Measure Theorey and Functional Analysis

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MeasureTheoreyandFunctional Analysis DMTH505

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SYLLABUS Measure Theory and Functional Analysis

Objectives: This course is designed for the of analysis of various types of spaces like Banach Spaces, Hilbert Space, etc. and also

Sr. No Description 1 Differentiation and Integration: Differentiation of monotone functions, Functions of bounded variation, 2 Differentiation of an integral, Absolute continuity 3 Spaces, Holder, Minkowski inequalities, Convergence and Completeness 4 Bounded linear functional on the Lp spaces, Measure spaces, Measurable Functions, Integration 5 General Convergence Theorems, Signed Measures, Radon-Nikodym theorem. 6 Banach spaces: Definition and some examples, Continuous linear transformations, The Hahn-Banach theorem 7 The natural imbedding of N in N**, The open mapping theorem, The closed graph theorem, 8 The conjugate of an operator, The uniform boundedness theorem, The uniform boundedness theorem, Hilbert spaces : The definition and some simple properties 9 Orthogonal complements, Orthonormal Sets, The conjugate space H*, The Adjoint of an Operator, Self Adjoint Operators 10 Normal and Unitary Operators, Projections, Finite dimensional spectral theory : the spectrum of an operator on a finite dimensional Hilbert space, the Spectral theorem

CONTENTS

Unit 1: Differentiation and Integration: Differentiation of Monotone Functions 1 Unit 2: Functions of Bounded Variation 11 Unit 3: Differentiation of an Integral 27 Unit 4: Absolute Continuity 36 Unit 5: Spaces, Hölder 50 Unit 6: Minkowski Inequalities 65 Unit 7: Convergence and Completeness 72 Unit 8: Bounded Linear Functional on the Lp-spaces 82 Unit 9: Measure Spaces 100 Unit 10: Measurable Functions 106 Unit 11: Integration 121 Unit 12: General Convergence Theorems 141 Unit 13: Signed Measures 158 Unit 14: Radon-Nikodym Theorem 166 Unit 15: Banach Space: Definition and Some Examples 174 Unit 16: Continuous Linear Transformations 183 Unit 17: The Hahn-Banach Theorem 201 Unit 18: The Natural Imbedding of N in N** 211 Unit 19: The Open Mapping Theorem 217 Unit 20: The Closed Graph Theorem 225 Unit 21: The Conjugate of an Operator 231 Unit 22: The Uniform Boundedness Theorem 238 Unit 23: Hilbert Spaces: The Definition and Some Simple Properties 244 Unit 24: Orthogonal Complements 253 Unit 25: Orthonormal Sets 262 Unit 26: The Conjugate Space H* 273 Unit 27: The Adjoint of an Operator 283 Unit 28: Self Adjoint Operators 294 Unit 29: Normal and Unitary Operators 305 Unit 30: Projections 316 Unit 31: Finite Dimensional Spectral Theory 329

Unit 1: Differentiation and Integration: Differentiation of Monotone Functions

Unit 1: Differentiation and Integration: Differentiation Notes of Monotone Functions

CONTENTS

Objectives Introduction 1.1 Differentiation and Integration 1.1.1 Lipschitz Condition 1.1.2 Lebesgue Point of a Function 1.1.3 Covering in the Sense of Vitali 1.1.4 Four Dini's Derivatives 1.1.5 Lebesgue Differentiation Theorem 1.2 Summary 1.3 Keywords 1.4 Review Questions 1.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand differentiation and integration

 Describe Lipschitz condition and Lebesgue point of a function

 State Vitali’s Lemma and understand its proof.

 Explain four Dini’s derivatives and its properties

 Describe Lebesgue differentiation theorem.

Introduction

Differentiation and integration are closely connected. The fundamental theorem of the integral calculus is that differentiation and integration are inverse processes. The general principle may be interpreted in two different ways: 1. If f is a Riemann integrable function over [a, b], then its indefinite integral i.e.

x F : [a, b] R defined by F (x) = f (t) dt is continuous on [a, b]. Furthermore if f is

continuous at a point xo [a, b], then F is differentiable thereat and F (xo) = f (xo). 2. If f is Riemann integrable over [a, b] and if there is a differentiable function F on [a, b] such that F f(x) for x [a, b], then

x f (t) dt = F (x) – F (a) [a x b].

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Notes 1.1 Differentiation and Integration

1.1.1 Lipschitz Condition

Definition: A function f defined on [a, b] is said to satisfy Lipschitz condition (or Lipschitzian function), if a constant M > 0 s.t.

|f (x) – f (y)| M |x – y|, x, y [a, b].

1.1.2 Lebesgue Point of a Function

Definition: A point x is said to be a Lebesgue point of the function f (t), if

1 x n Lim |f(t) f(x)|dt 0. h 0 h x

1.1.3 Covering in the Sense of Vitali

Definition: A set E is said to be covered in the sense of Vitali by a family of intervals (may be open, closed or half open), M in which none is a singleton set, if every point of the set E is contained in some small interval of M i.e., for each x E, and > 0, an interval I M s.t. x I and (I) .

The family M is called the Vitali Cover of set E.

Example: If E = {q : q is a rational number in the interval [a, b]}, then the family Iqi 1 1 where I q , q , i N is a vitali cover of [a, b]. qi i i

Vitali's Lemma

Let E be a set of finite outer measure and M be a family of intervals which cover E in the sense of

Vitali; then for a given > 0, it is possible to find a finite family of disjoint intervals {Ik, k = 1, 2, … n} of M, such that

n m * E I  k < . k 1

Proof: Without any loss of generality, we assume that every interval of family M is a closed interval, because if not we replace each interval by its closure and observe that the set of end

points of I1, I2, …… In has measure zero. [Due to this property some authors take family M of closed intervals in the definition of Vitali’s covering]. Suppose 0 is an open set containing E s.t. m* (0) < m* (E) + 1 < we assume that each interval in M is contained in 0, if this can be achieved by discarding the intervals of M extending beyond 0 and still the family M will cover the set E in the sense of Vitali.

Now we shall use the induction method to determine the sequence of disjoint intervals of M as follows:

2 LOVELY PROFESSIONAL UNIVERSITY Unit 1: Differentiation and Integration: Differentiation of Monotone Functions

 Notes Let I1 be any interval in M and let 1 be the supremum (least upper bound of the lengths of the intervals in M disjoint from I1 (i.e., which do not have any point common with I1).

Obviously  1 < as  1 m (0) < .

1 Now we choose an interval I from M, disjoint from I , such that (I).  Let  be the 2 1 22 1 2 supremums of lengths of all those intervals of M which do not have any point common with I 2  or I2 obviously 2 < .

 In general, suppose we have already chosen r intervals I 1, I2, … Ir (mutually disjoint). Let r be the supremums of the length of those intervals of M which do not have any point in common with I (i.e., which do not meet any of the intervals I , I , … I . Then  m (0) < .  i 1 2 r r i 1

r r Now if E is contained in I , then Lemma established. Suppose IE . Then we can find  i  i i 1 i 1

1 interval I s.t. (I)  which is disjoint from I , I , … I . r+1 r 12 r 1 2 r

Thus at some finite iteration either the Lemma will be established or we shall get an infinite 1 sequence of disjoint intervals of M s.t. (I)  and  r < , n = 1, 2, 3 …. r r 12 r

Note that <  r > is a monotonically decreasing sequence of non-negative real numbers.

Obviously, we have that I 0 (  ) m(0) hence for any arbitrary > 0, we can  r r i 1 r 1 find an integer N s.t.

1  (I)r . r N 1 5

N I Let a set F =  r . r 1

N The lemma will be established if we show that m* (F) < . For, let x F, then x I x is an  r r 1 N element of E not belonging to the closed set I an interval I in M s.t. x I and (I) is so  r r 1 N small that I does not meet the I , i.e.  r r 1

I Ir = , r = 1, 2, … N.

Therefore we shall have (I) N 2  (I n 1 ) as by the method of construction we take 1 I  . n 12 N

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Notes   It also I IN+1 = , we should have I N 1 . Further if the interval I does not meet any of the

intervals in the sequence , we must have

I  r , r

which is not true as r2  (I r 1 ) 0 as r .

I must meet at least one of the intervals of the sequence . Let p be the least integer s.t. I

meets Ip. Then p > N and (I) p 1 2  (I p ). Further let x I as well x Ip, then the distance of x

from the mid point of Ip is at most

1 1 5 (I)  (  )2(I)   (I)  (I) 2p p 2 p 2 p

Thus if Ip is an interval having the same mid point as Ip but length 5 times the length of Ip, i.e.   (Jp ) 5 (I p ) . Then x Jp also.

Thus for every x F, an integer p > N s.t. x Jp

and (Jp ) 5  (I p ) . Also

FJ  p p N 1

m*(F) (J)5p  (I)5 p p N 1 p N 1 5

and hence the Lemma holds good.

1.1.4 Four Dini's Derivatives

The usual condition of differentiability of a function f (x) is too strong. Here we are studying the functions under slightly weaker condition (measurability). So why we define four quantities, called as Dini’s Derivatives, which may be defined even at the points where the function is not differentiable.

f(x h) f(x) 1. D+ f (x) = Lim , called upper right derivative n 0 h

f(x h) f(x) Lim 2. D+ f (x) = , called lower right derivative h 0 h

f(x h) f(x) 3. D– f (x) = Lim , h 0 h

f(x h) f(x) or Lim , called upper left derivative h 0 h

f(x h) f(x) 4. D f (x) = Lim – h 0 h f(x h) f(x) or Lim , called lower left derivative h 0 h

4 LOVELY PROFESSIONAL UNIVERSITY Unit 1: Differentiation and Integration: Differentiation of Monotone Functions

Notes

+ 1. D f (x) D+ f (x) and D f(x) D f(x)

+ If D f (x) = D+ f (x), then we conclude that right hand derivative of f (x) exists at the + – point x and denoted by f (x ). Similarly if D f (x) = D– f (x), we say that f (x) is left differentiable at x and denote this common value by f (x–). 2. The function is said to be differentiable at x if all the four Dini’s derivatives are equal but different than , i.e. if

+ – D f (x) = D+ f (x) = D f (x) = D– f (x) and their common value is denoted by f (x).

Properties of Dini's Derivatives

1. Dini’s derivatives always exist, may be finite or infinite for every function f. 2. D+ (f + g) D+ f + D+ g with similar properties for the other derivatives. 3. If f and g are continuous at a point ‘x’, then D+ (f . g) (x) f (x) D+ g (x) + g (x) D+ f (x).

+ 4. D+ f (x) = – D (– f (x))

– and D– f (x) = – D (– f (x)). 5. If f is a continuous function on [a, b] and one of its derivatives (say D +) is non-negative on (a, b). Then f is non-decreasing on [a, b] i.e. f (x) f (y) whenever x y, y [a, b]. 6. If f is any function on an interval [a, b], then the four derivatives if exist are measurable.

1.1.5 Lebesgue Differentiation Theorem

Statement: Let f : [a, b] R be a finite valued monotonically increasing function, then f is differentiable. Also f : [a, b] R is L-integrable and

b f (x) dx f(b) f(a) . a

Proof: Define a sequence of non-negative functions, where fn : [a, b] R such that, 1 f (x) = nfx f(x), x [a,b] … (1) n n and set f (x) = f (b), for x b.

By hypothesis, f : [a, b] R is an increasing function, therefore fn : [a, b] R is also an increasing function and hence integrable in the Lebesgue sense. Again from (i) we have

f {x (1/n)} f(x) Lim fn (x) = Lim , x [a, b] , n 1/n 0 (1/n)

= f (x), a.e.

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Notes Thus, the sequence converges to f (x), a.e. Using Fatou’s Lemma, we have

b b

f (x) dx Lim inf fn (x) dx … (ii) a n a

b b 1 Again Lim inf fn (x) dx = Lim inf n f x f(x) dx n a n a n

b1 b = Lim inf n f x dx f(x) dx n an a

Putting t = x + (1/n), we get

b 1 b(1/n) b(1/n) f x dx = f(t)dt f(x)dx a n a(1/n) a(1/n)

[By the first property of definite integrals]

b b (1/n) b

Lim inf fn (x) dx = Lim inf n f(x) dx f(x) dx n a n a (1/n) a

b(1/n) a(1/n) = Lim inf n f(x) dx f(x) dx … (iii) n b a

Now extend the definition of f by assuming

f (x) = f (b), x [b, b + 1/n].

b (1/n) b (1/n) 1 f(x) dx = f(b) dx f(b) b b n

1 Also f (a) f (x), for x a, a , therefore n

a (1/n) a (1/n ) 1 f(x) dx f(a) dx f(a) a a n

a (1/n) 1 – f(x) dx f(a) a n

b b(1/n) a(1/n) Lim inf f (x) dx Lim inf n f(b) dx f(x) dx (iii) n = n n a b a

1 1 Lim inf n f(b) f(a) f(b) f(a) n n n

Thus from (ii), we get

b f (x) dx f (b) – f (a) a

f (x) is integrable and hence finite a.e. thus f is differentiable a.e.

6 LOVELY PROFESSIONAL UNIVERSITY Unit 1: Differentiation and Integration: Differentiation of Monotone Functions

Notes Example: Let f be a function defined by f (0) = 0 and f (x) = x sin (1/x) for x 0. Find + – D f (0), D+ f (0), D f (0), D– f (0).

1 f(0 h) f(o) h sin 0 D+ f (0) = Lim Lim h h o hh o h

1 1 = Lim sin 1, as 1 sin 1 h o h h

f(0 h) f(0) 1 Also D f (0) = Lim Lim sin 1 + h oh h o h

1 ( h)sin 0 f(0 h) f(0) h D– f (0) = Lim Lim h o 0 hh o h

1 = Lim sin 1 h o h

f(0 h) f(0) 1 Lim Lim sin 1 and D– f (0) = h oh h o h

Theorem: Let x be a Lebesgue point of a function f (t); then the indefinite integral

x F (x) = F (a) + f(t) dt a is differentiable at each point x and F (x) = f (x). Proof: Given that x is a Lebesgue point of f (t), so that

1 x h Lim f(t) f(x) dt = 0 … (i) h o h x

x h x h 1 1 1 x h Now f(x) dt = f(x) 1 dt f(x)[t]x h x hx h

1 = f(x).h f(x) h

1 x h Thus f (x) = f(x) dt … (ii) h x

x h x Also F (x + h) – F (x) = f(t) dt f(t) dt a a

x x h x x h = f(t) dt f(t) dt f(t)dt f(t) dt a x a x

F(x h) F(x) 1 x h = f(t)dt … (iii) h h x

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Notes From (ii) and (iii) we have

F(x h) F(x) 1x h 1 x h f(x) = f(t)dt f(x)dt h hx h x

1 x h 1 x h = f(t) f(x) dt f(t) f(x) dt h x h x

F(x h) F(x) 1 x h Lim f(x) Lim f(t) f(x) dt 0 [Using (i)] h o h h o h x

F(x h) F(x) or Lim f(x) 0 … (iv) h o h

Since modulus of any quantity is always positive, therefore

F(x h) F(x) Lim f(x) 0 … (v) h o h

Combining (iv) and (v), we obtain

F(x h) F(x) Lim f(x) = 0 h o h

F(x h) F(x) Lim = f (x) h o h

F (x) = f (x). Theorem: Every point of continuity of an integrable function f (t) is a Lebesgue point of f (t). Proof: Let f (t) be integrable over the closed interval [a, b] and let f (t) be continuous at the

point xo.

f (t) is continuous at t = xo implies that > 0, a > 0 such that,

|f (t) – f (xo) | < , whenever |t – xo| < .

xo h x o h

|f (t) f(xo )|dt dt h whenever|h| . xo x o

1 xo h |f (t) f(xo )|dt … (i) h xo

Now h 0 0. So from (i), we have

1 xo h Lim |f (t) f(xo )|dt 0 h o … (ii) h xo

1 xo h Now Lim |f (t) f(xo )|dt h o h xo

1 xo h Lim |f (t) f(xo )|dt 0 h o [Using (ii)] h xo

8 LOVELY PROFESSIONAL UNIVERSITY Unit 1: Differentiation and Integration: Differentiation of Monotone Functions

Notes 1 xo h Lim |f (t) f(xo )|dt 0 or h o [ Modulus of any quantity is always non-negative] h xo

1 xo h i.e. Lim |f (t) f(xo )|dt 0 . h o h xo

This shows that xo is a Lebesgue point of f (t).

1.2 Summary

 A function f defined on [a, b] is said to satisfy Lipschitz condition if a constant M > 0 such that

| f (x) – f (y) M |x – y|, x, y [a, b].

 A point x is said to be a Lebesgue point of the function f (t), if

1 x h Lim f(t) f(x) dt 0 = 0 h o h x  Let E be a set of finite outer measure and M be a family of intervals which cover E in the sense of Vitali; then for a given > 0, it is possible to find a finite family of disjoint intervals

{Ik, k = 1, 2, …, n} of M, such that

n m * E I .  k k 1

 Lebesgue differentiation theorem: Let f : [a, b] R be a finite valued monotonically increasing function, then f is differentiable. Also f : [a, b] R is L-integrable and

b f (x) dx f(b) f(a) .

a 1.3 Keywords

Dinni’s Derivatives: These are the ways to define the quantities to judge the measurability of the functions even at the points where it is not differentiable.

Fundamental Theorem of the Integral: The fundamental theorem of the integral calculus is that differentiation and integration are inverse processes. Measurable functions: An extended real valued function f defined over a measurable set E is said to be measurable in the sense of Lebesgue if set E (f > a) = {x E : f (x) > a} is measurable for all extended real numbers a. Vitali's Lemma: Let E be a set of finite outer measure and M be a family of intervals which cover E in the sense of Vitali; then for a given > 0, it is possible to find a finite family of disjoint intervals {Ik, k = 1, 2, … n} of M, such that

n m * E I  k < . k 1

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Notes 1.4 Review Questions

1. If the function f assumes its maximum at c, show that D+ f (c) 0 and D– f (c) 0. 2. Give an example of functions such that D+ (f + g) D+ f + D+g. 3. Find the four Dini’s derivatives of function f : [0, 1]  R such that f (x) = 0, if x 0, if x Q and f (x) = 1, if x Q. 4. Evaluate the four Dini’s derivative at x = 0 of the function f (x) given below:

1 1 ax sin2 bx cos 2 , x 0 x x f (x) = 1 1 px sin2 qx cos 2 , x 0 x x

and f (0) = 0, given that a < b, p < q. 5. Every point of continuity of an integrable function f (t) is a Lebesgue point of f (t). Elucidate.

1.5 Further Readings

Books J. Yeh, Real Analysis: Theory of Measure and Integration Bartle, Robert G. (1976). The Elements of Real Analysis (second edition ed.)

Online links www.solitaryroad.com/c756.html www.public.iastate.edu/.../Royden_Real_Analysis_Solutions.pdf

10 LOVELY PROFESSIONAL UNIVERSITY Unit 2: Functions of Bounded Variation

Unit 2: Functions of Bounded Variation Notes

CONTENTS Objectives Introduction 2.1 Functions of Bounded Variation 2.1.1 Absolute Continuous Function 2.1.2 Monotonic Function 2.1.3 Functions of Bounded Variation – Definition 2.1.4 Theorems and Solved Examples 2.2 Summary 2.3 Keywords 2.4 Review Questions 2.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define absolute continuous function.

 Define monotonic function.

 Understand functions of bounded variation.

 Solve problems on functions of bounded variation.

Introduction

Functions of bounded variation is a special class of functions with finite variation over an interval. In Mathematical analysis, a function of bounded variation, also known as a BV function, is a real-valued function whose total variation is bounded: the graph of a function having this property is well behaved in a precise sense. Functions of bounded variation are precisely those with respect to which one may find Riemann – Stieltjes integrals of all continuous functions. In this unit, we will study about absolute continuous function, Monotonic function and functions of bounded variation.

2.1 Functions of Bounded Variation

2.1.1 Absolute Continuous Function

A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if for an arbitrary 0 , however small, a, 0, such that

n n

f br – f a r ,wherever b r a r , r 1 r 1

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Notes where a1 b 1 a 2 b 2 ... a n b n i.e., ai’s and bi’s are forming finite collection

ai ,b i : i 1,2,...,n of pair-wise disjoint (non-overlapping) intervals (or of disjoint closed intervals).

Obviously, every absolutely continuous function is continuous.

Notes

 If a function satisfies f br – f a r , even then it is absolutely continuous.

 The condition br a r , means that total length of all the intervals must be r 1 less than .

2.1.2 Monotonic Function

Recall that a function f defined on an interval I is said to be monotonically non-increasing, iff

x y fx fy,x,yI

and monotonically non-decreasing, iff

x y fx fy,x,yI

Also f is said to be strictly decreasing, iff

x y f x f y

and strictly increasing, iff

x y f x f y

2.1.3 Functions of Bounded Variation – Definition

Let f be a real-valued function defined on [a,b] which is divided by means of points

a x0 x 1 x 2 ... x n b.

Then the set P x0 ,x 1 ,x 2 ,...,x n is termed as subdivision or partition of [a,b].

b n 1 b b Let us take V f,P f xr 1 f x r , and V f,P sup V f,P for all possible subdivisions P of a r 0 a a

b [a,b]. (This is called total variation of f over [a,b] and also denoted by T f ). a

12 LOVELY PROFESSIONAL UNIVERSITY Unit 2: Functions of Bounded Variation

b Notes If V f is finite, then f is called a function of bounded variation or function of finite variation a over [a,b]. Set of all the functions of bounded variation on [a,b] is denoted by BV [a,b].

Notes

If f is defined on R, then we define

a V f lin V f . a a

Some important observations about the functions of bounded variations.

Let f: [a,b] R and P be any subdivision of [a,b]. Then:

x (i) fx fa Vf,x [a,b] a a (ii) V f 0 a

b b (iii) P1 P 2 Vf,P 1 Vf,P, 2 where P1 and P2 are any two subdivisions of [a,b]. a a

b b (iv) V f,P V f , for all subdivisions P of [a,b]. a a (v) For each 0, however small, at least one subdivision P’ of [a,b] such that

b b V f,P' V f . a a

b (vi) V f 0. a

b c (vii) a b c V f V f . a a

2.1.4 Theorems and Solved Examples

Theorem 1: A monotonic function on [a,b] is of bounded variation.

Proof: Divide the interval [a,b] by means of points

a x0 x 1 x 2 ... x n b. without any loss of generality, we can take f(x) as increasing function on [a,b]. Since if f is a decreasing function, –f is an increasing function and so by taking –f = g, we see that g is an increasing function and so we are allowed to consider only increasing functions. Thus

xr x r 1 f x r f x r 1

f xr 1 – f x r 0

f xr 1 – f x r f x r 1 – f x r ...(i)

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Notes n 1 n 1

Now V fxr 1 –fx r fx r 1 –fx r [using (i)] r 0 r 0

V f xn f x 0 f b f a .

Now f is monotonic f b and f a are finite quantities.

V a finite quantity independent of the mode subdivision. Hence f is of bounded variation.

Notes

If f is a monotonic function on [a,b], then

b T f f b f a a

Theorem 2: Let V, P, N denote total, positive and negative variations of a bounded function f on [a,b]; then prove that

V = P+N, and P–N= f(b)–f(a).

Proof: Let the interval [a,b] be divided by means of points

a x0 x 1 x 2 ... x n b.

n 1

v f xr 1 – f x r r 0

If P denotes the sum of those differences f xr 1 – f x r which are +n for positive and –n for negative, then obviously,

v = p + n, f(b) – f(a) = p – n ...(i)

Let P supp,V supv,N supn, ...(ii)

where suprema are taken over all subdivisions of [a,b]. From (i), we have

v = 2p + f(a) – f(b), ...(iii)

v = 2n + f(b) – f(a). ...(iv)

Taking supremum in (iii) and (iv) and using (ii), we get

V = 2P + f(a) – f(b), ...(v)

V = 2N + f(b) – f(a). ...(vi)

By adding and subtracting, (v) and (vi) give

V = P+N and f(b) – f(a) = P–N.

Theorem 3: If f1 and f2 are non-decreasing functions on [a,b], then f1–f2 is of bounded variation on [a,b].

Proof: Let f = f1 – f2 defined on [a,b].

Then for any partition P a x0 ,x 1 ,...,x n b , we have

14 LOVELY PROFESSIONAL UNIVERSITY Unit 2: Functions of Bounded Variation

Notes fx–fxi i1 fx–fx 1i1i1 fx–fx 2i2i1

f1 b – f 1 a f 2 b – f 2 a

as f1 and f2 are monotonically increasing.

b Vf fb1 fb 2 fa 2 fa, 1 which is a finite quantity. a

b V f and hence f is of bounded variation. a

Theorem 4: If f BV [a,b] and c a,b , then f BV [a,c] and f BV [c,b] . Also

b c b V f V f V f a a c

Proof: Since f BV[a,b] and [a,c] [a,b] we get

c b V f V f a a

f BV [a,c] and similarly f BV [c,b] .

Now if P1 and P2 are any subdivisions of [a,c] and [c,b] respectively, then PPP1 2 is a subdivision of [a,b].

c b b b Vf,P1 Vf,P 2 Vf,P Vf . a c a a

But P1 and P2 are any subdivisions. So taking supremums on P1 and P2, we get

c b b V f V f V f . ...(1) a c a

Now let P a x0 ,x 1 ,x 2 ,...,x n b be a subdivision of [a,b] and c xr 1 ,x r

P1 x 0 ,x 1 ,x 2 ,...,x r 1 ,c and

P2 c,x r ,x r 1 ,x r 2 ,...,x n are the subdivisions of [a,c] and [c,b] respectively.

b r 1 n Now Vf,P fxi fx i 1 fx r fx r 1 fx i fx i 1 a i 1 i r 1

r 1 n

fxi fx i 1 fx r fcfcfx r 1 fx i fx i 1 i 1 i r 1

r 1 n

fxfxi i 1 fcfx r 1 fx r fc fxfx i i 1 i 1 i r 1

c b c b Vf,P1 Vf,P 2 Vf Vf a c a c

b c b (i) and (ii) V f V f V f . ...(ii) a a c

LOVELY PROFESSIONAL UNIVERSITY 15 Measure Theory and Functional Analysis

Notes

 This theorem enables us to define a new function (called variation function) say

x V x V f , x [a,b] . a

yx y  If x > y in [a,b], then V f V f V f . a a x

y i.e. v(y) = v(x) + V f . x

v x is an increasing function.

 If a c1 c 2 ... c n b,then

bc1 c 2 b V f V f V f ... V f a a c1 c n

Corollary:

f BV[a,b] f BV[a,c],

f BV[c,b] for each c [a,b].

Theorem 5: If a function f of bounded variation in [a,b] is continuous at c [a,b], then the function

x defined by v(x) = V f , is also continuous at x = c and vice versa. a

Proof: Suppose f is continuous at x = c. Hence for arbitrary /2 0, we can find a 1 such that

a c1 x c or x c 1 f x f c /2 ...(i)

Also we know by remark (v) after the definition (2.1.3), for above , we can get a subdivision

P = a x0 ,x 1 ,x 2 ,...,x n c of [a,c]

c c s.t. V f V f,P ...(ii) a a 2

Now choosing positive min[1 ,c x n 1 ], we get that for any x such that c x c , we also

have xn 1 x x n .

c n 1 (ii) V f f xr f x r 1 f x n f x n 1 a r 1 2

n 1

fxr fx r 1 fx n fxfxfx n 1 r 1 2

16 LOVELY PROFESSIONAL UNIVERSITY Unit 2: Functions of Bounded Variation

n 1 Notes fxr fx r 1 fxfx n 1 fx n fx r 1 2

x V f + f c f x a 2

c x V f V f , [by (i)] a a

c x 0 V f V f , a a

x s.t. c x c,we have v c v x

lt v x = v c . x c 0

v x is continuous on the left at x = c.

Similarly considering the partition of [c,b], one can show that v(x) is right continuous also at x = c and hence v(x) is also continuous at x = c.

Converse of the above Theorem

If v(x) is continuous at x c [a,b] so is f also at x – c.

Proof: Since v(x) is continuous at x = c, for arbitrary small 0, a 0 such that

v x v c ,x c ,c ...(i)

Now let c x c . Then by Note (ii) of Theorem 4, we get

x c x V f V f V f a a c

x v x v c V f c

x v x v c V f f x f c c

f x f c v x v c [by (i)] ...(ii)

Similarly, we can show that f c f x ,if c x c. ...(iii)

(ii) and (iii) show that f(x) is also continuous at x = c.

Theorem 6: Let f and g be functions of bounded variation on [a,b] ; then prove that f+g, f-g, fg and f/g g x 0, x and cf are functions of bounded variation, c being constant.

Proof:

(i) Set f + g = h, then

hxr 1 hx r fx r 1 gx r 1 fxgx r r

where a = x0 < x1< x2<...

LOVELY PROFESSIONAL UNIVERSITY 17 Measure Theory and Functional Analysis

Notes f xr 1 f x r g x r 1 g x r

f xr 1 f x r g x r 1 g x r .

n 1 n 1 n 1

hxr 1 hx r hx r 1 hx r gx r 1 gx r . r 0 r 0 r 0

b b b or V h V f V g . a a a

Now by hypothesis, f, g are functions of bounded variations.

b b V f and V g are finite. a a

b V h a finite quantity. a

Hence h = f + g is of bounded variation in [a,b].

(ii) Let h = f – g. Then as above,

hxr 1 hx r fx r 1 fx r gx r 1 gx. r

b b b V h V f V g a a a

b V h a finite quantity. a Hence h = f – g is of bounded variation in [a,b].

(iii) Let h(x) = f(x).g(x). Then

hxr 1 hx r fx r 1 .gx r 1 fx.gx r r

fxr1 .gx r1 fxgx r r1 fxgx r r1 fxgx r r

gxr 1 fx r 1 fx r fxgx r r 1 gx r .

gxr 1 fx r 1 fx r fx.gx r r 1 gx r .

Let A = sup f x : x [a,b] ,

B = sup g x : x [a,b] ,

h xr 1 h x r B. f x r 1 f x r A. g x r 1 g x r .

n 1 n 1 n 1

hxr 1 hx r B.hx r 1 hx r A gx r 1 gx r . r 0 r 0 r 0

18 LOVELY PROFESSIONAL UNIVERSITY Unit 2: Functions of Bounded Variation

b b b Notes i.e. V h B.V f A.V g . a a a

= a finite quantity.

Hence h(x) = f(x).g(x) is of bounded variation in [a,b].

(iv) First, we shall show that 1/g is of bounded variation, where g x 0, x [a,b].

Now, g x 0, x [a,b]

1 1 0, x [a,b]. g x

Again, we observe that

1 1g xr g x r 1 1 2 g xr g x r 1 g xr 1 g x r g x r .g x r 1

n 11 1 1 n 1 2 g xr g x r 1 r 0g xr 1 g x r r 0

b1 1 b V V g a finite quantity. ag 2 a

1 Hence g is of bounded variation in [a,b].

1 Now f and g are of bounded variation in [a,b].

1 f. is of bounded variation in [a,b] [by case (iii)] g

f is of bounded variation in [a,b]. g

b b (v) Do yourself. Note that V cf c V f . a a

Notes

Since BV [a,b] is closed for all four algebraic operations, it is a linear space.

Theorem 7: Every absolutely continuous function f defined on [a,b] is of bounded variation.

Proof: Since f is absolutely continuous on [a,b]; for 1, a 0

LOVELY PROFESSIONAL UNIVERSITY 19 Measure Theory and Functional Analysis

Notes n

s.t. f bi f a i 1, i 1

whenever bi a i , i 1

and a a1 b 1 a 2 b 2 ... a n b n b.

Now consider another subdivision of [a,b] or say refinement of P by adjoining some additional points to P in such a way that all the intervals can be divided into r parts each of total length less than .

Let the r sub-intervals be [c0,c1], [c1,c2],...,[cr-1,cr] such that

a = c0, cr = b and (ck+1–ck) < , K 0,1,2,...,(r 1)

Obviously, f xi 1 – f x i 1 1,where x i 1 ,x i . [c k ,c k 1 ] i

ck 1 or V f 1, [Using (i)] ck

b c1 c 2 c r Hence V f V f V f ... V f 1 1 1 ... 1 r finite quantity. a c0 c1 c r 1 Hence, f is of bounded variation.

Notes

Converse of above theorem is not necessarily true. These exists functions of bounded variation but not absolutely continuous.

Theorem 8: Jordan Decomposition Theorem

A function f is of bounded variation, if and only if it can be expressed as a difference of two monotonic functions both non-decreasing.

Proof: Let f be the function of f :[a,b] R.

Case I. f BV[a,b]. Then we can write

f = v – (v – f), ...(i)

so that f x v x v x f x ,x [a,b].

Now if x, y [a, b] such that x < y, then by the remark (ii) of theorem 4, we get

yx y V f V f V f . a a x

y v y v x V f 0 x

v x v y and hence v is a non-decreasing function on [a,b].

20 LOVELY PROFESSIONAL UNIVERSITY Unit 2: Functions of Bounded Variation

Again, if x

y vy vx Vf fy fx fy fx x

vy fy vx fx vfy vfx

v f is also a non-decreasing function on [a,b].

Thus (i) shows that f is expressible as a difference of two monotonically non-decreasing functions.

Case II. Set g (x) and h (x) be increasing functions such that f(x) = g(x) – h(x).

Divide the closed interval [a,b] by means of points a = x0 < x1< x2<...< xn=b.

n 1

Let V f xr 1 f x r r 0

Now, we have that

fxr 1 fx r gx r 1 hx r 1 gx r hx r

g xr 1 g x r h x r h x r 1

g xr 1 g x r h x r 1 h x r

Now, g(x) and h(x) are monotonically increasing functions, so that g xr 1 g x r 0

and h xr 1 h x r 0

g xr 1 g x r g x r 1 g x r

and h xr 1 h x r h x r 1 h x r .

Hence fxr 1 fx r gx r 1 gx r hx r 1 hx r

n 1 n 1 n 1

fxr 1 fx r gx r 1 gx r hx r 1 hx r r 0 r 0 r 0

n 1

Now g xr 1 g x r g x 1 g x 0 g x 2 g x 1 ...... g x n g x n 1 r 0

g xn g x 0

g b g a  xn b,x 0 a

n 1

Similarly, h xr 1 h x r h b h a . r 0

n 1

Hence fxr 1 fx r gb ga hb ha. r 0

LOVELY PROFESSIONAL UNIVERSITY 21 Measure Theory and Functional Analysis

Notes since f is finite in [a,b] Now g b ,g a h b ,h a are finite numbers.

n 1

f xr 1 h x r r 0

b V f . a

f is a function of bounded variation. Alternatively, since g (x) and h(x) are both non-decreasing, so by theorem 3, g(x) – h(x) and hence f(x) is of bounded variation. Corollary: A continuous function is of bounded variation iff it can be expressed are as a difference of two continuous monotonically increasing functions. It follows from the results of Theorems 5 and 8.

Theorem 9: An indefinite integral is a function of bounded variation, i.e. if f L[a,b] and F x is

x indefinite integral of f x i.e. F (x) = f t dt, then F BV[a,b].Also show that a

x b V f f . a a

Proof: Since f L[a,b], also f L[a,b].

Let P = xi : i 0,1,2,...,n be a subdivision of the interval [a,b]. Then

x x n n i i 1

F xi F x i 1 f f r 0 i 1 a a

x x ni n i f f i 1 i 1 xi 1 x i 1

b f . a

b b f BV[a,b] and V f,p f . a a Further above result is true for any subdivision of P of [a,b]. Therefore taking supremum, we get

b b f f .

a a

22 LOVELY PROFESSIONAL UNIVERSITY Unit 2: Functions of Bounded Variation

Notes Example: A function f of bounded variation on [a,b] is necessarily bounded on [a,b] but not conversely.

x b Solution: If x [a,b], then f x f a V f V f a a

b b V f f x f a V f a a

b b f a V f f x V f f a a a

f x is bounded on [a,b]

For the converse, define the function f on [0,1] by

0,if x 0 f x x.sin ,if 0 x 1 x since 0 x 1 and 1 sin 1, the function f is obviously bounded. Now consider the x partition

2 2 2 2 2 P= 0, , ,..., , , ,1 of [0,1] 2n+1 2n–1 7 5 3

Where n N. Then we get

1 2 2 2 2 V f,P f f 0 ... f f f 1 f 0 2n 1 3 5 3

2n 2 2 2 1 0 ... 1 .1 0 1 2n 1 3 5 3

2 2 2 2 ... 2n 1 3 5 3

1 1 1 4. ... . 3 5 2n 1

1 But we know that series is divergent. Therefore letting we get that 2n 1 n

1 1 V f lt V f,P 0 n 0

f is not of bounded variation.

LOVELY PROFESSIONAL UNIVERSITY 23 Measure Theory and Functional Analysis

Notes Example: Show that the function

xsin if 0 x 2 f xx is continuous 0,if x 0

without being of bounded variation.

show that there exists a continuous function without being of bounded variation.

Solution: We know that lt f x 0 f 0 x 0

f x is continuous but not of bounded variation (see converse of above example.)

Hence the result.

Problem: Show that if f exists and is bounded on [a, b], then f BV [a, b].

Solution: According to given, let |f | M on [a, b].

Then for any Xi – 1, xi [a, b], we get

f(xi ) f(x i 1 ) M |f(xi ) f(x i 1 )| M(x i x i 1 ) xi x i 1

for any partition P of [a, b],

b V(f) M (xi x i 1 ) M(b a) a

f B V [a, b].

Problem: Show that the function f defined as

1 f(x) xp sin for 0 x 1, f(o) 0, p 2. x

is of bounded variation [0, 1].

1 (0 h)p sin 0 h Solution: Note that RF (0) = Lim h o h

1 = Lim h(p 1) sin 0 h o h

1 ( h)p sin 0 h and Lf (0) lim 0 h o h

24 LOVELY PROFESSIONAL UNIVERSITY Unit 2: Functions of Bounded Variation

Notes 1 1 1 f (0) = 0 and f (x) = xp cos px p 1 sin x x2 x

1 1 f (x) = xp–2 pxsin cos , for 0 < x 1 x x

f (x) is bounded for 0 x 1.

According to above problem, f BV [0, 1].

2.2 Summary

 A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if for an arbitrary 0, however small, a, 0,s.t.

n n

f br f a r whenever b r a r , r 1 r 1

where a1 b 1 a 2 b 2 ... a n b n

 A function f defined on an interval I is said to be monotonically non-increasing, iff

x y f x f y,,. x y I

and monotonically non-decreasing, iff x > y f(x), f(y) x, g I.

n 1 b b b  Let and for all possible subdivisions P of V f,P f xr 1 f x r , V f Sup V f,P a a a r 0

b [a,b]. If V f is finite, then f is called a function of bounded variation over [a,b]. a

2.3 Keywords

Absolute Continuous Function: A real valued function f defined on [a, b] is said to be absolutely continuous on [a, b], if for an arbitrary > 0, however small, a, > 0, such that

n n

|f(br ) f(a r )| , wherever (b r a r ) r 1 r 1

where a1 < b1 a2 < b2 … an < bn i.e. a1’s and b1’s are forming finite collection {(ai, bi) : i = 1, 2, …, n} of pair-wise disjoint intervals.

Continuous: A continuous function is a function f : X Y where the pre-image of every open set in Y is open in X.

Disjoint: Two sets A and B are said to be disjoint if they have no common element, i.e. AB.

LOVELY PROFESSIONAL UNIVERSITY 25 Measure Theory and Functional Analysis

Notes Monotonic Decreasing Function: A monotonic decreasing function is a function that either

decreases or remains the same, never increases i.e. a function f(x) such that f(x2) f(x1) for x2 > x1. Monotonic Function: A monotonic function is a function that is either a monotonic increasing or monotonic decreasing.

Monotonic Increasing Function: A monotonic increasing function is a function that either

increases or remains the same, never decreases i.e. a function f(x) such that f(x2) f(x1) for x2 > x1.

2.4 Review Questions

1. Show that sum and product of two functions of bounded variation is again a function of bounded variation.

2. Show that the function f defined on [0,1] by x xcos for 0 x 1 f x 2 0 for x 0 is continuous but not of bounded variation on [0,1].

3. Show that the function f defined on [0,1] as f(x) = xsin for x > 0, f(0)=0 is continuous but x is not of bounded variation on [0,1].

4. Define a function of bounded variation on [a,b]. Show that every increasing function on [a,b] is of bounded variation and every function of bounded variation on [a,b] is differentiable on [a,b].

5. Show that a continuous function may not be of bounded variation.

6. Show that a function of bounded variation may not be continuous.

7. If f is a function such that its derivative f’ exists and is bounded. Then prove that the function f is of bounded variation.

2.5 Further Readings

Books Halmos, Paul (1950), Measure Theory, Van Nostrand and Co.

Kolmogorov, Andrej N.; Fomin, Sergej V. (1969). Introductory Real Analysis, New York: Dovers Publications.

Online links www.ams.org

www.whitman.edu/mathematics/SeniorProjectArchive/.../grady.pdf

26 LOVELY PROFESSIONAL UNIVERSITY Unit 3: Differentiation of an Integral

Unit 3: Differentiation of an Integral Notes

CONTENTS

Objectives Introduction 3.1 Differentiation of an Integral 3.2 Summary 3.3 Keyword 3.4 Review Questions 3.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define differentiation of an integral.

 Solve problems related to it.

Introduction

If f is an integrable function on [a, b], we define its indefinite integral to be the function F defined on [a, b] by

x F (x) = f(t) dt

Here, it is shown that the derivative of the indefinite integral of an integrable function is equal to the integrand almost everywhere. We begin by establishing some lemmas.

3.1 Differentiation of an Integral

If f is an integrable function on [a, b] then f is integrable on any interval [a, x] [a, b]. The function F given by

x F (x) = f(t) dt c ,

a where c is a constant, called the indefinite integral of f. Lemma 1: If f is integrable on [a, b] then the indefinite integral of f namely the function F on

x [a, b] given by F (x) = f(t) is a continuous function of bounded variation on [a, b].

Proof: Let xo be any point of [a, b].

LOVELY PROFESSIONAL UNIVERSITY 27 Measure Theory and Functional Analysis

Notes x xo

Then F(x) F(xo ) = f(t) dt f(t) dt a a

x a = f(t) dt f(t) dt

a xo

a x = f(t) dt f(t) dt

xo a

x = f(t) dt

x |f(t)|dt

But f is integrable on [a, b] |f| is integrable on [a, b] [Since we know that measurable function f is integrable over iff |f| is integrable over E] Given > 0, > 0 such that for every measurable set A [a, b] with m (A) < , we have

|f| by theorem, “if f is a non-negative function which is integrable over a set E, then

given > 0, there is a > 0 such that for every set A E with m (A) < , f .”

|f(t)|dt < , for |x – xo| < .

x x f(t) dt |f(t)|dt |F(x) – F (xo)| = < xo x o

whenever |x – xo| < .

|F(x) – F(xo)| < wherever |x – xo| <

F is continuous at xo and hence in [a, b]. Now we shall show that F is a function of bounded variation.

Let P = {a = xo < x1 < x2 < … < xn = b} be a partition of [a, b]. Then

n n xi

F(xi ) F(x i 1 ) = f(t) dt i 1 i 1 xi 1

28 LOVELY PROFESSIONAL UNIVERSITY Unit 3: Differentiation of an Integral

Notes n xi |f(t)|dt i 1 xi 1

b = |f(t)|dt

b b T(F)a |f(t)|dt a

But |f| is integrable therefore.

b |f|dt

b T(F)a <

F BV [a, b] Hence the Proof. Theorem 1: Let f be an integrable on [a, b].

x If f(t)dt 0 x [a, b] then f = 0 a.e. in [a, b].

Proof: Let if possible, f 0 a.e. in [a, b]. Let f (t) > 0 on a set E of positive measure, then there exists a closed set F E with m (F) > 0. Let A = (a, b) – F. Then A is an open set.

b Now f(t) dt f(t) dt

a A F

b But f(t) dt 0

b f(t) dt 0

f(t) dt f(t) dt 0

f(t) dt f(t) dt 0 f(t) dt f(t) dt

AFAF

LOVELY PROFESSIONAL UNIVERSITY 29 Measure Theory and Functional Analysis

Notes But f (t) > 0 on F with m (F) > 0 implies

f(t) dt 0

Therefore f(t) dt 0

Now, A being as open set, it can be expressed as a union of countable collection {(an, bn)} of disjoint open intervals as we know that an open set can be expressed as a union of countable collection of disjoint open intervals.

bn Thus f(t) dt f(t) dt n A an

But f(t) dt 0

bn f(t) dt 0 n an

bn f(t) dt 0 for some n

an either f(t) dt 0

bn Or f(t) dt 0

In either case, we see that if f is positive on a set of positive measure, then for some x [a, b] we have

x f(t) dt 0 .

Similarly if f is negative on a set of positive measure we have

x f(t) dt 0 .

But it leads to the contradiction of the given hypothesis. Hence our supposition is wrong. f = 0 a.e. in [a, b]. Hence the proof.

30 LOVELY PROFESSIONAL UNIVERSITY Unit 3: Differentiation of an Integral

Theorem 2: First fundamental theorem of calculus statement: If f is bounded and measurable on Notes

x [a, b] and F (x) = f(t) dt + F (a), then F (x) = f (x) a.e. in [a, b].

Proof: Since every indefinite integral is a function of bounded variation, therefore F (x) is a function of bounded variation over [a, b]. Thus F (x) can be expressed as a difference of two monotonic functions and since every monotonic function has a finite differential coefficient at every point of a set of non-zero measure, therefore F (x) has a finite differential coefficient a.e. in [a, b]. Now F is given to be bounded; |f| M (say) … (1)

F(x h) F(x) Let f (x) = . n h

1 with h = . x

1 Then |f (x)| = (F(x h) F(x) n h

x h x 1 = f(t) dt f(t) dt h a a

x h a 1 = f(t) dt f(t) dt h a x

a x h 1 = f(t) dt f(t) dt h x a

x h 1 = f(t) dt h x

But |f| M

x h MM |f (x)| dt (x h x) n h h x

M |f (x)| (h) n h

|fn(x)| M

Since fn (x) F (x) a.e.,

LOVELY PROFESSIONAL UNIVERSITY 31 Measure Theory and Functional Analysis

Notes then the bounded convergence theorem implies that

x x F (x) dx = lim Fx (x) dx h a a

x 1 = lim [F(x h) F(x)]dx h 0 h a

x h a h 1 1 = lim [F(x)dx F(x) dx h 0 h h x a

= F (x) – F (a)

x = f(t) dt, by hypothesis a

x x F(x) f(t)dt F(a) F(x) F(a) f(t)dt

a a

x or [F (t) f(t)]dt 0, x

F (x) – f (x) = 0 a.e. in [a, b]

x Hence F (x) = f(x) a.e. in [a, b] by the theorem, “If f is integrable on [a, b] and f(t) dt 0, x [a, b]

a then f = 0 a.e. in [a, b]”. Hence F (x) = f (x) a.e. in [a, b]. Hence the proof.

x Theorem 3: If f is an integrable function on [a, b] and if F(x) = f(t) dt + F(a) then F (x) = f (x) a.e.

a in [a, b].

Proof: Without loss of generality, we may assume that f (x) 0 x

Let us define a sequence {fn} of functions

fn : [a, b] R, where

f(x) if f(x) n, f (x) = n n if f(x) n

Clearly, each fn is bounded and measurable function and so, by the theorem,

x Let f be a bounded and measurable function defined on [a, b]. If F(x) = f(t) dt + F(a), then F (x) a = f(x) a.e. in [a, b]”, we have

x d f f (x) a.e. dx n n a

32 LOVELY PROFESSIONAL UNIVERSITY Unit 3: Differentiation of an Integral

Notes Also, f – fn > 0 n, and therefore, the function Gn defined by

Gn (x) (f f n ) a is an increasing function of x, which must have a derivative almost everywhere by Lebesgue theorem and clearly, this derivative must be non-negative.

(f fn ) Since Gn (x) = a

x x

= f(t ) dt fn (t ) dt a a

x x

f(t ) dt = Gn (x) f n (t ) dt a a

Now the relation

x F (x) = f(t ) dt F(a) becomes a

F (x) = Gn (x) f n (t ) dt F(a) , a

F (x) = Gn (x) f n (x)a.e.

fn(x) a.e. n. since n is arbitrary, we have F (x) f(x) a.e.

b b F (x) dx f (x) dx … (1)

a a

Also by the Lebesgue’s theorem, i.e. “Let f be an increasing real-valued function defined on [a, b]. Then f is differentiable a.e. and the derivative f is measurable.

b and f (x) dx f (b) – f (a)”, we have

b F (x) dx F (b) – F (a) … (2)

LOVELY PROFESSIONAL UNIVERSITY 33 Measure Theory and Functional Analysis

Notes b But F (x) = f (t) dt F(a) a

b F(b) – F(a) = f (x) dx a

Therefore (2) becomes

b b F (x) dx f(x) dx … (3)

a a

From (1) and (3), we get

b b F (x) dx = f(x) dx

a a

b b F (x) dx f(x) dx = 0 a a

b F (x) f(x) dx = 0

since F (x) – f(x) 0 a.e., which gives that F (x) – f(x) = 0 a.e. and so F (x) = f(x) a.e.

3.2 Summary

 If f is an integrable function on [a, b] then f is integrable on any interval [a, x] [a, b]. The function F given by

x F (x) = f(t) dt c ,

where c is a constant, called the indefinite integral of F.

x  Let f be an integrable on [a, b]. If f(t) dt 0 x [a,b] then f = 0 a.e. in [a, b].

3.3 Keyword

Differentiation of an Integral: If f is an integrable function on [a, b] then f is integrable on any interval [a, x] [a, b]. The function F given by

x F (x) = f(t) dt c ,

where c is a constant, called the indefinite integral of f.

34 LOVELY PROFESSIONAL UNIVERSITY Unit 3: Differentiation of an Integral

3.4 Review Questions Notes

x 1. If f is an integrable function on [a, b] and if F (x) = f(t) dt F(a) then check whether

a F (x) = f (x) is absolute continuous function in [a, b] or not.

x 2. If F is an absolutely continuous function on [a, b], then prove that F (x) = f(t) dt C where

a f = F a.e. on [a, b] and C is constant.

3.5 Further Readings

Books Flanders, Harley. Differentiation under the Integral Sign Frederick S. Woods, Advanced Calculus, Ginn and Company David V. Widder, Advanced Calculus, Dover Publications Inc., New Edition (Jul 1990).

Online links www.physicsforums.com > Mathematics > Calculus & Analysis www.sp.phy.cam.ac.uk/~ alt 36/partial diff.pdf

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Notes Unit 4: Absolute Continuity

CONTENTS

Objectives

Introduction

4.1 Absolute Continuity

4.1.1 Absolute Continuous Function

4.1.2 Theorems and Solved Examples

4.2 Summary

4.3 Keywords

4.4 Review Questions

4.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define Absolute Continuous function.

 Solve problems on absolute continuity

 Understand the proofs of related theorems.

Introduction

It may happen that a continuous function f is differentiable almost everywhere on [0,1], its derivative f’ is Lebesgue integrable, and nevertheless the integral of f’ differs from the increment of f. For example, this happens for the Cantor function, which means that this function is not absolutely continuous. Absolute continuity of functions is a smoothness property which is stricter than continuity and uniform continuity.

4.1 Absolute Continuity

4.1.1 Absolute Continuous Function

A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if for an arbitrary 0, however small, a, 0, such that

n n f b f a r r whenever br a r , r 1 r 1

where a1 b 1 a 2 b 2 ... a n b n i.e. ai’s and bi’s are forming finite collection

ai ,b i : i 1,2,...,n of pair-wise disjoint intervals.

Obviously, every absolutely continuous function is continuous.

36 LOVELY PROFESSIONAL UNIVERSITY Unit 4: Absolute Continuity

Notes

1. If a function satisfied f br f a r , even then it is absolutely continuous.

2. The condition br a r means that total length of all the intervals must be less r 1

than .

4.1.2 Theorems and Solved Examples

Theorem 1: Every absolutely continuous function f defined on [a,b] is of bounded variation.

Proof: Since f is absolutely continuous on [a,b]; for 1, a 0 such that

f bi f a i 1, r 1

n whenever bi a i , r 1

and a a1 b 1 a 2 b 2 ... a n b n b.

Let the r-sub-intervals be c0 ,c 1 , c 1 ,c 2 ,..., c r 1 ,c r such that

a c0 ,c r b and c k+1 c k , k 0,1,2,..., r 1

Obviously, f xi 1 f x i 1, i

where xi 1 ,x i c k ,c k 1

ck 1 or V f 1, ck

c c c b 1 2 r Hence V f V f V f ... V f 1 1 ... 1 r finite quantity. a c0 c 1 c r 1

Hence f is of bounded variation.

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Notes

Notes Converse of above theorem is not necessarily true. There exists functions of bounded variation but not absolutely continuous.

Theorem 2: Let f(x) and g(x) be absolutely continuous functions, then prove that f x g x and

f x f x .g x are also absolutely continuous functions. Hence show that if g x 0, x is also g x

absolutely continuous function.

Proof: Given f x and g x are absolutely continuous functions on the closed interval [a,b],

therefore for each 0 , there exists 0 such that

f br f a r and r 1

g br g a r , r 1

whenever br a r , for all the points a1 ,b 1 ,a 2 ,b 2 ,...,a n ,b n such that r 1

a1 b 1 a 2 b 2 ... a n b n .

n n n

(i) We have, fbgbr r faga r r fbfa r r gbga. r r r 1 r 1 r 1

Now if br a r ,then r 1

n n

f br f a r and g b r g a r . r 12 r 1 2

f br g b r f a r g a r + = , r 1 2 2

whenever br a r . r 1

This show that f x g x are also absolutely continuous functions over [a,b].

38 LOVELY PROFESSIONAL UNIVERSITY Unit 4: Absolute Continuity

Notes n

(ii) We have f br g b r f a r g a r r 1

fbgbr r fbga r r fbga r r faga r r r 1

fbgbr r ga r ga r fb r fa r r 1

n n

fbgbr r ga r gafb r r fa r r 1 r 1

n n

fbgbr r ga r gafb r r fa. r r 1 r 1

Now every absolutely continuous function is bounded therefore f(x) and g(x) are bounded in the closed interval [a,b].

Let f x K1 , g x K 2 , x [a,b].

Then we have

fbgbfagar r r r K 1 K 2 K 1 K, 2 r 1

Whenever br a r . r 1

Setting K1 K 2 *,

We have f br g b r f a r g a r = *, r 1

Whenever br a r , r 1

where a1 b 1 a 2 b 2 ... a n b n ;

Product of two absolutely continuous functions is also absolutely continuous.

(iii) We have g x 0 x [a,b] ; therefore

g x ,where 0, x [a,b].

LOVELY PROFESSIONAL UNIVERSITY 39 Measure Theory and Functional Analysis

Notes n n 1 1 g ar g b r Now, = <2 , r 1g br g a r r 1 g b r g a r

Whenever br a r . Setting2 = *,we get r 1

n 1 1 < *. r 1 g br g a r

This show that 1 is absolutely continuous function over [a,b]. g x

Now f(x), 1 are absolutely continuous. g x

1 f x . is absolutely continuous. g x

f x is also absolutely continuous over [a,b]. g x

Hence the theorem is true.

Note By Theorem 1, its remark and above theorem it follows that set of all absolutely continuous functions on [a,b] is a proper subspace of the space BV [a,b] of all functions of bounded variation on [a,b].

Theorem 3: If BV[a,b], then f is absolutely continuous on [a,b], iff the variation function

x v x V f is absolutely continuous on [a,b]. a

Proof: Case I: Given v(x) is absolutely continuous.

For arbitrary >0, 0 s.t.

n n

v br v a r < ,whenever b r a r . r 1 r 1

x Also, we know that f x f a V f v x a

40 LOVELY PROFESSIONAL UNIVERSITY Unit 4: Absolute Continuity

Notes n n

f br f a r = f b r f a f a f a r r 1 r 1

f br f a f a f a r r 1

Now taking supremum over all collections of Pi of [ai, bi] for i = 2,...,n, we get

n bi V f . ai r 1

bi a i b i But V f V f V f a a ai

bi b i a i V f V f V f ai a a

bi V f v bi v a i ai

v bi v a i < i 1

v x is absolutely continuous.

Theorem 4: A necessary and sufficient condition that a function should be an indefinite integral is that it should be absolutely continuous.

Proof: Condition is sufficient.

Let f(x) be an absolutely continuous function over the closed interval [a,b].

Therefore f is of bounded variation and hence we can express f(x) as

f(x) = f1(x) – f2(x) where f1(x) and f2(x) are monotonically increasing functions and hence both are differentiable.

n n

v br v a r < whenever b r a r r 1 r 1

f is also absolutely continuous on [a,b]. Case II: Given f is absolutely continuous on [a,b].

for a given 0, a 0 s.t.

f bi f a i < , ...(i) i 1

LOVELY PROFESSIONAL UNIVERSITY 41 Measure Theory and Functional Analysis

Notes for every finite collection P ai ,b i , i 1,2,...,n of pairwise disjoint sub-intervals of [a,b] such

that bi a i . i 1

i i Now, let Pi x k 1 ,bx ki ,k 1,...,m i be a finite collection of non-overlapping intervals of the

interval [ai,bi].

i i Then the collection xk1 , bx ki : i 1,2,..., n , k 1,..., m i is a finite collection of non-overlapping sub-intervals of [a,b] such that

m ni n i i xk x k 1 b i a i . i 1 k 1 i 1

m n i i i and hence by (i), f xk f x k 1 . i 1 k 1

'' Hence f’(x) exists and f' x f1 x f 2 x

fx fbfbfafa1 2 1 2 , a

f x is integrable also.

Now let F(x) be an definite integral of f’(x) i.e.

x F(x) = F(a) + f t dt, x [a,b] ...(ii)

Using fundamental theorem of integral calculus,

We get

F’(x) = f’(x)

or F(x) = f(x) + constant (say c) ...(iii)

From (ii), we have F(a) = f(a),

Using this in (iii), we get c = 0 and hence F (x) = f(x).

Thus every absolutely continuous function f(x) is an indefinite integral of its own derivative.

Condition is necessary: Let f(x) be an indefinite integral of f(x) defined on the closed interval [a,b], so that

x Fx ftdt fa, x [a,b] and f(x) is integrable over [a,b]. a

Corresponding to arbitrary small 0, let >0 be such that if m(A) ,then f , A

42 LOVELY PROFESSIONAL UNIVERSITY Unit 4: Absolute Continuity

Now select 2n real numbers such that Notes

ai b i a 2 b 2 a 3 ... a n b n

n n such that A = U ai ,b i and b i a i . i 1 i 1

n n bi a i

Then F bi F a i f f i 1 i 1 a a

nbi n b i F f f . i 1 i 1 ai ai A

Thus, we have shown that for arbitrary small 0, a 0 s.t. bi a i . i 1

F bi F a i . i 1

F is absolutely continuous.

Thus every indefinite integral is absolutely continuous.

Theorem 5: If a function f is absolutely continuous in an interval [a,b] and if f’(x) = 0. a.e. in [a,b], then f is constant.

Proof: Let c [a, b] be arbitrary. If we show that f(c) = f(a), then the theorem will be proved.

Let E = x a,c : f '(x) 0 . since c is arbitrary, therefore set E a,c . This implies any x E f' x 0.

Let , > 0 arbitrary. Now f' x 0, x E an arbitrary small interval x,x h a,c

f x h f x such that f x h f x h. h

This implies that corresponding to every x E, an arbitrary small closed interval x,x h contained in [a,c] s.t.

f x h f x h.

Thus the interval x,x h , x E, over E in Vitali’s sense. Thus by Vitali’s Lemma, we can determine a finite number of non-overlapping intervals Ik, where

Ik x k ,y k k 1,2,3,...,n such that this collection covers all of E except for a set of measure less than 0 where is pre- assigned number which corresponds to occurring in the definition of absolute continuity of f.

LOVELY PROFESSIONAL UNIVERSITY 43 Measure Theory and Functional Analysis

Notes Suppose xk x k 1 ; then adjoining the points y0, xn+1.

We have a y0 x 1 y 1 x 2 y 2 ... x n y n x n 1 c.

Now since f is absolutely continuous, therefore for above subdivision of [a,c], we have

n n

f xk 1 f y k , whenever x k 1 y k . k 0 k 0

n n

(i) fyk fx k n yx k k nca. k 1 k 1

n n

Now fcfa fxk 1 fy k fyfx k k k 0 k 1

n n

f xk 1 f y k f y k f x k k 0 k 1

n c a

But ,n and hence n c a are arbitrary small positive numbers. So letting 0,n 0

We get f(c) = f(a)

f x is a constant function.

Corollary: If the derivatives of two absolutely continuous functions are equivalent, then the functions differ by a constant.

Proof: Let f and g be two absolutely continuous functions and f’ = g’ f g ' 0 by above theorem f – g = constant and hence the result.

Example: If f is an absolutely continuous monotone function on [a,b] and E a set of measure zero, then show that f (E) has measure zero.

Proof: Let the function f be monotonically increasing. By the definition of absolute continuity of

f, for 0, 0 and non-overlapping intervals In a n ,b n such that

bn a n f b n f a n

or f bn f a n

Now, E [a,b] E In

f E f In  f I n

m*fE m*fIn fx n fx n ,

44 LOVELY PROFESSIONAL UNIVERSITY Unit 4: Absolute Continuity

Notes f x where n and f xn are the maximum and maximum values of f(x) in the interval [an,bn].

Also note that xn xn b n a n

m * f E , being arbitrary.

m * f E 0 m f E 0.

Example: Give an example which is continuous but not absolutely continuous.

Solution: Consider the function f : F R, where F is the Cantor’s ternary set.

x Let x F x x x x ...k ,x 0 or 2 1 2 3k k K 1 3

r 1 Define f xk , where r x . k k k K 1 2 2

= 0. r 1, r2, r3..... This function is continuous but not absolutely continuous.

(i) Note that this function is constant on each interval contained in the complement of the Cantor’s ternary set.

For, let (a,b) be one of the countable open intervals contained in F c. Then in ternary notation,

a = 0.a1a2...an–1 0 2 2 2

and b = 0.a1a2...an-1 2 0 0 0,

where ai = 0 or 2, for i n 1.

a f a 0.r ,r ,...,r 0 1 1 1 1 ...,where ri , 1 2 n 1 i 2

f b 0.r1 ,r 2 ,...,r n 1 1 0 0 0 0 ...

But in binary notation

0.r1 ,r 2 ,...,r n 1 0 1 1 1 1 ... 0.r 1 ,r 2 ,...,r n 1 1 0 0 0 0 ...

f a f b .

Thus, we extend the function f overall of the set [0,1] instead of F by defining fx fb,x a,b F.c Thus, the Cantor’s function is defined over [0,1] and maps it onto [0,1].

It is clearly a non-decreasing function.

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Notes (ii) To show that f(x) is a continuous function. Note that if c',c'' F, then we have

c' 0. 2p1 2p 2 2p 3 ... each pi ,q i 0 or =1 c'' 0. 2q1 2q 2 2q 3 ...

1 If c' c'' , then p = q , for 1 i n 1 and hence 3n i i

1 f c' f c'' n ...(i) 2

as n ,c' c'',f c' f c'' ,

Hence if c0 F and c n is a sequence in F such that cn c 0 ,when n , then

f cn f c 0 ,when n .

Now let x0 [0,1] and let xn be a sequence in [0,1] such that xn x 0 as n .

c Case I: Let x0 F x 0 I,say a,b F

xn I and hence f x n f x f a

and hence f xn f x 0 as n .

Case II: Let x0 F. Now for each n such that xn F,set x n c n and hence f x n f x 0 .

c If xn F,then an open interval I F .

(i) if xn x 0 , then set cn as the upper end point of I.

(ii) If x0 x n , then set cn as the lower end point of I.

in any case f xn f x 0 as n .

But the sequence xn was any sequence satisfying the stated conditions.

f is a continuous function.

(iii) To show f(x) is not absolutely continuous. Note that f’(x) = 0 at each x Fc .

f' x exists and is zero on [0,1] and is summable on [0,1].

We know that for f(x) to be absolutely continuous, we must have

x f x f' x dx f 0 .

Particularly, we must have

1 f 1 f 0 f' x dx.

46 LOVELY PROFESSIONAL UNIVERSITY Unit 4: Absolute Continuity

x Notes But f(1) – f(0) = 1 and f' 1 dx 0 as f' x 0 0

x f 1 f 0 f' 1 dx 0 as f' x 0

f x is not absolutely continuous.

Theorem 6: Prove that an absolutely continuous function on [a, b] is an indefinite integral.

Proof: Let f(x) be an absolutely continuous function in a closed interval [a, b] so that f (x) &

x f (t) dt exists finitely x [a, b].

Let F(x) be an indefinite integral of f (x), so that

x F(x) = f(a) f (t) dt, x [a,b] … (1) a

We shall prove that F(x) = f(x).

Since an indefinite integral is an absolutely continuous function.

Therefore F(x) is absolutely continuous in [a, b].

Then from (1),

F (x) = f (x) a.e.

d [F(x) f(x)] = 0. dx

Integrating, we get

F(x) – f(x) = c (constant) … (2)

Taking x = a in (1), we get

a F(a) = f(a) f (t) dt a

F(a) – f(a) = 0 or F(x) – f(x) = 0 for x = a

Then from (2), we get c = 0.

Thus (2) reduces to

F(x) – f(x) = 0 a.e.

F(x) = f(x) a.e. which shows that f(x) is indefinite integral of its own derivative.

LOVELY PROFESSIONAL UNIVERSITY 47 Measure Theory and Functional Analysis

Notes 4.2 Summary

 A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if for an arbitrary 0, however small, a, 0, such that

n n

f br f a r , whenever b r a r . r 1 r 1  Every absolutely continuous function is continuous.

 Every absolutely continuous function f defined on [a,b] is of bounded variation.

4.3 Keywords

Absolute Continuity of Functions: Absolute continuity of functions is a smoothness property which is stricter than continuity and uniform continuity.

Absolute Continuous Function: A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if for an arbitrary 0, however small, a, 0, such that

n n f b f a r r whenever br a r , r 1 r 1

where a1 b 1 a 2 b 2 ... a n b n i.e. ai’s and bi’s are forming finite collection

ai ,b i : i 1,2,...,n of pair-wise disjoint intervals.

4.4 Review Questions

1. Define absolute continuity for a real variable. Show that f(x) is an indefinite integral, if F is absolutely continuous.

2. If f,g: [0,1] R are absolutely continuous, prove that f + g and fg are also absolutely continuous.

3. Show that the set of all absolutely continuous functions on an interval I is a linear space.

4. If g is a non-decreasing absolutely continuous function on [a,b] and f is absolutely continuous on [g(a), g(b)], show that fog is also absolutely continuous on [a,b].

5. If f is absolutely continuous on [a,b] and f' x 0 for almost all x [a,b], show that f is non-decreasing on [a,b].

4.5 Further Readings

Books Krishna B Athreya, N Soumendra Lahiri, Measure Theory and Probability Theory, Springer (2006).

48 LOVELY PROFESSIONAL UNIVERSITY Unit 4: Absolute Continuity

Aole Nielsen, An Introduction to Integration and Measure Theory, Wiley-Interscience, Notes (1997).

N.L. Royden, Real Analysis (third ed.), Collier MacMillan, (1988).

Online links dl.acm.org

mrich.maths.org

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Notes Unit 5: Spaces, Hölder

CONTENTS

Objectives Introduction 5.1 Spaces, Hölder 5.1.1 Lp-spaces 5.1.2 Conjugate Numbers 5.1.3 Norm of an Element of Lp-space 5.1.4 Simple Version of Hölder's Inequality 5.1.5 Hölder's Inequality 5.1.6 Riesz-Hölder's Inequality 5.1.7 Riesz-Hölder's Inequality for 0 < p < 1 5.2 Summary 5.3 Keywords 5.4 Review Questions 5.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand Lp-spaces, conjugate numbers and norm of an element of Lp-space

 Understand the proof of Hölder’s inequality.

Introduction

In this unit, we discuss an important construction, which is extremely useful in virtually all branches of analysis. We shall study about Lp-spaces and Hölder’s inequality.

5.1 Spaces, Hölder

5.1.1 LP-Spaces

The class of all measurable functions f (x) is known as Lp-spaces over [a, b], if Lebesgue – integrable over [a, b] for each p exists, 0 < p < , i.e.

b |f|p dx , (p 0)

and is denoted by Lp [a, b].

50 LOVELY PROFESSIONAL UNIVERSITY Unit 5: Spaces, Hölder

Notes

Note The symbol Lp is used for such classes when limits of integration are known and mentioning of interval is not necessary.

5.1.2 Conjugate Numbers

1 1 Let p, q be any two n on-negative extended real numbers s.t. 1 , then p, q are called p q (mutually) conjugate numbers. Obviously, 2 is self-conjugate number. Also if p 2, then q 2. Further, if p = , then q = 1 1, are conjugate numbers.

Note Non-negativity p 1, q 1.

5.1.3 Norm of an Element of LP-space

p The p-norm of any f L [a, b], denoted by f p, is defined as

1 b p p f p = |f| , 0 < p < . a

Theorem 1: If f Lp [a, b] and g f, then g Lp [a, b]. Proof: Let be any positive real number. {x [a, b] : g (x) > } = {x [a, b] : < g (x) f (x)} ( g f) = {x [a, b] : f (x) > } Again f Lp [a, b] f is measurable over [a, b]. {x [a, b] : f (x) > } is a measurable set. {x [a, b] : g (x) > } is a measurable set. g is a measurable function over [a, b]

Again since g (x) f (x), x [a, b]

b b |g|p dx |f| p dx ( |f|p L [a, b])

a a

b p or |g| dx a

Thus |g|p L [a, b].

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Notes Thus we have proved that g is a measurable function over [a, b] such that |g|p L [a, b] Hence g Lp [a, b] Theorem 2: If f Lp [a, b], p > 1, then f L [a, b] Proof: f Lp [a, b] f is measurable over [a, b]

Let A1 = {x [a, b] : |f (x) 1}

and A2 = {x [a, b] : |f (x) 1}

Then [a, b] = A1 A2 and A1 A2 = Using countable additive property of the integrals, we have

b |f|dx = |f|dx |f|dx … (i)

a AA1 2

 Now |f (x)| 1, x A1

p |f| |f| on A1 as p > 1

p |f|dx |f| dx as f Lp [a, b] … (ii) A A1 1

Now |f (x) |<|, x A2 Using first mean value theorem, we get

|f|dx m (A2 ) = A finite quantity … (iii)

Combining (ii) and (iii) and making use of (i), we get

b |f|dx <

Thus f is a measurable function over [a, b], such that

b |f|dx <

|f| L [a, b] and hence f L [a, b]. Theorem 3: If f Lp [a, b], g Lp [a, b]; then f + g Lp [a, b] Proof: Since f, g Lp [a, b] f, g are measurable over [a, b] f + g is measurable over [a, b]

Let A1 = {x [a, b] : |f (x)| |g (x)|}

and A2 = {x [a, b] : |f (x)| < |g (x)|}

Then [a, b] = A1 A2 and A1 A2 =

52 LOVELY PROFESSIONAL UNIVERSITY Unit 5: Spaces, Hölder

Notes Therefore |f g|p dx |f g| p |f g| p dx .

AA1 2

p p p p Again, |f + g| (|f| + |g|) (|g| + |g|) on A2 and (|f| + |f|) on A1

p p p 2P |g| on A2 and 2 |f| on A1 Integrating, we have

|f g|p 2p |f| p

A1 A1 and |f g|p 2p |g| p

A2 A2

Since f, g Lp [a, b] |f|p and |g| p

AA1 2

 |f g|p and |f g| p [by (i) and (ii)]

AA1 1

b |f g|p dx f + g Lp [a, b]

5.1.4 Simple Version of Hölder's Inequality

1 1 Lemma 1: Let p, q > 1 be such that 1 , and let u and v be two non-negative numbers, at p q least one being non-zero. Then the function f : [0, 1] R defined by

1 f (t) = ut + v(1 tq ) q , t [0, 1], has a unique maximum point at

1 up q s = … (1) up v p

The maximum value of f is

1 max f(t) = (up v p ) p … (2) t [0, 1]

Proof: If v = 0, then f (t) = tu, t [0, 1] (with u > 0), and in this case, the Lemma is trivial.

Likewise, if u = 0, then

1 f (t) = v(1 tq ) q , t [0, 1] (with v > 0), … (3)

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Notes and using the inequality

1 (1 tq ) q 1, t (0, 1],

We immediately get f (t) < f (0), t (0, 1],

and the Lemma again follows. For the remainder of the proof we are going to assume that u, v > 0. Obviously f is differentiable on (0, 1) and the solutions of the equation (3) f (t) = 0 Let s be defined as in (1), so under the assumption that u, v > 0, we clearly have 0 < s < 1. We are going to prove first that s is the unique solution in (0, 1) of the equation (3). We have

1 1 1 f (t) = u v (1 tq ) q q t q 1 … (4) q

tq = u v , t (0, 1) 1 tq

so the equation (3) reads

tq u v = 0. 1 tq

Equivalently, we have

1 tq p = u/v, 1 tq

tq = (u/v)p, 1 tq

(u/v)p u p t2 = , 1 (u/v)p u p v p

Having shown that the “candidates” for the maximum point are 0, 1 and s let us show that s is the only maximum point. For this purpose, we go back to (4) and we observe that f is also continuous on (0, 1).

Since lim f (t) = u > 0 and t 0

lim f (t) = – t 1

and the equation (3) has exactly one solution in (0, 1), namely s, this forces

f (t) > 0 t (0, s)

54 LOVELY PROFESSIONAL UNIVERSITY Unit 5: Spaces, Hölder

and f (t) < 0 t (s, 1). Notes This means that, f is increasing on [0, s] and decreasing on [s, 1], and we are done. The maximum value of f is then given by

max f(t) = f (s), t [0, 1] and the fact that f (s) equals the value in (2) follows from an easy computation.

5.1.5 Hölder's Inequality

Statement: Let a1, a2, …, an, b1, b2, …, bn be non-negative numbers. Let p, q > 1 be real number 1 1 with the property 1 . Then p q

1 1 n np n q p q aj b j aj b j … (5) j 1 j 1 j 1

p p q q Moreover, one has equality only when the sequences a1 , , a n and b , , bn are proportional.

Proof: The proof will be carried on by induction on n. The case n = 1 is trivial. Case n = 2.

Assume (b1, b2) (0, 0). (otherwise everything is trivial). Define the number

b1 r = 1 . q q q b1 b 2

Notice that r [0, 1] and we have

1 b2 q q 1 = 1 r q q q b1 b 2

1 q q q Notice also that, upon dividing by b1 b 2 , the desired inequality

1 1 q qp q q q a1 b1 + a2 b2 a1 a 2 b 1 b 2 … (6) reads

1 1 q q q q p a1 r + a2 1 r a1 a 2 … (7)

It is obvious that this is an equality when a1 = a2 = 0. Assume (a1, a2) (0, 0), and set up the function.

1 q q f (t) = a1 t a 2 1 t , t [0, 1].

LOVELY PROFESSIONAL UNIVERSITY 55 Measure Theory and Functional Analysis

Notes We now apply Lemma (1) stated above, which immediately gives us (7). Let us examine when equality holds.

If a1 = a2 = 0, the equality obviously holds, and in this case (a1, a2) is clearly proportional to (b1,

b2). Assume (a1, a2) (0, 0). Again by Lemma (1), we know that equality holds in (7), exactly when

1 p q a1 r = p p a1 a 2

1 p q b1 a1 that is 1 = p p , q q q a1 a 2 b1 b 2 or equivalently

q p b1 a1 q q = p p . b1 b 2 a1 a 2

Obviously this forces

q q b1 a1 q q = p p , b1 b 2 a1 a 2

p p q q so indeed a1 , a 2 and b1 , b 2 are proportional.

Having proven the case n = 2, we now proceed with the proof of:

The implication: Case n = k case n = k + 1, start with two sequences (a1, a2, …, ak, ak+1) and

(b1, b2, …, ak, bk+1). Define the numbers

1 1 k p k q p q a = a j and b = bj . j 1 j 1

Using the assumption that the case n = k holds, we have

1 1 k 1 kp k q a b p q + a b j j aj b j k+1 k+1 j 1 j 1 j 1

= ab + ak+1 bk+1 … (8) Using the case n = 2, we also have

1 1 p pp q q q ab + ak+1 bk+1 a ak 1 b b k 1

1 1 k 1p k 1 q p q = aj b j , … (9) j 1 j 1

so combining with (8) we see that the desired inequality (5) holds for n = k + 1.

56 LOVELY PROFESSIONAL UNIVERSITY Unit 5: Spaces, Hölder

Assume now we have equality. Then we must have equality in both (8) and in (9). Notes

p p p q q q On one hand, the equality in (8) forces a1 , a 2 , , a k and b 1 , b 2 , , b k to be proportional (since

p p q q we assume the case n = k). On the other hand, the equality in (9) forces a , ak 1 and b , bk 1 to be proportional (by the case m = 2). Since

k k p p q q a aj and b b j , j 1 j 1

p p p p q q q q it is clear that a1 , a 2 , , a k , a k 1 and b1 , b 2 , , b k , b k 1 are proportional.

5.1.6 Riesz-Hölder's Inequality

Statement: Let p and q be conjugate indices or exponents (numbers) and f Lp [a, b], g Lq [a, b]; then show that

(i) f g L[a, b]

fg f g i.e. (ii) p q

1 1 p q |fg| |f|p |g| q with equality only when |f|p = |g|q a.e. for some non-zero constants and . Lemma: If A and B are any two non-negative real numbers and 0 < < 1, then A B1– A + (1 – ) B, with equality when A = B. Proof: If either A = 0 or B = 0, then the result is trivial. Let A > 0, B > 0 Consider the function (x) = x – x, where 0 x < and 0 < < 1

d d2 x 1 and ( 1)x 1 . dx dx2

d Now solving = 0, we get x = 1. dx

d2 Also at x = 1, < 0 as 0 < < 1. dx2

By calculus, (x) is maximum at x = 1, so (x) (1) i.e. x – x l – . … (1)

A Now, putting x = , we get B

AA A 1 or A B 1 BB B

LOVELY PROFESSIONAL UNIVERSITY 57 Measure Theory and Functional Analysis

Notes Or A B1– – A (1 – ) B or A B1– A + B (1 – ) … (2) Obviously equality holds good only for x = 1, i.e. only when A = B.

Proof of Theorem

Note that when p = 1, q = , the proof of theorem is obvious. Let us assume that 1 < p < and 1 < q < .

1 Now set = ; p > 1 < 1 p

1 1 1 Therefore = 1 –  1 q p q

Putting these values of and 1 – in (2), we get

1 1 AB ABp q … (3) p q

If one of the functions f (x) and g (x) is zero a.e. then the theorem is trivial. Thus, we assume that f 0, g 0 a.e. and hence the integrals

b b |f|p dx and |g| q dx

a a

are strictly positive and hence f p > 0, g q > 0.

f(x) g(x) Set f (x) = , g(x) f g p q

1 1 and A p = |f(x)|, Bq g(x) .

Then (3) gives

p q f(x) g(x) |f(x) g(x)| p q

Integrating, we get

b b b 1 1 |f(x) g(x)|dx |f(x)|p dx |g(x)| p dx p q a a a

b b 1 |f(x)|p 1 |g(x)|q = dx dx pb q b a|f|p dx a |g| q dx

a a

58 LOVELY PROFESSIONAL UNIVERSITY Unit 5: Spaces, Hölder

b b Notes |f|p dx |g| q dx 1 1 = a a pb q b |f|p dx |g| q dx

a a 1 1 = 1 . p q

b Hence |f(x) g(x)|dx 1.

Putting the values of f (x) and g (x), we get

b |f(x) g(x)|dx 1 or fg f g … (4) f g p q a p q

Now f Lp [a, b], g Lq [a, b]

b b |f|p dx and |g| q dx

a a

f and g p q

Therefore, from (4), we have

fg 1 < fg L [a, b] Also the equality will hold when A = B i.e. |f (x)|p = |g (x)|q, a.e.

|f|p |g|q i.e. if p = q , a.e. f g p q

q p or if g |f|p = f |g|q , a.e. q p or if we have got some non-zero constants , |f|p = |g|q, a.e. Hence the theorem.

5.1.7 Riesz-Hölder's Inequality for 0 < p < 1

If 0 < p < 1 and p and q are conjugate exponents, and f Lp and g Lq, then

q |fg| fp g q , provided |g| 0 .

(In this case, the inequality is reversed than that of the case for 1 p < .)

1 1 Proof: Conjugacy of p, q 1 p q

LOVELY PROFESSIONAL UNIVERSITY 59 Measure Theory and Functional Analysis

Notes 1 1 1 q p

p 1 p q

p p 1 q

1 p 1 1 1 1 1 If we take p and , then 1 and since 0 p 1 1 P 1 , P PQ p Q 0 P

1 p 1 i.e. 1< P < < and also 1 p 0 1 as 0 < p < 1 Q > 1. Q q Q

P, Q are conjugate numbers with 1 < P < . If we take |fg| = Fp and |g|q = GQ.

1 q 1 q p p Q q p Q |f|p |g| Then fg = |fg | |g| =

= |f|p. f, g are non-negative measurable functions s.t. Also f Lp and g Lq.

Applying the Hölder’s inequality for P, Q to the functions f and g, we get

|FG| F P G Q

1 1 PQ |f|p |F|PQ |G| as |fg| = fg = |f|p

p p q |f|p |fg| |g|q

1 1 p q |f|p |fg| |g|q

1 |fg| p p q |f| 1 , provided |g| 0 q |g|q

1 1 pp q q |fg| |f| |q| fp g q

60 LOVELY PROFESSIONAL UNIVERSITY Unit 5: Spaces, Hölder

Theorem 4: SCHWARZ or CAUCHY-SCHWARZ INEQUALITY statement: Let f and g be square Notes integrable, i.e.

f, g 2 L [a, b]; then fg L [a, b] and fg f 2 g 2. Proof: Let x [a, b] be arbitrary, then [|f(x)| – |g(x)|]2 0 or 2|f(x)|.|g(x)| |f(x)2 + |g(x)|2. On integrating, we get

b b b 2 2 2 |f(x)g(x)|dx |f(x) dx |g(x)| dx … (i)

a a a

b b Now f, g L2[a, b] f and g are measurable over [a, b] and |f(x)|2 dx , |g(x)| 2 dx .

a a

b Using in (i), we get |f(x) g(x)|dx

Thus fg L [a, b]. Let a R be arbitrary. Then ( |f| + |g|)2 0

b ( |f| |g|)2 0

b b b or 2|f| 2 dx 2 |fg|dx |g| 2 dx 0

a a a

b b b Write A = |f|2 dx, B 2 |fg|dx, C |g| 2 dx

a a a

Then we have 2A + B + C 0 … (ii) Now, if A = 0, then f(x) = 0 a.e. in [a, b] and hence B = 0 and both sides of the inequality to be proved are zero. Thus when A = 0, the inequality is trivial.

B Again, let A 0. Writing = – in (ii), we get 2A

BB2 A B C 0 . 2A 2A which gives B2 4AC.

LOVELY PROFESSIONAL UNIVERSITY 61 Measure Theory and Functional Analysis

Notes Now putting the values of A, B, C in last inequality, we have

2 b b b 4 |fg|dx 4 |f|2 |g| 2

a a a

b b1/2 b 1/2 or |f(x) (g(x)|dx |f(x)|2 |g(x)| 2

a a a

or fg f 2 g 2. Note: The above theorem is a particular case of Hölder’s inequality.

Example: Let f, g be square integrable in the Lebesgue sense then prove f + g is also

square integrable in the Lebesgue sense, and f + g 2 f 2 + g 2. Solution: By hypothesis f2 L [a, b], g2 L [a, b]. f2, g2 L [a, b] fg L [a, b]. [by Schwarz inequality] Again (f + g)2 = f2 + g2 + 2fg L [a, b]. Hence (f + g) is square integrable, again, we have

b b b b 2 (f g) = f2 g 2 2 fg a a a a

b b b1 /2 b 1 /2 f2 g 2 2 f 2 g 2 (by Schwarz inequality)

a a a a

2 b1 /2 b 1 /2 = f2 g 2 a a

b 1 /2 b1 /2 b 1 /2 2 (f g) f2 g 2 a a a

or f + g 2 f 2 + g 2.

Example: Prove that f + g 1 f 1 + g 1. Solution: We know that |f + g| |f| + |g|. Integrating both the sides.

|f g| |f| |g|

f + g 1 f 1 + g 1.

62 LOVELY PROFESSIONAL UNIVERSITY Unit 5: Spaces, Hölder

5.2 Summary Notes

 The class of all measurable functions f (x) is known as L p – space over [a, b], if Lebesgue- integrable over [a, b] for each p exists, 0 < p < , i.e.

b |f|p dx , (p 0)

 p The p-norm of any f L [a, b], denoted by f p, is defined as

1 b p f |f|p , 0 p p a

1 1  Let p, q > 1 be such that 1 , and let u and v be two non-negative numbers, at least p q one being non-zero. Then the function f : [0, 1] R defined by

1 f(t) ut v 1 tq q , t [0, 1],

has a unique maximum point at

1 up q s up v p

 Let p and q be conjugate indices or exponents and f Lp [a, b], g Lq [a, b], then it is evident that (i) f, g L [a, b]

(ii) fg f p g q i.e.

1 1 p q |fg| |f|p |g| q

5.3 Keywords

1 1 Conjugate Numbers: Let p, q be any two n on-negative extended real numbers s.t. 1 , then p q p, q are called (mutually) conjugate numbers.

Hölder's Inequality: Let a1, a2, …, an, b1, b2, …, bn be non-negative numbers. Let p, q > 1 be real 1 1 number with the property 1 . Then p q

1 1 n np n q p q aj b j aj b j j 1 j 1 j 1

p p q q Moreover, one has equality only when the sequences a1 , , a n and b , , bn are proportional.

LOVELY PROFESSIONAL UNIVERSITY 63 Measure Theory and Functional Analysis

Notes LP-Spaces: The class of all measurable functions f (x) is known as Lp-spaces over [a, b], if Lebesgue – integrable over [a, b] for each p exists, 0 < p < , i.e.

b |f|p dx , (p 0)

and is denoted by Lp [a, b].

p p-norm: The p-norm of any f L [a, b], denoted by f p, is defined as

1 b p p f p = |f| , 0 < p < . a

5.4 Review Questions

1. If f and g are non-negative measurable functions, then show that in Hölder’s inequality, equality occurs iff some constants s and t (not both zero) such that sfp + tgq = 0. 2. State and prove Hölder’s Inequality.

5.5 Further Readings

Books G.H. Hardy, J.E. Littlewood, G. Polya, Inequalities, Cambridge University Press, (1934) L.P. Kuptsov, Hölder inequality, Springer (2001) Kenneth Kuttler, An Introduction of Linear Algebra, BRIGHAM Young University, 2007

Online links www.m–hiKari.com www.math.Ksu.edu www.tandfonline.com

64 LOVELY PROFESSIONAL UNIVERSITY Unit 6: Minkowski Inequalities

Unit 6: Minkowski Inequalities Notes

CONTENTS

Objectives Introduction 6.1 Minkowski Inequalities 6.1.1 Proof of Minkowski Inequality Theorems 6.1.2 Minkowski Inequality in Integral Form 6.2 Summary 6.3 Keywords 6.4 Review Questions 6.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define Lp-space, conjugate numbers and norm of an element of Lp-space.

 Understand Minkowski inequality.

 Solve problems on Minkowski inequality.

Introduction

In mathematical analysis, the Minkowski inequality establishes that the L p spaces are normed vector spaces. Let S be a measure space, let 1 p and let f and g be elements of Lp (s). Then f + g is in Lp (s), we have the triangle inequality

f + g p f p + g p with equality for 1 < p < if and only if f and g are positively linearly dependent, i.e. f = g for some 0. In this unit, we shall study Minkowski’s inequality for 1 p < and for 0 < p < 1. We shall also study almost Minkowski’s inequality in integral form.

6.1 Minkowski Inequalities

Here, the norm is given by:

1/p p f p = |f| d if p < , or in the case p = by the essential supremum

f = ess supx S |f (x)|. The Minkowski inequality is the triangle inequality in L p(S). In fact, it is a special case of the more general fact

LOVELY PROFESSIONAL UNIVERSITY 65 Measure Theory and Functional Analysis

Notes

f p = sup |fg|d , 1/p + 1/q = 1 g 1 q

where it is easy to see that the right-hand side satisfies the triangular inequality. Like Holder's inequality, the Minkowski inequality can be specialized to sequences and vectors by using the counting measure:

n1/p n 1/p n 1/p p p p |xk y k | |x k | |y k | k 1 k 1 k 1

for all real (or complex) numbers x1, ..., xn, y1, ..., yn and where n is the cardinality of S (the number of elements in S). Thus, we may conclude that If p > 1, then Minkowski's integral inequality states that

b1 /p b 1 /p b 1 /p |f(x) g(x)|p dx |f(x)| p dx |g(x)| p dx a a a

Similarly, if p >1 and ak, bk > 0, then Minkowski's sum inequality states that

n1/p n 1/p n 1/p p p p |ak b k | |a k | |b k | k 1 k 1 k 1

Equality holds iff the sequences a1, a2, ... and b1, b2, ... are proportional.

6.1.1 Proof of Minkowski Inequality Theorems

Theorem 1: State and prove Minkowski inequality. If f and g Lp (1 p < ), then f + g Lp and

f + g p f p + g p. or Let 1 p . Prove that for every pair f, g Lp {0, 1}, the function f + g Lp {0, 1} and that

f + g p f p + g p. When does equality occur? Or Suppose 1 p . Prove that for any two functions f and g in Lp [a, b]

1 1 1 bp b p b p |f g|p |f| p dx |g| p dx

a a a

Proof: When p = 1, the desired result is obvious. If p = , then |f| f a.e. |g| g a.e.

66 LOVELY PROFESSIONAL UNIVERSITY Unit 6: Minkowski Inequalities

|f + g| f + |g| Notes f + g a.e. f + g f + g Hence the result follows in this case also. Thus, we now assume that 1 < p < . Since Lp is a linear space, f + g Lp.

1 1 Let q be conjugate to p, then 1 . p q

Now (f + g) Lp (f + g)p/q Lq

1 1 1 1 1 p 1 1 1 Since p q q p q p

q p 1q p (p 1) p, |f g| |f g|

p and therefore |f + g|p–1 Lp (f + g)p/q Lp because p – 1 = . q

On applying Hölder’s inequality for f and (f + g)p/q, we get

1 p 1 p p q p |f||f g|q dx |f|p ) dx |f g| q dx

1 1 p p q or |f||f g|q dx |f|p ) dx |f g| p dx … (1)

Since g Lp, therefore interchanging f and g in (1), we get

1 1 p p q |g||f g|q dx |g|p ) dx |f g| p dx … (2)

Adding, we get

1 1 1 p p p p q |f||f g|q dx |g||f g| q dx |f|p dx |g| p dx |f g| p dx … (3)

p 1 Now |f + g|p = f g f g

1 1 p p But 1 1 p p 1 p q q q

p |f + g|p = f g f g q

p |f| |g| |f g| q

LOVELY PROFESSIONAL UNIVERSITY 67 Measure Theory and Functional Analysis

Notes p p or |f + g|p |f||f g|q |g||f g| q

Integrating, we get

p p p |f g| dx |f||f g|q |g||f g| q dx … (4)

Using (3), relation (4) becomes

1 1 p p p 1 |f g| dx |f|p dx |g| p dx |f g| p dx q

1 Dividing each term by |f g|p dx q , we get

1 1 1 1 q p p |f g|p dx |f|p dx |g| p dx

1 1 1 1 But 1 1 p q q p

1 1 1 p p p So |f g|p dx |f|p dx |g| p dx

or f + g p f p + g p Hence the proof.

Note Equality hold in Minkowski’s inequality if and only if one of the functions f and g is a multiple of the other. Theorem 2: Minkowski’s inequality for 0 < p < 1. If 0 < p < 1 and f, g are non-negative functions in Lp, then

f + g p f p + g p. Proof: For this proceed as in theorem Minkowski’s inequality and applying the Hölder’s inequality for 0 < p < 1 for the functions f Lp and (f + g)p/q Lq, we get

1 p 1 q |f||f g|p/q |f|p |f g| p/q ) q

1 p 1 q |f||f g|p/q |f|p |f g| p … (i)

Also g Lp, proceeding as above, we get

1 p 1 q p p |g||f g|p/q |g| dx |f g| … (ii)

68 LOVELY PROFESSIONAL UNIVERSITY Unit 6: Minkowski Inequalities

Adding these two, Notes

1 1 1 p/q |f g| (|f| |g|) |f|pp |g| p p |f g| p q … (iii)

1 1 p Also 1 1 p p q q

p p |f + g|p = |f g||f g|q |f| |g| |f g| q , as f 0, g 0

1 p 1 1 q |f g| = |f|pp |g| p p |f g| p

1 q Dividing by |f g|p , we get

1 (1/q) |f g|p f g p p

1/q p |f g| f p + g p

f + g p f p + g p

6.1.2 Minkowski Inequality in Integral Form

Statement: Suppose f :  ×   is Lebesgue measurable and 1 p < . Then

1/p 1/p h(x, y) dy dx h(x, y)p dx dy

Proof: By an approximation argument we need only consider h of the form

h(x,y) f(x)1f(y),(x,y)j j   , j 1

where N is a positive integer, fj is Lebesgue measurable, and Fj Ln, j = 1, … Ni and Fi Fj = if 1 i < j N. We use Minkowski’s inequality to estimate

p 1/p 1/p NN 1/p p h(x,y) dy dx = Fj f j (x) F j f i (x) dx j 1 j 1 dx

But

p 1/p NN1/p 1/p p p h(x,y) dx dy = |h(x,y)| dx |fj (x)| dx FF j 1i j 1 j

LOVELY PROFESSIONAL UNIVERSITY 69 Measure Theory and Functional Analysis

Notes 2 2 Example: If is a sequence of functions belonging to L (a, b) and also f L (a, b) and

Lim fn – f 2 = 0, then prove that

b b 2 2 f dx Lim fn dx a a

Solution: By Minkowski’s inequality, we get

fn 2 f 2 fn – f 2

Lim f f n 2 Lim fn – f 2 = 0

Lim f f n 2 = 0 Lim fn 2 = f 2

1/2 1/2 b b b b 2 2 2 2 Lim (fn ) dx f dx Lim fn dx f dx . a a a a

6.2 Summary

 The class of all measurable function f (x) is known as L p space over [a, b], if Lebesgue integrable over [a, b] for each p exists, 0 < p < .

 p p If f and g L (1 p ), then f + g L and f + g p f p + g p.

6.3 Keywords

Lp-space: The class of all measurable functions f (x) is known as Lp-space over [a, b], if Lebesgue- integrable over [a, b] for each exists, 0 < p < , i.e.,

b |f|p dx , (p 0)

and is denoted by Lp [a, b]. Minkowski Inequality in Integral Form: Suppose f :  ×   is Lebesgue measurable and 1 p < . Then

1/p 1/p h(x, y) dy dx h(x, y)p dx dy

Minkowski Inequality: Minkowski inequality establishes that the Lp spaces are normed vector spaces. Let S be a measure space, let 1 p and let f and g be elements of Lp (s). Then f + g is in Lp (s), we have the triangle inequality

f + g p f p + g p with equality for 1 < p < if and only if f and g are positively linearly dependent.

70 LOVELY PROFESSIONAL UNIVERSITY Unit 6: Minkowski Inequalities

6.4 Review Questions Notes

1. If f, g are square integrable in the Lebesgue sense, prove that f + g is also square integrable and

f + g 2 f 2 + g 2. 2. If | < p < , then show that equality can be true, iff there are non-negative constants and , such that f = g.

6.5 Further Readings

Books Books: Stein, Elias (1970). Singular Integrals and Differentiability Properties of Functions. Princeton University Press. Hardy, G.H.; Littlewood, J.E.; Polya, G. (1952). Inequalities, Cambridge Mathematical Library (second ed.). Cambridge: Cambridge University Press.

Online links Mathworld.wolfram.com>Calculus and Analysis>Inequalities Planet math.org/Minkowski In-equality.html

LOVELY PROFESSIONAL UNIVERSITY 71 Measure Theory and Functional Analysis

Notes Unit 7: Convergence and Completeness

CONTENTS

Objectives Introduction 7.1 Convergence and Completeness 7.1.1 Convergent Sequence 7.1.2 Cauchy Sequence 7.1.3 Complete Normed Linear Space 7.1.4 Banach Space 7.1.5 Summable Series 7.1.6 Riesz-Fischer Theorem 7.2 Summary 7.3 Keywords 7.4 Review Questions 7.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand convergence and completeness.

 Understand Riesz-Fischer theorem.

 Solve problems on convergence and completeness.

Introduction

Convergence of a sequence of functions can be defined in various ways, and there are situations in which each of these definitions is natural and useful. In this unit, we shall start with the definition of convergence and Cauchy sequence and proceed with the topic completeness of Lp.

7.1 Convergence and Completeness

7.1.1 Convergent Sequence

. Definition: A sequence in a normal linear space X with norm is said to converge to an

element x X if for arbitrary > 0, however small, n0 N such that xn – x < , n > n0.

Then we write lim xn x . n

72 LOVELY PROFESSIONAL UNIVERSITY Unit 7: Convergence and Completeness

7.1.2 Cauchy Sequence Notes

. Definition: A sequence in a normal linear space (X, ) is said to be a Cauchy sequence if for arbitrary > 0, n0 N s.t.

xn – xm < , n, m n0.

7.1.3 Complete Normed Linear Space

. Definition: A normed linear space (X, ) is said to be complete if every Cauchy sequence in it converges to an element x X.

7.1.4 Banach Space

Definition: A complete normed linear space is also called Banach space.

7.1.5 Summable Series

Definition: A series un in N is said to be summable to a sum u if u N and lim Sn u , where 1 1 n n 1

Sn = u1 + u2 + … + un In this case, we write

u un . n 1

Further, the series un is said to be absolutely summable if un . n 1 n 1

7.1.6 Riesz-Fischer Theorem

Theorem: The normed Lp-spaces are complete for (p 1). Proof: In order to prove the theorem, we shall show that every Cauchy sequence in Lp [a, b] space p p converges to some element f in L -space. Let be one of such sequences in L -space. Then for given > 0, a natural number n0, such that

m, n n0 fm – fn p < ,

1 since is arbitrary therefore taking , we can find a natural number n such that 2 1

1 for all m, n n f – f < 1 m n p 2

1 Similarly, taking , k N, we can find a natural number n , such that 2k k

1 for all m, n n f – f < k m n p 2k

LOVELY PROFESSIONAL UNIVERSITY 73 Measure Theory and Functional Analysis

Notes 1 In particular, m > n f – f < . k m nk 2k

Obviously n1 < n2 < n3 … < nk < …

i.e. is a monotonic increasing sequence of natural numbers.

Set gk = fnk , then from above, we have

1 g – g = fn f n , 2 1 p 2 2 p 2

1 g3 – g2 p = fn f n 2 , 3 2 p 2

… … … … … … … … … …

1 gk + 1 – gk p = f f . nk 1 n k p 2k

… … … … … … … … … … Adding these inequalities, we get

1 g g < 1 … (i) k 1 k p k k 1 k 1 2

Thus g g is convergent. Define g such that k 1 k p k 1

g (x) = g (x) g g if R.H.S. is convergent … (ii) 1k 1 k p k 1

and g (x) = , if right hand side is divergent.

1 b b n p |g(x)|p dx p Now, = lim |g1 (x) g g k n k 1 dx a a k 1

or g = lim g1 g g k (By Minkowski’s inequality) p n p k 1 p k 1

g gk = g1 p + k 1 p < g1 p + 1, [by (i)] k 1

p g p < g L [a, b]. Let E = {x [a, b] : g (x) = }.

74 LOVELY PROFESSIONAL UNIVERSITY Unit 7: Convergence and Completeness

Now we define a function f such that Notes

f (x) = 0, x E

and f (x) = g1 (x) + gk 1 gk , for x [a, b] but x E, k 1

m 1 or f (x) = lim g1 g g k , for x E m k 1 k 1

= lim gm (x) m

Thus f (x) = 0, for x E and

f (x) = lim gm (x) for x E. m

f (x) = lim gm (x) a.e. in [a, b] m

or lim gm f = 0 a.e. in [a, b] … (iii) m

m 1

Also, gm (x) = g1 gk 1 g k k 1

m 1

|gm| |g1| + gk 1 gk k 1

|g1| + gk 1 gk = g, k 1

|gm| g, m N

lim gm (x) g m

(iii) |f| g.

Again, |gm – f| |gm| + |f| g + g = 2g.

|gm – f| 2 g. Thus there exists a function g Lp [a, b] s.t.

|gm – f| 2g, m

and lim gm f = 0 a.e. in [a, b] … (iv) m

Applying Lebesgue dominated convergence theorem,

b b b p p lim gm f dx = lim gm f dx 0 dx 0 [Using (iv)] m m a a a

LOVELY PROFESSIONAL UNIVERSITY 75 Measure Theory and Functional Analysis

Notes 1 b p p = 0 lim gm f dx m a

lim gm f = 0 m p

lim fn f = 0 as g = fn m m p m m

f f < . nm p

Also f f < . m nm p

f f f f fm – f p = m nm n m p

f f f f m nmp n m p

< ( + ) = .

Hence lim fm f = 0 m p

p or lim fm = f L [a, b]. m

This proves the theorem.

Alternative Statement of this Theorem

p p A convergent sequence in L -spaces has a limit in L -space. Or

p p Every Cauchy sequence in the L -space converges to a function in L -space. Theorem: Prove that a normed linear space is complete iff every absolutely summable sequence is summable. Proof: Necessary part . Let X be a complete normed linear space with norm and be an absolutely summable sequence of elements of X

fn = M < , n 1

For arbitrary > 0, however small, no N

s.t. fn < , … (i) n no

n f Now, if Sn = i , then n m no, we get i 1

76 LOVELY PROFESSIONAL UNIVERSITY Unit 7: Convergence and Completeness

Notes n n f Sn – Sm p = i fi i m 1 i m 1

fi i no

Sequence of partial sums is a Cauchy sequence

converges.

Sequence is summable to some element S X.

But X is a complete space. Therefore will converge to some element S X. Sufficient part: Given that every absolutely summable sequence in the space X is summable. To show that X is complete.

Let be a Cauchy sequence in X.

For each positive integer k, we can choose a number nk N such that

1 f – f < , n, m n … (ii) n m 2k k

We can choose these nk’s such that nk+1 > nk.

Then f is a subsequence of . nk 1 n

th Setting g1 = fn1 and gk = fnk f n k 1 , (k > 1), we get a sequence s.t. its k partial.

Sum = Sk = g1 + g2 + … + gk = fn1 (f n 2 f n 1 ) (f n k f n k 1 ) f n k .

1 Now, g = f f , [by (ii)], k > 1 k nk n k 1 2k 1

1 gk g 1 k 1 = g1 + 1 (a finite quantity) k 1 k 2 2

The sequence is absolutely summable and hence by the hypothesis, it is a summable sequence. The sequence of partial sums of this sequence converges to some S X.

The sequence converges and hence fnk converges to some f X.

Now, we shall show that the limit fn = f.

Again, since is a Cauchy sequence, we get that for each > 0, however small, n N s.t. n, m > n .

f – f < . n m 2

Also since fnk f, n N such that k n ,

LOVELY PROFESSIONAL UNIVERSITY 77 Measure Theory and Functional Analysis

Notes f f . nk 2

Choosing a number k, as large that k > n and nk > n , we get

f f f f f f , n n nk n k 2 2

n > n , we obtain fn – f < , where is an arbitrary quantity.

fn f X and hence X is a complete space.

p p Theorem: Let {fn} be a sequence in L , 1 p < , such that fn f a.e. and that f L .

If lim fn = f , then n p p

lim fn f = 0. n p

Proof: Without any loss of generality, we may assume that each fn 0 a.e. so that f is also 0 a.e. since the result in general case follows by considering f = f + – f–. Now, let a and b be any pair of non-negative real numbers, we have |a – b|p 2p (|a|p + |b|q), 1 p < So, we get

p p p p 2 (|fn| + (|f| ) – |fn – f| 0 a.e. Thus, by Fatou’s Lemma and by the given hypothesis, We get

p p p 1 p p p 2 |f| = lim 2 fn f f n f n

p pp p lim inf 2 fn f f n f n

p 1p p p p = 2 lim fn 2 f lim inf f n f n n

p 1 p p = 2 f limsup fn f . n

p Since f it follows that

p limsup fn f 0. n

p p Therefore limsup fn f liminf f n f 0 , n n

p So that lim fn f = 0 n

78 LOVELY PROFESSIONAL UNIVERSITY Unit 7: Convergence and Completeness

1 Notes p p lim fn f dt = 0 n

lim fn f = 0. n p

Theorem: In a normed linear space, every convergent sequence is a Cauchy sequence.

Proof: Let the sequence in a normed linear space N, converges to a point xo N1. We shall show that it is a Cauchy sequence.

Let > 0 be given. Since the sequence converges to xo a positive integer mo s.t.

n mo xn – xo < /2 … (1)

Hence for all m, n mo, we have

xm – xn = xm – xo + xo – xn

xm – xo + xo – xn

< = by (1) 2 2

It follows that the convergent sequence is a Cauchy sequence. Theorem: Prove that L [0, 1] is complete.

Proof: Let (fn) be any Cauchy sequence in L , and let

Ak = {x : |fk (x)| > fk },

Bm, n = {x : |fk (x) – fm (x)| > fk – fm }.

Then m (Ak) = 0 = m (Bm, n) (k, m, n = 1, 2, 3, …), So that if E is the union of these sets, we have m (E) = 0. Now, if x F = [0, 1] – E, then

|fk (x) fk

|fn (x) – fm (x) fn – fm 0 as n, m .

Hence the sequence (fn) converges uniformly to a bounded function on F. Define f : [0, 1] R by

lim fn (x) if x F f (x) = n 0, if x E

Then f L and fn – f 0 as n . Thus L is Hence proved.

7.2 Summary

 . A sequence in a normal linear space X with norm is said to converge to an element

x X if for arbitrary > 0, however small, no N s.t. xn – x < , n > no. Then we write

lim xn x n .

LOVELY PROFESSIONAL UNIVERSITY 79 Measure Theory and Functional Analysis

Notes  A complete normed linear space is also called Banach space.

 The normed Lp-spaces are complete for (p 1).

 p p A convergent sequence in L -spaces has a limit in L -space.

 A normed linear space is complete iff every absolutely summable sequence is summable.

 In a normed linear space, every convergent sequence is a Cauchy sequence.

7.3 Keywords

Banach Space: A complete normed linear space is also called Banach space. . Cauchy Sequence: A sequence in a normal linear space (X, ) is said to be a Cauchy

sequence if for arbitrary > 0, n0 N s.t.

xn – xm < , n, m n0. Complete Normed Linear Space: A normed linear space (X, . ) is said to be complete if every

Cauchy sequence in it converges to an element x X.

Convergence almost Everywhere: Let be a sequence of measurable functions defined over a

measurable set E. Then is said to converge almost everywhere in E if there exists a subset Eo of E s.t.

(i) fn (x) f (x), x E – Eo.

and (ii) m (Eo) = 0. . Convergent Sequence: A sequence in a normal linear space X with norm is said to

converge to an element x X if for arbitrary > 0, however small, n0 N such that xn – x <

, n > n0.

Then we write lim xn x . n

Normed Linear Space: A linear space N together with a norm defined on it, i.e., the pair (N, ) is called a normed linear space.

Summable Series: A series un in N is said to be summable to a sum u if u N and lim Sn u , 1 1 n n 1 where

Sn = u1 + u2 + … + un

7.4 Review Questions

n 1. Prove that  p is complete.

2. Prove that the vector space L equipped with is a complete vector space. L

3. Suppose f L is supported on a set of finite measure. Then f Lp for all p < , and

f f as p . LL

80 LOVELY PROFESSIONAL UNIVERSITY Unit 7: Convergence and Completeness

p p Notes If f L (p > 0), f 0 and f = min (f, n) (n N), show that f L and lim fn f = 0. n n n p

7.5 Further Readings

Books H.L. Royden, Real analysis. Walter Rudin, Real and Complex Analysis, Third, McGraw-Hill Book Co., New York, 1987.

Online links www.public.iastate.edu www.scribd.com/doc/49732162/103.Convergence-and-Completeness.

LOVELY PROFESSIONAL UNIVERSITY 81 Measure Theory and Functional Analysis

Notes Unit 8: Bounded Linear Functional on the Lp-spaces

CONTENTS Objectives Introduction 8.1 Bounded Linear Functionals on Lp-spaces 8.1.1 Linear Functional 8.1.2 Bounded Linear Functional 8.1.3 Bounded Linear Functional on Lp-spaces 8.1.4 Norm 8.1.5 Continuous Linear Functional 8.1.6 Theorems 8.2 Summary 8.3 Keywords 8.4 Review Questions 8.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand bounded linear functional on Lp-spaces

 Understand related theorems.

 Solve problems on bounded linear functionals.

Introduction

In this unit, we obtain the representation of bounded linear functionals on L p-space. We shall * also study about linear functional, continuous linear functionals and norm of f  p . Further, we shall prove important theorems on bounded linear functionals.

8.1 Bounded Linear Functionals on Lp-spaces

8.1.1 Linear Functional

Definition: Let N1 be a normed space over a field R (or C). A mapping f : N1 R (or C) is called a

linear functional on N1 if f ( x + y) = f (x) + f (y), x, y N1 and , R (or C).

8.1.2 Bounded Linear Functional

Definition: A linear functional f on a normed space N 1 is said to be bounded if there is a constant k > 0 such that

|f (x)| k x , x N1 … (1)

82 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

The smallest constant k for which (1) holds is called the norm of f, written f . Notes

|f(x)| Thus f = sup : x 0 and x N or equivalently x 1

f = sup {|f (x)| : x X and x = 1}.

Also |f (x)| f x x N1. Definition: Let p R, p > 0. We define Lp = Lp [0, 1] to be the set of all real-valued functions on [0, 1] such that

1 1 p p (i) f is measurable and (ii) f p = |f| < .

Note L1 or simply L denotes the class of measurable function f (x) which are also L-integrable.

8.1.3 Bounded Linear Functional on Lp-spaces

If x  p and f is bounded linear functional on  p , then f has the unique representation of the form as an infinite series

f (x) = xk f(e k ) k 1 8.1.4 Norm

* The norm of f  p is given by

1 q q f = |f(ek )| k 1

Likewise in finite dimensional case, the bounded linear functionals are characterised by the values they assume on the set ek, k = 1, 2, 3, … .

8.1.5 Continuous Linear Functional

A linear functional f is continuous if given > 0 there exists > 0 so that |f (x) – f (y)| whenever x – y .

8.1.6 Theorems

1 1 Theorem 1: Suppose 1 p < , and 1 , then, with  = Lp we have p q

* = Lq ,

LOVELY PROFESSIONAL UNIVERSITY 83 Measure Theory and Functional Analysis

Notes in the following sense: For every bounded linear functional  on Lp there is a unique g Lq so that

 (f) = f(x)g(x)d (x), for all f Lp

Moreover,  = g . * Lq

This theorem justifies the terminology where by q is usually called the dual exponent of p. The proof of the theorem is based on two ideas. The first, as already seen, is Hölder’s inequality; to which a converse is also needed. The second is the fact that a linear functional  on Lp, 1 p < , leads naturally to a (signed) measure . Because of the continuity of  the measure is absolutely continuous with respect to the underlying measure , and our desired function g is then the density function of in terms of . We begin with: Lemma: Suppose 1 p, q , are conjugate exponents.

q g sup fg (i) If g L , then Lq . f Lp 1

(ii) Suppose g is integrable on all sets of finite measure and

sup fg = M < f Lp 1 f simple

Then g Lq, and g = M. Lq

For the proof of the lemma, we recall the signum of a real number defined by

1 if x 0 sign (x) = 1 if x 0 0 if x 0

Proof: We start with (i). If g = 0, there is nothing to prove, so we may assume that g is not 0 a.e., and hence g 0 . By Hölder’s inequality, we have that Lq

g q sup fg . L f Lp 1

To prove the reverse inequality we consider several cases.

 f 1 First, if q = 1 and p = , we may take f (x) = sign g (x). Then, we have L and

clearly fg g . L1

q 1  If 1 < p, q < , then we set f (x) = |g (x)|q–1 sign g(x) g . We observe that Lq

p p(q1) p(q1) f g(x) d g 1 since p (q – 1) = q, and that fg g . LLp q Lq

84 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

 Finally, if q = and p = 1, let > 0, and E a set of finite positive measure, where |g (x)| Notes g . (Such a set exists by the definition of g and the fact that the measure is - L L finite). Then, if we take f (x) =  E (x) sign g (x)/ (E), where E denotes the characteristic function of the set E, we see that f 1 , and also L1

1 fg |g| g . (E) E

This completes the proof of part (i).

To prove (ii) we recall that we can find a sequence {gn} of simple functions so that |gn (x) |g (x)| q 1 while g (x) g (x) for each x. When p > 1 (so q < ), we take f (x) = |g (x)|q–1 sign g(x) g . n n n n Lq As before, f 1 . However n Lp

q gn (x) , fn g q 1 g n q g L n Lq and this does not exceed M. By Fatou’s Lemmas if follows that |g|q M q , so g Lq with

g M . The direction gq M is of course implied by Hölder’s inequality. When p = 1 the Lq L  argument is parallel with the above but simpler. Here we take fn (x) = (sign g (x)) En (x), where

En is an increasing sequence of sets of finite measure whose union is X. The details may be left to the reader. With the lemma established we turn to the proof of the theorem. It is simpler to consider first the case when the underlying space has finite measure. In this case, with  the given functional on Lp, we can then define a set function by

(E) =  ( E),

 p where E is any measurable set. This definition make sense because E is now automatically in L since the space has finite measure. We observe that | (E)| C ( (E))1/p … (1)

1/p where C is the norm of the linear functional, taking into account the fact that  (E) . E Lp

Now the linearity of  clearly implies that is finitely-additive. Moreover, if {En} is a countable collection of disjoint measurable sets, and we put EE,EE* , then obviously n 1 n N  n N 1 n

E *  E n . EN  n 1

N * * Then (E)E(E)N n . However EN 0 , as N because of (1) and the n 1 assumption p < . This shows that is countably additive and moreover (1) also shows us that is absolutely continuous with respect to . We can now invoke the key result about absolutely continuous measures, the Lebesgue-Radon

– Nykodin theorem. It guarantees the existence of an integrable function g so that (E) = g du E

LOVELY PROFESSIONAL UNIVERSITY 85 Measure Theory and Functional Analysis

Notes for every measurable set E. Thus we have ( E)  Eg d . The representation (f) fg d

then extends immediately to simple function f, and by a passage to the limit, to all f Lp since the p g  simple functions are dense in L , 1 p < . Also by lemma, we see that Lq .

To pass from the situation where the measure of X is finite to the general case, we use an

increasing sequence {E } of sets of finite measure that exhaust X, that is, X = E . According n  n 1 n

to what we have just proved, for each n there is an integrable function gn on En (which we can set c to be zero in En ) so that

(f) = fgn d … (2)

whenever f is supported in E and f Lp. Moreover by conclusion (ii) of the lemma g  . n n Lp

Now it is easy to see because of (2) that gn = gm a.e. on Em , whenever n m. Thus limn gn (x) = g (x) exists for almost every x, and by Fatou’s lemma, g Lq  . As a result we have that

 p (f) f g du for each f L supported in En, and then by a simple limiting argument, for all f

p  q L supported in En. The fact that g L is already contained in Hölder’s inequality and therefore the proof of the theorem is complete. Theorem 2: Let f be a linear functional defined on a normed linear space N, then f is bounded f is continuous. Proof: Let us first show that continuity of f boundedness of f. If possible let f is continuous but not bounded. Therefore, for any natural number n, however

large, there is some point xn such that

|f (xn)| n xn … (1)

xn Consider the vector yn so that n xn

1 y = . n n

yn 0 as n .

yn 0 in the norm.

Since any continuous functional maps zero vector into zero, and f is continuous f (yn) f (0) = 0.

1 f(x ) But |f (yn)| = n … (2) n xn

It now follows from (1) and (2) that

|f (yn)|> 1, a contradiction to the fact that

f (yn) 0 as n . Thus if f is bounded then f is continuous.

86 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

Notes Conversely, let f is bounded. Then for any sequence (xn), we have

|f (xn)| k xn n = 1, 2, … and k 0.

Let xn 0 as n then f (xn) 0 f is continuous at the origin and consequently it is continuous everywhere. This completes the proof of the theorem. Theorem 3: If L is a linear space of all n-tuples, then

n n (i) p*  q

n n (ii) 1 * 

n n (iii) *  1

Proof: Let (e1, e2, …, en) be a standard basis for L so that any x = (x 1, x2, …, xn) L can be written as x = x1e1 + x2e2 + … + xnen. If f is a scalar valued linear function defined on L, then we get

f (x) = x1 f (e1) + x2f (e2) + … + xnf (en) … (1)

f determines and is determined by n scalars yi = f (ei). Then the mapping

n x y y = (y1, y2, … yn) f where f (x) = i i is an isomorphism of L onto the linear space L of all i 1 function f. We shall establish (i) – (iii) by using above given facts.

n th (i) If we consider the space L =  p (1 p < ) with the p norm, then f is continuous and L

* n n represents the set of all continuous linear functionals on  p so that L =  p .

Now for y f as an isometric isomorphism we try to find the norm of y’s. For 1 < p < , we show that

n n p * =  p .

n For x  p , we have defined,

1 n p p x = |xi | i 1

n n

Now |f (x)| = xi y i |x i ||y i | i 1 i 1

By using Holder’s inequality, we get

1 1 n np n q  p q so that |xi y i | |xi | |y i | i 1 i 1 i 1

LOVELY PROFESSIONAL UNIVERSITY 87 Measure Theory and Functional Analysis

Notes 1 1 nq n p  q p |f (x)| |yi | |x i | i 1 i 1

Using the definition of norm for f, we get

1 n q q f  |yi | … (2) i 1

Consider the vector, defined by

|y |q i , y 0 xi = i and xi = 0 if yi = 0 … (3) yi

1 1 p np n q p p |yi | Then, x = |xi | … (4) i 1 i 1 |yi |

Since q = p (q – 1) we have from (4),

1 n q q x = |yi | … (5) i 1

n n q |yi | Now f (x) = xi y i y i … (4) i 1 i 1 yi

n q = |yi | (By (3)) i 1

So that

n q |yi | = |f (x) f x … (6) i 1

From (5) and (6) we get,

1 1 n p q f |yi | i 1

1 n q q f … (7) |yi | i 1

Also from (2) and (7) we have

1 n q q f = |yi | , so that i 1

88 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

y f is an isometric isomorphism. Notes

n n Hence  p * =  q .

n (ii) Let L =  1 with the norm defined by

x = |xi | i 1

Now f defined in (1), above is continuous as in (i) and L here represents the set of continuous

n linear functional on  1 so that

n L =  1 * .

We now determine the norm of y’s which makes y f an isometric isomorphism.

Now, f (x) = xi y i i 1

|xi ||y i | i 1

n n

But |xi ||y i | max. |yi | |x i | so that i 1 i 1

f (x) max. |yi | |x i |. i 1

From the definition of the norm for f, we have

f = max. {|yi| : i = 1, 2, …, n} … (8) Now consider the vector defined as follows:

If |y | = max |yi | , let us consider the vector x as i 1 i n

|yi | x = when|yi | max |y i | and x i 0 … (9) i 1 i n yi

otherwise

From the definition, xk = 0 k i, so that we have

y (x) = i = 1 y

Further | f (x)| = (xi y i ) = |yi|. i 1

Hence |yi| = |f (x)| f x

LOVELY PROFESSIONAL UNIVERSITY 89 Measure Theory and Functional Analysis

Notes  |yi| f or max {|yi|} [ x = 1] f … (10) From (8) and (10), we obtain

f = max. {|yi|} so that

n y f is an isometric isomorphism of L to 1 *.

n n Hence 1 *  .

(iii) Let L = n with the norm

x = max {|xi| : 1, 2, 3, …, n}. Now f defined in (1) above is continuous as in (1). Let L represents the set of all continuous linear functionals on n so that

L = n * .

Now we determine the norm of y’s which makes y f as isometric isomorphism.

n n

|f (x)| = xi y i |x i ||y i |. i 1 i 1

n n

But |xi ||y i | max (|xi |) |y i | i 1 i 1

Hence we have

n n

|f (x)| |yi | ( x ) so that f |y i | … (11) i 1 i 1

Consider the vector x defined by

|yi | xi = when yi 0 and xi = 0 otherwise. … (12) yi

|y | Hence x = maxi 1 . |yi |

n n

and |f (x)| = |xi y i | |y i | i 1 i 1

Therefore

|yi | = |f (x)| f x = f . i 1

|yi | f … (13) i 1

90 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

It follows now from (11) and (13) that Notes

f = |yi | so that y f is an isometric isomorphism. i 1

n n Hence, *  1 .

This completes the proof of the theorem.

Note We need the signum function for finding the conjugate spaces of some infinite dimensional space which we define as follows: If is a complex number, then

sgn if 0 | | 0 if 0

(i) |sgn | = 0 if = 0 and | sgn | = 1 if 0

(ii) sgn = 0 if = 0 and sgn = =| |, if 0. | |

Theorem 4: The conjugate space of  p is  q , where

1 1 = 1 and 1 < p < . p q

* or p  q .

 p Proof: Let x = (xn) p so that |xn | … (1) n 1

th Let en = (0, 0, 0, …, 1, 0, 0, …) where 1 is in the n place.  en p for n = 1, 2, 3, … .

* We shall first determine the form of f and then establish the isometric isomorphism of  p onto

 q .

By using (en), we can write any sequence

(x1, x2, … xn, 0, 0, 0, …) in the form xk e k and k 1

x xk e k (0, 0, 0, , x n 1 , x n 2 , ). k 1

LOVELY PROFESSIONAL UNIVERSITY 91 Measure Theory and Functional Analysis

Notes 1 n p p Now x xk e k |x k | … (2) k 1 k n 1

The R.H.S. of (2) gives the remainder after n terms of a convergent series (1).

1 p Hence p 0 as n . … (3) |xk | k n 1

From (2) and (3) if follows that

x = xk e k … (4) k 1

n  Let f *p and Sn = xk e k then k 1

Sn x as n (Using (4)) Since f is linear, we have

f (Sn) = xk f(e k ) . k 1

Also f is continuous and Sn x, we have

f (Sn) f (x) as n .

f (x) = xk f(e k ) … (5) k 1

which gives the form of the functional on  p .

 Now we establish the isomeric isomorphism of  *p onto q , for which proceed as follows:

Let f (ek) = k and show that the mapping  T :  *p q given by

  T (f) = ( 1, 2, …, k, …) is an isomeric isomorphism of *p onto q .

First, we show that T is well defined.

 For let x p , where x = ( 1, 2, …, n, 0, 0, …) where

g 1 | | sgnk , 1 k n = k k 0 n k

q–1 | k| = | k| for 1 k n.

p p (q 1) q 1 1 | | = |k | = | | .  q p(q 1) q k k p q

92 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

q 1 q 1 Notes Now k k k| k | sgnk | k | sgn k

q p k k = | k| = | k| … (7) (Using property of sgn function)

1 n p p x = |k | k 1

1 n q q = |k | … (8) k 1

Since we can write

x = ke k , we get k 1

n n

f (x) = kf(e k ) k k k 1 k 1

n q f (x) = |k | (Using (7)) … (9) k 1

We know that for every x  p

| f (x) | f x , which upon using (8) and (9), gives

1 n n p |f (x)| q q |k | f | k | k 1 k 1 which yields after simplification.

1 n p q |k | f … (10) k 1

Since the sequence of partial sum on the L.H.S. of (10) is bounded; monotonic increasing, it converges. Hence

1 n p q |k | f … (11) k 1

 So the sequence ( k) which is the image of f under T belongs to q and hence T is well defined.

LOVELY PROFESSIONAL UNIVERSITY 93 Measure Theory and Functional Analysis

Notes We next show that T is onto  q .

 * Let ( k) q , we shall show that is a g p such that T maps g into ( k).

Let x  p so that

x = xk e k k 1 We shall show that

g (x) = xk k is the required g. k 1

Since the representation for x is unique, g is well defined and moreover it is linear on  p . To prove it is bounded, consider

n n

|g (x)| = kx k k x k k 1 k 1

1 1 np n q p q xk k (Using Hölder’s inequality) k 1 k 1

1 n q q |g (x)| x |k | k 1

g is bounded linear functional on  p .

 Since ek p for k = 1, 2, …, we get

g (ek) = k for any k so that   Tg = ( k) and T is on *p onto q .

We next show that Tf = f so that T is an isometry.

Since Tf  q , we have from (6) and (10) that

1 q q = Tf f k k 1

Also, x  p x = xk e k . Hence k 1

f (x) = xk (e k ) x k k k 1 k 1

94 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

Notes

|f (x)| |xk || k | k 1

1 1 n q p q p |k | x k (Using Hölder’s inequality) k 1 k 1

1 n q q or |f (x)| |k | x x  p . k 1

Hence we have

1 |f(x)| q sup | |q = Tf (Using (6)) x 0 x k k 1 which upon using definition of norm yields f Tf … (13) Thus f = Tf (Using (12) and (13)) From the definition of T, it is linear. Also since it is an isometry, it is one-to-one and onto

(already shown). Hence T is an isometric isomorphism of  *p onto  q , i.e.,

*p  q .

This completes the proof of the theorem.

1 1 Theorem 5: Let p > 1 with 1 and let g L (X). Then the function defined by p q p

F (f) = fg d for f Lp (X) X

is a bounded linear functional on Lp (X) and

F = g q … (1) Proof: We first note that

F is linear on Lp (X). For if f1, f2 Lp (X), then we get

(f f)gd fgd fgd F (f1 + f2) = 1 2 1 2 XXX

= F (f1) + F (f2) So that

F (f1 + f2) = F (f1) + F (f2) and F ( F) = fg d F(f) .

Now |F (f)| = fg d |fg|d … (2) XX

LOVELY PROFESSIONAL UNIVERSITY 95 Measure Theory and Functional Analysis

Notes Making use of Hölder’s inequality, we get

1 1 p q |fg|d |f|p d |g| q d

X XX

= f p g q … (3) From (2) and (3) it follows that

|F (f)| f p g q.

F|f| Hence sup : f Lp(X) and f 0 g f q p

F g q (Using definition of the norm) … (4)

Further, let f = |g|q–1 sgn g … (5)

Since sgn g = 1, we get

|f|p = |g|p(q–1) = |g|q ( p (q – 1) = q)

1 1 p q p q Thus, f Lp (X) and |f| d = |g| d … (6)

X X

1 p q /p But q p = g |g| d |g| d q

which implies on using (6) that

q /p f = g … (7) p q

Now F (f) = fg d |g|q 1 g sgng d

q = |g|q d g q X

q Hence g g = F (f) F f . q p

and this on using (7) yields that

q q /p g = F (f) F g q q

q q /p g = g F … (8) q q

( g 0)

96 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

From (4) and (8) it finally follows that Notes

F = g q. This completes the proof of the theorem.

Approximation by Continuous Function

Theorem 6: If f is a bounded measurable function defined on [a, b], then for given > 0, a continuous function g on [a, b], such that

f – g 2 <

x Proof: Let F (x) = f(t) dt where x [a, b].

x h x Then |F (x + h) – F (x)| = f(t) dt f(t) dt

a a

x h x h = f(t) dt f(t) dt

x x

Mh, where |f (x)| M, x [a, b].

Taking h < , and Mh < 1, we get

|x + h – x| < | F (x + h) – F (x) | < 1 F (x) is continuous on [a, b].

x h n f(t) dt : x [a, b] and n N; Let Gn (x) = x

1 then G (x) = n F x F(x) ( F (x) is continuous on [a, b] n n

Gn (x) is continuous on [a, b] n)

x Again, since F (x) = f(t) dt, x [a, b].

F (x) = f (x) a.e. in [a, b].

F(x (1/n) F(x) Now, Lim Gn (x) = Lim n n 1/n

F(x h) F(x) 1 = Lim , h h 0 h n

= F (x) = f (x) a.e. in [a, b]

2 and hence Lim Gn (x) f(x) = 0. n

LOVELY PROFESSIONAL UNIVERSITY 97 Measure Theory and Functional Analysis

Notes x(1/n) x(1/n)

Also |Gn (x)| = f(t) dt n |f(t)|dt M . x x

Hence |Gn (x)| M, n N and x [a, b].

2 2 2 [Gn(x) – f (x)] (M + M) = 4M , x [a, b]. On applying Lebesgue bounded convergence theorem, we get

b b 2 2 Lim (Gn f) = Lim (Gn f) 0 n a a

2 Lim Gn f = 0 n 2

Lim Gn f n 2

or Lim f Gn = 0 n

for given > 0, no N, such that n no

f – Gn 2 <

Particularly for n = no.

f G < no 2

f – g 2 < (Taking Gno = g)

Thus there exists a continuous function Gno (x) g(x)

x (1 /no )

= no f(t) dt, x [a, b], x which satisfies the given condition.

8.2 Summary

 A linear functional f on a normed space N1 is said to be bounded if there is a constant k > 0 such that

|f (x)| k x , x N1

 If x  p and f is bounded linear functional on  p , then f has the unique representation of the form as an infinite series.

f (x) = xk f(e k ) k 1

*  The norm of f  p is given by

1 q q f = |f(ek )| k 1

98 LOVELY PROFESSIONAL UNIVERSITY Unit 8: Bounded Linear Functional on the Lp-spaces

8.3 Keywords Notes

p Bounded Linear Functional on L -spaces: If x  p and f is bounded linear functional on  p , then f has the unique representation of the form as an infinite series

f (x) = xk f(e k ) k 1

Bounded Linear Functional: A linear functional f on a normed space N1 is said to be bounded if there is a constant k > 0 such that

|f (x)| k x , x N1 Continuous Linear Functional: A linear functional f is continuous if given > 0 there exists > 0 so that |f (x) – f (y)| whenever x – y .

Linear Functional: Let N1 be a normed space over a field R (or C). A mapping f : N1 R (or C) is called a linear functional on N1 if f ( x + y) = f (x) + f (y), x, y N1 and , R (or C).

* Norm: The norm of f  p is given by

1 q q f = |f(ek )| k 1

8.4 Review Questions

1. Account for bounded linear functionals on Lp-space. 2. State and prove different continuous linear functional theorems. 3. Describe approximation by continuous function. 4. How will you explain norms of bounded linear functional on L p-space? 5. What is Isometric Isomorphism?

8.5 Further Readings

Books Rudin, Walter (1991), Functional Analysis, Mc-Graw-Hill Science/Engineering/ Math Kreyszig, Erwin, Introductory Functional Analysis with Applications, WILEY 1989. T.H. Hilderbrandt, Transactions of the American Mathematical Society. Vol. 36, No. = 4, 1934.

Online links www.math.psu.edu/yzheng/m597k/m597kLIII4.pdf www.public.iastate.edu/…/Royden_Real_Analysis_Solutions.pdf

LOVELY PROFESSIONAL UNIVERSITY 99 Measure Theory and Functional Analysis

Notes Unit 9: Measure Spaces

CONTENTS

Objectives Introduction 9.1 Measure Space 9.1.1 Null Set in a Measure Space 9.1.2 Complete Measure Space 9.1.3 Measurable Mapping 9.2 Summary 9.3 Keywords 9.4 Review Questions 9.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define measure space.

 Define null set in a measure space.

 Understand theorems based on measure spaces.

 Solve problems on measure spaces.

Introduction

A measurable space is a set S, together with a non-empty collection, S, of subsets of S, satisfying the following two conditions: 1. For any A, B in the collection S, the set1 A – B is also in S.

2. For any A1, A2, … S, Ai S. The elements of S are called measurable sets. These two conditions are summarised by saying that the measurable sets are closed under taking finite differences and countable unions.

9.1 Measure Space

Measurable Space: Let  be a -algebra of subsets of set X. The pair (X, ) is called a measurable space. A subset E of X is said to be -measurable if E . (a) If is a measure on a -algebra  of subsets of a set X, we call the triple (X, , u) a measure space. (b) A measure on a -algebra  of subsets of a set X is called a finite measure if m (X) < . In this case (X, , ) is called a finite measure space.

100 LOVELY PROFESSIONAL UNIVERSITY Unit 9: Measure Spaces

(c) A measure on a -algebra  of subsets of a set X is called a -finite measure if there exists Notes     a sequence (En : n ) in such that n  En = X and (En) < for every n . In this case (X, , ) is called a -finite measure space. (d) A set D  in an arbitrary measure space (X, , ) is called a -finite set if there exists a     sequence (Dn : n ) in such that Un  Dn = D and (Dn) < for every n . Lemma 1: (a) Let (X, , ) be a measure space. If D  is a -finite set, then there exists an increasing sequence (F : n ) in  such that lim Fn D and (F ) < for every n  and there n n n     exists a disjoint sequence (Gn : n ) in such that n  Gn = D and (Gn) < for every n . (b) If (X, , ) is a -finite measure space then every D  is a -finite set. Proof 1: Let (X, , ) be a measure space. Suppose D  is a -finite set. Then there exists a      sequence (Dn : n ) in such that Un  Dn = D and (Dn) < for every n . For each n , n let FUDn k 1 k . Then (F : n ) is an increasing sequence in  such that lim Fn Un  F n n n

n n and (F)D(D) for every n . UDDn  n n k k k 1 k 1

n 1   Let G1 = F1 and Gn = Fn\ UFk 1 k for n 2. Then (Gn: n ) is a disjoint sequence in such that  UGUFDnn n  n as in the proof of Lemma “let (En : n ) be an arbitrary sequence in an    algebra of subsets of a set X. Then there exists a disjoint sequence (Fn: n ) in such that

NN (1) EF for every N , n  n n 1 n 1

and

(2) EF ”. n  n n n 

n 1

(G1) = (F1) < and (G)FF(F)n n k n for n 2. This proves (a).  k 1

   2. Let (X, , ) be a -finite measure space. Then there exists a sequence (En : n ) in such     that Un  En = X and (En) < for every n . Let D . For each n , let Dn = D En.   Then (Dn : n ) is a sequence in such that Un  Dn = D and m (Dn) (En) < for every n . Thus D is a -finite set. This proves (b).

9.1.1 Null Set in a Measure Space

Definition: Given a measure on a -algebra  of subsets of a set X . A subset E of X is called a null set with respect to the measure if E  and (E) = 0. In this case we say also that E is a null set in the measure space (X, , ). (Note that is a null set in any measure space but a null set in a measure space need not be .) Theorem 1: A countable union of null sets in a measure space is a null set of the measure space.   Proof: Let (En : n ) be a sequence of null sets in a measure space (X, , ). Let E = Un  En. Since  is closed under countable unions, we have E .

LOVELY PROFESSIONAL UNIVERSITY 101 Measure Theory and Functional Analysis

Notes By the countable subadditivity of on ,

we have (E) n  (En) = 0. Thus (E) = 0. This shows that E is a null set in (X, , ).

9.1.2 Complete Measure Space

Definition: Given a measure on a -algebra  of subsets of a set X. We say that the -algebra 

is complete with respect to the measure if an arbitrary subset E0 of a null set E with respect to    is a member of (and consequently has (E0) = 0 by the Monotonicity of ). When is complete with respect to , we say that (X, , ) is a complete measure space.

Example: Let X = {a, b, c}. Then = { , {a}, {b, c}, X} is a -algebra of subsets of X. If we define a set function on by setting ( ) = 0, ({a}) = 1, ({b, c}) = 0, and (X) = 1, then is a measure on . The set {b, c} is a null set in the measure space (X, , ), but its subset {b} is not a member of . Therefore, (X, , ) is not a complete measure space.

9.1.3 Measurable Mapping

Let f be a mapping of a subset D of a set X into a set Y. We write D (f) and R (f) for the domain of definition and the range of f respectively. Thus D (f) = D X, R (f) = {y Y : y = f (x) for some x D (f)} Y. For the image of D (f) by f, we have f (D (f)) = R (f). For an arbitrary subset E of y we define the preimage of E under the mapping f by F–1 (E) : = {x X : f (x) E} = {x D (f) : f (x) E}.

Notes

1. E is an arbitrary subset of Y and need not be a subset of R (f). Indeed E may be disjoint from  (f), in which case f–1 (E) = . In general, we have f (f–1 (E)) E. 2. For an arbitrary collection C of subsets of Y, we let f–1 (C) : = {f–1 (E) : E C}.

Theorem 2: Given sets X and Y. Let f be a mapping with D (f) X and  (f) Y. Let E and E be arbitrary subsets of Y. Then 1. f–1 (Y) = D (f), 2. f–1 (EC) = f–1 (Y\E) = f–1 (Y)\f–1 (E) = D (f) \ f–1 (E),

–1 –1 3. f (U  E ) = U  f (E ),

–1 –1 4. f (  E ) = f (E ). Theorem 3: Given sets X and Y. Let f be a mapping with D (f) X and R (f) Y. If  is a -algebra of subsets of Y then f–1 () is a -algebra of subsets of the set D (f). In particular, if D (f) = X then f–1 () is a -algebra of subsets of the set X.

102 LOVELY PROFESSIONAL UNIVERSITY Unit 9: Measure Spaces

Proof: Let  be a -algebra of subsets of the set Y. To show that f–1 () is a -algebra of subsets of Notes the set D (f) we show that D (f) f–1 (); if A f–1 () then D (f)\A f–1 (); and for any sequence  –1  –1 (An : n ) in f ( ) we have Un An f (B). 1. By (1) of above theorem, we have D (f) = f–1 (Y) f–1 (B) since Y . 2. Let A f–1 (). Then A = f–1 () for some B . Since BC  we have f–1 (BC) f–1 (). On the other hand by (2) of above theorem, we have f–1 (BC) = D (f)\f–1 (B) = D (f)\A. Thus D (f)\A f–1 ().

 –1  –1   3. Let (An : n ) be a sequence in f ( ). Then An = f (Bn) for some Bn for each n . Then by (3) of above theorem, we have

A f1 (B ) f 1 B f 1 ( ) , n  n  n n n  n 

since B() .  n n 

Measurable Mapping

Definition: Given two measurable spaces (X, ) and (Y, ). Let f be a mapping with D (f) X and  (f) Y. We say that f is a / measurable mapping if f–1 (B) for every B , that is, f–1 () . Theorem 4: Given two measurable spaces (X, ) and (Y, ). Let f be a /-measurable mapping.      (a) If , is a -algebra of subsets of X such that , , then f is 1/ -measurable.      (b) If 0 is a -algebra of subsets of Y such that 0 , then f is / 0-measurable. –1    –1 –1   Proof: (a) Follows from f ( ) 1 and (b) from f (B0) f ( ) . Composition of two measurable mappings is a measurable mapping provided that the two measurable mappings from a chain. Theorem 5: Given two measurable spaces (X, ) and (Y, ), where = () and  is arbitrary collection of subsets of Y. Let f be a mapping with D (f) and  (f) Y. Then f is a /- measurable mapping of D (f) into Y if and only if f–1 () . Proof: If f is a /-measurable mapping of D (f) into Y, then f–1 ()  so that f–1 () . Conversely if f–1 () , then (f–1 () () = . Now by theorem, “Let f be a mapping of a set X into a set Y. Then for an arbitrary collection C of subsets of Y, we have (f–1 ()) = f–1 ( ().” (f–1 () = f–1 ( ()) = f–1 (). Thus f–1 ()  and f is a /- measurable mapping of D (f).

Theorem 6: If X0 is a thick subset of a measure space (X, S, ), if S0 = S X0, and if, for E in S, 0 (E

X0) = (E), then (X0, S0, 0) is a measure space.

Proof: If two sets, E1 and E2, in S are such that E1 X0 = E2 X0, then (E1 E2) Xo = 0, so that

(E1 E2) = 0 and therefore (E1) = (E2). In other words 0 is indeed unambiguously defined on

S0.

Suppose next that {Fn} is a disjoint sequence of sets in S0, and let En be a set in S such that

Fn = En X0, n = 1, 2, … .

LOVELY PROFESSIONAL UNIVERSITY 103 Measure Theory and Functional Analysis

Notes  If En E n {E i : 1 i n}, n 1, 2, , then

 EEXn n 0 = Fn F i : 1 i n F n

= Fn Fn = 0,

 So that EEn n = 0, and therefore

  n 1 0(F) n = n10n(E)(E)E n10n n1n

= EFn 0 n n 1  n 1

In other word 0 is indeed a measure, and the proof of the theorem is complete.

9.2 Summary

 Let  be a -algebra of subsets of a set X. The pair (X, ) is called a measurable space. A subset E of X is said to be -measurable if E .

 If is a measure on a -algebra of subsets of a set X, we call the triple (X, , ) a measure space.

 A subset E of X is called a null set with respect to the measure if E  and (E) = 0.

 Two measurable spaces (X, ) and (Y, ). Let f be a mapping with D (f) X and  (f) Y. We say that f is a /-measurable mapping if f–1 (B)  for every B , that is f–1 () .

9.3 Keywords

Complete Measure Space: Given a measure on a -algebra  of subsets of a set X. We say that  the -algebra is complete with respect to the measure if an arbitrary subset E0 of a null set   E with respect to is a member of (and consequently has (E0) = 0 by the Monotonicity of ). When  is complete with respect to , we say that (X, , ) is a complete measure space. Measurable Mapping: Given two measurable spaces (X, ) and (Y, ). Let f be a mapping with D (f) X and  (f) Y. We say that f is a / measurable mapping if f–1 (B) for every B , that is, f–1 () . Measurable Space: A measurable space is a set S, together with a non-empty collection, S, of subsets of S. Null Set in a Measure Space: A subset E of X is called a null set with respect to the measure if E  and (E) = 0. In this case we say also that E is a null set in the measure space (X, , ). Sigma Algebra:  is sigma algebra which establishes following relations:

(i) A  for all k implies A  k  k k 1 (ii) A implies AC  (iii) 

104 LOVELY PROFESSIONAL UNIVERSITY Unit 9: Measure Spaces

9.4 Review Questions Notes

1. Let  be a -algebra of subsets of a set X and let Y be an arbitrary subset of X. Let  = {A Y : A }. Show that is a -algebra of subsets of Y.   2. Let (X, , ) be a measure space. Show that for any E 1, E2 we have the equality:

(E1 E2) + (E1 E2) = (E1) + (E2).

9.5 Further Readings

Books Paul Halmos, (1950). Measure Theory. Van Nostrand and Co. Bogachev, V.I. (2007), Measure Theory, Berlin : Springer

Online links planetmath.org/measurable space.html mathworld.wolfram.com > Calculus and Analysis > Measure Theory

LOVELY PROFESSIONAL UNIVERSITY 105 Measure Theory and Functional Analysis

Notes Unit 10: Measurable Functions

CONTENTS

Objectives Introduction 10.1 Measurable Functions 10.1.1 Lebesgue Measurable Function/Measurable Function 10.1.2 Almost Everywhere (a.e.) 10.1.3 Equivalent Functions 10.1.4 Non-Negative Functions 10.1.5 Characteristic Function 10.1.6 Simple Function 10.1.7 Step Function 10.1.8 Convergent Sequence of Measurable Function 10.1.9 Egoroff's Theorem 10.1.10 Riesz Theorem 10.2 Summary 10.3 Keywords 10.4 Review Questions 10.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand measurable functions.

 Define equivalent functions and characteristic function.

 Describe Egoroff's theorem and Riesz theorem.

 Define simple function and step function.

Introduction

In this unit, we shall see that a real valued function may be Lebesgue integrable even if the function is not continuous. In fact, for the existence of a Lebesgue integral, a much less restrictive condition than continuity is needed to ensure integrability of f on [a, b]. This requirement gives rise to a new class of functions, known as measurable functions. The class of measurable functions plays an important role in Lebesgue theory of integration.

106 LOVELY PROFESSIONAL UNIVERSITY Unit 10: Measurable Functions

10.1 Measurable Functions Notes

10.1.1 Lebesgue Measurable Function/Measurable Function

Definition: Let E be a measurable set and R* be a set of extended real numbers. A function f : E R* is said to be a Lebesgue measurable function on E or a measurable function on E iff the set

E (f > ) = {x E : f (x) > } = f–1 { , )} is a measurable subset of E R.

Notes

1. The definition states that f is a measurable function if for every real number , the inverse image of ( , ) under f is a measurable set. 2. The measure of the set E (f > ) may be finite or infinite. 3. A function whose values are in the set of extended real numbers is called an extended real valued function. 4. If E = R, then the set E (f > ) becomes an open set.

Example: A constant function with measurable domain is measurable.

Sol: Let f be a constant function defined over a measurable set E so that f (x) = x E.

Then for any real number ,

E, if c E (f > ) = , if c

The sets E and are measurable and hence E (f > ) is measurable i.e. the function f is measurable. Theorem 1: Let f and g be measurable real valued functions on E, and c is a constant. Then each of the following functions is measurable on E. (a) f c (b) c f (c) f + g (d) f – g (e) |f| (f) f2 (g) fg (h) f/g (g vanishes no where on E) Proof: Let be an arbitrary real number. (a) Since f is measurable and

E (f ± c > ) = E (f >  c), the function f ± c is measurable. (b) To prove c f is measurable over E. If c = 0, then cf is constant and hence measurable because a constant function is measurable.

LOVELY PROFESSIONAL UNIVERSITY 107 Measure Theory and Functional Analysis

Notes Consider the case in which c 0, then

E f if c 0 c E (cf > ) = E f if c 0 c

Both the sets on R.H.S. are measurable.

Hence E (cf > ) is measurable and so cf is measurable c R. (c) Before proving f + g is measurable, we first prove that if f and g are measurable over E then the set E (f > g) is also measurable. Now f > g a rational number r such that f (x) > r > g (x)

Thus E (f > g) =  [(E (f r) (E(g r))] r Q

= an enumerable union of measurable sets. = measurable set, since Q is an enumerable set. Now, we shall prove that f + g is measurable over E. Let a be any real number. Now E (f + g > a) = E (f > a – g) … (1) Again, g is measurable

cg is measurable, c is constant. ( We know that if f is a measurable function and c is constant then cf is measurable)

a + cg is measurable a, c R a – g is measurable by taking c = –1, since f and a – g are measurable E (f > a – g) is measurable. E (f + g > a) is a measurable set. f + g is a measurable function. (d) To prove that f – g is measurable. Before proving f – g is measurable, we first prove that if f and g are measurable over E then the set E (f > g) is also measurable. Now f > g a rational number r, such that f (x) > r > g (x).

Thus E (f > g) =  [(E (f r) (E(g r))] r Q

= an enumerable union of measurable sets. = measurable sets, since Q is an enumerable set. Now we shall prove that f – g is measurable over E. Let a be any real number.

108 LOVELY PROFESSIONAL UNIVERSITY Unit 10: Measurable Functions

Now E (f – g > a) = E (f > a + g) Notes since g is measurable. cg is measurable, c is constant.

a + cg is measurable a, c R a + g is measurable by taking c = 1, since f and a + g are measurable E (f > a + g) is measurable. E (f – g > a) is a measurable set. f – g is a measurable function. (e) To prove |f| is measurable. We have

E if 0 E (|f| > ) = [E (f )] [E(f )] if 0

[because we know that |x| > a x > a or x < –a] since f is measurable therefore E (f > ) and E (f < – ) are measurable by definition. Also we know that finite union of two measurable sets is measurable. E (f > ) E (f < – ) is measurable. E (|f| > ) is measurable.

|f| is measurable. (f) To prove f2 is measurable.

E if 0 We have E (f2 > ) = E (|f| )] if 0

But E (|f| > ) = [E(f )] [E(f )], if 0 ( |x| > a x > a or x < – a)

E if 0 E (f2 > ) = [E (f )] [E(f )] if 0

But f is measurable over E.

E (f ) and E (f ) are measurable sets.

[E (f )] [E(f )] is measurable.

( union of two measurable sets is measurable) E (f2 > ) is measurable because both the sets on RHS are measurable. f2 is measurable over E. (g) To prove fg is measurable.

Clearly, f + g and f – g are measurable functions over E.

LOVELY PROFESSIONAL UNIVERSITY 109 Measure Theory and Functional Analysis

Notes (f + g)2, (f – g)2 are measurable functions over E. (f + g)2 – (f – g)2 is a measurable function over E.

1 (f g)2 (f g) 2 is a measurable function over E. 4

fg is a measurable function over E. (h) To prove f/g is measurable. Let g vanish nowhere on E, so that

g (x) 0 x E.

1 exists. g

1 Now we shall show that is measurable. g

We have

E(g 0) if 0 1 1 E [E(g 0)] E g if 0 g 1 [E(g 0)] E(g 0) E g

1 Also finite union and intersection of measurable sets are measurable. Hence E is g measurable in every case.

1 Since f and are measurable. g

1 (f) is measurable over E. g

f is measurable over E. g

10.1.2 Almost Everywhere (a.e.)

Definition: A property is said to hold almost everywhere (a.e.) if the set of points where it fails to hold is a set of measure zero.

Example: Let f be a function defined on R by

0, if x is irrational f (x) = 1, if x is rational

Then f (x) = 0 a.e.

110 LOVELY PROFESSIONAL UNIVERSITY Unit 10: Measurable Functions

10.1.3 Equivalent Functions Notes

Definition: Two functions f and g defined on the same domain E are said to be equivalent on E, written as f ~ g on E, if f = g a.e. on E, i.e. f (x) = g (x) for all x E – E1, where E1 E with m (E1) = 0. Theorem 2: If f, g : E R (E M) such that g (E).

Proof: Let be any real number and let E1 = E (f > ) and E2 = E (g > )

Then E1 E2 = (E1 – E2) (E2 – E1) = {x E : f (x) g (x)} so that by given hypothesis we have

m (E1 E2) = 0.

This together with the fact that E1 is measurable

E2 is measurable.

Hence g (E).

10.1.4 Non-negative Functions

Definition: Let f be a function, then its positive part, written f+ and its negative part, written f–1, are defined to be the non-negative functions given by f+ = max (f, 0) and f–1 = max (–f, 0) respectively.

Note f = f+ – f–1 and |f| = f+ + f–1

Theorem 3: A function is measurable iff its positive and negative parts are measurable. Proof: For every extended real valued function f, we may write

1 f+ = [f + |f|] 2

1 and f–1 = [|f| – f] 2

But f is measurable then |f| is measurable and hence positive and negative parts of f i.e. f + and f– are measurable. Conversely, let f+ and f–1 be measurable. Since f = f+ – f–1 Since we know that if f and g are measurable functions defined on a measurable set E then f – g is measurable on E. Here f+ – f–1 is measurable. and hence f is measurable.

LOVELY PROFESSIONAL UNIVERSITY 111 Measure Theory and Functional Analysis

Notes Theorem 4: If f is a measurable function and f = g a.e. then g is measurable. Proof: Let E = {x : f (x) g (x)}. Then m (E) = 0 Let be a real number. {x : g (x) > } = {x : f (x) > } {x E : g (x) > } – {x E : g (x) } since f is measurable, the first set on the right is measurable i.e. {x : f (x) > } is measurable. The last two sets on the right are measurable since they are subsets of E and m (E) = 0. Thus, {x : g (x) > } is measurable. So, g is measurable.

Example: Give an example of function for which f is not measurable but |f| is measurable. Sol: Let k be a non-measurable subset of E = [0, 1). Define a function f : E R by

1 if x k f (x) = 1 if x k The function f is not measurable, since E (f > 0) (=k) is a non-measurable set. But |f| is measurable as the set

E if 1 E (|f| > ) = if 1 is measurable

10.1.5 Characteristic Function

Definition: Let A be subset of real numbers. We define the characteristic function A of the set A as follows:

1 if x A (x) = A 0 if x A

Note The characteristic function A of the set A is also called the indicator function of A.

Theorem 5: Show that the characteristic function A is measurable iff A is measurable.

Proof: Let A be measurable.

Since A = {x : A (x) > 0} is measurable.

But A is measurable, therefore the set {x : A (x) > 0} is measurable. A is measurable. Conversely, let A be measurable and be any real number. if 1 A if 0 1 then E ( A > ) = E if 0

112 LOVELY PROFESSIONAL UNIVERSITY Unit 10: Measurable Functions

Every set on R.H.S. is measurable. Notes

Therefore E ( A > ) is measurable.

Hence A is measurable.

Note The above theorem asserts that the characteristic function of non-measurable sets are non-measurable even though the domain set is measurable.

10.1.6 Simple Function

A real valued function is called simple if it is measurable and assumes only a finite number of values.

If is simple and has the values 1, 2, … n, then

= i Ai i 1

where Ai = {x : (x) = i} and Ai Aj is a null set. Thus we can always express a simple function as a linear combination of characteristic function.

Notes

(i) is simple A is are measurable. (ii) sum, product and difference of simple functions are simple. (iii) the representation of as given above is not unique.

But if is simple and { 1, 2, ……, n} is the set of non-zero values of f, then

= i Ai i 1

where Ai = {x : (x) = i}

This representation of is called the Canonical representation. Here A is are disjoint

and is are distinct and non-zero. (iv) Simple function is always measurable.

10.1.7 Step Function

A real valued function S defined on an interval [a, b] is said to be a step function if these is a partition a = xo < x1 … < xn = b such that the function assumes one and only one value in each interval.

LOVELY PROFESSIONAL UNIVERSITY 113 Measure Theory and Functional Analysis

Notes

(i) Step function also assumes finite number of values like simple functions but the sets

{x : S (x) = Ci} are intervals for each i. (ii) Every step function is also a simple function but the converse is not true.

1, x is rational e.g. f : R R such that f (x) = 0, x is irrational

is a simple function but not step as the sets of rational and irrational are not intervals.

Theorem 6: If f and g are two simple functions then f + g is also a simple function. Proof: Since f and g are simple functions and we know that every simple function can be expressed as the linear combination of characteristic function. f and g can be expressed as the linear combination of characteristic function.

f = i Ai i 1

and g = j Bj j 1

where A is and B js are disjoint.

Ai = {x : f (x) = i}

Bj = {x : g (x) = j}

The set Ek obtained by taking all intersections Ai Bj from a finite disjoint collection of measurable sets and we may write

f = ak Ek k 1

and g = bk Ek k 1

where n = mm .

n n f + g = ak Ek b k E k k 1 k 1

= ( ak bk ) E k k 1

which is a linear combination of characteristic functions, therefore it is simple.

114 LOVELY PROFESSIONAL UNIVERSITY Unit 10: Measurable Functions

n Notes

Similarly fg = ak b k Ek k 1 which is again a linear combination of characteristic function, therefore fg is simple.

Theorem 7: Let E be a measurable set with m (E) < and {fn} a sequence of measurable functions converging a.e. to a real valued function defined on E. Then, given > 0 and > 0, there is a set

A E with m (A) < and an integer N such that |fn (x) – f (x)| < for all x E – A and all n N.

Proof: Let F be the set of points of E for which fn f. Then m (F) = 0 and fn (x) f (x) for all x

E – F = E1 (say). Then by the previous theorem for the set E1, we get A1 E1 with m (A1) < and an integer N such that

|fn (x) – f (x) | < for all n N and x E1 – A1. We get the required result by taking

A = A1 F since m (F) = 0 and E – A = E1 – A1

Note Before proving this theorem first prove the previous theorem.

10.1.8 Convergent Sequence of Measurable Function

Definition: A sequence {fn} of measurable functions is said to converge almost uniformly to a measurable function f defined on a measurable set E if for each > 0 there exists a measurable set A E with m (A) < such that {fn} converges to f uniformly an E – A.

10.1.9 Egoroff's Theorem

Statement: Let E be a measurable set with m (E) < and {fn} a sequence of measurable functions which converge to f a.e. on E. Then, given > 0 there is a set A E with m (A) < with that the sequence {fn} converges to f uniformly on E – A.

Proof: Applying the theorem, “Let E be a measurable set with m (E) < and {fn} a sequence of measurable function converging a.e. to real valued function f defined on E. Then given > 0 and > 0 there is a set A E with m (A) < and an integer N such that

|fn (x) – f (x) | < for all x E – A and all n N” with = 1, = /2, we get a measurable set

A1 E with m (A1) < /2 and a positive integer N1, such that

|fn (x) – f (x)|< 1 for all x N1 and x E1, where E1 = E – A1. Again, taking = 1/2 and = n/22,

2 we get a measurable set A2 E1 with m (A2) < /2 , and a positive integer N2 such that

1 |f (x) – f (x)| < n N and x E where E = E – A , and so on. n 2 2 2 2 1 2

LOVELY PROFESSIONAL UNIVERSITY 115 Measure Theory and Functional Analysis

Notes At the pth stage, we get a measurable set

n A E with m (A ) < and a positive integer N such that p p–1 p 2p p

1 |f (x) – f (x)| <  n N and x E where n p p p

Ep = Ep – 1 – Ap.

Let A = Ap , p 1

then m (A) m(Ap ) p 1

n But m (A ) < p 2p

n m (A) < p p 1 2

1 1 1 a But is a G.P. series so S = 2 2 = 1 2p 1 r 1 1 p 1 1 2 2

m (A) <

Also, E – A = E – A  p p

= (EA)  p p

= (EA)  p 1 p p

= E  p p

Let x E – A. Then x Ep p and so

1 |f (x) – f (x)| < , n N . n p p

1 Let us choose p such that < , we get p

|fn (x) – f (x)| < x E – A and n Np = N

116 LOVELY PROFESSIONAL UNIVERSITY Unit 10: Measurable Functions

Notes

Note Egoroff’s theorem can be stated as: almost every where convergence implies almost uniform convergence.

10.1.10 Riesz Theorem

Let {fn} be a sequence of measurable functions which converges in measure to f. Then there is a subsequence fnk which converges in measure to f a.e.

Proof: Let { n} and { n} be two sequences of positive real numbers such that n 0 as n and

n . n 1

Let us now choose an increasing sequence {nk} of positive integers as follows. Let n be a positive integer such that

m x Ef(x)n f(x)

Since fn f in measure for a given 1 > 0 and 1 > 0, a positive integer n1 such that

mxE,f(x)f(x)n1 1 1 , n 1 n

Similarly, let n2 be a positive number such that mxE,f(x)f(x)n2 2 2 , n 2 n 1 and so on.

In general let nk be a positive number such that

m x : x E, fnk (x) f(x) k k and that n k n k 1 .

We shall now prove that the subsequence fnk converges to f a.e.

Let A = x : x E, f (x) f(x) ,k N and A A . k ni i  k i k k 1

Clearly, {Ak} is a decreasing sequence of measurable sets and m (A1) < . Therefore, we have

m (A) = limm(Ak ) k

But m (Ak) i 0 as k . i k

Hence m (A) = 0.

Now it remains to show that {fn} converges to f on E – A. Let xo E – A.

LOVELY PROFESSIONAL UNIVERSITY 117 Measure Theory and Functional Analysis

Notes A Then xo ko for some positive integer ko.

or xo {x E : |fn (x) – f (x)| k}, k ko

which gives |fn (xo) – f (xo)| < k, k ko

But k 0 as k .

Hence lim fn (x o ) f(x o ) . k k

10.2 Summary

 Let E be a measurable set and R* be a set of extended real numbers. A function f : E R* is said to be a Lebesgue measurable function on E or a measurable function on E iff the set E (f > ) = {x E : f (x) > } = f–1 {( , )} is a measurable subset of E R.

 A property is said to hold almost everywhere (a.e.) if the set of points where it fails to hold is a set of measure zero.

 Two functions f and g defined on the same domain E are said to be equivalent on E, written

as f ~ g on E, if f = g a.e. on E, i.e. f (x) = g (x) for all x E – E1, where E1 E with m (E1) = 0.

 f+ = max (f, 0) and f–1 = max (–f, 0) |f| = f+ + f–1

 Let A be subset of real numbers. We define the characteristic function A of the set A as follows:

1, if x A (x) = A 0, if x A

 A real valued function is called simple if it is measurable and assumes only a finite number of values.

10.3 Keywords

Almost Everywhere (a.e.): A property is said to hold almost everywhere (a.e.) if the set of points where it fails to hold is a set of measure zero. Characteristic Function: Let A be subset of real numbers. We define the characteristic function

A of the set A as follows:

1 if x A (x) = A 0 if x A

Egoroff’s Theorem: Let E be a measurable set with m (E) < and {fn} a sequence of measurable functions which converge to f a.e. on E. Then given n > 0 there is a set A E with m (A) < n such

that the sequence {fn} converges to f uniformly on E – A. Equivalent Functions: Two functions f and g defined on the same domain E are said to be

equivalent on E, written as f ~ g on E, if f = g a.e. on E, i.e. f (x) = g (x) for all x E – E1, where E1

E with m (E1) = 0.

118 LOVELY PROFESSIONAL UNIVERSITY Unit 10: Measurable Functions

Notes Haracteristic Function: Let A be subset of real numbers. We define the characteristic function A of the set A as follows: 1 if x A (x) = A 0 if x A

Lebesgue Measurable Function: A function f : E R* is said to be a Lebesgue measurable function on E or a measurable function on E iff the set

E (f > ) = {x E : f (x) > } = f–1 { , )} is a measurable subset of E R. Measurable Set: A set E is said to be measurable if for each set T, we have m* (T) = m* (T E) + m* {T Ec) Non-negative Functions: Let f be a function, then its positive part, written f+ and its negative part, written f–1, are defined to be the non-negative functions given by f+ = max (f, 0) and f–1 = max (–f, 0) respectively.

Riesz Theorem: Let {fn} be a sequence of measurable functions which converges in measure to f.

Then there is a subsequence fnk which converges to f a.e.

Simple Function: A real valued function is called simple if it is measurable and assumes only a finite number of values.

If is simple and has the values 1, 2, … n, then

= i Ai i 1

where Ai = {x : (x) = i} and Ai Aj is a null set. Step Function: A real valued function S defined on an interval [a, b] is said to be a step function if these is a partition a = xo < x1 … < xn = b such that the function assumes one and only one value in each interval.

Subsequence: If (xn) is a given sequence in X and (nk) is an strictly increasing sequence of positive integers, then xnk is called a subsequence of (xn).

10.4 Review Questions

1. If f is a measurable function and c is a real number, then is it true to say that cf is measurable? 2. A non-zero constant function is measurable if and only if X is measurable comment. 3. Let Q be the set of rational number and let f be an extended real-valued function such that {x : f (x) > } is measurable for each Q. Then show that f is measurable. 4. Show that if f is measurable then the set {x : f (x) = } is measurable for each extended real number . 5. If f is a continuous function and g is a measurable function, then prove that the composite function fog is measurable.

LOVELY PROFESSIONAL UNIVERSITY 119 Measure Theory and Functional Analysis

Notes 6. Show that

(i) ABAB

(ii) ABABAB

1 (iii) Ac A

10.5 Further Readings

Books Dudley, R.M. (2002). Real Analysis and Probability (2 ed.). Cambridge University Press Strichartz, Robert (2000). The Way of Analysis. Jones and Bortlett.

Online links mathworld.wolfram.com>calculus and Analysis > Measure theory planetmath.org/Measurable functions.html zeta.math.utsa.edu/~ mqr 328/class/real2/mfunct.pdf

120 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

Unit 11: Integration Notes

CONTENTS

Objectives Introduction 11.1 Integration 11.1.1 The Riemann Integral 11.1.2 Lebesgue Integral of a Bounded Function over a Set of Finite Measure 11.1.3 The Lebesgue Integral of a Non-negative Function 11.1.4 The General Lebesgue Integral 11.2 Summary 11.3 Keywords 11.4 Review Questions 11.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define the Riemann integral and Lebesgue integral of bounded function over a set of finite measure.

 Understand the Lebesgue integral of a non-negative function.

 Solve problems on integration.

Introduction

We now come to the main use of measure theory: to define a general theory of integration. The particular case of the integral with respect to the Lebesgue measure is not, in any way, simpler the general case, which will give us a tool of much wider applicability.

11.1 Integration

11.1.1 The Riemann Integral

Let f be a bounded real valued function defined on the interval [a, b] and let a = x0 < x1 < … < xn= b be a sub-division of [a, b]. Then for each sub-division we can define the sums

S = (xi x i 1 ) M i i 1

n and s = (xi x i 1 ) m i i 1

LOVELY PROFESSIONAL UNIVERSITY 121 Measure Theory and Functional Analysis

Notes sup f(x) where Mi = , xi 1 x x i

mi = inf f(x) xi 1 x x i

We then define the upper Riemann integral of f by

b R f(x) dx inf S

where the infimum is taken over all possible sub-divisions of [a, b]. Similarly, we define the lower Riemann integral

b R f(x) dx sup S

The upper integral is always at least as large as the lower integral, and if the two are equal, we say that f is Riemann integrable and we call this common value the Riemann integral of f. It will be denoted by

b R f(x) dx

Note By a step function we mean function s.t.

(x) = i x [xi–1, xi]

for some sub-division of [a, b] and some set of constant i then

b x1 x n (x) dx = (x) dx (x) dx

a xo x n 1

x1 x 2 x n

= 1dx 2 dx n dx

xo x 1 x n 1

= 1 (x1 – x0) + 2 (x2 – x1) + … + n (xn–1 – xn)

= i(x i x i 1 ) … (1) i 1

with this in mind, we see that

R f (x) dx = infp (f)

122 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

n Notes

= inf Mi (x i x i 1 ) i 1

b = inf (x) dx for all step functions

(x) f (x)

Similarly R f (x) dx = sup Lp (f) a

= sup mi (x i x i 1 ) i 1

b = sup (x) dx for all step function a

(x) f (x).

11.1.2 Lebesgue Integral of a Bounded Function over a Set of Finite Measure

Characteristic Function

The function E defined by

1 if x E (x) = E 0 if x E is called the characteristic function of E.

Simple Function

A linear combination (x) = i Ei (x) is called a simple function if the sets Ei are measurable. i 1

This representation of is not unique. However, a function is simple if and only if it is measurable and assumes only a finite number of values.

Canonical Representation

If is simple function and { 1, 2, …, n} the set of non-zero values of , then

= i Ei , i 1

where Ei = {x : (x) = i}.

LOVELY PROFESSIONAL UNIVERSITY 123 Measure Theory and Functional Analysis

Notes This representation of is called the canonical representation. Here Ei’s are disjoint and i’s are finite in number, distinct and non-zero.

Elementary Integral

Definition: If vanishes outside a set of finite measure, we define the elementary integral of by

(x) dxi mE i when has the canonical representation. i 1

= i Ei . i 1

We sometimes abbreviate the expression for this integral . If E is any measurable set, we

define the elementary integral of over E by E . E

b If E = [a, b], then the integral will be denoted by .

[a, b] a

Theorem 1: Let and be simple functions which vanish outside a set of finite measure, then

(a b ) = a b and if a.e., then

Proof: Since and are simple functions. Therefore these can be written in the canonical form

= i Ai

and = Bj Bj j 1

where {Ai} and {Bj} are disjoint sequences of measurable sets and

Ai = {x : (x) = i}

and Bj = {x : (x) = j}

The set Ek obtained by taking all intersections Ai Bj form a finite disjoint collection of measurable sets. We may write

= ak Ek and k 1

= bk Ek (where N = mm ) k 1

124 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

NN Notes

Now a + b = a ak Ek b b k E k k 1 k 1

= (aak bb k ) Ek k 1 which is again a simple function.

Since = ai m E i i

(a b ) = (aak bb k ) m E k (by definition) k 1

= a ak mE k b b k mE k k 1 k 1

= a b

Now since a.e. – 0 a.e. We have proved that

(a b ) = a b

Put a = 1, b = –1 in the first part, we get

() =

Since – 0 a.e. is a simple function, by the definition of the elementary integral, we have

() 0

Theorem 2: Riemann integrable is Lebesgue integrable. Proof: Since f is Riemann integrable over [a, b], we have

b b b

inf1 (x) dx sup 1 (x) dx R f(x) dx 1 f f a a a

where 1 and 1 vary over all step functions defined on [a, b].

LOVELY PROFESSIONAL UNIVERSITY 125 Measure Theory and Functional Analysis

Notes Since we know that every step function is a simple function,

b b

sup1 (x) dx sup (x) dx 1 f f a a

b b and inf (x) dx inf1 (x) dx f 1 f a a

where and vary over all the simple functions defined on [a, b]. Thus from the above relation, we have

b b b b R f(x) dx sup (x) dx inf (x) dx R f(x) dx f f a a a a

b b sup (x) dx inf (x) dx f f a a

b b f(x) dx R f(x) dx

a a

Note The converse of this theorem is not true i.e. A Lebesgue integrable function may not be Riemann integrable e.g. Let f be a function defined on the interval [0, 1] as follows:

1, if x is rational f (x) = 0, if x is irrational

Let us consider a partition p of an interval [0, 1].

U (p, f) = Mi x i i 1

= 1 x1 + 1 x2 + …… + 1 xn = 1 – 0 = 1.

1 f dx = inf U (p, f) = 1 – 0 = 1.

1 f dx = sup L (p, f)

= sup {0 x1 + 0 x2 + …… + 0 xn} = 0

126 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

Notes Thus f dx f dx

The function is not Riemann integrable.

Now for Lebesgue Integrability

Let A1 be the set of all irrational numbers and A2 be the set of all rational numbers in [0, 1].

The partition P = {A1, A2} is a measurable partition of [0, 1] and mA1 = 0, mA2 = 1.

L (p, f) = inf f(x) mA1 inf f(x) mA 2 AA1 2

= 0 mA1 1 mA 2 = 1.

U (p, f) = sup f(x) mA1 sup f(x) mA 2 AA1 2

= 0 mA1 1 mA 2 = 1.

sup L (p, f) = 1 inf U(p, f) p p

f is Lebesgue integrable over [0, 1]. Theorem 3: If f and g are bounded measurable functions defined on the set E of finite measure, then

(1) (af bg) a f b g

EEE

(2) If f = g a.e., then f g

(3) If f g a.e., then f g

Hence f |f|

(4) If A and B are disjoint measurable set of finite measure, then

f f f

ABAB

Proof of 1: Result is true if a = 0 Let a 0. If is a simple function then so is a and conversely. Hence for a > 0

af = inf a a af E E

LOVELY PROFESSIONAL UNIVERSITY 127 Measure Theory and Functional Analysis

Notes = inf a ( a > 0) f E

= inf a f E

= a inf f E

= a f

Again if a < 0,

af inf a = a af E E

= sup a ( a < 0) f E

= sup a f E

= a sup f E

= a f

Therefore in each case

af = a f … (i)

E E

Now we prove that

(f g) = f g

E EE

Let 1 and 2 be two simple functions such that 1 > f and 2 g, then 1 + 2 is a simple

function and 1 + 2 f + g.

or f + g = 1 + 2

(f g) ()1 2 E E

But 1 2 = 1 2 E EE

128 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

Notes

(f g) 1 2 E EE

Since inf 1 = f f 1 E E

and inf 2 = g g 2 E E

(f g) f g … (2)

E EE

On the other hand if 1 and 2 are two simple functions such that 1 < f and 2 g. Then 1 + 2 is simple function and

1 + 2 f + g, or f + g 1 + 2

(f g) ()1 2 E E

But ()1 2 = 1 2 E EE

(f g) 1 2 E EE

Since sup 1 = f f 1 E E

and sup 2 = g f 2 E E

(f g) f g … (3)

E EE

From (2) and (3), we get

(f g) = f g

E EE

(af bg) = af bg

E EE

= a f b g from (i)

Proof of 2: Since f = g a.e.

f – g = 0 a.e.

LOVELY PROFESSIONAL UNIVERSITY 129 Measure Theory and Functional Analysis

Notes Let F = {x : f (x) g (x)} Then by definition of a.e., we have mF = 0 and F E.

(f g) = (f g) (f g) (f g)

E F(EF)FEF

= (f – g) mF + (f – g) m (E – F) = (f – g) . 0 + 0 . m (E – F) [ mF = 0 and f – g = 0 over E – F] = 0

(f g) = 0

f g = 0 f g

EE EE

Note Converse need not be true e.g. Let the functions f : [–1, 1] R and g : [–1, 1] R be defined by

2 if x 0 f (x) = 0 if x 0

and g (x) = 1 x.

1 1 Then f(x) dx = 2 = g(x) dx

1 1

But f g a.e. In other words, they are not equal even for a single point in [–1, 1].

Proof of 3: f g a.e. f – g 0 a.e. Let be simple function, = f – g 0 [ f – g = 0 a.e.]

(f g) 0

f g 0

130 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

Notes f g

E E

Since f |f|

f |f| … (1)

E E

Again – f |f|

f |f|

E E or |f| f … (2)

E E

From (1) and (2) we get

|f| f |f|

E E E

f |f|

E E

Proof of 4: It follows from (3) and the fact that 1 mE .

Proof of 5: f = f AB AB

Now AB = ABAB where A and B are disjoint measurable sets i.e. A B =

f = f(ABAB ) f AB

= fAB f 0 [ A B = and m ( ) = 0]

= f f

11.1.3 The Lebesgue Integral of a Non-negative Function

Definition: If f is a non-negative measurable function defined on a measurable set E, we define

f = sup h , h f E E

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Notes where h is a bounded measurable function such that m {x : h (x) 0} < 

Theorem 4: Let f be a non-negative measurable function. Show that f 0 implies f = 0 a.e.

Proof: Let be any measurable simple function such that f. Since f = 0 a.e. on E 0 a.e.

(x) dx 0

Taking supremum over all those measurable simple functions f, we get

f dx 0 … (1)

Similarly let be any measurable simple function such that f Since f = 0 a.e. 0 a.e.

(x) dx 0

Taking infimum over all those measurable simple functions f, we get

f dx 0 … (2)

From (1) and (2), we get

f dx = 0.

Conversely, let f dx = 0

1 If E = x : f(x) , then n n

1 f dx > (x) dx n En E E

1 1 But E (x) dx = mE (By definition) n n n n E

132 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

Notes 1 f dx > mE n n E

1 Or mE < f dx n n E

But f dx < 0

1 mE < 0 n n

m En < 0

But m En 0 is always true

m En = 0

But {x : f (x) > 0} =  En n 1

and m En = 0

m En = 0 n 1

m {x : f (x) > 0} = 0 f = 0 a.e. on E Theorem 5: Let f and g be two non-negative measurable functions. If f is integrable over E and g (x) < f (x) on E, then g is also integrable over E, and

(f g) = f g .

E EE

Proof: Since we know that if f and g are non-negative measurable functions defined on a set E, then

(f g) = f g

E EE

Since f = (f – g) + g, therefore we have

f = (f g g) (f g) g … (1)

E EEE

Since the functions f – g and g are non-negative and measurable. Further, f being integrable over

E, f < (by definition)

LOVELY PROFESSIONAL UNIVERSITY 133 Measure Theory and Functional Analysis

Notes Therefore, each integral on the right of (1) is finite.

In particular, g < ,

which shows that g is an integrable function over E.

Since f = (f g) g

E EE

f g = (f g) .

EE E

11.1.4 The General Lebesgue Integral

For the positive part f+ of a function f, we define f+ = max (f, 0)

and negative part f– by f– f– = max (–f, 0) and that f is measurable if and only if both f+ and f– are measurable.

Note f = f+ – f– and |f| = f+ + f–

Definition: A measurable function f is said to be Lebesgue integrable over E if f+ and f– are both

Lebesgue integrable over E. In this case, we define f f f .

EEE

Theorem 6: Let f and g be integrable over E, then

(a) The function of f is integrable over E, and cf c f .

(b) Sum of two integrable functions is integrable i.e. the function f + g is integral over E, and

(f g) = f g

E EE

(c) If f g a.e., then f g .

(d) If A and B are disjoint measurable sets contained in E, then

f = f f .

AB AB

134 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

Proof: (a) If c 0, then Notes (cf)+ = cf+ (cf)– = cf and if c < 0, then

(cf)* = (–c) . f (cf)– = (–c) . f+ Since f is integrable so f+ and f– are also integrable and conversely. Hence the result follows.

(b) In order to prove the required result first of all we show that if f1 and f2 are non-negative

integrable functions such that f = f1 – f2, then

f = f1 f 2 … (1) E EE

Since f = f+ – f–,

Also f = f1 – f2,

+ – then f – f = f1 – f2

+ – f + f2 = f1 + f … (2) Also we know that if f and g are non-negative measurable functions defined on a set E, then

(f g) = f g

E EE

Then from (2), we get

f f2 = f1 f EE EE

f f = f1 f 2 … (3) EE EE

But f is integrable so f+ and f– are integrable i.e.

f = f f

Therefore (3) becomes

Hence f = f1 f 2 E EE

which proves (1). Now, if f and g are integrable functions over E, then f+ + g+, f– + g– and f + g = (f+ + g+) – (f– + g–) and also integrable functions over E.

LOVELY PROFESSIONAL UNIVERSITY 135 Measure Theory and Functional Analysis

Notes Furthermore f and g are integrable, it implies that |f| and |g| are integrable. ( A measurable function f is integrable over E if and only if |f| is integrable over E.) Thus |f| + |g| is integrable over E.

( (f g) f g and by the definition of integrable)

EEE

Since |f + g| |+|f| + |g| which shows that f + g is integrable. Hence sum of two integrable functions is integrable.

Thus (f g) = (f g ) (f g )

E EE

= f g f g

EEEE

= f f g g

EEEE

= f g

(c) f g a.e. f – g 0 a.e. g – f 0 a.e.

(g f) 0

Since g = f + (g – f) and f, g – f are integrable over E. Then by the given hypothesis (g – f)– = 0 a.e.

then (g f) = 0,

(Since we know that if f = 0 a.e. then f = 0)

g f (gf) f (gf) (gf)

EEEEEE

becomes

g = f (gf) 0 f (gf) ( (g – f) 0)

E EEEE

136 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

Notes f

g f f g

E EEE

(d) f = f AB AB

= fAB f

= f f .

Example: Let f be a non-negative integrable function. Show that the function F defined by

x F (x) = f(t) dt is continuous on R.

Solution: Since f is a non-negative integrable function, then given > 0 there is a > 0 such that for every set A R with m (A) < , we have

f <

If xo R, then x R with |x – xo| < , we have

x f (t) dt <

x f (t) dt f (t) dt <

x f (t) dt f (t) dt < xo

x xo f (t) dt f (t) dt <

|F (x) – F (xo)| <

Hence F is continuous at xo. Since xo R is arbitrary, F is continuous on R.

LOVELY PROFESSIONAL UNIVERSITY 137 Measure Theory and Functional Analysis

Notes 11.2 Summary

 Let and be simple functions which vanish outside a set of finite measure, then

(a b ) a b and if a.e., then

 A Lebesgue integrable function may not be Riemann integrable.

 Let A1 be the set of all irrational numbers and A2 be the set of all rational numbers in [0, 1].

 If f is a non-negative measurable function defined on a measurable set E, we define

f sup h h f , EE

where h is a bounded measurable function such that m {x : h (x) 0} <

 Let f and g be two non-negative measurable functions. If f is integrable over E and g (x) < f (x) on E, then g is also integrable over E, and

(f g) f g . EEE

11.3 Keywords

Canonical Representation: If is simple function and { 1, 2, …, n} the set of non-zero values of , then

= i Ei , i 1

where Ei = {x : (x) = i}.

Characteristic Function: The function E defined by

1 if x E (x) = E 0 if x E

is called the characteristic function of E. Elementary Integral: If vanishes outside a set of finite measure, we define the elementary

integral of by (x) dxi mE i when has the canonical representation. i 1

= i Ei . i 1

Lebesgue Integrable: A measurable function f is said to be Lebesgue integrable over E if f+ and f–

are both Lebesgue integrable over E. In this case, we define f f f .

EEE

138 LOVELY PROFESSIONAL UNIVERSITY Unit 11: Integration

n Notes

Simple Function: A linear combination (x) = i Ei (x) is called a simple function if the sets i 1

Ei are measurable.

Simple Function: A linear combination (x) = i Ei (x) is called a simple function if the sets Ei i 1 are measurable. This representation of is not unique. However, a function is simple if and only if it is measurable and assumes only a finite number of values. The Lebesgue Integral of a Non-negative Function: If f is a non-negative measurable function defined on a measurable set E, we define

f = sup h , h f E E where h is a bounded measurable function such that m {x : h (x) 0} <  The Riemann Integral: Let f be a bounded real valued function defined on the interval [a, b] and let a = x0 < x1 < … < xn= b be a sub-division of [a, b]. Then for each sub-division we can define the sums

S = (xi x i 1 ) M i i 1

n and s = (xi x i 1 ) m i i 1

sup f(x) where Mi = , xi 1 x x i

mi = inf f(x) xi 1 x x i

11.4 Review Questions

1. Prove that af a f real number a.

2. If f is bounded real valued measurable function defined on a measurable set E of finite

measure such that a f (x) b, then show that amE f bmE.

3. If f and g are non-negative measurable functions defined on E  then prove that

(a) cf c f, c 0

LOVELY PROFESSIONAL UNIVERSITY 139 Measure Theory and Functional Analysis

Notes (b) (f g) f g

EEE

(c) f 0 f 0 a.e.

(d) If f g a.e. then f g

4. If f is integrable over E, then show that |f| is integrable over E, and f |f|.

5. Show that if f is a non-negative measurable function then f = 0 a.e. on E iff f = 0.

6. If f = 0 and f (x) 0 on E, then f = 0 a.e.

11.5 Further Readings

Books Erwin Kreyszig, Introductory Functional Analysis with Applications, John Wiley & Sons Inc., New York, 1989 Walter Rudin, Real and Complex Analysis, Third McGraw Hill Book Co., New York, 1987 R.G. Bartle, The Elements of Integration and Lebesgue Measure, Wiley Interscience, 1995

Online links www.maths.manchester.ac.uk www.uir.ac.za

140 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

Unit 12: General Convergence Theorems Notes

CONTENTS Objectives Introduction 12.1 General Convergence Theorems 12.1.1 Convergence almost Everywhere 12.1.2 Pointwise Convergence 12.1.3 Uniform Convergence, Almost Everywhere (a.e.) 12.1.4 Bounded Convergence Theorem 12.1.5 Fatou’s Lemma 12.1.6 Monotone Convergence Theorem 12.1.7 Lebesgue Dominated Convergence Theorem 12.2 Summary 12.3 Keywords 12.4 Review Questions 12.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand bounded convergence theorem.

 State and prove monotone convergence theorem and Lebesgue dominated convergence theorem.

 Solve related problems on these theorems.

Introduction

Convergence of a sequence of functions can be defined in various ways and there are situations in which each of these definitions is natural and useful. In this unit, we shall study about convergence almost everywhere, pointwise and uniform convergence. We shall also prove bounded convergence theorem and monotone convergence theorem which are so useful in solving problems on convergence. The dominated convergence theorem is one of the most important results of Lebesgue’s integration theory. It gives a general sufficient condition for the validity of proceeding to the limit of a sequence of functions under the integral sign. It is an invaluable tool to study functions defined by integrals.

12.1 General Convergence Theorems

12.1.1 Convergence almost Everywhere

Let be a sequence of measurable functions defined over a measurable set E. Then is said to converge almost everywhere in E if there exists a subset E0 of E s.t.

LOVELY PROFESSIONAL UNIVERSITY 141 Measure Theory and Functional Analysis

Notes (i) fn(x) f (x), x E – E0,

and (ii) m (E0) = 0.

12.1.2 Pointwise Convergence

Let be a sequence of measurable functions on a measurable set E. Then is said to converge “pointwise” in E, if a measurable function f on E such that

fn(x) f (x) x E or

lt fn (x) = f (x) n

12.1.3 Uniform Convergence, Almost Everywhere (a.e.)

Let be a sequence of measurable functions defined over a measurable set E. Then the

sequence is said to converge uniformly a.e. to f, if a set E0 E s.t.

(i) m (E0) = 0 and

(ii) converges uniformly to f on the set E – E0.

12.1.4 Bounded Convergence Theorem

Theorem 1: State and Prove: Bounded Convergence Theorem

Statement: Let {fn} be a sequence of measurable functions defined on a set E of finite measure, and

suppose that there is a real number M such that |f (x)| M n and x. If f (x) = lim fn (x) for each n n x in E, then

f = lim fn n E E

Proof: Since f (x) = lim fn (x) and f is measurable on E n n E

f is also measurable on E Let > 0 be given

Then measurable set A E with mA < and a positive integer N such that 4M

|f (x) – f (x)| < n N and x E – A n 2mE

fn f = (fn f) EE E

(f f) n E

142 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

Notes

= (fn f) (f n f) as (E – A) A = EAA

1 f f 2mE n EAA

m(E A) 2M 1 2mE A

mE 2M mA as m (E – A) mE 2mE

< 2M 2 4M

= 2 2

Thus fn f < EE

But was arbitrary

lim fn = f n E E

12.1.5 Fatou’s Lemma

If {fn} is a sequence of non-negative measurable functions and fn(x) f (x) almost everywhere on a set E, then

f lim fn n E E

i.e. f lim inf fn n E E

Proof: Since integrals over sets of measure zero are zero. Without loss of generality, we may assume that the convergence is everywhere. Let h be a bounded measurable function with h f and h (x) = 0 outside a set E E of finite measure.

Define a function hn by

hn(x) = Min. {h(x), fn(x)}

then hn(x) h(x) and hn(x) fn(x)

hn is bounded by the boundedness of h and vanishes outside E as x E – E h(x) = 0

hn(x) = 0 because

LOVELY PROFESSIONAL UNIVERSITY 143 Measure Theory and Functional Analysis

Notes Since hn = h or hn = fn

hn is measurable function on E

If hn = h, then hn h

If hn = fn < h f

then fn h as fn f

hn h

Thus hn h

Since hn (x) h (x) for each x E and {hn} is a sequence of bounded measurable functions on E By Bounded Convergence Theorem

h h h h lim h n n EEEEEE

as E = (E – E ) E & (E – E ) E =

lim h n = n E

lim fn as h n f n n E

lim fn as E E n E

h lim fn n EE

Taking supremum over all h f, we get

sup h h lim fn n f n E EE

f f lim fn n E EE

Remarks:

(1) If in Fatou’s Lemma, we take

1, n x n 1 f (x) = n 0, otherwise

with E = R

then f lim fn n EE

Thus in Fatou’s Lemma, strict inequality is possible.

144 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

Notes (2) fn 0 x E is essential for Fatou’s Lemma However, if we take

1 2 n, x fn(x) = n n 0, otherwise

with E = [0, 2]

Then f lim fn . n EE

12.1.6 Monotone Convergence Theorem

Statement: Let {fn} be an increasing sequence of non-negative measurable functions and let f = lim fn . Then n

f lim fn n

Proof: Let h be a bounded measurable function with h f and h (x) = 0 outside a set E E of finite measure

Define a function hn by

hn(x) = Min. {h(x), fn(x)}

then hn(x) h(x) and hn(x) fn(x)

hn is bounded by the boundedness of h and vanishes outside E as

x E – E h(x) = 0 hn(x) = 0 because fn(x) 0

Since hn = h or hn = fn

hn is measurable function on E

If hn = h, then hn h

then fn h as fn f

hn h

Thus hn h

Since hn(x) h(x) for each x E and {hn} is a sequence of measurable functions on E By Bounded Convergence Theorem

h h h h lim h n n EEEEEE

as E = (E – E ) E & (E – E ) E = .

= lim h n n E

LOVELY PROFESSIONAL UNIVERSITY 145 Measure Theory and Functional Analysis

Notes

= lim fn n E

lim fn as E E n E

h lim fn n E E

Taking supremum over all h f, we get

sup h lim fn n E E

f lim fn … (1) n E E

Since {fn} is monotonically increasing sequence and fn f

fn f

lim fn f … (2) n

From (1) and (2), we have

f lim fn lim f n f n n

f lim fn n

Theorem 2: Let {un} be a sequence of non-negative measurable functions, and let f = un . h 1

Then f un h 1

Proof: Let fn = u1 + u2 + … + un = u j j 1

then fn f

i.e. lim fn f n

Let h be a bounded measurable function with h f and h(x) = 0 outside a set E E of finite measure.

Define a function hn by

hn(x) = Min. {h (x), fn(x)}

146 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

Notes then hn (x) h(x) and hn(x) fn(x)

hn is bounded by the boundedness of h and vanishes outside E as

x E – E h(x) = 0 hn(x) = 0 because fn(x) 0.

Since hn = h or hn = fn

hn is measurable function on E

If hn = h, then hn h

If hn = fn < h < f

then fn h as fn f

hn h

Thus hn h

Since hn(x) h(x) for each x E and {hn} is a sequence of measurable function on E By Bounded Convergence Theorem

h h h h lim h n n EEEEEE

as E = (E – E ) E & (E – E ) E =

= lim h n n E

= lim fn n E

lim fn as E E n E

h lim fn n E E

Taking supremum over all h f, we get

sup h lim fn h f n … (1) E E

f lim fn n E E

Since {fn} is monotonically increasing sequence and fn f

fn f

lim fn f … (2) n

LOVELY PROFESSIONAL UNIVERSITY 147 Measure Theory and Functional Analysis

Notes From (1) and (2), we have

f lim fn lim f n f n n

f lim fn n

= lim fn n

= lim u j n j 1

lim u j = n j 1

= u j j 1

Hence f = un n 1

Theorem 3: Let f be a non-negative function which is integrable over a set E. Then given > 0 there is a > 0 such that for every set A E with mA < , we have

f <

Proof: If f is bounded function on E Then positive real number M such that

|f (x)| M x E

For given > 0 = such that for every set A E with mA < , we have M

f M M mA M. M M AA

i.e. f

Thus the result is true if f is a bounded function. So assume that f is not a bounded function on E.

Define a function fn on E by

f(x) if f(x) n f (x) n n otherwise

148 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

Notes Then each fn is bounded and

fn f at each point

Since {fn} is an increasing sequence of bounded functions such that fn f on E By the monotone convergence theorem

lim fn f n EE

For given > 0 a positive integer N such that

f f for n N n 2 EE

f f N 2 EE

f f 2N 2 EE

(f f ) N 2 E

Choose 2N

If mA < , then we have

f (f f ) f = NN A A

(f f ) f = NN AA

(f f ) N as f N NN EA

< NmA 2

< N 2

< N 2 2N

= 2 2

LOVELY PROFESSIONAL UNIVERSITY 149 Measure Theory and Functional Analysis

Notes f

12.1.7 Lebesgue Dominated Convergence Theorem

Theorem 4: State and prove Lebesgue dominated convergence theorem

Statement: Let g be an integrable function on E and let {fn} be a sequence of measurable functions

such that |f | g on E and lim fn = f a.e. on E. Then n n

f lim fn n . EE

Proof: Since we know that if f is a measurable function over a set E and there is an integrable

function g such that |f| g, then f is integrable over E. So clearly, each fn is integrable over E.

Also lim fn = f a.e. on E. n

and |fn| g a.e. on E |f| g a.e. on E. Hence f is integrable over E.

Let { n} be a sequence of functions defined by n = fn + g. Clearly, n is a non-negative and integrable function for each n. Therefore, by Fatou’s Lemma, we have

(f g) lim (fn g) n E E

f lim fn … (1) n E E

Similarly, let { n} be a sequence of functions defined by n = g – fn. Clearly n is a non-negative and integrable function for each n. So, given by Fatou’s Lemma, we have

(g f) lim (g fn ) n E E

g f g lim fn n EE EE

f lim fn n E E

f lim fn … (2) E E

Hence from (1) & (2), we get

f = lim fn lim f n E EE

150 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

Notes

But lim fn = lim fn lim f n E EE

Hence f = lim fn . E E

Corollary: Let {un} be a sequence of integrable functions on E such that un converges a.e. on n 1

E. Let g be a function which is integrable on E and satisfy ui g a.e. on E for each n. Then i 1

un is integrable on E and un u n . n 1 EEn 1 n 1

Proof: Let ui f n . i 1

Applying Lebesgue Dominated Convergence Theorem for the sequence {fn}, we get

un u n EEn 1 n 1

Corollary: If f is integrable over E and {Ei} is a sequence of disjoint measurable sets such that

EEi , then i 1

f f i 1 EEi

Proof: Since {Ei} is a sequence of disjoint measurable sets, we may write.

f = f Ei i 1

The function f. Ei is integrable over E since fEi |f| and |f| is integrable over E. Moreover

fEi |f|, n N i 1

Thus the conditions of above corollary are satisfied and hence

= f f Ei E E i 1

LOVELY PROFESSIONAL UNIVERSITY 151 Measure Theory and Functional Analysis

Notes

= f Ei i 1 E

= f i 1 Ei

nx Example: Show that the theorem of bounded convergence applies to f (x) = , 0 n 1 n2 x 2 x 1.

nx Sol: f (x) = n 1 n2 x 2

1 = 1 nx nx

1 = 2 1 nx 2 nx

1 2

1 1 Thus a number such that |f (x)| . 2 n 2

Hence it satisfies the conditions of bounded convergence theorem. Now

1 1 nx lim fn (x) dx = lim dx n 1 n2 x 2 0 0

1 = lim log(1 n2 x 2 ) form n 2n

[1/(1 n2 x 2 )] 2nx 2 = lim [Using L’Hospital Rule] n 2

nx2 = lim n 1 n2 x 2

1 x2 = limn 0 n 1 x2 n2

1 1 nx and lim fn (x) dx = lim dx n n 1 n2 x 2 0 0

152 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

1 Notes = (0) dx 0

1 1

lim fn (x) dx = lim fn (x) dx n n 0 0

This verifies the result of bounded convergence theorem.

Example: Use Lebesgue dominated convergence theorem to evaluate lim fn (x) dx , n 0 where

n3/2 x f (x) = , n = 1, 2, 3, … 0 x 1. n 1 n2 x 2

n3/2 x Solution: f (x) = n 1 n2 x 2

1 n3/2 x 2 = x 1 n2 x 2

1 g(x), (say) x

fn(x) g (x) and g (x) L (0, 1], Hence by Lebesgue Dominated Convergence Theorem.

1 1

lim fn (x) dx = lim fn (x) dx n n 0 0

1 n3/2 x = lim dx n 1 n2 x 2 0

1 1 x = lim dx n n 1 2 0 x n2

1 = 0 dx = 0.

Example: If (fn) is a sequence of non-negative function s.t. fn f and fn f for each n, show that

f lim fn

LOVELY PROFESSIONAL UNIVERSITY 153 Measure Theory and Functional Analysis

Notes Solution: From the given hypothesis it follows that

lim fn f … (1)

Also by Fatou’s Lemma, we have

f lim fn … (2)

Then from (1) and (2), we get

f lim fn lim f n f .

Hence f = lim fn lim f n lim f n .

n x n Example: If > 0, prove that Lim 1 x1 dx e x .x 1 dx , where the integrals are n n o o taken in the Lebesgue sense.

n n x x x Solution: If f (x) = 1 .x1 0 , then f (x) g(x), where g(x) = e–x.x –1 recall Lim 1 e n n n n n

Also g(x) L[0, ], hence by Lebesgue dominated convergence theorem, we get

Lim fn (x) dx = Lim fn (x) dx n n o o

x n = Lim 1 .x1 dx n n o

x 1 = e .x dx o

Example: Show that if > 1,

1 xsin x dx 0(n1 ) as n . 1 (nx) o

Solution: Consider the sequence s.t.

nx sin x f (x) = , n = 1, 2, ……… n 1 (nx)

Obviously since > 1, and x [0, 1]

154 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

Notes nx sin x 1 1 (nx)

If (x) = 1, x, then | fn (x) | (x), x. Hence by dominated convergence theorem, we get

1 1 1 nx sin x nx sin x Lim dx Lim dx (0) dx 0 n1 (nx) n 1 (nx) 0 0 0

1 x sin x Lim n dx = 0 n 1 (nx) 0

1 x sin x dx = 0 (n–1). 1 (nx) 0

2 2 n2 xe n x Example: Show that Lim dx = 0, if a > 0, but not for a = 0. n 1 x2 a

putting nx u Solution: If a > 0, and du ndx , we get

2 2 2 n2 xe n x 2 ue u du ue u dx du 1 x2 1 u 2 /n 2(na, ) 1 u 2 /n 2 a na o

u2 u.e 2 Also u.eu L [0, ] 1 (u2 /n 2 ) (na, )

2 u.e u and Lim(ne, ) 0 as ( , ) 0 . n 1 u2 /n 2

Hence by Lebesgue dominated convergence theorem, we obtain

2 2 2 n2 xe n x u.e u Lim dx Lim(na, ) du n1 x2 n 1 u 2 /n 2 a a

2 u.e u = Lim(na, ) du 0 dx 0 . n 1 u2 /n 2 a o

Now when a = 0,

2 21 2 2 n2 xe n x n 2 xe n x dx dx 1 x2 1 x 2 o 0

LOVELY PROFESSIONAL UNIVERSITY 155 Measure Theory and Functional Analysis

Notes 1 1 2 2 n2 x.e n x dx (putting 1 in place of x2) 2 0

12 2 1 1 = e n x 40 4

12.2 Summary

 Bounded Convergence Theorem: Let {fn} be a sequence of measurable functions defined on

a set E of finite measure, and suppose that there is a real number M such that |fn(x)| < M

n and all x. If f (x) = lim fn (x) for each x in E, then n

f lim fn n EE

 Monotone Convergence Theorem: Let {fn} be an increasing sequence of non-negative

measurable functions and let f = lim fn . Then n

f lim fn n

 Lebesgue Dominated Convergence Theorem: Let g be an integrable function on E and let {fn}

be a sequence of measurable functions such that |f | g on E and lim fn = f a.e. on E. Then n n

f lim fn n EE

12.3 Keywords

Convergence almost Everywhere: Let be a sequence of measurable functions defined over a

measurable set E. Then is said to converge almost everywhere in E if there exists a subset E0 of E s.t.

(i) fn(x) f (x), x E – E0,

and (ii) m (E0) = 0. Convergence: Refers to the notion that some functions and sequence approach a limit under certain conditions.

Fatou’s Lemma: If {fn} is a sequence of non-negative measurable functions and fn(x) f (x) almost everywhere on a set E, then

f lim inf fn n EE

Pointwise Convergence: Let be a sequence of measurable functions on a measurable set E.

Then is said to converge “pointwise” in E, if a measurable function f on E such that

fn(x) f (x) x E or

lt fn (x) = f (x) n

156 LOVELY PROFESSIONAL UNIVERSITY Unit 12: General Convergence Theorems

Notes Uniform Convergence, Almost Everywhere (a.e.): Let be a sequence of measurable functions defined over a measurable set E. Then the sequence is said to converge uniformly a.e. to f, if a set E0 E s.t.

(i) m (E0) = 0 and

(ii) converges uniformly to f on the set E – E0.

12.4 Review Questions

1. Show that we may have strict inequality in Fatou’s Lemma.

2. Let be an increasing sequence of non-negative measurable functions, and let f = lim fn.

Show that f lim fn .

Deduce that f un , if un is a sequence of non-negative measurable functions and n 1

f un . n 1

3. State the Monotone Convergence theorem. Show that it need not hold for decreasing sequences of functions.

4. Let {gn} be a sequence of integrable functions which converge a.e. to an integrable function

g. Let {fn} be a sequence of measurable functions such that |fn| gn and {fn} converges to f a.e.

If g lim gn n

then prove that f lim fn . n

5. State and prove monotone convergence theorem.

12.5 Further Readings

Books G.F. Simmons, Introduction to Topology and Modern Analysis, New York: McGraw Hill, 1963. H.L. Royden, Real analysis, Prentice Hall, 1988.

Online links dl.acm.org math.stanford.edu www.springerlink.com

LOVELY PROFESSIONAL UNIVERSITY 157 Measure Theory and Functional Analysis

Notes Unit 13: Signed Measures

CONTENTS

Objectives Introduction 13.1 Signed Measures 13.1.1 Signed Measure: Definition 13.1.2 Positive Set, Negative Set and Null Set 13.1.3 Hahn Decomposition Theorem 13.1.4 Hahn Decomposition: Definition 13.2 Summary 13.3 Keywords 13.4 Review Questions 13.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define signed measure.

 Describe positive and negative and null sets.

 Solve problems on signed measure.

Introduction

We have seen that a measure is a non-negative set function. Now we shall assume that it takes both positive and negative values. Such assumption leads us to a new type of measure known as signed measure. In this unit, we shall start with definition of signed measure and we shall prove some important theorems on it.

13.1 Signed Measures

13.1.1 Signed Measure: Definition

Definition: Let the couple (X, ) be a measurable space, where  represents a -algebra of subsets of X. An extended real valued set function :  [– , ] defined on  is called a signed measure, if it satisfies the following postulates: (i) assumes at most one of the values – or + . (ii) ( ) = 0.

158 LOVELY PROFESSIONAL UNIVERSITY Unit 13: Signed Measures

Notes (iii) If is any sequence of disjoint measurable sets, then

A(A),  n n n 1 n 1

i.e., is countably additive. From this definition, it follows that a measure is a special case of a signed measure. Thus, every measure on  is a signed measure but the converse is not true in general, i.e. every signed measure is not a measure in general. If – < (A) < , for very A , then we say that signed measure is finite.

13.1.2 Positive Set, Negative Set and Null Set

Definition

(a) Positive Set: Let (X, ) be a measurable space and let A be any subset of X. Then A X is said to be a positive set relative to a signed measure defined on (X, ), if (i) A , i.e. A is measurable.

(ii) (E) 0, E A s.t. E is measurable.

Obviously, it follows from the above definition that: (i) every measurable subset of a positive set is a positive set, (ii) is a positive set w.r.t. every signed measure. Also for A to be positive. (A) 0 is the necessary condition, but not in general sufficient for A to be positive. (b) Negative Set: Let (X, ) be a measurable space. Then a subset A of X is said to be a negative set relative to a signed measure defined on measurable space (X, ) if (i) A  i.e., A is measurable.

(ii) (E) 0, E A s.t. E is measurable. set A is negative w.r.t. , provided it is positive w.r.t. – . (c) Null Set: A set A X is said to be a null set relative to a signed measure defined on measurable space (X, ) is, A is both positive and negative relative to . Thus, measure of every null set is zero. Now, we know that a measurable set is a set of measure zero, iff every measurable subset of it has measure zero. Thus, if A X is a null set relative to then (E) = 0, measurable subsets E A. In other words.

A is a null set (E) = 0, measurable subsets E A. Theorem 1: Countable union of positive sets w.r.t. a signed measure is positive.   Proof: Let (X, ) be a measurable space and let be a signed measure defined on (X, ). Let n A be a sequence of positive subsets of X, let A =  i and let B be any measurable subset of A. i 1

 B A ACC A , n N. Set n = n n 1 1

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Notes C  where An (n 1,2,3 n 1) denotes complement of An(n = 1, 2, 3, … n – 1) with respect to X.

Now, we know that complement of a measurable set is also measurable so that each

C An (n 1,2,3 n 1) is measurable relative to . Again, intersection of countable collection of

measurable sets is also measurable. Hence Bn is a measurable subset of the positive set An. Thus

(Bn) 0 (by the definition of positive set) … (i)

Obviously, the set Bn are disjoint and

B if B =  n , we get … (ii) n 1

(B) = (B)n … (iii) n 1

In view of (i). (B) 0. Thus, we have (1) A is measurable for

An is a positive set An is a measurable set

countable union A is measurable,  n n 1

A A =  n is measurable n 1

(2) (B) 0, B A s.t. B is a measurable set.

Hence A is a positive set, by definition. Theorem 2: Let (X, ) be a measurable space and let be a signed measure defined on (X, A). If B is a measurable set with finite negative measure i.e., – < (B) < 0, then prove that B contains a negative set A B with the property (A) < 0. Proof: If B is itself a negative set, then we may take A = B and theorem is done. Therefore

consider the case when B is not a negative set. Then there must exist a measurable subset E 1 B

and a smallest positive integer n1, s.t.

1 (E1) > n1

 B = (B – E1) E1 and (B – E1) E1 = ,

(B) = (B – E1) + (E1) … (i)

or (B – E1) = (B) – (E1) … (ii)

Since (B) is finite, (i) implies that (B – E1) and (E1) are finite. Again (B) < 0, (ii) implies that

(B – E1) < 0.

160 LOVELY PROFESSIONAL UNIVERSITY Unit 13: Signed Measures

Notes Now, the set B – E1 is either negative or contains a subset of positive measure. If the set B – E1 is a negative set, then we may take A = B – E1 and the theorem is done. So, suppose that B – E1 is not a negative set. Then there must exist a measurable subset E2 of B – E1 and a smallest positive number n2 with a property

1 (E2) > . n2

Since B = (B – E1 E2) (E2 E2), and (B – E1 E2) (E1 E2) = , we have (B) = (B – E1 E2) + (E2 E2) or (B – E1 E2) = (B) – (E2 E2)

= (B) – (E1) – (E2).  As before, (B – E1 E2) > 0 [ (B) < 0, (Er) > 0 for r = 1, 2]

Thus, B – E1 E2 is a set of negative measure, which is either a negative set or contains a subset of positive measure. If B – E1 E2 is a negative set, then the theorem is done by taking B = A – E1

E2. Otherwise we repeat the above process. On repeating this process, at some stage we shall get either a negative subset A B s.t. (A) < 0 or a sequence of disjoint measurable sets and a sequence of positive integers s.t.

r 1 1 E B – E and < (E ) < r  n n r n 1 r

In first case, we have nothing to do. In the latter case, let

E E A = B –  n or B = A  n … (iii) n 1 n 1

Then as before, it follows that

(B) = (A) + (E)n . n 1

1 > (A) + … (iv) k 1 nk

[ change of suffix is in material] Since (B) is finite and assumes at most one of the values – and , it follows from (iv) that

1 (A) is finite and the series is convergent. k 1 nk

1 Then (A) < (B) – k 1 nk

= a finite negative number ( (B) is a finite negative number)

LOVELY PROFESSIONAL UNIVERSITY 161 Measure Theory and Functional Analysis

Notes or (A) < 0. Again, we know that difference of two measurable sets is measurable and enumerable union of measurable sets is measurable therefore it follows from (iii) that A is a measurable set. Now we shall prove that A is a negative set. Let E A be an arbitrary measurable set.

E Since A = B –  n , n 1

E = B – E .  n n 1

Since nk , we can choose k so large that

1 (E) nk 1

Letting nk , we obtain (E) 0. Thus we have (1) A is measurable.

(2) (E) 0, E A s.t. E is measurable.

Hence A is a negative set.

13.1.3 Hahn Decomposition Theorem

Theorem 3: Let be a signed measure on a measurable space (X, ). Then there exists a positive set P and a negative set Q s.t. P Q = and P Q = X. Proof: Let (X, ) be a measurable space and let be a signed measure defined on a measurable space (X, ). Since, by definition, assumes at most one of the values + or collection of all negative subsets of X w.r.t. and let  be a collection of all negative subsets of X w.r.t. and let k = inf { (E) : E )  (i) that there exists a sequence in such that

Lim (En ) = k. n

E Let Q =  n . n 1

 Since is a family of negative sets, < En> is a sequence of negative sets. Again, we know by remark of theorem 1 that countable union of negative sets is negative, it follows that Q is a negative subset of X so that (Q) K.

162 LOVELY PROFESSIONAL UNIVERSITY Unit 13: Signed Measures

Notes Now, Q – En is a subset of Q, it follows that

(Q – En) 0.

Since (Q – En) En = and Q = (Q – En) En, we have (Q) = (Q – En) + (En)

(Q) (En), n N and En B. Therefore (Q) K. … (iii) (ii) and (iii) (Q) = K – < k. … (iv) Now we shall show that p = QC, the complement of Q w.r.t. is a positive subset of X. Suppose not, i.e. P is negative. Then E P s.t. E is measurable and (E) < 0. Now we know that if – < (E) < 0, we get a negative set A E s.t. (A) < 0. A, Q are distinct negative subsets of X A Q is negative set (A Q) K [using (i)] (A) + (Q) K, (A) + K K, [using (iv)] (A) 0, a contradiction, for (A) < 0 P = QC is a positive subset of X Q is a negative subset of X. Thus X = P Q, P Q = .

13.1.4 Hahn Decomposition: Definition

A decomposition of a measurable space X into two subsets s.t. X = P Q, P Q = , where P and Q are positive and negative sets respectively relative to the signal measure , is called as Hahn decomposition for the signed measure . P and Q are respectively called positive and negative components of X. Notice that Hahn decomposition is not unique.

13.2 Summary

 Let the couple (X, ) be a measurable space, where represents a -algebra of subsets of X. An extended real-valued set function :  [– , ] defined on  is called a signed measure, if it satisfies the following postulates: (i) assumes at most one of the values – or + . (ii) ( ) = 0.

(iii) If is any sequence of disjoint measurable sets, then is countably additive.

LOVELY PROFESSIONAL UNIVERSITY 163 Measure Theory and Functional Analysis

Notes  Let (X, ) be a measurable space and then A X is said to be a positive set relative to a signed measure defined on (X, ) if (i) A is measurable

(ii) (E) 0, E A s.t. E is measurable.

 Let (X, ) be a measurable space. Then A X is said to be negative set relative to a signed measure if (i) A is measurable

(ii) (E) 0, E A s.t. E is measurable.

 A X is said to be a null set relative to a signed measure defined on measurable space (X, ) is: A is both positive and negative relative to .

13.3 Keywords

Hahn Decomposition: Definition: A decomposition of a measurable space X into two subsets s.t. X = P Q, P Q = . Negative Set: Let (X, ) be a measurable space. Then a subset A of X is said to be a negative set relative to a signed measure defined on measurable space (X, ) if (i) A  i.e., A is measurable.

(ii) (E) 0, E A s.t. E is measurable. Null Set: A set A X is said to be a null set relative to a signed measure defined on measurable space (X, ) is, A is both positive and negative relative to .

Positive Set: Let (X, ) be a measurable space and let A be any subset of X. Then A X is said to be a positive set relative to a signed measure defined on (X, ), if (i) A , i.e. A is measurable.

(ii) (E) 0, E A s.t. E is measurable. Signed Measure: Let the couple (X, ) be a measurable space, where  represents a -algebra of subsets of X. An extended real valued set function :  [– , ] defined on  is called a signed measure, if it satisfies the following postulates: (i) assumes at most one of the values – or + . (ii) ( ) = 0.

13.4 Review Questions

2 1. If (E) = xex dx, then find positive, negative and null sets w.r.t. . Also give a Hahn

E decomposition of R w.r.t. . 2. State and prove Hahn decomposition theorem for signed measures.

3. If is a measure and 1, 2 are the signed measures given by 1 (E) = (A E), 2 (E) = (B

E), where (A B) = 0, show that 1 2.

164 LOVELY PROFESSIONAL UNIVERSITY Unit 13: Signed Measures

Notes 4. Show that if 1 and 2 are two finite signed measures, then so is a 1 + b 2 where a, b are real numbers.

13.5 Further Readings

Books Bartle, Robert G., The Elements of Integration, New York – London – Sydney: John Wiley and Sons Cohn, Donald L. (1997) [1980], Measure Theory (reprint ed.), Boston – Based – Stuttgart: Birkhauser Verlag

Online links www.maths.bris.ac.uk www.planetmath.org/signedmeasure.html

LOVELY PROFESSIONAL UNIVERSITY 165 Measure Theory and Functional Analysis

Notes Unit 14: Radon-Nikodym Theorem

CONTENTS

Objectives Introduction 14.1 Radon-Nikodym Theorem 14.1.1 Absolutely Continuous Measure Function 14.2 Summary 14.3 Keywords 14.4 Review Questions 14.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define Absolutely continuous measure function

 State Radon-Nikodym theorem

 Understand the proof of Radon-Nikodym theorem

 Solve problems on this theorem

Introduction

In mathematics, the Radon-Nikodym theorem is a result in measure theory that states that given a measurable space (X, ), if a -finite measure on (X, ) is absolutely continuous with respect to a -finite measure on (X, ), then there is a measurable function f on X and taking values in [0, ], such that for any measurable set A. The theorem is named after Johann Radon, who proved the theorem for the special case where the underlying space is RN in 1913, and for Otto Nikodym who proved the general case in 1930. In 1936 Hans Freudenthal further generalised the Radon-Nikodym theorem by proving the Freudenthal spectral theorem, a result in Riesz space theory, which contains the Radon-Nikodym theorem as a special case. If Y is a Banach space and the generalisation of the Radon-Nikodym theorem also holds for functions with values in Y, then Y is said to have the Radon-Nikodym property. All Hibert spaces have the Radon-Nikodym property.

14.1 Radon-Nikodym Theorem

14.1.1 Absolutely Continuous Measure Function

Let (X, ) be a measurable space and let and be measure functions defined on the space (X, A). The measure is said to be absolutely continuous w.r.t. if (A) = 0 or | | (A) = 0, A  (A) = 0, and is denoted by  .

166 LOVELY PROFESSIONAL UNIVERSITY Unit 14: Radon-Nikodym Theorem

Notes

 If is -finite, the converse is also true.

 If and are signed measures on (X, ), then  if | |  | |.

Radon-Nikodym Theorem

Let (X, , ) be a -finite measure space. If be a measure defined on A s.t. is absolutely continuous w.r.t. , then there exists a non-negative measurable function f on s.t.

(A) = f d , A . A The function f is unique in the sense that if g is any measurable function with the property defined as above, then f = g almost everywhere with respect to . Proof: To establish the existence of the function f, we shall use the following two Lemmas:  Lemma 1: Let E be a countable set of real numbers. Let for each a E there is a set Fa s.t.  Fa Fb, whenever b < a i.e. is a monotonic decreasing sequence of subsets of corresponding to the sequence of real numbers in E. Then a measurable extended real valued function f on X s.t.

f (x) a, x Fa, and f (x) a, x (X – Fa).

Proof: Let f (x) = inf {a : x Fa} x X and let, conventionally inf {empty collection of real numbers} =

Now, x Fa f(x) a

x Fa x Fa for every b < a f (x) a

Now, f (x) < a x Fb for some b < a

[F] or {x : f (x) < a} =  b . b a

Also x Fb f (x) b < a for some b < a. Hence f is measurable. Again, by definition of f, we observe that

f (x) a, x Fa, and f (x) a, x Fa. Thus f is the required function. Lemma 2: Let E be a countable set of real numbers. Let corresponding to each a E, there is a set  Fa s.t.

(Fa – Fb) = 0 whenever b > a. Then there exists a measurable function f with the property

x Fa f (x) a a.e. and x (X – Fa) f (x) > a a.e.

LOVELY PROFESSIONAL UNIVERSITY 167 Measure Theory and Functional Analysis

Notes {FF} Proof: Let P =  a b . b a

Evidently (P) = 0.

Let Fa = Fa P.

This FFa b = (Fa – Fb) – P = for a < b. In view of Lemma 1, it follows that a measurable function f s.t.

f (x) a, x Fa

and f (x) a, x X – Fa Thus we have

x F f(x) a a.e., a except for x P. x X Fa f(x) a a.e.,

Proof of the main theorem

At first, suppose that is finite. ( – a ) is a signed measure on  for each rational number a.

Let (Pa, Qa) be a Hahn decomposition for the measure ( – a ).

Let PO = X and QO = . By the definition of Hahn decomposition theorem,

Pa Qa = X,

and Pb Qb = X.

Therefore, Qa – Qb = Qa – (X – Pb)

= Qa Pb.

Thus, ( – a ) (Qa – Qb) 0 … (i) Similarly, we can prove that

( – b ) (Qa – Qb) 0 … (ii) Let a < b, then from (i) and (ii), we have

(Qa – Qb) = 0. Therefore, by Lemma (ii)

f (x) a, a.e. x Pa

and f (x) a, a.e. x Qa, where f is measurable

Since QO = , it follows that f is non-negative Again, let A  be arbitrary.

QQr 1 r Define Ar = A no n o

168 LOVELY PROFESSIONAL UNIVERSITY Unit 14: Radon-Nikodym Theorem

Notes Q A = A – r . no

A Evidently, A = A  r , r o where A is disjoint union of measurable sets.

(A) = (A ) + (A)r . r o

QQr 1 r Obviously Ar no n o

r r 1 f(x) , x Ar no n o

r r 1 (Ar ) fd (A r ) [by first mean value theorem] no n o Ar

r r 1 Again (A)(A)(A)r r r , we have no n o

1 1 (A ) (A ) fd (A ) (A ) r r r r … (iii) no n o Ar

Now, if (A ) > 0, then g (A ) = 0, [ ( – a ) A is positive, a] and (A ) = 0 if (A ) = 0 [  )]

In either case, (A ) = f d . Ar

Adding the inequalities (iii) over r, we get

1 1 (A) (A) fd (A) (A).r no n o A

Since no is arbitrary and (A) is assumed to be finite, it follows that

(A) f d A . A

To show that the theorem is true for -finite measure , decompose X into a countable union of

Xi of finite -measure. Applying the same argument as above for each Xi, we get the required function. To show the second part, let g be any measurable function satisfying the condition,

(A) = f d A . A

LOVELY PROFESSIONAL UNIVERSITY 169 Measure Theory and Functional Analysis

Notes For each n N, define

1 A = x X:f(x) g(x)  n n

1 and B = x X:g(x) f(x) . n n

1 Since f (x) – g (x) , x A , we have by first mean value theorem n n

1 (f g)d (A ) n n An

1 fd gd (A ) n n AAn n

1 1 (A ) – (A ) (A ) or 0 (A ) n n nn n n

(An) 0.

Since (An) is always greater than equal to zero, we have (An) = 0. Similarly, we can show that

(Bn) 0. If C = {x X : f (x) g (x)}

(A B ), =  n n n 1

then (C) = 0 f = g a.e. on X w.r.t. . Hence the theorem.  Theorem 1: If 1, 2 are -finite signed measures on (X, A) and 1 , 2 , then

d( ) d d d d( ) 1 2 1 2and 1 1 d d d d d

  Proof: Since 1, 2 are -finite and 1 , 2 , we have that 1 + 2 is also -finite and  1 + 2 . Now for any A ,

( 1 + 2) (A) = 1 (A) + 2 (A)

d d d d = 1d 2 d 1 2 d d d d d AAA

d d d 1 2d 1 2 d d d d AA

170 LOVELY PROFESSIONAL UNIVERSITY Unit 14: Radon-Nikodym Theorem

Notes d(1 2 ) d 1 d 2 d d d

Prove the other result yourself.

Theorem 2: If is a -finite signed measures and is a -finite measure s.t.  , show that

d| | d . d d

Proof: Let = + – – with Hahn decomposition A, B.

d d d d Then on A, and on B, d d d d

d d d d( ) d| | d d d d d .

Theorem 3: If be a -finite signed measure and , be -finite measures on (X, A) s.t.  ,  : then show that

d d d d d d Proof: Since we may write = + – – and

d d( ) d d( ) , . d d d d we need to prove the above result for measures only.

d d If f and g , (f, g are non-negative functions as obtained in Radon-Nikodym Theorem), d d then we need to prove that

(F) = fg d . F

Let be a measurable simple function s.t.

= ai Ei , i 1

n then d ai (E i F) F i 1

= ai gd g d . i 1 EFFi

Let < n> be a sequence of measurable simple function which converges to f, then

(F) = fd limn d . FF

LOVELY PROFESSIONAL UNIVERSITY 171 Measure Theory and Functional Analysis

Notes

= limn gd f g d as n g fg FF

d d du = fg . d d d

14.2 Summary

 Let (X, ) be a measurable space and let r and m be measure functions, defined on the space (X, A). The measure is said to be absolutely continuous w.r.t. if

(A) = 0 or | | (A) = 0, A  (A) = 0, and is denoted by  .

 Let (X, , ) be a -finite measure space. If Y be a measure defined on A s.t. is absolutely continuous w.r.t. , then there exists a non-negative measurable function f s.t.

(A) = fd , A 

The function f is unique in the sense that if g is any measurable function with the property defined as above, then f = g almost everywhere with respect to .

14.3 Keywords

Absolutely Continuous Measure Function: Let (X, ) be a measurable space and let and be measure functions defined on the space (X, A). The measure is said to be absolutely continuous w.r.t. if (A) = 0 or | | (A) = 0, A  (A) = 0, and is denoted by  . Radon-Nikodym Theorem: Let (X, , ) be a -finite measure space. If be a measure defined on A s.t. is absolutely continuous w.r.t. , then there exists a non-negative measurable function f on s.t.

(A) = f d , A . A The function f is unique in the sense that if g is any measurable function with the property defined as above, then f = g almost everywhere with respect to .

14.4 Review Questions

1 d d 1. Show that , d d where and are -finite signed measures and  ,  .

2. If (E) = fd , , where fd exists, then find | | (E).

E E

3. State and prove Radon-Nikodym theorem.

172 LOVELY PROFESSIONAL UNIVERSITY Unit 14: Radon-Nikodym Theorem

14.5 Further Readings Notes

Book G.E. Shilov and B.L. Gurevich, Integral, Measure and Derivative: A Unified Approach, Richard A. Silverman, trans. Dover Publications, 1978.

Online links www.math.ksu.edu/nnagy/real-an/4-04-rn.pdf mathworld.woltram.com www.csun.edu pioneer.netserv.chula.ac.th

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Notes Unit 15: Banach Space: Definition and Some Examples

CONTENTS

Objectives Introduction 15.1 Banach Spaces 15.1.1 Normed Linear Space 15.1.2 Convergent Sequence in Normed Linear Space 15.1.3 Subspace of a normed Linear Space 15.1.4 Complete Normed Linear Space 15.1.5 Banach Space 15.2 Summary 15.3 Keywords 15.4 Review Questions 15.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Know about Banach spaces.

 Define Banach spaces.

 Solve problems on Banach spaces.

Introduction

Banach space is a linear space, which is also, in a special way, a complete metric space. This combination of algebraic and metric structures opens up the possibility of studying linear transformations of one Banach space into another which have the additional property of being continuous. The concept of a Banach space is a generalization of Hilbert space. A Banach space assumes that there is a norm on the space relative to which the space is complete, but it is not assumed that the norm is defined in terms of an inner product. There are many examples of Banach spaces that are not Hilbert spaces, so that the generalization is quite useful.

15.1 Banach Spaces

15.1.1 Normed Linear Space

Definition: Let N be a complex (or real) linear space. A real valued function n : N R is said to define, a norm on N if for any x, y N and any scalar (complex number) , the following conditions are satisfied by n: (i) n (x) 0, n (x) = 0, x = 0; (ii) n (x + y) n (x) + n (y); and

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(iii) n ( x) = | | n (x) Notes It is customary to denote n (x) by n (x) = x (read as norm x) With this notation the above conditions (i) – (iii) assume the following forms: (i) x 0, x = 0 x = 0; (ii) x + y x + y ; and (iii) x = x . A linear space N together with a norm defined on it, i.e., the pair (N, ) is called a normed linear space and will simply be denoted by N for convenience.

Notes

1. The condition (ii) is called subadditivity and the condition (iii) is called absolute homogeneity. 2. If we drop the condition viz. x = 0 x = 0, then is called a semi norm (or pseudo norm) or N and the space N is called a semi-normed linear space.

Theorem 1: If N is a normed linear space and if we define a real valued function d : N × N R by d (x, y) = x – y (x, y N), then d is a metric on N. Proof: We shall verify the conditions of a metric (i) d (x, y) 0, d (x, y) = 0 x – y = 0 x = y; (ii) d (x, y) = x – y = (–1) (y – x) = |–1| y – x = y – x = d (y, x); (iii) d (x, y) = x – y = x – z + z – y (z = N) x – z + z – y = d (x, z) + d (z, y) Hence, d defines a metric on N. Consequently, every normed linear space is automatically a metric space. This completes the proof of the theorem.

Notes

1. The above metric has the following additional properties: (i) If x, y, z N and is a scalar, then d (x + z, y + z) = (x + z) – (y + z) = x – y = d (x, y). (ii) d ( x, y) = x – y = (x – y) = | | x – y = | | d (x, y). 2. Since every normed linear space is a metric space, we can rephrase the definition of convergence of sequences by using this metric induced by the norm.

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Notes 15.1.2 Convergent Sequence in Normed Linear Space

Definition: Let (N, ) be a normed linear space. A sequence (xn) is N is said to converge to an

element x in N if given > 0, there exists a positive integer no such that

xn – x < for all n no.

If x converges to x, we write Lim xn x . n n

or xn x as n It follows from the definition that

xn x xn – x 0 as n Theorem 2: If N is a normed linear space, then

x y x – y for any x, y N

Proof: We have x = (x – y) + y x – y + y x – y x – y … (1) Using (1), we have – ( x – y ) = y – x y – x But y – x = (–1) (x – y) = |–1| x – y Therefore – ( x – y ) x – y so that x – y – x – y … (2) From (1) and (2) we get

x y x – y

This completes the proof of the theorem.

15.1.3 Subspace of a Normed Linear Space

Definition: A subspace M of a normed linear space is a subspace of N consider as a vector space with the norm obtain by restricting the norm of N to the subset M. This norm on M is said to be induced by the norm on N. If M is closed in N, then M is called a closed subspace of N.

Theorem 3: Let N be a normed linear space and M is a subspace of N. Then the closure M of M is also a subspace of N.

(Note that since M is closed, M is a closed subspace).

Proof: To prove that M is a subspace of N, we must show that any linear combination of element in M is again in M. That is if x and y M , then x + y M for any scalars and .

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Notes Since x, y M , there exist sequences (xn) and (yn) in M such that

xn x and yn y, By joint continuity of addition and scalar multiplication in M.

xn + yn x + y for every scalars and .

Since xn + yn M, we conclude that

x + y M and consequently M is a subspace of N.

This completes the proof of the theorem.

Notes

1. The scalars , can be assumed to be non-zero. For if = 0 = , then

x + y = 0 M M

2. In a normed linear space, the smallest closed subspace containing a given set of vectors S is just the closure of the subspace spanned by the set S. To see this, let S be the subset of a normed linear space N and let M be the smallest closed subspace of N, containing S. We show that M = [S] , where [S] is the subspace spanned by S.

By theorem, [S] is a closed subspace of N and it contains S.

Since M is the smallest closed subspace containing S, we have

M [S] .

But [S] M and M = M , we must have

[S] M = M so that [S] M.

Hence [S] = M.

15.1.4 Complete Normed Linear Space

Definition: A normed linear space N is said to be complete if every Cauchy sequence in N converges to an element of N. This means that if xm – xn 0 as m, n , then there exists x N such that

xn – x 0 as n .

15.1.5 Banach Space

Definition: A complete normed linear space is called a Banach space. OR A normed linear space which is complete as a metric space is called a Banach space.

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Notes In the definition of a Banach space completeness means that if

xm – xn 0 as m, n , where (xn) N, then a x N such that

xn – x 0 as n .

Note A subspace M of a Banach space B is a subspace of B considered as a normed linear space. We do not require M to be complete.

Theorem 4: Every complete subspace M of a normed linear space N is closed. Proof: Let x N be any limit point of M. We have to show that x M.

Since x is a limit point of M, there exists a sequence (xn) in M and xn x as n .

But, since (xn) is a convergent sequence in M, it is Cauchy sequence in M.

Further M is complete (xn) converges to a point of M so that x M. Hence M is closed. This completes the proof of the theorem. Theorem 5: A subspace M of a Banach space B is complete iff the set M is closed in B. Proof: Let M be a complete subspace of a Banach space M. They be above theorem, M is closed (prove it). Conversely, let M be a closed subspace of Banach space B. We shall show that M is complete.

Let x = (xn) be a Cauchy sequence in M. Then

xn x in B as B is complete. We show that x M.

Now x M x M ( M being closed M = M )

Thus every Cauchy sequence in M converges to an element of M. Hence the closed sequence M of B is complete. This completes the proof of the theorem.

Example 1: The linear space R of real numbers or C of complex numbers are Banach spaces under the norm defined by x = |x|, x R (or C) Solution: We have x = |x| > 0 and x = 0 |x| = x = 0

1 2 Further, let z1, z2 C and let z and z be their complex conjugates, then

2 |z1 + z2| = (z1 + z2) (z1 z 2 )

= z1 z1 z 1 z 2 z 2 z 1 z 2 z 2

2 2 z1 2 z 1 z2 z 2

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2 2 Notes = z1 2 z 1 z 2 z 2  z1 , z 2 z 1 , z 2 z 1 z 2

2 = (|z1| + |z2|)

|z1 + z2| |z1| + |z2|  or z1 + z2 z1 + z2 ( x = |x|) Also x = | x| = | | |x| = | | x Hence all the conditions of normed linear space are satisfied. Thus both C or R are normed linear space. And by Cauchy general principle of convergence, R and C are complete under the matrices induced by the norm. So R and C are Banach spaces.

n n Example 2: Euclidean and Unitary spaces: The linear space R and C of all n-tuples (x1, x2 …, xn) of real and complex numbers are Banach spaces under the norm

1/2 n 2 x = |xi | i 1

[Usually called Euclidean and unitary spaces respectively].

Solution: (i) Since each |xi| 0, we have x 0

n 2 and x = 0 |xi | = 0 xi = 0, i = 1, 2, …, n i 1

(x1, x2, … xn) = 0 x = 0

(ii) Let x = (x1, x2, …, xn)

n n and y = (y1, y2, … yn) be any two numbers of C (or R ). Then

2 2 x + y = (x1, x2, …, xn) + (y1, y2, … yn)

2 = (x1+ x1), (x2 + y2), …, (xn + yn)

n 2 = |xi y i | i 1

|xi y i |(|x i | |y i |) i 1

n n

|xi y i ||x| i |x i y i ||y| i i 1 i 1

Usually Cauchy inequality for each sum, we get

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Notes = x + y x + x + y y = ( x + y ) ( x + y ). If x + y = 0, then the above inequality is evidently true. If x + y 0, we can divide both sides by it to obtain x + y x + y .

1 1 n2 n 2 2 2 (iii) x = | xi | | | |x i | i 1 i 1

= | | x . This proves that Rn or Cn are normed linear spaces. Now we show the completeness of Cn (or Rn).

n n Let < x1, x2, … xn > be a Cauchy sequence in C (or R ). Since each xm is an n-tuple of complex (or real) numbers, we shall write

x(m) , x (m) , , x (m) xm = 1 2 n

(m) th So that xk is the k coordinate of xm.

Let > 0 be given, since is a Cauchy sequence, there exists a positive integer mo, such that

 x x , m mo m 

2 2 xm x

n (m) ( ) 2 xi x i … (1) i 1

(m) ( ) 2 xi x i (i = 1, 2, ……, n)

(m) ( ) xi x i

Hence x(m ) is a Cauchy sequence of complex (or real) numbers for each fixed but i m 1 arbitrary i.

Since C (or R) is complete, each of these sequences converges to a point, say 2i in C (or R) so that

(m) Lim xi = z (i = 1, 2, …, n) … (2) m i

n Now we show that the Cauchy sequence converges to the point z = (z1, z2, ……, zn) C (or Rn).

To prove this let  in (1). Then by (2) we have

n (m) 2 2 xi z i i 1

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2 2 Notes xm – z <

xm – z <

n n It follows that the Cauchy sequence converges to z C (or R ). Hence Cn or Rn are complete spaces and consequently they are Banach spaces.

15.2 Summary

 A linear space N together with a norm defined on it, i.e. the pair (N, ) is called a normed linear space.

 Let (N, ) be a normed linear space. A sequence (xn) in N is said to converge to an element

x in N if given > 0, there exists a positive integer no such that

xn – x < for all n no.

 If N is a normed linear space, then

x y x y for any x, y N.

 A normed linear space N is said to be complete if every Cauchy sequence in N converges to an element of N.

 A complete normed linear space is called a Banach space.

15.3 Keywords

A Subspace M of a Normed Linear Space: A subspace M of a normed linear space is a subspace of N consider as a vector space with the norm obtain by restricting the norm of N to the subset M. If norm on M is said to be induced by the norm on N. If M is closed in N, then M is called a closed subspace of N. Banach Space: A complete normed linear space is called a Banach space. Complete Normed Linear Space: A normed linear space N is said to be complete if every Cauchy sequence in N converges to an element of N. This means that if xm – xn 0 as m, n , then there exists x N such that

xn – x 0 as n . Normed Linear: A linear space N together with a norm defined on it, i.e., the pair (N, ) is called a normed linear space and will simply be denoted by N for convenience.

15.4 Review Questions

1. Let N be a non-zero normed linear space, prove that N is a Banach space {x : x = 1} is complete. 2. Let a Banach space B be the direct sum of the linear subspaces M and N, so that B = M N. If z = x + y is the unique expression of a vector z in B as the sum of vectors x and y in M and N, then a new norm can be defined on the linear space B by z = x + y . Prove that this actually is a norm. If B symbolizes the linear space B equipped with this new norm, prove that B is a Banach space of M and N are closed in B.

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Notes 15.5 Further Readings

Books Bourbaki, Nicolas (1987), Topological Vector Spaces, Elements of Mathematics, Berlin: Springer-Verlag. Beauzamy, Bernard (1985), Introduction to Banach Spaces and their Geometry (Second revised ed.), North-Holland.

Online links mathword.wolfram.com?Calculus and Analysis>Functional Analysis homepage.ntlworld-com/ivan.wilde/notes/fal/fal.pdf www.math.ucdavis.edu/~ hunter/book.chs.pdf

182 LOVELY PROFESSIONAL UNIVERSITY Unit 16: Continuous Linear Transformations

Unit 16: Continuous Linear Transformations Notes

CONTENTS

Objectives Introduction 16.1 Continuous Linear Transformation 16.1.1 Continuous Linear Functionals Definition 16.1.2 Bounded Linear Functional 16.1.3 Norm of a Bounded Linear Functional 16.1.4 Equivalent Methods of Finding F 16.1.5 Representation Theorems for Functionals 16.2 Summary 16.3 Keywords 16.4 Review Questions 16.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand continuous linear transformation

 Define bounded linear functional and norm of a bounded linear functional

 Understand theorems on continuous linear transformations.

Introduction

In this unit, we obtain the representation of continuous linear functionals on some of Banach spaces.

16.1 Continuous Linear Transformation

16.1.1 Continuous Linear Functionals Definition

 Let N be a normed linear space. Then we know the set R of real numbers and the set C of complex numbers are Banach spaces with the norm of any x R or x C given by the absolute value of x. Thus with our previous notations, (N, R) or (N, C) denote respectively the set of all continuous linear transformations from N into R or C.

 We denote the Banach space (N, R) or (N, C) by N* and call it by the conjugate space (or dual space or adjoint space) of N.

 The elements of N* will be referred to as continuous linear functionals or simply functionals on N.

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Notes

Note The conjugate space (N*)* of N* is called the second conjugate space of N and shall be denoted by N**. Also note that N** is complete too.

Theorem 1: The conjugate space N* is always a Banach space under the norm

f(x) f = sup : x N, x 0 … (i) x

= sup f(x) : x 1

= inf k, k 0 and f(x) k x x

Proof: As we know that if N, N are normed linear spaces, (N, N ) is a normed linear space. If N is a Banach space, (N, N ) is Banach space. Hence (N, R) or (N, C) is a Banach space because R and C are Banach spaces even if N is not complete. This completes the proof of the theorem.

Theorem 2: Let f be a linear functional on a normed linear space. If f is continuous at x o N, it must be continuous at every point of N.

Proof: If f is continuous at x = xo, then

xn xo f (xn) f (x) To show that f is continuous everywhere on N, we must show that for any y N,

yn y f (yn) f (y)

Let yn y as n

Now f (yn) = f (yn – y + xo + y – xo) since f is linear.

f (yn) = f (yn – y + xo) + f (y) – f (xo) … (1)

As yn y yn – y + xo xo by hypothesis

Also f is continuous, f (yn – y + xo) f (xo) … (2) From (1) and (2), it follows that

f (yn) f (y) as n . f is continuous at y N and consequently as it is continuous everywhere on N. Hence proved.

16.1.2 Bounded Linear Functional

A linear functional on a normed linear space N is said to be bounded, if there exists a constant k such that

f (x) K x x N .

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Notes

Note

We may find many K’s satisfying the above condition for a given bounded function. If it

is satisfied for one K, it is satisfied for a K1 > K.

Theorem 3: Let f be a linear functional defined on a normed linear space N, then f is bounded f is continuous. Proof: Let us first show that continuity of f boundedness of f. If possible let f is continuous but not bounded. Therefore, for any natural number n, however large, there is some point xn such that

|f (xn)| n || xn|| … (1)

xn Consider the vector, yn = so that n xn

1 y = . n n

yn 0 as n

yn 0 in the norm.

Since any continuous functional maps zero vector into zero and f is continuous f (y n) f (0) = 0.

1 But |f (yn)| = f (xn) … (2) n xn

It now follows from (1) & (2) that |f (yn)| > 1, a contradiction to the fact that f (yn) 0 as n . Thus if f is bounded, then f is continuous.

Conversely, let f is bounded. Then for any sequence (xn), we have

|f (xn)| K || xn||  n = 1, 2, …, and K 0.

Let xn 0 as n then

f (xn) 0 f is continuous at the origin and consequently it is continuous everywhere. This completes the proof of the theorem.

Note The set of all bounded linear function on N is a vector space denoted by N*. As in the case of linear operators, we make it a normed linear space by suitably defining a norm of a functional f.

16.1.3 Norm of a Bounded Linear Functional

If f is a bounded linear functional on a normed space N, then the norm of f is defined as: f(x) || f|| = sup … (1) x 0 x

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Notes We first note that the above norm is well defined. Since f is bounded, we have |f (x)| M || x||, M 0.

f(x) Let M be the set of real numbers M satisfying this relation. Then the set ; x 0 is x

bounded above so that it must possess a supremum. Let it be f . So f is well defined and we must have

f(x) f x 0. x

or |f (x)| f x .

Let us check that defined by (1) is truly a norm on N*:

If f, g N*, then

f(x) g(x) f g = sup x 0 x

f(x) g(x) sup sup x 0x x 0 x

f g f g .

Similarly, we can see that f f .

16.1.4 Equivalent Methods of Finding F

If f is a bounded linear functional on N, then

|f (x)| M x , M 0.

(I) f = inf {M : M M } where M is the set of all real numbers satisfying

|f (x)| M x ,

Since f M and M is the set of all non-negative real numbers, it is bounded below by zero so that it has an infimum. Hence

f inf {M : M M } … (2)

f(x) For x 0 and M M we have M. Since M is the only upper bound then from x definition (2), we have

f(x) M sup = f for any M M . x 0 x

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Notes Since M is bounded below by f , it has an infimum so that we have

inf M = inf {M : M M } f … (3) MM

From (2) and (3), it follows that

f = inf {M : M M }

(II) f = sup f (x) x 0

Let us consider x 1. Then

f(x) f x f .

Therefore, we have

sup f(x) f . … (4) x

Now by definition, f(x) f = sup x 0 x

It follows from the property of the supremum that, given > 0, an x N such that

f(x ) > ( f ) … (5) x

Define

x x . Then x is a unit vector. x

Since x 1 x 1 , we have

1 sup f(x) f(x) f(x) (f ) [by (2)] x 1 x

Hence > 0 is arbitrary, we have

sup f(x) > f … (6) x 1

From (4) and (6), we obtain

sup f(x) = f . x 1

(III) f = sup f(x) . x 1

Consider x = 1, we have

f (x) f x f

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Notes So that

sup f(x) f … (7) x 1

Now consider

f(x) f = sup x 0 x

By supremum property, given > 0, x 0 Such that |f (x )| > (||f|| – ) ||x ||

x Define x . x

Since f is continuous in ||x|| 1 and reaches its maximum on the boundary ||x|| = 1, We get

1 sup f(x) f (x) f (x ) f . x 1 x

sup f(x) f . x 1

The arbitrary character of yields that

sup f(x) f x 1 … (8)

Hence from (7) and (8), we get

f = sup f(x) . x 1

Note If N is a finite dimensional normed linear space, all linear functions are bounded and hence continuous. For, let N be of dimension n so that any x N is of the form

ix i , where x1, x2, …, xn is a basis of N and 1, 2, …, n are scalars uniquely determined i 1 by the basis. Since f is linear, we have

f (x) = i f (xi) so that i 1

| f (x)| i |f (xi)| … (1) i 1

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We have from (1) by using the notation of the Zeroth norm in a finite dimensional space, Notes

n x f(x ) | f (x)| 0 i … (2) i 1

If f(xi ) = M, then from (2), we have i 1

x | f (x)| M 0 .

Hence f is bounded with respect to 0 .

, Since any norm on N is equivalent to 0 f is bounded with respect to any norm on N. Consequently, f is continuous on N.

16.1.5 Representation Theorems for Functionals

We shall prove, in this section, the representation theorems for functionals on some concrete Banach spaces.

n n Theorem 4: If L is a linear space of all n-tuples, then (i) p*  q .

Proof: Let (e1, e2, …, en) be a standard basis for L so that any x = (x 1, x2, …, xn) L can be written as

x = x1e1 + x2e2 + … + xnen. If f is a scalar valued linear function defined on L, then we get

f (x) = x1 f (e1) + x2 f (e2) + … + xn f (en) … (1) f determines and is determined by n scalars

yi = f (ei). Then the mapping

y = (y1, y2, …, yn) f

n where f (x) = xi y i is an isomorphism of L onto the linear space L of all function f. We shall i 1 establish (i) – (iii) by using above given facts. (i) If we consider the space

n th L =  p (1 p < ) with the p norm, then f is continuous and L represents the set of all

n continuous linear functionals on  p so that

n L =  p * .

Now for y f as an isometric isomorphism we try to find the norm for y’s. For 1 < p < , we show that

n n  p * =  q .

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Notes n For x  p , we have defined

1 n p p x = xi i 1

n n

Now |f (x)| = xi y i x i y i i 1 i 1

By using Hölder’s inequality, we get

1 1 n np n q p q xi y i xi y i i 1 i 1 i 1

so that

1 1 nq n p q p |f (x)| yi x i i 1 i 1

Using the definition of norm for f, we get

1 n q q f yi … (2) i 1

Consider the vector, defined by

q yi xi = , yi 0 and xi = 0 if yi = 0 … (3) yi

Then

1 1 q p p np n p yi x = xi … (4) i 1 i 1 yi

Since q = p (q – 1) we have from (4),

1 n p q x = yi … (5) i 1

Now

n n q yi |f (x)| = xi y i y i i 1 i 1 yi

n q = yi , (By (3)) i 1

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So that Notes

n q yi = |f(x)| f x … (6) i 1

From (5) and (6), we get

1 1 n p q yi f i 1

1 n q q yi f … (7) i 1

Also from (2) and (7), we have

1 n q q f = yi , so that i 1 y f is an isometric isomorphism.

n n Hence p*  q .

n n (ii) Let L =  1 with the norm defined by x xi . i 1

Now f defined in (1), above is continuous as in (i) and L here represents the set of continuous

n linear functional on  1 so that

n L =  1 * .

We now determine the norm of y’s which makes y f an isometric isomorphism. Now,

|f (x)| = xi y i i 1

xi y i i 1

n n n

But xi y i max. y i x i so that f(x) max. yi x i . i 1 i 1 i 1

From the definition of norm for f, we have

f = max. yi : i 1,2, ,n … (8) Now consider the vector defined as follows:

If |y | = max yi , let us consider vector x as i 1 i n

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Notes yi x = when |y | = max yi i i 1 i n yi

and xi = 0 otherwise.

From the definition, xk = 0 k i. So that we have y x = i = 1 y

n (x y ) Further |f (x)| = i i = |yi| i 1

f x Hence |yi| = |f (x)|

f  |yi| or max. {|yi|} [ ||x|| = 1]

||f|| … (10) From (8) and (10), we obtain

||f|| = max. {|yi|} so that

n y f is an isometric isomorphism of L to  1 * .

n n Hence 1 *  .

(iii) Let L = n with the norm

x = max {|xi| : i = 1, 2, 3, …, n}.

Now f defined in (1) above is continuous as in (1).

Let L represents the set of all continuous linear functionals on n so that

L = n * .

Now we determine the norm of y’s which makes y f as isometric isomorphism

n n

|f (x)| = xi y i x i y i . i 1 i 1

n n

But xi y i max( xi ) y i i 1 i 1

Hence we have

| f (x) | yi x so that i 1

f yi … (11) i 1

192 LOVELY PROFESSIONAL UNIVERSITY Unit 16: Continuous Linear Transformations

Consider the vector x defined by Notes

yi xi = when yi 0 and xi = 0 otherwise. … (12) yi

y Hence x maxi 1 . yi

n n

and |f (x)| = xi y i y i . i 1 i 1

Therefore yi f(x) f x f . i 1

yi f … (13) i 1

It follows now from (11) and (13) that f yi so that y f is an isometric i 1 isomorphism.

n n Hence, * 1 .

This completes the proof of the theorem.

Theorem 5: The conjugate space of  p is  q , where

1 1 1 and 1 < p < . p q

* or p  q .

 p Proof: Let x = (xn) p so that xn . … (1) n 1

th Let  n = (0, 0, 0, …, 1, 0, 0, …) where 1 is in the m place.

 en p for n = 1, 2, 3, …

* We shall first determine the form of f and then establish the isometric isomorphism of  p onto

 q .

By using (en), we can write any sequence (x1, x2, …, xn, 0, 0, 0, …) in the form xk e k and k 1

n x x e k k = (0, 0, 0, …, xn+1, xn+2, …). k 1

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Notes 1 n p p Now x xk e k x k … (2) k 1 k n 1

The R.H.S. of (2) gives the remainder after n terms of a convergent series (1).

1 p p Hence xk 0 as n … (3) k n 1

From (2) and (3), it follows that

x = xk e k . … (4) k 1

n * Let f  p and sn x k e k then k 1

sn x as n . (Using (4)) since f is linear, we have

f (sn) = xk f(e k ) . k 1

Also f is continuous and sn x, we have

f (sn) f (x) as n

f (x) = xk f(e k ) … (5) k 1

which gives the form of the functional on  p .

* Now we establish the isometric isomorphism of  p onto  q , for which we proceed as follows:

Let f (ek) = k and show that the mapping

* T :  p  q given by … (6)

*  T (f) = ( 1, 2, …, k, …) is an isometric isomorphism of p onto q .

First, we show that T is well defined.

 For let x p , where x = ( 1, 2, …, n, 0, 0, …)

g 1 sgn , 1 k n where = k k k 0 n k

q – 1 | k| = | k| for 1 k n.

p 1 1 | |p = (q 1) = | |q.  q p(q 1) q k k k p q

194 LOVELY PROFESSIONAL UNIVERSITY Unit 16: Continuous Linear Transformations

q 1 q 1 Notes sgn sgn Now k k = k k k k k k

q p k k = | k| = | k| (Using property of sgn function) … (7)

1 n p p x = k k 1

1 n q q = k … (8) k 1

Since we can write

x = ke k , we get k 1

n n

f (x) = kf (e k ) k k k 1 k 1

n q

f (x) = k ( Using (7)) … (9) k 1

We know that for every x  p

| f (x)| f x , which upon using (8) and (9), gives

1 n n p q q |f (x)| kf k k 1 k 1 which yields after simplification.

1 n p q k f … (10) k 1 since the sequence of partial sums on the L.H.S. of (10) is bounded, monotonic increasing, it converges. Hence

1 n q q k f … (11) k 1

 so the sequence ( k) which is the image of f under T belongs to q and hence T is well defined.

We next show that T is onto  q .

 * Let ( k) q , we shall show that there is a g p such that T maps g into ( k).

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Notes Let x  p so that

x = xk e k . k 1 We shall show that

g (x) = xk k is the required g. k 1

Since the representation for x is unique, g is well defined and moreover it is linear on  p . To prove it is bounded, consider

|g (x) | = kx k k x k k 1 k 1

1 1 p q p q xk k (Using Hölder’s inequality) k 1 k 1

1 q q |g (x)| x k . k 1

g is bounded linear functional on  p .

 since ek p for k = 1, 2, …, we get

g (ek) = k for any k so that

*  Tg = ( k) and T is on p onto p .

We next show that

Tf f so that T is an isometry.

Since Tf  q , we have from (6) and (10) that

1 q q k = || Tf || || f || … (12) k 1

Also, x p x x k e k . Hence k 1

f (x) = xk (e k ) x k k . k 1 k 1

|f (x)| xk k k 1

196 LOVELY PROFESSIONAL UNIVERSITY Unit 16: Continuous Linear Transformations

1 1 Notes q p q p kx k (Using Hölder’s inequality) k 1 k 1

1 q q or |f (x)| kx x  p . k 1

Hence, we have

1 q f(x) q sup k = Tf (Using (6)) x 0 x k 1 which upon using definition of norm yields. f Tf … (13) Thus f = Tf (Using (12) and (13)) From the definition of T, it is linear. Also since it is an isometry, it is one-to-one and onto.

* Hence T is an isometric isomorphism of  p onto  p , i.e.

* p  q

Theorem 6: Let N and N be normed linear and let T be a linear transformation of N into N . Then the inverse T–1 exists and is continuous on its domain of definition if and only if there exists a constant m > 0 such that

m x T (x) x N. … (1) Proof: Let (1) holds. To show that T–1 exists and is continuous. Now T–1 exists iff T is one-one.

Let x1, x2, N. Then

T(x1) = T(x2) T (x1) – T(x2) = 0

T (x1 – x2) = 0

x1 – x2 = 0 by (1)

x1 = x2 Hence T is one-one and so T–1 exists. Therefore to each y in the domain of T –1, there exists x in N such that T (x) = y T–1 (y) = x … (2) Hence (1) is equivalent to

1 m T–1(y) y T–1(y) y m

T–1 is bounded T–1 is continuous converse. Let T–1 exists and be continuous on its domain T(N). Let x be an arbitrary element in N. Since T–1 exists, there is y T(N) such that T–1 (y) = x T(x) = y.

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Notes Again since T–1 is continuous, it is bounded so that there exists a positive constant k such that T–1 (y) k y x k T(x)

1 m x T(x) where m = > 0. k This completes the proof of the theorem. Theorem 7: Let T : N N be a linear transformation. Then T is bounded if and only if T maps bounded sets in N onto bounded set in N . Proof: Since T is a bounded linear transformation, T (x) k x for all x N. Let B be a bounded subset of N. Then

x k1k x B. We now show that T(B) is bounded subset of N . From above we see that

T (x) k1 x B. T (B) is bounded in N . Conversely, let T map bounded sets in N into bounded sets in N . To prove that T is a bounded linear transformation, let us take the closed unit sphere S [0, 1] in N as a bounded set. By hypothesis, its image T (S[1, 0]) must be bounded set in N .

Therefore there is a constant k1 such that

T (x) k1 for all x S [0, 1]

x Let x be any non-zero vector in N. Then S[0,1] and so we get x

x T k x 1

T (x) k1 x . Since this is true for x = 0 also, T is a bounded linear transformation. This completes the proof of the theorem.

16.2 Summary

 Let N be a normed linear space. Then we know the set R of real numbers and the set C of complex numbers are Banach spaces with the norm of any x R or x C be the absolute value of X. (N, R) or (N, C) denote respectively the set of all continuous linear transformations from N into R or C.

 A linear functional on a normed linear space N is said to be bounded, if there exists a constant k such that

|f (x)| k x x N.

198 LOVELY PROFESSIONAL UNIVERSITY Unit 16: Continuous Linear Transformations

 If f is a bounded linear functional on a normed space N, then the norm of f is defined as: Notes

f(x) f = sup x 0 x

16.3 Keywords

Bounded Linear Functional: A linear functional on a normed linear space N is said to be bounded, if there exists a constant k such that

f (x) K x x N .

Continuous Linear Transformations: Let N be a normed linear space. Then we know the set R of real numbers and the set C of complex numbers are Banach spaces with the norm of any x R or x C given by the absolute value of x. Thus with our previous notations, (N, R) or (N, C) denote respectively the set of all continuous linear transformations from N into R or C. Norm of a Bounded Linear Functional: If f is a bounded linear functional on a normed space N, then the norm of f is defined as:

f(x) || f|| = sup x 0 x

Second Conjugate: The conjugate space (N*)* of N* is called the second conjugate space of N .

16.4 Review Questions

1. Prove that the conjugate space of  1 is  ,

* i.e. 1  .

 2. Prove that the conjugate space of co is 1 .

* or co 1

1 1 3. Let p > 1 with = 1 and let g L (X). p q q Then prove that the function defined by

F (f) = fg d for f Lp (X) X is a bounded linear functional on Lp (X) and

F = g q

4. Let N be any n dimensional normed linear space with a basis B = {x1, x2, ..., xn}. If

(r1, r2, ..., rn) is any ordered set of scalars, then prove that, there exists a unique continuous linear functional f on N such that

f (xi) = ri for i = 1, 2, …, n 5. If T is a continuous linear transformation of a normed linear space N into a normed linear space N , and if M is its null space, then show that T induces a natural linear transformation T of N/M into N and that T = T .

LOVELY PROFESSIONAL UNIVERSITY 199 Measure Theory and Functional Analysis

Notes 16.5 Further Readings

Books JB Conway (1990), A Course in Functional Analysis. E Hille (1957), Functional Analysis and Semigroups.

Online links pt.scribd.com/doc/86559155/14/Continuous-Linear-Transformations www.math.psu.edu/bressan/PSPDF/fabook.pdf

200 LOVELY PROFESSIONAL UNIVERSITY Unit 17: The Hahn-Banach Theorem

Unit 17: The Hahn-Banach Theorem Notes

CONTENTS

Objectives Introduction 17.1 The Hahn-Banach Theorem 17.1.1 Theorem: The Hahn-Banach Theorem – Proof 17.1.2 Theorems and Solved Examples 17.2 Summary 17.3 Keywords 17.4 Review Questions 17.5 Further Readings

Objectives

After studying this unit, you will be able to:

 State the Hahn-Banach theorem

 Understand the proof of the Hahn-Banach theorem

 Solve problems related to it.

Introduction

The Hahn-Banach theorem is one of the most fundamental and important theorems in functional analysis. It is most fundamental in the sense that it asserts the existence of the linear, continuous and norm preserving extension of a functional defined on a linear subspace of a normed linear space and guarantees the existence of non-trivial continuous linear functionals on normed linear spaces. Although there are many forms of Hahn-Banach theorem, however we are interested in Banach space theory, in which we shall first prove Hahn-Banach theorem for normed linear spaces and then prove the generalised form of this theorem. In the next unit, we shall discuss some important applications of this theorem.

17.1 The Hahn-Banach Theorem

17.1.1 Theorem: The Hahn-Banach Theorem – Proof

Let N be a normed linear space and M be a linear subspace of N. If f is a linear functional defined on M, then f can be extended to a functional fo defined on the whole space N such that

fo = f . Proof: We first prove the following lemma which constitutes the most difficult part of this theorem.

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Notes Lemma: Let M be a linear subspace of a normed linear space N let f be a functional defined on M.

If xo N such that xo M and if Mo = M + [xo] is the linear subspace of N spanned by M and xo, then

f can be extended to a functional fo defined on Mo s.t.

fo = f . Proof: We first prove the following lemma which constitutes the most difficult part of this theorem. Lemma: Let M be a linear subspace of a normed linear space N let f be a functional defined on M.

If xo N such that xo M and if Mo = M + [xo] is the linear subspace of N spanned by M and xo, then

f can be extended to a functional fo defined on Mo s.t.

fo = f . Proof: The lemma is obvious if f = o. Let then f 0. Case I: Let N be a real normed linear space.

Since xo M, each vector y in Mo is uniquely represented as

y = x + xo, x M and R. This enables us to define

fo : Mo R by

fo (y) = fo (x + xo) = f (x) + ro,

where ro is any given real number … (1)

We show that for every choice of the real number ro, fo is not only linear on M but it also extends

f from M to Mo and

fo = f .

Let x1, y1 Mo. Then these exists x and y M and real scalars and such that

x1 = x + xo and y1 = y + xo,

Hence, fo (x1 + y1) = fo (x + xo + y + xo)

= fo (x + y + ( + ) xo)

= f (x + y) + ( + ) ro, ro is a real scalar … (2) Since f is linear M, f (x + y) = f (x + y) … (3) From (2) and (3) it follows after simplification that

fo (x1 + y1) = f (x) + ro + f (y) + ro

= fo (x + xo) + fo (y + xo)

= fo (x1) + fo (y1)

fo (x1 + y1) = fo (x1) + fo (y1) … (4)

Let k be any scalar. Then if y Mo, we have

fo (ky) = fo [k (x + xo)]

= fo (kx + k xo)

But fo (kx + k xo) = f (kx) + k ro

= k f (x) + k ro

202 LOVELY PROFESSIONAL UNIVERSITY Unit 17: The Hahn-Banach Theorem

Notes Hence fo (ky) = k [f (x) + ro] = k fo (y) … (5)

From (4) and (5) it follows that fo is linear on Mo. If y M, then = 0 in the representation for y so that y = x.

Hence fo (x) = f (x) x M.

fo extends f from M to Mo. Next we show that

fo = f .

If = 0 this is obvious. So we consider when 0. Since M is a subspace of Mo we then have

fo = sup {|fo (x)| : x Mo, x 1}

sup {|fo (x)| : x M, x 1}  = sup {|f (x)| : x M, x 1} ( fo = f on M) = f .

Thus, fo f … (A)

So our problem now is to choose ro such that fo f .

Let x1, x2 M. Then we have

f (x2) – f (x1) = f (x2 – x1)

|f (x2 – x1)|

f (x2 + xo) – (x1 + xo)

f ( x2 + xo + – (x1 + xo) )

= f x2 + xo + f x1 + xo

Thus – f (x1) – f x1 + xo – f (x2) + f x2 + xo … (6)

Since this inequality holds for arbitrary x1, x2 M, we see that

sup f(y) f y x int f(y) f y xo o y M y M

Choose ro to be any real number such that

sup f(y) f y xo ro y M

inf f(y) f y xo y M

From this, we get for all y M

sup {– f (y) – f ( y + xo )} ro

inf {– f (y) + f ( y + xo )}

x Let us take y = in the above inequality, we have

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Notes x x r sup f f xo o y M

x x inf f f xo … (7) y M

If > 0 the right hand side of (7) becomes

1 1 f(x) f x x ro o which implies that

f (x) + ro = fo (x + xo) f x + xo

If z = x + xo Mo then we get from above

|fo (z)| f z … (8) If < 0, then from L.H.S. of (7) we have

x x f f xo ro

1 1 1 1 f(x) f x xo ro, since < 0, .

f (x) + ro f x + xo

fo (z) f z for every z Mo … (9) Replacing z by –z in (9) we get

– fo (z) f z , since fo is linear on Mo … (10) Hence we get from (9) and (10).

|fo (z)| f z … (11) Since f is functional on M, f is bounded.

Thus ( ) shows that fo is a functional on Mo.

Since fo = sup {|fo (z)| : z Mo, z |}, it follows from ( ) that

fo f We finally obtain from (A) and (B) that

fo = f This power the lemma for real normed linear space. Case II: Let N be a complex normed linear space. Let N be a normed linear space over C and f be a complex valued functional on a subspace M of N. Let g = Re (f) and h = Im (f) so that we can write f (x) = g (x) + i h (x). We show that g (x) and h (x) are real valued functionals. Since f is linear, we have f (x + y) = f (x) + f (y)

204 LOVELY PROFESSIONAL UNIVERSITY Unit 17: The Hahn-Banach Theorem

g (x + y) + i h (x + y) = g (x) + i h (x) + g (y) + i h (y) Notes = g (x) + g (y) + i (h (x) + h (y)) Equating the real and imaginary parts, we get g (x + y) = g (x) + g (y) and h (x + y) = h (x) + h (y) If R, then we have f ( x) = g ( x) + i h ( x) Since f is linear f ( x) = f (x) = g (x) + i h (x) f ( x) = g (x) and h ( x) = h (x) (equating real and imaginary parts) g, h are real linear functions on M. Further |g (x)| |f (x)| f x If f is bounded on M, then g is also bounded on M. Similarly h is also bounded on M. Since a complex linear space can be regarded as a real linear space by restricting the scalars to be real numbers, we consider M as a real linear space. Hence g and h are real functional on real space M. For all x in M we have f (i x) = i f (x) = i {g (x) + i h (x)} or g (i x) + i h (i x) = – h (x) + i g (x) Equating real and imaginary parts, we get g (i x) = – h (x) and h (i x) = g (x) Therefore we can express f (x) either only by g or only h as follows: f (x) = g (x) – i g (i x) = h (i x) + i h (x).

Since g is a real functional on M, by case I, we extend g to a real functional g o on the real space Mo such that go = g . For x Mo, we define

fo (x) = go (x) – i go (i x)

First note that fo is linear on the complex linear space Mo. Such that fo = f on M.

Now fo (x + y) = go (x + y) – i go (i x + i y)

= go (x) + go (y) – i go (i x) – i go (i y)

= fo (x) + fo (y). Now for a, b R, we have

fo ((a + i b) x) = go (ax + i bx) – i go (– bx + i ax)

= a go (x) + b go (i x) – i (–b) go (x) – i a go (i x)

= (a + ib) {go (x) – i go (i x)}

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Notes So that

fo ((a + i b) x) = (a + i b) fo (x)

fo is linear on Mo and also go = g on M.

fo = f on M.

Now have to show that fo = f on Mo.

i Let x Mo and fo (x) = re

–i i –i |fo (x) | = r = e re = e fo (x) … (12)

Since fo (x) is linear,

–i –i e fo (x) = fo (e x) … (13) So we get from (12) and (13) that

–i |fo (x) | = r = fo (r e x).

Thus the complex valued functional fo is real and so it has only real part so that

–i –i |fo (x) | = go (e x) |go (e x)|

–i –i But |go (e x)| go e x

= go x ,

We get |fo (x)| go x

Since go is the extension of g, we get

go x = g x f x Therefore

|fo (x) f x so that from the definition of the norm of fo, we have

fo f

As in case I, it is obvious that f fo

Hence fo = f . This completes the proof of the theorem.

17.1.2 Theorems and Solved Examples

Theorem: The generalized Hahn-Banach Theorem for Complex Linear Space.

Let L be a complex linear space. Let p be a real valued function defined on L such that p (x + y) p (x) + p (y)

and p ( x) = | | p (x) x L and scalar .

If f is a complex linear functional defined on the subspace M such that |f (x)| p (x) for x M,

then f can be extended to a complex linear functional to be defined on L such that |fo (x)| p (x) for every x L. Proof: We have from the given hypothesis that f is a complex linear functional on M such that

| f (x) p (x) x M.

206 LOVELY PROFESSIONAL UNIVERSITY Unit 17: The Hahn-Banach Theorem

Let g = Re (f) then g (x) |f (x)| p (x). Notes So by the generalised Hahn-Banach Theorem for Real Linear space, can be extended to a linear functional go on L into R such that go = g on M and go (x) p (x) x L.

Define fo (x) = go (x) – i go (i x) for x L as in the Hahn-Banach Theorem, fo is linear functional on

L such that fo = f on M. To complete the proof we have to prove that

|fo (x)| p (x) x L.

i Let x L and fo (x) = r e , r > o and real. Then

– i –i |fo (x)| = r = e re = e fo (x)

–i = fo (e x).

–i Since r = fo (e x), fo is real so that we can take

–i –i |fo (x)| = r = fo (e x) = go (e x) … (1)

–i –i Since go (x) p (x), go (e x) p (e x) for x L.

–i –i –i But p (e x) = |e | p (x) so that go (e x) p (x) … (2) It follows from (1) and (2) that

|fo (x)| p (x) This completes the proof of the theorem. Corollary 1: Deduce the Hahn-Banach theorem for normed linear spaces from the generalised Hahn-Banach theorem. Proof: Let p (x) = f x for x N. We first note that p (x) 0 for all x N. Then for any x, y N, we have p (x + y) = f x + y f ( x + y ) = f x + f y = p (x) + p (y) p (x + y) p (x) + p (y)

Also p ( x) = f x = | | f x = | | p (x). Hence p satisfies all the conditions of the generalized Hahn-Banach Theorem for

Complex Linear space. Therefore a functional fo defined on all of N such that fo = f on M and

|fo (x)| p (x) = f x x N.

fo f … (3)

Since fo is the extension of f from a subspace M, we get

f fo … (4) From (3) and (4) it follows that

fo = f

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Notes

Notes Let L be a linear space. A mapping p : L R is called a sub-linear functional on L if it satisfies the following two properties namely,

(i) p (x + y) p (x) + p (y) x, y L (sub additivity)

(ii) p ( x) = p (x), 0 (positive homogeneity) Thus p defined on L in the above theorems is a sub-linear functional on L.

Some Applications of the Hahn-Banach Theorem

Theorem: If N is a normed linear space and xo N, xo 0 then there exists a functional fo N* such that

fo (xo) = xo and fo = 1.

Proof: Let M denote the subspace of N spanned by xo, i.e.,

M = { xo : any scalar}. Define f : M F (R or C) by

f ( xo) = xo . We show that f is a functional on M with f = 1.

Let x1, x2 M so that

x1 = 1 xo and x2 = 2 xo. Then

f (x1 + x2) = f ( 1 xo + 2 xo)

= ( 1 + 2) xo

But ( 1 + 2) xo = 1 xo + 2 xo

= f (x1) + f (x2)

Hence f (x1 + x2) = f (x1) + f (x2) … (1)

Let k be a scalar (real or complex). Then if x M, then x = xo.

Now f (kx) = f (k xo) = k xo = k f (x) … (2) If follows from (1) and (2) that f is linear.

Further, we note that since xo M with = 1, we get

f (xo) = xo .

For any x M, we get, | f (x)| = | | || xo|| = xo = x |f (x) | = x f is bounded and we have

|f(x)| sup = 1 for x M and x 0. x

So by definition of norm of a functional, we get f = 1.

208 LOVELY PROFESSIONAL UNIVERSITY Unit 17: The Hahn-Banach Theorem

Notes Hence by Hahn-Banach theorem, f can be extended to a functional fo N* such that fo (M) = f (M) and fo = f = 1, which in particular yields that

fo (xo) = f (xo) = xo and fo = 1. This completes the proof of the theorem. Corollary 2: N* separates the vector (points) in N. Proof: To prove the cor. it suffices to show that if x, y N with x y, then there exists a f N* such that f (x) f (y). Since x y x – y 0. So by above theorem there exists a functional f N* such that f (x – y) = f (x) – f (y) 0 and hence f (x) f (y). This shows that N* separates the point of N. Corollary 3: If all functional vanish on a given vector, then the vector must be zero, i.e. if f (x) = 0 f N* then x = 0.

Proof: Let x be the given vector such that f (x) = 0 f N*.

Suppose x 0. Then by above theorem, there exists a function f N* such that f (x) = x > 0, which contradicts our supposition that f (x) = 0 f N*. Hence we must have x = 0.

17.2 Summary

 The Hahn-Banach Theorem: Let N be a normed linear space and M be a linear subspace of

N. If f is a linear functional defined on M, then f can be extended to a functional f o defined on the whole space N such that

fo = f

 If f is a complex linear functional defined on the subspace M such that |f (x)| p (x) for

x M, then f can be extended to a complex linear function fo defined on L such that

| fo (x) | p (x) for every x L.

17.3 Keywords

Hahn-Banach theorem: The Hahn-Banach theorem is one of the most fundamental and important theorems in functional analysis. It is most fundamental in the sense that it asserts the existence of the linear, continuous and norm preserving extension of a functional defined on a linear subspace of a normed linear space and guarantees the existence of non-trivial continuous linear functionals on normed linear spaces. Sub-linear Functional on L: Let L be a linear space. A mapping p : L R is called a sub-linear functional on L if it satisfies the following two properties namely,

(i) p (x + y) p (x) + p (y) x, y L (sub additivity) (ii) p ( x) = p (x), 0 (positive homogeneity)

LOVELY PROFESSIONAL UNIVERSITY 209 Measure Theory and Functional Analysis

Notes Thus p defined on L in the above theorems is a sub-linear functional on L. The Generalized Hahn-Banach Theorem for Complex Linear Space: Let L be a complex linear space. Let p be a real valued function defined on L such that p (x + y) p (x) + p (y)

and p ( x) = | | p (x) x L and scalar .

17.4 Review Questions

1. Let M be a closed linear subspace of a normed linear space N and x o is a vector not in M.

Then there exists a functional fo N* such that

fo (M) = 0 and fo (xo) 0

2. Let M be a closed linear subspace of a normed linear space N, and let x o be a vector not in

M. If d is the distance from xo to M, then these exists a functional fo N* such that

fo (M) = 0, fo (xo) = d, and fo = 1.

3. Let M is a closed linear subspace of a normed linear space N and x o N such that xo M.

If d is the distance from xo to M, then there exists a functional fo N* such that fo (M) = 0, fo 1 (x ) = 1 and f = . o o d

4. Let N be a normed linear space over R or C. Let M N be a linear subspace. Then M = N f N* is such that f (x) = 0 for every x M, then f = 0.

5. A normed linear space is separable if its conjugate (or dual) space is separable.

17.5 Further Readings

Books Walter Rudin, Functional Analysis (2nd ed.). McGraw-Hill Science/Engineering/ Math 1991. Eberhard Zeidler, Applied Functional Analysis: Main Principles and their Applications, Springer, 1995.

Online links mat.iitm.ac.in www.math.ksu.edu mizar.uwb.edu.pl/JFW/Vol5/hahnban.html

210 LOVELY PROFESSIONAL UNIVERSITY Unit 18: The Natural Imbedding of N in N**

Unit 18: The Natural Imbedding of N in N** Notes

CONTENTS

Objectives Introduction 18.1 The Natural Imbedding of N into N** 18.1.1 Definition: Natural Imbedding of N into N** 18.1.2 Definition: Reflexive Mapping 18.1.3 Properties of Natural Imbedding of N into N** 18.1.4 Theorems and Solved Examples 18.2 Summary 18.3 Keywords 18.4 Review Questions 18.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define the natural imbedding of N into N**.

 Define reflexive mapping.

 Describe the properties of natural imbedding of N into N*.

Introduction

As we know that conjugate space N* of a normed linear space N is itself a normed linear space. So, we can find the conjugate space (N*)* of N*. We denote it by N** and call it the second conjugate space of N. Likewise N*, N** is also a Banach space. The importance of the space N** lies in the fact that each vector x in N given rise to a functional Fx in N** and that there exists an isometric isomorphism of N into N**, called natural imbedding of N into N**.

18.1 The Natural Imbedding of N into N**

18.1.1 Definition: Natural Imbedding of N into N**

The map J : N N** defined by

J (x) = Fx x N, is called the natural imbedding of N into N**. Since J (N) N**, N can be considered as part of N** without changing its basic norm structure. We write N N** in the above sense.

LOVELY PROFESSIONAL UNIVERSITY 211 Measure Theory and Functional Analysis

Notes 18.1.2 Definition: Reflexive Mapping

If the map J : N N** defined by

J (x) = Fx x N, is onto also, then N (or J) is said to be reflexive (or reflexive mapping). In this case we write N = N**, i.e., if N = N**, then N is reflexive.

Note Equality in the above definition is in the sense of isometric isomorphism under the natural imbedding. Since N** must always be a complete normed linear space, no incomplete space can be reflexive.

18.1.3 Properties of Natural Imbedding of N into N**

I. Let N be a normed linear space. If x N, then x = sup {|f (x)|: f N* and f = 1}. Using natural imbedding of N into N**, we have for every x N,

Fx (f) = f (x) and Fx = x .

Now, Fx = sup {|Fx (f)|} sup {|f(x)|, f N*} f 1 f 1

therefore, x = sup {|f (x)|: f N*, f = 1}. II. Every normed linear space is a dense linear subspace of a Banach space. Let N be a normed linear space. Let J : N N** be the natural imbedding of N into N**.

The image of the mapping is linear subspace J (N) N**. Let J(N) be the closure of N(N) in N**.

Since N** is a Banach space, its closed subspace J(N) is also a Banach space. Hence if we identity N with J(N), then J(N) is a dense subspace of a Banach space.

18.1.4 Theorems and Solved Examples

Theorem 1: Let N be an arbitrary normal linear space. Then each vector x in N induces a functional Fx on N* defined by

Fx(f) = f(x) for every f N* such that Fx = x .

Further, the mapping J : N N** : J (x) = Fx for every x N defines and isometric isomorphisms of N into N**.

Proof: To show that Fx is actually a function on N*, we must prove that Fx is linear and bounded (i.e. continuous).

We first show Fx is linear.

212 LOVELY PROFESSIONAL UNIVERSITY Unit 18: The Natural Imbedding of N in N**

Let f, g N* and , be scalars. Then Notes

Fx ( f + g) = ( f + g) x = f (x) + g (x)

= Fx (f) + Fx (g)

Fx is linear

Fx is bounded. For any f N*, we have

|Fx (f)| = |f (x)| f x … (1)

Thus the constant x is bounded (in the sense of a bounded linear functional) for Fx. Hence Fx is a functional on N*.

We now prove Fx = x

We have Fx = sup {|Fx (f)| : f 1} sup { F x : f 1 } (Using (1)) x

Hence Fx x … (2) To prove the reverse inequality we consider the case when x = 0. In this case (2) gives

Fo = 0 = 0.

But Fx = 0 always. Hence Fo = 0 i.e. Fx = x for x = 0.

Not let x 0 be a vector in N. Then by theorem (If N is a normal linear space and x o N, xo 0, then there exists a functional fo N* such that

fo (xo) = xo and fo = 1.) a functional f N* such that f (x) = x and f = 1.

But Fx = sup {|Fx (f)| : f 1} = sup { f (x) : f = 1} and since x = f (x) sup {|f (x)| : f = 1} we conclude that Fx x … (3) [Note that since f (x) = x 0 we have f (x) = |f (x)|] From (2) and (3); we have

Fx = x … (4) Finally, we show that J is an isometric isomorphism of N into N**. For any x, y N and scalar.

Fx+y (f) = f (x + y) = f (x) + f (y)

= Fx (f) + Fy (f)

Fx+y (f) = (Fx + Fy) f … (5)

Fx+y = Fx + Fy … (6)

Further, F x (f) = f ( x) = f (x) = ( Fx) (f)

LOVELY PROFESSIONAL UNIVERSITY 213 Measure Theory and Functional Analysis

Notes Hence F x = Fx … (7) Using definition of J and equations (6) and (7) we get

J(x + y) = Fx+y = Fx + Fy = J (x) + J (y) … (8)

and J( x) = F x = Fx = J (x) … (9) (8) and (9) J is linear and also (4) shows that J is norm preserving. For any x and y in N, we have

J (x) – J (y) = Fx – Fy = Fx–y = x – y … (10) Thus J preserve distances and it is an isometry. Also (10) shows that J (x) – J (y) = 0 J (x – y) = 0 x – y = 0 i.e. J (x) = J (y) x = y so that J is one-one. Hence J defines an isometric isomorphism of N into N**. This completes the proof of the theorem.

n Example 1: The space  p (1 p < ) are reflexive. Solution: We know that if 1 p < , then

n * n p =  p .

n * n But q =  p

n ** n Hence p =  p

n ** n Similarly we have 1 =  1 for p = 1

and n ** = n for p =

n So that  p spaces are reflexive for 1 p < .

Example 2: The space  p for 1 < p < are reflexive.

  Sol: We know that if *p p and *q  p

**q  p .

 p are reflexive for 1 < p < .

A similar result can be seen to hold for Lp (X).

Example 2: If N is a finite dimensional normed linear space of dimension m, then N* also has dimension m.

Solution: Since N is a finite dimensional normed linear space of dimension m then {x 1, x2, …, xm}

is a basis for N, and if ( 1 2 … m) is any set of scalars, then there exists a functional f on N such

that f (xi) = i, i = 1, 2, …m.

214 LOVELY PROFESSIONAL UNIVERSITY Unit 18: The Natural Imbedding of N in N**

To show that N* is also of dimension m, we have to prove that there is a uniquely determined Notes basis (f1, f2, …, fm) in N*, with fj (xi) = y.

By the above fact, for each i = 1, 2, …, m, a unique fj in N* exists such that fj (xi) = ij. We show now that {f1, f2, …, fn} is a basis in N* to complete our proof.

Let us consider 1 f1 + 2 f2 + …… m fm = 0 … (1)

For all x N, we have 1 f1 (x) + 2 f2 (x) + …… + m fm (x) = 0.

We have jf j (x j ) 0 j i j i for i = 1, 2, …, m, when x = xi. j j

f1, f2, …… fm are linearly independent in N*.

Now let f (xi) = i.

Therefore if x = ix i , we get d

f (x) = 1f (x1) + 2f (x2) + …… + mf (xm) … (2)

Further fj (x) = 1fj (x1) + …… + ifj (xi) + …… + mfj (xm)

fj (x) = j From (1) and (2), it follows that

f (x) = 1 f1 (x) + 2 f2 (x) + …… + m fm (x)

= ( 1f + 2f2 + …… + m fm) (x)

(f1, f2, …… fm) spans the space. N* is m-dimensional.

18.2 Summary

 The map J : N N** defined by

J (x) = Fx x N, is called the natural imbedding of N into N**.

 If the map J : N N** defined by

J (x) = Fx x N, is onto also, then N (or J) is said to be reflexive. In this case we write N = N**, i.e., if N = N**, then N is reflexive.

 Let N be an arbitrary normal linear space. Then each vector x in N induces a functional F x

on N* defined by Fx (f) = f (x) for every f N* such that Fx = x .

18.3 Keywords

Natural Imbedding of N into N**: The map J : N N** defined by

J (x) = Fx x N, is called the natural imbedding of N into N**.

LOVELY PROFESSIONAL UNIVERSITY 215 Measure Theory and Functional Analysis

Notes Reflexive Mapping: If the map J : N N** defined by

J (x) = Fx x N, is onto also, then N (or J) is said to be reflexive (or reflexive mapping).

18.4 Review Questions

1. Let X be a compact Hausdorff space, and justify the assertion that C (X) is reflexive if X is finite. 2. If N is a finite-dimensional normed linear space of dimension n, show that N* also has dimension n. Use this to prove that N is reflexive. 3. If B is a Banach space, prove that B is reflexive B* is reflexive. 4. Prove that if B is a reflexive Banach space, then its closed unit sphere S is weakly compact. 5. Show that a linear subspace of a normed linear space is closed it is weakly closed.

18.5 Further Readings

Books G.F. Simmons, Introduction to topology and Modern Analysis. McGraw-Hill, Kogakusha Ltd. J.B. Conway, A Course in Functional Analysis. Springer-Verlag.

Online links www.mathoverflow.net/…/natural-embedding www.tandfonline.com

216 LOVELY PROFESSIONAL UNIVERSITY Unit 19: The Open Mapping Theorem

Unit 19: The Open Mapping Theorem Notes

CONTENTS

Objectives Introduction 19.1 The Open Mapping Theorem 19.1.1 Lemma 19.1.2 Proof of the Open Mapping Theorem 19.1.3 Theorems and Solved Examples 19.2 Summary 19.3 Keywords 19.4 Review Questions 19.5 Further Readings

Objectives

After studying this unit, you will be able to:

 State the open mapping theorem.

 Understand the proof of the open mapping theorem.

 Solve problems on the open mapping theorem.

Introduction

In this unit, we establish the open mapping theorem. It is concerned with complete normed linear spaces. This theorem states that if T is a continuous linear transformation of a Banach space B onto a Banach space B , then T is an open mapping. Before proving it, we shall prove a lemma which is the key to this theorem.

19.1 The Open Mapping Theorem

19.1.1 Lemma

Lemma 1: If B and B are Banach spaces and T is a continuous linear transformation of B onto B , then the image of each sphere centered on the origin in B contains an open sphere centered on the origin in B .

Proof: Let Sr and Sr respectively denote the open sphere with radius r centered on the origin in B and B .

We one to show that T (Sr) contains same Sr .

However, since T (Sr) = T (r S1) = r T (S1), (by linearity of T).

2 It therefore suffices to show that T (S1) contains some Sr for then S , where = r , will be contained in T (Sr). We first claim that T(S)1 (the closure of T (S1)) contains some Sr .

LOVELY PROFESSIONAL UNIVERSITY 217 Measure Theory and Functional Analysis

Notes If x is any vector in B we can by the Archimedean property of real numbers find a positive

integer n such that n > x , i.e., x Sn, therefore

B = S  n n 1

and since t is onto, we have B = T (B)

S = T  n n 1

= T(S)  n n 1

Now B being complete, Baire’s theorem implies that some TSn0 possesses an interior point

Z0. This in turn yields a point y0 TSn0 such that y0 is also an interior point of TSn0 .

Further, maps j : B B and g : B B

defined respectively by j (y) = y = y – y0 and g (y) = 2 n0 y

where no is a non-zero scalars, are homeomorphisms as shown below f is one-to-one and onto. –1 To show f, f are continuous, let yn B and yn y in B.

Then f (yn) = yn – y0 y – y0 = f (y)

–1 –1 and f (yn) = yn + y0 y + y0 = f (y) Hence f and f–1 are both continuous so that is a homeomorphism.

Similarly g : B B : g (x) = 2n0y is a homeomorphism for, g is one-to-one, onto and bicontinuous

for n0 0. Therefore we have

(i) f (y0) = 0 = origin in B is an interior point of f T Sn .

f T Sn (ii) 0 = f T Sn0

= T Sn0 y 0

TS2  y0 T S n n0 0

(iii) TS = T 2n S 1 2n 0 T S 1 2n0 0

= g (T(S1 )) g (T(S 1 ))

= 2n0 T(S 1 )

218 LOVELY PROFESSIONAL UNIVERSITY Unit 19: The Open Mapping Theorem

Notes Combining (i) – (iii), it follows that origin is also an interior point of (T(S1 )) . Consequently, there exists > 0 such that

S T(S)1

This justifies our claims. We conclude the proof of the lemma by showing that

S /3 T(S1 ) , i.e., S T(S3 )

Let y B such that y < . Then y T(S)1 and therefore there exists a vector x1 B such that

x1 < 1, y – y1 < /2 and y1 = T (x1) We next observe that

S S /2 T(S1/2 ) and y – y1 /2

Therefore there exists a vector x2 B such that

1 x , y y y and y = T (x ) 32 1 2 22 2 2

Continuing this process, we obtain a sequence (xn) in B such that

1 x , y T(x ) and y (y y y ) n2n 1 n n 1 2 n 2 2

Let sn = x1 + x2 + … + xn, then

sn = x1 + x2 + … + xn

x1 + x2 + … + xn

1 1 1 < 1  2 22 2 n 1

1 2 1 2n < 2 Also for n > m, we have

sn – sm = sm+1 + … + xn

xm+1 + … + xn

1 1 <  2m 2 n 1

1 1 1 2m 2 n m = 1 (summing the G.P.) 1 2

1 1 = 2m 1 2 n m 1 0 as m, n

LOVELY PROFESSIONAL UNIVERSITY 219 Measure Theory and Functional Analysis

Notes Thus (sn) is a Cauchy sequence in B and since B is complete, a vector x B such that

sn x and therefore

x = lim sn lim s n 2 3 , n n

i.e., x S3. It now follows by the continuity of T that

T (x) = T lim s n n

= lim T(sn ) n

= lim (y1 y 2 y n ) n

= y

Hence y T (S3)

Thus y S y T (S3). Accordingly

S T (S3)

This completes the proof of the lemma.

Note If B and B are Banach spaces, the symbol S (x; r) and S (x; r) will be used to denote

open spheres with centre x and radius r in B and B respectively. Also Sr and Sr will denote these spheres when the centre is the origin. It is easy to see that

S (x; r) = x + Sr and Sr = r S1 For, we have y S (x; r) y – x < r

z < r and y – x = z, z Sr y = x + z and z < r

y x + Sr

Thus S (x; r) = x + Sr

x and S = {x : x < r} = x : 1 r r

= {r . y y < 1}

= r S1

Thus Sr = r S1 Now we prove an important lemma which is key to the proof of the open mapping theorem.

220 LOVELY PROFESSIONAL UNIVERSITY Unit 19: The Open Mapping Theorem

19.1.2 Proof of the Open Mapping Theorem Notes

Statement: If T is a continuous linear transformation of a Banach space B onto a Banach space B , then T is an open mapping. Proof: Let G be an open set in B. We are to show that T (G) is an open set in B i.e. if y is any point of T (G), then there exists an open sphere centered at y and contained in T (G). y T (G) y = T(x) for some x G. x G, G open in B there exists an open sphere S (x; r) with centre x and radius r such that S (x; r) G.

But as remarked earlier we can write S (x; r) = x + Sr, where Sr is open sphere of radius r centered at the origin in B.

Thus x + Sr G … (1) By lemma (2) (prove it),

T (Sr) contains some Sr1 . Therefore

S (y; r1) = y + Sr1

y + T (Sr)

= T (x) + T (Sr)

= T (x + Sr) T (G), (Using (1)) since x + Sr = S (x; r) G. Thus we have shown that to each y T (G), there exists an open sphere in B centered at y and contained in T (G) and consequently T (G) is an open set. This completes the proof of the theorem.

19.1.3 Theorems and Solved Examples

Theorem 1: Let B and B be Banach spaces and let T be an one-one continuous linear transformation of B onto B . Then T is a homeomorphism. In particular, T–1 is automatically continuous. Proof: We know that a one-to-one continuous open map from B onto B is a homeomorphism. By hypothesis T : B B is a continuous one-to-one onto mapping. By the open mapping theorem, T is open. Hence T is a homeomorphism. Since T is homeomorphism, T– exists and continuous from B to B so that T– is bounded and hence T–1 (B , B). This completes the proof of the theorem. Cor. 1: Let B and B be Banach spaces and let T (B, B ). If T : B B is one-to-one and onto, there are positive numbers m and M such that m x T (x) M x .

LOVELY PROFESSIONAL UNIVERSITY 221 Measure Theory and Functional Analysis

Notes Proof: By the theorem, T : B B is a homeomorphism. So that T and T–1 are both continuous and hence bounded. Hence by theorem, Let N and N be normed linear spaces. Then N and N are topologically isomorphic if and only if there exists a linear transformation T of N onto N and positive constants m and M such that m x T (x) M x , for every x N. constants m and M such that m x T (x) M x . Theorem 2: If a one-to-one linear transformation T of a Banach space B onto itself is continuous, then its inverse T–1 is continuous. Proof: T is a homeomorphism (using theorem of B onto B. Hence T–1 is continuous. This completes the proof of the theorem.

Note The following examples will show that the completeness assumption in the open mapping theorem and theorem can neither be omitted in the domain of definition of T nor in the range of T.

Example 1: Let C [0, 1] be the set of all continuous differentiable function on [0, 1]. We know that C [0, 1] is an incomplete space with the norm f = sup {|f (x)| : 0 x 1} But it is complete with respect to the norm f = f + f . Now let us choose B = [C [0, 1], ] and N = [C [0, 1], ]. Consider the identity mapping I : B N. The identity mapping is one-to-one onto and continuous. I–1 is not continuous. For, if it were continuous, then it is a homeomorphism. Mapping of a complete space into an incomplete space which cannot be. Hence I does not map open sets into open sets. Thus the open mapping theorem fails if the range of T is not a Banach space.

Example 2: Let B be an infinite dimensional Banach space with a basis { i : i I} with

i = 1 for each i I. Let N be the set of all functions from I to C which vanish everywhere except a finite member of points in I. Then N is a linear space under addition and scalar multiplication. We can define the norm on N as f = |f (i)|, i I. Then N is an incomplete normed linear space. Now consider the transformation T : N B defined as follows.

For each f N, let T (f) = f (i) i.

222 LOVELY PROFESSIONAL UNIVERSITY Unit 19: The Open Mapping Theorem

Then T is linear and Notes

T (f) |f(i)| i i I

= |fi | = f for every f N. i I

Hence T is bounded transformation from N to B . It is also one-to-one and onto. But T does not map open subsets of N onto B . For, if it maps, it is a linear homeomorphism from N onto B which cannot be since N is incomplete. Theorem 3: Let B be a Banach space and N be a normed linear space. If T is a continuous linear open map on B onto N, then N is a Banach space.

Proof: Let (yn) be a Cauchy sequence in N. Then we can find a sequence of positive integer (n k) such that nk < nk + 1 and for each k 1 y y nk 1 n k 2k

Hence by theorem: “Let N and N be normed linear spaces. A linear map T : N N is open and onto if and only if there is a M > 0 such that for any y N , there is a x N such that Tx = y and x M y .”

For ynk 1 y n N , there is a nk B and a constant M such that

y y x y y T (xk) = nk 1 n and k nk 1 n .

By on choice is convergent so that is convergent. Since B is a Banach ynk 1 y n k xnk k 1 k 1 space there is a x B such that

x = Lim xn n k k 1

Since T is continuous T(xk ) T (x) as n k 1

But so that T(xk ) y nk 1 y n 1 k 1

ynk 1 y n 1 T(x) y n k 1 y n 1 T(x)

Since (yn) is a Cauchy sequence such that every subsequences is convergent, (yn) itself converges y and yn n1 + T (x) in N.

Hence N is complete. Consequently, N is a Banach space. This completes the proof of the theorem.

Example: Let N be complete in two norms 1 and 2 respectively. If there is a number a > 0 such that x 1 a x 2 for all x N, then show that the two norms are equivalent. Solution: The identity map i : (N, 2) (N, 1) is an one-one onto map.

LOVELY PROFESSIONAL UNIVERSITY 223 Measure Theory and Functional Analysis

Notes Also x 1 a x 2 i is bounded … (1) i is continuous.

Hence by open-mapping theorem, i is open and so it is homeomorphism of (N, x 2) onto

(N, x 1). Consequently i is bounded as a map from (N, x 1) (N, x 2)

–1 Since i (x) = x, a, b s.t. x 2 b 1 … (2) (1) & (2) imply that the norms are equivalent.

19.2 Summary

 If B and B are Banach spaces and T is a continuous linear transformation of B onto B , then the image of each sphere centered on the origin in B contains an open sphere centered on the origin in B .

 The open mapping theorem : If T is a continuous linear transformation of a Banach space B onto a Banach space B , then T is an open mapping.

19.3 Keywords

Banach Space: A normed space V is said to be Banach space if for every Cauchy sequence

n V then there exists an element V such that lim n . n 1 n Homeomorphism: A map f : (X, T) (Y, U) is said to be homeomorphism if (i) f is one-one onto. (ii) f and f–1 are continuous.

+ Open Sphere: Let xo X and r R . Then set {x X : p (xo, x) < r} is defined as open sphere with

centre xo and radius r.

19.4 Review Questions

1. If X and Y are Banach spaces and A : X Y is a bounded linear transformation that is bijective, then prove that A–1 is bounded.

2. Let X be a vector space and suppose 1 , and 2 are two norms on X and that T1 and T2

are the corresponding topologies. Show that if X is complete in both norms and T 1 T2,

then T1 = T2.

19.5 Further Readings

Books Walter Rudin, Functional Analysis, McGraw-Hill, 1973. Jean Diendonne, Treatise on Analysis, Volume II, Academic Press (1970).

Online links euclid.colorado.edu/ngwilkin/files/math 6320…/OMT_CGT.pdf people.sissa.it/nbianchin/courses/…/lecture05.banachstein.pdf

224 LOVELY PROFESSIONAL UNIVERSITY Unit 20: The Closed Graph Theorem

Unit 20: The Closed Graph Theorem Notes

CONTENTS

Objectives Introduction 20.1 The Closed Graph Theorem 20.1.1 Graph of Linear Transformation 20.1.2 Closed Linear Transformation 20.1.3 The Closed Graph Theorem - Proof 20.2 Summary 20.3 Keyword 20.4 Review Questions 20.5 Further Readings

Objectives

After studying this unit, you will be able to:

 State the closed graph theorem.

 Understand the proof of the closed graph theorem

 Solve problems based on the closed graph theorem.

Introduction

Though many of the linear transformations in analysis are continuous and consequently bounded, there do exist linear transformation which are discontinuous. The study of such kind of transformation is much facilitated by studying the graph of transformation and using the graph of the transformation as subset in the Cartesian product space to characterise the boundedness of such transformations. The basic theorem in this regard is the closed graph theorem.

20.1 The Closed Graph Theorem

20.1.1 Graph of Linear Transformation

Definition: Let N and N be a normed linear space and let T : N N be a mapping with domain N and range N . The graph of T is defined to be a subset of N × N which consists of all ordered pairs (x, T (x)). It is generally denoted by GT. Therefore the graph of T : N N is

GT = {(x, T (x) : x N}.

Notes GT is a linear subspace of the Cartesian product N × N with respect to coordinate- wise addition and scalar multiplications.

LOVELY PROFESSIONAL UNIVERSITY 225 Measure Theory and Functional Analysis

Notes Theorem 1: Let N and N be normed linear spaces. Then N × N is a normed linear space with coordinate-wise linear operations and the norm.

1 p p (x, y) = x x p , where x N, y N

and | p < . Moreover, this norm induces the product topology on N × N , and N × N is complete iff both N and N are complete. Proof:

(i) It needs to prove the triangle inequality since other conditions of a norm are immediate. Let (x, y) and (x , y ) be two elements of N × N . Then (x, y) + (x , y ) = (x + x , y + y )

1 p p x x y y p

1 p p = x x y y p

1 1 p pp p p p = x y x y

(By Minkowski’s inequality) = (x, y) + (x , y ) . This establishes the triangular inequality and therefore N × N is a normed linear space.

Furthermore (xn, yn) (x, y) xn = x and yn = y. Hence theorem on N × N induces the product topology. (ii) Next we show that N × N is complete N, N are complete.

Let (xn, yn) be a Cauchy sequence in N × N . Given > 0, we can find a no such that

(xn, yn) – (xm, ym) < m, n no. … (1)

(xn – xm) < and yn – ym < m, n no

(xn) and (yn) are Cauchy sequences in N and N respectively. Since N, N are complete, let

xn xo N and yn yo N in their norms,

i.e. (xn – xo) < and yn – ym < m, n no. … (2)

since xo N, yo N , (xo, yo) N × N .

Further (xn, yn) – (xo, yo) < n no (using (2))

(xn, yn) (xo, yo) in the norm of N × N and (xo, yo) N × N . N × N is complete. The converse follows by reversing the above steps. This completes the proof of the theorem.

226 LOVELY PROFESSIONAL UNIVERSITY Unit 20: The Closed Graph Theorem

Notes

Notes The following norms are equivalent to above norm (i) (x, y) = max { x , y } (ii) (x, y) = x + y (p = 1 in the above theorem)

20.1.2 Closed Linear Transformation

Definition: Let N and N be normed linear spaces and let M be a subspace of N. Then a linear transformation T : M N is said to be closed iff xn M, xn x and T (xn) y imply x M and y = T (x). Theorem 2: Let N and N be normed linear spaces and B be a subspace of N. Then a linear transformation T : M N is closed its graph GT is closed.

Proof: Let T is closed linear transformation. We claim that its graph GT is closed i.e. GT contains all its limit point.

Let (x, y) be any limit point of GT. Then a sequence of points in GT, (xn, T (xn), xn M, converging to (x, y). But

(xn, T (xn)) (x, y)

xn, T (xn) – (x, y) 0

(xn – x), T (xn) – y 0

xn – x + T (xn) – y 0

xn – x 0 and T (xn) – y 0  xn x and T (xn) y ( T is closed)

(x, y) GT. (By def. of graph)

Thus we have shown that every limit point of GT is in GT and hence GT is closed.

Conversely, let the graph of T, GT is closed. To show that T is closed linear transformation.

Let xn M, xn x and T (xn) y.

Then it can be seen that (x, y) is an adherent point of GT so that  (x, y) GT . But GT = GT ( GT is closed)

Hence (x, y) GT and so by the definition of GT we have x M and y = T (x). Consequently, T is a closed linear transformation. This completes the proof of the theorem.

20.1.3 The Closed Graph Theorem – Proof

If B and B are Banach spaces and if T is linear transformation of B into B , then T is continuous

Graph of T (GT) is closed. Proof: Necessary Part:

Let T be continuous and let GT denote the graph of T, i.e.

GT = {(x, T (x) : x B} B × B .

LOVELY PROFESSIONAL UNIVERSITY 227 Measure Theory and Functional Analysis

Notes We shall show that GT = GT.

Since GT GT always, it suffices to show that GT GT.

Let (x, y) GT . Then there exists a sequence (xn, T (xn)) in GT such that

(xn, T (xn)) (x, y)

xn x and T (xn) y.

But T is continuous T (xn) T (x) and so y = T (x)

(x, y) = (x, T (x)) GT

GT GT

Hence GT = GT i.e. GT is closed. Sufficient Part:

Let GT is closed. Then we claim that T is continuous. Let B1 be the given linear space B renormed

by 1 given by

x 1 = x + T (x) for x B.

Now T (x) x + T (x) = x 1.

T is bounded (continuous) as a mapping from B1 to B .

So if B and B1 have the same topology then T will be continuous from B to B . To this end, we have

to show that B and B1 are homeomorphic. Consider the identity mapping

I : B1 B defined by

I (x) = x for every x B1. Then I is always one-one and onto.

Further I (x) = x x + T (x) = x 1

I is bounded (continuous) as a mapping from B1 onto B.

Therefore if we show that B1 is complete with respect to 1, then B1 is a Banach space so by theorem. “Let B and B be Banach spaces and let T be one-one continuous linear transformation of B onto B . Then T is a homeomorphism. In particular, T–1 is automatically continuous.”

I is homeomorphism. Therefore to complete the proof, we have to show that B 1 is complete

under the norm 1.

Let (xn) be a Cauchy sequence in B1. Then

xn – xm 1 = xn – xm + T (xn – xm) 0 as m, n

(xn) and (T (xn)) are Cauchy sequences in B and B respectively. Since B and B are complete, we have

xn x in B and T (xn) T (x) in B … (1)

228 LOVELY PROFESSIONAL UNIVERSITY Unit 20: The Closed Graph Theorem

Notes Since GT is closed, we have

(x1 T (x)) GT and if we take y = T(x); then (x, y) GT.

Now xn – x 1 = xn – x + T (xn – x)

= xn – x + T (xn) – T(x)

= xn – x + T (xn – y) 0 as n . (Using (1))

Hence, the sequence (xn) in B1 x B1 and consequently B1 is complete. This completes the proof of the theorem.

Theorem 3: Let B and B be Banach spaces and let T : B B be linear. If GT is closed in B × B and if T is one-one and onto, then T is a homeomorphism from B onto B . Proof: By closed graph theorem, T is continuous. Let T = T–1 : B B. Then T is linear.

Further (x, y) GT (y, x) GT .

GT is closed in B × B.

T is continuous (By closed graph theorem) T is a homeomorphism on B onto B . This completes the proof of the theorem. Theorem 4: Let a Banach space B be made into a Banach space B by a new norm. Then the topologies generated by these two norms are the same if either is stronger than the other. Proof: Let the new norm on B be . Let is stronger than . Then a constant k such that x k x for every x B. Consider the identity map I : B B .

We claim that G1 is closed.

Let xn x in B and xn y in B .

Then x k x x B, I (xn) = xn y in also.

Since a sequence cannot converge to two distinct points in , y = x. Consequently G1 is closed. Hence closed graph theorem, I is continuous. Therefore a k s such that x = I(x) k x for every x B. Hence is stronger than . Hence two topologies are same.

20.2 Summary

 Let N and N be a normal linear space and let T : N N be a mapping with domain N and range N . The graph of T is defined to be a subset of N × N which consist of all ordered

pairs (x, T (x)). It is generally denoted by GT.

LOVELY PROFESSIONAL UNIVERSITY 229 Measure Theory and Functional Analysis

Notes  Let N and N be normed linear spaces and let M be a subspace of N. Then a linear

transformation T : M N is said to be closed iff xn M, xn x and T (xn) y imply x M and y = T (x).

 If B and B are Banach spaces and if T is linear transformation of B into B , then T is

continuous Graph of T (GT) is closed.

20.3 Keyword

Closed Linear Transformation: Let N and N be normed linear spaces and let M be a subspace of N. Then a linear transformation T : M N is said to be closed

iff xn M, xn x and T (xn) y imply x M and y = T (x).

20.4 Review Questions

1. If X and Y are normed spaces and A : X Y is a linear transformation, then prove that

graph of A is closed if and only if whenever x n 0 and Axn y, it must be that y = 0. 2. If P is a projection on a Banach space B, and if M and N are its range and null space, then prove that M and N are closed linear subspaces of B such that B = M N.

20.5 Further Readings

Books Folland, Gerald B, Real Analysis: Modern Techniques and their Applications (1st ed.), John Wiley & Sons, (1984). Rudin, Walter, Functional Analysis, Tata McGraw-Hill (1973).

Online links euclid.colorado.edu/ngwilkin/files/math6320.../OMT_CGT.pdf mathworld.wolfram.com

230 LOVELY PROFESSIONAL UNIVERSITY Unit 21: The Conjugate of an Operator

Unit 21: The Conjugate of an Operator Notes

CONTENTS

Objectives Introduction 21.1 The Conjugate of an Operator 21.1.1 The Linear Function 21.1.2 The Conjugate of T 21.2 Summary 21.3 Keywords 21.4 Review Questions 21.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand the definition of conjugate of an operator.

 Understand theorems on it.

 Solve problems relate to conjugate of an operator.

Introduction

We shall see in this unit that each operator T on a normed linear space N induces a corresponding operator, denoted by T* and called the conjugate of T, on the conjugate space N*. Our first task is to define T* and our second is to investigate the properties of the mapping T T*.

21.1 The Conjugate of an Operator

21.1.1 The Linear Function

Let N* be the linear space of all scalar-valued linear functions defined on N. Clearly the conjugate space N* is a subspace of N*. Let T be a linear transformation T of N* into itself as follows:

If f N+, then T (f) is defined as [T (f)]x = f (T (x)) Since f (T (x)) is well defined, T is a well-defined transformation on N+. Theorem 1: Let T : N+ N+ be defined as [T (j)] x = f (T (x)), f N+, then (a) T (j) is a linear junction defined on N. (b) T is a linear mapping of N+ into itself.

LOVELY PROFESSIONAL UNIVERSITY 231 Measure Theory and Functional Analysis

Notes (c) T (N*) N* T is continuous, where T is a linear transformation of N into itself which is not necessarily continuous. Proof:

(a) x, y N and , be any scalars. Then [T (f)] ( x + y) = f (T ( x + y)) Since T and f are linear, we get f (T ( x + y)) = f (T (x) + f (T (y)) = [T (f)] (x) + [T (f)]y part (a). (b) Let f, g N+ and , be any scalars. Then [T ( f + g) (x)] = ( f + g) (T (x)) = [T (f)] (x) + (T (g)] (x) T is linear on N+ part (b) (c) Let S be a closed unit sphere in N. Then we know that T is continuous T (S) is bounded f (T (S)) is bounded for each f N*. By definition of T , f (T (S)) is bounded if and only if [T (f)] (S) is bounded for each f in N* = T (f) is in N* for each f in N*. T (N) N* part (c) This completes the proof of the theorem. Note: Part (c) of the above theorem enables us to restrict T to N* iff T is continuous. Hence by making T continuous we define an operation called the conjugate of T by restricting T to N*. We see it below.

21.1.2 The Conjugate of T

Definition: Let N be normed linear space and let T be a continuous linear transformation of N into itself (i.e. T is an operator). Define a linear transformation T* of N* into itself as follows: If f N*, then, T* (f) is given by

[T* (f)] (x) = f (T (x)) We call T* the conjugate of T. Theorem 2: If T is a continuous linear transformation on a normed linear space N, then its conjugate T* defined by T* : N* N* such that T* (f) = f.T where

[T* (f)] (x) = f (T (x)) f N* and all x N

is a continuous linear transformation on N* and the mapping T T* given by : (N) (N*) such that

232 LOVELY PROFESSIONAL UNIVERSITY Unit 21: The Conjugate of an Operator

(T) = T* for every (N) Notes is an isometric isomorphism of b (N) into b (N*) reverses products and preserves the identify transformation. Proof: We first show that T* is linear Let f, g N* and , be any scalars then [T* ( j + g)] (x) = ( j + g) (T (x)) = ( j) T (x) + ( g) T (x) = [j (T (x))] + [g (T (x))] = [T* (j)] (x) + [T* (g)] (x)

= [ T* (j) + T* (g)] (x) x N Hence T* ( f + g) = T* (f) + T* (g) T* is linear on N*. To show that T* is continuous, we have to show that it is bounded on the assumption that T is bounded. T* = sup { T* (f) : f 1} = sup { [T* (f)] (x) : f 1 and x } = sup { f (T (x))|: f 1 and x } = sup { f T x : x 1 and x } … (1)

T T* is a bounded linear transformation on N* into N*. Hence by application of Hahn- Banach theorem, for each non-zero x in N, a functional f N* such that f = 1 and f (T (x)) = T (x) … (2)

T(x) Hence T = sup : x 0 x

f T(x) = sup : f 1, x 0 (by (2)) x

T * (f) (x) = sup : f 1, x 0 (by (1)) x

T * (f) x sup : f 1, x 0 x

= sup T *(f) : f 1 T * … (3)

From (1) and (3) it follows that T = T* . … (4)

LOVELY PROFESSIONAL UNIVERSITY 233 Measure Theory and Functional Analysis

Notes Now we show that

: (N) (N*) given by (T)T* … (5)

for every T (N) is an isometric isomorphism which reverses the product and preserve the identity transformation. The isometric character of follows by using (5) as seen below: (T) = T* = T .

Next we show that is linear and one-to-one. Let T, T1 (N) and , be any scalars. Then

( T + T1) = ( T + T)* by (3)

But [( T + T1)* (f)] (x) = f ( T + T1) (x)

= f ( T (x) + T1 (x) ) Since f is linear, we get

[( T + T1)* (f)] = f (T (x)) + f (T1 (x))

= [T* (f) (x) + T *1 (f) (x)]

= [T * (f)] [T *1 (f)](x)

x N. Hence we get

[( T + T1)* (f)] = [T* (f)] + [T *1 (f) ]

= T * T *1 (f)

Hence ( T + T1)* = T* + T1* … (6)

Therefore ( T + T1) = ( T + T1)* = T* + T*1 = (T) + (T1) is linear.

To show is one-to-one, let (T) = (T1)

Then T* = T*1

T*T*1 = 0

Using (6) by choosing = 1, = – 1 we get

(T – T1)* = 0 T – T1 = 0 or T = T1. is one-to-one. Hence is an isometric isomorphism on (N) onto (N*). Finally we show that reverses the product and preserves the identity transformation.

Now [(T T1)* (f)] (x) = f ((T T1) (x))

= f (T (T1 (x))

= [T* (f)] [T1 (x)], since T1 (x) N and T* (f) N*.

234 LOVELY PROFESSIONAL UNIVERSITY Unit 21: The Conjugate of an Operator

Notes = [ T*1 (T* (f))] (x)

= [( T*1 T*) (f)] (x)

Hence, we get

(T T1)* = T*1 T* so that

(T T1) = (T T1)* = T*1 T.

reverses the product. Lastly if I is the identity operator on N, then [I* (f)] (x) = f (I (x)) = f (x) = (I f) (x). I* = I so that (I) = I* = I preserves the identity transformation. This completes the proof of the theorem. Theorem 3: Let T be an operator on a normal linear space N. If N N* in the natural imbedding, then T** is an extension of T. If N is reflexive, then T** = T. Proof: By definition, we have (T*)* = T** Using theorem 2, we have T* = T . Hence T** = T* = T . By definition of conjugate of an operator T : N N, T* : N* N*, T** : N** N**.

Let J ; x Fx be the natural imbedding of N onto N** so that

Fx (f) = f (x) and J (x) = Fx. Further, since T** is the conjugate operator of T*, we get T** (x ) x = x (T* (x )) where x N* and x N** T** (x ) x = T** (J (x)) x . Using the definition of conjugate, we get T** (J (x)) x = J (x) (T* (x )). By definition of canonical imbedding J (x) (T* (x )) = T* (x ) x. Again T* (x ) (x) = x (T (x)) (By definition of conjugate) Now x (T (x)) = J (T (x))x (By natural imbedding) Hence T** (J (x))x = J (T (x))x . T** . J = J T and so T** is the norm preserving extension of T. If N is reflexive, N = N** and so T** coincides with T.

LOVELY PROFESSIONAL UNIVERSITY 235 Measure Theory and Functional Analysis

Notes This completes the proof of the theorem. Theorem 4: Let T be an operator on a Banach space B. Then T has an inverse T–1 T* has an inverse (T*)–1, and (T*)–1 = (T–1)* Proof: T has inverse T–1 TT–1 = T–1 T = I By theorem 2, the mapping : T T* reverse the product and preserves the identity (TT–1)* = (T–1 T)* = I* (T–1)*T = T* (T–1)* = I (T*)–1 exists and it is given by (T*)–1 = (T–1)*. This completes the proof of the theorem.

21.2 Summary

 Let N+ be the linear space of all scalar-valued linear functions defined on N. Clearly the conjugate space N* is a subspace of N+. Let T be a linear transformation T of N into itself as follows:

If f N+, then T (f) is defined as T (f) x = f (T (x))

 Let N be a normed linear space and let T be a continuous linear transformation of N into itself. Define a linear transformation T* of N* into itself as follows: If f N+, then T (f) is given by T (f) x = f (T (x)) We call T* the conjugate of T.

21.3 Keywords

The Conjugate of T: Let N be normed linear space and let T be a continuous linear transformation of N into itself (i.e. T is an operator). Define a linear transformation T* of N* into itself as follows: If f N*, then, T* (f) is given by [T* (f)] (x) = f (T (x)) We call T* the conjugate of T. The Linear Function: Let N* be the linear space of all scalar-valued linear functions defined on N. Clearly the conjugate space N* is a subspace of N*. Let T be a linear transformation T of N* into itself as follows: If f N+, then T (f) is defined as [T (f)]x = f (T (x)) Since f (T (x)) is well defined, T is a well-defined transformation on N+.

21.4 Review Questions

1. Let B be a Banach space and N a normed linear space. If {Tn} is a sequence in B (B, N) such

that T(x) = lim Tn (x) exists for each x in B, prove that T is a continuous transformation.

236 LOVELY PROFESSIONAL UNIVERSITY Unit 21: The Conjugate of an Operator

2. Let T be an operator on a normed linear space N. If N is considered to be part of N** by Notes means of the natural imbedding. Show that T** is an extension of T. Observe that if N is reflexive, then T** = T. 3. Let T be an operator on a Banach space B. Show that T has an inverse T–1 T* has an inverse (T*)–1, and that in this case (T*)–1 = (T–1)*.

21.5 Further Readings

Books James Wilson Daniel, The Conjugate Gradient Method for Linear and Non-linear Operator Equations. G.O. Okikiolu, Special Integral Operators: Poisson Operators, Conjugate Operators and related Integrals. Vol. Okikiolu Scientific and Industrial Organization, 1981.

Online links epubs.siam-org/sinum/resource/1/sjnamm/v9/i|/p165_s| www.ima.umm.edu

LOVELY PROFESSIONAL UNIVERSITY 237 Measure Theory and Functional Analysis

Notes Unit 22: The Uniform Boundedness Theorem

CONTENTS Objectives Introduction 22.1 The Uniform Boundedness Theorem 22.1.1 The Uniform Boundedness Theorem – Proof 22.1.2 Theorems and Solved Examples 22.2 Summary 22.3 Keywords 22.4 Review Questions 22.5 Further Readings

Objectives

After studying this unit, you will be able to:

 State the uniform boundedness theorem.

 Understand the proof of this theorem.

 Solve problems related to uniform boundedness theorem.

Introduction

The uniform boundedness theorem, like the open mapping theorem and the closed graph theorem, is one of the cornerstones of functional analysis with many applications. The open mapping theorem and the closed graph theorem lead to the boundedness of T –1 whereas the uniform boundedness operators deduced from the point-wise boundedness of such operators. In uniform boundedness theorem we require completeness only for the domain of the definition of the bounded linear operators.

22.1 The Uniform Boundedness Theorem

22.1.1 The Uniform Boundedness Theorem – Proof

If (a) B is a Banach space and N a normed linear space,

(b) {Ti} is non-empty set of continuous linear transformation of B into N, and

(c) {Ti (x)} is a bounded subset of N for each x B, then { Ti } is a bounded set of numbers, i.e.

{Ti} is bounded as a subset of (B, N) Proof: For each positive integer n, let

Fn = {x B : Ti (x) n i}.

Then Fn is a closed subset of B. For if y is any limit point of Fn, then a sequence (xk) of points of

Fn such that

xk y as k

238 LOVELY PROFESSIONAL UNIVERSITY Unit 22: The Uniform Boundedness Theorem

Notes Tixk Tiy as k (By continuity of Ti)

Tixk Tiy as k (By continuity of norm)

T y = lt T x i k i k

 n i ( xk Fn)

y Fn.

Thus Fn contains all its limit point and is therefore closed. Further, if x is any element of B, then by hypothesis (c) of the theorem a real number k 0 s.t.

Tix k i Let n be a positive integer s.t. n > k. Then

Tix < n i so that x Fn.

Consequently, we have B = F .  n n 1

n Since B is complete, it therefore follows by Baire’s theorem that closure of some Fn, say FFo no , possesses an interior point xo. Thus we can find a closed sphere So with centre xo and radius ro such that So Fno .

Now if y is any vector in Ti (So), then

y = Ti so

where so So Fno .

y = Tsi o no.

Thus norm of every vector in Ti (So) is less than or equal to no. We write this fact as Ti (So) no.

S x Let S = o o . Then S is a closed unit sphere centred at the origin in B and ro

S x T o o Ti (S) = i ro

1 = Ti (S o ) T i (x o ) ro

1 Ti (S o ) T i (x o ) ro

2n o i ro

2no Hence Ti i ro This completes the proof of the theorem.

LOVELY PROFESSIONAL UNIVERSITY 239 Measure Theory and Functional Analysis

Notes 22.1.2 Theorems and Solved Examples

Theorem 1: If B is a Banach space and (fi (x)) is sequence of continuous linear functionals on B such

that (|fi (x)|) is bounded for every x B, then the sequence ( fi ) is bounded. Proof: Since the proof of the theorem is similar to the theorem (1), however we briefly give its proof for the sake of convenience to the readers.

For every m, let Fm B be the set of all x such that |fn (x)| m n

|fn (x)| m n

Now Fm is the intersection of closed sets and hence it is closed. As in previous theorem, we have

F B =  m . Since B is complete. It is of second category. Hence by Baire’s theorem, there is a m 1

xo Fm and a closed sphere S [xo, ro] such that |fn (x)| m n.

Let x be a vector with x ro.

Now fn (x) = fn (x + xo – xo)

= fn (x + xo) – fn (xo)

|fn (x)| |fn (x + xo)| + |fn (xo)| … (1)

Since x + xo – xo = x < ro, we have (x + ro) S [xo, ro]

|fn (x + xo) | m … (2)

Also we have |fn (xo)| k n … (3)

From (1), (2) and (3), we have for x S [xo, r].

|fn (x)| < (m + k) n.

r x Now for x B, consider the vector o . x

f (x) xro x x n m k Then |fn (x)| f n (m k) so that . ro x r o x ro In other words,

m k f . ro This completes the proof of the theorem.

Example 1: Show that the completeness assumption in the domain of (Ti) in the uniform boundedness theorem cannot be dropped. Solution: Consider N= space of all polynomial x

a t = x (t) = n n , an 0 n 0

for finitely many n’s.

240 LOVELY PROFESSIONAL UNIVERSITY Unit 22: The Uniform Boundedness Theorem

It we define the norm on N as Notes

x = max {|an|, n = 0, 1, 2, …} then N is an incomplete normed linear space.

n 1

Now define fn (x) = ak , n = 1, 2, … k 0

The functions {fn} are continuous linear functional on N.

m If we take x = a0 + a1t + …… + am t then

|fn (x)| (m + 1) max {|ak|} = (m + 1) x , so that (|fn (x)|) is point-wise bounded.

2 n–1 Now consider x = 1 + t + t + … + t . Then x = 1 and from the definition of |fn (x)| = n.

fn (x) Hence f = n. n x

( fn ) is unbounded.

Thus if we drop the condition of completeness in the domain of (Ti), the uniform boundedness theorem is not true anymore.

Theorem 2: Let N be a normed linear space and B be a Banach space. If a sequence (Tn) (B, N) such that T (x) = lim Tn (x) exists for each x in B, then T is a continuous linear transformation. Proof: T is linear.

T ( x + y) = lim Tn ( x + y)

= lim {Tn ( x) + Tn ( y)|}

= lim Tn (x) + lim Tn (y) = T (x) + T (y) for x, y B and for any scalars and . since lim Tn (x) exists, (Tn (x)) is a convergent sequence in N. Since convergent sequences are bounded, (Tn (x)) is point-wise bounded.

Hence by uniform bounded theorem, ( Tn ) is bounded so that a positive constant such that

Tn n.

Now Tn(x) Tn x x .

Since Tn (x) T (x), we have

T (x) x T is bounded (continuous) linear transformation. This completes the proof of the theorem.

Corollary 1: If f is a sequence in B* such that f (x) = lim fn (x) exists for each x B, then f is n continuous linear functional on B.

Example 2: Let (an) be a sequence of real or complex numbers such that for each x = (xn)

co, an x n converges. Prove that |an | . n 1 n 1

LOVELY PROFESSIONAL UNIVERSITY 241 Measure Theory and Functional Analysis

Notes n n a x Solution: For every x co, let fn a i x i . Since each i i is a finite sum of scalars, (fn) is i 1 i 1

sequence of continuous linear functional on c . Let f (x) = lim fn (x) lim a i x i . By cor. 1, f (x) o n n i 1

exists and bounded. f = |an |. Since f is bounded, |an | < . n 1 n 1

Theorem 3: A non-empty subset X of a normed linear space N is bounded f (X) is a bounded set of numbers for each f in N*.

Proof: Let X be a bounded subset of N so that a positive constant 1 such that

x 1 x X … (1) To show that f (X) is bounded for each f N*. Now f N* f is bounded.

2 > 0 such that |f (x)| 2 x x N … (2) It follows from (1) & (2) that

|f (x)| 1 2 x X. f (X) is a bounded set of real numbers for each f N*. Conversely, let us assume that f (X) is a bounded set of real numbers for each f N*.

To show that X is bounded. For convenience, we exhibit the vectors in X by writing X = {x i}. We now consider the natural imbedding J from N to N** given by

J : xi Fxi

From the definition of this natural imbedding, we have

Fx (f) = f (x) for each x N.

Hence our assumption f (X) = {f (xi)} is bounded for each f N* is equivalent to the assumption

that Fxi (f) is bounded set for each f N*.

Since N* is complete Fxi is bounded subset of N** by uniform boundedness theorem.

That is, is a bounded set of numbers. Since the norms are preserved in natural imbedding, Fxi

we have Fxi = xi for every xi X.

Therefore ( xi ) is a bounded set of numbers. Hence is bounded subset of Ni. This completes the proof of the theorem. Theorem 4: Let N and N be normed linear space A linear transformation. T : N N is continuous for each f N*, f o T N*. Proof: We first note that f o T is linear. Also f o T is well defined, since T (x) N for every x N and f is a functional on N so that f (T (x)) is well defined and f o T N*. Since T is continuous and f is continuous, f o T is continuous on N.

242 LOVELY PROFESSIONAL UNIVERSITY Unit 22: The Uniform Boundedness Theorem

Conversely, let us assume that f o T is continuous for each f N*. To show that T is continuous Notes it suffices to show that T(B) = {Tx : x N, B = x 1} is bounded in N . For each f N*, f o T is continuous and linear on N and so (f o T) B = f (T(B)) is bounded set for every f N*, where we have considered the unit sphere B with centre at the origin and radius 1. Since any bounded set in N can be obtained from B, T (B) is bounded by a non-empty subset X of a normed linear space N of bounded f (X) is a bounded set of number for each f in N*.

22.2 Summary

 Uniform Boundedness Theorem: If (a) B is Banach space and N a normed linear space,

(b) {Ti} is non-empty set of continuous linear transformation of B into N and (c) {Ti (x)} is a

bounded subset of N for each x B, then { Ti } is a bounded set of numbers, i.e. {Ti} is bounded as a subset of (B, N).

 If B is a Banach space and (fi (x)) is sequence of continuous linear functionals on B such that

(|fi (x)|) is bounded for every x B, then the sequence ( Ti ) is bounded.

22.3 Keywords

Imbedding: Imbedding is one instance of some mathematical structure contained within another instance, such as a group that is a subgroup. Uniform Boundedness Theorem: The uniform boundedness theorem, like the open mapping theorem and the closed graph theorem, is one of the cornerstones of functional analysis with many applications.

22.4 Review Questions

1. If X is a Banach space and A X*, then prove that A is a bounded set if and only if for every x in X, Sup {|f (x)| : f A} < .    2. Let be a Hilbert space and let be an orthonormal basis for . Show that a sequence {hn}   in satisfies 0 for every h in if and only if sup { hn : n 1} < and 0 for every e in .

22.5 Further Readings

Books Bourbaki, Nicolas, Topological vector spaces, Elements of mathematics, Springer (1987). Diendonne, Jean, Treatise on Analysis, Volume 2, Academic Press, (1970). Rudin, Walter, Real and Complex Analysis, McGraw-Hill, 1966.

Online links www.jstor.org/stable/2035429 www.sciencedirect.com/science/article/pii/S0168007211002004

LOVELY PROFESSIONAL UNIVERSITY 243 Measure Theory and Functional Analysis

Notes Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

CONTENTS

Objectives Introduction 23.1 Hilbert Spaces 23.1.1 Inner Product Spaces 23.1.2 Hilbert Space and its Basic Properties 23.1.3 Hilbert Space: Definition 23.1.4 Examples of Hilbert Space 23.2 Summary 23.3 Keywords 23.4 Review Questions 23.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define inner product spaces.

 Define Hilbert space.

 Understand basic properties of Hilbert space.

 Solve problems on Hilbert space.

Introduction

Since an inner product is used to define a norm on a vector space, the inner product are special normed linear spaces. A complete inner product space is called a Hilbert space. We shall also see from the formal definition that a Hilbert space is a special type of Banach space, one which possesses additional structure enabling us to tell when two vectors are orthogonal. From the above information, one can conclude that every Hilbert space is a Banach space but not conversely in general. We shall first define Inner Product spaces and give some examples so as to understand the concept of Hilbert spaces more conveniently.

23.1 Hilbert Spaces

23.1.1 Inner Product Spaces

Definition: Let X be a linear space over the field of complex numbers C. An inner product on X is a mapping from X × X C which satisfies the following conditions:

244 LOVELY PROFESSIONAL UNIVERSITY Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

(i) ( x + y, z) = (x, z) + (y, z), x, y, z X and , C. (Linearity property) Notes

(ii) (x, y) = (y, x) (Conjugate symmetry)

(iii) (x, x) 0, (x, x) = 0 x = 0 (Non-negativity) A complex inner product space X is a linear space over C with an inner product defined on it.

Notes

1. We can also define inner product by replacing C by R in the above definition. In that case, we get a real inner product space. 2. It should be noted that in the above definition (x, y) does not denote the ordered pair of the vectors x and y. But it denotes the inner product of the vectors x and y.

Theorem 1: If X is a complex inner product space then (a) ( x – y, z) = (x, z) – (y, z)

(b) (x, y + z) = (x, y) + (x, z)

(c) (x, y – z) = (x, y) – (x, z)

(d) (x, 0) and (0, x) = 0 for every x X. Proof: (a) ( x – y, z) = ( x + (– ) y, z) = (x, z) + (– ) (y, z) = (x, z) – (y, z).

(b) (x, y + z) = (y z,x) (y,x) (z,x)

= (y, x) (z,x)

= (x, y) (x,z)

(c) (x, y – z) = (x, y + (– ) z) = (x,y) ( )(x,z)

= (x, y) (x,z)

(d) (0, x) = (0 , x) = 0 ( , x) = 0, where is the zero

element of x and (x,0) (0,x) 0 0 .

Further note that (x, y + z) = (x, |y + 1|z) = 1 (x, y) + 1 (x, z) Hence (x, y + z) = (x, y) + (x, z). This completes the proof of the theorem.

LOVELY PROFESSIONAL UNIVERSITY 245 Measure Theory and Functional Analysis

Notes

1. Part (b) shows an inner product is conjugate linear in the second variable.

2. If (x, y) = 0 x X, then y = 0. If (x, y) = 0 x X, it should be true for x = y also, so that (y, y) = 0 y = 0.

n Example 1: The space  2 is an inner product space.

n Solution: Let x = (x1, x2, ……, xn), y = (y1, y2, ……, yn) 2 .

n Define the inner product on  2 as follows:

(x, y) = xi yi i 1 Now n

i (i) ( x + y, z) = xi yi z i 1

n n

= xi zi y i z i i 1 i 1

= (x, z) + (y, z)

(ii) (x,y) = xi yi i 1

 = (x1 y1 x 2 y 2 x n y n )

= x1 y1 x 2 y 2 x n y n = (y, x)

(iii) (x, x) = xi xi i 1

n 2 = xi 0 i 1

Hence (x, x) 0 and (x, x) = 0 xi = 0 for each i, i.e. (x, x) = 0 x = 0.

n (i) – (iii)  2 is a inner product space.

23.1.2 Hilbert Space and its Basic Properties

By using the inner product, on a linear space X we can define a norm on X, i.e. for each x X, we define x = (x,x) . To prove it we require the following fundamental relation known as Schwarz inequality.

246 LOVELY PROFESSIONAL UNIVERSITY Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

Theorem 2: If x and y are any two vectors in an inner product space then Notes |(x, y)| x y … (1) Proof: If y = 0, we get y = 0 and also theorem 1 implies that |(x, y)| = 0 so that (1) holds. Now, let y 0, then for any scalar C we have 0 x – y 2 = (x – y,x – y) But (x – y, x – y) = (x, x) – (x, y) – ( y, x) + ( y, y)

= (x, x) – (x, y) – (y, x) + (y, y) = x 2 – (y, x) – (x, y) + | |2 y 2

x 2 – (y, x) – (x, y) +| |2 y 2 = (x, y, x – y) = x – y 2 0 … (2)

(x,y) Choose = 2 , y 0, y 0 . y We get from (2)

2 2(x,y)(xy) (x,y) (x,y) 2 x2 2 (x,y) 4 y 0 y y y

2 2 2 2 |(x,y)| |(x,y)| (x,y) x2 2 2 0 y y y

2 2 |(x,y)| x2 0 y x 2 y 2 |(x, y)|2 or |(x, y)| x y . This completes the proof of the theorem.

Theorem 3: If X is an inner product space, then (x, x) has the properties of a norm, i.e.

x = (x, x) is a norm on X.

Proof: We shall show that satisfies the condition of a norm.

(i) x = (x, x) x 2 = (x, x) 0 and x = 0 x = 0.

(ii) Let x, y X, then x + y 2 = (x + y, x + y) = (x, x) + (x, y) + (y, x) + (y, y) … (1)

= x 2 + (x, y) + (x, y) + y 2

= x 2 + 2Re (x, y) + y 2 [ (x,y) (x, y) 2Re(x,y)]

x 2 + 2|(x, y)| + y 2 [ Re (x, y) |(x, y)|]

LOVELY PROFESSIONAL UNIVERSITY 247 Measure Theory and Functional Analysis

Notes x 2 + 2 x y + y 2 [using Schwarz inequality] = ( x + y )2 Therefore x + y x + y (iii) x 2 = ( x, x) = (x, x) = | |2 x 2 , x = x

(i)-(iii) imply that x = (x, x) is a norm on X. This completes the proof of the theorem.

Note Since we are able to define a norm on X with the help of the inner product, the inner product space X consequently becomes a normed linear space. Further if the inner product space X is complete in the above norm, then X is called a Hilbert space.

23.1.3 Hilbert Space: Definition

A complete inner product space is called a Hilbert space. Let H be a complex Banach space whose norm arises from an inner product which is a complex function denoted by (x, y) satisfying the following conditions:

H1 : ( x + y, 2) = (x, 2) + (y, 2),

H2 : (x, y) = (y, x), and

2 H3 : (x, x) = x , for all x, y, z H and for all , C.

23.1.4 Examples of Hilbert Space

n 1. The space  2 is a Hilbert space.

n n We have already shown in earlier example that  2 is an inner product space. Also  2 is a

n n Banach space. Consequently  2 is a Hilbert space. Moreover  2 , being a finite dimensional,

n hence  2 is a finite dimensional Hilbert space.

2.  2 is a Hilbert space.

 Consider the Banach space 2 consisting of all infinite sequence x = (xn), n = 1, 2, … of

2 complex numbers such that xn with norm of a vector x = (xn) defined by x = n 1

2 xn . n 1

248 LOVELY PROFESSIONAL UNIVERSITY Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

Notes We shall show that if the inner product of two vectors x = (x n) and y = (yn) is defined by

 (x, y) = xn yn , then 2 is a Hilbert space. n 1

We first show that inner product is well defined. For this we are to show that for all x, y in  2 the

infinite series xn yn is convergent and this defines a complex number. n 1

By Cauchy inequality, we have

1 1 n n2 n 2 2 2 x y xi yi i i i 1 i 1 i 1

1 1 2 2 2 2 xn y n . n 1 n 1

n 2 2 Since xn and yn are convergent, the sequence of partial sum xi yi is a monotonic n 1 n 1 i 1

increasing sequence bounded above. Therefore, the series xi yi is convergent. Hence n 1

xn yn is absolutely convergent having its sum as a complex number. n 1

Therefore (x, y) = xn yn is convergent so that the inner product is well defined. The condition n 1 of inner product can be easily verified as in earlier example. Theorem 4: If x and y are any two vectors in a Hilbert space, then (x + y) 2 + x – y 2 = 2 ( x 2 + y 2) Proof: We have for any x and y (x + y) 2 = (x + y, x + y) (By def. of Hilbert space) = (x, x + y) + (y, x + y) = (x, x) + (x, y) + (y, x) + (y, y) = x 2 + (x, y) + (y, x) + y 2 … (1) x – y 2 = (x – y, x – y) = (x, x – y) – (y, x – y) = (x, x) – (x, y) – (y, x) + y 2 … (2) Adding (1) and (2), we get x + y 2 + (x – y) 2 = 2 x 2 + 2 y 2 = 2 ( x 2 + y 2) This completes the proof of the theorem.

LOVELY PROFESSIONAL UNIVERSITY 249 Measure Theory and Functional Analysis

Notes Theorem 5: In a Hilbert space the inner product is jointly continuous i.e.,

xn x, yn y (xn, yn) (x, y) Proof: We have

|(xn, yn) – (x, y)| = |(xn, yn) – (xn, y) + (xn, y) – (x,, y)|

= |(xn, yn – y) + (xn – x, y)| (by linearity property of inner product)  |(xn, yn – y)| + |(xn – x, y)| [ | + | | | + | |]

xn + yn – y + xn – x y [By Schwarz inequality]

Since xn x and yn y as n .

Therefore yn – yn 0 and xn – x 0 as h . Also (xn) is a continues sequence, it is bounded

so that xn M n. Therefore

| (xn, yn) – (x, y) | 0 as n .

Hence (xn, yn) (x, y) as n . This completes the proof of the theorem. Theorem 6: A closed convex set E in a Hilbert space H continuous a unique vector of smallest norm. Proof: Let = inf { e ; e E}

To prove the theorem it suffices to show that there exists a unique x E s.t. x = .

Definition of yields us a sequence (xn) in E such that

Lim xn = … (1) n

x x Convexity of E implies that m n E . Consequently 2

x x m n x + x 2 … (2) 2 m n

Using parallelogram law, we get

2 2 2 2 xm + xn + xm – xn = 2 xn + 2 xn

2 2 2 2 or xm – xn = 2 xm + 2 xn – xm – xn

2 2 2 2 xm + 2 xn – d (Using (2)) 0 as m, n (Using (1))

2 xm – xn 0 as m, n

(xn) is a CAUCHY sequence in E. x E such that , since H is complete and E is a closed subset of H, therefore Lim xn x n

E is also complete and consequently (xn) is in E is a convergent sequence in E.

250 LOVELY PROFESSIONAL UNIVERSITY Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

Notes Now x = Lim xn n

= Lim xn ( norm is continuous mapping) n

= . Uniqueness of x. Let us suppose that y E, y x and y = .

x y Convexity of E E 2

x y … (3) 2

Also by parallelogram law, we have

2 2 2 x y 2 x y x y = 2 2 2 2 2

2 2 2 2 x y2 x y = 2 2 2 2

< 2. So that

x y 2 < , a result contrary to (3). 2

Hence we must have y = x. This completes the proof of the theorem.

Example: Give an example of a Banach space which is not an Hilbert space. Solution: C [a, b] is a Banach space with supremum norm, i.e. if x C [a, b] then x = Sup {|x(t)| : t [a, b]}. Then this norm does not satisfy parallelogram law as shown below:

t a Let x(t) = 1 and y (t) = . Then x = 1, y = 1 b a

t a Now x (t) + y (t) = 1 + so that x + y = 2 b a

t a x (t) – y (t) = 1 – so that x – y = 1 b a

Hence 2 ( x 2 – y 2) = 4, and x + y 2 + x – y 2 = 5 So that x + y 2 + x – y 2 2 x 2 + 2 y 2. C [a, b] is not a Hilbert space.

LOVELY PROFESSIONAL UNIVERSITY 251 Measure Theory and Functional Analysis

Notes 23.2 Summary

 Let X be a linear space over the field of complex numbers C. An inner product on X is a mapping from X × X C which satisfies the following conditions:

(i) ( x + y, z) = (x, z) + (y, z) x, y, z X and , C.

(ii) (x, y) = (y, x)

(iii) (x, x) 0, (x, x) = 0 x = 0

 A complete inner product space is called a Hilbert space.

23.3 Keywords

Hilbert Space: A complete inner product space is called a Hilbert space. Inner Product Spaces: Let X be a linear space over the field of complex numbers C. An inner product on X is a mapping from X × X C which satisfies the following conditions:

(i) ( x + y, z) = (x, z) + (y, z), x, y, z X and , C. (Linearity property)

(ii) (x, y) = (y, x) (Conjugate symmetry)

(iii) (x, x) 0, (x, x) = 0 x = 0

23.4 Review Questions

n 1. For the special Hilbert space  2 , use Cauchy’s inequality to prove Schwarz’s inequality.

n 2. Show that the parallelogram law is not true in  2 (n > 1).

3. If x, y are any two vectors in a Hilbert space H, then prove that 4 (x, y) = x + y 2 – x – y 2 + i x + iy 2 – i x – iy 2. 4. If B is complex Banach space whose norm obeys the parallelogram law, and if an inner product is defined on B by 4 (x, y) = x + y 2 – x – y 2 + i x + iy 2 – i x – iy 2, then prove that B is a Hilbert space.

23.5 Further Readings

Books Bourbaki, Nicolas (1987), Topological vector Spaces, Elements of Mathematics, BERLIN: Springer – Verlag. Halmos, Paul (1982), A Hilbert space Problem Book, Springer – Verlag.

Online links www.math-sinica.edu.tw/www/file_upload/maliufc/liu_ch04.pdf mathworld.wolfram.com>Calculus and Analysis > Functional Analysis

252 LOVELY PROFESSIONAL UNIVERSITY Unit 24: Orthogonal Complements

Unit 24: Orthogonal Complements Notes

CONTENTS

Objectives Introduction 24.1 Orthogonal Complement 24.1.1 Orthogonal Vectors 24.1.2 Pythagorean Theorem 24.1.3 Orthogonal Sets 24.1.4 Orthogonal Compliment: Definition 24.1.5 The Orthogonal Decomposition Theorem or Projection Theorem 24.2 Summary 24.3 Keywords 24.4 Review Questions 24.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define Orthogonal complement

 Understand theorems on it

 Understand the Orthogonal decomposition theorem

 Solve problems related to Orthogonal complement.

Introduction

In this unit, we shall start with orthogonality. Then we shall move on to definition of orthogonal complement. Let M be a closed linear subspace of H. We know that M is also a closed linear subspace, and that M and M are disjoint in the sense that they have only the zero vector in common. Our aim in this unit is to prove that H = M M , and each of our theorems is a step in this direction.

24.1 Orthogonal Complement

24.1.1 Orthogonal Vectors

Let H be a Hilbert space. If x, y H then x is said to be orthogonal to y, written as x y, if (x, y) = 0. By definition, (a) The relation of orthogonality is symmetric, i.e., x y y x

LOVELY PROFESSIONAL UNIVERSITY 253 Measure Theory and Functional Analysis

Notes For, x y (x, y) = 0

(x,y) 0

(y, x) = 0 y x (b) If x y then every scalar multiple of x is orthogonal to y i.e. x y x y for every scalar C. For, let be any scalar, then ( x, y) = (x, y) = . 0 = 0 x y x y. (c) The zero vector is orthogonal to every vector. For every vector x in H, we have (0, x) = 0 0 x for all x H. (d) The zero vector is the only vector which is orthogonal to itself. For, if x x (x, x) = 0 x 2 = 0 x = 0 Hence, if x x, then x must be a zero vector.

24.1.2 Pythagorean Theorem

Statement: If x and y are any two orthogonal vectors in a Hilbert space H, then x + y 2 = x – y 2 = x 2 + y 2. Proof: Given x y (x, y) = 0, then we must have y x i.e. (y, x) = 0 Now x + y 2 = (x + y, x + y) = (x, x) + (x, y) + (y, x) + (y, y) = x 2 + 0 + 0 + y 2 = x 2 + y 2 Also, x – y 2 = (x – y, x – y) = (x, x) – (x, y) – (y, x) + (y, y) = x 2 – 0 – 0 – y 2 = x 2 + y 2 x + y 2 = x – y 2 = x 2 + y 2

24.1.3 Orthogonal Sets

Definition: A vector x is to be orthogonal to a non-empty subset S of a Hilbert space H, denoted by x S if x y for every y in S.

254 LOVELY PROFESSIONAL UNIVERSITY Unit 24: Orthogonal Complements

Notes Two non-empty subsets S1 and S2 of a Hilbert space H are said to be orthogonal denoted by

S1 S2, if x y for every x S1 and every y S2.

24.1.4 Orthogonal Compliment: Definition

Let S be a non-empty subset of a Hilbert space H. The orthogonal compliment of S, written as S and is read as S perpendicular, is defined as

S = {x H : x y y S} Thus, S is the set of all those vectors in H which are orthogonal to every vectors in H which are orthogonal to every vector in S.

Theorem 1: If S, S1, S2 are non-empty subsets of a Hilbert space H, then prove the following: (a) {0} = H (b) H = {0} (c) S S {0}

(d) S1 S2 SS2 1 (e) S S

Proof:

(a) Since the orthogonal complement is only a subset of H, {0} H. It remains to show that H {0} . Let x H. Since (x, 0) = 0, therefore x {0} . Thus x H x {0} . H {0} . Hence {0} = H

(b) Let x H. Then by definition of H, we have

(x, y) = 0 y H Taking y = x, we get (x, x) = 0 x 2 = 0 x = 0 Thus x H x = 0 H = {0} (c) x S S . Then x S and x S Since x S , therefore x is orthogonal to every vector in S. In particular, x is orthogonal to x because x S. Now (x, x) = 0 x 2 = 0 x = 0. 0 is the only vector which can belong to both S and S . S S {0} If S is a subspace of H, then 0 S. Also S is a subspace of H. Therefore 0 S . Thus, if S is a subspace of H, then 0 S S . Therefore, in this case S S = {0}.

LOVELY PROFESSIONAL UNIVERSITY 255 Measure Theory and Functional Analysis

Notes (d) Let S1 S2, we have

x S2 x is orthogonal to every vector in S2

x is orthogonal to every vector in S1 because S1 S2.

x S1

SS2 1

(e) Let x S. Then (x, y) = 0 y S . by definition of (S ) , x (S ) . Thus x S x S . S S . This completes the proof of the theorem. Theorem 2: If S is a non-empty subset of a Hilbert space H, then S is a closed linear subspace of H and hence a Hilbert space.

Proof: We have

S = {x H : (x, y) = 0 y S} by definition. Since (0, y) = 0 y S, therefore at least 0 S and thus S is non-empty.

Now let x1, x2 S and , be scalars. Then (x1, y) = 0, (x2, y) = 0 for every y S. For every y S, we have

( x1 + x2, y) = (x1, y) + (x2, y) = (0) + (0) = 0

x1 + x2 S S is a subspace of H. Next we shall show that S is a closed subset of H.

Let (xn) S and xn x in H. Then we have to show that x S . For this we have to prove (x, y) = 0 for every y S.

Since xn S , (xn, y) = 0 for every y S and for n = 1, 2, 3, … Since the inner product is a continuous function, we get

(xn, y) (x, y) as n

Since (xn, y) = 0 n, (x, y) = 0 x S . Hence S is a closed subset of H.

256 LOVELY PROFESSIONAL UNIVERSITY Unit 24: Orthogonal Complements

Now S is a closed subspace of the Hilbert space H. Notes So, S is complete and hence a Hilbert space. This completes the proof of the theorem. Theorem 3: If M is a linear subspace of a Hilbert space H, then M is closed M = M . Proof: Let us assume that M = M , M being a subspace of H. by theorem (2), (M ) is closed subspace of H. Therefore M = M is a closed subspace of H. Conversely, let M be a closed subspace of H. We shall show that M = M . We know that M M . Now suppose that M M .

Now M is a proper closed subspace of Hilbert space M . a non-zero vector zo in M such that zo M or zo M .

Now zo M and M gives zo M M … (1) Since M is a subspace of H, we have M M = {0} … (2) (by theorem 1 (iii))

From (1) and (2) we conclude that z = 0, a contradiction to the fact that zo is a non-zero vector. M M can be a proper inclusion. Hence M = M . This completes the proof of the theorem. Cor. If M is a non-empty subset of a Hilbert space H, then M = M . Proof: By theorem (2), M is a closed subspace of H. So by theorem (3), M = (M ) = M . Theorem 4: If M and N are closed linear subspace of a Hilbert space H such that M N, then the linear subspace M N is closed. Proof: To prove: M + N is closed, we have to prove that it contains all its limit point. Let z be a limit point of M + N,

a sequence (zn) in M + N such that zn z in H.

Since M N, M N = {0} and M + N is the direct sum of the subspace M and N, zn can be written uniquely as

zn = xn + yn where xn M and yn N.

Taking two points zm = xm + ym and zn = xn + yn, we have

zm – zn = (xm – xn) + (ym – yn).

Since xm – xn M and ym – yn N, we get

(xm – xn) (ym – yn)

LOVELY PROFESSIONAL UNIVERSITY 257 Measure Theory and Functional Analysis

Notes So, by Pythagorean theorem, we have

2 2 2 (xm – xn) + (ym – yn) = xm – xn + ym – yn .

But (xm – xn) + (ym – yn) = zm – zn so that

2 2 2 zm – zn = xm – xn + ym – yn … (1)

Since (zn) is a convergent sequence in H, it is a Cauchy sequence in H.

2 Hence zm – zn 0 as m, n … (2) Using (2) in (1), we see that

2 2 xm – xn 0 and ym – yn 0

So that (xn) and (yn) are Cauchy sequence in M and N. Since H is complete and M and N are closed subspace of a complete space H, M and N are complete.

Hence, the Cauchy sequence (xn) in M converges to x in M and the Cauchy sequence (yn) in N converges to y in N.

Now z = lim zn = lim (xn + yn)

= lim xn + lim yn

But lim xn + lim yn = x + y M + N Thus, z = x + y M + N M + N is closed.

24.1.5 The Orthogonal Decomposition Theorem or Projection Theorem

Theorem 5: If M is a closed linear subspace of a Hilbert space H, then H = M M . Proof: If M is a subspace of a Hilbert space H, then we know that M M = {0}. Therefore in order to show that H = M M , we need to verify that H = M + M . Since M and M are closed subspace of H, M + M is also a closed subspace of H by theorem 4. Let us take N = M + M and show that N = H. From the definition of N, we get M N and M N. Hence by theorem (1), we have

N M and N M . Hence N M M = {0}. N = {0} N = {0} = H … (1) Since N = M + M is a closed subspace of H, we have by theorem (3), N = N … (2) From (1) and (2), we have N = M + M = H.

258 LOVELY PROFESSIONAL UNIVERSITY Unit 24: Orthogonal Complements

Since M M = {0} and Notes H = M + M , we have from the definition of the direct sum of subspaces, H = M M . This completes the proof of the theorem. Theorem 6: Let M be a proper closed linear sub space of a Hilbert space H. Then there exists a non-zero vector zo in H such that zo M. Proof: Since M is a proper subspace of H, there exists a vector x in H which is not in M. Let d = d (x, M) = inf { x – y } : y M}. Since x M, we have d > 0. Also M is a proper closed subspace of H, then by theorem: “Let M be a closed linear subspace of a Hilbert space H. Let x be a vector not in M and let d = d (x, m) (or d is the distance from x to M).

Then there exists a unique vector yo in M such that x – yo = d.”

There exists a vector yo in M such that

x – yo = d.

Let zo = x – yo. We then here

zo = x – yo = d > 0.

zo is a non-zero vector.

Now we claim that Zo M.

Let y be an arbitrary vector in M. We shall show that zo y. For any scalar , we have

zo – y = x – yo – y = x – (yo + y). since M is a subspace of H and yo, y M,

yo + M M. Then by definition of d, we have

x – (yo + y) d

Now zo – y = x – (yo + y) d = zo

2 2 zo – y zo or (zo – y, zo – y) – (zo, zo) 0 or (zo, zo) – (zo, y) – (y, zo) + (y, y) – (zo, zo) 0

or (z,y)o (z,y) o (y,y) 0 … (1)

The above result is true for all scalars .

Let us take (zo ,y) .

Putting the value of , in (1), we get

2 2 (z,y)(z,y)o o (z,y)(z,y) o o (z,y)(z,y) o o y 0

2 2 2 2 or –2 |(zo, y)| + |(zo, y)| y 0

LOVELY PROFESSIONAL UNIVERSITY 259 Measure Theory and Functional Analysis

Notes 2 2 or |(zo, y)| { y – 2} 0 … (2)

The above result is true for all real suppose that (zo, y) 0. Then taking positive and so small 2 2 2 that y < 2, we see from (2) that |(zo, y)| { y – 2} < 0. This contradicts (2).

Hence we must have (zo, yo) = 0 zo y, y M.

zo M. This completes the proof of the theorem.

24.2 Summary

 Let H be a Hilbert space. If x, y H then x is said to be orthogonal to y, written as x y, if (x, y) = 0.

 If x and y are any two orthogonal vectors in a Hilbert space H, then x + y 2 = x – y 2 = x 2 + y 2.

 Two non-empty subsets S1 and S2 of a Hilbert space H are said to be orthogonal denoted by

S1 S2, if x y for every x S1 and every y S2.

 Let S be a non-empty subsets of a Hilbert space H. The orthogonal compliment of S, written as S and is read as S perpendicular, is defined as

S = {x H : x y y S}

 The orthogonal decomposition theorem: If M is a closed linear subspace of a Hilbert space H, then H = M M .

24.3 Keywords

Orthogonal Compliment: Let S be a non-empty subset of a Hilbert space H. The orthogonal compliment of S, written as S and is read as S perpendicular, is defined as

S = {x H : x y y S}

Orthogonal Sets: A vector x is to be orthogonal to a non-empty subset S of a Hilbert space H, denoted by x S if x y for every y in S.

Two non-empty subsets S1 and S2 of a Hilbert space H are said to be orthogonal denoted by S1

S2, if x y for every x S1 and every y S2. Orthogonal Vectors: Let H be a Hilbert space. If x, y H then x is said to be orthogonal to y, written as x y, if (x, y) = 0. Pythagorean Theorem: If x and y are any two orthogonal vectors in a Hilbert space H, then x + y 2 = x – y 2 = x 2 + y 2.

24.4 Review Questions

1. If S is a non-empty subset of a Hilbert space, show that S = S . 2. If M is a linear subspace of a Hilbert space, show that M is closed M = M . 3. If S is a non-empty subset of a Hilbert space H, show that the set of all linear combinations of vectors in S is dense in H S = {0}.

260 LOVELY PROFESSIONAL UNIVERSITY Unit 24: Orthogonal Complements

4. If S is a non-empty subset of a Hilbert space H, show that S is the closure of the set of all Notes linear combinations of vectors in S. 5. If M and N are closed linear subspace of a Hilbert space h such that M N, then the linear subspace M + N is closed.

24.5 Further Readings

Books Halmos, Paul R. (1974), Finite-dimensional Vector Spaces, Berlin, New York Paul Richard Halmos, A Hilbert Space Problem Book, 2nd Ed.

Online links Itcconline.net/green/courses/203/…/orthogonal complements.html www.math.cornell.edu/~andreim/Lec33.pdf www.amazon.co.uk

LOVELY PROFESSIONAL UNIVERSITY 261 Measure Theory and Functional Analysis

Notes Unit 25: Orthonormal Sets

CONTENTS

Objectives Introduction 25.1 Orthonormal Sets 25.1.1 Unit Vector or Normal Vector 25.1.2 Orthonormal Sets, Definition 25.1.3 Examples of Orthonormal Sets 25.1.4 Theorems on Orthonormal Sets 25.2 Summary 25.3 Keywords 25.4 Review Questions 25.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand orthonormal sets

 Define unit vector or normal vector

 Understand the theorems on orthonormal sets.

Introduction

In linear algebra two vectors in an inner product space are orthonormal if they are orthogonal and both of unit length. A set of vectors from an orthonormal set if all vectors in the set are mutually orthogonal and all of unit length. In this unit, we shall study about orthonormal sets and its examples.

25.1 Orthonormal Sets

25.1.1 Unit Vector or Normal Vector

Definition: Let H be a Hilbert space. If x H is such that x = 1, i.e. (x, x) = 1, then x is said to be a unit vector or normal vector.

25.1.2 Orthonormal Sets, Definition

A non-empty subset { ei } of a Hilbert space H is said to be an orthonormal set if

(i) i j ei ej, equivalently i j (ei, ej) = 0

(ii) ei = 1 or (ei, ej) = 1 for every i.

262 LOVELY PROFESSIONAL UNIVERSITY Unit 25: Orthonormal Sets

Thus a non-empty subset of a Hilbert space H is said to be an orthonormal set if it consists of Notes mutually orthogonal unit vectors.

Notes

1. An orthonormal set cannot contain zero vector because 0 = 0. 2. Every Hilbert space H which is not equal to zero space possesses an orthonormal set.

x Since 0 x H. Then x 0. Let us normalise x by taking e = , so that x

x 1 e = x = 1. x x

e is a unit vector and the set {e} containing only one vector is necessarily an orthonormal set.

xi 3. If {xi} is a non-empty set of mutually orthogonal vectors in H, then {ei} = is an xi orthonormal set.

25.1.3 Examples of Orthonormal Sets

n th 1. In the Hilbert space 2 , the subset e1, e2, …, en where ei is the i-tuple with 1 in the i place and O’s elsewhere is an orthonormal set.

n n For (ei, ej) = 0 i j and (ei, ej) = 1 in the inner product xi yi of 2 . i 1

 th 2. In the Hilbert space 2 , the set {e1, e2, …, en, …} where en is a sequence with 1 in the n place and O’s elsewhere is an orthonormal set.

25.1.4 Theorems on Orthonormal Sets

Theorem 1: Let {e1, e2, …, en} be a finite orthonormal set in a Hilbert space H. If x is any vector in H, then

n 2 2 (x, ei ) x ; … (1) i 1 further,

x (x, ei ) e i ej for each j … (2) i 1

Proof: Consider the vector

y = x (x, ei ) e i i 1

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Notes We have y 2 = (y, y)

n n

= x (x,e)e,xi i (x,e)e i i i 1 i 1

n n

= (x,x) (x,e)(e,x)i i (x,e)(x,e) j j i 1 j 1

n n

(x,ei )(x,e i )(e i ,e) j i 1 j 1

n n n 2 = x (x,e)(x,e)i i (x,e)(x,e) i j (x,e)(x,e) j i i 1 j 1 i 1

On summing with respect to j and remembering that (ei, ej) = 1, i = j and (ei, ej) = 0, i j

n n n 2 2 2 2 = x x,ei x,e i (x,e) i i 1 i 1 i 1

n 2 2 = x (x, ei ) i 1

n 2 2 2 Now y 0, therefore x – (x, ei ) 0 i 1

n 2 2 (x, ei ) x i 1 result (1). Further to prove result (2), we have for each j (1 j n),

n n

x (x, ei ) e i , e j = (x,e)j (x,e)e,e i i j i 1 i 1

= (x,e)j (x,e)(e,e) i i j i 1  = (x, ej) – (x, ej) [ (ei, ej) = 1, i j 0, i = j] = 0

Hence x (x, ei ) e i e j for each j. i 1

This completes the proof of the theorem.

Note The result (1) is known as Bessel’s inequality for finite orthonormal sets.

264 LOVELY PROFESSIONAL UNIVERSITY Unit 25: Orthonormal Sets

Notes Theorem 2: If {ei} is an orthonormal set in a Hilbert space H and if x is any vector in H, then the set S = {ei : (x, ei) 0} is either empty or countable. Proof: For each positive integer n, consider the set

2 2 x S = e : (x, e ) . n i i n

If the set Sn contains n or more than n’ vectors, then we must have

2 2 x (x, e ) > n x … (1) i n ei S n By theorem (1), we have

2 2 (x, ei ) x … (2) ei S n which contradicts (1).

Hence if (2) were to be valid, Sn should have at most (n – 1) elements. Hence for each positive n, the set Sn is finite.

2 Now let ei S. Then (x, ei) 0. However small may be the value of |(x, ei)| , we can take n so large that

2 x |(x, e )|2 > . i n

Therefore if e S, then e must belong to some S . So, we can write S = S . i i n  n n 1 S can be expressed as a countable union of finite sets. S is itself a countable set.

If (x, ei) = 0 for each i, then S is empty. Otherwise S is either a finite set or countable set. This completes the proof of the theorem.

2 Theorem 3: Bessel’s Inequality: If {ei} is an orthonormal set in a Hilbert space H, then |(x, ei)| x 2 for every vector x in H.

Proof: Let S = {ei : (x, ei) 0}. By theorem (2), S is either empty or countable.

If S is empty, then (x, ei) = 0 i.

2 So if we define |(x, ei)| = 0, then

2 2 |(x, ei)| = 0 x . Now let S is not empty, then S is finite or it is countably infinite.

If S is finite, then we can write S = {e1, e2, …, en} for some positive integer n. In this case, we have

n 2 2 2 |(x, ei)| = |(x, ei )| x … (1) i 1 which represents Bessel’s inequality in the finite case.

LOVELY PROFESSIONAL UNIVERSITY 265 Measure Theory and Functional Analysis

Notes If S is countable infinite, let S be arranged in the definite order such as {e1, e2, …, en, …}. In this case we can write

2 2 |(x, ei)| = |(x, en )| … (2) n 1

The series on the R.H.S. of (2) is absolutely convergent. Hence every series obtained from this by rearranging the terms is also convergent and all such series have the same sum.

2 2 Therefore, we define the sum |(x, ei)| to be |(x, en )| . n 1

2 Hence the sum of |(x, ei)| is an extended non-negative real number which depends only on S and not on the rearrangement of vectors. Now by Bessel’s inequality in the finite case, we have

n 2 2 |(x, ei )| x … (3) i 1

For various values of n, the sum on the L.H.S. of (3) are non-negative. So they form a monotonic increasing sequence. Since this sequence is bounded above by x 2, it converges. Since the sequence is the sequence of partial sums of the series on the R.H.S. of (2), it converges and we

have ei S,

2 2 2 |(x, ei)| = |(x, ei )| x n 1

This completes the proof of the theorem.

2 Note: From the Bessel’s inequality, we note that the series |(x, en )| is convergent series. n 1

Corollary: If en S, then (x, en) 0 as n .

2 Proof: By Bessel’s inequality, the series |(x, en )| is convergent. n 1

2 Hence |(x, en)| 0 as n .

(x, en) 0 as n .

Theorem 4: If {ei} is an orthonormal set in a Hilbert space H and x is an arbitrary vector in H, then

{x – (x, ei) ei} ej for each j.

Proof: Let S = {ei : (x, ei) 0} Then S is empty or countable. [See theorem (2)]

If S is empty, then (x, ei) = 0 for every i.

In this case, we define (x, ei) ei to be a zero vector and so we get

x – (x, ei) ei = x – 0 = x.

266 LOVELY PROFESSIONAL UNIVERSITY Unit 25: Orthonormal Sets

Notes Hence in this case, we have to show x ej for each j.

Since S is empty, (x, ej) = 0 for every j.

x ej for every j. Now let S is not empty. Then S is either finite or countably infinite. If S is finite, let

S = {e1, e2, …, en} and we define

n (x, e ) e , (x, ei) ei = i i i 1

and prove that x (x, ei ) e i ej for each j. This result follows from (2) of theorem (1). n 1 Finally let S be countably infinite and let

S = {e1, e2, …, en, …}

(x, e ) e Let sn = i i i 1

2 m 2 For m > n, sm – sn = (x, ei )e i i n 1

m 2 = (x, ei ) i n 1

2 By Bessel’s inequality, the series (x, en ) is convergent. n 1

2 Hence (x, ei ) as m, n . i n 1

2 sm – sn 0 as m, n .

(sn) is a Cauchy sequence in H.

Since H is complete sn s H. Now s H can be written as

s = (x, en ) e n n 1

(x, e ) e Now we can define (x, ei) ei = n n . n 1

Before, completing the proof, we shall show that the above sum is well-defined and does not depend upon the rearrangement of vectors. For this, let the vector in S be arranged in a different manner as

S = {f1, f2, f3, …, fn, …}

LOVELY PROFESSIONAL UNIVERSITY 267 Measure Theory and Functional Analysis

Notes n

Let sn = (x, fi ) f i i 1

As shown for the case above for (sn), let

sn s in H where we can take

s = (x, fn ) f n . n 1

We prove that s = s . Given > 0 we can find n0 such that n n0.

2 x, e 2 s s i < , sn – s < , n < … (1) i no 1

For some positive integer m0 > n0, we can find all the terms of sn in smo also.

Hence smo s n o contains only finite number of terms of the type (x, ei) ei for i = n0 + 1, n0 + 2, …

2 2 Thus, we get smo s n o x, e i so that we have i no 1

smo s n o < … (2)

Now s – s = s smo s m o s n o s n o s

s sm0 s m 0 s n 0 s n 0 s

< + + = 3 (Using (1) and (2)) Since > 0 is arbitrary, s – s = 0 or s = s . Now consider

(x – (x, ei) ei, ej) = (x – s, ej)

But (x – s, ej) = (x, ej) – (s, ej)

= (x, ej) – (lim sn, ej) … (3) By continuity of inner product, we get

(lim sn, ej) = lim (sn, ej) … (4) Using (3) in (4), we obtain

(x – (x, ei) ei, ej) = (x, ej) – lim (sn, ej)

If ej S, then

(sn, ej) = (x, ei ) e i , e j = 0 i 1

lim (sn , e j ) = 0 n

268 LOVELY PROFESSIONAL UNIVERSITY Unit 25: Orthonormal Sets

Notes Hence (x – (x, ei) ei, ej) = (x, ej) = 0 since ej S.

If ej S, then (sn, ej) = (x, ei ) e i , e j … (5) i 1

If n > j, we get (x, ei ) e i , e j = (x, ej) … (6) i 1

From (5) & (6), we get

lim (sn , e j ) = (x, e ). n j

So, in this case

(x – (x, ei) ei, ej) = (x, ej) – (x, ej) = 0

Thus (x – (x, ei) ei, ej) = 0 for each j.

Hence x – (x, ei) ei ej for each j. This completes the proof of the theorem. Theorem 5: A Hilbert space H is separable every orthonormal set in H is countable.

Proof: Let H be separable with a countable dense subset D so that H = D . Let B be an orthonormal basis for H. We show that B is countable.

For x, y B, x y, we have

x – y 2 = x 2 + y 2 = 2 Hence the open sphere

1 1 S x; z : z x = as x B are all disjoint. 2 2

1 Since D is dense, D must contain a point in each S x, . 2

Hence if B is uncountable, then B must also be uncountable and H cannot be separable contradicting the hypothesis. Therefore B must be countable.

Conversely, let B be countable and let B = {x1, x2, …}. Then H is equal to the closure of all finite linear combinations of element of B. That is H = L(B) . Let G be a non-empty open subset of H.

Then G contains an element of the form ai x i with ai C. We can take ai C. We can take ai i 1 to be complex number with real and imaginary parts as rational numbers. Then the set

D = ai x i , n 1, 2, , a i rational i 1 is a countable dense set in H and so H is separable. This completes the proof of the theorem.

LOVELY PROFESSIONAL UNIVERSITY 269 Measure Theory and Functional Analysis

Notes Theorem 6: A orthonormal set in a Hilbert space is linear independent. Proof: Let S be an orthonormal set in a Hilbert space H. To show that S is linearly independent, we have to show that every finite subset of S is linearly independent.

Let S1 = {e1, e2, …, en} be any finite subset of S. Now let us consider

1e1 + 2e2 + … + nen = 0 … (1)

Taking the inner product with ek (1 k n),

n n

ie i , e k = i(e i , e k ) … (2) i 1 i 1

Using the fact that (ei, ek) = 0 for i k and (ek, ek) = 1, we get

i(e j , e k ) = k … (3) i 1

It follows from (2) on using (1) and (3) that

(0, ek) = k

k = 0 k = 1, 2, …, n.

S1 is linearly independent. This completes the proof of the theorem.

Example: If {ei} is an orthonormal set in a Hilbert space H, and if x, y are arbitrary vectors

in H, then (x,e)(y,e)i i x y .

Solution: Let S ei :(x, e i )(y, e i ) 0

Then S is either empty or countable. If S is empty, then we have

(x, ei )(y, e i ) = 0 i

and in this case we define

2 2 (x, ei )(y, e i ) to be number 0 and we have 0 x y .

If S is non-empty, then S is finite or it is countably infinite. If S is finite, then we can write

S = {e1, e2, …, en} for some positive integer n. In this case we define

(x, ei )(y, e i ) = (x, ei ) (y, e i ) i 1

270 LOVELY PROFESSIONAL UNIVERSITY Unit 25: Orthonormal Sets

1 1 Notes n2 n 2 2 2 (x, ei ) (y, e i ) (By Cauchy inequality) i 1 i 1

x 2 y 2 (by Bessel’s inequality for finite case)

(x, ei ) (y, e i ) x y … (1) i 1

Finally let S is countably infinite. Let the vectors in S be arranged in a definite order as

S = {e1, e2, …, en, …}. Let us define

(x, ei ) (y, e i ) = (x, en ) (y, e n ) . i 1

But this sum will be well defined only if we can show that the series (x, en ) (y, e n ) is n 1 convergent and its sum does not change by rearranging its term i.e. by any arrangement of the vectors in the set S. Since (1) is true for every positive integer n, therefore it must be true in the limit. So

(x, en ) (y, e n ) x y … (2) n 1

From (2), we see that the series (x, en ) (y, e n ) is convergent. Since all the terms of the series n 1 are positive, therefore it is absolutely convergent and so its sum will not change by any rearrangement of its terms. So, we are justified in defining

(x, ei ) (y, e i ) = (x, en ) (y, e n ) n 1 and from (2), we see that this sum is x y .

25.2 Summary

 Two vectors in an inner product space are orthonormal if they are orthogonal and both of unit length. A set of vectors from an orthonormal set if all vectors in the set are mutually orthogonal and all of unit length.

 Examples of orthonormal sets are as follows:

n th (i) In the Hilbert space 2 , the subset e1, e2, …, en where ei is the i-tuple with 1 in the i place and O’s elsewhere is an orthonormal set.

n n For (ei, ej) = 0 i j and (ei, ej) = 1 in the inner product xi yi of 2 . i 1

LOVELY PROFESSIONAL UNIVERSITY 271 Measure Theory and Functional Analysis

Notes  (ii) In the Hilbert space 2 , the set {e1, e2, …, en, …} where en is a sequence with 1 in the nth place and O’s elsewhere is an orthonormal set.

25.3 Keywords

Orthonormal Sets: A non-empty subset { ei } of a Hilbert space H is said to be an orthonormal set if

(i) i j ei ej, equivalently i j (ei, ej) = 0

(ii) ei = 1 or (ei, ej) = 1 for every i. Unit Vector or Normal Vector: Let H be a Hilbert space. If x H is such that x = 1, i.e. (x, x) = 1, then x is said to be a unit vector or normal vector.

25.4 Review Questions

1. Let {e1, e2, …, en} be a finite orthonormal set in a Hilbert space H, and let x be a vector in H.

If 1, 2, …, n are arbitrary scalars, show that xi e i attains its minimum value i 1

i = (x, ei) for each i. 2. Prove that a Hilbert space H is separable every orthonormal set in H is countable.

25.5 Further Readings

Book Sheldon Axler, Linear Algebra Done Right (2nd ed.), Berlin, New York (1997).

Online links www.mth.kcl.ac.uk/~jerdos/op/w3.pdf mathworld.wolfram.com www.utdallas.edu/dept/abp/PDF_files

272 LOVELY PROFESSIONAL UNIVERSITY Unit 26: The Conjugate Space H*

Unit 26: The Conjugate Space H* Notes

CONTENTS

Objectives

Introduction

26.1 The Conjugate Space H*

26.1.1 Definition

26.1.2 Theorems and Solved Examples

26.2 Summary

26.3 Keywords

26.4 Review Questions

26.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define the conjugate space H*.

 Understand theorems on it.

 Solve problems related to conjugate space H*.

Introduction

Let H be a Hilbert space. A continuous linear transformation from H into C is called a continuous linear functional or more briefly a functional on H. Thus if we say that f is a functional on H, then f will be continuous linear functional on H. The set H,C of all continuous linear functional on H is denoted by H* and is called the conjugate space of H. The elements of H* are called continuous linear functional or more briefly functionals. We shall see that the conjugate space of a Hilbert space H is the conjugate space H* of H is in some sense is same as H itself. After establishing a correspondence between H and H*, we shall establish the Riesz representation theorem for continuous linear functionals. Thereafter we shall prove that H * is itself a Hilbert space and H is reflexive, i.e. has a natural correspondence between H and H** and this natural correspondence is an isometric isomorphism of H onto H**.

26.1 The Conjugate Space H*

26.1.1 Definition

Let H be a Hilbert space. If f is a functional on H, then f will be continuous linear functional on H. The set H,C of all continuous linear functional on H is denoted by H* and is called the conjugate space of H. The conjugate space of a Hilbert space H is the conjugate space H* of H is in some sense is same as H itself.

LOVELY PROFESSIONAL UNIVERSITY 273 Measure Theory and Functional Analysis

Notes 26.1.2 Theorems and Solved Examples

Theorem 1: Let y be a fixed vector in a Hilbert space H and let fy be a scalar valued function on H defined by

fy x x,y x H.

Then fy is a functional in H* i.e. fy is a continuous linear functional on H and y fy .

Proof: From the definition

fy : H C defined as fy x x,y x H.

We prove that fy is linear and continuous so that it is a functional.

Let x1 ,x 2 H and , be any two scalars. Then for any fixed y H,

fy x1 x 2 x 1 x 2 ,y

x1 ,y x 2 ,y

fy x1 fy x 2

fy is linear.

To show fy is continuous, for any x H

fy x x,y x . y ...(1)

(Schwarz inequality)

Let y M. Then for M > 0

fy x M x so that fy is bounded and hence fy is continuous.

Now let y = 0, y 0 and from the definition fy = 0 so that fy y .

Sup fy x Further let y 0. Then from (1) we have y . x

Hence using the definition of the norm of a functional,

we get fy y ...(2)

Further fy sup fy x : x 1 ...(3)

y Since y 0, is a unit vector. y

From (3), we get

y fy fy y ...(4)

274 LOVELY PROFESSIONAL UNIVERSITY Unit 26: The Conjugate Space H*

Notes y y 1 But fy , y y, y y ...(5) y y y

Using (5) in (4) we obtain

fy y

From (2) and (6) it follows that

fy y

This completes the proof of the theorem.

Theorem 2: (Riesz-representation Theorem for Continuous Linear Functional on a Hilbert Space): Let H be a Hilbert space and let f be an arbitrary functional on H*. Then there exists a unique vector y in H such that f = fy, i.e. f(x) = (x,y) for every vector x H and f y .

Proof: We prove the following three steps to prove the theorem.

Step 1: Here we show that any f H * has the representation f = fy. If f = 0 we take y = 0 so that result follows trivially.

So let us take f 0.

We note the following properties of y in representation if it exists. First of all y 0, since otherwise f = 0.

Further (x,y) = 0 x for which f(x) = 0. This means that if x belongs to the null space N(f) of f, then (x,y) = 0.

y N f .

So let us consider the null space N(f) of f. Since f is continuous, we know that N(f) is a proper closed subspace and since f 0,N f H and so N f 0 .

Hence by the orthogonal decomposition theorem, ay0 0 in N f . Let us define any arbitrary x H.

z f x y0 f y 0 x

Now fz fxfy0 fyfx 0 0

z N f .

Since y0 N f , we get

0 z,y0 fxy 0 fyx,y 0 0

f x y0 , y 0 f y 0 x,y 0

Hence we get

fxy,y0 0 fy 0 x,y 0 0 ...(3)

LOVELY PROFESSIONAL UNIVERSITY 275 Measure Theory and Functional Analysis

Notes 2 Noting that y0 , y 0 y 0 0, we get from (3),

f y0 f x2 x,y0 ...(4) y0

We can write (4) as

f y0 f x x,2 y0 y0

f yo Now taking yo as y, we have established that there exists a y such that f(x) = (x,y) for x H. yo

Step 2: In this step we know that

f y

If f =0, then y = 0 and f y hold good.

Hence let f 0. Then y 0.

From the relation f(x) = (x,y) and Schwarz inequality we have

f x x,y x y .

f x sup y . x 0 x

Using definition of norm of f, we get from above

f y ...(5)

Now let us take x = y in f(x) = (x,y), we get

2 y y,y f y f y

y f ...(6)

(5) and (6) implies that

f y .

Step 3: We establish the uniqueness of y in f(x) = (x,y). Let us assume that y is not unique in f(x) = (x,y).

Let for all x H, y1 ,y 2 such that

f(x) = (x,y1) = (x,y2)

Then (x,y1) – (x,y2) = 0

(x,y1 – y2) = 0 x H.

276 LOVELY PROFESSIONAL UNIVERSITY Unit 26: The Conjugate Space H*

Notes Let us choose x to be y1 – y2 so that

2 y1 y 2 ,y 1 y 2 y 1 y 2 0

y1 y 2 0

y1 y 2

y is unique in the representation of f(x) = (x,y)

This completes the proof of the theorem.

Note The above Riesz representation theorem does not hold in an inner product space which is not complete as shown by the example given below. In other words the completeness assumption cannot be dropped in the above theorem.

Example: Let us consider the subspace M of l2 consisting of all finite sequences. This is the set of all scalar sequence whose terms are zero after a finite stage. It is an incomplete inner product space with inner product

x,y xn y n x,y M n 1

Now let us define

xn f x as x xn M. n 1 n

Linearity of f together with Hölder’s inequality yields

2 2 xn 1 2 f x2 xn n 1n n 1 n n 1

2 2 x,x x2 , 6 6

1 2 since 2 . n 1 n 6

f is a continuous linear functional on M.

We now prove that there is no y M such that

f x x,y x M.

th Let us take x = en = (0,0,....,1,0,0,.....) where 1 is in n place.

LOVELY PROFESSIONAL UNIVERSITY 277 Measure Theory and Functional Analysis

Notes 1 Using the definition of f we have f(x) = . n

Suppose y yn M satisfying the condition of the theorem, then

f x x,y xn y n y n as x e n .

1 Thus Riesz representation theorem is valid if and only if y 0 for every n. n n

Hence y yn M.

no y M such that f(x,y) = (x,y) for every x H.

the completeness assumption cannot be left out from the Riesz-representation theorem.

Theorem 3: The mapping :HH* defined by :HH* defined by y fy where fy(x) = (x,y) for every x H is an (i) additive, (ii) one-to-one, (iii) onto, (iv) symmetry, (v) not linear.

Proof:

(i) Let us show that is additive, i.e.,

y1 y 2 y 1 y 2 for y 1 ,y 2 H.

Now from the definition y1 y 2 f y1 y 2

Hence for every x H, we get

f x x,y y x,y x,y y1 y 2 1 2 1 2

f x f x y1 y2

f y y f f y y y1 y 2 1 2 y 1 y 2 1 2

(i) is one-to-one. Let y1 ,y 2 H

Then y =f and y =f . Then 1 y1 2 y 2

y = (y ) f = f 1 2 y1 y 2

fy x = f y x x H. ...(1) 1 2 f x = x,y and f x x,y y1 1 y 2 2 from (1), we get

x,y1 x,y 2 x,y 1 x,y 2 0

x,y1 y 2 0 x H ...(2)

Choose x = y1 – y2 then from (2) if follows that (y1 – y2,y1 – y2) = 0

278 LOVELY PROFESSIONAL UNIVERSITY Unit 26: The Conjugate Space H*

2 Notes y1 y 2 0

y1 y 2 is one-to-one.

(iii) is onto: Let f H* . Then y H such that

f(x) = (x,y)

since f(x) = (x,y) we get

f = fy so that y = fy = f.

Hence for f H* , a pre-image y H. Therefore is onto.

(iv) is isometry; let y1 ,y 2 H, then

y y f f 1 2 y1 y 2

fy f 1 y2

But f f f y y (By theorem (1)) y1 y 2 y 1 y 2 1 2

Hence y1 y 2 y 1 y 2 .

(v) To show is not linear, let y H and be any scalar. Then ,y f y. Hence for any x H, we get

fy (x) (x, y) (x,y) f(x) y

fy f y y y is not linear. Such a mapping is called conjugate linear.

This completes the proof of the theorem.

Note: The above correspondence is referred to as natural correspondence between H and H*.

Theorem 4: If H is a Hilbert space, then H* is also an Hilbert space with the inner product defined by

(fx, fy) = (y,x) … (1) Proof: We shall first verify that (1) satisfies the condition of an inner product.

Let x,y H and , be complex scalars.

(i) We know (see Theorem 3) that

fy f y

fy f y f y .

Now fx f y ,f z f x fy ,f z ... (2)

LOVELY PROFESSIONAL UNIVERSITY 279 Measure Theory and Functional Analysis

Notes But fx fy ,fz z, x, y (by (1))

Now z, x y z,x z,y

fx ,f z f y ,f z … (3)

From (2) and (3) it follows that fx f y ,f z f x ,f z f y ,f z

(ii) fx ,f y y,x x,y f x ,f y .

2 2 (iii) fx ,f x x,x x f x so f x ,f x 0 and f x 0 f x 0.

i iii implies that (1) represents an inner product. Now the Hilbert space H is a complete normed linear space. Hence its conjugate space H* is a Banach space with respect to the norm defined on H*. Since the norm on H* is induced by the inner product, H* is a Hilbert space with

the inner product (fx, fy) = (y,x) This completes the proof of the theorem.

Cor. The conjugate space H** of H* is a Hilbert space with the inner product defined as follows:

* ** If f,g H ,let Ff and F g be the corresponding elements of H obtained by the Riesz representation theorem.

** Then (Ff,Fg) = (g,f) defines the inner product of H . Theorem 5: Every Hilbert space is reflexive.

Proof: We are to show that the natural imbedding on H and H** is an isometric isomorphism.

* Let x be any fixed element of H. Let Fx be a scalar valued function defined on H by Fx(f) = f(x) for * ** every f H . We have already shown in the unit of Banach spaces that FH.x Thus each vector ** * x H gives rise to a functional Fx in H . Fx is called a functional on H induced by the vector x.

** Let J:HH be defined by J x Fx for every x H. We have also shown in chapter of Banach spaces that J is an isometric isomorphism of H into H **. We shall show that J maps H onto H**.

int o * Let T1 : H H defined by

T1 x f x ,f x y y,x for every y H.

***int o and T2 : H H defined by

* T2 f x F fx ,F f x f f,f x for f H .

** Then T2.T1 is a composition of T2 and T1 from H to H . By Theorem 3, T1,T2 are one-to-one and onto.

Hence T2.T1 is same as the natural imbedding J.

For this we show that J(x) = (T2.T1)x for every x H.

Now (T2.T1)x = T2(T1(x)) = T2(fx) = Ffx .

280 LOVELY PROFESSIONAL UNIVERSITY Unit 26: The Conjugate Space H*

Notes By definition of J, J(x) = Fx. Hence to show T2.T1 = J, we have to prove that Fx = Ffx .

* For this let f H . Then f = fy where f corresponds to y in the representation

Ffx (f) (f,f) x (f,f) y x (x,y) .

But (x,y) = fy(x) = f(x) = Fx(f).

Thus we get for every * Ffx (f) F x (f) f H .

Hence the mapping and F are equal. Ffx x

** T2.T1 = J and J is a mapping of H onto H , so that H is reflexive. This completes the proof of the theorem.

Notes 1. Since (From above theorem) Fx F fx x H

by using def. of inner product on H** and by the F,Fx y F,F fy f y f,f y x x,y def. of inner product on H*.

2. Since an isometric isomorphism of the Hilbert space H onto Hilbert space H**, therefore we can say that Hilbert space H and H** are congruent i.e. they are equivalent metrically as well as algebraically. We can identify the space H ** with the space H.

26.2 Summary

 Let H be a Hilbert space. If f is a functional on H, then f will be continuous linear functional on H. The set H,C of all continuous linear functional on H is denoted by H* and is called conjugate space of H. Conjugate space of a Hilbert space H is the conjugate space H * of H.

 Riesz-representation theorem for continuous linear functional on Hilbert space:

Let H be a Hilbert space and let f be an arbitrary functional on H*. Then there exists a unique vector y in H such that f = fy, i.e. f(x) = (x,y) for every vector x H and f y .

26.3 Keywords

Continuous Linear Functionals: Let N be a normal linear space. Then we know that the set R of real numbers and the set C of complex numbers are Banach spaces with the norm of any x R or x C given by the absolute value of x. We denote the BANACH space N,R or N,C by N*.

The elements of N* will be referred to as continuous linear functionals on N.

Hilbert space: A complete inner product space is called a Hilbert space.

Let H be a complex Banach space whose norm arises from an inner product which is a complex function denoted by (x,y) satisfying the following conditions:

LOVELY PROFESSIONAL UNIVERSITY 281 Measure Theory and Functional Analysis

Notes H1: x y,z x,z y,z

H2: x,y y,x

2 H3: x,x x

for all x,y,z H and for all , C.

Inner Product: Let X be a linear space over the field of complex numbers C. An inner product on X is a mapping from X × X C which satisfies the following conditions:

(i) ( x + y, z) = (x, z) + (y, z) x, y, z X and , C.

(ii) (x, y) = (y, x)

(iii) (x, x) 0, (x, x) = 0 x = 0

Riesz-representation Theorem for Continuous Linear Functional on a Hilbert Space: Let H be a Hilbert space and let f be an arbitrary functional on H*. Then there exists a unique vector y in H such that

f = fy, i.e. f(x) = (x,y) for every vector x H and f y .

The Conjugate Space H*: Let H be a Hilbert space. If f is a functional on H, then f will be continuous linear functional on H. The set H,C of all continuous linear functional on H is denoted by H* and is called the conjugate space of H. The conjugate space of a Hilbert space H is the conjugate space H* of H is in some sense is same as H itself.

26.4 Review Questions

1. Let H be a Hilbert space, and show that H* is also a Hilbert space with respect to the inner

product defined by (fx, fy) = (y, x). In just the same way, the fact that H* is a Hilbert space

implies that H** is a Hilbert space whose inner product is given by (Ff, Fg) = (g, f). 2. Let H be a Hilbert space. We have two natural mappings of H onto H**, the second of

which is onto: the Banach space natural imbedding x Fx, where fx (y) = (y, x) and

Ffx (f) (F, f x ). Show that these mappings are equal, and conclude that H is reflexive. Show

that (Fx, Fy) = (x, y).

26.5 Further Readings

Books Hausmann, Holm and Puppe, Algebraic and Geometric Topology, Vol. 5, (2005)

K. Yosida, Functional Analysis, Academic Press, 1965.

Online links www.spot.colorado.edu

www.arvix.org

282 LOVELY PROFESSIONAL UNIVERSITY Unit 27: The Adjoint of an Operator

Unit 27: The Adjoint of an Operator Notes

CONTENTS

Objectives

Introduction

27.1 Adjoint of an Operator

27.2 Summary

27.3 Keywords

27.4 Review Questions

27.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define the adjoint of an operator.

 Understand theorems on adjoint of an operator.

 Solve problems on adjoint of an operator.

Introduction

We have already proved that T gives rise to an unique operator T* and H* such that (T*f) (x) = f(Tx) f H * and x H. The operator T* on H* is called the conjugate of the operator T on H.

In the definition of conjugate T* of T, we have never made use of the correspondence between H and H*. Now we make use of this correspondence to define the operator T* on H called the adjoint of T. Though we are using the same symbol for the conjugate and adjoint operator on H, one should note that the conjugate operator is defined on H*, while the adjoint is defined on H.

27.1 Adjoint of an Operator

Let T be an operator on Hilbert space H. Then there exists a unique operator T* on H such that

(Tx,y)= (x,T*y) for all x, y H

The operator T* is called the adjoint of the operator T.

Theorem 1: Let T be an operator on Hilbert space H. Then there exists a unique operator T* on H such that

(Tx,y)= (x,T*y) for all x, y H ...(1)

The operator T* is called the adjoint of the operator T.

Proof: First we prove that if T is an operator on H, there exists a mapping T* on H onto itself satisfying

(Tx,y)= (x,T*y) for all x,y H. ...(2)

LOVELY PROFESSIONAL UNIVERSITY 283 Measure Theory and Functional Analysis

Notes Let y be a vector in H and fy its corresponding functional in H*. Let us define

T* : Hint o H * by

T* : fy f z ...(3)

Under the natural correspondence between H and H*, let z H corresponding to fz H *. Thus starting with a vector y in H, we arrive at a vector z in H in the following manner:

y fy T * f y f z z, ...(4)

where T* : H* H * and y fy and z f z are on H to H*

under the natural correspondence. The product of the above three mappings exists and it is denoted by T*.

Then T* is a mapping on H into H such that

T * y z.

We define this T* to be the adjoint of T. We note that if we identify H and H* by the natural

correspondence y fy , then the conjugate of T and the adjoint of T are one and the same.

After establishing, the existence of T*, we now show (1). For x H, by the definition of the conjugate T* on an operator T,

T * fy x f y Tx ...(5)

By Riesz representation theorem,

y fy so that

fy Tx Tx,y ...(6)

Since T* is defined on H*, we get

T * fy x f z x x,z ...(7)

But we have from our definition T*y = z ...(8)

From (5) and (6) it follows that

T * fy x Tx,y ...(9)

From (7) and (8) it follows that

T * fy x x,T * y ...(10)

From (9) and (10), we thus obtain

Tx,y x,T*y x,y H.

This completes the proof of the theorem.

284 LOVELY PROFESSIONAL UNIVERSITY Unit 27: The Adjoint of an Operator

Notes

Note The relation Tx,y x,T * y can be equivalently written as

T * x,y x,Ty since

T * x,y y,T * x = Ty,x = x,Ty = x,Ty

T * x,y x,Ty .

Example: Find adjoint of T if T is defined on  2 as Tx 0,x1 ,x 2 ,... for every x xn 2 .

Let T* be the adjoint of T. Using inner product in  2 , we have

T * x,y = x,Ty

since Ty 0,y1 ,y 2 ,... , we have

T * x,y x,Ty = xn+1 yn Sx,y , n=1

where S x x2 ,x 3 ,...

Hence T * x,y Sx,y for every x in  2 .

Since T* is unique, T*=S so that we have

T * x x2 ,x 3 ,x 4 ,... .

Theorem 2: Let H be the given Hilbert space and T* be adjoint of the operator T. Then T* is a bounded linear transformation and T determine T* uniquely.

Proof: T* is linear.

Let y1 ,y 2 H and , be scalars. Then for x H, we have

x,T* y1 y 2 Tx,y 1 y 2

But Tx,y1 y 2 Tx,y 1 Tx,y 2

Tx,y1 x,T * y 2

x, T*y1 x,T*y 2 .

Hence for any x H,

x,T* y1 y 2 x,T*y 1 x,T*y 2

LOVELY PROFESSIONAL UNIVERSITY 285 Measure Theory and Functional Analysis

Notes x, T * y1 T * y 2 .

T* is linear.

T* is bounded

for any y H, let us consider

2 T * y T * y,T * y

TT * yy

TT * y y using Schwarz inequality

T T * y y

2 Hence T * y T T * y y 0 ...(1)

If T * y 0 then T * y T y because T y 0

Hence let T * y 0.

Then we get from (1)

T * y T y .

since T is bounded,

T M so that

T * y M y for every y H.

T* is bounded.

T* is continuous.

Uniqueness of T*.

Let if T* is not unique, let T’ be another mapping of H into H with property

Tx,y = x,T * y x,y H.

Then we have

Tx,y = x,T'y ...(2)

and Tx,y = x,T * y ...(3)

From (2) and (3) it follows that

x,T'y=x,T*y x,y H

x, T'y T * y =0

286 LOVELY PROFESSIONAL UNIVERSITY Unit 27: The Adjoint of an Operator

Notes x,T' T*y=0 x H

T' T * y = 0 for every y H

Hence T'y = T * y for every y H.

TT*. This completes the proof of the theorem.

Notes 1. We note that the zero operator and the identity operator I are adjoint operators. For,

(i) x,0 * y 0x,y 0,y 0 x,0 = x,0y

so from uniqueness of adjoint 0* = 0.

(ii) (x,Iy) = (Ix,y) = (x,y) = (x,Iy)

so from uniqueness of adjoint I*=I.

2. If H is only an inner product space which is not complete, the existence of T* corresponding to T in the above theorem is not guaranteed as shown by the following example.

Example: Let M be a subspace of L2 consisting of all real sequences, each one containing only finitely many non-zero terms. M is an incomplete inner product space with the same inner product for  2 given by

x,y = xn y n ...(1) n=1

For each x M, define

x T xn ,0,0,...... (2) n=1 n

Then from the definition, for x,y M,

xn T x,y y1 . n=1 n

th Now let en 0,0,...,1,0,... where 1 is in the n place. Then using (3) we obtain

e (j) 1 T ,e 1.n 1. . en 1 j n

LOVELY PROFESSIONAL UNIVERSITY 287 Measure Theory and Functional Analysis

Notes Now we check whether there is T* which is adjoint of T. Now e,T*en 1 T*e 1 .e, n where the

R.H.S. gives the component wise inner product. Since T * e1 M,T * e 1 .e n cannot be equal to

1 n 1,2,... n

there is no T* on M such that

T en ,e 1 e n ,T * e 1

Hence completeness assumption cannot be ignored from the hypothesis.

Notes

1. The mapping TT* is called the adjoint operation on H. 2. From Theorem (2), we see that the adjoint operation is mapping TT* on H into itself.

Theorem 3: The adjoint operation T T* on (H) has the following properties:

(i) TT*T*T*1 2 1 2 (preserve addition)

(ii) TT*T*T*1 2 2 1 (reverses the product)

(iii) T*T* (conjugate linear)

(iv) T*T

(v) T*TT 2

Proof: (i) For every x, y H, we have

(x, (T1 + T2)*y = ((T1 + T2) x, y) (By def. of adjoint)

= (T1x + T2x, y)

= (T1x, y) + (T2x, y)

= (x, T1*y) + (x, T2*y)

= (x, T1*y + T2*y)

= (x, (T1* + T2*)y)

(T1 +T2)* = T1* + T2* by uniqueness of adjoint operator

(ii) For every x,y H, we have

x T1 T 2 * y T 1 T 2 x,y

T1 T 2 x ,y

288 LOVELY PROFESSIONAL UNIVERSITY Unit 27: The Adjoint of an Operator

Notes T2 x,T 1 * y

** x,T2 , T 1 y

** x, T2 T 1 ,y

** Therefore from the uniqueness of adjoint operator, we have TT*TT.1 2 2 1

(iii) For every x,y H, we have

x, T*y Tx,y Tx,y

Tx,y

x,T * y x, T * y

x T * y .

Therefore from the uniqueness of adjoint operator, we have

T*T*.

(iv) For every y H we have

2 T * y T * y,T * y

TT * y,y

2  T * y TT * y,y is a real number 0 TT * y,y

TT * y y By Schwarz inequality

T T * y y  Tx T x

2 Thus T * y T T * y y y H

T * y T y y H. ...(1)

Now T * Sup T * y : y 1

from (1), we see that if y 1 then T * y T

T*T ...(2)

Now applying (2) from the operator T* in place of operator T, we get

T**T*

LOVELY PROFESSIONAL UNIVERSITY 289 Measure Theory and Functional Analysis

Notes T**T*

TT* T**T ...(3)

From (2) and (3) it follows that

TT*.

(v) We have T*TT*T

TT  T*T

T 2 ...(4)

Further for every x H, we have

Tx2 Tx,Tx

T * Tx,x

T * T x,x

T * T x x (By Schwarz inequality)

T * T x 2

Then we have

Tx2 T * T x 2 x H ...(5)

Now T sup Tx : x 1

2 T2 sup Tx : x 1

sup Tx2 : x 1

From (5) we see that

if x 1 , then Tx2 T * T .

Therefore, Sup Tx2 :x 1 T*T

TT*T.2 ...(6)

From (5) and (6) it follows that

T*TT.2

This completes the proof of the theorem.

290 LOVELY PROFESSIONAL UNIVERSITY Unit 27: The Adjoint of an Operator

Notes Cor: If Tn is a sequence of bounded linear operators on a Hilbert space and

Tn T, then T n* T *.

We have

T*T*TT*n n

TTn (By properties of T*)

Since Tn T as n

** Tn T as n .

Theorem 4: The adjoint operation on H is one-to-one and onto. If T is a non-singular operator on H, then T* is also non-singular and

T*T*.1 1

Proof: Let :HH is defined by

T T * for every T H .

To show is one-to-one, let T,TH.1 2 Then we shall show that TTTT.1 2 1 2

Now TT1 2

T*T*1 2

T**T**1 2 (using Theorem 4. prop (iv))

TT1 2

is one-to-one.

is onto:

For T*H,we have on using Theorem 4 (iv),

T * = T* * =T.

Thus for every T*H,there is a T*H such that

T * T is onto.

Next let T be non-singular operator on H. Then its inverse T 1 exists on H and

TT1 T 1 T I.

LOVELY PROFESSIONAL UNIVERSITY 291 Measure Theory and Functional Analysis

Notes Taking the adjoint on both sides of the above, we obtain

TT1 * T 1 T * I *.

By using Theorem 4 and note (2) under Theorem 2, we obtain

T1 * T* T * T 1 * I.

T* is invertible and hence non-singular. Further from the above, we conclude

T*T*.1 1

This completes the proof of the theorem.

Note From the properties of the adjoint operation T T * on H discussed in Theorems (3) and (4), we conclude that the adjoint operation TT* is one-to-one conjugate linear mapping on H into itself.

Example: Show that the adjoint operation is one-to-one onto as a mapping of (H) into itself.

Solution: Let : (H) (H) be defined

(T) = T* T (H)

We show is one-to-one and onto.

is one-one:

Let T1, T2 (H). Then

(T1) = (T2) T1* = T2*

(T1*)* = (T2*)*

T1** = T2**

T1 = T2 is one-to-one.

is onto:

Let T be any arbitrary member of (H). Then T* (H) and we have (T*) = (T*)* = T** = T. Hence, the mapping is onto.

27.2 Summary

 Let T be an operator on Hilbert Space H. Then there exists a unique operator T* on H such that Tx,y x,T * y for all x,y H.The operator T* is called the adjoint of the operator T.

292 LOVELY PROFESSIONAL UNIVERSITY Unit 27: The Adjoint of an Operator

 The adjoint operation T T * on H has the following properties: Notes

(i) TT*T*T*1 2 1 2

(ii) TT*T*T*1 2 2 1

(iii) T*T*

(iv) T*T

(v) T*TT 2

27.3 Keywords

Adjoint of the Operator T: Let T be an operator on Hilbert space H. Then there exists a unique operator T* on H such that

(Tx,y)= (x,T*y) for all x, y H

The operator T* is called the adjoint of the operator T.

Conjugate of the Operator T on H: T gives rise to an unique operator T* and H* such that (T*f) (x) = f(Tx) f H * and x H. The operator T* on H* is called the conjugate of the operator T on H.

27.4 Review Questions

1. Show that the adjoint operation is one-to-one onto as a mapping of H into itself.

2. Show that TT * T2 .

3. Show that O*=O and I*=I. Use the latter to show that if T is non-singular, then T* is also

1 non-singular, and that in this case T*T*.1

27.5 Further Readings

Books N.I. Akhiezer and I.M. Glazman, Theory of Linear Operators in Hilbert Space, Vol. II, Pitman, 1981.

K. Yosida, Functional Analysis, Academic Press, 1965.

Online links www.math.osu.edu/ gerlach.1/math.BVtypset/node 78.html.

sepwww.standford.edu/sep/prof/pvi/conj/paper_html/node10.html.

LOVELY PROFESSIONAL UNIVERSITY 293 Measure Theory and Functional Analysis

Notes Unit 28: Self Adjoint Operators

CONTENTS

Objectives

Introduction

28.1 Self Adjoint Operator

28.1.1 Definition: Self Adjoint

28.1.2 Definition: Positive operator

28.2 Summary

28.3 Keywords

28.4 Review Questions

28.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define self adjoint operator.

 Define positive operator.

 Solve problems on self adjoint operator.

Introduction

The properties of complex number with conjugate mapping z z motivate for the introduction of the self-adjoint operators. The mapping z z of complex plane into itself behaves like the adjoint operation in H as defined earlier. The operation z z has all the properties of the adjoint operation. We know that the complex number is real iff z z . Analogue to this characterization in H leads to the motion of self-adjoint operators in the Hilbert space.

28.1 Self Adjoint Operator

28.1.1 Definition: Self Adjoint

An operator T on a Hilbert space H is said to be self adjoint if T*=T.

We observe from the definition the following properties:

(i) O and I are self adjoint O* O and I* I

(ii) An operator T on H is self adjoint if

Tx,y x,Ty x,y H and conversely.

294 LOVELY PROFESSIONAL UNIVERSITY Unit 28: Self Adjoint Operators

If T* is an adjoint operator T on H then we know from the definition that Notes

Tx,y x,T * y x,y H .

If T is self-adjoint, then T = T*.

Tx,y x,Ty x,y H.

Conversely, if Tx,y x,Ty x,y H then we show that T is self-adjoint.

If T* is adjoint of T then (Tx, y) = (x, T*y)

We have x,Ty x,T * y

x, T T * y 0 x, y H

But since x 0 T T * y 0 y H

T = T*

T is self adjoint.

(iii) For any T H ,T T * and T * T are self adjoint. By the property of self-adjoint operators, we have

TT**T*T**

T*T

TT*

T T * * T T*,

and T*T*T*T**T*T

T*T**T*T.

Hence T T * and T * T are self adjoint.

Theorem 1: If An is a sequence of self-adjoint operators on a Hilbert space H and if An converges to an operator A, then A is self adjoint.

Proof: Let An be a sequence of self adjoint operators and let An A.

An is self adjoint An * A n for n 1,2,...

We claim that AA*

Now AA*AAAA*A*A*n n n n

AA*AAAA*AA*n n n n

AAAAAAn n n n  A*An n

An A 0 A n A

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Notes = 2 An – A 0 as n

A A * 0 or A A* 0 A A *

A is self-adjoint operator.

This completes the proof of the theorem.

Theorem 2: Let S be the set of all self-adjoint operators in (H). Then S is a closed linear subspace of (H) and therefore S is a real Banach space containing the identity transformation.

Proof: Clearly S is a non-empty subset of (H), since O is self adjoint operator i.e. O S.

H, since O is self adjoint operator i.e. O S.

Let A,AS,1 2 We prove that A1 A 2 S.

A1 ,A 2 S A *1 A 1 and A 2* A 2 ...(1)

For ,R, we have

AA*A*A*1 2 1 2

A*A*1 2

 AA1 2 , are real numbers, ,

AA1 2 is also a self adjoint operator on H.

A1 ,A 2 S A 1 A 2 S.

S is a real linear subspace of H.

Now to show that S is a closed subset of the Banach space H. Let A be any limit point of S.

Then a sequence of operator An is such that AA.n We shall show that A S i.e. A A * .

Let us consider

AA*AAAA*n n

AAAA*n n

AAAA*A*A*n n n n

AAAAAA*n n n n ASA*An n n

An A 0 A n A

296 LOVELY PROFESSIONAL UNIVERSITY Unit 28: Self Adjoint Operators

Notes  2 An A 0 0, T * T and T T

0 as An A

A A * 0 A A* 0

AA*A is self adjoint

S is closed.

Now since S is a closed linear subspace of the Banach space H, therefore S is a real Banach space. ( S is a complete linear space)

Also I* I the identity operator I S. This completes the proof of the theorem.

Theorem 3: If A,A1 2 are self-adjoint operators, then their product A,A1 2 is self adjoint

A,AA,A1 2 2 1(i.e. they commute)

Proof: Let A,A1 2 be two self adjoint operators in H.

Then A*A,A*A.1 1 2 2

Let A,A1 2 commute, we claim that A,A1 2 is self-adjoint.

A,A*A*A*AAAA1 2 2 1 2 1 1 2

A,A*AA1 2 1 2

AA1 2 is self adjoint.

Conversely, let AA1 2 is self adjoint, then

AA*AA1 2 1 2

A*A*AA2 1 1 2

A1 ,A 2 commute

This completes the proof of the theorem.

Theorem 4: If T is an operator on a Hilbert space H, then T = T 0 Tx,y 0 x,y H.

Proof: Let T = 0 (i.e. zero operator). Then for all x and y we have

Tx,y Ox,y O,y O.

Conversely, Tx,y O x,y H

Tx,Tx O x,y H (taking y = Tx)

Tx O x,y H

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Notes TO i.e. zero operator. This completes the proof of the theorem.

Theorem 5: If T is an operator on a Hilbert space H, then

Tx,x 0 x in H T O.

Proof: Let T = O. Then for all x in H, we have

Tx,x Ox,x 0,x 0.

Conversely, let Tx,x 0 x,y H. Then we show that T is the zero operator on H.

If , any two scalars and x,y are any vectors in H, then

Tx y,x y Tx Ty,x y

T x , x y T y , x y

Tx,x Tx,y Ty,x Ty,x

2 2 Tx,x Tx,y Ty,x Ty,x

2 2 Tx y,x y Tx,x Ty,y Tx,y Ty,x ...(1)

But by hypothesis Tx,x 0 x H.

L.H.S. of (1) is zero, consequently the R.H.S. of (1) is also zero. Thus we have

Tx,y Ty,x 0 ...(2)

for all scalars , and x,y H.

Putting 1, 1 in (2) we get

Tx,y Ty,x 0 ...(3)

Again putting i, 1 in (2) we obtain

iTx,y iTy,x 0 ...(4)

Multiply (3) by (i) and adding to (4) we get

2iTx,y 0 x,y H

Tx,y 0 x,y H

Tx,Tx 0 x,y H (Taking y = Tx)

Tx 0 x,y H

298 LOVELY PROFESSIONAL UNIVERSITY Unit 28: Self Adjoint Operators

T = 0 (zero operator) Notes

This completes the proof of the theorem.

Theorem 6: An operator T on a Hilbert space H is self-adjoint.

Tx,x is real for all x. Proof: Let T* =T (i.e. T is self adjoint operator)

Then for every x H,we have

Tx,Tx x,T * x x,Tx Tx,x

Tx,x equals its own conjugate and is therefore real.

Conversely, let Tx,x is real x H. We claim that T is self adjoint i.e. T*=T. since Tx,x is real x H,

Tx,x Tx,x x,T * x T * x,x

Tx,x T*x,x 0 x H

Tx T * x,x 0 x H

T T * x,x 0 x H

T – T* = 0 [ if (Tx, x) = 0 T = 0]

T = T*

T is self adjoint.

This completes the proof of the theorem.

Cor. If H is real Hilbert space, then A is self adjoint

Ax,y Ay,x x,y H.

A is self adjoint for any x,y H.

Ax,y x,A * y A * y,x . since H is real Hilbert space A * y,x A * y,x so that Ax,y Ay,x A*A

Theorem 7: The real Banach space of all self-adjoint operators on a Hilbert space H is a partially ordered set whose linear and order structures are related by the following properties:

(a) If A1 A 2 then A 1 +A A 2 +A for every A S;

(b) If A1 A 2 and 0, then A 1 A 2 .

Proof: Let S represent the set of all self-adjoint operators on H. We define a relation on S as follows:

If A1 A 2 S, we write A 1 A 2 if A 1 x,x A 2 ,x x in H.

We shall show that '' is a partial order relation on S. '' is reflexive.

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Notes Let A S. Then

Ax,x Ax,x x H

By definition A A.

'' on S is reflexive.

'' is transitive.

Let A1 A 2 and A 2 A 3 then

Ax,x1 Ax,x 2 x H.

and Ax,x2 Ax,x 3 x H.

From these we get

Ax,x1 Ax,x 3 x H.

and Ax,x2 Ax,x 3 x H.

From these we get

Ax,x1 Ax,x 3 x H.

Therefore by definition AA1 3 and so the relation is transitive.

'' is anti-symmetric.

Let A1 A 2 and A 2 A 1 then to show that A 1 A 2 .

We have A1 A 2 Ax,x 1 Ax,x 2 x H.

Also A2 A 1 A 2 x,x A 1 x,x x H.

From these we get

Ax,x1 Ax,x 2 x H.

A1 x A 2 x,x 0 x H.

A1 A 2 x,x 0 x H.

A1 A 2 0

AA1 2

'' on anti-symmetric.

Hence '' is a partial order relation on S.

Now we shall prove the next part of the theorem.

(a) We have A1 A 2 Ax,x 1 Ax,x 2 x H.

300 LOVELY PROFESSIONAL UNIVERSITY Unit 28: Self Adjoint Operators

Notes A1 x,x Ax,x A 2 x,x Ax,x x H.

A1 Ax,x 2 A 1 Ax,x x H.

A1 A 2 A 2 A, by def. of .

(b) We have A1 A 2 Ax,x 1 Ax,x 2 xH

A1 x,x A 2 x,x x H  0

A1 x,x A 2 x,x x H

A1 x,x A 2 x,x x in H

A1 A 2 ,by def. of ' '. This completes the proof of the theorem.

28.1.2 Definition – Positive Operator

A self adjoint operator on H is said to be positive if A 0 in the order relation. That is if Ax,x 0 x H.

We note the following properties from the above definition.

(i) Identity operator I and the zero operator O are positive operators.

Since I and O are self adjoint and

Ix, x x, x x2 0

also (Ox, x) = (0, x) = 0

I,O are positive operators. (ii) For any arbitrary T on H, both TT* and T*T are positive operators. For, we have

TT * * T * * T* TT *

TT * is self adjoint

Also T*T*T*T**T*T

T*T is self adjoint Further we see that

TT * x,x T * x,T * x T * x2 0

and T * Tx,x Tx,T * *x (Tx,Tx) Tx2 0

Therefore by definition both TT* and T*T are positive operators.

Theorem 8: If T is a positive operator on a Hilbert space H, then I+T is non-singular.

Proof: To show I+T is non-singular, we are to show that I+T is one-one and onto as a mapping of H onto itself.

I+T is one-one.

First we show I T x 0 x 0

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Notes We have I T x 0 Ix Tx 0 x Tx 0

Tx x

Tx,x x,x x 2

x2 0  Tx,x 0

x2 0

x2 0  x2 is always 0

x 0 I T x 0 x 0.

Now ITx ITy ITxy 0

x y 0 x y

Hence I+T is one-one.

I+T is onto.

Let M = range of I+T. Then I+T will be onto if we prove that M=H.

We first show that M is closed.

For any x H, we have

2 I T x x Tx 2

x Tx,x Tx

x,x x,Tx Tx,x Tx,Tx

x2 Tx 2 Tx,x Tx,x

x2 Tx 2 2 Tx,x T is positive T is self-adjoint Tx,x real

x 2 T is positive Tx,x 0

Thus x I T x x H

Now let I T xn be a CAUCHY sequence in M. For any two positive integers m,n we have

xm x n I T x m x n

I T xm I T x n 0,

since I T x is a CAUCHY sequence.

xm x n 0

xn is a CAUCHY sequence in H. But H is complete. Therefore by CAUCHY sequence xn in H converges to a vector, say x in H.

302 LOVELY PROFESSIONAL UNIVERSITY Unit 28: Self Adjoint Operators

 Notes Now Lim I T xn I T lim x n I T is a continuous mapping

I T x M range of I T

Thus the CAUCHY sequence I T xn in M converges to a vector I T x in M.

every CAUCHY sequence in M is a convergent sequence in M.

M is complete subspace of a complete space is closed.

M is closed.

Now we show that M = H. Let if possible M H.

Then M is a proper closed subspace of H.

Therefore, a non-zero vector x0 in H s.t. x0 is orthogonal in M.

Since I T x0 M, therefore

x0 M I T x 0 ,x 0

x0 Tx 0 ,x 0 0

x0 ,x 0 Tx 0 ,x 0 0

2 x Tx0 ,x 0 0

2 x Tx0 ,x 0

2 x 0 T positive Tx0 ,x 0 0

x2 0

2 x 0  x 0

x 0

a contradiction to the fact that x0 0. Hence we must have M = H and consequently I+T is onto. Thus I+T is non-singular.

This completes the proof of the theorem.

Cor. If T is an arbitrary operator on H, then the operator I+TT* and I+T*T are non-singular.

Proof: We know that for an arbitrary T on H, T*T and TT* are both positive operators.

Hence by Theorem (8) both the operators I+TT* and I+T*T are non-singular.

28.2 Summary

 An operator T on a Hilbert space H is said to be self adjoint if T*=T.

 A self adjoint operator on H is said to be positive if A 0 in the order relation. That is if Ax,x 0 x H.

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Notes 28.3 Keywords

Positive Operator: A self adjoint operator on H is said to be positive if A 0 in the order relation. That is

if Ax,x 0 x H.

Self Adjoint: An operator T on a Hilbert space H is said to be self adjoint if T*=T.

28.4 Review Questions

1. Define a new operation of “Multiplication” for self-adjoint operators by

AAAA1 2 2 1 AA, and note that AA is always self-adjoint and that it equals AA1 2 1 2 2 1 2

whenever A1 and A2 commute. Show that this operation has the following properties:

AAAA,1 2 2  1

AAAAAAA,1 2 3 1  2 1  3

AAAAAA,1 2 1  2 1  2

and A I I  A A. Show that

    AAAAAA1 2 3 1 2 3 whenever A1 and A3 commute.

2 2. If T is any operator on H, it is clear that Tx,x Tx x T x ; so if H 0 ,we have

2 sup Tx,x / x : x 0 T . Prove that if T is self-adjoint, then equality holds here.

28.5 Further Readings

Books Akhiezer, N.I.; Glazman, I.M. (1981), Theory of Linear Operators in Hilbert Space

Yosida, K., Functional Analysis, Academic Press

Online links www.ams.org/bookstore/pspdf/smfams-14-prev.pdf-UnitedStatesmath world.wolfram.com

304 LOVELY PROFESSIONAL UNIVERSITY Unit 29: Normal and Unitary Operators

Unit 29: Normal and Unitary Operators Notes

CONTENTS

Objectives

Introduction

29.1 Normal and Unitary Operators

29.1.1 Normal Operator

29.1.2 Unitary Operator

29.1.3 Isometric Operator

29.2 Summary

29.3 Keywords

29.4 Review Questions

29.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand the concept of Normal and Unitary operators.

 Define the terms Normal, Unitary and Isometric operator.

 Solve problems on normal and unitary operators.

Introduction

An operator T on H is said to be normal if it commutes with its adjoint, that is, if TT*=T*T. We shall see that they are the most general operators on H for which a simple and revealing structure theory is possible. Our purpose in this unit is to present a few of their more elementary properties which are necessary for our later work. In this unit, we shall also study about Unitary operator and Isometric operator.

29.1 Normal and Unitary Operators

29.1.1 Normal Operator

Definition: An operator T on a Hilbert space H is said to be normal if it commutes with its adjoint i.e. if TT* = T*T

Conclusively every self-adjoint operator is normal. For if T is a self adjoint operator i.e. T*=T then TT* =T*T and so T is normal.

Note A normal operator need not be self adjoint as explained below by an example.

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Notes Example: Let H be any Hilbert space and I:HH be the identity operator.

Define T = 2iI. Then T is normal operator, but not self-adjoint. Solution: Since I is an adjoint operator and the adjoint operation is conjugate linear, T* = –2iI* = –2iI so that TT* =T*T =4I. T is a normal operator on H. But T = T* T is not self-adjoint.

Note If TH is normal, then T* is normal.

since if T* is the adjoint of T; then T**= T. T is normal TT*=T*T Hence T*T**=T*T=TT*=T**T* so that T*T**=T**T

T* is normal if TH.

Theorem 1: The limit T of any convergent sequence (Tk) of normal operators is normal.

Proof: Now T*T*TT*TTk k k

T *k T * as k since T k T as k .

Now we prove TT* T * T so that T is normal.

**** TT*T*T TT* TT*kkkkkkk TT* T*T T*T TT* TT TT kkkkkk TT TT

T *k T k TT *

* TT* T*T TT* TT*k k T*T k k TT ...(1)

 Tk is normal i.e. T k T* k T *k T k

since Tk T as T k* T*, R.H.S. of (1) 0

TT * T * T 0

TT* T * T T is normal. This completes the proof of the theorem.

Theorem 2: The set of all normal operators on a Hilbert space H is a closed subspace of H which contains the set of all set-adjoint operators and is closed under scalar multiplication. Proof: Let M be the set of all normal operators on a Hilbert space H. First we shall show that M is closed subset of H .

306 LOVELY PROFESSIONAL UNIVERSITY Unit 29: Normal and Unitary Operators

Let T be any limited point of M. Then to show that TM i.e. to show that T is a normal operator Notes on H.

Since T is a limited point of M, therefore a sequence Tn of distinct point of M such that

Tn T. We have

Tn * T * T n T * T n T 0.

Tn * T * 0 T n* T * .

Now, TT * T * T TT * Tn T * T n T *n T * T TT * T n T n* T n T *n T * T

TT * Tn T * T n T n* T n* T n T n* T n T * T

TT * Tn T *n T n T n* T *n T n T *n T n T * T

TT* TT*n n 0 T*T n n T*T

Tn M T n is a normal operator on H i.e. T n T* n T* nT n and 0 0

TT * Tn T *n T *n T n T * T 0 since T n T and T n* T *

Thus, TT * T * T 0 TT * T * T 0

TT* T * T T is normal operator on H.

TM and so M is closed. Now every self adjoint operator is normal. Therefore the set M contains the set of all self-adjoint operators on H.

Finally, we show that M is closed with respect to scalar multiplication i.e. TM

T M, is any scalar.

In other words, we are to show that if T is a normal operator on H and is any scalar, then T is normal operator on H. Since T is normal, therefore TT* =T*T.

We have T*T*.

Now T T * T T * TT * .

Also T * T T T T * T ( ) (TT*)

TT*T*T

T is normal. This completes the proof of the theorem.

Theorem 3: If N1, N2 are normal operators on a Hilbert space H with the property that either commutes with the adjoint of the other, then N1 + N2 and N1N2 are also normal operators.

Proof: Since N1, N2 are normal operators, therefore

N1 N* 1 = N* 1 N 1 and N 2 N* 2 N* 2 N 2 ...(1)

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Notes Also by hypothesis, we have N1 N* 2 =N* 2 N 1 and N 2 N* 1 N* 1 N 2 … (2)

we claim that N1 + N2 is normal.

i.e. (N1 + N2)(N1 + N2)* = (N1 + N2)*(N1 + N2) ...(3) since adjoint operation preserves addition, we have

(N1 + N2)(N1 + N2)* = (N1 + N2) N*1 +N* 2

...(4) NN*1 1 NN 1 2 NN* 2 1 NN* 2 2

N1 N* 1 N* 2 N 1 N* 1 N 2 N* 2 N 2

= N*1 N* 2 N 1 N 2

= NN*NN1 2 1 2 (using (1) and (2))

NNNN*NN*NN1 2 1 2 1 2 1 2

NN1 2 is normal.

Now we show that N1N2 is normal i.e.

NNNN*NN*NN.1 2 1 2 1 2 1 2

L.H.S.= NNNN*NNN*N*1 2 1 2 1 2 2 1

NNN*N*1 2 2 1

NN*NN*1 2 2 1

N*NN*N2 1 1 2

N*N*NN2 1 1 2

NN*NN1 2 1 2

NNNN*NN*NN1 2 1 2 1 2 1 2

N1N2 is normal. This completes the proof of the theorem.

Theorem 4: An operator T on a Hilbert space H is normal T * x Tx for every x H.

Proof: We have T is normal TT* T * T

TT * T * T 0

308 LOVELY PROFESSIONAL UNIVERSITY Unit 29: Normal and Unitary Operators

Notes TT * T * T x,x 0 x

TT * x;x T * Tx,x x

T * x,T * x Tx,T * *x x

T * x2 Tx 2 x T**T

T * x Tx x.

This completes the proof of the theorem.

2 Theorem 5: If N is normal operator on a Hilbert space H, then NN.2

Proof: We know that if T is a normal operator on H then

Tx T * x x ...(1)

Replacing T by N, and x by Nx we get

NNx N * Nx x

N2 x N * Nx x ...(2)

Now N2 Sup N 2 x : x 1

Sup N * Nx : x 1 (by (2))

N*N

N 2

This completes the proof of the theorem.

Theorem 6: Any arbitrary operator T on a Hilbert space H can be uniquely expressed as

T T1 iT 2 where T 1 ,T 2 are self-adjoint operators on H.

T T * 1 Proof: Let T and T T T * 12 2 2i

Then T1 iT 2 T ...(1)

1 * Now T*TT* 1 2 1 TT* * 2 1 T*T** 2 1 1 T*TTT*T 2 2 1

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Notes T*T1 1

is self-adjoint. T1

1 * Also T*TT* 2 2i

1 TT* * 2i

1 T*T** 2i

1 1 T*TTT*T 2 2i 2

T*T2 2

is self-adjoint. T2

Thus T can be expressed in the form (1) where T1,T2 are self adjoint operators. To show that (1) is unique.

Let T = U1 + iU2, U1,U2 are both self-adjoint

We have T* U1 iU 2 *

U *1 iU 2 *

U *1 iU *2

U *1 iU *2 U 1 iU 2

T T* U1 iU 2 U 1 iU 2 2U,

1 UTTT* 12 1

and T T* U1 iU 2 U 1 iU 2 2iU 2

1 UTT*T 22i 2

expression (1) for T is unique.

This completes the proof of the theorem.

Note The above result is analogous to the result on complex numbers that every complex number z can be uniquely expressed in the form z = x + iy where x, y are real. In the above

theorem T =T1 + T2, T1 is called real part of T and T2 is called the imaginary part of T.

310 LOVELY PROFESSIONAL UNIVERSITY Unit 29: Normal and Unitary Operators

Theorem 7: If T is an operator on a Hilbert space H, then T is normal its real and imaginary Notes parts commute.

Proof: Let T1 and T2 be the real and imaginary parts of T. Then T 1, T2 are self-adjoint operators and T = T1 + i T2.

We have T* = (T1 + iT2)* = T1* + (iT2)* i = Ti* + T2*

= Ti* – iT2*

= T1 – iT2

Now TT* = (T1 + iT2) (Ti – iT2)

2 2 = T1 + T2 + i (T2T1 – T1T2) … (1) and T*T = (Ti – iT2) (T1 – iT2)

2 2 = T1 + T2 + i (T1T2 – T2T1) … (2) Since T is normal i.e. TT* = T*T.

Then from (1) and (2), we see that

2 2 2 2 T1 + T2 + i (T2T1 – T1T2) = T1 + T2 + i (T1T2 – T2T1)

T2T1 – T1T2 = T1T2 – T2T1

2T2T1 = 2T1T2

T2T1 = T1T2 T1, T2 commute.

Conversely, let T1, T2 commute i.e. T1T2 = T2T1, then from (1) and (2) We see that

TT* = T*T T is normal.

Example: If T is a normal operator on a Hilbert space H and is any scalar, then T – I is also normal.

Solution: T is normal TT* = T*T

Also (T – I)* = T* – ( I)*

= T* – I*

= T* – I.

Now (T – I) (T – I)* = (T – I) (T* – I)

= TT* – I – T* + | |2I … (1)

Also (T – I)* (T – I) = (T* – I) (T – I)

= T*T – I* – T + | |2I … (2)

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Notes Since TT* = T*T, therefore R.H.S. of (1) and (2) are equal. Hence their L.H.S. are also equal. (T – I) (T – I)* = (T – I)* (T – I)

T – I is normal.

29.1.2 Unitary Operator

An operator U on a Hilbert space H is said to be unitary if UU* =U*U =I.

Notes (i) Every unitary operator is normal.

(ii) U* = U-1 i.e. an operator is unitary iff it is invertible and its inverse is precisely equal to its adjoint.

Theorem 8: If T is an operator on a Hilbert space H, then the following conditions are all equivalent to one another.

(i) T*T = I.

(ii) (Tx,Ty) = (x,y) for all x,y H.

(iii) Tx x x H.

Proof: (i) (ii)

(Tx,Ty) = (x,T*Ty) = (x,Iy) = (x,y) x and y.

(ii) (iii)

We are given that

Tx,Ty x,y x,y H.

Taking y = x, we get

2 2 (Tx,Tx) = (x,x) Tx x

Tx x x H.

(iii) (i)

Given Tx x x

Tx2 x 2

Tx,Tx x,x

T * Tx,x x,x

312 LOVELY PROFESSIONAL UNIVERSITY Unit 29: Normal and Unitary Operators

Notes T * T I x,x O x H

T*TIO

T*TI

This completes the proof of the theorem.

29.1.3 Isometric Operator

Definition: An operator T on H is said to be isometric if Tx Ty x y x,y H.

Since T is linear, the condition is equivalent to Tx x for every x H.

For example: let e1 ,e 2 ,...,e n ,... be an orthonormal basis for a separable Hilbert space H and

TH be defined as T x1 e 1 x 2 e 2 ... x 1 e 2 x 2 e 3 ... where x x n .

22 2 Then Tx xn x n 1 T is an isometric operator.

The operator T defined is called the right shift operator given by Te n = en+1. Theorem 9: If T is any arbitrary operator on a Hilbert space H then H is unitary it is an isometric isomorphism of H onto itself.

Proof: Let T is a unitary operator on H. Then T is invertible and therefore T is onto.

Further TT* = I.

Hence Tx x for every x H. [By Theorem (7)]

T preserves norms and so T is an isometric isomorphism of H onto itself.

Conversely, let T is an isometric isomorphism of H onto itself. Then T is one-one and onto. Therefore T–1 exists. Also T is an isometric isomorphism.

Tx x x

T*T = I [By Theorem (7)]

T * T T1 IT 1

T * TT1 T 1

T*IT 1

TT* I T * T and so T is unitary.

This completes the proof of the theorem.

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Notes

Note If T is an operator on a Hilbert space H such that Tx x x H and T is definitely an isometric isomorphism of H onto itself. But T need not be onto and so T need not be unitary. The following example will make the point more clear.

Example: Let T be an operator on l2 defined by T x1 ,x 2 ,... 0,x 1 ,x 2 ,...

Tx x x I2 .

T is an isometric isomorphism of l2 into itself.

However T is not onto. If y1 ,y 2 ,... is a sequence in l2 such that y1 0, then no sequence in l2

whose T-image is y1 ,y 2 ,... . Therefore T is not onto and so T is not unitary.

29.2 Summary

 An operator T on a Hilbert space H is said to be normal if it commutes with its adjoint i.e. if TT* = T*T. Conclusively every self adjoint operator is normal.

 The set of all normal operators on a Hilbert space H is a closed subspace of H which contains the set of all set-adjoint operators and is closed under scalar multiplication.

 An operator U on a Hilbert space H is said to be unitary if UU* = U*U =I.

 An operator T on H is said to be isometric if Tx Ty x y x,y H , since T is linear,

the condition is equivalent to Tx x for every x H.

29.3 Keywords

Normal Operator: An operator T on a Hilbert space H is said to be normal if it commutes with its adjoint i.e. if TT* = T*T.

Unitary Operator: An operator U on a Hilbert space H is said to be unitary if UU* = U*U = I.

Isometric Operator: An operator T on H is said to be isometric if Tx Ty x y x,y H. Since T is linear, the condition is equivalent to Tx x for every x H.

29.4 Review Questions

1. If T is an operator on a Hilbert space H, then T is normal its real and imaginary part commute.

2. An operator T on H is normal T * x Tx for every x.

3. The set of all normal operators on H is a closed subset of H which contains the set of all self-adjoint operators and is closed under scalar multiplication.

4. If H is finite-dimensional, show that every isometric isomorphism of H into itself is unitary.

5. Show that the unitary operators on H form a group.

314 LOVELY PROFESSIONAL UNIVERSITY Unit 29: Normal and Unitary Operators

29.5 Further Readings Notes

Books Arch W. Naylor, R Sell George, Linear Operator Theory in Engineering and Sciences, New York Springer, (1982).

Paul Garret, Operators in Hilbert Spaces, 2005.

Online link www.maths.leeds.ac.uk/nkisilv/

LOVELY PROFESSIONAL UNIVERSITY 315 Measure Theory and Functional Analysis

Notes Unit 30: Projections

CONTENTS

Objectives

Introduction

30.1 Projections

30.1.1 Perpendicular Projections

30.1.2 Invariance

30.1.3 Orthogonal Projections

30.2 Summary

30.3 Keywords

30.4 Review Questions

30.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Define perpendicular projections.

 Define invariance and orthogonal projections.

 Solve problems on projections.

Introduction

We have already defined projections both in Banach spaces and Hilbert spaces and explained how Hilbert spaces have plenty of projection as a consequence of orthogonal decomposition theorem or projection theorem. Now, the context of our present work is the Hilbert space H, and not a general Banach space, and the structure which H enjoys in addition to being a Banach space enables us to single out for special attention those projections whose range and null space are orthogonal. Our first theorem gives a convenient characterisation of these projections.

30.1 Projections

30.1.1 Perpendicular Projections

A projection P on a Hilbert space H is said to be a perpendicular projection on H if the range M and null space N of P are orthogonal.

Theorem 1: If P is a projection on a Hilbert space H with range M and null space N then MNP is self-adjoint and in this case NM.

Proof: Let MN and z be any vector in H. Then since HMN, we can write z uniquely as

z x y,x M,y N.

316 LOVELY PROFESSIONAL UNIVERSITY Unit 30: Projections

Thus Pz P x y Notes

Px Py

Px x y N

Pz,z x,z  Pz = P x+y x,P being projection on H

x,x y

x,x x,y

x 2 and Pz*,z z,Pz

x y,x x,x x,y

x2 .

Hence Pz,z Pz*,z z H

P P * z,z 0 z H

P P* 0 i.e. P P *

P is self adjoint.

Further, MNNM

If NM, then N is a proper closed linear subspace of the Hilbert space M and therefore a vector z0 0 M s.t. z 0 N.

Now z0 M and z0 N and H = M N.

z0 H z 0 0, a contradiction.

Hence NM

Conversely, let P* P,x,y be any vectors in M and N respectively. Then

x,y Px,y

x,P * y x,Py

x,0 0

M N. This completes the proof of the theorem.

Theorem 2: If P is the projection on the closed linear subspace M of H, then

x M Px x Px x .

Proof: We have, P is a projection on H with range M then, to show x M Px x.

LOVELY PROFESSIONAL UNIVERSITY 317 Measure Theory and Functional Analysis

Notes Let Px = x. Then X is in the range of P because Px is in the range of P.

Px x x M. Conversely, let x M. Then to show Px = x. Let Px = y. Then we must show that y = x.

We have

Px y PPx Py Px2 Py

Px Py  P2 =P

P x y 0

x y is a in null space of P.

x y M .

x y z,z M .

x y z.

Now y Px y is in the range of P.

i.e. y is in M. Thus we have expressed

x y z,y M,z M .

But x is in M. So we can write x = x+0, x M,0 M

But HMM. Therefore we must have y = x, z = 0

Hence x M Px x.

Now we shall show that Px = x Px x .

If Px = x then obviously Px x .

Conversely, suppose that Px x .

We claim that Px = x. We have

2 x2 Px I P x ...(1)

Now Px is in M. Also P is the projection on M.

I P is the projection on M .

I P x in M .

Px and I P x are orthogonal vectors.

Then by Pythagorean theorem, we get

2 2 Px I P x Px2 I P x ...(2)

318 LOVELY PROFESSIONAL UNIVERSITY Unit 30: Projections

From (1) and (2), we have Notes

2 x2 Px 2 I P x

2 I P x 0  by hypothesis Px x

I P x 0

x Px 0

Px 0 This completes the proof of the theorem.

Theorem 3: If P is a projection on a Hilbert space H, then

(i) P is a positive operator i.e. P 0

(ii) 0 P 1

(iii) Px x x H.

(iv) P 1.

Proof: P, projection on H P2 P,P * P.

Let M = range of P.

(i) Let x H. Then

Px,x PPx,x

Px,Px* Px,Px Px2 0

Px,x 0 x H.

P is a positive operator i.e. P 0.

(ii) P is a projection on H I – P is also a projection on H.

I – P 0. (by (i))

But P 0, consequently 0 P 1.

(iii) Let x H. If M is the range of P, then M is the range of (I – P).

Now Px is in M and (I – P)x is in M .

Therefore Px and (I – P)x are orthogonal vector. So by Pythagorean theorem, we have

2 2 Px I P x Px2 I P x

2 x2 Px 2 I P x  Px+ I–P x x

LOVELY PROFESSIONAL UNIVERSITY 319 Measure Theory and Functional Analysis

Notes 2 2 x Px

Px x .

(iv) We have P sup Px : x 1

But Px x x H (by(iii))

sup Px : x 1 1

Hence P 1

This completes the proof of the theorem.

Example: If P and Q are the projections on closed linear subspaces M and N of H. Show that PQ is a projection PQ = QP. In this case, show that PQ is the projection on M N.

Solution: Since P and Q are projections on H, therefore P2 = P, P* = P, Q2 = Q, Q* = Q. Also it is given that M is range of P and N is the range of Q.

Now suppose PQ is projection on H. Then to prove that PQ = QP.

Since PQ is a projection on H.

(PQ)* = PQ

Q* P* = PQ

QP = PQ ( Q* = Q, P* = P)

Conversely, let PQ = QP. We shall show that PQ is a projection on H.

We have (PQ)* = Q*P* = QP = PQ.

Also (PQ)2 = (PQ) (PQ) = (PQ) (QP)

= PQ2P = PQP

= QPP = QP2

= QP = PQ

Thus (PQ)* = PQ and (PQ)2 = PQ.

PQ is a projection on H.

Finally we are to show that PQ is the projection on M N, i.e. we are to show that range of PQ is M N.

Let R (PQ) = range of PQ.

Let x M N x M, x N we have

(PQ) (x) = P (Qx) = Px [ N is range of Q and x N Qx = x]

= x [ M is range of P and x P]

(PQ)x = x

x R (PQ)

x M N x R (PQ)

320 LOVELY PROFESSIONAL UNIVERSITY Unit 30: Projections

M N R (PQ) Notes

Now let x R (PQ). Then (PQ)x = x

Now (PQ) x = x

P [(PQ) x] = Px

[P (PQ)] x = Px

(P2Q) x = Px

(PQ) x = Px

But (PQ) x = x.

We have Px = x x M i.e. the range of P.

Also PQ = QP

x R (PQ) (PQ) x = x

(QP)x = x Q [(QP)x] = Qx

(Q2P)x = Qx (QP)x = Qx

But (QP)x = x, Qx = x x N.

Thus x R (PQ) x M and x N

x M N

R(PQ) M N

Hence R(PQ) = M N.

Example: Show that an idempotent operator on a Hilbert space H is a projection on H it is normal.

Solution: P is an idempotent operator on H i.e. P2 = P.

Let P be a projection on H. Then P* = P. We have

PP* = P* P* [taking P* in place of P in L.H.S.]

= P* P [ P* = P]

P is normal.

Conversely, let PP* = P*P.

Then to prove that P* = P.

For every vector y H, we have

(Py, Py) = (y, P* Py) = (y, PP*y) [ P*P = PP*]

= (P*y, P*y) [ (P*)* = P]

From this we conclude that

Py = 0 P*y = 0.

Now let x be any vector in H.

Let y = x – Px. Then

LOVELY PROFESSIONAL UNIVERSITY 321 Measure Theory and Functional Analysis

Notes Py = P(x – Px) = Px – P2x = Px – Px = 0

0 = P*y = P*(x – Px) = P*x – P*Px

P*x = P*Px x H

P* = P*P

Now P = (P*)* = (P*P)* = P*P = P*

P is a self adjoint operator.

Also P2 = P.

Hence P is a projection on H.

30.1.2 Invariance

Definition: Let T be an operator on a Hilbert space H and M be a closed subspace of H. Then M is said to be invariant under T if T M M. If we do not take into account the action of T on vectors outside M, then T can be regarded as an operator on M itself. The operator T on H induces

on operator TM on M such that TM(x) = T(x) for every x M. This operator TM is called the restriction of T on M.

Further, let T be an operator on Hilbert space H. If M is a closed subspace of H and if M and M are both invariant under T, then T is said to be reduced by M. If T is reduced by M, we also say that M reduces T.

Theorem 4: A closed linear subspace M of a Hilbert space H is invariant under the operation TM is invariant under T*.

Proof: Let M is invariant under T, we show M is invariant under T*.

Let y be any arbitrary vector in M . Then to show that T*y is also in M i.e. T*y is orthogonal to every vector in M.

Let x be any vector in M. Then Tx M because M is invariant under T.

Also y M y is orthogonal to every vector in M.

Therefore y is orthogonal to Tx i.e.

(Tx,y) = 0

(x,Ty) = 0

T*y is orthogonal to every vector x in M.

T * y is in M and so M is invariant under T*.

Conversely, let M is invariant under T*. Thus to show that M is invariant under T. Since M is

a closed linear subspace of H invariant under T*, therefore by first case M is invariant under T.

But M M M and T * * T * * T.

322 LOVELY PROFESSIONAL UNIVERSITY Unit 30: Projections

Hence M is invariant under T. Notes

This completes the proof of the theorem.

Theorem 5: A closed linear subspace M of a Hilbert space H reduces on operator M is invariant under both T and T*.

Proof: Let M reduces T, then by definition both M and M are invariant under T*. But by theorem 4, if M is invariant under T then M i.e. M is invariant under T*. Thus M is invariant under T and T*.

Conversely, let M is invariant under both T and T*. Since M is invariant under T*, therefore M is invariant under T** = T (by theorem 4). Thus both M and M are invariant under T. Therefore M reduces T.

Theorem 6: If P is the projection on a closed linear subspace M of a Hilbert space H, then M is invariant under an operator T TP PTP.

Proof: Let M is invariant under T.

Let x H. Then Px is in the range of T, Px M TPx M.

Now P is projection and M is the range of P. Therefore TPx M TPx will remain unchanged under P. So, we have

PTPx = TPx

PTP = TP (By equality of mappings)

Conversely, let PTP = TP. Let x M. Since P is a projection with range M and x M , therefore

Px = x

TPx = Tx

PTPx =Tx PTPTP

PTPx = TPx TPx Tx But P is a projection with range M.

P TPx TPx TPx M Tx M Since TPx = Tx.

Thus x M Tx M M is invariant under T.

Theorem 7: If P is the projection on a closed linear subspace of M of a Hilbert space H, then M reduces an operator TP PT.

Proof: M reduces TM is invariant under T and T*.

TP PTP and T * P PT * P

TP PTP and T * P * PT * P *

TP PTP and P * T * * P * T * *P *

TP PTP and PT PTP P is projection P* P. AlsoTT* T

LOVELY PROFESSIONAL UNIVERSITY 323 Measure Theory and Functional Analysis

Notes Thus M reduces T.

TP PTP and PT PTP ...(1)

Now suppose M reduces T. Then from (1), TP = PTP and PT = PTP. This gives TP = PT.

Conversely, let TP = PT

PTP =P2T (Multiplying both sides on left by P.)

or PTP = PT PP2

similarly multiplying both sides of TP = PT on the right of P, we get

TP2 = PTP or TP = PTP. Thus

TP = PT TP = PTP and PT = PTP.

Therefore from (1), we conclude that M reduces T.

Theorem 8: If M and N are closed linear subspace of a Hilbert space H and P and Q are the projections on M and N respectively, then

(i) M N PQ O. and

(ii) PQ O QP O.

Proof: Since P and Q are projections on a Hilbert space H, therefore P* = P, Q* = Q.

We first observe that

PQ O PQ * O * Q * P* O *

QP O.

Therefore to prove the theorem it suffices to prove that

MNPQO.

First suppose M N. If y is any vector in N, then M N y is orthogonal to every vector in M.

so y M .Consequently N M .

Now, let z be any vector in H. Then Qz is the range of Q i.e. Qz is in N.

Consequently Qz is in M which is null space of P.

Therefore P(Qz) = O.

Thus PQz = O z H

PQ = O

Conversely, let PQ = O and x M and y N.

since M is the range of P, therefore Px = x. Also N is the range of Q. Therefore

Qy = y

We have (x,y) = (Px, Qy) = (x,P*Qy)

324 LOVELY PROFESSIONAL UNIVERSITY Unit 30: Projections

Notes = (x,PQy) P*P

= (x,Oy) PQ O

= (x,O) = O

M and N are orthogonal i.e. M N.

30.1.3 Orthogonal Projections

Definition: Two projections P and Q on a Hilbert space H are said to be orthogonal if PQ = O.

Note: By theorem 8, P and Q are orthogonal iff their ranges M and N are orthogonal.

Theorem 9: If P1, P2, ... Pn are projections on closed linear subspaces M1, M2, ... Mn of a Hilbert space H, then P = P1 + P2 + ... + Pn is a projection the P'si are pair-wise orthogonal (in the sense that Pi P j 0,i j).

Also then P is the projection on M = M1 + M2 + ... + Mn.

Proof: Given that P1, P2, ... Pn are projections on H.

2 * Therefore PPPi i i for each i = 1, 2, ...,n.

* * ** Let P = P1 + P2+ ... + Pn. Then P = (P1 + P2+ ... + Pn) = P1 ... P n

= P1 + P2 + ... + Pn = P. Sufficient Condition:

Let Pi P j O,i j. Then to prove that P is a projection on H. We have

2 P = PP = (P1 + P2+ ... + Pn) (P1 + P2+ ... + Pn)

2 2 2 = P1 P 2 ... P n Pi P j 0,i j

= P1 + P2+ ... + Pn = P

Thus, P* = P = P*.

Therefore P is a projection on H.

Necessary Condition:

Let P is a projection on H.

Then P2 = P = P*.

We are to prove that Pi P j 0 if i j.

We first observe that if T is any projection on H and z is any vector in H, then

(Tz, z) = (T Tz, z) = (Tz, T*z)

= (Tz, Tz)

= Tz 2 ...(1)

LOVELY PROFESSIONAL UNIVERSITY 325 Measure Theory and Functional Analysis

Notes Now let x belongs to the range of some Pi so that Pix = x. Then

2 2 x Pi x

n 2 2 2 Pj x P 1 x ... P n x j 1

Pj x,x [Using (i)] j 1

Px,x1 ... Px,x n

P1 P 2 ... P n x

Px,x

Px 2 [by (1)]

x 2 ...(2)

Thus we conclude that sign of equality must hold throughout the above computation. Therefore we have

n 2 2 Pi x P j x j 1

2 Pj x O if j i

Pj x O, j i

x is in the null space of Pj ,i j

x Mj ,if j i

x is orthogonal to the range Mj of every Pj with j i.

j i. Thus every vector x in the range Pi(i = 1,...,n) is orthogonal to the range of every Pj with

Therefore the range of Pi is orthogonal to the range of every Pj with j i. Hence

Pi P j O, i j [By theorem (8)]

Finally in order to show that P is the projection on M M1 M 2 ... M n

We are to show that R(P) = M where R(P) is the range of P.

Let x M. Then x x1 x 2 ... x n

x2 Px 2 where xi M i , 1 i n. Now from (2), we observe that if x is the range of some Pi.

xi M, i.e. the range of P i .

326 LOVELY PROFESSIONAL UNIVERSITY Unit 30: Projections

2 2 Notes Pi x x i P i x x i

Pxi x i

xi the range of P.

xi R P , for each i 1,2,...,n

x1 x 2 ... x n R P .

x R P .

Then x M x R P

MRP ...(3)

Now suppose that x R P . Then

Px = x

P1 P 2 ... P n x x

P1 x P 2 x ... P n x x

But P1 x M,P 1 2 x M,...,P 2 n x M. n

x M1 M 2 ... M n and so RP M ...(4)

Hence from (3) and (4), we get

M = R(P)

This completes the proof of the theorem.

30.2 Summary

 A projection P on a Hilbert space H is said to be a perpendicular projection on H if the range M and null space N of P are orthogonal.

 Let T be an operator on a Hilbert space H and M be a closed subspace of H. Then M is said to be invariant under T if TMM.

 Let T be an operator on Hilbert space H, if M is closed subspace of H and if M and M are both invariant under T, then T is said to be reduced by M.

 Two projections P and Q on a Hilbert space H are said to be orthogonal if PQ = O.

30.3 Keywords

Invariance: Let T be an operator on a Hilbert space H and M be a closed subspace of H. Then M is said to be invariant under T if T M M. Orthogonal Projections: Two projections P and Q on a Hilbert space H are said to be orthogonal if PQ = O.

Perpendicular Projections: A projection P on a Hilbert space H is said to be a perpendicular projection on H if the range M and null space N of P are orthogonal.

LOVELY PROFESSIONAL UNIVERSITY 327 Measure Theory and Functional Analysis

Notes 30.4 Review Questions

1. If P and Q are the projections on closed linear subspaces M and N of H, prove that PQ is a projection PQ QP. In this case, show that PQ is the projection on MN.

2. If P and Q are the projections on closed linear subspaces M and N of H, prove that the following statements are all equivalent to one another:

(a) P Q;

(b) Px Qx for every x;

(c) MN;

(d) PQ P; (e) QP = P.

3. If P and Q are the projections on closed linear subspaces M and N of H, prove that QP is a projection P Q. In this case, show that Q – P is the projection on NM.

30.5 Further Readings

Books Borbaki, Nicolas (1987), Topological Vector Spaces, Elements of mathematics, Berlin: Springer – Verlag

Rudin, Walter (1987), Real and Complex Analysis, McGraw-Hill.

Online links www.math.Isu.edu/~ sengupta/7330f02/7330f02proiops.pdf

Planetmath.org/....OrthogonalProjections OntoHilbertSubspaces.html.

328 LOVELY PROFESSIONAL UNIVERSITY Unit 31: Finite Dimensional Spectral Theory

Unit 31: Finite Dimensional Spectral Theory Notes

CONTENTS

Objectives Introduction 31.1 Finite Dimensional Spectral Theory 31.1.1 Linear Operators and Matrices on a Finite Dimensional Hilbert Space 31.1.2 Similar Matrices 31.1.3 Determinant of an Operator 31.1.4 Spectral Analysis 31.1.5 Spectrum of an Operator 31.1.6 Spectral Theorem 31.1.7 Spectral Resolution of an Operator 31.1.8 Compact Operators 31.1.9 Properties of Compact Operators 31.2 Summary 31.3 Keywords 31.4 Review Questions 31.5 Further Readings

Objectives

After studying this unit, you will be able to:

 Understand finite dimensional spectral theory.

 Describe spectral analysis and spectral resolution of an operator.

 Define compact operators and understand properties of compact operators.

 Solve problems on spectral theory.

Introduction

The generalisation of the matrix eigenvalue theory leads to the spectral theory of operators on a Hilbert space. Since the linear operators on finite dimensional spaces are determined uniquely by matrices, we shall study to some extent in detail the relationship between linear operators in a finite dimension Hilbert spaces and matrices as a preliminary step towards the study of spectral theory of operators on finite dimensional Hilbert spaces.

LOVELY PROFESSIONAL UNIVERSITY 329 Measure Theory and Functional Analysis

Notes 31.1 Finite Dimensional Spectral Theory

31.1.1 Linear Operators and Matrices on a Finite Dimensional Hilbert Space

Let H be the given Hilbert space of dimension n with ordered basis B = {e1, e2, …, en} where the ordered of the vector is taken into consideration. Let T (H) (the set of all bounded linear operators). Since each vector in H is uniquely expressed as linear combination of the basis, we

can express Tej as Tej = ije i , where the n-scalars 1j, 2j, … nj are uniquely determined by Tej. i 1

2 2 Then vectors Te1, Te2, …, Tej determine uniquely the n scalars ij, i, j = 1, 2, … n. These n scalars th th determine matrix with ( i1, i2, …, in) as the i row and ( 1j, 2j, … nj) as its j column. We denote this matrix by {T} and call this matrix as the matrix of the operator T with respect to the ordered basis B.

11 12 1n  Hence  = [ ] = 21 22 2n ij 

n1 n2 nn

We note that (i) [0] = 0, which is the zero matrix.

(ii) [I] = I = [ ij], which is a unit matrix of order n. Here ij is the Kronecker delta.

Definition: The set of all n × n matrices denoted by An is complex algebra with respect to addition, scalar multiplication and multiplication defined for matrices. This algebra is called the total matrices algebra of order n. Theorem 1: Let B be an ordered basis for a Hilbert space of dimension n. Let T (H) with (T) = –1 –1 [ ij], then T is singular [ ij] is non-singular and we have [ ij] = [T ]. Proof: T is non-singular iff there exists an operator T –1 on H such that T–1 T = T T–1 = I … (1) Since there is one-to-one correspondence between T and [T–1], (1) is true [T–1 T] = [TT–1] = [I]

–1 –1 from (2) [T ] [T] = [T] [T ] = [I] = [ ij]

–1 –1 so that [T ] [ ij] = [ ij] [T ] = [ ij], [T] = [ ij].

–1 –1 [ ij] is a non-singular and [ ij] = [T ]. This completes the proof of the theorem.

31.1.2 Similar Matrices

Let A, B are square matrix of order n over the field of complex number. Then B is said to be similar to A if there exists a n × n non-singular matrix C over the field of complex numbers such that B = C–1 AC.

330 LOVELY PROFESSIONAL UNIVERSITY Unit 31: Finite Dimensional Spectral Theory

This definition can be extended similarly for the case when A, B are operators on a Hilbert space. Notes

Notes

1. The matrices in An are similar iff they are the matrices of a single operator on H relative to two different basis H. 2. Similar matrices have the same determinant.

31.1.3 Determinant of an Operator

Let T be an operator on an n-dimensional Hilbert space H. Then the determinant of the operator T is the determinant of the matrix of T, namely [T] with respect to any ordered basis for H. Following we given properties of a the determinant of an operator on a finite dimensional Hilbert space H. (i) det (I) = 1, I being identity operator.

Since det (I) = det ([I]) = det ([ ij]) = 1.

(ii) det (T1 T2) = (det T1) (det T2) (iii) det (T) 0 [T] is non-singular det ([T]) 0. Hence det (T) 0 [T] is non-singular.

31.1.4 Spectral Analysis

Definition: Eigenvalues

Let T be bounded linear operator on a Hilbert space H. Then a scalar is called an eigenvalue of T if there exists a non-zero vector x in H such that Tx = x. Eigenvalues are sometimes referred as characteristic values or proper values or spectral values.

Definition: Eigenvectors

If is an eigenvalue of T, then any non-zero vector x H such Tx = x, is called a eigenvector (characteristic vector or proper vector or spectral vector) of T.

Properties of Eigenvalues and Eigenvectors

Notes If the Hilbert space has no non-zero vectors then T cannot have any eigenvectors and consequently the whole theory reduces to triviality. So we shall assume that H 0 throughout this unit. 1. If x is an eigenvector of T corresponding to the eigenvalue and is a non-zero scalar, then x is also an eigenvector of T corresponding to the same eigenvalue .

Contd...

LOVELY PROFESSIONAL UNIVERSITY 331 Measure Theory and Functional Analysis

Notes Since x is an eigenvector of T, corresponding to the eigenvalue 0 and Tx = x. 0 x 0 Hence T ( x) = T(x) = ( x) Therefore corresponding to an eigenvalue there are more than one eigenvectors. 2. If x is an eigenvector of T, then x cannot correspond to more than one eigenvalue of T.

If possible let 1, 2 be two eigenvalues of T, ( 1 2) for eigenvector x. Then

Tx = 1x and Tx = 2x

1x = 2x

( 1 – 2)x = 0  1 – 2 = 0 ( x 0)

l = 2 3. Let be an eigenvalue of an operator T on H. If M is the set consisting of all eigenvectors of T corresponding to together with the vector 0, then M is a non- zero closed linear subspace of H invariant under T. By definition x M Tx = x … (1) By hypothesis 0 M and 0 vector satisfies (1). M = {x H : Tx = x} = {x H : (T – I) x = 0} Since T and I are continuous, M is the null space of continuous transformation T – I. Hence M is closed. Next we show that if x M , then Tx M . If x M then Tx = x. Since M is a linear subspace of H, x M x = Tx M . M is invariant under T.

Definition: Eigenspace

The closed subspace M is called the eigenspace of T corresponding to the eigenvalue . From property (3), we have proved that each eigenspace of T is a non-zero linear subspace of H invariant under T.

Note It is not necessary for an operator T on a Hilbert space H to possess an eigenvalue.

Example: Consider the Hilbert space  2 and the operator T on  2 defined by

T (x1, x2, …) = (0, x1, x2, …) Let be a eigenvalue of T. Then a non zero vector

y = (y1, y2, …) in  2 such that Ty = y.

332 LOVELY PROFESSIONAL UNIVERSITY Unit 31: Finite Dimensional Spectral Theory

Notes Now Ty = y T (y1, y2, …) = (y1, y2, …)

(0, y1, y2, …) = ( y1, y2 …)

y1 = 0, y2 = y1, ……

Now y is a non-zero vector y1 0

y1 = 0 = 0.

Then y2 = y1 y1 = 0 and this contradicts the fact that y is a non-zero vector. Therefore T cannot have an eigenvalue.

31.1.5 Spectrum of an Operator

The set of all eigenvalues of an operator is called the spectrum of T and is denoted by (T). Theorem 1: If T is an arbitrary operator on a finite dimensional Hilbert space H, then the spectrum of T namely (T) is a finite subset of the complex plane and the number of points in (T) does not exceed the dimension n of H. Proof: First we shall show that an operator T on a finite dimensional Hilbert space h is singular if and only if there exists a non-zero vector x H such that Tx = 0. Let a non-zero vector x H s.t. Tx = 0. We can write Tx = 0 as Tx = T0. Since x 0, the two distinct elements x, 0 H have the same image under T. Therefore T is not one-to-one. Hence T–1 does not exist. Hence it is singular. Conversely, let T is singular. Let no non-zero vector such that Tx = 0. This means Tx = 0 x = 0. Then T must be one-to-one. Since H is finite dimensional and T is one-to-one, T is onto, so that T is a non-singular, contradicting the hypothesis that T is singular. Hence there must be non- zero vector x s.t. Tx = 0. Now if T is an operator on a finite dimensional Hilbert space H of dimension n. Then A scalar (T), if there exists a non-zero vector x such that (T – I)x = 0. Now (T – I)x = 0 (T – I) is a singular. (T – I) is singular det (T – I) = 0. Thus (T) satisfies the equation det (T – I) = 0. Let B be an ordered basis for H. Thus det (T – I)

= det ([T – I]B)

But det ([T – I]B) = det ([T]B – [I]B)

Thus det (T – I) = det ([T]B – [ ij]).

So det (T – I) = 0 det ([T]B – [ ij]) = 0 … (1)

If [T]B = [ ij] is a matrix of T then (1) gives

11 12 1n

22 22 2n     … (2)

n1  nn

The expression of determinant of (2) gives a polynomial equation of degree n in with complex coefficients in the variable . This equation must have at least one root in the field of complex number (by fundamental theorem of algebra). Hence every operator T on H has eigenvalue so

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Notes that (T) . Further, this equation in has exactly n roots in complex field. If the equation has repeated roots, then the number of distinct roots are less than n. So that T has an eigenvalue and the number of distinct eigenvalue of T is less than or equal to n. Hence the number of elements of (T) is less than or equal to n. This completes the proof of the theorem.

Example: For a two dimensional Hilbert space H, let B = {e1, e2} be a basis and T be an operator on H given by the matrix

A = 11 12 … (1) 21 22

(i) If T is given by Te1 = e2 and Te2 = – e2, find the spectrum T. (ii) If T is an arbitrary operator on H with the same matrix representation, then

2 T ( 11 + 22) T + ( 11 22 – 12 21) I = 0 Sol:

(i) Using the matrix A of the operator T, we have

Te1 = 11 e1 + 21 e2 = e2 so that 11 = 0 and 21 = 1

Te2 = 12 e1 + 22 e2 = –e1 so that 12 = –1 and 22 = 0

0 1 Hence [T] = 11 12 . 21 22 1 0

For this matrix, the eigenvalue are given by the characteristic equation

1 = 0 1

2 + 1 = 0 = i so (T) = { i}. (ii) Let us consider the eigenvalues of A, which are given by

11 12 = 0 21 22

2 – ( 11 + 22) + ( 11 22 – 12 21) = 0 … (2) Since (2) is true for , we can take T = I … (3) From (2) and (3) we get

2 T – ( 11 + 22) T + ( 11 22 – 12 21) I = 0 … (4) The operator T on H having as an eigenvalue satisfies equation (4). Theorem 2: If T is an operator on a finite dimension Hilbert space, then the following statements are true. (a) T is singular 0 (T) (b) If T is non-singular, then (T) –1 (T–1) (c) If A is non-singular, then (ATA–1) = (T) (d) If (T) and if P is polynomial then P ( ) (P (T)).

334 LOVELY PROFESSIONAL UNIVERSITY Unit 31: Finite Dimensional Spectral Theory

Proof: Notes

(a) T is singular a non-zero vector x H such that Tx = 0 or Tx = 0. Hence T is singular 0 is the eigenvalue of T i.e. 0 (T). (b) Let T be non-singular and (T). Hence 0 by (a) so that –1 exists. Since is an eigenvalue of T, a non-zero vector x H s.t. Tx = x. Premultiplying by T–1 we get T–1 Tx = T–1 ( x) 1 T–1 (x) = x for x 0

Hence –1 ( (T–1)) Conversely, if –1 is an eigenvalue of T–1 then ( –1)–1 = is an eigenvalue of (T–1)–1 = T. Hence (T). (c) Let S = ATA–1. Then we find S – I. Now S – I = ATA–1 – A ( I) A–1 = A (T – I) A–1 det (S – I) = det (A(T – I) A–1) = det (T – I) is an eigenvalue of T det (T – I) = 0. Hence det (T – I) = 0 det (S – I) = 0 S and T have the some eigenvalues so that (ATA–1) = (T). (d) If = (T), is an eigenvalue of T. Then a non-zero vector x such that Tx = x. Hence T (Tx) = T ( x) = Tx = 2x. Hence if is an eigenvalue of T, then 2 is an eigenvalue of T2. Repeating this we get that if is an eigenvalue of T, then n is an eigenvalue of Tn for any positive integer n.

m Let P (t) = a0 + a1t + … + amt , a0, a1, ……, am are scalars.

m Then [P (T)]x = (a0I + a2T + …… + amT )x

m = a0x + a1 ( x) + …… + am ( x)

m = [a0 + a1 ( ) + …… + am ]x

m Hence P( ) = a0 + a1 + …… + am is an eigenvalue of P (T). This if (T), then P ( ) (P (T)). This completes the proof of the theorem.

31.1.6 Spectral Theorem

Statement: Let T be an operator on a finite dimensional Hilbert space H with 1, 2, …, m as the eigenvalues of T and with M1, M2, …, Mm be then corresponding eigenspaces. If P1, P2, …, Pm are the projections on the spaces, then the following statements are equivalent.

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Notes (a) The Mi’s are pairwise orthogonal and span H:

(b) The Pi’s are pairwise orthogonal and P 1 + P2 + … + Pm = I and T = 1P1 + 2P2 + … + mPm. (c) T is normal operator on H. Proof: We shall show that (i) (ii) (iii) (i) (i) (ii)

Assume that Mi’s are pairwise orthogonal and span H. Hence every x H can be represented uniquely as

x = x1 + x2 + … + xm … (1)

where xi Mi for i = 1, 2, …, m

by hypothesis Mi’s are pairwise orthogonal. Since P i’s are projections in Mi’s Pi’s are pairwise

orthogonal, i.e. i j PiPj = 0. If x is any vector in H, then from (1) for each i,

Pi(x) = Pi (x1 + x2 + … + xm)

= Pix1 + Pix2 + … + Pixm … (2)

Since Pi is the range of Mi, Pi xi = xi.

For i j Mj Mi since xj Mj for each j we have

xj Mi for j i.

Hence xj Mi (null space of Pi)

xj Mi Pixj = xi

from (2) we get

Pix = xi … (3) Since I is the identity mapping on H, we get

Ix = x1 + x2 + … + xm … (by (1))

= P1x + P2x + … + Pmx … (by (3))

= (P1 + P2 + … + Pm) x x H.

This show that I = P1 + P2 + …… + Pm. For every x H, we have from (1)

T (x) = T (x1 + x2 + … + xm)

= Tx1 + Tx2 + … + Txm

Since xi Mi Txi = xi

Tx = 1x1 + 2x2 + …… + mxm … (4)

= 1P1x + 2P2x + …… + mPmx … (5)

T = 1P1 + 2P2 + …… + mPm (ii) (iii)

336 LOVELY PROFESSIONAL UNIVERSITY Unit 31: Finite Dimensional Spectral Theory

Notes Let T = 1P1 + 2P2 + …… + mPm, where Pi’s are pairwise orthogonal projections and to show that T is normal.

2 Since Pi’s are projection and P *i P i and P i P i … (6)

Further we have PiPj = 0 for i j Since adjoint operation is conjugate linear, we get

T* = ( 1P1 + 2P2 + …… + mPm)*

= 1P*P*P*1 2 2 m m

= 1PPP1 2 2 m m .

Now TT* = (PPP)(PPP)1 1 2 2 m m1 1 2 2 m m 2 2 2 2 2 2  = |1 | P 1 | 2 | P 2 | 1 | P m ( PiPj = 0, i j)

2 2 2 = | 1| P1 + | 2| P2 + …… + | m| Pm … (by (6)) Similarly T*T can be found s.t.

2 2 2 T*T = | 1| P1 + | 2| P2 + …… + | m| Pm Hence T*T = TT* T is normal. (iii) (i)

Let T is normal operator on H and prove that Mi’s are pairwise orthogonal and M i’s span H. We know that if

T is normal on H its eigenspaces Mi’s are pairwise orthogonal.

So it suffices to show that Mi’s span H.

Let M = M1 + M2 + …… + Mm and P = P1 + P2 + …… + Pm

Since T is normal on H, each eigenspace Mi reduces T. Also Mi reduces T PiT = TPi for each Pi.

TP = T (P1 + P2 + …… + Pm)

= TP1 + TP2 + …… + TPm

= P1T + P2T + …… + PmT

= (P1 + P2 + …… + Pm) T = PT Since P is projection on M and TP = PT, M reduces T and so M is invariant under T. Let U be the restriction of T to M . Then U is an operator on a finite dimensional Hilbert space M and Ux = Tx x M . If x is an eigenvector for U corresponding to eigenvalue then x M and Ux = x. Tx = x and so x is also eigenvector for T. Hence each eigenvector of U is also an eigenvector for T. But T has no eigenvector in M . Hence M M = {0}. So U is an operator on a finite dimensional Hilbert space M and U has no eigenvector and so it has no eigenvalue.

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Notes M = {0}. For if M {0}, then every operator on a non-zero finite dimensional Hilbert space must have an eigenvalue. Now M = {0} M = H.

Thus M = M1 + M2 + …… + Mm = H and so Mi’s span H. This complete the proof of the theorem.

31.1.7 Spectral Resolution of an Operator

Let T be an operator on a Hilbert space H. If there exist distinct complex numbers 1, 2, …, m

and non-zero pairwise orthogonal projections p 1, p2, …, pm such that

T = 1p1 + 2p2 + … + mpm

and p = p1 + p2 + … + pm, then the expression

T = 1p1 + 2p2 + … + mpm for T is called the spectral resolution for T.

Note We note that the spectral theorem coincides with the spectral resolution for a normal operator on a finite dimensional Hilbert space.

Theorem: The spectral resolution of the normal operator on a finite dimensional non-zero Hilbert space is unique.

Proof: Let T = 1p1 + 2p2 + … + mpm be a spectral resolution of a normal operator on a non-zero finite dimensional Hilbert space H.

Then 1, 2, …, m are distinct complex numbers and pi s are non-zero pairwise orthogonal

projections such that p1 + p2 + … + pm = 1. We establish that 1 + 2, …, m are precisely the distinct eigenvalues of T.

To this end we show first that for each i, i is an eigenvalue of T. Since pi 0 is a projection, a

non-zero x in the range of p1 such that pix = x Let us consider

Tx = ( 1p1 + 2p2 + … + mpm)x

2 = ( 1p1pi + 2p2pi + … + ipi + … + mpmpi)x

2 So pi’s are pairwise orthogonal pipj = 0 for i j and pi = pi, we have Tx = ipix = ix by pix = x.

i is an eigenvalue of T.

Next we show that each eigenvalue of T is an element of the set ( 1, 2, …, m). Since T is an operator on a finite dimensional Hilbert space, T must have an eigenvalue. If is an eigenvalue of T then Tx = x = Ix.

( 1 p1 + 2 p2 + … + mpm) x = (p1 + p2 + … + pm) x

{( 1 – ) p1 + ( 2 – ) p2 + … + ( m – ) pm} x = 0 … (2)

2 Since pi = pi and pi pj = 0 for i j operating with pi throughout (2), we get

( i – ) pix = 0 for i = 1, 2, …, m.

338 LOVELY PROFESSIONAL UNIVERSITY Unit 31: Finite Dimensional Spectral Theory

Notes If i for each i, pix = 0 for each i.

Hence pix + p2x + … + pmx = 0

(p1 + p2 + … + pm) x = 0 Ix = 0

x = 0, a contradiction to the fact that x 0. Hence must be equal to i for some i. This in the spectral resolution (1) of T, the scalar i are the precisely the eigenvalue of T. If the spectral resolution is not unique.

Let T = 1Q1 + 2Q2 + … + nQn … (3) be another revolution of T. Then 1, 2, … n is the same set of eigenvalues of T written in different order. Hence writing the eigenvalues in the same order as in (1) and renaming the projections, we can write (3) as

T = 1Q1 + 2Q2 + … + mQm … (4)

To prove uniqueness, we shall show that pi in (1) and Qi in (4) are some.

2 Using the fact pi = pi, pipj = 0 i j, we can have

0 T = I = p1 + p2 + … + pm

1 T = 1p1 + 2p2 + … + mpm

2 2 2 T = 1p1 + … + mpm

n n n and T = 1 p1 + … + m pm for any positive integer n. … (5) Now if g (t) is a polynomial with complex coefficient in the complex variable t, we can write g (T) as

g (T) = g ( 1) p1 + g ( 2) p2 + … + g ( m) pm

= g (j ) p j … (by 5) j 1

Let i be a polynomial such that i ( i) = 1 and i ( j) = 0 if i j

Taking i in place of g, we get

m m ( ) p p p i (T) = i j j ij j i j 1 j 1

Hence for each i, let pi = i (T) which is a polynomial in T. The proof is complete if we show the existence of i over the field of complex number.

(t1 ) (t i 1 ) (t i 1 )...(t m ) Now i (t) (i 1 )....( i i 1 ) ( i i 1 ) ( i m ) satisfies our requirements i.e. i ( i) = 1 and i ( j) = 0 if i j

Repeating the above discussion for Qi’s we get in a similar manner Qi = i (T) for each i.

pi = Qi for each i. This completes the proof of the theorem.

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Notes 31.1.8 Compact Operators

Definition: A subset A in a normed linear space N is said to be relatively compact if its closure A is compact. A linear transformation T of a normed linear space N into a normed linear space N is said to be a compact operator if it maps a bounded set of N into a relatively compact set in N , i.e.

T : N N is compact of every bounded set B N, T(B) is compact in N .

31.1.9 Properties of Compact Operators

1. Let T : N N be a compact operator. Then T is bounded (continuous) linear operator. For, let B be a bounded set in N. Since T is compact, T(B) is compact in N . So T(B) is complete

and totally bounded in N . Since a totally bounded set is always bounded, T(B) is bounded and consequently T(B) is bounded, since a subset of a bounded set is bounded. T is a bounded linear transformation and it is continuous. 2. Let T be a linear transformation on a finite dimensional space N. Then T is compact operator. For, N is finite dimensional and T is linear T(N) is finite dimensional. Since any linear transformation on a finite dimensional space is bounded. T(B) is bounded subset of T(N) for every bounded set B N. Now if T(B) is bounded so is T(B) and is closed. T(N) is

finite dimensional, any closed and bounded subset of T(N) is compact, so that T(B) is compact, being closed and bounded subset of T(N). 3. The operator O on any normed linear space N is compact. 4. If the dimension of N is infinite, then identity operator I : N N is not compact operator. For consider a closed unit sphere. S = {x N : x 1} then S is bounded. Since N is a infinite dimensional.

I (S) = S = S is not necessarily compact.

Hence I : N N is not compact operator. But I is a bounded (continuous) operator. Theorem: A set A in a normed linear space N is relatively compact every sequence of points in A contains a convergent sub sequence. Proof: Let A is relatively compact.

Since A A , every sequence in A is also sequence in A . Since A is compact, such a sequence in A contains a convergent subsequence. Hence every sequence in A has a convergent subsequence. Conversely, let every sequence in A has a convergent subsequence.

Let (yn) be a sequence of points in A . Since A is dense in A , a sequence (xn) of points of A s.t.

1 xn y n … (1) n

340 LOVELY PROFESSIONAL UNIVERSITY Unit 31: Finite Dimensional Spectral Theory

Notes and xnk x A … (2)

(y ) we can find a nk of (yn) s.t.

ynk x = ynk x n k x n k x

ynk x n k x n k x

0 as n .

A is compact. This completes the proof of the theorem.

31.2 Summary

 If T is an arbitrary operator on a finite dimensional Hilbert space H, then the spectrum of T namely (T) is a finite subset of the complex plane and the number of points in (T) does not exceed the dimension n of H.

 Let T be bounded linear operator on a Hilbert space H. Then a scalar is called an eigenvalue of T if there exists a non-zero vector x in H such that Tx = x.

 The closed subspace M is called the eigenspace of T corresponding to the eigenvalue .

 The set of all eigenvalues of an operator is called the spectrum of T. It is denoted by (T).

 The spectral resolution of the normal operator on a finite dimensional non-zero Hilbert space is unique.

 A subset A in a normed linear space N is said to be relatively compact if its closure A is compact.

31.3 Keywords

Eigenspace: The closed subspace M is called the eigenspace of T corresponding to the eigenvalue . Eigenvalues: Let T be bounded linear operator on a Hilbert space H. Then a scalar is called an eigenvalue of T if there exists a non-zero vector x in H such that Tx = x. Eigenvalues are sometimes referred as characteristic values or proper values or spectral values. Eigenvectors: If is an eigenvalue of T, then any non-zero vector x H such Tx = x, is called a eigenvector. Similar Matrices: Let A, B are square matrix of order n over the field of complex number. Then B is said to be similar to A if there exists a n × n non-singular matrix C over the field of complex numbers such that B = C–1 AC. Spectrum of an Operator: The set of all eigenvalues of an operator is called the spectrum of T and is denoted by (T).

Total Matrices Algebra: The set of all n × n matrices denoted by An is complex algebra with respect to addition, scalar multiplication and multiplication defined for matrices. This algebra is called the total matrices algebra of order n.

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Notes 31.4 Review Questions

1. If T (H) is a self-adjoint operator, then (T) = {m, M} where m, M are spectral values. 2. If T is self-adjoint operator then (T) is the subset of the real line [– T , T ]. 3. Let R (T) = (T – I)–1 for a T B (X, X). Prove that R T 0 as . 4. Prove that the projection of a Hilbert space H onto a finite dimensional subspace of H is compact.

31.5 Further Readings

Books Walter Rudin, Real and complex analysis, Third, McGraw-Hill Book Co., New York, 1987. Erwin Kreyszig, Introductory functional analysis with applications, John Wiley & Sons Inc., New York, 1989.

Online links www.math.washington.edu chicago.academia.edu www.math.ethz.ch/~ Kowalski/spectral-theory.pdf

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