Functional Analysis

Marco Di Francesco, Stefano Spirito 2 Contents

I Introduction 5

1 Motivation 7 1.1 Someprerequisitesofsettheory...... 10

II Lebesgue integration 13

2 Introduction 15 2.1 Recalling Riemann integration ...... 15 2.2 A new way to count rectangles: Lebesgue integration ...... 17

3 An overview of Lebesgue measure theory 19

4 Lebesgue integration theory 23 4.1 Measurablefunctions...... 23 4.2 Simplefunctions ...... 24 4.3 Deﬁnition of the Lebesgue integral ...... 25

5 Convergence properties of Lebesgue integral 31 5.1 Interchanging limits and integrals ...... 31 5.2 Fubini’sTheorem...... 34 5.3 Egorov’s and Lusin’s Theorem. Convergence in measure ...... 36

6 Exercises 39

III The basic principles of functional analysis 43

7 Metric spaces and normed spaces 45 7.1 Compactnessinmetricspaces...... 45 7.2 Introduction to Lp spaces ...... 48 7.3 Examples: some classical spaces of sequences ...... 52

8 Linear operators 57

9 Hahn-Banach Theorems 59 9.1 The analytic form of the Hahn-Banach theorem ...... 59 9.2 The geometric forms of the Hahn-Banach Theorem ...... 61

10 The uniform boundedness principle and the closed graph theorem 65 10.1 The Baire category theorem ...... 65 10.2 Theuniformboundednessprinciple ...... 66 10.3 The open mapping theorem and the closed graph theorem ...... 67

3 4 CONTENTS

11 Weak topologies 69 11.1 The inverse limit topology of a family of maps ...... 69 11.2 The weak topology σ(E, E ∗) ...... 69 11.3 The weak ∗ topology σ(E∗, E )...... 73 11.4 Reﬂexive spaces and separable spaces ...... 75 11.5 Uniformly Convex Space ...... 78

12 Exercises 79

IV Lp spaces and Hilbert spaces 87

13 Lp spaces 89 p p 13.1 Density properties and separability of L spaces. Lloc spaces ...... 89 13.2 Dual space of Lp and reﬂexivity of Lp spaces ...... 92

14 Hilbert spaces 97 14.1 Deﬁnition of Hilbert spaces and geometrical properties ...... 97 14.2 Dual of a Hilbert space and reﬂexivity of Hilbert spaces ...... 100 14.3 Hilbert sums. Orthonormal bases ...... 102

15 Exercises 109

V Operator theory 115

16 Introductory concepts 117 16.1 Basic concepts and examples ...... 117 16.2 Adjoint operators ...... 120

17 Compact operators 123 17.1 Deﬁnitions and basic properties ...... 123 17.2 TheRiesz-Fredholmtheory ...... 124 17.3 The spectrum of a compact operator ...... 128

18 Self-adjoint operators on Hilbert spaces 131

19 Exercises 133 Part I

Introduction

Chapter 1

Motivation

Why functional analysis?

Recall a typical problem from linear algebra. Problem 1.0.1 . Let A be an n n matrix with real entries, and let y Rn be a given vector. Find all the vectors x Rn such that × ∈ ∈ Ax = y. (1.1) The vector x has to be determined. Borrowing geometrical terminology we call the vectors x, y points . The right to treat this problem is that of vector spaces (Rn) and linear mappings or operators (A). Now note that a point x = ( x ,...,x ) Rn can be viewed as a function x : 1,...,n R, i.e. 1 n ∈ { } → Rn = x : 1,...,n R . { { } → } So our points are actually functions on a ﬁnite set .

Functional analysis is the right setting to treat problems (equations) where the sought-after object (unknown) is a function on an inﬁnite set .

Let us look at a more speciﬁc example.

Problem 1.0.2 . Let α, y 0 R and assume we are given a continuous function g : [0 , + ) R. Find all the diﬀerentiable functions y :∈ [0 , + ) R such that ∞ → ∞ →

y′(t) = αy (t) + g(t) for all t 0 ≥ (y(0) = 0

This is a Cauchy problem for a linear differential equation. Integrating the differential equation on the interval [0 , t ] for an arbitrary t > 0 we get t t y(t) α y(s)ds = y + g(s)ds. (1.2) − 0 Z0 Z0 The unknown ‘point’ is now a differentiable function y on the half line [0 , + ). Can we see the set of differentiable functions on [0 , + ) as a linear space? Can we see the mapping ∞ ∞ t y = y(t) A[y] = A[y]( t) A[y]( t) = y(t) α y(s)ds 7→ − Z0 as a linear operator acting on the ‘point’ y? If so, then the expression (1.2) could be written as

. t A[y]( t) = h(t) , h (t) = y0 + g(s)ds, Z0 and the latter expression could be reminiscent of (1.1).

7 8 CHAPTER 1. MOTIVATION

Functional analysis provides the best setting to extend notions such as linear space and linear operator to functions on inﬁnite sets .

In the example of problem 1.0.2, the candidate linear space to work with is the space of diﬀerentiable functions on the half line [0 , + ). This is actually a subset of a wider set, the set of continuous functions on [0 , + ). Let us focus on this set for∞ a moment, and change the half line [0 , + ) to [0 , 1] for simplicity. ∞ ∞ Fact 1.0.1 . The set of continuous functions on [0 , 1] is a linear space . In order to see this we must equip said set (that we name C([0 , 1])) with two operations, namely sum between points and product of a point by a real number. Given f, g C([0 , 1]), we introduce f + g C([0 , 1]) deﬁned by ∈ ∈ (f + g)( x) = f(x) + g(x).

Given λ R and f C([0 , 1]), we introduce λf C([0 , 1]) deﬁned by ∈ ∈ ∈ (λf )( x) = λf (x).

Note that in the above two definitions we are using two well known operations (sum and product between real numbers) to define operations between functions. Every linear space needs its zero point. In our case, the zero point is obviously the function f(t) = 0 for all t [0 , 1]. ∈ Fact 1.0.2 . The space C([0 , 1]) has infinite dimension . To see this, recall from linear algebra that the dimension of a linear space is the largest possible integer number N of linearly independent vectors, i.e. the largest possible number of vectors v1,...,v N for which α1v1 + . . . + αN vN = 0 implies α1 = . . . = αN = 0, i.e. the largest possible number of vectors in which none of them can be written as a linear combination of the others. Now, assume that the dimension of our linear space C([0 , 1]) is given by some (possibly very large) integer N N. Consider the set of points ∈ . . . 2 . N v0(t) = 1 , v 1(t) = t, v 2(t) = t , ...,v N (t) = t . We claim that the above N + 1 points are linearly independent. Assume

N i αit = 0 (1.3) i=0 X for some α , α ,...,α R. Note that (1.3) must hold for all t [0 , 1], so in particular for t = 0. This implies 0 1 N ∈ ∈ α0 = 0. Now divide (1.3) by t > 0, which implies

N i 1 α1t − = 0 i=1 X for all t > 0. Now t can be clearly sent to zero in the above expression, and this easily implies α1 = 0. Iterating the procedure we prove that α2 = . . . = αN = 0. Hence, the N + 1 monomials are linearly independent. But that contradicts the fact that the dimension of the space was N, and N was an arbitrary integer. Hence, the dimension is infinite. So, here is the big issue with functional analysis. Linear spaces of functions such as C([0 , 1]) do not fit the classical background of linear algebra that we learned in our bachelor studies, because there linear spaces were always finite dimensional . This had a lot of implications, most importantly linear mappings could be expressed via matrices with a finite number of entries.

Functional analysis is the extension of linear algebra to linear spaces with inﬁnite dimension . Such spaces are typically spaces of functions on inﬁnite sets, often functions on subsets of R or subsets of the Euclidean space Rn.

It is known that one can measure distances in a ﬁnite dimensional linear space. In the most intuitive case, n namely the Euclidean space R , the distance between two points x = ( x1,...,x n) and y = ( y1,...,y n) is given by the canonical formula d(x, y ) = (x y )2 + . . . + ( x y )2. 1 − 1 n − n p 9

Other non-canonical ways could be d (x, y ) = x y + . . . x y , 1 | 1 − 1| | n − n| or d (x, y ) = max xi yi . ∞ i=1 ,...,n | − | We could measure distances between points using all of those distances, and of course they might (except in the case n = 1) give rise to different concepts of distance. However, here is an important fact. + n n Fact 1.0.3 . Given a sequence of points xk k=1∞ R and a point x R . Then d1(xk, x ) 0 as k + if and { } ⊂ ∈ → → ∞ n only if the same holds with d2 or d replacing d1. In other words, the notion of convergence of a sequence in R ∞ is not affected by the choice of the ‘distance’ ( d1, d, or d ). Another way to see it: ‘being close’ in the d1 sense is equivalent to being close in the d sense of in the d sense.∞ Proving this fact is not too difficult, but we shall omit ∞ the details at this stage. The main idea is the following: think of x = ( x1,...,x n) and xk = (( xk)i,..., (xk)n) as n vectors in R ; having xk and x very close in any of the three distances under consideration is equivalent to having the i-th components ( xk)i and xi very close (as real numbers!) for all i = 1 ,...,n .

In finite dimensional spaces, different ways of measuring distances generate the same notion of closeness . If two points are close to each other in a finite dimensional space, this does not depend on the type of distance we are using. This is, in general, not true in infinite dimensional spaces . Therefore, having to deal with infinite dimensional spaces, functional analysis has to explore many possible ways to measure distances between points (functions).

Before Fact 1.0.3, the only subject we invoked as a background was linear algebra. All of a sudden, in Fact 1.0.3 we started dealing with sequences, limits, etc, things we were used to in calculus and analysis . Functional analysis is indeed a lot about analysis, not just linear algebra and linear mappings. In fact,

functional analysis is a subject in which analysis and linear algebra merge together .

So far we provided a partial answer to the question ‘Why functional analysis?’. We tried somehow to justify functional analysis as a natural continuation of a pedagogical path based on linear algebra and analysis. But is all that of any use? Problem 1.0.2 is a good start, as it involves differential equations, an undoubtedly useful tool in science and engineering. Fact 1.0.4 . In finite dimensions, a continuous function f : K R defined on a bounded and closed set K Rn has a maximum and a minimum. This is a famous theorem in→ analysis due to Weierstrass. It is quite useful⊂ while looking for the solution to an optimisation problem depending on finitely many variable, since it guarantees under quite general assumptions that - no matter how good we are in finding a solution to the problem - there actually is a solution. The key issue behind Weierstrass theorem is the fact that a bounded and closed set in finite dimension is always compact , i.e. every sequence in K has a convergent subsequence. Example 1.0.1 . In classical mechanics, the trajectory of a material body subject to conservative forces can be found by minimising the quantity T L = m x˙(t) 2 + U(x(t)) dt, | | Z0 in which the first term is the kinetic energy and second one the potential energy. The unknown of the problem is a curve [0 , T ] t x(t) R3 standing for the optimal trajectory, i. e. the unknown is a vector in which each component is a∋ differentiable7→ ∈ function. Hence, the minimiser of the quantity L above lives in an infinite dimensional space. Weierstrass Theorem no longer holds in infinite dimensions, and ensuring that a quantity of the form of L attains the minimum on some curve t x(t) is far from being trivial. This is due to the fact that in infinite dimension is much more difficult to prove7→ that a set is compact .

One of the most important concepts developed in a functional analysis course is that of compactness . 10 CHAPTER 1. MOTIVATION

Mathematical modelling in applied sciences often relies on numerical calculus , in which the solution to a differential equation is approximated via some finite elements method. Example 1.0.2 . A differential equation

y˙(t) = f(y(t), t ), t [0 , T ], ∈ can be approximated via ﬁnite diﬀerence methods e.g. by the set of equations

yi+1 = yi + (∆ t)f(yi+1 , t ), (1.4) with i 0,...,n 1 and (∆ t)n = T . The unknown of the system of equations in (1.4) is the ( n + 1)-dimensional ∈ { − } vector ( y0, y 1,...,y n), whereas the unknown of the differential equation above is a function on [0 , T ], which lives in an infinite dimensional space. In order to make sure that the numerical scheme (1.4) works, one has to prove that a suitable interpolation (piecewise constant, or piecewise linear) of the solution ( y0, y 1,...,y n) to (1.4) gets closer and closer to the solution y(t) of the above differential equation as ∆ t goes to zero. Once again, it’s a matter of measuring a distance between functions. What is the correct distance to use? How can one prove that our numerical scheme converges ? Functional analysis provides tools to answer to these questions. Why has functional analysis become so fashionable? Despite the above examples, we admit that it is rarely the case that functional analysis leads to a result completely unobtainable with other traditional techniques. The strength and appeal of functional analysis is that it is a convenient way of examining the mathematical behavior of various mathematical models. More precisely, functional analysis clarifies , rigorises , and unifies the underlying concepts. It clarifies because - as already said - functional analysis is a generalisation and combination of linear algebra, analysis, and geometry (yes, there is a bit of geometry too when you measure distances), expressed in a simple mathematical notation which allows these three aspects of the problem to be easily seen. It rigorises, because it has the back up of a vast mathematical machinery which subsumes many of the classical results on differential equations, numerical methods, calculus of variations, and applied mathematical techniques. It unifies, because the simple notation does away with many of the complicating details leaving the essential standing out clearly, so that problems from many different fields have the same functional analytical symbolism.

1.1 Some prerequisites of set theory

We introduce here one of the fundamental concepts of modern mathematics, namely the Axiom of Choice 1 Axiom 1.1.1 (Axiom of choice) . Let E be an arbitrary family of sets indexed by α A. Then, there exists a α ∈ function mapping each set α A into one of the elements xα Eα. Therefore, it is always possible to choose an element x E for all α. ∈ ∈ ∈ α Remark 1.1.1 . In the language of infinite Cartesian products, the Axiom of choice basically states that every Cartesian product indexed by an arbitrary set of indexes is non empty (check!). A relation s A A is called an order relation if it satisfies the following properties: ∈ × s is reflexive, i. e. ( a, a ) s for all a A. • ∈ ∈ s is anti-symmetric, i. e. if ( a, b ) s and ( b, a ) s, then a = b. • ∈ ∈ s is transitive, i. e. if ( a, b ) s and ( b, c ) s, then ( a, c ) s. • ∈ ∈ ∈ A typical notation for an order relation is , i. e. ( a, b ) s if and only a b. We also use the expression ‘ a precedes b’ for a b. An order relation in A≺is called a total∈ order if for all a,≺ b A, either a b or b a. A set X equipped with≺ an order relation is often called a partially ordered set . ∈ ≺ ≺ Example 1.1.1 . Examples of order relations are: Let X be a set. Consider the power set (X) with the relation A B if and only if A B, for all • A, B (X). Prove that is an order relation.P Is a total order? Why?≺ ⊆ ∈ P ≺ ≺ Consider the real line R with the relation a b if and only if a b. Prove that is an order relation. Is a • total order? Why? ≺ ≤ ≺ ≺

1The axiom of choice was formulated for the ﬁrst time by Ernst Zermelo [5] in 1904. 1.1. SOME PREREQUISITES OF SET THEORY 11

Let A, B be two sets. Set • P = (f, C ) : C A, f : C B . { ⊂ → } Define the relation ( f, C ) (g, D ) if and only if C D and g C = f (g restricted on the set C equals f). Prove that is an order relation.≺ is called extension-restriction⊂ | order relation. ≺ ≺ Definition 1.1.1. Let ( P, ) be a partially ordered set (i. e. a set P and an order relation on P ). A chain in P is a subset C P such that,≺ for all a, b C, either a b or b a. Hence, a chain in P is a≺ subset of P which is a total order with⊆ the relation defined in P∈. Let A P .≺ An upper≺ bound for A is an element m P such that a m for all a A. A maximal element for A P is an⊂ element m A such that, for all a A, m ∈ a implies m =≺a. ∈ ⊆ ∈ ∈ ≺ Exercise 1.1.1. Let (R, ) be the real line equipped with the order relation. Prove that in this case, the definition of maximal element for a≤ subset A R coincides with the≤ definition of maximum for A. To prove that the two concepts of maximum and maximal element⊂ in general are not equivalent, find an example of a partially ordered set in which they are not.

Deﬁnition 1.1.2. A partially ordered set ( P, ) is called inductive if every chain in P has an upper bound. ≺ The next theorem is of paramount importance in functional analysis. It is a direct consequence of the Axiom of choice. Theorem 1.1.2 (Zorn’s lemma) . Every inductive partially ordered set P has a maximal element.

We shall not prove Zorn’s lemma here. We refer to [2, Volume 1, Theorem 1.2.7] for the detailed proof. Example 1.1.2 . We show some example of applications of Zorn’s lemma in the following examples.

Let X be a set. Consider the partially ordered set ( (X), ). Every chain in (X) has X as upper bound. • Hence ( (X), ) is inductive. Indeed, this is not surprising,P ⊆ as X is a maximalP element for ( (X), ). P ⊆ P ⊆ Let V a vector space, and let P be the set of families of pairwise linearly independent vectors in V . Let • vα α , vβ β P , and let us set vα α vβ β if and only if vα α vβ β. Clearly is an order {{relation,} } hence{{ } ( P,} ∈ ) is a partially ordered{{ } set.} ≺ Let {{ C} be} a chain in P .{ Then,} ⊆ the { union} of all elements≺ in C ≺ is clearly an upper bound for C. Therefore, Zorn’s lemma implies that P has a maximal element = vα α. Such statement means the following: if is contained in a family of pairwise linearly independentV vectors,{ } then such family must coincide with .V Hence, let v V be a vector in V . Consider the family v . Assume such family is made up by pairwiseV linearly independent∈ vectors. Since v , then{ Zorn’s} ∪ V lemma implies that the two families must coincide, which is a contradiction. The wrongV ⊆ {statement} ∪ V in the ﬁrst place was assuming that v is a family of pairwise linearly independent vectors, which is therefore not true. Note that v V is totally{ } ∪ V arbitrary, hence has the property that every arbitrary vector v V can be written as a linear∈ combination of elements in .V Therefore, is a base for the vector space V . ∈ V V (R, ) is not an inductive set, since R itself is a chain and has no upper bound. Hence, R cannot have a • maximal≤ element.

Consider the extension-restriction order relation on the maps A B, i. e. • → P = (f, C ) : C A, f : C B , { ⊂ → }

with ( f, C ) (g, D ) if and only if C D and g C = f. Let = (fα, C α) α be a chain in P . has an upper bounded constructed≺ very naturally.⊂ Consider C| = C ,C and {f : C B} defined as follows.C Let x C, let α α → ∈ α such that x Cα, set f(x) = fα(x). It is immediately seen that ( fα, C α) (f, C ) for all α, hence ( f, C ) is an upper bound∈ for the chain . Then, Zorn’s lemmaS implies that P has a maximal≺ element, i. e. there exists C a function f defined on a subset A′ A such that no functions defined on subsets of A can be extensions of f. ⊆ 12 CHAPTER 1. MOTIVATION Part II

Lebesgue integration

Chapter 2

Introduction

2.1 Recalling Riemann integration

The concept of integral in the majority of introductory analysis and calculus courses is centered on the Riemann integral 1. The intuition behind the Riemann integral is based on dividing the domain of integration D Rd of N ∈ a given function f into a ﬁnite number of ‘simple’ disjoint subsets Aj, j = 1 , . . . , N with D = j=1 Aj, and sampling the function f on arbitrary points cj Aj . The domains Aj are typically intervals, or rectangles, or multi-rectangles, depending on the dimension d ∈of the domain. Computing the Riemann integral thenS relies on d+1 computing the d + 1-dimensional ‘measure’ of multi-rectangles Aj f(cj) in R , taking the sum over j, and hoping that the output is a good approximation of the desired result× as the number of sets N gets larger and larger. Example 2.1.1 . As an example, consider the one-dimensional case. Let D = [ a, b ]. Take an N-points partition

= a = x0, x 1,...,x N 1, b = xN P { − } of D, with a < x 1 < ... < x N 1 < b . For a given function f : D R, deﬁne the Riemann sum on − → P N (f, ) = f(cj )( xj xj 1), (2.1) R P − − j=1 X where cj [xj 1, x j ). Roughly speaking, the Riemann integral of f on D exists if (f, ) has a limit as N + ∈ − R P → ∞ (and the size of all intervals [ xj 1, x j) tends to zero) which depends neither on the choice of the partitions nor − P on the sampling points cj. The Riemann integral is quite handy and easy to use in most situations. In particular, it is a successful method e. g. when f is continuous on D (or piecewise continuous, i. e. continuous on each set in a pair-wise disjoint partition of the domain) and bounded.

Example 2.1.2 . Let [ a, b ] R, and let a = x0 < x 1 < ... < x N 1 < x N = b be a partition of [ a, b ]. Let f : [ a, b ] R be continuous on each interval⊂ [ x , x ) for all j = 0 , . . . , N −1, f having jump discontinuities on all points x ,→ and j j+1 − j bounded on the whole interval [ a, b ], see Figure 2.1. Clearly, f is Riemann integrable on each interval [ xj, x j+1 ), and the integral on the whole interval [ a, b ] is easily computed by taking the sum over j of xj+1 f(x)dx , see the exercise xj 4.4 of the introductory material. An alternative way to see that f is Riemann integrable is the following. Assume R we are sampling f on a partition = a < y 1 < ...,y k 1 < y k = b , namely we are considering the Riemann sum P { − } k f(c )( y y ) (2.2) j j+1 − j j=1 X for cj Ij = [ yj, y j+1 ). Inevitably, some intervals Ij will contain the discontinuity points xj, and near a point ∈ − xj the value f(cj) may oscillate between the left and right limits lim x x f(x) and lim x x+ f(x). This may arise ր j ց j the doubt that the Riemann sum (2.2) does not converge in the limit (because it may have more than one limit point). However, each value f(cj) near the discontinuity is multiplied by the size of the interval [ yj, y j+1 ] in the Riemann sum (2.2), and the size of [ yj, y j+1 ] goes to zero in the limit! Provided f is bounded near those points xj ,

1We assume the reader is familiar with the Riemann integral. See Chapter the preliminary material

15 16 CHAPTER 2. INTRODUCTION

it won’t matter how much (and fast) f will oscillate: its value on the sampling points cj will be multiplied by the inﬁnitesimal (in the limit) quantity yj+1 yj. Now, since f only has a ﬁnite number of jump discontinuities, the contribution near those points in the Riemann− sum (2.2) will be negligible in the limit .

Figure 2.1: A discontinuous Riemann integrable function

On the other hand, there are examples of functions for which the Riemann integral procedure does not produce a uniquely deﬁned output in the limit. With the previous example 2.1.2 in mind, when the set of discontinuities of the integrand is inﬁnite and accumulates on an interval, the set in which the values f(cj ) in the Riemann sum are oscillating is no longer negligible, and the Riemann sums simply do not converge. A famous example in this sense is the following. Example 2.1.3 . Consider the Dirichlet function d : [0 , 1] R → 1 if x [0 , 1] Q d(x) = ∈ ∩ 0 if x [0 , 1] Q, ( ∈ \ 2.2. A NEW WAY TO COUNT RECTANGLES: LEBESGUE INTEGRATION 17

where Q is the set of rational numbers in R. For an arbitrary partition of [0 , 1], the values f(cj) in the Riemann sum (2.1) can be arbitrarily chosen to be either 0 or 1 because the set PQ is dense in R 2. Therefore, there is no hope that the Riemann sum converges in the limit. One may simply live with the idea that the above example is just a very unlikely case in our daily work. First of all, this is far from being true. The same phenomenon will occur anytime a function is discontinuous on a set which accumulates on a nontrivial interval. Moreover, the above example reveals one of the technical faults of the Riemann integral, namely that the existence of a Riemann integral is not (in general) preserved under pointwise convergence 3. This is explained in the next example.

Example 2.1.4 . Let Q = rn n N be the set of rational numbers in [0 , 1] expressed via a sequence. Consider the sequence of functions { } ∈

1 if x [0 , 1] r1,...,r n dn(x) = ∈ ∩ { } 0 if x [0 , 1] r ,...,r . ( ∈ \ { 1 n} 1 Since dn is continuous except on a ﬁnite number of points in [0 , 1], dn is Riemann integrable, and 0 dn(x)dx = 0 in the sense of Riemann. Now, the pointwise limit as n + of dn is the Dirichlet function in Example 2.1.3, and 1 → ∞ R we know from Example 2.1.3 that 0 d(x)dx is not well deﬁned in the sense of Riemann integral.

Exercise 2.1.1. Show that the sequenceR dn : [0 , 1] R in the Example 2.1.4 converges pointwise to the Dirichlet function d in the Example 2.1.3. →

2.2 A new way to count rectangles: Lebesgue integration

An equally intuitive, but (for some reason) long in coming method of integration, was introduced by Henri Lebesgue [3] in 1902. Rather than taking a partition of the domain of the integrand f, Lebesgue proposed to consider a partition of the range of f. Hence, instead of trying to determine the value of f in an interval of the partition of the domain, he asked how much of the domain is mapped by the function to an interval of the partition of the range.

Figure 2.2: Riemann integration: partition of the domain Figure 2.3: Lebesgue integration: partition of the range

2See Example 6.2 of the introductory material 3The notion of pointwise convergence is a prerequisite. It is recalled in the preliminary material 18 CHAPTER 2. INTRODUCTION

This method relies on being able to measure the inverse image through f of open intervals in R. In order to perform this task, we have to provide an eﬃcient method to measure the largest possible number of sets in a coherent and consistent form which somehow extends Riemann’s method. Going back to the example 2.1.3, Lebesgue’s measure theory will give as a result that the measure of the set Q equals zero, so that the integral of the Dirichlet function on [0 , 1] equals zero. Chapter 3

An overview of Lebesgue measure theory

In this chapter we recall the main properties of Lebesgue’s measure theory on Rd. Notation 3.0.1 (Extended nonnegative real axis) . The extended nonnegative real axis [0 , + ] is the nonnegative real axis [0 , + ) with the additional element labelled by + . We shall work with this system∞ because many sets in Rd have inﬁnite∞ measure. Of course + is not a real number,∞ but we think of it as an extended real number. We extend the basic operations on the extended∞ real axis as follows:

+ + x = x + (+ ) = + , for all x [0 , + ], • ∞ ∞ ∞ ∈ ∞ + x = x (+ ) = + for all x (0 , + ], • ∞ · · ∞ ∞ ∈ ∞ + 0 = 0 (+ ) = 0, • ∞ · · ∞ x < + for all x [0 , + ) • ∞ ∈ ∞ Most of the laws of algebra for addition, multiplication, and order continue to hold in this extended number system. However, we caution that the laws of cancellation do not apply once some of the variables are allowed to be infinite. We also mention that, with the convention 0 (+ ) = 0 multiplication on [0 , + ] is upward continuous, but · ∞ ∞ not downward continuous. More precisely, if xn and yn are sequences in [0 , + ], with xn increasing and converging to x [0 , + ] and y increasing and converging to y [0 , + ], then x y converges∞ to xy . On the other hand, if ∈ ∞ n ∈ ∞ n n xn = 1 /n 0 = x and yn = + = y, then xnyn = + = 0 = xy . We note→ that infinite sums∞ in [0 , + ] are always well∞ 6 defined in [0 , + ]. ∞ ∞ The goal of a suitable measure theory is that of defining a convenient setting for measuring sets in Rd in a way to fix the problems raised at the end of the introductory chapter (2), matching the intuition in measuring elementary sets (rectangles, cubes, etc.), and extending the set of measurable sets consistently with Riemann’s theory (and with its multidimensional measure counterpart, called Peano–Jordan theory ). A very interesting discussion, with historic introduction to the subject, is developed in [4, Section 1.1].

Definition 3.0.1. Let A Rd be an open set. We define the measure of A as the extended real number ⊂ m(A) = sup m(Y ) : Y elementary , Y A . { ⊆ } Let K Rd be a compact set. We define the measure of K as the real number ⊂ m(K) = inf m(Y ) : Y elementary , Y K . { ⊇ } d Definition 3.0.2 (Lebesgue measure) . Let E R be a bounded set. We define m∗(E) the outer measure of E and m (E) the inner measure of E as follows: ⊂ ∗ m (E) = sup m(K) : K E compact ∗ { ⊆ } m∗(E) = inf m(A) : A E open . { ⊇ }

The set E is said Lebesgue measurable if m∗(E) = m (E), and the number m(E) = m∗(E) = m (E) is called the (d-dimensional) Lebesgue measure of E. An unbounded∗ set E Rd is said Lebesgue measurable∗ if the sets ⊆ E BR(0) are measurable for all R > 0, and we set m(E) = lim R + m(E BR(0)) = sup R> 0 m(E BR(0)). ∩ → ∞ ∩ ∩ 19 20 CHAPTER 3. AN OVERVIEW OF LEBESGUE MEASURE THEORY

Proposition 3.0.1 (Characterisation of measurable sets in terms of open and compact sets) . Let E Rd a bounded set. Then, E is Lebesgue measurable if and only if for all ǫ > 0 there exist an open set A and a compact⊂ set K with K E A and m(A) m(K) < ǫ . ⊆ ⊆ − Let us list here some basic property of Lebesgue measure. Proposition 3.0.2. The measurable sets E Rd and the Lebesgue measure m satisfy the following properties: ⊂ (a) The empty set and Rd are measurable. m( ) = 0 , m(Rd) = + . ∅ ∅ ∞ (b) Let E F be measurable subsets of Rd. Then m(E) m(F ). ⊆ ≤ (c) Let E Rd be measurable and x Rd. Then m(E) = m(x + E). ⊆ ∈ (d) If E Rd is a measurable set with m(E) = 0 , then all the subsets of E are measurable and have zero measure. ⊆ (e) All bounded subsets of Rd have ﬁnite measure. In the next theorem, we collect the main properties of Lebesgue measure.

Theorem 3.0.3. Let Ek be a sequence of Lebesgue measurable sets. Then, k Ek and k Ek are also measurable. Moreover, S T m ( E ) m(E ) (countable subadditivity of Lebesgue measure). • k k ≤ k k If ES E = Pfor all i = j, then m ( E ) = m(E ) (countable additivity of Lebesgue measure). • i ∩ j ∅ 6 k k k k If E E , then m(E ) m(E ). S P • k ↑ 0 k ր 0 If E E and m(E ) < + , then m(E ) m(E ). • k ↓ 0 1 ∞ k ց 0 Remark 3.0.1 . Every Peano–Jordan measurable set is also Lebesgue measurable. Definition 3.0.3. We say that a given property holds almost everywhere (a. e.) on Rd if there exists a set E Rd with m(E) = 0 such that the given property holds on Rd E. ⊂ \ Exercise 3.0.1. Let E Rd be a measurable set with m(E) > 0. Then, the set ⊂ E E := x y : x, y E − { − ∈ } contains an open ball centered at zero Bǫ(0) . Solution. Let σ (0 , 1/3) , and let K compact and A open such that K E A, m(A) (1 + σ)m(E), and m(K) (1 σ)m∈(E). Let ǫ = min k a : k K, a Ac . Since K⊂is closed⊂ and A ≤open, we have ǫ > 0 (exercise!).≥ − By definition of ǫ we have{k v +−Kk= v +∈ k : k ∈ K } A for all v Rd with v ǫ. Moreover, with the same assumption on v we claim that K (v +{K) = . Assume∈ } ⊂ by contradiction∈ that suchk k intersection ≤ is empty. Then, ∩ 6 ∅ m(A) m(K) + m(v + K) = 2 m(K), ≥ but then (1 + σ)m(E) m(A) 2m(K) 2(1 σ)m(E), ≥ ≥ ≥ − which yields σ 1/3, a contradiction. The last claim implies that for all v Rd with v ǫ there exists z K (v + K)≥, which means v = z k for some z, k K, i. e. v K K. With∈ the assumptionsk k ≤ on v, we have proven∈ ∩B (0) K K, which proves− the assertion. ∈ ∈ − ǫ ⊂ − We are now almost ready to describe Vitali’s set. We observe that the construction of such set will require the Axiom of choice introduced in 1.1.1. Example 3.0.1 (Vitali’s set) . The relation x y x y Q defines an equivalence relation in R. Let us consider the quotient set [0 , 1] / . Using the axiom∼ of choice⇔ − we∈ can construct a set E R such that E contains one representant of each equivalence∼ class x + Q, x [0 , 1], and that such representant⊂ is unique in E. Clearly, the union of the family of classes q + E : q [0 , 1] ∈Q covers [0 , 1] (for every real number r [0 , 1] consider its class r + Q, and let e E be its representant{ ∈ in E. This∩ } means that r e = q Q, i. e. r q∈+ E). Now we want to prove that the classes∈ q + E and r + E are disjoint for two given rational− numbers∈ r = q∈. Assume by contradiction that x (q + E) (r + E). Then, x = q + e = r + e for some e , e E. But this6 implies q r = e e = 0, ∈ ∩ 1 2 1 2 ∈ − 2 − 1 6 21

and therefore e1 and e2 are distinct elements in E, which is a contradiction because they belong to the same class (e2 = ( q r)+ e1!). Now we observe that all the sets in q+E : q [0 , 1] Q have the same measure because of the translation− invariance of Lebesgue measure. Such family{ of sets is∈ countab∩le,} they are disjoint, they cover [0 , 1], and they all have the same measure. Such measure cannot be zero (because otherwise [0 , 1] would have measure zero, a contradiction!). On the other hand, if the measure would be positive, m(E) > 0, the exercise 3.0.1 implies that E E contains a non empty open ball. In particular, there exists a (small enough) rational number r [0 , 1] Q − ∈ ∩ such that r E E, which means r = e1 e2 for two distinct e1, e 2 E. This, once again, a contradiction, because e = r + e ∈with −r Q implies that e and− e are in the same class.∈ Hence, E is not Lebesgue measurable. 1 2 ∈ 1 2 22 CHAPTER 3. AN OVERVIEW OF LEBESGUE MEASURE THEORY Chapter 4

Lebesgue integration theory

Notation 4.0.1 (Extended real axis) . In this chapter we shall consider the extended real axis [ , + ], similarly to what we did in the notation 3.0.1. We add two points to the real line R, namely and−∞ + , ordered∞ in the most obvious way ( < r < + for all r R). All the operations are defined similarly−∞ to notation∞ 3.0.1: −∞ ∞ ∈ + + = r + = + for all r R, • ∞ ∞ ∞ ∞ ∈ = r = for all r R, • −∞ − ∞ − ∞ −∞ ∈ + a = + if a > 0, + a = if a < 0, • ∞ · ∞ ∞ · −∞ a = if a > 0, a = + if a < 0, • −∞· −∞ −∞ · ∞ + 0 = 0, 0 = 0. • ∞ · −∞ · The difference + is not defined . ∞ − ∞ 4.1 Measurable functions

Deﬁnition 4.1.1 (Measurable function) . A function f : Rd [ , + ] is called a measurable function if, for all 1 → −∞ ∞ open subsets U R, the set f − (U) is Lebesgue measurable. ⊂ We immediately remark what follows. 1 Remark 4.1.1 . In order to check whether or not f is measurable, one doesn’t need to check that f − (U) is measurable for all open sets U [ , + ]. It suﬃces to test that condition only on open sets of the form ( a, + ], i. e. we need to check that the⊂ −∞ sets f∞ > a are measurable for all a R. This is due to the following two facts:∞ { } ∈ Every open interval ( a, b ) [ , + ] we have • ⊂ −∞ ∞

1 1 f − (( a, b )) = f > a f > b , { } \ − k k 1 [≥ 1 and therefore f − (( a, b )) is measurable if all the sets of the form f > r are. { } Since every open set U R can be written as a countable union of open intervals ( a, b ), one can use that • ∈ 1 countable unions of measurable sets are measurable to obtain f − (U) measurable assuming that sets of the 1 form f − (( a, b )) are measurable. Exercise 4.1.1. Let f : Rd [ , + ]. Then, the following conditions are equivalent: → −∞ ∞ f is measurable, • 1 all the sets of the form f − ([ , a ]) with a R are measurable, • −∞ ∈ 1 all the sets of the form f − ([ , a )) with a R are measurable, • −∞ ∈ 1 all the sets of the form f − ([ a, + ]) with a R are measurable. • ∞ ∈ 23 24 CHAPTER 4. LEBESGUE INTEGRATION THEORY

Exercise 4.1.2. We can use the arguments in remark 4.1.1 to prove what follows. If f and g are measurable functions and λ R, then f +λg is a measurable function provided f +λg is always • well deﬁned on [ , + ]. ∈ −∞ ∞ + + If f and g are measurable functions, then the functions f = max f, 0 , f − = max f, 0 , f = f + f −, • f g = sup f, g , f g = inf f, g , f g are measurable. { } {− } | | ∨ { } ∧ { } ·

If fn n is a sequence of measurable functions, then sup n fn, inf n fn, lim sup n + fn, lim inf n + fn, • { } → ∞ → ∞ lim n + fn (if it exists), n 1 fn (if it exists) are measurable. → ∞ ≥ Moreover, one can easily proveP that if f : Rd [ , + ] is continuous, then f is measurable. Finally, one can easily prove that if f : Rd [ , + ] is→ measurable−∞ ∞ and φ : [ , + ] R is continuous, then φ f is measurable. → −∞ ∞ −∞ ∞ → ◦

4.2 Simple functions

Notation 4.2.1 . In what follows we use the expression f zero outside a compact set for a given function f : Rd [ , + ] if there exists a compact set K Rd such that f(x) = 0 for all x K. → −∞ ∞ ⊂ 6∈ Deﬁnition 4.2.1 (Simple functions) . Let A Rd. We denote by 1 the indicator function of A ⊂ A 1 if x A 1A(x) = ∈ 0 if x A. ( 6∈ A simple function on Rd is a function φ : Rd R measurable, zero outside a compact set, attaining only a ﬁnite number of values in R. We denote by (Rd) the→ set of all simple functions on Rd. S Exercise 4.2.1. Let A, B Rd. Prove that • ⊂ 1A B = 1A1B. ∩ d + Let φ, ψ (R ) and λ R. Prove that the functions λφ + ψ, φ ψ, ψ , ψ−, φ , φ ψ, ψ φ all belong to • (Rd). ∈ S ∈ · | | ∨ ∧ S d Remark 4.2.1 . We observe that if φ (R ), then let c1,...,c N be the real values attained by φ, and let d ∈ S { } Ej = x R : φ(x) = cj , we can write { ∈ } N

φ = cj1Ej . (4.1) j=1 X Note that all the sets Ej above are measurable, bounded, and pairwise disjoint (check!). We observe that one can represent φ as a linear combination of indicator functions of measurable sets in infinitely many way. What makes the representation (4.1) unique is the property of having the sets Ej ’s pair-wise disjoint and the constant cj without repetitions. The student is asked to verify that. This is indeed a tricky aspect. Some textbooks use arbitrary representations for simple functions, but such approach has the disadvantage of having to prove that all the quantities defined from simple functions (including integrals, see below) are independent of the representation. Lemma 4.2.1. Let f : Rd R be measurable. If f is bounded and zero outside a compact set. Then, there → exist two sequences φk k and ψk k of simple functions such that φk f and ψk f as k + uniformly. If d { } { } ↑ ↓ → ∞ f : R [0 , + ] is not necessarily bounded, then there exists a sequence of nonnegative simple functions φk k → ∞ d { } such that φk f pointwise. If f : R R is not necessarily bounded, then there exists a sequence of simple functions φ such that↑ φ f pointwise. → { k}k k → Proof. Consider the case of f bounded first. Let M > 0 be large enough such that f M and f(x) = 0 for k| | ≤ x BM (0). For a given positive integer k, let us partition the interval [ M, M ] into 2 parts with equal size, and let6∈ us set − j 1 j Ek := x Rd : M + 2 M − f(x) < M + 2 M B (0) , j = 1 ,..., 2k. j ∈ − 2k ≤ − 2k ∩ M 4.3. DEFINITION OF THE LEBESGUE INTEGRAL 25

Clearly, all the sets Ej are measurable, pairwise disjoint, and bounded. Let us set

k 2 j 1 φk(x) = M + 2 M − 1Ek , − 2k j j=1 X k 2 j ψk(x) = M + 2 M 1Ek . − 2k j j=1 X Both φ and ψ are simple functions, and by construction we have φ f ψ . Moreover, φ is non-decreasing k k k ≤ ≤ k k w.r.t. k, and ψk is non-increasing w.r.t. k (check!). Finally, 2M ψk(x) φk(x) 0, | − | ≤ 2k −−−−−→k + → ∞ which proves the assertion. We now consider the case of f nonnegative and measurable (not necessarily bounded). For a ﬁxed M > 0, let fM := ( f M) 1BM (0) . Due to the previous case, for all ǫ > 0 there exists a simple function φ such that 0 φ f and∧ f · φ < ǫ . Therefore, we choose ǫ = 1 /M and call φ the corresponding simple function. We ≤ ≤ M M − M have φM fM 1/M f as M + , which proves the assertion. ≥ − ↑ → ∞ + + Finally, the case of a sign changing f can be sorted by decomposing f = f f −, f = max f, 0 , f − = max f, 0 , and use the steps above (exercise). − { } {− } 4.3 Deﬁnition of the Lebesgue integral

We are now ready to start deﬁning integrals of measurable functions in the Lebesgue sense. As we pointed out in the introduction of this part (see Section 2.2), Lebesgue integration relies upon partitioning the image of a function rather than its domain. For simple function such action is very simple. All we have to do is measuring the set at which f attains each of its (ﬁnite) values.

d N Definition 4.3.1 (Lebesgue integral of a simple function) . Let φ (R ), φ = j=1 cj1Ej with Ej measurable, bounded, and pairwise disjoint. Then, the Lebesgue integral of φ is∈ the S number P N φ(x)dx = cj m(Ej). j=1 Z X Remark 4.3.1 . Due to additivity and linearity properties of the Lebesgue measure, the Definition 4.3.1 is well posed. The proof is left as an exercise. Remark 4.3.2 (Linearity and monotonicity of the integral of simple functions) . Let φ, ψ (Rd) and λ R. Then, we have (λφ + ψ)dx = λ φdx + ψdx , and if φ ψ we have φdx ψdx . Consequently∈ S we have the∈ formula φdx φ dx . All these formula are easy to check≤ by using the definitions≤ of simple function and integral of a simple| | function. ≤ R | | They areR left to theR student as an exercise. R R R R We are now ready for the definition of the Lebesgue integral for a measurable function. Definition 4.3.2 (Lebesgue integral of a measurable function) . Let f : Rd [0 , + ] measurable and nonnegative. The Lebesgue integral of f if the number (finite of + ) • → ∞ ∞ f(x)dx = sup φdx : 0 φ f , φ (Rd) . ≤ ≤ ∈ S Z Z Let f : Rd [ , + ] be measurable. Assume that one of two integrals • → −∞ ∞ + f (x)dx, f −(x)dx, Z Z are finite. Then, f is Lebesgue integrable , and the Lebesgue integral of f is defined as

+ f(x)dx = f (x)dx f −(x)dx. (4.2) − Z Z Z 26 CHAPTER 4. LEBESGUE INTEGRATION THEORY

Let E Rd be a measurable set and let f : Rd [ , + ]. Then, the Lebesgue integral of f on E, in • formulas⊂ f(x)dx , is deﬁned as f(x)dx = f→1 −∞dx . ∞ E E · E Remark 4.3.3 . RWe immediately notice twoR things: R

First of all, once a given function f is measurable, the only property we have to check to hold in order to be • able to perform Lebesgue integration is that the difference + does not occur in (4.2). ∞ − ∞ In particular (and this is the second thing that pops up from the previous definition), in order to compute f • we do not need a procedure such as the one for the Riemann/Darbous integral (see the Introductory Material), in which f is well defined only if we can approximate it well from both below and above with elementaryR integrals. It really seems like the notion of measurability is enough to produce an integrable function , provided the + R issue above is avoided. But in particular, every nonnegative measurable function is integrable . The reason∞ − ∞ behind this remark is contained in the next remark.

d Remark 4.3.4 . Let f : R R be measurable, bounded, zero outside a compact set, say BM (0). In this case we can deﬁne the upper integral→ and the lower integral of f as

∗ fdx = inf φdx ; φ (Rd), φ f , fdx = sup φdx ; φ (Rd), φ f ∈ S ≥ ∈ S ≤ Z Z Z∗ Z (note that both sets above are non empty because f is bounded). Now, since f is bounded and zero outside a compact set, both ∗ fdx and fdx are ﬁnite. Now, due to Lemma 4.2.1 there exists two sequences of simple ∗ functions φk and ψk such that φk f ψk such that φk f and ψk f uniformly. Therefore, we have the following estimate R R ≤ ≤ ↑ ↓

0 ψkdx φkdx = (ψk φk)dx m(BM (0)) sup φk(x) ψk(x) 0. ≤ − − ≤ x Rd | − | −−−−−→k + Z Z Z ∈ → ∞ Therefore, the lower and upper integrals of f must coincide. In other words, if f is measurable, bounded, and zero outside a compact set, then the integral of f can be approximated from above and below by integrals of simple functions. Proposition 4.3.1. Let f, g : Rd [ , + ] measurable and λ R. Then the formulas below hold whenever all the integrals below are well-deﬁned:→ −∞ ∞ ∈

(λf + g)dx = λ fdx + gdx, Z Z Z f g fdx gdx, ≤ ⇒ ≤ Z Z fdx f dx. ≤ | | Z Z

Proof. If f and g are nonnegative, the these formulas are direct consequences of taking the supremum on simple + + functions and using remark 4.3.2. In the general case, write f = f f − and use the ﬁrst part for f and f −. We omit the details. −

Exercise 4.3.1 (Integral of an almost everywhere null function) . Let f : Rd [0 , + ] be measurable, and assume there exists a measurable set E with m(E) = 0 such that f(x) = 0 for all x →E. Then,∞ fdx = 0 . Solution. Let φ (Rd) be a nonnegative simple function such that φ 6∈ f. Then, clearly φ = 0 on Rd E. ∈ S ≤ R \ Therefore, φ = N c 1 for some measurable sets E E. Hence, it is easily seen that φdx = 0 . Since φ is j=1 j Ej j ⊂ arbitrary f, by deﬁnition of Lebesgue integral as a supremum, fdx = 0 . ≤ P R Deﬁnition 4.3.3 (Summable function) . Let f : Rd [ , + R ] be measurable. If f(x) dx < + then f is called summable . We also use the notation f L1(Rd→). −∞ ∞ | | ∞ ∈ R Exercise 4.3.2. Prove that summable functions satisfy the following properties:

For every summable function f : Rd [ , + ], f(x)dx exists and it is ﬁnite. • → −∞ ∞ If f is measurable, bounded, zero outside a compactR set, then f is summable (Hint: f dx = f +dx + • | | f −dx ). R R R 4.3. DEFINITION OF THE LEBESGUE INTEGRAL 27

Remark 4.3.5 (Compatibility with the Riemann integral) . For simplicity we shall restrict to the one-dimensional case. Let f : R R be measurable, bounded, and zero outside a compact set. Recall (exercise) that f is Riemann integrable if and→ only if the lower Riemann integral

fdx = sup hdx : h is piecewise constant , h f ≤ Z Z and the upper Riemann integral

fdx = inf hdx : h is piecewise constant , h f ≥ Z Z coincide. We claim that if f is Riemann integrable, then its Lebesgue integral coincide with its Riemann integral. To see this, we remark that piecewise constant functions are a special case of simple functions (because ﬁnite intervals are measurable sets). Hence, the lower Riemann integral of f is controlled from above by the Lebesgue lower integral fdx deﬁned in remark 4.3.4. Similarly, the upper Riemann integral is controlled from below by the upper Lebesgue∗ integral. Therefore, we have R

∗ fdx fdx fdx fdx. ≤ ≤ ≤ Z Z∗ Z Z Hence, if f is Riemann integrable, then fdx = fdx , and therefore all the four above integrals coincide, which means that the Riemann and Lebesgue integrals of f coincide. R R The above remark is indeed very important. In order to compute a Lebesgue integral, if the function is also Riemann integrable we can use all the rules on Riemann integration we learned in basic calculus. Example 4.3.1 . The function 1 if x Q [0 , 1] d(x) = ∈ ∩ (0 otherwise is known to be Lebesgue integrable. Indeed, it is a simple function . In the Lebesgue sense, one has

d(x)dx = 0 . R Z On the other hand, the Darboux upper integral of d is 1 whereas the lower Darboux integral of d is 0 (Exercise). Hence, d is not Riemann integrable. We record a basic inequality, known as Markov’s inequality, that asserts that the Lebesgue integral of a measurable function controls how often that function can be large. Lemma 4.3.2 (Markov’s inequality) . Let f : Rd [0 , + ] be measurable. Then, for any 0 < λ < + one has → ∞ ∞ 1 m( x Rd : f(x) λ ) f(x)dx. { ∈ ≥ } ≤ λ Rd Z Proof. We have the trivial pointwise inequality

f(x) λ1 x Rd : f(x) λ (x), ≥ { ∈ ≥ } and since the function on the left-hand-side is a simple function, we have

d f(x)dx λ1 x Rd : f(x) λ (x)dx = λm ( x R : f(x) λ ), ≥ { ∈ ≥ } { ∈ ≥ } Z Z which proves the assertion.

Markov’s lemma has the following three important applications. Lemma 4.3.3. Let f : Rd [0 , + ] be a nonnegative measurable function. → ∞ (i) If f(x)dx < + then f is ﬁnite almost everywhere. ∞ R 28 CHAPTER 4. LEBESGUE INTEGRATION THEORY

(ii) f(x)dx = 0 if and only if f is zero almost everywhere.

Moreover,R

(iii) let f : Rd [ , + ] be measurable such that f(x)dx = 0 for all measurable sets E Rd. Then f = 0 → −∞ ∞ E ⊂ almost everywhere on Rd. R d Proof. Statement (i) can be proven as follows. For all M N let EM = x R : f(x) M , and let d ∈ { ∈ ≥ } E = x R : f(x) = + . Clearly, EM E . From Markov’s lemma one has ∞ { ∈ ∞} ↓ ∞ 1 m(E ) f(x)dx. M ≤ M Z From the last statement in Theorem 3.0.3, we have

m(E ) = lim m(EM ) = 0 , ∞ M + → ∞ where the last identity follows form the fact that fdx is ﬁnite. For statement (ii), assume ﬁrst that f = 0 almost everywhere. Then the exercise 4.3.1 gives that fdx = 0. R Vice versa, assume fdx = 0. Then, Markov’s inequality gives for all integers k > 0 R R 1 1 0 = f(x)dx m x Rd : f(x) , ≥ k ∈ ≥ k Z d 1 which implies m x R : f(x) k = 0. By using Theorem 3.0.3 as in step (i), since the family of sets d 1 ∈ ≥ x R : f(x) k k N is increasing, we get ∈ ≥ ∈ 1 1 m( x Rd : f(x) > 0 ) = m x Rd : f(x) = lim m x Rd : f(x) = 0 . { ∈ } ∈ ≥ k k + ∈ ≥ k !k N " → ∞ [∈ d Statement (iii) can be proven as follows. For all k N set Ek = x R : f(x) 1/k . Then, from the assumption on f we have ∈ { ∈ ≥ } 1 0 = f(x)dx m(E ), ≥ k k ZEk and that implies m(Ek) = 0. Since the family of sets Ek k N is increasing and { } ∈ x Rd : f(x) > 0 = E , { ∈ } k k N [∈ then Theorem 3.0.3 implies d m( x R : f(x) > 0 ) = lim m(Ek) = 0 . { ∈ } k + → ∞ d Now, similarly we set F = x R : f −(x) 1/k . Then we have k { ∈ ≥ } 1 0 = f −(x)dx m(F ), ≥ k k ZFk which implies m(F ) = 0 for all k N. With the same argument as above this implies k ∈ d m( x R : f(x) < 0 ) = lim m(Fk) = 0 . { ∈ } k + → ∞ Therefore,

m( x Rd : f(x) = 0 ) = m( x Rd : f(x) > 0 x Rd : f(x) < 0 ) { ∈ 6 } { ∈ } ∪ { ∈ } m( x Rd : f(x) > 0 ) + m( x Rd : f(x) < 0 ) = 0 , ≤ { ∈ } { ∈ } which proves the assertion. 4.3. DEFINITION OF THE LEBESGUE INTEGRAL 29

Remark 4.3.6 (Extension to complex valued functions) . Lebesgue integration theory can be trivially extended to functions f : Rd C as follows. We say that f = Re( f) + iIm( f) is measurable if and only if Re( f) and Im( f) are measurable (as real→ valued functions). All the basic properties of measurable functions are easily generalised. In particular,

If f, g are measurable, so are f + g, f g, f/g if g = 0, f (here f = Re( f)2 + Im( f)2). • · 6 | | | |

If fk is a sequence of measurable functions, then inf k fk , sup k fk , liminfp k + fk , lim sup k + fk are • measurable. | | | | → ∞ | | → ∞ | | The integral of f is deﬁned as follows

f(x)dx = Re( f(x)) dx + i Im( f(x)) dx Z Z Z provided the two real valued integrals above are well deﬁned. The linearity property is trivially satisﬁed. The monotonicity property is replaced by f(x)dx f(x) dx. ≤ | | Z Z

A function f is said to be summable if

f(x) dx < + . | | ∞ Z 30 CHAPTER 4. LEBESGUE INTEGRATION THEORY Chapter 5

Convergence properties of Lebesgue integral

As explained partly in Chapter 2, the Lebesgue integral provides a better tool than Riemann integration to preserve the integral in a limiting procedure. As far as Riemann integration is concerned, Proposition 6.12 of the Introduc- tory Material provides a useful result in which integral and limit can be interchanged, but this requires uniform convergence of the sequence. We shall see below that Lebesgue integral allows for much more ﬂexibility in this sense. d We recall from Deﬁnition 3.0.3 that a sequence fn : R R of measurable functions converges almost everywhere d → d to a measurable function f : R R if there exists a set A R with md(A) = 0 such that fn(x) f(x) for all x Rd A (f converges pointwise→ to f outside a set of null⊂ measure). → ∈ \ n 5.1 Interchanging limits and integrals

The next theorem is due to the Italian-Argentinian mathematician Beppo Levi (1875-1961) and it is known in the literature as Beppo Levi’s lemma or Lebesgue’s monotone convergence theorem.

d Theorem 5.1.1 (Monotone Convergence theorem) . Let fn : R [0 , + ] be a sequence of measurable functions, with the monotonicity property → ∞

f (x) f (x), for almost every x Rd and for all n N. n ≤ n+1 ∈ ∈ Then, we have

lim fn(x) dx = lim fn(x)dx. Rd n + n + Rd Z → ∞ → ∞ Z

Proof. The statement in the Theorem implicitly asserts that fn has a limit almost everywhere. Let us set

f(x) = sup fn(x). n N ∈

Clearly, fn converges to f almost everywhere. Now, let

A = x Rd : f converges to f as n + . { ∈ n → ∞} We set f˜n(x) = fn(x)1A(x) , f˜(x) = f(x)1A(x).

Then it is clear that f˜n converges to f˜ pointwise. Now, let

B = x Rd : f (x) f (x) for all n N . { ∈ n ≤ n+1 ∈ } Clearly, B A, and then ⊂ m(Rd A) m(Rd B) = 0 , \ ≤ \ 31 32 CHAPTER 5. CONVERGENCE PROPERTIES OF LEBESGUE INTEGRAL

where the last equality follows by the assumptions. Therefore, fñ and f˜ differ from fn and f respectively by a null set, which means that the Lebesgue integrals of fn and f are not affected by that change. Hence, it is not restrictive to assume that f f pointwise everywhere and f f everywhere . n → n ≤ n+1 Since f f, we have k ≤ lim fk(x)dx = sup fk(x)dx f(x)dx. k + ≤ → ∞ Z k Z Z In order to finish the proof, we need to prove that

sup f (x)dx f(x)dx. k ≥ k Z Z Let φ be a simple function such that 0 φ f, say ≤ ≤ N

φ(x) = cj1Bj (x). j=1 X Fix t (0 , 1), and let us set ∈ E = x Rd : f (x) tφ (x) . k { ∈ k ≥ } Since f is increasing, E is an increasing family of sets. Moreover, E = Rd. To see this, let x Rd, we have k k k k ∈ tφ (x) < φ (x) f(x), and since fk(x) f(x) one has that fk(x) tφ (x) for k larger than some K N. Therefore, ≤ → ≥ S ∈ N f (x)dx f (x)dx t φ(x)dx = t c m(E B ). k ≥ k ≥ j k ∩ j Ek Ek j=1 Z Z Z X

Now, let k + in the above inequalities. Since the family of sets Ek Bj k is increasing to Bj , we can apply the monotonicity→ ∞ of Lebesgue measure to get { ∩ }

N lim fk(x)dx t cjm(Bj) = t φ(x)dx. k + ≥ → ∞ j=1 Z X Z Since φ and the left hand side above do not depend on t, we can send t 1 and get ր

lim fk(x)dx φ(x)dx. k + ≥ → ∞ Z Z Since φ is an arbitrary simple function 0 φ f, taking the sup with respect to φ we obtain the assertion. ≤ ≤ d Corollary 5.1.2 (Tonelli) . Let un n N be a sequence of nonnegative measurable functions deﬁned on R . Then, { } ∈ + + ∞ ∞ un(x)dx = un(x)dx. Rd Rd n=1 n=1 Z X X Z n Proof. This corollary is an immediate consequence of Theorem 5.1.1, by deﬁning sn(x) = k=1 uk(x) and by observing that sn is monotone increasing with respect to n. P

In general, when the sequence fn is not monotonic with respect to n, one can only obtain the following result, which is however of paramount importance.

Lemma 5.1.3 (Fatou’s Lemma) . Let f : Rd [0 , + ] be a sequence of measurable functions, then n → ∞

lim inf fn(x)dx lim inf fn(x)dx. Rd + ≤ n + Rd Z → ∞ → ∞ Z Proof. Let us set gn(x) = inf fk(x). k n ≥ 5.1. INTERCHANGING LIMITS AND INTEGRALS 33

Then, by deﬁnition of liminf we have liminf n + fn(x) = lim n + (inf k n fk(x)) = lim n + gn(x). Since gn is → ∞ → ∞ ≥ → ∞ monotone increasing, theorem 5.1.1 applied to gn implies

lim inf fn(x)dx = lim gn(x)dx = lim gn(x)dx. Rd + Rd n + n + Rd Z → ∞ Z → ∞ → ∞ Z

Now, for all k n one has gn(x) fk(x), and hence Rd gn(x)dx Rd fk(x)dx . By taking the inf k n on the last term we get ≥ ≤ ≤ ≥ R R gn(x)dx inf fk(x)dx. Rd ≤ k n Rd Z ≥ Z Therefore, from above we get

lim inf fn(x)dx = lim gn(x)dx lim inf fk(x)dx = lim inf fn(x)dx. Rd + n + Rd ≤ n + k n Rd n + Rd Z → ∞ → ∞ Z → ∞ ≥ Z → ∞ Z

A natural question arises at this stage. What makes a non monotonic sequence fn fail in satisfying the limit interchange property stated in Theorem 5.1.1? In particular, are there examples of (non monotonic) sequences for which the strict inequality in Fatou’s Lemma holds? Here are two important examples, showing the two typical phenomena that may produce a sudden loss of mass (i. e. a sudden loss of fn) in the limit n + : concentrations and travelling waves . → ∞ R Example 5.1.1 (Concentration) . This example anticipates the concept of Dirac’s delta , which is usually covered in more advanced units. Here the mass (integral) of the sequence is concentrating on a point (the origin) and to disappear in the limit. Consider fn : R R, fn(x) = n1[0 ,1/n ]. One has lim n + fn(x) = 0 when x = 0, and → → ∞ 6 lim n + fn(0) = + . Hence, fn converges to f(x) 0 almost everywhere. Moreover, Rd fn(x) = 1 for all n N. Now,→ ∞ ∞ ≡ ∈ R lim inf fn(x)dx = lim fn(x)dx = 0dx = 0 , Rd n + Rd n + Rd Z → ∞ Z → ∞ Z whereas

lim inf fn(x)dx = lim inf 1 = 1 , n + Rd n + → ∞ Z → ∞ and therefore

lim inf fn(x)dx lim inf fn(x)dx. Rd n + n + Rd Z → ∞ → ∞ Z Example 5.1.2 (Travelling wave) . The other phenomenon that may arise is that the mass of the sequence concentrates at inﬁnity . Here is an example. Let f : R R, f (x) = 1 . Clearly f (x)dx = n+1 dx = 1. Moreover, n → n [n,n +1] Rd n n lim + fn(x) = 0 = f(x) for all x R, and therefore f(x)dx = 0. This shows once again that the integral of the limit→ is∞ strictly less than the limit of∈ the integrals. R R R Remark 5.1.1 (Nonnegativity assumptions) . It may have passed as a hidden information that so far we have always requested the sequence to be nonnegative in both Beppo Levi’s and Fatou’s lemmas. Such assumption cannot be 1 completely removed. Indeed, let fn(x) = . Clearly lim n + fn(x) = 0 = f(x), and therefore f(x)dx = 0 − n → ∞ whereas fn(x)dx = . This example shows that the thesis of Fatou’s lemma is not true for this example. On the other hand, the−∞ nonnegativity assumption can be relaxed by assuming that f g a. e. forR some (non R n ≥ necessarily nonnegative) measurable function g such that g(x)dx > . Indeed, one can set hn = fn g and apply Fatou’s lemma to h (exercise!). −∞ − n R A possible way to prevent concentrations and travelling waves, in a general set of assumptions to guarantee the interchange of limits and integrals, is the following result, known as the dominated convergence theorem , which deals with the interchange between limits and integrals in the case of a.e. converging sequences. The dominated convergence theorem does not hold for the Riemann integral because the limit of a sequence of Riemann-integrable functions is in many cases not Riemann-integrable, see example 2.1.4. d Theorem 5.1.4 (Dominated convergence theorem) . Let fn : R R be a sequence of measurable functions, and f : Rd R be a measurable function, such that → → (i) f (x) f(x) almost everywhere in Rd, n → (ii) f (x) g(x) almost everywhere in Rd for some measurable function g : Rd [0 , + ] with g(x)dx < + . | n | ≤ → ∞ Rd ∞ R 34 CHAPTER 5. CONVERGENCE PROPERTIES OF LEBESGUE INTEGRAL

Then,

f(x)dx = lim fn(x)dx. Rd n + Rd Z → ∞ Z Proof. First of all, as in theorem 5.1.1, we can assume without restriction that all the properties stated above hold everywhere and not just almost everywhere. Moreover, note that the property (ii) allows to deduce that all the fn have ﬁnite integral. We now apply Fatou’s lemma to the nonnegative functions h (x) := g(x) f (x). We get n − n

(g(x) f(x)) dx lim inf (g(x) fn(x)) dx. Rd − ≤ n + Rd − Z → ∞ Z This shows

f(x)dx lim sup fn(x)dx. Rd ≥ n + Rd Z → ∞ Z We now re-apply Fatou’s lemma to the nonnegative functions Hn(x) = g(x) + fn(x) and get

(g(x) + f(x)) dx lim inf (g(x) + fn(x)) dx, Rd ≤ n + Rd Z → ∞ Z which gives

f(x)dx lim inf fn(x)dx lim sup fn(x)dx f(x)dx, Rd ≤ n + Rd ≤ n + Rd ≤ Rd Z → ∞ Z → ∞ Z Z which proves the assertion. Remark 5.1.2 . A seemingly obvious remark to be added here is that a uniform control on f dx is not enough | n| to apply dominated convergence theorem. One has to control the integrand fn pointwise , uniformly in n, by a summable function. | | R

5.2 Fubini’s Theorem

Typically one expects that, while integrating a function f : Rn+m [ , + ], we can integrate with respect to x Rn ﬁrst, and then with respect to y Rm, at least under reasonable→ −∞ ∞ assumptions. This is the case for ∈ ∈ Lebesgue’s integration theory. For simplicity we shall restrict to the case m = n = 1. In what follows, md denotes the Lebesgue measure on Rd. The results in this section are not misplaced (this section concerns with convergence properties of the Lebesgue integral). Actually, they are consequence of the Monotone Convergence’s theorem. Lemma 5.2.1. Let E R2 be a measurable set. Let ⊂ E = y R : ( x, y ) E x { ∈ ∈ } be the section of E in x R. Then Ex is measurable for almost every x R, the function x m1(Ex) is measurable, and ∈ ∈ 7→

m2(E) = m1(Ex)dx. R Z Proof. The assertion is obvious if E is an elementary set (exercise). Let E = A be an open set, and let Yj be a sequence of elementary sets such that Y A. We observe that ( Y ) is a sequence of elementary sets in R, A is an j ↑ j x x open set in R, and ( Yj)x A. Therefore, m1(( Yj)x) m1(Ax) and m2(Yj ) m2(A). Therefore, due to Beppo-Levi’s theorem and to the continuity↑ of Lebesgue measure,↑ we have ↑

m2(A) = lim m2(Yj) = lim m1(( Yj)x)dx = m1(Ax)dx. j + j + → ∞ → ∞ Z Z Here the function g(x) = m1(Ax) is measurable since it is the almost everywhere limit of the simple functions φj(x) = m1(( Yj )x) (simple exercise: check that φj is simple!). The case E = K compact can be proven in a similar way by approximating with elementary sets from outside the set. Now let E be a bounded measurable set. Choose ǫ = 1 /j , we obtain two sequences K E A , K compact j ⊂ ⊂ j j and Aj open such that m (K ) m(E) , m (A ) m (E). 2 j ↑ 2 j ↓ 2 5.2. FUBINI’S THEOREM 35

Set fj(x) = m1(( Kj)x), gj(x) = m1(( Aj)x). The sequence gj fj is nonnegative and decreasing. Let h 0 be its limit function. Due to the previous steps and to elementary proper− ties of Lebesgue integration, we have ≥

0 hdx (g f )dx = m (A ) m (K ) 0. ≤ ≤ j − j 2 j − 2 1 → Z Z

Hence, fdx = 0, which implies h = 0 almost everywhere. Now, let φ (x) = m1 (Ex), φ∗(x) = m1∗(Ex). Clearly we ∗ ∗ have fj φ φ∗ gj, but gj fj 0, therefore φ = φ∗ = lim j fj = lim j gj almost everywhere, and therefore R≤ ∗ ≤ ≤ − → ∗ Ex is measurable for almost every x. Finally, Beppo Levi’s Theorem implies

m2(E) = lim m2(Kj) = lim fjdx = m1(Ex)dx. j j Z Z

The case of an unbounded measurable set easily follows by taking intersections with balls BR(0), and is therefore left as an exercise.

Remark 5.2.1 . The statement in Lemma 5.2.1 can be easily proven also in higher dimensions, say if E Rn+k is measurable, then ⊂

mn+k(E) = mn(Ex)dx = mk(Ey)dy, Z Z where E = y Rk : ( x, y ) E and E = x Rn : ( x, y ) E . We omit the proof. x { ∈ ∈ } y { ∈ ∈ } Exercise 5.2.1. Let A Rn and B Rk be measurable. Then A B Rn+k is measurable and ⊂ ⊂ × ⊂ m (A B) = m (A)m (B). n+k × n k Lemma 5.2.2. Let f : Rn [0 , + ] be a nonnegative measurable function, and let E be the subgraph of f, i. e. → ∞ E = (x, t ) Rn R : 0 t f(x) . { ∈ × ≤ ≤ } Then, E is measurable and

f(x)dx = mn+1 (E). Z

Proof. Let f = c1A with c 0 and A measurable. We have E = A [0 , c ], therefore mn+1 (E) = cm n(A) = fdx . Therefore, the statement is≥ trivially obtained in the case of f simple× function. Now let f 0 me measurable. From ≥ R Lemma 4.2.1, we know that there exists a sequence φj of simple functions with 0 φj f. Beppo Levi’s Theorem then implies ≤ ↑ φ dx fdx. k ↑ Z Z Let Ek be the subgraph of φk. We clearly have Ek E. Hence, E is measurable and the continuity of Lebesgue measure implies ↑

m(E) = lim m(Ek) = lim φkdx = fdx. k k Z Z

Theorem 5.2.3 (Fubini’s Theorem) . Let f : R2 R be a summable function. Then, for almost every x, the function f(x, ) is summable, moreover the function → ·

R x f(x, y )dy R ∋ 7→ ∈ Z is summable, and we have the formula

f(x, y )dxdy = f(x, y )dy dx = f(x, y )dx dy. R2 R R R R Z Z Z Z Z 36 CHAPTER 5. CONVERGENCE PROPERTIES OF LEBESGUE INTEGRAL

+ Proof. We only prove the statement in the case f 0. The general case easily follows by writing f = f f −, and is left as an exercise. Let E be the subgraph of≥f. Lemma 5.2.1 (and its generalisation to higher dimensions)− implies

m3(E) = m2(Ex)dx, R Z but E is the subgraph of f(x, ), hence x · m2(Ex) = f(x, y )dy, R Z which implies that f(x, ) is summable for almost every x, and that ·

f(x, y )dx = m3(E) = m2(Ex)dx = f(x, y )dy dx. R2 R Z Z Z Z The same identity with reversed roles for x and y is proven in the same way.

Corollary 5.2.4 (Tonelli’s Theorem) . Let R2 [0 , + ], f measurable. Then → ∞

f(x, y )dxdy = f(x, y )dy dx = f(x, y )dx dy. R2 R R R R Z Z Z Z Z

Proof. Apply Fubini’s Theorem to the function fR = ( f R)1BR(0) , and then pass to the limit with Beppo Levi’s Theorem. The details are left as an exercise. ∧

Fubini’s Theorem and Tonelli’s Theorem can be easily proven in higher dimensions. We omit the details. The importance of Tonelli’s Theorem lies on the fact that it is not necessary to check a priori that f is summable in order to use the reduction formula for integrals. In case the function is nonnegative, one can compute the integral with the reduction formula, and then deduce that the function is summable in case the result is ﬁnite.

5.3 Egorov’s and Lusin’s Theorem. Convergence in measure

The following result establishes a surprising property of almost everywhere converging sequences on a set of finite measure. d Theorem 5.3.1 (Severini–Egorov) . Let A R be a set with finite Lebesgue measure, and let fn : A R be a sequence of measurable functions which converges⊂ almost everywhere to a function f : A R. Then, for→ all ǫ > 0 there exists a measurable set A A such that m(A A ) < ǫ and f f uniformly on A→. ǫ ⊂ \ ǫ n → ǫ Proof. For all n N we set ∈ + ∞ 1 Ek = x A : f (x) f(x) < . n ∈ | j − | k j=n \ k k k k We observe that, for fixed k, n N, En En+1 , which implies A En A En+1 , i. e. the family of sets k ∈ k ⊂ + k \ ⊃ \ A En n N is decreasing. Let us set E = n=1∞ En. Since fn f almost everywhere, then there exists N A { \ } ∈ → k ⊂ such that m(N) = 0 and fn(x) f(x) for all x A N. We observe that A N E for all k N. Indeed, let → S ∈ \ \ ⊂ 1 ∈ x A N, then fn(x) f(x), hence there exists N N such that fn(x) f(x) k for all n N, and hence ∈ k\ k → k ∈ | − | ≤ k ≥ x EN E . Now, this implies A (E ) N. Therefore, since m(N) = 0 then A (E ) is measurable with m(∈A (E⊂k)) = 0. From the third property\ in⊂ Theorem 3.0.3 we have \ \ k k k lim m(A En) = m( (A En)) = m(A E ) = 0 . n + \ \ \ → ∞ n N \∈ k k Hence, for a given ǫ > 0 there exists an integer nk N such that m(A Enk ) < ǫ 2− . This is true for all k N. + k ∈ \ ∈ We now set A = ∞ E , and claim that m(A A ) < ǫ . Indeed, we have ǫ k=1 nk \ ǫ T + + + + ∞ k ∞ k ∞ k ∞ k m(A A ) = m(A E ) = m( (A E )) m(A E ) < ǫ 2− = ǫ. \ ǫ \ nk \ nk ≤ \ nk k\=1 k[=1 kX=1 kX=1 5.3. EGOROV’S AND LUSIN’S THEOREM. CONVERGENCE IN MEASURE 37

Now we claim that fn f uniformly on Aǫ. Let η > 0, and let k N such that 1 /k < η . Let x Aǫ. This means that x Ek for all k,→ and hence ∈ ∈ ∈ nk 1 f (x) f(x) < < η, for all j n = n(η), | j − | k ≥ k which implies the desired assertion because nk does not depend on x. The following is an another important property of measurable functions, sometimes refereed to as the function being nearly continuous . Theorem 5.3.2 (Lusin) . Let A Rd be a measurable set with m(A) < + , and let f : A R be a function. Then the following are equivalent: ⊂ ∞ →

(i) f is measurable

(ii) For all ǫ > 0 there exists a compact set K A such that m(A K ) < ǫ and f is continuous on K . ǫ ⊂ \ ǫ ǫ Step 1. 1 Proof. (ii) implies (i). For all n N we set ǫ = n and let Kn A be the corresponding compact set from the assumptions in (ii), such that f is continuous∈ on K . Now, let α ⊂R, and set A = x A : f(x) α . We n ∈ α { ∈ ≥ } set Kñ = x Kn : f(x) α . We observe that Kñ is a closed set since f is continuous on Kn. Hence, since Kn { ∈ ˜ ≥ } ˜ is compact, then Kn is compact (as a closed subset of a compact set). Now set K = n N Kn, and compute ∈ S 1 m∗(A K) = m∗ A K˜ m∗(A K˜ ) = m∗(A K ) m∗(A K ) , for all n N, α \ α \ n ≤ α \ n α \ n ≤ \ n ≤ n ∈ ! n N " [∈ and this proves that m∗(Aα Kn) = 0, and therefore Aα Kn is measurable with Lebesgue measure zero. We observe that A = K (A \K). Since A K is measurable,\ and since K is measurable (as it is the countable α ∪ α \ α \ union of measurable sets), then Aα is measurable. Since α R is arbitrary, this shows that f is a measurable function. ∈ Step 2. n (i) implies (ii). Let us first consider the case of f = φ simple function. Hence, φ = i=1 αi1Ai with n A = i=1 Ai and Ai Aj = for i = j. Let ǫ > 0. By definition of measurable sets, for all i there exists Ki ∩ ∅ 6 ǫ n P compact such that Ki Ai and m(Ai Ki) < . Set K = Ki. Clearly K is compact, and we have S ⊂ \ n i=1 n n S n n ǫ m(A K) = m A K m (A K ) m(A K ) = ǫ. \ \ i ≤ i \ i ≤ i \ i ≤ n ! i=1 " !i=1 " i=1 i=1 [ [ X X Since all the K are disjoint, and φ is constant on all of them, then φ is continuous on K, and m(A K) ǫ. i \ ≤ Step 3. General case f measurable. Let ǫ > 0. Let φn be a sequence of simple functions such that φn f almost everywhere (see Lemma 4.2.1). Theorem 5.3.1 implies there exists a set E A such that m(A E) →< ǫ ⊂ \ 2 and φn f uniformly in E. Now, step 2 implies that for all n N there exists a compact set Kn such that → n 1 + ∈ m(E K ) < ǫ 2− − and φ is continuous on K . Set K = ∞ K . We have \ n n n n=1 n T + + ∞ ∞ n 1 ǫ m(E K) = m(E K ) = m( (E K )) m(E K ) ǫ2− − = . \ \ n \ n ≤ \ n ≤ 2 n n=1 n=1 \ [ X X ǫ ǫ Therefore, m(A K) m(A E)+ m(E K) 2 + 2 = ǫ. Moreover, φn is continuous on K Kn, and φn converges uniformly to f on\ K≤. Therefore,\ f is continuous\ ≤ on K (the uniform limit of continuous functions⊂ is continuous). Finally, K is compact as it is the countable intersection of compact sets. The assertion (i) implies (ii) is proven.

We now introduce a new notion of convergence of function sequences.

d Deﬁnition 5.3.1 (Convergence in measure) . A sequence of measurable functions fn : R R converges in measure to a measurable function f : Rd R if and only if, for all ǫ > 0 one has → → d lim m x R : fn(x) f(x) > ǫ = 0 . n + → ∞ ∈ | − | We make immediately clear that neither the sentence ‘convergence in measure implies almost everywhere convergence’ nor the converse statement are true. 38 CHAPTER 5. CONVERGENCE PROPERTIES OF LEBESGUE INTEGRAL

Example 5.3.1 . Almost everywhere convergence does not necessarily imply convergence in measure. Indeed, let fn : R R be defined as fn(x) = 1[n, + ). Then, fn 0 pointwise (and hence almost everywhere) but for all ǫ (0 , 1)→ one has ∞ → ∈ m ( x R : f (x) > ǫ ) = m([ n, + )) = + . { ∈ | n | } ∞ ∞ Example 5.3.2 (The typewriter sequence) . Let f : [0 , 1] [0 , 1] defined as follows. f = 1 , f = 1 , n → 1 [0 ,1) 2 [0 ,1/2) f3 = 1[1 /2,1) , f4 = 1[0 ,1/4) , f5 = 1[1 ,4,1/2) , f6 = 1[1 /2,3/4) , f7 = 1[3 /4,1) , f8 = 1[0 ,1/8) , and so on... For a given ǫ > 0 one clearly has m ( x [0 , 1] : fn(x) > ǫ ) 0 as n + , hence fn converges to zero in measure. On the other hand, f does not{ converge∈ to zero| almost| } everywhere.→ → Indeed,∞ let η = 1 /2 and let x [0 , 1). For all n N there n ∈ ∈ exists m > n such that fm(x) = 1 > 1/2, and this violates the convergence of fn(x) to zero as n + for x in a set of measure 1. | | → ∞ The example 5.3.2 clarified that convergence in measure does not imply almost everywhere convergence in general. However, the example itself reveals that the interplay between these two notions of convergence is not totally empty, as shown in the next theorem. Theorem 5.3.3. Let f : Rd R be a sequence of measurable functions which converges in measure to f : Rd R. n → → Then, there exists a subsequence of fn which converges to f almost everywhere.

d Proof. For all ǫ > 0, let Aǫ,n = x R : fn(x) f(x) > ǫ . Then, the assumptions imply that m(Aǫ,n ) 0 ∈ | − | 1 1 → as n + . Let ǫ = 1. Then, there exists n1 N such that m(A1,n 1 ) < 2 . Let ǫ = 2 . Then, there exists → ∞ 1 ∈ n2 N, n2 > n 1, such that m(A 1 ) < . By induction, we generate an increasing sequence of natural numbers ∈ 2 ,n 2 4 k + + n1 < n 2 < n 3 < ... , that we denote by nk k N such that m(A 1 ,n ) < 2− . Let us set E = ∞ ∞ A 1 ,n . ∈ k k j=1 k=j k k { } + Let x E. Then, there exists j N such that x ∞ A 1 , which means that x A 1 for all k j, or k=j k ,n k k ,n kT S 6∈ 1 ∈ 6∈ 6∈ ≥ equivalently f (x) f(x) for all k j. This implies that the sequence f (x) N converges to f(x) as nk k S nk k k + , and| this is− true for| ≤ all x E. The≥ assertion follows if we can prove that {E has null} ∈ measure. By deﬁnition of →E, for∞ all j N we have 6∈ ∈ + + + ∞ ∞ ∞ k j+1 m(E) m A 1 ,n m(A 1 ,n ) < 2− = 2 − . ≤  k k  ≤ k k k[=j Xk=j Xk=j   Since j is arbitrary, this shows that m(E) = 0. Chapter 6

Exercises

Exercise 6.1. Find a measurable function f : R R which is not Lebesgue integrable. →

Exercise 6.2. Let a > 0 and f : [ a, ) R. Prove that ∞ → 1. If f 0 a.e. on [a, ) and f is integrable in the improper sense, then f is Lebesgue integrable and the two integrals≥ coincide. ∞

2. There exists f : [ a, ) R such that f is integrable in the improper sense but f is not Lebesgue integrable. ∞ →

Exercise 6.3. Let f and g be two functions on Rd, and assume that f = g almost everywhere. Show that if f is measurable than g is measurable.

Exercise 6.4. Let f and g be two measurable functions on Rd. Prove that the set x Rd : f(x) g(x) is measurable. { ∈ ≥ }

d Exercise 6.5. Show that the indicator function 1A of a set A R is measurable if and only if the set A is Lebesgue measurable. ⊂

Exercise 6.6. Suppose fn n is a sequence of measurable, nonnegative functions. Assume fn f almost everywhere on Rd. Prove that {f is} almost everywhere nonnegative. →

Exercise 6.7. Suppose fn n is a sequence of measurable, nonnegative functions. Assume fn f almost every- d { } → where on R and fdx = lim n + fndx < + . Prove that → ∞ ∞ R R fdx = lim fndx, n + ZE → ∞ ZE for all Lebesgue measurable sets E (Hint: set gn = fn fn1E and hn = fn + fn1E and apply Fatou’s lemma for both). −

Exercise 6.8. Let f : Rd [0 , + ] be a summable function. Show that for every ǫ > 0 there exists a measurable set E Rd such that → ∞ ⊂ f(x) dx < ǫ Rd E | | Z \ (Hint: use the dominated convergence theorem).

39 40 CHAPTER 6. EXERCISES

Exercise 6.9. Suppose fk is a monotone function on (0 , 1) for each k N and fk converges in measure to f. Prove that f is equal almost everywhere to a monotone function. ∈

Exercise 6.10. Prove that a bounded function f on [a, b ] is Riemann integrable if and only if the set of points at which f is discontinuous has Lebesgue measure zero.

Exercise 6.11. Let fn n and gn n be sequences of summable functions such that fn f and gn g almost everywhere on Rd. Moreover,{ } assume{ } that f g and → → | n| ≤ n

g dx g dx. n → Z Z Prove that f dx f dx. n → Z Z

Exercise 6.12. Show that there exists a sequence of functions f such that f 0 almost everywhere on R, { n}n n →

fn dx 0, R → Z and there exists no summable function g such that f g. | n| ≤

d Exercise 6.13. Let fn n and be a sequences of summable functions such that fn f almost everywhere on R and f is summable. Moreover,{ } assume that →

fn dx f dx. R | | → R | | Z Z Show that

fn f dx 0. R | − | → Z

Exercise 6.14. Assume that f : R R R is such that x f(t, x ) is measurable for any t R and t f(t, x ) is continuous for any x R. Assume also× that→ there is a summable→ function g : R R such that∈ for any (t,→ x ) R R it holds f(t, x ) g(x∈). Prove that the function t f(t, x ) is summable and the→ function ∈ × | | ≤ →

F (t) = f(t, x ) dx R Z is continuous.

Exercise 6.15. Assume that f : R R R is such that x f(t, x ) is measurable for any t R and for some t R the function x f(t , x ) is summable.× → Moreover, assume→ that ∂ f(t, x ) exists for any (∈t, x ) R R and 0 ∈ → 0 t ∈ × there is a summable function g : R R such that for any (t, x ) R R it holds ∂tf(t, x ) g(x). Prove that the function x f(t, x ) is summable for→ any t R and the function ∈ × | | ≤ → ∈

F (t) = f(t, x ) dx R Z is diﬀerentiable with derivative d ∂ F ′(t) = f(t, x ) dx = f(t, x ) dx. dt ∂t ZR Z 41

Exercise 6.16. Let f be a sequence of measurable functions on a set A R, which has with ﬁnite measure. { n}n ⊂ Assume that fn n are uniformly bounded on A and fn f almost everywhere on A. Use Egorov’s Theorem to prove that { } →

f(x) dx = lim fn(x) dx. n ZA →∞ ZA

Deﬁnition 6.0.1. Let fn n be a sequence of summable functions on R. The sequence fn n is said equi-integrable is for any ε > 0 there exists{ } δ > 0 such that for any measurable set A R with A < δ { } ⊂ | |

f ε for any n N. | n| ≤ ∈ ZA Exercise 6.17. Let f be a sequence of summable functions. Then the following statement are equivalent. { n}n 1. f is equi-integrable. { n}n

2. lim sup fn = 0 . M n fn >M | | →∞ Z{| | }

Exercise 6.18. Suppose f is a summable function. Prove that for any ε > 0 there exists δ > 0 such that for any A R with A δ it holds ⊂ | | ≤ f dx ε. | | ≤ ZA

Exercise 6.19. Let f , f : [0 , 1] R be summable functions. Prove that the following statements are equivalent. n →

1. f is bounded in L1([0 , 1]) , f is equi-integrable, f f in measure on [0 , 1] . { n}n { n}n n →

1 2. f f dx 0. | n − | → Z0

Exercise 6.20. Let f : [0 , 1] R and τ > 0 such that n → 1 sup f 1+ τ < . | n| ∞ n Z0 Prove that f is equi-integrable. { n}n

Exercise 6.21. Let f(x) = 1 /√x for x (0 , 1) and f(x) = 0 otherwise. Let rn n be an enumeration of the rationals and set ∈ { } ∞ n g(x) = 2− f(x r ). − n n=1 X Prove that 1. g is summable and in particular g < almost everywhere. ∞ 2. g is discontinuous at every point and unbounded on every interval, and it remains so after any modiﬁcation on a Lebesgue null set. 3. g2 is ﬁnite almost everywhere, but g2 is not summable. 42 CHAPTER 6. EXERCISES Part III

The basic principles of functional analysis

Chapter 7

Metric spaces and normed spaces

We assume the student is familiar with the following preliminary concepts:

Vector spaces • Metric spaces • Normed spaces • Topology in metric spaces • Continuity on metric spaces • Completeness, Banach spaces • Spaces of continuous functions equipped with the uniform norm • Proof that all norms are equivalent in ﬁnite dimensional spaces • Completeness of spaces of continuous functions C(X), C (X), where X is a metric space. • b Banach’s ﬁxed point theorem. • 7.1 Compactness in metric spaces

The concepts of compactness, sequential compactness, and local compactness introduced for general topological spaces also apply to metric spaces. Our first goal is to prove that, on metric spaces, compactness and sequential compactness are equivalent. An additional notion, named total boundedness , is essential for that. Definition 7.1.1 (Totally bounded space) . Let ( X, d ) be a metric space. A subset K X is said to be totally bounded if for any ǫ > 0 there exists finitely many balls of radius ǫ which cover K. ⊂ Theorem 7.1.1. Let (X, d ) be a complete metric space and let K X. Then the following are equivalent: ⊂ (i) K is sequentially compact.

(ii) K is totally bounded and complete.

(iii) K is compact.

Proof. Step 1 . Assume (i) is satisfied, and let us prove (ii). First of all we prove that K is complete. Let xn be a Cauchy sequence in K with a limit x X (by completeness of X). By sequential compactness, all the limit points ∈ of xn are in K, and therefore, the unique limit point x is in K. Now, assume that K is not totally bounded. For a fixed point x K we could find x K such that d(x , x ) > ǫ , then we could find x K with d(x , x ) > ǫ 1 ∈ 2 ∈ 1 2 3 ∈ 1 3 and d(x2, x 3) > ǫ , and for a general n we could define a sequence xn such that d(xn, x m) > ǫ for all n, m N. Clearly, from such a sequence we cannot extract any convergent sub-sequence, which contradicts the assumption∈ of K sequentially compact. Hence, K is totally bounded.

45 46 CHAPTER 7. METRIC SPACES AND NORMED SPACES

Step 2 . Assume (ii) is satisfied, and let us prove (iii). Assume by contradiction that K is not compact, then there exists an open cover ( Eα)α A such that K α A Eα for which no finite sub-family Eα1 , . . . , E αm satisfies m ∈ ⊂ ∈ n K Eα . Now, from (ii) there exists n balls B1, . . . , B n of radius 1 such that K Bj . If Bi were ⊂ j=1 j S ⊂ j=1 coverable with a finite number of sets of the family ( Eα)α A , then the same would be true for K, which contradicts S ∈ S our assumption. Hence, there exists a ball Bj1 which is not covered by a finite sub-family of ( Eα)α A. Let us call X := K B . Now, from (ii) we know that X can be covered by a finite number of balls of radius∈ 1 /2. As 1 ∩ j1 1 above, one of these balls cannot be covered by a finite subfamily of ( Eα)α A. Let us denote such ball by B2 and set X = B K. Let us continue this procedure inductively. We obtain a∈ nested family of subsets of K 2 2 ∩ X X X . . . 1 ⊃ 2 ⊃ 3 ⊃ such that diam( Xi) < 2/i (the diameter of a set in a metric space can be easily defined as the supremum distance between arbitrary pairs of points on that set), and such that none of the Xj can be covered by a finite subfamily of ( Eα)α A. For all i = 1 , 2,... , let us choose xk Xk. This defines a sequence ( xk)k K. For m > n we have x , x ∈X , and hence d(x , x ) diam( X ) < ∈2 . Therefore, the sequence x is a Cauchy⊂ sequence. Since X is n m ∈ n n m ≤ n n n complete, xn converges to some x0 X. Since K is complete, we have x0 K. Now, let x Xk and let n > k . Then x X and the triangular inequality∈ yields ∈ ∈ n ∈ k 2 d(x, x ) d(x, x ) + d(x , x ) + d(x , x ), 0 ≤ n n 0 ≤ k n 0 so that, in the n + limit, → ∞ 2 d(x, x ) , for all x X . 0 ≤ k ∈ k 2 Now, let Eα¯ (Eα)α such that x0 Eα¯. Since Eα¯ is open, let BR(x0) Eα¯. If k > R , then d(x, x 0) < R for all x X , which∈ means X B (x ∈) E . This means that X is covered⊂ by just one set of the family ( E ) , ∈ k k ⊂ R 0 ⊂ α¯ k α α whereas by assumption no Xj is covered by a finite subfamily of ( Eα)α. This proves that K is compact. Step 3. Assume (iii) is satisfied. Let us prove (i). Let xk k be a sequence in K and assume by contradiction that no subsequence of x converges to a point of K. Now,{ } we claim that for all x K there exists an open { k}k ∈ ball Br(x) containing at most a finite number of elements of xk k. This is easily seen as follow. Assume the statement is false; then, for some x K and for all r > 0 there are{ } infinitely many elements of x in B (x). By ∈ { k}k r choosing r = 1 /n , this implies that we can extract a subsequence of xk k which is a Cauchy sequence, and hence converges in K. This contradicts the initial assumption. Now, for every{ }x K let B be the open ball centered in ∈ x x containing at most finitely many xk. The family Bx x K covers K. On the other hand, no finite subfamily of { } ∈ Bx x K covers K, as otherwise K would contain only a finite number of elements of the sequence xk k, which is a{ contradiction.} ∈ { }

The compact sets of the Euclidean space Rd are characterised as those sets which are closed and bounded, according to Heine-Borel’s theorem. This property is valid for all finite dimensional normed spaces. Indeed, this property characterises finite dimensional normed spaces, i. e. if the dimension of the space is infinite such property is no longer true . This is seen more precisely in the next theorem. First we prove the following Lemma. Lemma 7.1.2 (Riesz’s lemma) . Let (E, ) be a normed space. Let M E be a proper closed linear subspace of E. Let α (0 , 1) . Then, there exists a vectork · k x E such that x = 1 ⊂and inf x x , x M > α . ∈ α ∈ k αk {k − αk ∈ }

Proof. For a given y E M, let d = inf x M x y . Since M is closed, d > 0. Indeed, if d = 0 there would be no open balls centered on∈ x entirely\ contained∈ in kE −M,k which would contradict E M being an open set. Hence, since \ d y x0\ α (0 , 1), there exists a vector x0 M with d y x0 . Set xα = − . Clearly xα = 1. Moreover, for ∈ ∈ ≤ k − k ≤ α y x0 k k all x M we have k − k ∈ y x y x x + x y α x x = x − 0 = k − 0k 0 − y x x + x y . k − αk − y x y x ≥ d kk − 0k 0 − k 0 0 k − k k − k

Since y x0 x + x0 M, the quantity y x 0 x + x0 y is bigger than d, and therefore x xα α. k − k ∈ kk − k − k k − k ≥ We are now ready to prove the next important theorem.

Theorem 7.1.3. Let (E, ) be a normed space, and let B1(0) be the closed unit ball of E. Then, B1(0) is compact if and only if the dimensionk·k of E is ﬁnite. 7.1. COMPACTNESS IN METRIC SPACES 47

d Proof. Assume first that E is finite dimensional, let e1,...,e d be a basis for E. Let us set for x = i=1 xiei, d {d } T (x) = ( x1,...,x d) R . The linear map T : E R is a homeomorphism. Then sup x B1(0) x 2 < + , where ∈1/2 → ∈ k k P∞ x = d x 2 . The norm is the Euclidean norm on Rd. Hence, we can apply Heine-Borel theorem k k2 i=1 | i| k · k 2 d on R : since the set T (x) x B1(0) is closed and bounded in the Euclidean norm, Heine-Borel theorem gives that P ∈ { } 1 the set is compact. Since T − is continuous, B1(0) is also compact. Now, assume that B1(0) is compact. Assume by contradiction that E is not finite dimensional. Pick an element x1 E with x1 = 1, and denote by S1 the linear subspace generated by x1. According to Riesz’s lemma 7.1.2, ∈ k k 1 there exists an element x2 E with x2 = 1 and inf x S1 x x2 > . Now let S2 be the linear subspace ∈ k k ∈ k − k 2 generated by x1 and x2. Since E is not finite dimensional, S2 is a proper subspace, and hence there exists x3 E 1 ∈ with x3 = 1 and inf x S2 x x3 > 2 . If we proceed inductively, we can construct a sequence xn with xn = 1 whichk satisfiesk x x∈ >k 1−fork all n = m. Therefore, no convergent subsequence can be extracted from k xk , k n − mk 2 6 { n}n i. e. B1(0) is not compact, a contradiction.

Theorem 7.1.4. Every compact metric space is complete.

Proof. Let x be a Cauchy sequence, let us extract a subsequence x converging to ξ X. We have { n}n nk ∈ d(ξ, x ) d(ξ, x ) + d(x , x ). n ≤ nk nk n Hence, for a ﬁxed ǫ > 0, d(x , x ) < ǫ if n and n are larger than a suitable n . Therefore, for n n we have nk n k ǫ ≥ ǫ d(ξ, x n) < 2ǫ, which implies the assertion.

Exercise 7.1.1. Prove that every compact space X is bounded. Solution. For a ﬁxed R > 0, the family of open sets BR(x) x X covers the space X, therefore there exists a ﬁnite m { } ∈ number of points x1,...,x m such that X i=1 BR(xi). The last property implies in particular that X is contained in a large enough open ball, and it is therefore⊂ bounded. S Exercise 7.1.2. Prove that a continuous function on a compact metric space is uniformly continuous.

We shall often use the following terminology.

Deﬁnition 7.1.2. A subset K in a metric space ( X, d ) is pre-compact if it has compact closure, i. e. if K is compact.

Deﬁnition 7.1.3. A subset D in a metric space ( X, d ) is dense in X if D = X. Moreover, if D is countable, then X is separable.

Exercise 7.1.3. Show that a compact metric space is separable.

Discussion 7.1.1 . The result in Theorem 7.1.3 marks a basic difference between finite and infinite dimensional normed spaces: closed and bounded sets are compact in finite dimensions, whereas they are not in infinite dimensions. This poses a big problem for instance with function spaces, which typically have infinite dimension. Often (e. g. in calculus of variations or in the theory of partial differential equations), the solution to a given problem is computed by approximation and by compactness . More precisely, a family of approximations of the solution is constructed, and then one hopes to extract a converging sequence from it, which should converge to the actual solution of the problem. If we would be in a finite dimensional metric space ( X, ), a uniform control of the norm for the approximating family would be enough to extract a convergent subksequence. · k As we saw above, this is kno · k longer true in infinite dimension. However, one can recover the compactness of the sequence by requiring some additional property on the family (further to a mere uniform control of the norm). The next theorem constitutes the first example of compactness result for functions spaces, and is therefore of paramount importance in analysis.

Theorem 7.1.5 (Arzel´a-Ascoli) . Let (X, d ) be a compact metric space, and let be a subset of the Banach space F C(X) of functions f : X R equipped with the norm f = sup x X f(x) . The set is pre-compact if and only if the following two→ conditions hold: k · k ∞ k k∞ ∈ | | F

(i) is uniformly bounded, i. e. there exists M > 0 such that f M for all f . F k k∞ ≤ ∈ F (ii) is equi-continuous, i. e. for all ǫ > 0 there exists δ > 0 such that d(x, y ) < δ implies f(x) f(y) < ǫ for Fall f . | − | ∈ F 48 CHAPTER 7. METRIC SPACES AND NORMED SPACES

Proof. Step 1. Assume first that is compact. Exercise 7.1.1 shows that is bounded, which gives (i). Moreover, let ǫ > 0. By Theorem 7.1.1, is totallyF bounded. Hence, we can cover Fwith finitely many balls with radius ǫ/ 3. F F Let f1,...,f N be the centers of those balls. By the metric version of Heine–Cantor theorem (see exercise 7.1.2), the functions f are uniformly continuous. Therefore, there exists δ > 0 such that d(x, y ) < δ implies f (x) f (y) < ǫ i | i − i | 3 for all i = 1 , . . . , N . Now, let f , we have f B ǫ (f ) for some k 1, . . . , N . Hence, ∈ F ∈ 3 k ∈ { } ǫ ǫ ǫ f(x) f(y) f(x) f (x) + f (x) f (y) + f (y) f(y) + + = ǫ, | − | ≤ | − k | | k − k | | k − | ≤ 3 3 3 which yields (ii). Step 2. Assume (i) and (ii) are satisfied. We want to show that is pre-compact. Let fn n N be a F { } ∈ ⊂ F sequence in . It suffices to prove that fn has a subsequence which is a Cauchy sequence. This will then imply that the subsequenceF will converge to some f C(X), as C(X) is a complete metric space. Let ǫ > 0, and let δ > 0 ∈ given as in assumption (ii). Since X is compact, we can cover X with finitely many balls Bδ(x1), . . . , B δ(xN ) with radius δ (X is totally bounded). Hence, for a given x X, x Bδ(xj) for some j 1, . . . , N . Hence, we can use (ii) and get ∈ ∈ ∈ { }

f (x) f (x) f (x) f (x ) + f (x ) f (x ) + f (x ) f (x) 2ǫ + f (x ) f (x ) . | n − m | ≤ | n − n j | | n j − m j | | m j − m | ≤ | n j − m j | Now, the sequence v := ( f (x ),...,f (x )) RN is bounded in view of (i). Therefore, in view of Heine-Borel n n 1 n N ∈ theorem there exists a convergent subsequence vnk . This means that there exists M N such that for all h, k M we have ∈ ≥ f (x ) f (x ) < ǫ, for all j 1, . . . , N . | nh j − nk j | ∈ { }

Therefore, fnh (x) fnk (x) < 3ǫ for all x X. We have therefore shown that, for a given ǫ > 0 we can extract a | (1) − | ∈ subsequence fn such that { } (1) (1) fn fm 3ǫ, for all n, m Mǫ, k − k∞ ≤ ≥ (1) for a suitable Mǫ N. Now, ﬁx ǫ = 1 and extract the subsequence fn as above. Then ﬁxe ǫ = 1 /2 and extract (1) ∈ (2) { } (n) from fn another subsequence fn , and so on. The diagonal sequence fn will satisfy { } { } { }

(n) (m) 3 fn fm k − k∞ ≤ k

(n) for n, m M for some suitable M depending on k. Hence, fn is a Cauchy sequence, and the assertion follows. ≥ k k 1 Example 7.1.1 . A very useful application of Theorem 7.1.5 is the following. Let fn : [ a, b ] R be a function of C functions such that →

(i) fn C for all n N, k k∞ ≤ ∈

(ii) fn′ C for all n N. k k∞ ≤ ∈

Then, the sequence fn admits a convergent subsequence in the space ([ a, b ], ). k · k ∞ To see this, we observe that for all n N we have ∈ y y f (x) f (y) = f ′ (z)dz f ′ (z) dz C x y , | n − n | n ≤ | n | ≤ | − | Zx Zx and this implies that for a given ǫ > 0 one has f (x) f (y) < ǫ provided x y ǫ , which means that the | n − n | | − | ≤ C sequence fn is equi-continuous.

7.2 Introduction to Lp spaces

In this chapter we introduce one of the main classes of Banach spaces used in functional analysis, i. e. the Lp spaces. They are constructed as function spaces on Rd, and their theory makes use of the Lebesgue measure-integration theory developed before in the course. The theory we develop in this chapter will be deﬁned for functions on Rd with values on R, but everything can be easily generalised to the case of functions with values on C. 7.2. INTRODUCTION TO LP SPACES 49

Definition 7.2.1. Let p [1 , + ), and let E Rn be a Lebesgue measurable set. For a measurable function f : E R we define the L∈p norm∞ of f on Ω as the⊂ (finite or infinite) number → 1/p . p f p = f(x) dx . k kL (E) | | ZE Moreover, we set . C = α R : f(x) α almost everywhere on Ω . { ∈ | | ≤ } The L∞ norm of f (also called the essential supremum of f) is defined as

f ∞ = inf C. k kL (Ω) The essential supremum of f is the minimum essential upper bound for f , namely the minimum α such that f α almost everywhere. It is| easily| seen that, in general | | | | ≤

f L∞ sup f(x) , k k ≤ x E | | ∈ and very simple examples can be constructed in which the strict inequality holds above (basically we get a strict inequality anytime the f achieve its supremum f on a set of measure zero, and it is bounded above by a value strictly less than f elsewhere).

d Exercise 7.2.1. Prove that if f : R R is continuous then f L∞(Rd) = sup x Rd f(x) . → k k ∈ | | Remark 7.2.1 . A simple consequence of the above deﬁnition is that

f(x) f ∞ almost everywhere on E. | | ≤ k kL (E) To see this, for all k N let ∈ 1 C = x E : f(x) f ∞ + . k { ∈ | | ≤ k kL k } 1 Clearly, m(E C ) = 0 for all k N, because, for all k N, the value f ∞ + is an upper bound for f almost \ k ∈ ∈ k kL k | | everywhere. Hence, m k N(E Ck) = 0, and ∈ \ S (E C ) = x E : f(x) > f ∞ . \ k { ∈ | | k kL } k N [∈ p Clearly, we would like to define a norm by means of the L and L∞ norms. The problem is that, in general, f p = 0 does not imply f 0, which is one of the axioms of a norm. Indeed, for all p 1 the statement k kL ≡ ≥ f Lp(E) = 0 only implies that f(x) = 0 almost everywhere on E, but f could still be nonzero on a set of null measure.k k We will see in a few pages| how| 6 to bypass this problem. Definition 7.2.2. Let p [1 , + ]. The conjugate of p is the extended real number q [1 , + ] such that ∈ ∞ ∈ ∞ 1 1 + = 1 , p q with the convention that 1 / = 0. ∞ Theorem 7.2.1 (Hölder inequality) . Let f, g : E R be measurable functions, and let p, q [1 , + ] be conjugate. Then, → ∈ ∞ fg 1 f p g q . k kL (E) ≤ k kL (E)k kL (E) Proof. If p = 1 and q = + , we have ∞

fg 1 = f(x)g(x) dx g ∞ f(x) dx = g ∞ f 1 , k kL (E) | | ≤ k kL | | k kL k kL ZE Z where the first inequality is justified by the fact that g can be redefined on a set of measure zero in a way that g g ∞ everywhere, and this does not affect the integral. ≤ k kL 50 CHAPTER 7. METRIC SPACES AND NORMED SPACES

In the general case p > 1, the statement is trivial if either f or g are zero almost everywhere. Otherwise, we clearly have f p > 0 and g q > 0. For a ﬁxed α > 0 we have k kL k kL f(x) 1 f(x) p 1 f(x)g(x) = αg (x) + αg (x) q , | | α | | ≤ p α q | |

where we have used Young’s inequality (Exercise 7.3.1). By integrati ng the above inequality on E we get

1 1 p 1 q q fg 1 f p + α g q . k kL (E) ≤ p αp k kL (E) q k kL (E) We now choose α such that the two terms on the above right hand side are equal, namely

1 q f Lp α := k k 1 , p g q k kL which yields 1 g Lq p 1 f Lp q 1 fg L (E) k kp/q f Lp + k kq/p g Lq , k k ≤ p f p k k q g q k k k kL k kL and the deﬁnition of p and q implies the last term above equals f p g q . k kL k kL Remark 7.2.2 . H¨older inequality can be also rephrased as follows:

α β f(x) α g(x) βdx f(x) dx g(x) dx , | | | | ≤ | | | | ZE ZE ZE provided α + β = 1. Theorem 7.2.2 (Minkowski’s inequality) . Let f, g : E R be measurable functions. Let p [1 , + ]. Then, → ∈ ∞

f + g p f p + g p . k kL (E) ≤ k kL (E) k kL (E) Proof. The case p = + is trivial, since for all x E one has ∞ ∈ f(x) + g(x) f(x) + g(x) , | | ≤ | | | | and the right hand side is controlled by f L∞ + g L∞ almost everywhere on E. This implies the assertion. The case p = 1 is straightforward. k k k k Let p (1 , + ). The statement is trivially satisfied if either f Lp or g Lp equals + . Therefore, assume the are both∈ finite. First∞ of all we notice that f(x) ( f(x) p + g(kx)kp)1/p andk k g(x) ( f∞(x) p + g(x) p)1/p , which implies | | ≤ | | | | | | ≤ | | | | f(x) + g(x) p ( f(x) + g(x) )p 2p( f(x) p + g(x) p), | | ≤ | | | | ≤ | | | | p and integrating on E gives that f(x) + g(x) dx is finite, hence f + g p < + . We observe that | | k kL ∞ p p 1 p 1 p 1 f(x) + g(x) f(x) + g(xR) − ( f(x) + g(x) ) f(x) + g(x) − f(x) + f(x) + g(x) − g(x) . (7.1) | | ≤ | | | | | | ≤ | | | | | | | | Integrating on E yields, via the Hölder inequality,

p−1 1 p p p 1 p p p 1 f(x) + g(x) − f(x) dx f(x) + g(x) dx f(x) dx = f + g −p f p . | | | | ≤ | | | | k kL k kL ZE ZE ZE By performing the same manipulation on the last term in (7.1), we get

p p p 1 f(x) + g(x) dx = f + g p f + g −p ( f p + g p ) , | | k kL ≤ k kL k kL k kL ZE which proves the assertion.

Now, clearly the Lp norm veriﬁes 7.2. INTRODUCTION TO LP SPACES 51

f p 0, • k kL ≥ λf p = λ f p , • k kL | |k kL f + g p f p + g p , • k kL ≤ k kL k kL in the class of measurable functions on which the above quantities are finite. But this is not enough to make p k · k L a norm on such space, because in general f p = 0 does not imply f 0. k kL ≡ Definition 7.2.3. Let E Rd be a measurable set and p [1 , + ]. We call p(E) the set of measurable functions ⊂ ∈ ∞ L f : E R such that f Lp < + . Such set is a vector space with the usual operations of sum between functions and multiplication→ byk ak scalar (because∞ of Minkowski’s inequality). Now, we set the following relation on p(E). Let f, g p(E). We say that f g if f(x) = g(x) for almost every x E. We set L ∈ L ∼ ∈ Lp(E) = p(E)/ , L ∼ i. e. Lp(E) is the quotient (vector) space of p(E) through the relation 1. For a given equivalence class p L ∼ [f] L (E), we define the norm of [ f] as f Lp(E) for an arbitrary representant f [f]. From now on, by abuse of notation,∈ we shall confuse [ f] and its representantk k f. The space Lp(E) is the Lp space∈ on E. Exercise 7.2.2. Prove that the norm of f Lp(E) is well defined. ∈ Remark 7.2.3 . Clearly, the Lp norm on Lp(E) is now a norm, since all the ‘good’ properties proven above are easily inherited by the norm on the quotient space, and furthermore one has f Lp = 0 implies f = 0 almost everywhere, hence [ f] = 0. Therefore, Lp(E) is a normed space. k k p p p We shall say that a sequence fn L (E) converges in L to f L if fn f Lp 0 as n + . Lp spaces can be proven to be Banach∈ spaces. This is the content∈ of thek ne−xtk theorem.→ → ∞ Theorem 7.2.3 (Riesz-Fisher theorem) . Let E Rd be a measurable set and p [1 , + ]. Then the space Lp(E) is ⊂ ∈p ∞ Banach space. Moreover, if p [1 , + ) and fn n N is a Cauchy sequence in L (E), then there exist two functions f, h Lp(E) and a subsequence∈ f ∞of f such{ that} ∈ ∈ nk n (a) f (x) h(x) almost everywhere on E, | nk | ≤ (b) f f almost everywhere on E. nk → Proof. Consider first the case p = + . Let fn be a Cauchy sequence in L∞. This means that for all k N there ∞ 1 ∈ exists n(k) N such that, for all n, m n(k) we have f f ∞ < , i. e. ∈ ≥ k n − mkL k 1 f (x) f (x) < , for all n, m > n (k) and for all x V with m(V ) = 0 . | n − m | k 6∈ n,m,k n,m,k

Let V = n,m,k Vn,m,k . Clearly m(V ) = 0. We set S ˜ fn(x) = fn(x)1E V . \ Therefore, f˜ (x) f˜ (x) < 1 if n, m n(k) for all x E, i. e. f˜ (x) is a Cauchy sequence in R. Let | n − m | k ≥ ∈ n

f(x) = lim fñ(x). n + → ∞ By sending m + we get f˜ (x) f(x) < 1 for all x E, and hence → ∞ | n − | k ∈ 1 f˜ f ∞ < for all n n(k), k n − kL k ≥ i. e. fñ f in L∞. By definition of L∞ norm this clearly implies fn f in L∞. Let us→ now consider the case p [1 , + ). First of all, we claim that→ if (a) and (b) holds for some subsequence ∈ ∞ p fnk , for some h, and for some f, we have that L (E) is a complete space. To see this, let fn n be a Cauchy sequence in Lp. As a consequence of (a) and (b), we get { }

0 f (x) f(x) p 2p( h(x) p + f(x) p), ≤ | nk − | ≤ | | | | 1All the vector space operations on Lp(E) are inherited by the quotient space by considering operations between representant. For instance, given [ f], [g] ∈ Lp, [ f] + [ g] is the equivalence class of [ f + g]. Prove that such operation is well deﬁned. Deﬁne similarly λ[f] for some λ ∈ R. 52 CHAPTER 7. METRIC SPACES AND NORMED SPACES with f f almost everywhere. From Lebesgue’s dominated convergence theorem 5.1.4, we have nk →

p lim fnk (x) f(x) dx = 0 , k + | − | → ∞ ZE but then f f p f f p + f f p 0, k n − kL ≤ k n − nk kL k nk − kL → which proves that Lp is complete. p p Hence, we only have to prove (a) and (b). Let fn n L (E) be a Cauchy sequence in L . It is easily seen that for all k N there exists an integer n such that { } ∈ ∈ k 1 f f p . k nk − nk+1 kL ≤ 2k+1

For simplicity in the notation we shall write fk = fnk . We set

n g (x) = f (x) f (x) . n | k+1 − k | kX=1 Clearly, n n 1 g p f f p 1. k nkL ≤ k k+1 − kkL ≤ 2k+1 ≤ kX=1 kX=1 Clearly, gn is monotone increasing with respect to n. Therefore we can use Beppo Levi’s theorem 5.1.1, which implies p p g(x) dx = lim gn(x) dx 1, n + ≤ ZE → ∞ ZE where + ∞ g(x) = f (x) f (x) , | k+1 − k | kX=1 hence g p 1 and g(x) < + almost everywhere in E. Now, for all n > m N we have k kL ≤ ∞ ∈

fn(x) fm(x) fn(x) fn 1(x) + . . . + fm+1 (x) fm(x) = gn 1(x) gm(x) g(x) gm(x). | − | ≤ | − − | | − | − − ≤ − Now, since g g almost everywhere, for ǫ > 0 there exists N N such that for all n, m N n → ǫ ∈ ≥ ǫ f (x) f (x) g(x) g (x) < ǫ. | n − m | ≤ − m

Therefore, fn(x) n is Cauchy in R, and hence fn(x) f(x) almost everywhere for some measurable function f. Moreover, { } → f(x) f (x) g(x) g (x) almost everywhere , | − m | ≤ − m which implies f(x) f(x) f (x) + f (x) g(x) + f (x) , | | ≤ | − m | | m | ≤ | m | and this proves that f Lp(E) and (b) is proven. Moreover, ∈ f (x) f(x) + f (x) f(x) g(x) + f(x) , | m | ≤ | | | m − | ≤ | | and (a) follows by setting h(x) := g(x) + f(x) . Clearly h Lp(E). The proof is complete. | | ∈ 7.3 Examples: some classical spaces of sequences

Deﬁnition 7.3.1 (ℓp spaces) . Let p [1 , + ) be a real number. We say that a sequence x = xk k N of real ∈ ∞ { } ∈ numbers is in ℓp if + ∞ x p < + . | k| ∞ kX=1 7.3. EXAMPLES: SOME CLASSICAL SPACES OF SEQUENCES 53

The space of real sequences with the above property is called ℓ . For all x ℓ , the quantity p ∈ p + 1/p ∞ p x ℓp := xk k k # | | % kX=1 is called the ℓp-norm of x. The space ℓ is the space of bounded sequences, i. e. ∞

sup xk < + . k N | | ∞ ∈ For all x ℓ , the quantity ∈ ∞ x ℓ∞ := sup xk k k k N | | ∈ is called the ℓ -norm of x. ∞ The space ℓp can be seen as a subset of the vector space of all sequences of real numbers, with the obvious operations x = x ℓ , y = y ℓ , x + y = x + y • { k}k ∈ p { k}k ∈ p { k k}k x = x ℓ , λ R, λx = λx . • { k}k ∈ p ∈ { k}k Proving that the sum between two vectors in well defined, as well as proving that x is an actual norm, is k kℓp not immediate. The first two properties of a norm (i. e. x ℓp = 0 implies x = 0, and λx ℓp = λ x ℓp ) are trivial. For the third one, i. e. the triangular inequality, wek havek to struggle a bit more. Oncek k we have| |k that,k the sum between vectors will also be well defined, and we shall have a nice family of normed spaces to work with.

Exercise 7.3.1 (Young’s inequality) . Let p [1 , + ) and let p′ be its conjugate. Let a, b 0 be two positive numbers. Then, ∈ ∞ ≥ ′ ap bp ab + . ≤ p p′ ′ Solution. If ab = 0 there is nothing to prove. Assume a > b > 0. Set A = ap and B = bp . We need to prove that ′ A B A1/p B1/p + . ≤ p p′ Multiplication by 1/B makes the above inequality equivalent to

1 A 1 A 1/p + p B p ≥ B ′ 1 1 A where we have used ′ 1 = . Now, set t = 1. The above becomes equivalent to proving that p − − p B ≥ 1 1 t + t1/p for all t 1. p p′ ≥ ≥

1 1 1 1/p 1 1 p 1 But the function φ(t) = t + ′ t satisﬁes φ(1) = 0 , φ′(t) = t − , which is 0 for t 1. Therefore, p p − p − p ≥ ≥ φ(t) 0 for all t 1, which proves the assertion. ≥ ≥ Exercise 7.3.2 (Discrete H¨older’s inequality) . Let x = xk k ℓp and y = yk k ℓp′ with p, p ′ [1 , + ) conjugate numbers. Prove that { } ∈ { } ∈ ∈ ∞ + ∞ x y x y ′ . | k|| k| ≤ k kℓp k kℓp kX=1 Solution. Set xk yk Xk = , Y k = , for all k 1. x y ′ ≥ k kℓp k kℓp ′ p p + Xk Yk ∞ ′ We need to prove that k=1 Xk Yk 1. From Young’s inequality, Xk Yk | p | + | p| for all k 1, and taking the sum over k we get | || | ≤ | || | ≤ ≥ P + + + ′ ∞ 1 ∞ p 1 ∞ p 1 1 Xk Yk Xk + Yk = + = 1 . | || | ≤ p | | p′ | | p p′ kX=1 kX=1 kX=1 54 CHAPTER 7. METRIC SPACES AND NORMED SPACES

We are now ready to prove the triangular inequality on ℓp.

Exercise 7.3.3 (Discrete Minkowski’s inequality) . Let x, y ℓp [1 , + ]. Prove that x + y ℓp and x + y ℓp x + y . ∈ ∈ ∞ ∈ k k ≤ k kℓp k kℓp Solution. For p < + , compute ∞ p p 1 x + y = x + y x + y − | k k| | k k|| k k| k 1 k 1 X≥ X≥ p 1 p 1 x x + y − + y x + y − , ≤ | k|| k k| | k|| k k| k 1 k 1 X≥ X≥ p where we have used the obvious inequality xk + yk xk + yk . Now, since p′ = p 1 is conjugate of p, the above discrete H¨older’s inequality implies | | ≤ | | | | −

p−1 p−1 1/p p 1/p p x + y p x p x + y p + y p x + y p , | k k| ≤  | k|   | k k|   | k|   | k k|  k 1 k 1 k 1 k 1 k 1 X≥ X≥ X≥ X≥ X≥         which yields

− 1 p 1 1/p 1/p − p x + y p x p + y p ,  | k k|  ≤  | k|   | k|  k 1 k 1 k 1 X≥ X≥ X≥       which proves the assertion. The case p = + is a trivial exercise. ∞ Exercise 7.3.4 (Completeness of the ℓp spaces) . Let p [1 , + ]. Let xn = xn,k k be a Cauchy sequence in ℓp. Then, x x as n + in for some x ℓ . ∈ ∞ { } n → → ∞ k · k ℓp ∈ p Solution. Let us ﬁrst consider the case p = 1. Since

x x 0 as n, m + , | n,k − m,k | → → ∞ k 1 X≥ for all k 1 we have that xn,k n is a Cauchy sequence in R, and hence there exists some xk R such that x x≥as n + . We{ need} to prove that x = x ℓ and that x x 0 as n +∈ . Let ǫ > 0. n,k → k → ∞ { k} ∈ 1 k n − kℓ1 → → ∞ The Cauchy condition on the sequence xn reads

Since Nǫ does not depend on K, we can take the supremum with respect to K above and get

xn,k xk = sup xn,k xk < ǫ for n Nǫ, | − | K N | − | ≥ k 1 ∈ k=1 X≥ X 7.3. EXAMPLES: SOME CLASSICAL SPACES OF SEQUENCES 55 which shows that x x 0 as n + . By triangular inequality, we then get k n − kℓ1 → → ∞ x x x + x ǫ + x , k kℓ1 ≤ k − Nǫ kℓ1 k Nǫ kℓ1 ≤ k Nǫ kℓ1 and the last term above is ﬁnite. Now, let p (1 , + ). Assume x ℓ is a Cauchy sequence. Hence, as above we can easily show that ∈ ∞ { n}n ∈ p there exists a sequence xk k of real numbers such that xn,k xk as n + for all k 1. To prove that x = x ℓ , for a given{ ǫ}let N , k N such that, for all n, m → N , → ∞ ≥ { k}k ∈ p ǫ ǫ ∈ ≥ ǫ x x p < ǫ, for m n N . (7.2) | n,k − m,k | ≥ ≥ ǫ k 1 X≥ As before, for all K N we have ∈ K x x p < ǫ for m n N , | n,k − m,k | ≥ ≥ ǫ kX=1 and the assertion xn x ℓp 0 follows similarly as in the case p = 1. The triangular inequality then proves once again that x ℓ .k − k → ∈ p Finally, let us consider the case p = + . Assume xn n ℓ is a Cauchy sequence. For a given ǫ > 0, there exists N N such that, for all n, m N ∞one has { } ∈ ∞ ǫ ∈ ≥ ǫ

sup xn,k xm,k < ǫ. k N | − | ∈

This implies that each sequence xn,k n (for all k 1) converges in n to some xk R. Let x = xk k. We can set m > n and send m + {above} and get x ≥ x ǫ for all n N . This∈ holds for all{ k } 1, hence → ∞ | n,k − k| ≤ ≥ ǫ ≥ xn x ℓ∞ ǫ for all n Nǫ. Moreover, x ℓ∞ xNǫ x ℓ∞ + xNǫ ℓ∞ , which proves that x ℓ k − k ≤ ≥ k k ≤ k − k k k ∈ ∞ 56 CHAPTER 7. METRIC SPACES AND NORMED SPACES Chapter 8

Linear operators

In this section we explore more in detail the relation between linearity and continuity of an operator between two normed spaces. We have already proved that a linear operator between two ﬁnite dimensional spaces is continuous. In this chapter (as in the whole part) we won’t impose any restriction on the dimension.

Proposition 8.0.1. Let (X, X ) and (Y, Y ) be two normed spaces. Let T : X Y be a linear map. Then, the following are equivalent: k · k k · k → (i) T is continuous at x = 0 X. ∈ (ii) T is continuous on all points x X. ∈ (iii) T maps bounded sets of X into bounded sets of Y . (iv) There exists M > 0 such that T (x) M x . (8.1) k kY ≤ k kX

Proof. (i) implies (ii): Assume T is continuous at 0 X. Let x X. Let xn n be a sequence in X which converges to x as n + . Consider ∈ ∈ { } → ∞ T (x ) T (x) = T (x x) 0, k n − kY k n − kY → because x x converges to zero and T is continuous at zero. n − (ii) implies (iii): Let A X be a bounded set. This means A BR(0) for some R 0. We claim there exists S 0 such that T (A) B⊂(0) Y . Assume by contradiction that⊆ for every n N there≥ exists x A such that ≥ ⊆ S ⊂ ∈ n ∈ T (x ) n. Set v := xn . Since x R, then v 0 as n + . On the other hand, k n kY ≥ n n k nkX ≤ n → → ∞ T (x ) 1 T (v ) = n = T (x ) 1. k n kY n nk n kY ≥ Y

This is a contradiction with the fact that T is continuous at zero. (iii) implies (iv): By contradiction, assume that for all n N there exists x X with T (x ) n x . ∈ n ∈ k n kY ≥ k nkX Set y = xn . Clearly y = 1. Moreover, n xn X n X k k k k 1 T (y ) = T (x ) n, k n kY x k n kY ≥ k nkX which implies the set T ( y ) is unbounded, whereas y is bounded, a contradiction. { n}n { n}n (iv) implies (i): Let xn n converge to zero as n + . Then, the condition T (x) Y M x X easily implies T (x ) 0, which gives{ } the continuity of T at zero.→ ∞ k k ≤ k k k n kY → Remark 8.0.1 . Since the translation X x x+x0 E is clearly a continuous function, the condition of the linear operator T being continuous at zero in∋ (i) above7→ is equivalent∈ to state that T is continuous at one point x E. 0 ∈ Definition 8.0.1. Let X and Y be normed spaces. A linear operator T : X Y is called a bounded operator if it is continuous. The (operator) norm of T is the number → T (x) T = sup k kY , k k x=0 x X 6 k k 57 58 CHAPTER 8. LINEAR OPERATORS i. e. T is the infimum of all M such that condition (8.1) is satisfied. The space of all bounded linear operators from kX kto Y is denoted by (X, Y ). L Exercise 8.0.1. With the notation in the previous definition, prove that T (x) T x . k kY ≤ k kk kX Exercise 8.0.2. Prove that the operator norm is a norm on the linear space (X, Y ). k · k L Proposition 8.0.2. Let X be a normed space and Y be a Banach space. Then the space (X, Y ) is a Banach space L

Proof. We only need to prove that (X, Y ) is complete. Let T be a Cauchy sequence on (X, Y ) with respect L n L to the operator norm . This means that, for all ǫ > 0, there exists an N(ǫ) such that Tn Tm ǫ for all n, m N(ǫ). Hence, fork · all k x X, k − k ≤ ≥ ∈ T (x) T (x) = (T T )( x) T T x ǫ x . k n − m kY k n − m kY ≤ k n − mkk kX ≤ k kX

The above shows that the sequence Tn(x) is a Cauchy sequence in Y , which is complete, i. e. there exists an element y Y such that Tn(x) y. Since y depends on x via T , we name y = T (x). It is easily shown that T is a linear operator.∈ Indeed, for x ,→ x X, we have 1 2 ∈

T (x1 + x2) = lim Tn(x1 + x2) = lim (Tn(x1) + Tn(x2)) = T (x1) + T (x2). n + n + → ∞ → ∞ Moreover, T is a bounded operator. To see this, let ǫ > 0. Then, one easily sees that there exists N N such that T T ǫ (as a consequence of the Cauchy condition on T ). Therefore, ∈ k − N k ≤ n T (x) = (T T )( x) + T (x) T T x + T x ǫ x + M x , k kY k − N kY k N kY ≤ k − N kk kX k N kk kX ≤ k kX N k kX where M = T . This implies T (x) (ǫ + M ) x , i. e. T is a bounded operator. N k N k k kY ≤ N k kX Deﬁnition 8.0.2. We say that a sequence Tn (X, Y ) is norm-convergent to T (X, Y ) if Tn T 0 as n + . We say that T converges to T pointwise∈ L if T (x) T (x) 0 as n ∈+ L for all xk X−. k → → ∞ n k n − kY → → ∞ ∈ Chapter 9

Hahn-Banach Theorems

9.1 The analytic form of the Hahn-Banach theorem: extension of linear functionals

The main result in this section concerns the extension of a linear functional defined on a linear subspace of E by a linear functional defined on all of E. Theorem 9.1.1 (Hahn-Banach analytic form) . Let p : E R be a function satisfying → (i) p(λx ) = λp (x) for all x E and all λ > 0, ∈ (ii) p(x + y) p(x) + p(y) for all x, y E. ≤ ∈ Let G E be a linear subspace of E and let g : G R be a linear functional defined on G such that ⊂ → g(x) p(x) for all x G. ≤ ∈ Then, there exists a linear functional f defined on all of E that extends g, i. e. g(x) = f(x) for all x G, and such that f(x) p(x) for all x E. ∈ ≤ ∈ The proof uses Zorn’s lemma, i. e. Theorem 1.1.2. Proof. Consider the set

P = h : D(h) E R : D(h) is a linear subspace of E, h is linear, G D(h), { ⊂ → ⊂ h extends g, and h(x) p(x) for all x D(h) . ≤ ∈ } On P we deﬁne the order relation

h h D(h ) D(h ) and h extends h . 1 ≺ 2 ⇔ 1 ⊂ 2 2 1 It is clear that P is non empty, since g P . We claim that P is inductive. Let Q P be a chain. We write ∈ ⊂ Q = ( hi)i I for some set of indexes I. Let us set ∈ D(h) = D(h ), h (x) = h (x) if x D(h ) for some i I. i i ∈ i ∈ i I [∈ First of all let us check that the deﬁnition of h is well posed. Let x D(h ) D(h ) for some i, j I, i = j. ∈ i ∩ j ∈ 6 Then either D(hi) D(hj) or D(hi) D(hj). In either cases hi extends hj or viceversa and therefore h(x) = h (x) = h (x). It is⊂ easily checked that ⊃h P . Indeed, since all h extend g so does h, and moreover h is linear and i j ∈ i satisﬁes the estimate h(x) p(x) (the same estimate is true for all hi). Finally, one easily sees that hi h for all i I. Therefore, h is an upper≤ bound for the chain Q. Hence, Zorn’s lemma (Theorem 1.1.2) implies that≺ P has a maximal∈ element, that we call f P . The proof will be completed once we prove that D(f) = E. Assume by contradiction that∈D(f) = E. Let x D(f). Set D(h) = D(f) + Rx , and for every y D(h), 6 0 6∈ 0 ∈ y = x + tx 0, set h(y) = f(x) + tα , where the constant α R will be chosen in such a way that h P . We must ensure that ∈ ∈ f(x) + tα p(x + tx ), for all x D(f) and for all t R. ≤ 0 ∈ ∈ 59 60 CHAPTER 9. HAHN-BANACH THEOREMS

In view of (i), it suﬃces to check that

f(x) + α p(x + x ) for all x D(f) ≤ 0 ∈ (9.1) f(x) α p(x x ) for all x D(f). ( − ≤ − 0 ∈ Indeed, if e. g. the first inequality above is satisfied for all x D(f), since f is linear, the two above inequalities in (9.1) yield for all t 0 ∈ ≥ x x f(x) + tα = t f + α tp + x = p(x + tx ) t ≤ t 0 0 as desired, and for t 0 ≤ x x f(x) + tα = t f α t p x = p(x + tx ), | | t − ≤ | | t − 0 0 | | | | as desired once again. Now, in order to find α as in (9.1), we need to find some α R satisfying ∈

sup f(y) p(y x0) α inf p(x + x0) f(x) . y D(f){ − − } ≤ ≤ x D(f){ − } ∈ ∈ Now, the property (ii) implies

f(x) + f(y) = f(x + y) p(x + y) p(x + x ) + p(y x ), ≤ ≤ 0 − 0 and therefore f(y) p(y x ) p(x + x ) f(x), − − 0 ≤ 0 − for all y, x D(f), which shows it is possible to ﬁnd α as desired. Therefore, f h, and this contradicts the fact that f is a∈ maximal element. Therefore D(f) = E and the proof is complete. ≺ The above theorem works on an arbitrary vector space. The reference to a function p, however, suggests that the theorem can be applied to normed vector spaces.

Deﬁnition 9.1.1. Let ( E, ) be a normed space. We denote by E∗ the dual space of E, that is, the space of all k · k continuous linear functionals on E. The (dual) norm on E∗ is deﬁned by

f E∗ = sup f(x) . k k x 1, x E | | k k≤ ∈

When there is no confusion we shall also write f instead of f ∗ . Given f E∗ and x E, we shall often write k k k kE ∈ ∈ f, x instead of f(x). We often call the operation , the duality product on E∗ E. h i h· ·i × As a consequence of the fact that R is a Banach space, and due to Proposition 8.0.2, E∗ is a Banach space even if E is not complete. Corollary 9.1.2. Let E be a normed space. Let G E be a linear subspace. If g : G R is a continuous linear ⊂ → functional, then there exists f E∗ that extends g and such that ∈

f E∗ = sup g(x) = g G∗ . k k x G, x 1 | | k k ∈ k k≤ Proof. Use Theorem 9.1.1 with p(x) = g ∗ x . k kG k k Corollary 9.1.3. Let E be a normed space. For every x E there exists f E∗ such that 0 ∈ 0 ∈ f = x , and f , x = x 2. k 0k k 0k h 0 0i k 0k Proof. Use Corollary 9.1.2 with G = tx : t R and g(tx ) = t x 2, so that g ∗ = x . { 0 ∈ } 0 k 0k k kG k 0k In general, the element f0 E∗ in Corollary 9.1.3 is not unique for a given x0 E. It is unique in special cases, ∈ p ∈ e. g. E a Hilbert space or an L space p (1 , + ), as we shall see later on. In general, for an element x0 E we set ∈ ∞ ∈ 2 F (x ) = f E∗ : f = x and f , x = x . 0 { 0 ∈ k 0k k 0k h 0 0i k 0k } The multi-valued map E x F (x ) (E∗) is called the duality map from E into E∗. ∋ 0 7→ 0 ∈ P 9.2. THE GEOMETRIC FORMS OF THE HAHN-BANACH THEOREM 61

Corollary 9.1.4. Let x E. If for all f E∗ one have f(x ) = 0 then x = 0 . 0 ∈ ∈ 0 0 Proof. Assume by contradiction that x0 = 0. Then, by Corollary 9.1.3 there exists f0 E∗ with f0 = 0 such that f = x and f , x = x 2 = 0, and6 this is a contradiction. ∈ 6 k 0k k 0k h 0 0i k 0k 6 Corollary 9.1.5. Let E be a normed space. For every x E we have ∈

x = sup f, x = max∗ f, x . k k f E∗, f 1 |h i| f E , f 1 |h i| ∈ k k≤ ∈ k k≤ Proof. The assertion is trivial if x = 0. If x = 0, it is clear that 6 sup f, x x . f E∗, f 1 |h i| ≤ k k ∈ k k≤ 2 On the other hand, we know from Corollary 9.1.3 that there is some f0 E∗ such that f0 = x and f0, x = x . f0(x) ∈ k k k k h i k k Set f1(x) := x . We have f1 = 1 and f1, x = x . k k k k h i k k 9.2 The geometric form of the Hahn-Banach Theorem: separation of convex sets

Deﬁnition 9.2.1. An aﬃne hyperplane is a subset H of a linear space E of the form H = x E : f(x) = α , { ∈ } where f is a linear functional 1 that does not vanish identically and α R is a given constant. We write H = f = α and say that f = α is the equation of H. ∈ { } Proposition 9.2.1. Let E be a normed space. The hyperplane H = f = α is closed if an only if f is continuous. { } Proof. If f is continuous then the pre-image of closed sets if closed, hence H is closed because α is closed in R. Conversely, let us assume that H is closed. The complement Hc of H is open and nonempty{ (since} f is not c identically constant, otherwise f is trivially continuous). Let x0 H , so that f(x0) = α, for example f(x0) < α . Fix r > 0 such that B (x ) Hc where ∈ 6 r 0 ⊂ B (x ) = x E : x x < r . r 0 { ∈ k − 0k } We claim that f(x) < α for all x Br(x0). Suppose by contradiction that f(x1) > α for some x1 Br(x0). The segment ∈ ∈ x = (1 t)x + tx : t [0 , 1] { t − 0 1 ∈ } is contained in Br(x0) (indeed, xt x0 = t x1 x0 < tr r) and thus f(xt) = α for all t [0 , 1]. On the other hand, let k − k k − k ≤ 6 ∈ α f(x ) t = t¯= − 0 . f(x ) f(x ) 1 − 0 We compute (by linearity of f) α f(x ) f(x ) = (1 t)f(x ) + tf (x ) = f(x ) + t(f(x ) f(x )) = f(x ) + − 0 (f(x ) f(x )) = α. t − 0 1 0 1 − 0 0 f(x ) f(x ) 1 − 0 1 − 0 Hence, we have found an element of Hc at which f attains the value f = α, a contradiction. Hence, f(x) < α for all x B (x ). Now, this implies f(x + rz ) < α for all z E with z < 1. In particular ∈ r 0 0 ∈ k k 1 1 α f(x ) f(z) = f(rz ) = f(x + rz ) f(x ) − 0 , | | r | | r | 0 − 0 | ≤ r and the r.h.s. above does not depend on z. Hence, α f(x ) sup f(z) − 0 , z 1 | | ≤ r k k≤ which means that f is continuous. 1We do not assume necessarily here that f is continuous 62 CHAPTER 9. HAHN-BANACH THEOREMS

Deﬁnition 9.2.2. Let E be a normed space. Let A and B be two subsets of E. We say that the hyperplane H = f = α separates A and B if { } f(x) α for all x A and f(x) α for all x B. ≤ ∈ ≥ ∈ We say that H strictly separates A and B if there exists some ǫ > 0 such that

f(x) α ǫ for all x A and f(x) α + ǫ for all x B. ≤ − ∈ ≥ ∈ Geometrically, the separation means that A lies in one of the half-spaces determined by H, ad B lies in the other one. We recall that a subset A E is convex if ⊂ tx + (1 t)y A for all x, y A and for all t [0 , 1] . − ∈ ∈ ∈ Theorem 9.2.2 (Hahn-Banach, first geometric form) . Let E be a normed space. Let A and B be two non empty and convex subsets of E such that A B = . Assume that one of them is open. Then there exists a closed hyperplane that separates A and B. ∩ ∅ The proof of the above theorem relies on the following two lemmas, in which we shall always assume E is a normed space. Lemma 9.2.3. Let C E be an open convex set with 0 C. For every x E set ⊂ ∈ ∈ 1 p(x) = inf α > 0 : α− x C { ∈ } (p is called the gauge of C or the Minkowski functional of C). Then p satisfies (i) and (ii) in Theorem 9.1.1 and the following properties: (a) there is a constant M such that 0 p(x) M x for all x E ≤ ≤ k k ∈ (b) C = x E : p(x) < 1 . { ∈ } Proof. The proof of the property (i) is an easy exercise. Let us prove (a). Since C is open and 0 C, let B (0) C. For a given x E, choose α = x /r . We get ∈ r ⊂ ∈ k k 1 x 1 α− x = r , which implies α− x = r, x k k k k and this clearly shows that x p(x) k k ≤ r x x 1 1 (otherwise, assuming p(x) > krk would imply there exists α′ > krk for which ( α′)− x C, i. e. α− x Br(0), and x 6∈ 6∈ this implies x rα ′ > r krk = x , a contradiction). Since clearly p 0, (a) is proven. Let us provek k (b):≥ Assume firstk thatk x C. Since C is open, there exists≥ ǫ > 0 such that (1+ ǫ)x C. Therefore, 1 ∈ ∈ 1 p(z) 1+ ǫ < 1, and hence x x : p(x) < 1 . Conversely, if p(x) < 1 there exists α (0 , 1) such that α− x C, and thus≤ ∈ { } ∈ ∈ 1 x = α(α− x) + (1 α)0 C − ∈ because C is convex. Finally, let us prove (ii). Let x, y E, and let ǫ > 0. Using (i) and (b) we get x C and y C. Thus, ∈ p(x)+ ǫ ∈ p(y)+ ǫ ∈ since C is convex, tx (1 t)y + − C for all t [0 , 1] . p(x) + ǫ p(y) + ǫ ∈ ∈ Choosing the value p(x) + ǫ t = , p(x) + p(y) + 2 ǫ we find that x+y C, and using (i) and (b) once again we get p(x)+ p(y)+2 ǫ ∈ p(x + y) < p (x) + p(y) + 2 ǫ, for all ǫ > 0, which proves the assertion. 9.2. THE GEOMETRIC FORMS OF THE HAHN-BANACH THEOREM 63

Lemma 9.2.4. Let C E be a nonempty open convex set and let x0 E with x0 C. Then, there exists f E∗ such that f(x) < f (x )⊂for all x C. In particular, the hyperplane f ∈= f(x ) separates6∈ C and x . ∈ 0 ∈ { 0 } { 0} Proof. After a translation we may always assume that 0 C. We may this introduce the gauge functional p of C as in Lemma 9.2.3. Consider the linear subspace G = Rx ,∈ and the linear functional g : G R defined by g(tx ) = t 0 → 0 for all t R. We claim that g(x) p(x) for all x G. Let us first consider the case x = tx 0 with t > 0. Then, p(x) = p∈(tx ) = tp (x ) t = g(tx ≤), where we have∈ used p(x ) 1 since x C. Let us then consider the case 0 0 ≥ 0 0 ≥ 0 6∈ t 0. Then, tp (x0) < 0 p(tx 0), whereas g(tx 0) = t 0, therefore g(x) p(x). ≤Now, from theorem 9.1.1≤ there exists a linear functional≤ f on E that≤ extends g and satisfies

f(x) p(x) for all x E. ≤ ∈

In particular, f(x0) = 1 and f is continuous because of the property (a) in the previous lemma 9.2.3. Moreover, the property (b) in lemma 9.2.3 ensures that f(x) < 1 for all x C, which proves the assertion. ∈ Proof of Theorem 9.2.2. Let A be open. Set C = A B = x y, x A, y B . It is easily seen that C is convex − { − ∈ ∈ } (exercise!). Moreover, C is open since C = y B(A y), i. e. C is the arbitrary union of open sets. Since 0 C ∈ − 6∈ (A and B are disjoint!), from lemma 9.2.4 there is some f E∗ such that f(z) < 0 for all z C, i. e. f(x) < f (y) for all x A and y B. Fix a constant α SR satisfying ∈ ∈ ∈ ∈ ∈ sup f(x) α inf f(y). x A ≤ ≤ y B ∈ ∈ The hyperplane f = α separates A and B as desired. { } Theorem 9.2.5 (Hahn-Banach, second geometric form) . Let A E and B E be two nonempty convex subsets such that A B = . Assume that A is closed and B is compact. Then,⊂ there exists⊂ a closed hyperplane that strictly separates A∩and B∅. Proof. Set C = A B as before. C can be proved to be convex and closed (exercise). Moreover, 0 C. Hence, there is some r > 0− such that B (0) C = . By Theorem 9.2.2, there is a closed hyperplane that separates6∈ B (0) r ∩ ∅ r and C. Therefore, there is some f E∗, f 0, such that ∈ 6≡ f(x y) f(rz ), for all x A, y B, z B (0) . − ≤ ∈ ∈ ∈ 1 This implies f(x y) r f for all x A and y B. Let ǫ = 1 r f . We obtain − ≤ − k k ∈ ∈ 2 k k 1 f(x) + ǫ = f(x y) + r f + f(y) f(y) ǫ, − 2 k k ≤ − for all x A and y B. Choosing α such that ∈ ∈ sup f(x) + ǫ α inf f(y) ǫ, x A ≤ ≤ y B − ∈ ∈ we see that the hyperplane f = α strictly separates A and B. { } Corollary 9.2.6. Let F E be a linear subspace of the normed space E. Assume F = E. Then, there exists some ⊂ 6 f E∗, f = 0 , such that ∈ 6 f, x = 0 for all x F . h i ∈ Proof. Let x0 E, with x0 F . Using Theorem 9.2.5 with A = F and B = x0 , we ﬁnd a closed hyperplane f = α that strictly∈ separates6∈ F and x . Thus, we have { } { } { 0} f, x < α < f, x , for all x F . h i h 0i ∈ Since F is a linear subspace, if f, x would be non zero for some x F , then it would be positive for some h i ∈ x1 F , and then it would be arbitrarily large for a given x2 = λx 1 with λ > 0 large enough, which contradicts f, x∈ < α . h 2i The above corollary is used very often in proving that a linear subspace F E is dense in E. It suﬃces to show that every continuous linear functional that vanishes on F must vanish everywhere⊂ on E. 64 CHAPTER 9. HAHN-BANACH THEOREMS Chapter 10

The uniform boundedness principle and the closed graph theorem

10.1 The Baire category theorem

We recall that a subset E of a metric space X is called nowhere dense if X E is dense in X, or equivalently if E has empty interior. \

Definition 10.1.1. A metric space X is called a Baire first category space if X can be written as the union of a countable family of nowhere dense closed sets. The space X is called a Baire second category space if it is not a + first category space, i. e. if given a sequence closed subsets Fn n N in X, if X = ∞ Fn this implies that at { } ∈ n=1 least one of the Fn has a nonempty interior. S The following classical result plays an essential role in this chapter.

Theorem 10.1.1 (Baire) . Let X be a nonempty complete metric space. Then X is a Baire second category space.

◦ Proof. Let Xn n be a sequence of closed subsets of X with the property that Xn = . We claim that n Xn has empty interior.{ } If this is true, than in particular we have proven that if X is the countable∅ union of closed subsets, one of them must have a non empty interior, because otherwise we would get the contradiction that XShas a non empty interior, and therefore X is empty. c Let us prove the claim. Let On = Xn, so that On is an open set for all n. Moreover, On is dense in X. We c claim that G = n 1 On is dense in X, which would imply that G = n 1 Xn has empty interior. To see that, let ω be a non-empty≥ open set in X. We shall prove that ω G = (this would≥ imply there exists no open set in the complementT of G, i. e. the complement of G has empty interior,∩ 6 ∅ i. e.SG is dense). As usual, we set B (x) = y X : d(x, y ) < r . r { ∈ } Pick an arbitrary point x ω and r > 0 such that 0 ∈ 0 B (x ) ω. r0 0 ⊂

Since O1 is dense, choose x1 Br0 (x0) O1, and an r1 > 0 such that Br1 (x1) Br0 (x0) O1. We can always ∈r0 ∩ ⊂ ∩ choose r1 in a way that r1 < 2 . Inductively, we can construct two sequences xn n X and rn n (0 , + ) rn { } ⊂ { } ⊂ ∞ such that 0 < r n+1 < 2 and B (x ) B (x ) O . rn+1 n+1 ⊂ rn n ∩ n+1

It follows that xn is a Cauchy sequence. Since X is complete, let ℓ = lim n + xn. Since xn+p Brn (xn) for all n, p 0, we obtain in the p + limit that → ∞ ∈ ≥ → ∞ ℓ B (x ) for all n 0. ∈ rn n ≥

In particular, ℓ n 1 On = G, and ℓ ω. Therefore ℓ G ω = . ∈ ≥ ∈ ∈ ∩ 6 ∅ T 65 66 CHAPTER 10. THE UNIFORM BOUNDEDNESS PRINCIPLE AND THE CLOSED GRAPH THEOREM 10.2 The uniform boundedness principle

Theorem 10.2.1 (Banach-Steinhaus, uniform boundedness principle) . Let E and F be two Banach spaces and let Ti i I be a family (not necessarily countable) of continuous linear operators from E into F . Assume that { } ∈

sup Ti(x) < + for all x E. (10.1) i I k k ∞ ∈ ∈ Then, sup Ti (E,F ) < + . (10.2) i I k kL ∞ ∈ In other words, there exists a constant c such that

T (x) c x for all x E and for all i I. k i k ≤ k k ∈ ∈ Proof. For every n 1 let ≥ X = x E : T x n for all i I . n { ∈ k i k ≤ ∈ } Clearly, Xn is closed, and from the assumption (10.1) we have

Xn = E. n 1 [≥

From the Baire category theorem 10.1.1, the interior of Xn0 is non empty for at least one n0 N. Therefore there exists a ball B (x ) X for some x X and some r > 0. This implies ∈ r 0 ⊂ n0 0 ∈ n0 T (x + rz ) n for all i I and for all z B (0) . k i 0 k ≤ 0 ∈ ∈ 1 Therefore, for all x E with x = 0 we have ∈ 6 r rx rx T (x) = T T x + + T (x ) n + T (x ) , x k i k i x ≤ i 0 x k i 0 k ≤ 0 k i 0 k k k k k k k and this proves the assertion.

Corollary 10.2.2. Let E and F be two Banach spaces. Let Tn n be a sequence of continuous linear operators from E into F such that for every x E, T (x) converges to a{ limit} denoted by T (x) as n + . Then we have ∈ n → ∞

(a) sup n Tn (E,F ) < + , k kL ∞ (b) T (E, F ), ∈ L (c) T (E,F ) lim inf n + Tn (E,F ). k kL ≤ → ∞ k kL Proof. (a) follows directly from Theorem 10.2.1, and thus there exists a constant c such that

T (x) c x for all n N and for all n E. k n k ≤ k k ∈ ∈ Hence, for all n N, ∈ T (x) (T T )( x) + T (x) c x + (T T )( x) , k k ≤ k − n k k n k ≤ k k k − n k and taking the limit as n + we get → ∞ T (x) c x for all x E. k k ≤ k k ∈ Since T is clearly linear, we obtain (b). Finally, we have Tn(x) Tn (E,F ) x . k k ≤ k kL k k By taking the lim inf on both sides, recalling that Tn(x) converges to T (x), we get

T (x) lim inf Tn (E,F ) x , n + L k k ≤ → ∞ k k k k which proves (c). 10.3. THE OPEN MAPPING THEOREM AND THE CLOSED GRAPH THEOREM 67 10.3 The open mapping theorem and the closed graph theorem

This section contains two basic results due to Banach 1 Theorem 10.3.1 (Open mapping theorem) . Let E and F be two Banach spaces and let T be a continuous linear operator from E into F that is surjective. Then, there exists a constant c > 0 such that T (B (0)) B (0) . (10.3) 1 ⊃ c Remark 10.3.1 . The statement in the above theorem implies that the image under T of any open set in E is an open set in F , which justiﬁes the name given to this theorem. Indeed, let us suppose U is an open set in E and let us prove that T (U) is open. Fix a point y T (U), so that y T (x ) for some x U. Let r > 0 such that 0 ∈ 0 ∈ 0 0 ∈ Br(x0) U, i. e. x0 + Br(0) U. It follows that y0 + T (Br(0)) T (U). Using (10.3) and the fact that T is linear, we obtain⊂ T (B (0)) B (0)⊂ and therefore ⊂ r ⊃ rc y + B (0) T (U), 0 rc ⊂ which proves that T (U) contains an open neighborhood of y0, as desired. Proof of Theorem 10.3.1. Step 1. Assume that T is a linear surjective operator from E onto F . Then, we claim that there exists a constant c > 0 such that T (B (0)) B (0) . (10.4) 1 ⊃ 2c

Set Xn = T (Bn(0)) = nT (B1(0)). Since T is surjective, we have n 1 Xn = F , and by the Baire category theorem ≥ 10.1.1 there exists some n such that X has non empty interior. It follows that T (B (0)) has non empty interior. 0 n0 S 1 Pick a point y F and a constant c > 0 such that 0 ∈ B (y ) T (B (0)) . (10.5) 4c 0 ⊂ 1 In particular, y0 T (B1(0)), which reads y0 = lim n + T (xn) for some xn B1(0). By takingx ˜ n = xn B1(0), ∈ → ∞ ∈ − ∈ we get lim n + T (˜xn) = lim n + T (xn) = y0, which implies y0 T (B1(0)). Adding the latter formula to (10.5) we get→ ∞ − → ∞ − − ∈ B (0) T (B (0)) + T (B (0)) . 4c ⊂ 1 1 On the other hand, since T (B1(0)) is convex, we have

T (B1(0)) + T (B1(0)) = 2 T (B1(0)) , which proves the claim by linearity. Step 2. Assume T is a continuous linear operator from E into F that satisfies (10.4). Then, we claim that T (B (0)) B (0) . 1 ⊃ c To prove that, choose any y F with y < c . The aim is to find some x E such that x < 1 and T (x) = y. From (10.4) we know that for∈ all ǫ > 0k therek exists z E with z < 1/2 and∈ y T (z) k n we compute ∈ m m m k xm xn = zk zk 2− 0 as n, m + . k − k ≤ k k ≤ → → ∞ kX=n kX=n kX=n n Since E is complete, let xn x E . By sending n + in y T (xn) c2− we get y = T (x) since T is continuous. Since → ∈ → ∞ k − k ≤ + + + ∞ ∞ ∞ n x = zn zn < 2− = 1 , k k ≤ k k n=1 n=1 n=1 X X X we have written an arbitrary y Bc(0) as y = T (x ) for some x E with x < 1, which proves the assertion. ∈ ∈ k k 1Stefan Banach (1892 - 1945) was a Polish mathematician. He is generally considered to have been one of the 20th century’s most important and influential mathematicians. Banach was one of the founders of modern functional analysis and one of the original members of the Lwów School of Mathematics. His major work was the 1932 book, Théorie des opérations linéaires (Theory of Linear Operations), the first monograph on the general theory of functional analysis. 68 CHAPTER 10. THE UNIFORM BOUNDEDNESS PRINCIPLE AND THE CLOSED GRAPH THEOREM

Some important consequences of Theorem 10.3.1 are the following. Corollary 10.3.2. Let E an F be two Banach spaces and let T be a continuous linear operator from E into F that 1 is bijective (i. e. one-to-one and surjective). Then T − is also continuous from F into E. Proof. Property (10.3) and the assumption that T is bijective imply that if E is chosen so that T (x) < c then ∈ 1 k k x < 1. Now, let y F , set z = cy/ 2 y , which implies z < c . Hence, T − (z) < 1, which gives k k ∈ k k k k k k

1 2 T − y y . k k ≤ c k k

1 Therefore T − is continuous.

Corollary 10.3.3. Let E be a vector space provided with two norms 1 and 2. Assume that E is a Banach space with both norms and that there exists a constant C 0 such thatk · k k · k ≥ x C x for all x E. k k2 ≤ k k1 ∈ Then the two norms are equivalent, i. e. there is a constant c > 0 such that

x c x for all x E. k k1 ≤ k k2 ∈ Proof. Apply Corollary 10.3.2 with

E = ( E, ), F = ( E, ), T = I. k · k 1 k · k 2

Theorem 10.3.4 (Closed graph theorem) . Let E and F be two Banach spaces. Let T be a linear operator from E into F . Assume that the graph of T

G(T ) = (x, T (x)) E F : x E { ∈ × ∈ } is closed in E F . Then T is continuous. × Proof. Consider, on E, the two norms

x = x + T (x) , x = x k k1 k kE k kF k k2 k kE (the norm is called the graph norm ). We claim that E is a Banach space with the norm . To see this, let k · k 1 k · k 1 xn n be a Cauchy sequence in the norm 1. This means that xn is a Cauchy sequence in E and T (xn) is a Cauchy{ } sequence in . Since both E andk · k F are complete, x x in and T (x ) ykfor · k some x E and k · k F n → k · k E n → ∈ y F . Now, since G(T ) is closed, the element ( x, y ) E F belongs to G(T ), i. e. y = T (x). This proves that xn converges∈ to x in the graph norm. ∈ × Now, E is also a Banach space for the norm . Hence, we can apply Corollary 10.3.3, which implies that k · k 2 the two norms are equivalent, and thus there exists a constant c > 0 such that x 1 c x 2. This implies T (x) c x as desired. k k ≤ k k k kF ≤ k kE Remark 10.3.2 . The converse of the above statement is obviously true, since the graph of a any continuous map (linear or not) is closed. Chapter 11

Weak topologies

11.1 The inverse limit topology of a family of maps

Let ( E, ) be a normed space. Let us, for the moment, ignore the usual topology on E induced by the norm . k · k k · k We recall from the Introductory Material that, for a given family of maps f : E R, one can consider the coarsest topology τ that makes all maps f continuous. Here R is consideredF as topological→ space with the usual Euclidean topology. We recall that τ is constructed∈ F as the topology generated by the inverse images of all open sets 1 in R via all maps f . More in detail, one considers the family = f − (O) : O R open, f and take first the family of∈ the F intersection of finitely many elements in .C Finally,{ one takes ⊂τ as the union∈ F} of all sets in . Such topologyI is denoted by τ = τ( ), and is called the inverseC limit topology of . Recall that is a sub-base Ifor τ, i. e. τ is the coarsest topology containingC . Moreover, recall that the family F of finite intersectionsC in of sets in is a base for the topology τ. Finally, forC a given point x E, the family I C ∈ 1 1 = f − (O ) . . . f − (O ) : O , . . . , O open in R, f ,...,f , with f (x) O for all j = 1 ,...,k Ux { 1 1 ∩ ∩ k k 1 k 1 k ∈ F j ∈ j } is a base of neighborhoods for x in the topology τ, i. e. every open neighborhood of x in τ contains an open set of the family . Ux

Exercise 11.1.1. With the notation above, prove that a sequence xn n E converges to x E in τ( ) if and only if f(x ) f(x) for all f . { } ⊂ ∈ C n → ∈ F Solution: If x x in τ, then f(x ) f(x) for all f because all f are continuous in the topology n → n → ∈ F ∈ F τ. Vice versa, assume f(xn) f(x) for all f . Let Ux x be a basic neighborhood of x, in particular, 1 1 k → ∈ F ∈ U Ux = f1− (O1) . . . fk− (O ) for some f1,...,f k and some open sets O1, . . . , O k R with fi(x) Oi for all i = 1 ,...,k . By∩ assumption,∩ f (x ) f (x), hence∈ there F exist integers N N such that ⊂f (x ) O for∈ all n N , i n → i i ∈ i n ∈ i ≥ i for all i = 1 ,...,k . Set N := max N1, . . . , N k . Then, for all n N one has fi(xn) Oi for all i = 1 ,...,k , i. e. 1 1 { } ≥ ∈ x f − (O ) . . . f − (O ), i. e. x U . This implies that x x in τ. n ∈ 1 1 ∩ ∩ k k n ∈ x n → Exercise 11.1.2. With the notation above, let Z be a topological space and let ψ : Z E. Then ψ is continuous if and only if f ψ is continuous from Z into R for every f . → ◦ ∈ F Solution: If ψ is continuous then f ψ is also continuous for all f (all the f are continuous in τ, and the composition of continuous functions◦ is continuous). Vice versa,∈ assume F f ψ is∈ Fcontinuous for every 1 ◦ f . We have to prove that ψ− (U) is open for all sub-basic open sets U E in the τ topology. But we know ∈ F 1 ⊂ that sub basic open sets in τ are all of the form U = f − (O) for some f and some O R open. Hence, 1 1 1 1 ∈ F ∈ ψ− (U) = ψ− (f − (O)) = ( f ψ)− (O), and the latter is an open set by the continuity of f ψ. ◦ ◦

11.2 The weak topology σ(E, E ∗)

Let E be a normmed space and let f E∗. We denote by ϕ : E R the linear functional ϕ (x) = f, x . As ∈ f → f h i f runs through E∗ we obtain a collection φf f E∗ of maps from E into R. We now ignore the usual topology induced by on E and deﬁne a new topology{ } ∈ on the set E as follows. k · k

Deﬁnition 11.2.1. The weak topology σ(E, E ∗) on E is the inverse limit topology of the family of maps φf f E∗ . { } ∈ 69 70 CHAPTER 11. WEAK TOPOLOGIES

Remark 11.2.1 . Note that all f E∗ are continuous functionals on ( E, ). Since σ(E, E ∗) is the coarsest topology ∈ k·k that makes all f E∗ continuous, we deduce that the weak topology σ(E, E ∗) on E is weaker than the usual topology induced on E by∈ the norm . k · k Proposition 11.2.1. The weak topology σ(E, E ∗) is Hausdorﬀ.

Proof. Given x1, x 2 E with x1 = x2 we have to ﬁnd two open sets O1 and O2 in the weak topology σ(E, E ∗) such that x O , x O∈ and O O6 = . By Theorem 9.2.5 there exists a closed hyperplane strictly separating x 1 ∈ 1 2 ∈ 2 1 ∩ 2 ∅ { 1} and x2 (these two sets are closed and compact in the usual topology). Thus, there exists some f E∗ and some α R{ such} that ∈ ∈ f, x < α < f, x . h 1i h 2i Set

1 O = x E : f, x < α = ϕ− (( , α )) 1 { ∈ h i } f −∞ 1 O = x E : f, x > α = ϕ− (( α, + )) . 2 { ∈ h i } f ∞

Clearly, O1 and O2 are open sets for σ(E, E ∗) and they satisfy the required properties.

Proposition 11.2.2. Let x E. Given ǫ > 0 and a ﬁnite set f , f ,...,f E∗, consider 0 ∈ { 1 2 k} ⊂ V = V (f ,...,f ; ǫ) = x E : f , x x < ǫ, for all i = 1 ,...,k . 1 k { ∈ |h i − 0i| }

Then V is a neighborhood of x0 for the topology σ(E, E ∗). Moreover, we obtain a basis of neighborhoods of x0 by varying ǫ, k, and the fi’s in E∗. Proof. From the discussion in Section 11.1 we already know that the sets V in the statement are a basis of neighborhoods for x0 if the statement fi, x x0 < ǫ is replaced by fi Oi for some Oi open neighborhood of fi(x0). The statement follows by recalling|h that− thei| open sets of R are characterised∈ as the sets O such that for all y O there exists an open interval ( y ǫ, y + ǫ) O for some ǫ > 0. And this proves the assertion. ∈ − ⊂ Notation 11.2.1 . Let xn n be a sequence on E. If xn converges to x E in the σ(E, E ∗) topology we shall use the notation { } ∈ xn ⇀ x.

We shall sometimes say xn ⇀ x weakly in σ(E, E ∗). The convergence of xn to x in the usual topology will be sometimes emphasised by saying x x strongly , meaning x x 0. n → k n − k → We recall that the dual space E∗ is a Banach space equipped with the norm

f E∗ = sup f(x) . k k x 1 | | k k≤ Proposition 11.2.3. Let x E be a sequence in E. Then { n}n ⊂ (i) x ⇀ x weakly in σ(E, E ∗) is and only if f, x f, x for all f E∗. n h ni → h i ∈ (ii) If x x strongly, then x ⇀ x weakly in σ(E, E ∗). n → n (iii) If x ⇀ x weakly in σ(E, E ∗), then x is bounded and n {k nk} n

x lim inf xn . n + k k ≤ → ∞ k k

(iv) If x ⇀ x weakly in σ(E, E ∗) and if f f strongly in E∗ (i. e. f f ∗ 0), then f , x f, x . n n → k n − kE → h n ni → h i Proof. (i) is a consequence of the Exercise 11.1.1 and the deﬁnition of weak topology σ(E, E ∗) on E. (ii) is a consequence of (i), since

f, x f, x f ∗ x x . |h ni − h i| ≤ k kE k n − k Alternatively, it is a consequence of the fact that the weak topology is weaker than the norm topology. (iii) follows from Theorem 10.2.1. Indeed, for every n N deﬁne the map T : E∗ R as T (f) = f, x . ∈ n → n h ni Since x converges weakly to x, for every f E∗ the real sequence f, x is convergent, and hence bounded. n ∈ h ni 11.2. THE WEAK TOPOLOGY σ(E, E ∗) 71

Therefore, for all f E∗ the family T (f) is uniformly bounded in E∗ with respect to n N. Hence, from the ∈ n ∈ Banach-Steinhaus theorem 10.2.1 we have sup n N Tn (E∗,R) < + and there exists a constant c R such that ∈ k kL ∞ ∈ T (f) = f, x c f ∗ , for all n N. | n | |h ni| ≤ k kE ∈ Hence, Corollary 9.1.5 implies xn = sup f, x n c, k k f E∗ , f ∗ 1 |h i| ≤ ∈ k kE ≤ which proves that xn is a bounded sequence. Now, taking the liminf n + in the inequality f, x n f E∗ xn we obtain k k → ∞ |h i| ≤ k k k k f, x f E∗ lim inf xn , n + |h i| ≤ k k → ∞ k k which implies, once again by Corollary 9.1.5,

x = sup f, x lim inf xn . k k f 1 |h i ≤ n + k k k k≤ → ∞ (iv) follows from the inequality

f , x f, x f f, x + f, x x f f x + f, x x . |h n ni − h i| ≤ |h n − ni| |h n − i| ≤ k n − kk nk |h n − i|

Now, due to (iii) xn is uniformly bounded and the ﬁrst term in the r.h.s. above goes to zero. The second term goes to zero becausek ofk (i).

Proposition 11.2.4. When E is finite-dimensional, the weak topology σ(E, E ∗) and the usual topology are the same. In particular, a sequence x converges weakly if and only if it converges strongly. { n}n Proof. Since the weak topology is always weaker than the strong topology, it suffices to check that every strongly open set is weakly open. Let x E and let U be a neighborhood of x in the strong topology. We have to 0 ∈ 0 find a neighborhood V of x in the weak topology σ(E, E ∗) such that V U. In other words, we have to find 0 ⊂ f ,...,f E∗ and ǫ > 0 such that 1 k ∈ V = x E : f , x x < ǫ , for all i = 1 ,...,k U. { ∈ |h i − 0i| } ⊂

Fix r > 0 such that Br(x0) U. Pick a basis e1, e 2,...,e k E such that ei = 1 for all i. Hence, every x E can k ⊂ ∈ k k ∈ be written as x = i=1 xiei, and the maps x xi are continuous linear functionals on E denoted by fi. Choosing those functionals for the neighborhood V we have,7→ for all x V , P ∈ k x x f , x x < kǫ. k − 0k ≤ |h i − 0i| i=1 X Choosing ǫ = r/k , we obtain V B (x ) U as desired. ⊂ r 0 ⊂ Remark 11.2.2 . Open (resp. closed) sets in the weak topology σ(E, E ∗) are always open (resp. closed) in the strong topology. In any inﬁnite dimensional space the weak topology is strictly coarser than the strong topology: i. e., there exist open (resp. closed) sets in the strong topology that are not open (resp. closed) in the weak topology. Here are two examples. Example 11.2.1 . The unit sphere S = x E : x = 1 , with E inﬁnite dimensional, is never closed in the weak { ∈ k k } topology σ(E, E ∗). More precisely, we have

∗ σ(E,E ) S = B (0) = x E : x 1 , 1 { ∈ k k ≤ } ∗ σ(E,E ) where S denotes the closure of S in the topology of σ(E, E ∗). First, let us check that every x0 E with ∗ σ(E,E ) ∈ x0 < 1 belongs to S . Indeed, let V be a neighborhood of x0 in σ(E, E ∗). We have to prove that V S = . kIn viewk of Proposition 11.2.2, we may always assume that V has the form ∩ 6 ∅

V = x E : f , x x < ǫ for all i = 1 ,...,k , { ∈ |h i − 0i| } with ǫ > 0 and f ,...,f E∗. 1 k ∈ 72 CHAPTER 11. WEAK TOPOLOGIES

Now, we claim that there exists y0 E, y0 = 0, such that fi, y 0 = 0 for all i = 1 ,...,k . Assume by contradiction that the claim is false. We deﬁne∈ the6 following linearh map ϕi : E Rk, → ϕ(x) = ( f , x ,..., f , x ) Rk. h 1 i h k i ∈

If the claim is false, then for all x E there exists i 1,...,k such that fi(x) = 0, which means that for all x E the vector ϕ(x) Rk is always∈ non zero. Hence,∈ {ϕ is a linear} isomorphism6 from E onto ϕ(E), and hence dim(∈ E) = dim( ϕ(E)) ∈k, which contradicts the assumption that E has inﬁnite dimension. This proves the claim. Now, the function≤ [0 , + ) t g(t) := x + ty is continuous on [0 , + ). Moreover, g(0) < 1, and ∞ ∋ 7→ k 0 0k ∞ lim t + g(t) = + . Hence, there exists t0 > 0 such that x0 + ty 0 = 1, i. e. x0 + t0y0 S. Now, for all → ∞ ∞ k k ∈ i 1,...,k we have fi, (x0 + t0y0) x0 = t0 fi, y 0 = 0 < ǫ , clearly x0 + t0y0 V . Therefore, x0 + t0y0 V S. ∗ ∗ ∈ h − i h i σ(E,E ) ∈ σ(E,E ) ∈ ∩ The above argument proves that S B (0) S . To prove B (0) S it suﬃces to prove that ⊂ 1 ⊂ 1 ⊃ B1(0) is closed in the weak topology. Now, for a given x E with x 1 we have by Corollary 9.1.5 that ∈ k k ≤ ∗ sup f 1 f, x 1, hence x x E : f, x 1 for all f E∗ with f E 1, i. e. k k≤ |h i| ≤ ∈ { ∈ |h i| ≤ } ∈ k k ≤ B (0) x E : f, x 1 . 1 ⊂ { ∈ |h i| ≤ } f E∗, f 1 ∈ \k k≤

Moreover, again by Corollary 9.1.5 every element x f E∗, f 1 x E : f, x 1 satisﬁes ∈ ∩ ∈ k k≤ { ∈ |h i| ≤ } x = sup f, x 1, k k f 1 |h i| ≤ k k≤ and this proves that actually B (0) = x E : f, x 1 . 1 { ∈ |h i| ≤ } f E∗, f 1 ∈ \k k≤

Since the r.h.s. is the intersection of closed sets in the weak topology σ(E, E ∗), this implies that B1(0) is weakly closed. Example 11.2.2 . The open unit ball U = x E : x < 1 , { ∈ k k } with E infinite dimensional, is never open in the weak topology σ(E, E ∗). Suppose by contradiction that U is c weakly open. Then its complement U = x E : , x 1 is weakly closed. It follows that B1(0) U = S is weakly closed, which contradicts the example{ ∈ 11.2.1.| k k ≥ } ∩ Remark 11.2.3 . One can prove that the weak topology is never metrizable in infinite dimensions, i. e. there is no metric on E that induces the weak topology σ(E, E ∗). The proof is postponed. Remark 11.2.4 . Keep in mind that, in general, if two topological spaces have the same convergent sequences this does not automatically imply that they have the same topologies. Indeed, if two metric spaces have the same convergent sequences then they have the identical topologies. But in general, the set of convergent sequences is not enough to characterise a topology if this is not induced by a distance. This discussion implies in particular that, in principle, there might be examples of infinite dimensional spaces E for which every weakly converging sequence also converges strongly, whereas of course the two topologies (strong and weak) are distinct. Such examples are quite rare and pathological. The above examples have shown that in general every weakly closed set is strongly closed and the converse is false in infinite dimensions. However, such two classes become identical if we restrict to convex sets. We recall that a set C E is convex if for all x, y C and for all t [0 , 1], the vector tx + (1 t)y belongs to C (i. e. C contains all the segments⊂ connecting two arbitrary∈ points in C∈). Notice in particular that− S from the above example is not convex. Theorem 11.2.5. Let C be a convex subset of E. Then C is closed in the weak topology if and only if it is closed in the strong topology.

Proof. Assume C is closed in the strong topology and let us prove that C is closed in the weak topology. We shall check that the complement Cc is open in the weak topology. To perform this task, let x C. By the Hahn-Banach 0 6∈ theorem 9.2.5 there exists a closed hyperplane strictly separating x0 and C. Thus, there exists some f E∗ and some α R such that { } ∈ ∈ f, x < α < f, y , for all y C. h 0i h i ∈ 11.3. THE WEAK ∗ TOPOLOGY σ(E∗, E ) 73

Set V = x E : f, x < α , { ∈ h i } so that x V , V C = and V is open in the weak topology. This proves the assertion. 0 ∈ ∩ ∅

11.3 The weak ∗ topology σ(E∗, E )

We now turn our attention to the dual space E∗. As we know, E∗ is a normed space with the operator norm

f E∗ = sup f(x) . k k x 1 | | k k≤

Hence, one could consider the dual space of E∗, i. e. the space of all continuous linear functionals deﬁned on E∗. We denote such space by E∗∗ . As dual of the space E∗, the space E∗∗ is naturally equipped with the norm

ξ E∗∗ = sup ξ, f , ξ E∗∗ . k k f E∗, f ∗ 1 |h i| ∈ ∈ k kE ≤

1 There is a canonical injection J : E E∗∗ deﬁned as follows. Given x E, the map E∗ f f, x is a → ∈ ∋ 7→ h i continuous linear functional on E∗. This is due to the inequality

f, x x f ∗ . |h i| ≤ k kEk kE

Thus, such map is an element of E∗∗ . We denote such element as J(x). By deﬁnition, we have

J(x), f ∗∗ ∗ = f, x ∗ , for all f E∗. h iE ,E h iE ,E ∈

It is clear that J is linear and that J is an isometry 2, that is, J(x) ∗∗ = x . Indeed, we have k kE k kE

J(x) E∗∗ = sup J(x), f = sup f, x = x E, k k f ∗ 1 |h i| f ∗ 1 |h i| k k k kE ≤ k kE ≤ and the last step is due to Corollary 9.1.5. So far, we have two topologies on E∗:

(a) the usual strong topology associated to ∗ , k · k E

(b) the weak topology σ(E∗, E ∗∗ ), obtained by performing on E∗ the construction of section 11.2.

We are now going to define a third topology on E∗, called weak ∗ topology and denoted by σ(E∗, E ) (the is here to remind us that this topology is only defined on dual spaces). For every x E consider the linear functional∗ ∈ ϕx : E∗ R defined by E∗ f ϕx(f) = f, x . As x runs through E we obtain a collection ϕx x E of maps → ∋ 7→ h i { } ∈ from E∗ into R.

Deﬁnition 11.3.1. The weak ∗ topology σ(E∗, E ) on E∗ is the inverse limit topology of the family of maps ϕx x E. { } ∈ Remark 11.3.1 . Notice that ϕ is just another notation for J(x) above. Hence, for all x E, the linear functional x ∈ ϕ : E∗ R is continuous (as a function from the normed space ( E∗, ∗ ) to R), and therefore ϕ E∗∗ . This x → k · k E x ∈ fact implies that the weak ∗ topology σ(E∗, E ) on E∗ is coarser than the weak topology σ(E∗, E ∗∗ ) on E∗, i. e. σ(E∗, E ) has fewer open sets (resp. closed sets) than σ(E∗, E ∗∗ ), which in turn has fewer open sets than the strong topology induced on E∗ by the operator norm ∗ . k · k E One may probably wonder why there is such an interest into deﬁning weaker and weaker topologies. The reason is the following: a coarser topology has more compact sets , since there are less open covers to test the compactness condition.

Proposition 11.3.1. The weak ∗ topology is Hausdorﬀ.

1An injection of a set A into a set B is a one-to-one map from A into B. 2An isometry between two metric spaces is a map that preserves distances between points. 74 CHAPTER 11. WEAK TOPOLOGIES

Proof. Given f1, f 2 E∗ with f1 = f2 there exists x E such that f1, x = f2, x (this does not use Hahn-Banach, but just the fact that∈ f = f ). Assume6 for example∈ that f , x

1 O = f E∗ : f, x < α = ϕ− (( , α )) , 1 { ∈ h i } x −∞ 1 O = f E∗ : f, x > α = ϕ− (( α, + )) . 2 { ∈ h i } x ∞

Then O and O are open sets in σ(E∗, E ) such that f O , f O , and O O = . 1 2 1 ∈ 1 2 ∈ 2 1 ∩ 2 ∅

Proposition 11.3.2. Let f E∗. Given a ﬁnite set x ,...,x E and ǫ > 0, consider 0 ∈ { 1 k} ⊂

V = V (x ,...,x ; ǫ) = f E∗ : f f , x < ǫ, for all i = 1 ,...,k . 1 k { ∈ |h − 0 ii| }

Then V is a neighborhood of f0 for the topology σ(E∗, E ). Moreover, we obtain a basis of neighborhoods of f0 for σ(E∗, E ) by varying ǫ, k, and the xi’s in E.

Proof. Similar to the proof of Proposition 11.2.2, left as an exercise.

Notation 11.3.1 . Let fn n be a sequence on E∗. If fn converges to f E∗ in the σ(E∗, E ) topology we shall use the notation { } ∈ f ⇀∗ f. n − We shall sometimes say f ⇀∗ f weakly ∗ in σ(E∗, E ). The convergence of f to f in the weak topology σ(E∗, E ∗∗ ) n − n will be denoted by fn ⇀ f in σ(E∗, E ∗∗ ). The convergence of fn to f in the strong topology of E∗ will be sometimes emphasised by saying f f strongly , meaning f f ∗ 0. n → k n − kE → Exercise 11.3.1. By mimicking the proof of Proposition 11.2.3, prove the following statements for a general sequence f E∗: n ∈

(i) f ⇀∗ f in σ(E∗, E ) if and only if f , x f, x for all x E. n − h n i → h i ∈

(ii) If f f strongly, then f ⇀ f in σ(E∗, E ∗∗ ). If f ⇀ f in σ(E∗, E ∗∗ ), then f ⇀∗ f in σ(E∗, E ). n → n n n −

(iii) If fn ⇀∗ f in σ(E∗, E ) then fn n is bounded and f lim inf n + fn . − {k k} k k ≤ → ∞ k k

(iv) If f ⇀∗ f in σ(E∗, E ) and if x x strongly in E, then f , x f, x . n − n → h n ni → h i

Remark 11.3.2 . When dim( E) < + the three topologies (strong, weak, weak ∗) on E∗ coincide. Indeed, the ∞ canonical injection J : E E∗∗ is in this case surjective (since dim( E) = dim( E∗∗ ) and therefore σ(E∗, E ) = → σ(E∗, E ∗∗ ). We are now ready to prove one of the main results of this part. As we observed in Remark 11.3.1, weakening a topology implies having more compact sets. As seen in Theorem 7.1.3, one of the main points with the strong topology on a normed space of inﬁnite dimension is that the closed unit ball is not a compact set. Such a situation changes drastically when passing from the strong topology to weak topologies. The next result is a ﬁrst big step toward this direction. Theorem 11.3.3 (Banach - Alaoglu - Bourbaki) . The closed unit ball

B ∗ = f E∗ : f ∗ 1 E { ∈ k kE ≤ } is compact in the weak ∗ topology σ(E∗, E ).

Proof. Consider the Cartesian product Y = RE, which consists of all maps from E into R (or equivalently Y = Πx ER). We denote elements of Y by ω = ( ωx)x E with ωx R. The space Y is equipped with the standard product∈ topology, i. e. the coarsest topology on Y associated∈ to∈ the family of maps ω ω (as x runs through E). 7→ x In what follows, E∗ is systematically equipped with the weak ∗ topology σ(E∗, E ). Since E∗ consists of special maps from E into R (i. e. continuous linear maps), we may consider E∗ as a subset of Y . More precisely, let Φ : E∗ Y → 11.4. REFLEXIVE SPACES AND SEPARABLE SPACES 75

3 be the canonical injection from E∗ into Y , so that Φ( f) = ( ωx)x E with ωx = f, x . Clearly, Φ is continuous ∈ h i from E∗ into Y . Indeed, from exercise 11.1.2 we have to prove that the map E∗ f Φ( f) (Φ( f)) x R is continuous for all x R. But ∋ 7→ 7→ ∈ ∈ (Φ( f)) = f, x f x , | x| |h i| ≤ k kk k 1 and hence the above map is continuous. Now, we claim that the inverse map Φ − : Φ( E∗) E∗ is continuous (here → Φ( E∗) Y with the topology induced by the standard topology on Y ). To prove this, using once again exercise ⊂ 11.1.2 (recall E∗ is equipped with the σ(E∗, E ) topology, i. e the limit inverse topology of the family of linear maps E∗ f f, x R with x running through E), we only have to prove that, for every ﬁxed x E, the map ∋ 7→ h 1 i ∈ 1 ∈ Y ω Φ− (ω), x is continuous on Φ( E∗), which is obvious since Φ− (ω), x = ωx (this is seen by noticing ∋ 7→ h i 1 h i that ω = Φ( f) for some f E∗ and Φ− (ω), x = f, x = ω ). In other words, Φ is a homeomorphism from E∗ ∈ h i h i x onto Φ( E∗). Now, it is easily seen that Φ( BE∗ ) = K, where K is deﬁned by

K = ω Y : ω x , ω = ω + ω , ω = λω , for all λ R and for all x, y E . { ∈ | x| ≤ k k x+y x y λx x ∈ ∈ } The above formula, although quite scary at a ﬁrst sight, means nothing but the following: the image of the closed unit ball in E∗ via Φ is just the set of maps from E into R which are linear (see the properties ωx+y = ωx + ωy and ωλx = λω x) and have operator norm equal to 1 (see the property ωx x ). In order to complete the proof, it suﬃces to prove that K is compact in Y with the standard topology.| Write| ≤ k Kk as K = K K , where 1 ∩ 2 K = ω Y : ω x for all x E 1 { ∈ | x| ≤ k k ∈ } and K = ω Y : ω = ω + ω and ω = λω for all λ R and for all x E . 2 { ∈ x+y x y λx x ∈ ∈ } The set K1 may be also written as a Cartesian product of compact intervals

K1 = Π x E[ x , x ]. ∈ −k k k k

Due to Tychonoﬀ’s theorem, K1 is compact in Y . On the other hand, K2 is closed in Y . Indeed, for each ﬁxed λ R, x, y E, the sets ∈ ∈ A = ω Y : ω ω ω = 0 x,y { ∈ x+y − x − y } B = ω Y : ω λω = 0 , λ,x { ∈ λx − x } are closed in Y , since the maps ω ω ω ω and ω ω λω are continuous on Y . Therefore, we may 7→ x+y − x − y 7→ λx − x write K2 as

K = A B . 2  x,y  ∩  λ,x  x,y E x E, λ R \∈ ∈ \ ∈     Finally, K is compact since it is the intersection of a compact set K1 and a closed set K2.

11.4 Reﬂexive spaces and separable spaces

Definition 11.4.1. Let E be a Banach space and let J : E E∗∗ the canonical injection from E into E∗∗ defined → at the beginning of Section 11.3. The space E is said to be reflexive if J is surjective, i. e. J(E) = E∗∗ . When E is reflexive, E∗∗ is usually identified with E. Our next result describes a basic property of reflexive spaces. Theorem 11.4.1 (Kakutani) . Let E be a Banach space. Then E is reflexive if and only if

B = x E : x 1 E { ∈ k k ≤ } is compact in the weak topology σ(E, E ∗). 3This is not the canonical injection J described above, but just the canonical map which identiﬁes a continuous linear functional f ∈ E∗ with an element of the inﬁnite Cartesian product RE . 76 CHAPTER 11. WEAK TOPOLOGIES

Proof. We shall only prove one implication of the theorem, namely we prove that if E is reflexive than the closed unit ball in E is weakly compact. The other implication requires more work, and will be left to the interested reader, see [1, Lemma 3.3, Lemma 3.4, Theorem 3.17]. To prove the first implication, assume E is reflexive. Since J is an isometry, we have J(BE) = BE∗∗ , with BE∗∗ = ξ E∗∗ : ξ E∗∗ 1 . We already know from Banach-Alaoglu’s theorem that BE∗∗ is compact in the { ∈ k k ≤ } 1 topology σ(E∗∗ , E ∗). Therefore, it suffices to prove that J − is continuous from E∗∗ equipped with the σ(E∗∗ , E ∗) topology with values in E equipped with σ(E, E ∗). In view of exercise 11.1.2, we only have to prove that for every 1 f E∗ the map E∗∗ ξ f, J − ξ is continuous on E∗∗ equipped with the σ(E∗∗ , E ∗) topology. But by the definition∈ of J, ∋ 7→ h i 1 f, J − ξ = ξ, f , h i h i and the map ξ ξ, f is indeed continuous on E∗∗ for the topology σ(E∗∗ , E ∗). This proves the assertion. 7→ h i Here are some further properties of reflexive spaces. Proposition 11.4.2. Assume that E is a reflexive Banach space and let M E be a closed linear subspace of E. Then M is reflexive. ⊂ Proof. The space M (equipped with the norm of E) has a priori two distinct weak topologies: the topology induced by σ(E, E ∗), and the topology induced by σ(M, M ∗). In fact, these two topologies are the same. Indeed, by Hahn- Banach theorem every continuous linear functional on M is the restriction to M of a continuous linear functional on E, the details are left as an exercise). In view of Theorem 11.4.1, we have to check that B = x M : x 1 is M { ∈ k kE ≤ } compact in σ(M, M ∗) or equivalently in σ(E, E ∗). However, BE is compact in the topology σ(E, E ∗) and M is closed in the same topology (by Theorem 11.2.5). Therefore, BM is compact in σ(E, E ∗). Exercise 11.4.1. Let E and F two normed spaces with an isometry i : E F . If F is reflexive then E is reflexive. → Solution. Define the map τ : E∗ F ∗ → 1 τf, y = f, i − (y) , for all y F . h i h i ∈ 1 It is easily seen that τ is a linear invertible map from E∗ onto F ∗ (τ is 1 : 1 because τf = 0 implies f, i − (y) = 0 h i for all y F , and since i is an isometry this implies f = 0 . Moreover, for all g F ∗ one can see that g = τf with f = g i).∈ Moreover, τ is isometric: to see this, ∈ ◦ 1 τf F ∗ = sup f, i − (y) = sup f, x = f E∗ . k k y F 1 |h i| x E 1 |h i| k k k k ≤ k k ≤ Similarly, define the isometric linear invertible map ρ : E∗∗ F ∗∗ by → 1 ρξ, g = ξ, τ − g , for all ξ E∗∗ and for all g F ∗, h i h i ∈ ∈ it is an easy exercise to prove that such a map is linear, injective, surjective, and an isometry. Denote by JE and JF the canonical injections of E into E∗∗ and of F into F ∗∗ respectively. Pick an element ξ E∗∗ . Set 1 1 ∈ x = ( i− J − ρ)( ξ). Clearly x E, and for any f E∗ ◦ F ◦ ∈ ∈ 1 1 1 f, x = f, (i− J − ρ)( ξ) = (τf ), (J − ρ)( ξ) , h i h ◦ F ◦ i h F ◦ i ad by definition of JF we have 1 f, x = ρ(ξ), τ (f) = ξ, τ − τ(f) = ξ, f , h i h i h ◦ i h i which proves that f, x = ξ, f for all f E∗, i. e. ξ = J (x). Hence, E∗∗ = J(E) and E is reflexive. h i h i ∈ E Corollary 11.4.3. A Banach space E is reflexive if and only if its dual space E∗ is reflexive.

Proof. Let E be reflexive. Let us prove that E∗ is reflexive. Roughly speaking, since E∗∗ = E, we have that E∗∗∗ = E∗. More precisely, let J : E E∗∗ be the canonical isomorphism between E and E∗∗ . Let ϕ E∗∗∗ → ∈ be given. The map E x ϕ, J (x) R is a continuous linear functional on E. Call it f E∗, so that f, x = ϕ, J (x) for all ∋x E7→. But h we alsoi ∈ have ϕ, J (x) = J(x), f by definition of J, for all x ∈E. Since J is surjective,h i h we inferi that ∈ h i h i ∈ ϕ, ξ = ξ, f for all ξ E∗∗ , h i h i ∈ which means precisely that the canonical injection from E∗ into E∗∗∗ is surjective. Now assume E∗ is reflexive. From the above step we get that E∗∗ is reflexive. Since J(E) is a closed subspace of E∗∗ in the strong topology, we conclude by Proposition 11.4.2 that J(E) is reflexive. From the exercise 11.4.1, E is reflexive. 11.4. REFLEXIVE SPACES AND SEPARABLE SPACES 77

The next deﬁnition is a repetition of a property of topological spaces. Deﬁnition 11.4.2. A metric space E is separable if there exists a subset D E that is countable and dense. ⊂ Theorem 11.4.4. Let E be a Banach space such that E∗ is separable. Then, E is separable.

Proof. Let f be a countable dense set in E∗. Since { n}n fn = sup fn, x , k k x E 1h i k k ≤ we can ﬁnd some x E such that x = 1 and f , x 1 f . Let L be the vector space generated over Q n ∈ k nk h n ni ≥ 2 k nk 0 by the sequence xn n, i. e. { } N L = x = λ x : λ Q, N N . 0 { i i i ∈ ∈ } i=1 X We claim that L0 is countable. To see this, let Λ n be the linear space over Q generated by x1,...,x n. Clearly Λ n is countable, and hence L0 = n 1 Λn is countable. ≥ Let L denote the vector space over R generated by the xn n. Of course, L0 is a dense subset of L. We claim S { } that L is a dense subspace of E, which would conclude the proof. Let f E∗ be a continuous linear functional that vanishes on L. In view of Corollary 9.2.6, it suﬃces to prove that f =∈ 0. Given ǫ > 0, there is some integer N such that f f < ǫ . Since f vanishes on x L, we have k − N k N ∈ 1 f f , x = f f, x < ǫ. 2k N k ≤ h N N i h N − N i It follows that f f f + f < 3ǫ, thus f = 0. k k ≤ k − N k k N k The converse of the previous theorem is not true.

Corollary 11.4.5. Let E be a Banach space. Then, E is reﬂexive and separable if and only if E∗ is reﬂexive and separable.

Proof. We already know (from Corollary 11.4.3 and Theorem 11.4.4) that if E∗ is reflexive and separable, so is E. Conversely, if E is reflexive and separable, so is E∗∗ = J(E) (in view of exercise 11.4.1, and because the density property is preserved under isometries). Thus, E∗ is reflexive and separable. Separability properties are closely related to the metrizability of the weak topologies. Let us recall that a topological space X is called metrizable if there is a metric on X that induces the topology of X.

Theorem 11.4.6. Let E be a separable Banach space. Then B ∗ is metrizable in the weak topology σ(E∗, E ). E ∗ Conversely, if B ∗ is metrizable in the weak topology σ(E∗, E ) then E is separable. E ∗ Proof. Omitted. We refer to [1, Theorem 3.28] for the interested reader.

Theorem 11.4.7. Let E be a Banach space such that E∗ is separable. Then BE is metrizable in the weak topology σ(E, E ∗). Conversely, if BE is metrizable in the weak topology σ(E, E ∗), then E∗ is separable. Proof. Omitted. We refer to [1, Theorem 3.29] for the interested reader.

Corollary 11.4.8. Let E be a separable Banach space and let f be a bounded sequence in E∗. Then there { n}n exists a subsequence f that converges in the weak topology σ(E∗, E ). nk ∗ Proof. It follows from the previous Theorem 11.4.6 and from the fact that, on metric spaces the notion of topological compactness and sequential compactness coincide (Theorem 7.1.1). We can finally state what is probably the most important result of this part. Theorem 11.4.9 (Weak compactness on reflexive spaces) . Assume that E is a reflexive Banach space and let x { n}n be a bounded sequence in E. Then there exists a subsequence xnk that converges in the weak topology σ(E, E ∗).

Proof. Let M0 be the vector space generated by the xn’s and let M = M0. Clearly, M is separable (see the proof of theorem 11.4.4). Moreover, M is reﬂexive (by Proposition 11.4.2). By corollary 11.4.5, M ∗ is separable and reﬂexive. Hence, by Theorem 11.4.7 BM is metrizable in the σ(E, E ∗) topology. Hence, by Kakutani’s theorem BM is compact in a metrizable space, hence it is sequentially compact. Therefore, there is a convergent subsequence of x in the topology σ(M, M ∗), which coincides with the weak topology σ(E, E ∗). { n}n 78 CHAPTER 11. WEAK TOPOLOGIES 11.5 Uniformly Convex Space

Deﬁnition 11.5.1. A Banach space E is said to be em uniformly convex if for any ε > 0 there exists δ > 0 such that x + y [x, y E, x 1, y 1 and x y > ε ] = < 1 δ ∈ k k ≤ k k ≤ k − k ⇒ 2 −

Roughly speaking, uniform convexity asserts that if we consider a segme nt of length at least ε in the unit ball of E, then its midpoint must stay within the unit ball. 2 1 For example, if E = R and we consider x 2 = ( x1 + x2 ) 2 and x 1 = x1 + x2 we have that x 2 is uniformly convex while x is not. If E is a uniformlyk k | convex| | space| the followingk k | important| | | theorem holds.k k k k1 Theorem 11.5.1 (Milman-Pettis) . Every uniformly convex Banach space is reﬂexive.

Proof. We omit the proof and we refer to [1, Theorem 3.31].

Uniform convexity is a geometric property of the norm; an equivalent norm need not be uniformly convex. On the other hand, reflexivity is a topological property: a reflexive space remains reflexive for an equivalent norm. It is a striking feature of Theorem 11.5.1 that a geometric property implies a topological property. We stress that the contrary is ingeneral not true; there exist reflexive spaces that admit no uniformly convex equivalent norm. The following proposition is a useful property of uniformly convex space.

Proposition 11.5.2. Assume that E is a uniformly convex Banach space. Let (xn) be a sequence in E such that

xn ⇀ x weakly in σ(E, E ∗),

lim sup xn x . n k k ≤ k k →∞ Then x x strongly. n → Proof. We may always assume that x = 0 otherwise the conclusion is obvious. Set 6 1 1 λ = max( x , x ), y = λ− x , y = x − x. n k nk k k n n n k k

Then, λ x and y ⇀ y weakly in σ(E, E ∗). In particular, it follows Proposition ?? that n → k k n y + y y lim inf n . k k ≤ n 2 →∞

Since yn 1 and y = 1 we have that k k ≤ k k y + y n 1, 2 →

which in turn implies, by using the uniform convexity of E , that y y 0 and then x converge strongly to n n x. k − k → Chapter 12

Exercises

Exercise 12.1. Let (X, d ) be a metric space. Deﬁne

C(X) = f : X R : f is continuous { → } C (X) = f C(X) : f is continuous and bounded b { ∈ } C (X) = f C(X) : f is continuous and zero outside a compact subset of X . c { ∈ } For all f C(X), deﬁne the uniform norm of f as ∈ . f = sup f(x) . k k∞ x X | | ∈ (a) Prove that there are metric spaces X for which (C(X), ) is not a metric spaces. k · k ∞ (b) Prove that (C (X), ) is a Banach space. b k · k (c) Prove that there are metric spaces X for which Cc(X) is not closed in Cb(X). (d) Deﬁne C (X) = C (X). Find an example of a metric space X such that C (X) C (X) = . 0 c 0 \ c 6 ∅ (e) If X is compact, prove that (C(X), ) is a Banach space and that C(X) = Cb(X) = C0(X) = Cc(X). k · k ∞

Exercise 12.2. For x, y R let ∈ x y d(x, y ) = | − | . 1 + x y | − | Prove that d is a distance on R.

Exercise 12.3. Let a, b R, a < b . Let α : [ a, b ] R be a continuous function with α(x) β > 0 for some β > 0. For two functions f, g ∈C ([ a, b ]) , deﬁne → ≥ ∈ b

dα(f, g ) = sup [α(x) f(x) g(x) ] . x [a,b ] | − | ∈

Prove that dα is a metric. Prove that (Cb([ a, b ]) , d α) is a complete metric space. Is such distance induced by a norm? Which one? Is the eventual normed space obtained a Banach space? Motivate your answers.

d Exercise 12.4. Let Ω R be an open set. Let Ωm m N be an increasing sequence of bounded open sets such that + ⊂ { } ∈ Ωm Ωm+1 , Ω = m∞=1 Ωm. Prove that the space C(Ω) of continuous functions f : Ω R is a complete metric space⊂ with the metric → S + ∞ 1 max x Ω f(x) g(x) ρ(f, g ) = ∈ | − | . 2m 1 + max f(x) g(x) m=1 x Ω X ∈ | − | Prove also that ρ(f , f ) 0 if and only if f converges to f uniformly on compact subsets of Ω. m → m 79 80 CHAPTER 12. EXERCISES

+ Exercise 12.5. Let 1 p < q + . Prove that ℓ ℓ (Hint: use that is ∞ y < + then y 0 as ≤ ≤ ∞ p ⊂ q k=1 | k| ∞ k → k + ). Moreover, prove that if a sequence xn n N converges to zero in the ℓp norm then it converges to zero→ in∞ the norm. { } ∈ Pk · k k · k ℓq

Exercise 12.6. Let c = x ℓ : lim k + xk exists . Prove that c is closed in ℓ . Let c0 = x ℓ : { ∈ ∞ → ∞ } ∞ { ∈ ∞ lim k + xk = 0 equipped with the ℓ -norm. Prove that c0 is closed in c. → ∞ } ∞

Exercise 12.7. Consider the sequence n x = , for n, k N. n,k k2 + n2 ∈

For all n N, let xn := xn,k k N. Prove that xn ℓ1 ℓ for all n 1. Prove that xn 0 in ℓ but xn does ∈ { } ∈ ∈ ∩ ∞ ≥ → ∞ not converge to zero in ℓ1.

Exercise 12.8. Let (A, d ) be a complete metric space, and let Lip( A; R) be the space of Lipschitz continuous functions from A into R. For every f Lip( A; R), consider the number ∈ f(x) f(y) [f]Lip := sup | − |, x,y A d(x, y ) ∈ and the quantity f Lip = f + [ f]Lip , k k k k∞ where f is the usual uniform norm of f, i. e. f = sup x A f(x) . Prove that Lip is a norm on Lip( A; R). Provek thatk∞(Lip( A; R), ) is a Banach space. k k∞ ∈ | | k·k k · k Lip

Exercise 12.9. Let A = [ a, b ] R. Let C(A) the (Banach) space of continuous functions on A equipped with the usual uniform norm.⊂ Prove with an example that the linear subspace Lip( A) C(A) of all Lipschitz continuous functionsk · on k ∞A is not closed in (C(A), ). Explain why this is not a contradiction⊂ with the fact that the space of Lipschitz functions is a Banach spacek (and · k ∞ in particular a complete space) from the previous exercise.

Exercise 12.10. With the notation of the previous exercise, prove that Lip( A) is dense in (C(A), ) (Hint: for a given continuous function f C(A), construct an approximation of f in the uniform norm whichk · k ∞ is a Lipschitz function). ∈

Exercise 12.11. Let A = [ a, b ] R, and let α (0 , 1) . Let C0,α (A) be the space of α-H¨older continuous functions on A, i. e. the set of functions f⊂: A R such∈ that f(x) f(y) C x y α for all x, y A. For all f C0,α (A), set → | − | ≤ | − | ∈ ∈ f(x) f(y) 0,α [f]α := sup | − α |, f C = f + [ f]α. x,y A x y k k k k∞ ∈ | − | 0,α 0,α Prove that is a norm on C . Moreover, prove that the space C equipped with 0,α is a Banach space. k · k α k · k C

Exercise 12.12. Let A = [ a, b ] R, and let α (0 , 1) . Let C(A) be the space of continuous functions on A equipped with the uniform norm. Let⊂ ∈

M := f C(A) : f 0,α K . K { ∈ k kC ≤ }

Prove that MK is a compact subset of C(A). 81

Exercise 12.13. Let (l2, d ) be the complete metric space given by

2 ∞ 2 l = x = xn n N : xn R and xn < { } ∈ ∈ ∞ ( n=1 ) X and 1 2 ∞ d(x, y ) = x y 2 . | n − n| !n=1 " X Show that 1. The set Q = x l2 : x 1 for any n N { ∈ | n| ≤ ∈ } is not bounded. 2. The set ∞ B = x l2 : x2 1 { ∈ n ≤ } n=1 X is bounded but not totally bounded. 3. The set 1 H = x l2 : x for any n N { ∈ | n| ≤ n ∈ } is compact.

Exercise 12.14. Let fn n N C([0 , 1]) such that sup x [0 ,1] fn(x) 1 for any n N. Deﬁne Fn : [0 , 1] R by { } ∈ ⊂ ∈ | | ≤ ∈ → x Fn(x) = fn(t) dt. Z0

Show that the sequence Fn n N has a subsequence that converges uniformly on [0 , 1] . { } ∈

Exercise 12.15. Let G the following subset of C([0 , π ])

π G = g C([0 , π ]; g(x) = sin( xy )f(y) dy, f C([0 , π ]) , f 1 . ∈ ∈ k k∞ ≤ Z0 Prove the following statement. 1. G is relatively compact in in C([0 , π ]) . 2. G is relatively compact in L1([0 , π ]) . Exercise 12.16. Let f : R3 R be a continuous function and let A : C([0 , 1]) C([0 , 1]) the map deﬁned as follows → → 1 (Au )( x) = f(x, s, u (s)) ds, u C([0 , 1]) . ∈ Z0 Prove the following statement. 1. A is well deﬁned. 2. A is a continuous map. 3. If is a bounded subset of C([0 , 1]) , A( ) is relatively compact in C([0 , 1]) . U U

Exercise 12.17. Show that C([0 , 1]) , the space of continuous function from [0 , 1] to R, with the norm f = k k1 1 f(t) dt is not complete. 0 | | R 82 CHAPTER 12. EXERCISES

Exercise 12.18. Let X and Y be metric spaces and f : X Y be a continuous function. Show that if K X is compact then f(K) Y is compact as well. → ⊂ ⊂

Exercise 12.19. Prove that the set B = f C([0 , 1]) : f 1 { ∈ k k∞ ≤ } is not relatively compact with respect to the sup-norm. Hint: ﬁnd a sequence in B which admits no convergent subsequences.

2 Exercise 12.20. Let f : R R R be a continuous function in a neighborhood of the point (t0, y 0) R . Then, there exists ε > 0 such that× the ordinary→ diﬀerential equation ∈

y′(t) = f(t, y (t)) (y(t0) = x0 has a solution deﬁned on the interval [t ε, t + ε]. 0 − 0

Exercise 12.21. Let T > 0 and A : C([0 , T ]) C([0 , T ]) be the operator deﬁned as → t t s (Af )( t) = exp − f(s) ds. Z0 Show that is a linear continuos operator and compute and there exists T > 0 such that A < 1. k k

Exercise 12.22. Consider the space C([0 , 2]) with the maximum norm, f = sup x [0 ,2] f(x) . Let T be the functional deﬁned by k k∞ ∈ | | 1 2 T f = f(x) dx f(x) dx. − Z0 Z1 Show that T is a bounded linear functional and evaluate its norm.

Exercise 12.23. Consider the operator

T : ( C([0 , 1]) , ) (C([0 , 1]) , ), k · k ∞ → k · k ∞ deﬁned by 1 (T f )( s) = k(s, t )f(t) dt, s [0 , 1] . ∈ Z0 where k : [0 , 1] [0 , 1] R is the function × → (1 s)t if 0 t < s k(s, t ) = (1 − t)s if s ≤ t 1 − ≤ ≤ Show that T is a linear continuous operator and compute T . k k

Exercise 12.24. Let β > 0 and T : L2(0 , 1) L2(0 , 1) deﬁned as → (T f )( x) = f(xβ).

Show that 1. T is well deﬁned if and only if β (0 , 1] . ∈ 83

2. For β (0 , 1] T is a linear continuous operator and compute T . ∈ k k

Exercise 12.25. Let T : l2 l2 deﬁned as → (T x ) = a x n N. n n n+1 ∈ where an > 0 and lim an = 0 . n →∞ Show that T is a linear continuous operator and compute T . k k

Exercise 12.26. Let T : L∞(0 , 1) L∞(0 , 1) deﬁned as → 1 x (T f )( x) = g(t) dt. x Zx Show that T is a linear continuous operator and compute T . k k

Exercise 12.27. Deﬁne the right shift operator on ℓ2, p [1 , + ), ∈ ∞ 0 if n = 1 (Tra)n = a = an n. (an 1 if n 2, { } − ≥ Show that T is a linear continuous operator and compute T . r k rk

Exercise 12.28. Let C1([0 , 1]; R) be the space of diﬀerentiable function with continuous derivative equipped with the norm f = max f(x) . k k∞ x [0 ,1] | | ∈ Show that the operator T : X R deﬁned as → (T f ) = f ′(0) is not continuous.

Exercise 12.29. Consider the operator

A : ( C1([0 , 1]) , ) (C([0 , 1]) , ) k · k ∞ → k · k ∞ deﬁned by d (Af )( t) = f(t). dt Show that A is a linear operator but it is not continuous.

Exercise 12.30. Let X, Y be normed vector space, Tn n N (X, Y ) and T (X, Y ). We say that { } ∈ ⊂ L ∈ L Tn n N converges uniformly to T if • { } ∈ lim Tn T X Y = 0 . n → →∞ k − k

Tn n N converges strongly to T if for any x X • { } ∈ ∈

lim Tnx T x Y = 0 . n →∞ k − k 84 CHAPTER 12. EXERCISES

Prove the following statements.

1. If Tn n N converges uniformly to T then it converges strongly to T . { } ∈ 2. Let T : l1 l1 be the sequence of operator deﬁned by n → n n 1 T x = (T x)k k N = xk+n k N, x = xk k N l . { } ∈ { } ∈ { } ∈ ∈

Show that Tn are bounded, uniformly in n, operators, Tn n N converges strongly to T = 0 , but it does not converge uniformly. { } ∈

m Exercise 12.31. Let A = ( aij ) be an m n matrix with real coeﬃcients. With the notation y = ( y1,...,y m) R and x = ( x ,...,x ) Rn, consider the× linear map T : Rn Rm given by ∈ 1 n ∈ → y = T x = Ax.

Given p [1 , + ], consider the p-norms on Rn and Rm defined as usual for x Rn by ∈ ∞ ∈ p p 1/p x p = ( x1 + . . . + xn ) , x = max x1 ,..., xn , k k | | | | k k∞ {| | | |} m and a similar definition for y R . For a given p [1 , + ], define A p as the operator norm of T seen as a linear bounded operator from (R∈n, ) to (Rm, ∈). ∞ k k k · k p k · k p (a) Prove that A = max λ : λ is an eigenvalue of AT A . k k2 {| | } (b) Prove that n A = max 1 i m aij . k k∞ ≤ ≤  | | j=1 X  (c) Prove that   m A = max 1 j n a . k k1 ≤ ≤ | ij | (i=1 ) X Exercise 12.32. Show that the strong and weak convergence in l1 are equivalent.

Exercise 12.33. Let E be a Banach space and let A E be a subset that is compact in the weak topology σ(E, E ∗). Prove that A is bounded. ⊂

Exercise 12.34.

′ 1. Let xn lp with 1 p and assume that xn ⇀ x in the weak topology σ(lp, l p ). Prove that { }n ⊂ ≤ ≤ ∞ (a) xn is bounded in lp. { }n (b) xn x where xn = ( xn, x n, ..., x n, ... ) and x = ( x , x , ..., x , ... ). i → i 1 2 n 1 2 n 2. Suppose that xn lp with 1 < p . Assume that (a) and (b) hold for some limit x. Prove that x lp n ′ and xn ⇀ x in{ the} weak⊂ topology σ(l≤p, l ∞p ). ∈

n n n Exercise 12.35. For any n N let e be the inﬁnite sequence e = ei i N = δn,i i N. ∈ { } ∈ { } ∈ ′ 1. Show that, for 1 < p < + , en ⇀ 0 in the weak topology σ(lp, l p ). ∞ nk 1 1 2. Prove that there exists no subsequence e k N that converges in l with respect to the topology σ(l , l ∞). { } ∈ 85

3. Construct an example of a Banach space E and a sequence f in E∗ such that f = 1 for any n N { n}n k nk ∈ and f has no subsequence that converges in σ(E∗, E ). { n}n Hint: Take E = l∞.

Exercise 12.36. Let E = lp and F = lq with 1 < p, q < . Let a : R R be a continuous function such that ∞ → p a(t) C t q . | | ≤ | | Given x lp with components x . Set ∈ { i}i

Ax = ( a(x1), a (x2), a (x3), .... ).

1. Prove that Ax lq and the map x Ax is continuous from lp with the strong topology to lq with the strong topology. ∈ →

′ ′ 2. Prove that if xn lp such that xn ⇀ x in σ(lp, l p ) then Ax n ⇀ Ax in σ(lq, l q ). { }n ⊂ 3. Prove that A is continuous from BE equipped with the weak topology σ(E, E ∗) into F equipped with σ(F, F ∗). 86 CHAPTER 12. EXERCISES Part IV

Lp spaces and Hilbert spaces

Chapter 13

Lp spaces

p p 13.1 Density properties and separability of L spaces. Lloc spaces Definition 13.1.1 (Support of a continuous function) . Let Ω Rd be an open set. Let f C(Ω) 1. The support of f in Ω is the set ⊂ ∈ spt( f) = x Ω : f(x) = 0 . { ∈ 6 } If spt( f) is compact, we say that f is compactly supported . The space of compactly supported functions on Ω is denoted by C (Ω). We notice that C (Ω) C (Ω) C(Ω). c c ⊂ b ⊂ We now introduce the notion of distance between sets. Definition 13.1.2. Let x Rd, A, B Rd. We set ∈ ⊂ d(x, A ) = inf x y : y A , {k − k ∈ } and d(A, B ) = inf x y : x A, y B . {k − k ∈ ∈ } Here is the Euclidean norm on Rd. k · k Exercise 13.1.1. The distance function defined above has the following properties.

d(A, B ) = inf x A d(x, B ) = inf y B d(y, A ) (exercise). • ∈ ∈ The map Rd x d(x, A ) is continuous, indeed, the map is Lipschitz continuous with Lipschitz constant 1, • i. e. ∋ 7→ d(x, A ) d(y, A ) x y . | − | ≤ k − k Left as an exercise. Hint: take an arbitrary point z A and use the triangular inequality. ∈ If K, C Rd with K compact, C closed, and K C = , then d(K, C ) > 0 (exercise). • ⊂ ∩ ∅ Let A Rd and δ > 0. Set • ⊂ A = x Rd : d(x, A ) δ . δ { ∈ ≤ } d Prove that A Aδ and Aδ is closed. Moreover, if K R is compact then Kδ is compact. This is an easy exercise. ⊂ ⊂

As a consequence of the above exercises, let Ω Rd be open, and let K Ω be compact. Let δ = d(K, Rd Ω) . • ⊂ ⊂ 0 \ Clearly, δ0 > 0. Hence, for all δ < δ 0 one has

◦ K K K Ω. ⊂ δ ⊂ δ ⊂ The next proposition is a special case of a more general result in topology known as Urysohn’s lemma 2.

1We recall that C(Ω) is the space of continuous functions from Ω to R 2http://en.wikipedia.org/wiki/Urysohn’s_lemma

89 90 CHAPTER 13. LP SPACES

d Proposition 13.1.1. Let K Ω R , with K compact and Ω open. Then, there exists ϕ Cc(Ω) such that ϕ(x) = 1 for all x K and 0 ⊂ϕ(x⊂) 1 for all x Ω. ∈ ∈ ≤ ≤ ∈ Proof. Let δ such that 0 < δ < d (K, Rd Ω). Set \ d(x, Rd K ) ϕ(x) = \ δ . d(x, Rd K ) + d(x, K ) \ δ d Clearly, d(x, R Kδ) + d(x, K ) = 0 for all x Ω. Indeed, if d(x, K ) = 0 then x K, and therefore d(x, y ) δ d \ 6 ∈ ∈ ≥ for all y R Kδ. Moreover, ϕ(x) [0 , 1] for all x Ω, and x K implies d(x, K ) = 0 and ϕ(x) = 1. Finally, ϕ(x) = 0∈ only\ if d(x, Rd K ) = 0, which∈ is equivalent∈ to x K .∈ Hence, the support of ϕ is compact in Ω. 6 \ δ 6 ∈ δ Theorem 13.1.2 (Density of C in Lp). Let Ω Rd be an open set. c ⊂ (i) The space (Ω) of simple functions which are bounded and zero outside a compact set is dense in Lp if p [1 , + )S. ∈ ∞ (ii) C (Ω) is dense in Lp(Ω) if p [1 , + ). c ∈ ∞ (iii) C (Ω) is not dense in L∞(Ω) . (Ω) is dense in L∞(Ω) if Ω is bounded. c S p Proof. Proof of (i). Let f L (Ω). We have to construct a sequence of simple functions φj (Ω) with φj f Lp 0 as j + . Assume ﬁrst∈ that f 0. We know from Lemma 4.2.1 that there exists a∈ sequence S ofk nonnegative− k → → ∞ ≥ simple functions φj (Ω) with φj f almost everywhere in case f is zero outside a bounded set. In the general case of f 0, for a given∈ S n N thereր exists a sequence φ f1 . Consider the diagonal sequence φ (Ω). ≥ ∈ n,j ր Bn(0) j,j ∈ S Let ǫ > 0. For almost every x Ω one has x Bj (0), and hence f(x) φj,j ǫ for j large enough. Hence, the claim is true also for a general ∈f 0. Now, this∈ implies | − | ≤ ≥ 0 f φ p f p, ≤ | − j | ≤ | | almost everywhere on Ω. Therefore, we can apply Lebesgue dominated convergence theorem 5.1.4 to get

f(x) φ (x) pdx 0 as j + . | − j | → → ∞ Z The general case f sign changing can be solved by splitting f = f+ f , constructing sequences of simple functions − − as above for f+ and f , and applying the previous step. Proof of (ii). Let −f Lp(Ω) and let ǫ > 0. Due to (i) there exists a simple function φ on Ω such that ǫ ∈ ǫ f φ p . The proof will be completed once we ﬁnd a continuous function g on Ω such that g φ p . k − kL ≤ 2 k − kL ≤ 2 Assume ﬁrst that φ = α1F for some measurable bounded set F Ω and some α R. Fix σ > 0. Let K F A, K compact and A open, such that m(A) m(K) < σ . From Proposition⊂ 13.1.1, we∈ know that there exists a⊂ function⊂ g˜ C (A) such that 0 g˜ 1 andg ˜ =− 1 on K. Let g = αg˜. We have ∈ c ≤ ≤

p p p p p g φ p = g(x) φ(x) dx = αg˜ α1 dx α m(A K) α σ, k − kL | − | | − F | ≤ \ ≤ ZΩ ZΩ p ǫ and choosing σ = ( ǫ/ 2α) one has g φ p . Assume now that k − kL ≤ 2 N

φ = αj1Fj , j=1 X with Fj measurable and bounded sets. From the previous case we can ﬁnd continuous functions gj such that ǫ p gj 1Fj L . k − k ≤ 2N N α j=1 | j| N P Set g = j=1 αj gj, we have P N N ǫ g φ p = α (g 1 ) p α g 1 p , k − kL k j j − Fj kL ≤ k j| j − Fj |k L ≤ 2 j=1 j=1 X X P P 13.1. DENSITY PROPERTIES AND SEPARABILITY OF L SPACES. LLOC SPACES 91 and the assertion (ii) is proven. Proof of (iii). C (Ω) cannot be dense in L∞(Ω). Indeed, take f L∞(Ω) discontinuous at one point. The c ∈ density property would imply that there exists a sequence of continuous functions fj on Ω that converge in L∞ to f. But for continuous functions the convergence in L∞ is equivalent to the uniform convergence, and this is in contradiction with a well known convergence property of sequences of functions. If Ω is bounded, then the statement that simple functions are dense in L∞ is an immediate consequence of Lemma 4.2.1.

p p Theorem 13.1.3 (Separability of L ). L (Ω) is separable if p [1 , + ). L∞(Ω) is not separable. ∈ ∞ Proof. Case p < + . For simplicity let us assume Ω = Rd (the general case is a simple consequence of this special case). Let denote∞ the countable family of sets in Rd of the form R = Π d (a , b ) with a , b Q. Let denote R k=1 k k k k ∈ E the vector space over Q generated by the functions 1R R , that is, consists of all finite linear combinations with rational coefficients of indicator functions of sets{ in } .∈R It is easily seenE that is a countable set. We claim that is dense in Lp(Rd). Let f Lp(Rd) andR let ǫ > 0. From TheoremE 13.1.2 there exists f C (Rd) E ∈ 1 ∈ c such that f f p ǫ. Let R be any cube containing the support of f . Given δ > 0 it is easy to construct k − 1kL ≤ ∈ R 1 a function f2 such that f1 f2 Lp δ and f2 vanishes outside R. Indeed, it suffices to split R into small cubes of where∈ E the oscillationk of− f kis less≤ than δ (f is uniformly continuous on R!). Therefore we have R 1 1 1/p 1/p f f p f f ∞ m(R) < δm (R) . k 1 − 2kL ≤ k 1 − 2kL 1/p We conclude that f f p < 2ǫ provided δ > 0 is chosen so that δm (R) < ǫ . k − 2kL Case p = + . If we prove that there exists a family of open balls in L∞(Ω) which are pairwise disjoint and with an uncountable∞ cardinality, the proof will be completed. Indeed, if such property is satisfied, any dense subset S in L∞ should have at least one element in each of the above open balls, and this makes it impossible for S to be countable. Now, given two open balls B, B ′ Ω, assuming that B = B′, one has ⊂ 6 1 1 ′ ∞ = 1 , k B − B kL and the proof is an easy exercise. Now, for a given ball B Ω, set ⊂ 1 U = g L∞ : g 1 ∞ < . B { ∈ k − BkL 2} Clearly, the family = U : B is an open ball in Ω U { B } is more than countable, and every two distinct elements in are disjoint. This proves the assertion. U Remark 13.1.1 . Let Ω Rd be open, and let p, q [1 , + ] with p q. Is there any relation between Lp and Lq? More presicely, is one of⊂ the two spaces a subset of∈ the other∞ one? In≤ general the answer is negative. As an example, let p = 1 and q = 2, and let Ω = (0 , + ) R. Take f(x) = 1 . Clearly, f L1(Ω), where as f L2(Ω). Now, ∞ ⊂ 1+ x 6∈ ∈ let g(x) = 1 1 . Clearly, g L1(Ω) but g L2(Ω). Hence, L1(Ω) is not a subset of L2(Ω) and L2(Ω) is not a √x (0 ,1) ∈ 6∈ subset of L1(Ω). On the other hand, if m(Ω) < + , the Lp spaces are ordered. Indeed, let p q: then Lq(Ω) Lp(Ω). To see this, assume first q < + . We compute∞ ≤ ⊆ ∞ f pdx = f pdx + f pdx f qdx + m(Ω) f qdx + m(Ω) , Ω | | f 1 | | f <1 | | ≤ f 1 | | ≤ | | Z Z| |≥ Z| | Z| |≥ Z and hence f q < + implies f q < + . Now, let us consider the case q = + . We have k kL ∞ k kL ∞ ∞

p p f f ∞ m(Ω) , | | ≤ k kL ZΩ and this proves the assertion. d p Deﬁnition 13.1.3. Let Ω R be open. Let p [1 , + ]. The vector space Lloc (Ω) is the set of all measurable functions f : Ω R such that,⊂ for every compact∈ subset ∞K Ω, one has f1 Lp(Ω), or equivalently f Lp(K). → ⊂ K ∈ ∈ p p Exercise 13.1.2. Let p [1 , + ]. Prove that if f L than f Lloc . Show that the converse is not true in general. Prove that Lq ∈Lp if∞p q. ∈ ∈ loc ⊂ loc ≤ 92 CHAPTER 13. LP SPACES

1 d Proposition 13.1.4. Let u Lloc (Ω) , Ω R open. If Ω uφdx = 0 for all φ Cc(Ω) , then u = 0 almost everywhere on Ω. ∈ ⊂ ∈ R Proof. Let 1 Ω = x Ω : x < R , and d(x, Rd Ω) > . R { ∈ | | \ R} 1 For a given R 0, let E ΩR be a (bounded) measurable set. Clearly, 1E L (Ω R). Now, since Cc is dense in L1(Ω ), there≥ exists a sequence⊂ φ C (Ω ) with φ 1 as j + . From∈ Theorem 7.2.3, there exists a R j ∈ c R j → E → ∞ subsequence of φj that converges to 1E almost everywhere. By abuse of notation, we keep the notation φj for such a subsequence. Now, let us set ψ = ( φ 1) = max 0, min φ , 1 . Since 1 0, we have ψ 1 almost j j ∧ + { { j }} E ≥ j → E everywhere. Moreover, 0 ψj 1, and ψj Cc(Ω). The above implies that uψ j u1E almost everywhere as ≤ ≤ ∈1 1 → j + . Since uψ j u , and since u L (Ω R) ( u Lloc and ΩR is compact!), by the dominated convergence theorem→ ∞ 5.1.4 we| have| ≤ | | | | ∈ ∈ 0 = uψ dx u1 dx. i → E ZΩ Z This implies udx = 0. Since E is an arbitrary subset of Ω , the third property in Lemma 4.3.3 implies u 0 on E R ≡ ΩR. Therefore, u 0 on Ω = N N ΩN . R ≡ ∈ S 13.2 Dual space of Lp and reﬂexivity of Lp spaces

In this section Ω denotes an open subset of Rd and Lp := Lp(Ω). We want to study the reﬂexivity and the dual of Lp. In order to study reﬂexivity we start by proving that Lp is uniformly convex. Theorem 13.2.1. Lp is uniformly convex for any 1 < p < . ∞ Proof. We divide the proof in two cases. Case 1: 2 p < . Let a, b R, we claim that ≤ ∞ ∈ a + b p a b p a p b p + − | | + | | . (13.1) 2 2 ≤ 2 2

First, we note that for any α, β > 0 p αp + βp (α2 + β2) 2 . (13.2) ≤ Indeed, by homogeneity it is enough to assume that β = 1 and then (13.2) follows by observing that the function

p (x2 + 1) 2 xp 1 − − a+b a b is increasing on [0 , ). Choosing α = and β = − we obtain ∞ 2 2 p p 2 a + b p a b p a + b 2 a b 2 a2 b2 2 ap bp + − + − = + + , 2 2 ≤ 2 2 2 2 ≤ 2 2 ! "

p where in the last line we that the function x x 2 is convex for p 2. Let f, g Lp, by using (13.1) we have that→ | | ≥ ∈ f + g p f g p 1 + − f p + g p . (13.3) 2 2 ≤ 2 k kp k kp p p Let ε > 0, f 1, g 1 and f g > ε , then from (13.3) we have that k kp ≤ k kp ≤ k − kp f + g p ε p < 1 , 2 − 2 p 1 p p therefore Lp is uniformly convex for any p 2 with δ = 1 1 ε 2 . ≥ − − 2 Case 2: 1 < p 2. First we note that for any 1 < p < and any a, b R there exists C = C(p) > 0 such that ≤ ∞ ∈ p p p 2 p p p 1 p p a + b a b C ( a + b ) − 2 a + b 2 . (13.4) | − | ≤ | | | | | | | | − 2

13.2. DUAL SPACE OF LP AND REFLEXIVITY OF LP SPACES 93

p Indeed, again by homogeneity, it is enough to assume that b = 1. Let s = 2 , it is easy to prove that

p 1 s p t+1 s ( t + 1) − t + 1 2 inf | | | | − 2 > 0, t [ 1,1] t 1 p ∈ − ( ) | − | then(13.4) follows easily. Let f, g Lp, with 1 < p 2, then ∈ ≤ p p p 2 p p p 1 p p f + g f g C ( f + g ) − 2 f + g 2 dx k − kp ≤ | | | | | | | | − 2 Z

Let ε > 0, f p 1, g p 1 and f g p > ε , then, after using H¨older inequality and noticing that k k ≤p k k ≤ k − k f p + g p 2 f+g 0 due to the convexity of t tp, we get | | | | − 2 ≥ → p 2 f + g ε 1 2 . 2 ≤ − p p (2 C)

1 p p ε2 Thus, L is uniformly convex with δ = 1 1 2 . − − (2 C) p ! " Since Lp is uniformly convex from Proposition 11.5.2 follows the following theorem p p p p Theorem 13.2.2 (Radon-Riesz) . Let 1 < p < . Let fn n L and f L . If fn f in σ(L , (L )∗) as n and f f as n , then ∞ { } ⊂ ∈ → → ∞ k nkp → k kp → ∞ f f strongly in Lp. n → As a consequence of Theorem 11.5.1 and 13.2.1 we have the following theorem Theorem 13.2.3. Lp is reﬂexive for any 1 < p < . ∞ Next, we want to characterize the dual of Lp. We start with the following useful lemma. Lemma 13.2.4. Let f : Ω R be a measurable function. Let p, q (1 , + ) with p conjugate of q. → ∈ ∞ (a) If f Lp(Ω) , then ∈ f Lp = max fgdx = max fg dx. q q k k g L , g Lq =1 g L , g =1 | | ∈ k k Z ∈ k k Z

(b) If f Lp then 6∈ sup fg dx = + . g Lq , g =1 | | ∞ ∈ k k Z

Proof. To prove (a) we may clearly assume that f p = 0. By H¨older’s inequality, k kL 6

fg dx f p g q = f p if g q = 1 . | | ≤ k kL k kL k kL k kL Z We set 1 if f(x) > 0

θf (x) = 1 if f(x) < 0 − 0 if f(x) = 0 ,

p 1 p/q  and we deﬁne g = θ f − f −p . It is easily seen that θ is measurable. We have f | | k kL f

q (p 1) q p g dx = f − f −p dx = 1 . | | | | k kL Z Z Moreover, p p/q p p/q fgdx = f f −p dx = f −p = f p . | | k kL k kL k kL Z Z 94 CHAPTER 13. LP SPACES

This proves (a).

To prove (b), write Ω = n N(Ω Bn(0)). Introduce, for all n N, ∈ ∩ ∈ S f(x) if f(x) n and x Ω B (0) , | | ≤ ∈ ∩ n fn(x) = n if f(x) > n and x Ω B (0) ,  | | ∈ ∩ n 0 if x Ω Bn(0) . 6∈ ∩ p  p Then, fn L , fn f almost everywhere, and fn Lp + as n + (we are assuming f L !). Hence, from (a) we∈ have| | ≤ | | k k ր ∞ → ∞ 6∈

sup fg dx max fng dx = fn Lp + , q g q =1 | | ≥ g L =1 | | k k → ∞ k kL Z k k Z as n + . → ∞ Corollary 13.2.5. Let p (1 , + ) and let q be the conjugate of p. Let g Lq(Ω) . Then the functional ∈ ∞ ∈

G(f) = fgdx, for f Lp ∈ Z p is a continuous linear functional on L (Ω) , and G p ∗ = g q . k k(L ) k kL

Proof. G is clearly linear and continuous (exercise!). The equality G (Lp)∗ = g Lq follows from Lemma 13.2.4. Indeed, the latter implies k k k k

g Lq = max fgdx = max G(f) = G p ∗ . p p (L ) k k f L , f Lp =1 f L , f Lp 1 | | k k ∈ k k Z ∈ k k ≤

Theorem 13.2.6. Let 1 < p < + and let q be the conjugate of p. Then to every continuous linear functional p ∞ q F (L )∗ there corresponds a unique element g L such that ∈ ∈

F, f = fgdx, for all f Lp. h i ∈ Z Furthermore, F p ∗ = g q . (13.5) k k(L ) k kL ′ p p Proof. Let T : L (L )∗ deﬁned as →

′ p p T u, f (Lp)∗ Lp = ufdx, u L , f L . h i × ∈ ∈ ZΩ By corollary 13.2.5 it follows that ′ p T u p ∗ = u ′ for any u L . k k(L ) k kp ∈ ′ We claim that T is surjective. Let E = T (Lp ), since E is a closed subspace. it suﬃces to prove that E is dense in p p (L )∗. Let h (L )∗∗ satisfying ′ ∈ p h, T u (Lp)∗∗ (Lp)∗ = 0 for any u L . h i × ∈ Since Lp is reﬂexive, it follows that h Lp and ∈

′ h u dx = 0 for any u Lp . ∈ Zω p 2 Choosing u = h − h is follows that h = 0. | | Corollary 13.2.7. Let 1 < p < + and let q be the conjugate of p. Then there is an isometric isomorphism τ p q ∞ from (L )∗ onto L . 13.2. DUAL SPACE OF LP AND REFLEXIVITY OF LP SPACES 95

p Proof. It is a straightforward consequence of the previous Theorem. We have, for all F (L )∗, τ(F ) = g such that ∈ F, f = fgdx . h i Z

1 Theorem 13.2.8. To every continuous linear functional F (L )∗ there corresponds a unique element g L∞ such that ∈ ∈ F, f = fgdx, for all f L1. h i ∈ Z Furthermore, F 1 ∗ = g ∞ , (13.6) k k(L ) k kL 1 and the map τ : ( L )∗ L∞, τ(F ) = g is an isometric isomorphism. → d 1 1 Proof. Let Ω R measurable and L = L (Ω). Let us consider a sequence Ωn n such that Ω = n∞=1 Ωn and ⊂ 2 { } ∪ m(Ω n) < . It is easy to construct θ L (Ω) such that θ(x) εn > 0 on Ω n (The proof is left for exercise). Let 1 ∞ ∈ ≥ φ (L )∗, then ∈ 2 f φ, θf (L1)∗ L1 , for any f L (Ω) → h i × ∈ is a continuous linear functional on L2(Ω) and by Theorem 13.2.6 there exists v L2(Ω) such that ∈

φ, θf = vf dx for any f L2(Ω) . (13.7) h i ∈ Z Set u(x) = v(x)/θ (x) and note that u is well-deﬁned since θ > 0. Moreover, u is measurable and uχ L2(Ω) n ∈ where χ = 1 . It holds that for any g L∞(Ω) n Ωn ∈

φ, χ g = uχ g dx. (13.8) h n i n Z 2 Indeed, we just choose f = χ g/θ L (Ω) in (13.7). Then we prove that u L∞(Ω) and n ∈ ∈

u φ (L1)∗ . (13.9) k k∞ ≤ k k

Fix a constant C > φ 1 ∗ and set k k(L ) A := x Ω : u(x) > C . { ∈ | | } By choosing g = 1Asign u in (13.8) we have that

u φ (L1)∗ m(A Ωn). A Ωn | | ≤ k k ∩ Z ∩

Then, C m (A Ωn) φ (L1)∗ m(A Ωn). This implies that m(A Ωn) = 0 for any n N and then m(A) = 0. Finally, we claim∩ that≤ k k ∩ ∩ ∈ φ, h = uh dx for any h L1(Ω) . (13.10) h i ∈ Z 1 In order to prove (13.10) it is enough to choose gn = ( n) (n) h for h L (Ω) and noticing that gn h strongly 1 − ∨ ∧ ∈ → in L (Ω). By (13.10) it follows that φ (L1)∗ u and then by (13.10) φ (L1)∗ = u . The uniqueness is left for exercise. k k ≤ k k∞ k k k k∞

1 Corollary 13.2.9. The spaces L (Ω) and L∞(Ω) are not reﬂexive.

Proof. Since L∞ is not separable, from Theorem 11.4.4 we have that the dual of L∞ is not separable. Then, the 1 1 1 previous theorem implies that ( L )∗∗ is not separable. Since L is separable, this implies that L is not reﬂexive, otherwise one would have two separable spaces which are isometric, and this is a contradiction. Now, since L∞ is 1 1 the dual of L , Corollary 11.4.3 implies that L∞ is not reﬂexive since L is not either.

Remark 13.2.1 . As a consequence of the previous results, we can summarise the properties of Lp spaces w.r.t. to dual spaces and weak topologies as follows. 96 CHAPTER 13. LP SPACES

1 1 (1) The dual space (L )∗ of L is isometric to L∞. Hence, one can identify every linear and continuous functional 1 1 F (L )∗ with a measurable function g L∞(Ω). As a consequence of that, a sequence fn L (Ω) converges weakly∈ in L1 to f L1(Ω) if and only if ∈ ∈ ∈

f gdx fgdx, n → ZΩ ZΩ

as n + for all g L∞(Ω). → ∞ ∈ 1 (2) The space L (Ω) is separable , but not reﬂexive . Hence, the closed unit ball f L1 1 is not compact in the {k k ≤ } 1 weak topology . This is not surprising, as we know from the example fn(x) = nχ [0, 1/n ] L (R), fn L1 = 1 for all n N, but no subsequence of f converges weakly in L1 to a limit f L1, because∈ that wouldk k imply ∈ n ∈ 1/n n + n g(x)dx → ∞ f(x)g(x)dx, −−−−−→ R Z0 Z

for all g L∞(R). Now, take g continuous in L∞. One can easily see that ∈ 1/n n + n g(x)dx → ∞ g(0) , −−−−−→ Z0 and the identity g(0) = f(x)g(x)dx R Z cannot be satisﬁed for all continuous functions g in L∞(R). Indeed, if f is not zero almost everywhere on R 0 , then there exists a continuous function g compactly supported on R 0 such that R f(x)g(x)dx = 0 (due\{ to} Proposition 13.1.4), but g(0) = 0, which is a contradiction. On the\{ other,} hand, if f is zero almost6 everywhere, the above is contradicted every time g(0) = 0. R 6 p p q (3) Let p (1 , + ). Then, the dual space ( L )∗ of L is identiﬁed with L with q conjugate of p. Hence, a sequence∈ f ∞Lp(Ω) converges to f Lp(Ω) weakly in Lp if and only if n ∈ ∈

f gdx fgdx, n → ZΩ ZΩ as n + for all g Lq(Ω). → ∞ ∈ p (4) The space L (Ω) for p (1 , + ) is separable and reﬂexive . Hence, the closed unit ball f Lp 1 is compact ∈ ∞ p {k k ≤ } in the weak topology. As a consequence of that, every bounded sequence fn in L has a subsequence which converges weakly.

(5) The space L∞(Ω) is neither reﬂexive nor separable . Moreover, its dual is not easily detectable. This topic is postponed, as it goes beyond the purposes of our course. However, L∞ is itself the dual of another space, 1 1 namely L . Therefore, one can consider the weak ∗ topology σ(L∞, L ). One can easily prove that a sequence f L∞(Ω) converges in the L∞ weak ∗ topology to f L∞(Ω) if and only if n ∈ ∈

f gdx fgdx, n → ZΩ ZΩ as n + for all g L1(Ω). From the Banach-Alaoglu-Bourbaki theorem, we know that the closed unit → ∞ ∈ ball f ∞ 1 is compact in the weak ∗ topology. Since L∞ is the dual of a separable space, then the {k kL ≤ } closed unit ball is metrizable in the weak ∗ topology, and therefore every bounded sequence fn in L∞ has a subsequence which converges weakly ∗. Remark 13.2.2 . All the content of this chapter can be easily re-phrased for functions with values in the complex plane C. The absolute value is obviously replaced by the complex norm. Some proofs should be slightly modiﬁed to use the property that zz = z 2 for all z C, but the extension is a trivial (though tedious) exercise. | | ∈ Chapter 14

Hilbert spaces

In this Chapter we shall introduce Hilbert spaces as spaces with an inner product structure. In this chapter, we make the choice of producing the whole theory on linear spaces on C rather than on R. This choice has mostly a cultural motivation: the most classical applications of the Hilbert space theory occur in quantum physics, in which complex valued functions appear most often.

14.1 Deﬁnition of Hilbert spaces and geometrical properties

Deﬁnition 14.1.1. Let H be a vector space on C. A scalar product on H is a map

( , ) : H H C, · · × → such that (i) ( x, x ) 0 for all x H, and ( x, x ) = 0 if and only if x = 0. ≥ ∈ (ii) ( λx + y, z ) = λ(x, z ) + ( y, z ) for all x, y, z H and for all λ C. ∈ ∈ (iii) ( x, y ) = (y, x ) for all x, y H. ∈ We set x = (x, x ), for all x H, k kH ∈ and refer to it as the norm of x in H. Two nonp zero vectors x, y H are called orthogonal if ( x, y ) = 0. ∈ Exercise 14.1.1 (Cauchy-Schwartz inequality) . Let x, y H and B : H H R be a bilinear map such that ∈ × → i) B[x, x ] 0 for any x H. ≥ ∈ ii) B[x, y ] = B[y, x ] for any x, y H. ∈ 1 1 Then, B[x, y ] (B[x, x ]) 2 (B[y, y ]) 2 . In particular, if B[ , ] = ( , ) it holds that (x, y ) x y . | | ≤ · · · · | | ≤ k kk k Solution. For all λ C we have ∈ 0 B[x + λy, x + λy ] = B[x, x ] + 2Re( λB [x, y ]) + λ 2B[y, y ]. ≤ | | Now, for a given r R we choose ∈ rB [x, y ] λ = , B[x, y ] | | and get 0 B[x, x ] + 2 r B[x, y ] + r2B[y, y ]. ≤ | | Since the above right hand side is nonnegative for all r R, we must have ∈ B[x, y ] 2 B[x, x ]B[y, y ] 0, | | − ≤ which proves the assertion.

97 98 CHAPTER 14. HILBERT SPACES

Exercise 14.1.2. H is an actual norm on H. The only property that needs to be proven is the triangular inequality. This is ak simple · k consequence of Cauchy-Schwartz inequality (exercise). Deﬁnition 14.1.2. Let H be a vector space on C and let ( , ) be a scalar product on H. H is called a Hilbert space if the normed space ( H, ) is complete. · · k · k H Remark 14.1.1 . The scalar product ( , ) : H H C is continuous with respect to both variables. To see this, let · · × → xn x in H. This means xn x H 0. Then, (xn, y ) (x, y ) = (xn x, y ) xn x y 0 for all y H as n→ + . k − k → | − | | − | ≤ k − kk k → ∈ → ∞ Remark 14.1.2 (Parallelogram law) . For all x, y H there holds ∈ x + y 2 + x y 2 = 2 x 2 + 2 y 2. k k k − k k k k k To prove this, expand the left hand side

x + y 2 + x y 2 = x 2 + y 2 + 2Re( x, y ) + x 2 + y 2 2Re( x, y ), k k k − k k k k k k k k k − and this proves the assertion. Example 14.1.1 . The most important example of Hilbert space is the space L2(Ω) with Ω Rd open. The inner product on L2 is ⊂

(f, g )L2 = f(x)g(x)dx. ZΩ Remark 14.1.3 (Pythagoras’ Theorem) . Let x, y H be orthogonal. Then, ∈ x + y 2 = x 2 + 2Re( x, y ) + y 2 = x 2 + y 2, k k k k k k k k k k which is the famous Pythagoras’ Theorem in elementary geometry. Deﬁnition 14.1.3. Let M H be a subset of a Hilbert space. The set ⊂

M ⊥ = y H : ( y, x ) = 0 for all x M { ∈ ∈ } is called the orthogonal set of M.

Exercise 14.1.3. Prove that M ⊥ is a closed linear space (Hint: the linearity is trivial. For the closedness, write M ⊥ as the intersection of closed sets using the continuity of the scalar product). The next lemma is a stronger version than Riesz’ lemma 7.1.2 for normed space. Lemma 14.1.1 (Point of minimal distance) . Let H be a Hilbert space, let M H a non empty, closed, and convex subset of H. Then, for all x H there exists a unique y M such that ⊆ 0 ∈ 0 ∈

x0 y0 H = inf x0 y H . k − k y M k − k ∈ Proof. If x M then the statement is trivial. Assume x M. Set 0 ∈ 0 6∈

d = dist( x0, M ) = inf x0 y , y M k − k ∈ where d > 0 is a consequence of M being closed. Hence, there exists a sequence yn n M such that yn x0 d. Now, let us write the parallelogram law with the vectors x = x y and y ={x } y⊂, for n, m Nk: − k → 0 − n 0 − m ∈ 2x y y 2 + y y 2 = 2 x y 2 + 2 x y 2. k 0 − n − mk k n − mk k 0 − nk k 0 − mk Now, we have (y + y ) 2 2x y y 2 = 4 x m m , k 0 − n − mk 0 − 2

(ym+ym) and since M is convex, the mid-point of the segment connecting yn and ym, i. e. 2 , belongs to M. Therefore, 2x y y 2 4d2. k 0 − n − mk ≥ 14.1. DEFINITION OF HILBERT SPACES AND GEOMETRICAL PROPERTIES 99

Hence, we get 2 2 2 2 2 4d + yn ym 2 x0 yn + 2 x0 ym 4d , n,m + k − k ≤ k − k k − k −−−−−−→→ ∞ and this implies that yn ym 0, i. e. yn n is a Cauchy sequence. Since H is complete, let y0 = n,m + k − k −−−−−−→→ ∞ { } lim n + yn. By continuity of the norm, we get x0 y0 = d, and y0 is the desired point in M. → ∞ k − k To prove the uniqueness, assume y0, y 0′ are two points in M with the same property. We can use the parallelogram law as above to get 2 2 2 2 4d + y y′ 2 x y + 2 x y′ , k 0 − 0k ≤ k 0 − 0k k 0 − 0k and this implies y y′ = 0, i. e. y = y′ as desired. k 0 − 0k 0 0 Theorem 14.1.2 (Orthogonal decomposition) . Let H be a Hilbert space. Let M be a closed linear subspace of H. Then, every x H can be written uniquely as x = y + z with y M and z M ⊥, i. e. the vector space H 0 ∈ 0 0 0 0 ∈ 0 ∈ can be written as the direct sum of M and M ⊥, i. e. H = M M ⊥. The maps ⊕ H x P (x ) = y M, H x Q(x ) = z M ⊥ ∋ 0 7→ 0 0 ∈ ∋ 0 7→ 0 0 ∈ are projections , in the sense that they are linear operators on H, they satisfy P (H,H ) = Q (H,H ) = 1 and P 2 = P , Q2 = Q. k kL k kL

Proof. We set P (x0) to be the unique point of minimal distance from M given by Lemma 14.1.1. We then set Q(x ) = x P (x ), i. e. Q = I P . In other words, P (x ) = y M is the unique point in M with minimal 0 0 − 0 − 0 0 ∈ distance from x0. Now, we claim that Q(x0) is orthogonal to M. Let y M. For all λ C, the point y0 + λy is in M (M is a linear subspace). Hence, ∈ ∈ d2 = x y 2 x (y + λy ) 2 = x y 2 + λ 2 y 2 2Re( λ(y, x y )) , k 0 − 0k ≤ k 0 − 0 k k 0 − 0k | | k k − 0 − 0 which implies 2Re( λ(y, x y )) + λ 2 y 2 0. − 0 − 0 | | k k ≥ Now, choose λ = t (0 , + ), divide the above by t and send t 0+. We get ∈ ∞ ց Re( y, x y ) 0, for all y M. − 0 − 0 ≥ ∈ The above holds as well for y M, which gives Re( y, x y ) 0 for all y M, and therefore we get − ∈ − 0 − 0 ≤ ∈ Re( y, x y ) = 0 . 0 − 0 By choosing λ = it , once again ﬁrst with t > 0 and then with t < 0, dividing by t and sending t to zero we get similarly Im( y, x y ) = 0 , 0 − 0 and therefore ( y, x y ) = 0. Hence, ( y, Q (x )) = 0 for all y M, i. e. Q(x ) M ⊥. 0 − 0 0 ∈ 0 ∈ We now need to prove that y0 = P (x0) and z0 = Q(x0) are uniquely determined. Let

x = y + z = y + z , with y , y M and z , z M ⊥. 0 0 0 1 1 0 1 ∈ 0 1 ∈ Then, y y = z z M M ⊥. But M M ⊥ = 0 (easy exercise!). Therefore, y = y and z = z as desired. 0 − 1 0 − 1 ∈ ∩ ∩ { } 0 1 0 1 This finishes the proof of H = M M ⊥. ⊕ We now prove that P is linear. Let x0, x 1 H. Decompose uniquely xi = yi + zi, yi M, zi M ⊥, for i = 0 , 1. Then, ∈ ∈ ∈ x0 + x1 = ( y0 + y1) + ( z0 + z1), and since the decomposition on M M ⊥ is unique for x +x , this implies that P (x +x ) = y +y = P (x )+ P (x ). ⊕ 0 1 0 1 0 1 0 1 Similarly, P (λx 0) = λP (x0) (exercise!). As a consequence, Q = I P is also linear. The fact that P =P and Q2 = Q is obvious ( P (x) M for all− x H, hence P 2(x) = P (P (x) = P (x), and a similar argument holds for Q). ∈ ∈ Finally, let us prove that P (H,H ) = Q (H,H ) = 1. First of all, for all x H, from Pythagoras’ theorem we have k kL k kL ∈ x 2 = P (x) + Q(x) 2 = P (x) 2 + Q(x) 2, k k k k k k k k which implies P (x) x and Q(x) x . This implies P 1 and Q 1. Now, for m M one has k k ≤ k k k k ≤ k k k k ≤ k k ≤ ∈ P (m) = m , and for q M ⊥ one has Q(q) = q . This proves the assertion. k k k k ∈ k k k k Definition 14.1.4. Let M H be a closed linear subspace of the Hilbert space H. The linear operator P : H M defined in Theorem 14.1.2 is⊂ called the orthogonal projection of H onto M. → Exercise 14.1.4. Let z H such that (z, x ) = 0 for all x H. Prove that z = 0 . ∈ ∈ 100 CHAPTER 14. HILBERT SPACES 14.2 Dual of a Hilbert space and reflexivity of Hilbert spaces

Hilbert spaces feature a very important property with regard to taking their dual. The dual of a Hilbert space H is isometrically isomorphic to the Hilbert space H. We recall that, given a linear operator T : X Y with X and Y linear spaces, the set → Ker( T ) = x X : T (x) = 0 { ∈ } is a linear sub-space of X called the kernel of T . Moreover, if T is continuous then Ker( T ) is closed (easy exercise).

Theorem 14.2.1 (Riesz’ representation theorem) . Let H be a Hilbert space, and let f H∗ a linear and continuous functional on H. Then, there exists a unique z H such that ∈ ∈ f, x = ( x, z ), for all x H. (14.1) h i ∈

The map σ : H∗ f z H is a bijection of H∗ onto H, it is an isometry, i. e. σ(f) = f ∗ , and it is ∋ 7→ ∈ k kH k kH anti-linear, i. e. σ(f + λg ) = σ(f) + λσ (g) for all f, g H∗ and all λ C. ∈ ∈ Proof. Let N = Ker( f). If N = H, then f 0, and we set z = 0. If N = H then there exists z N ⊥ with ≡ 6 0 ∈ z0 = 0: indeed, for a given x0 H N, take z0 = Q(x0) with Q the orthogonal projection onto N ⊥. We observe that6 f, z = 0 as z Ker( f).∈ Moreover,\ for all x H one has h 0i 6 0 6∈ ∈ z x 0 f, x N. − f, z h i ∈ h 0i Indeed, z f, z f, x 0 f, x = f, x h 0i f, x = 0 . h − f, z h ii h i − f, z h i h 0i h 0i Therefore, z f, x f, x 0 = ( x 0 f, x , z ) = ( x, z ) h i (z , z ) = ( x, z ) z 2 h i , − f, z h i 0 0 − f, z 0 0 0 − k 0k f, z h 0i h 0i h 0i which implies f, z f, z f, x = h 0i(x, z ) = ( x, h 0iz ). h i z 2 0 z 2 0 k 0k k 0k f,z 0 Hence, h 2i z = σ(f) is the desired element in H for which 14.1. z0 0 k k Now, we claim that there is just one vector z H with the property (14.1). Assume f, x = ( x, z ′) for all ∈ h i x H. Then, ( x, z z′) = 0 for all x H, which implies z = z′. Moreover, σ is a bijection, because if σ(f1) = σ(f2) then∈ − ∈ f , x = ( σ(f ), x ) = ( σ(f ), x ) = f , x , h 1 i 1 2 h 2 i for all x H, i. e. f and f coincide. The anti-linearity is an easy exercise. Let us prove that σ is an isometry. It ∈ 1 2 suﬃces to prove that σ(f) = f ∗ . Let x H, we have (with z = σ(f)) k kH k kH ∈ f, x = (x, z ) x z , |h i| | | ≤ k kk k which implies f σ(f) . Choosing x = z we have f, z = z 2, which proves the assertion. k k ≤ k k |h i| k k Corollary 14.2.2. Every Hilbert space is reﬂexive.

Proof. Let J : H H∗∗ the canonical injection of H into H∗∗ . We claim that J(H) = H∗∗ . We use the map → σ : H∗ H from the previous theorem. We deﬁne the following scalar product on H∗: let f, g H∗, we set → ∈ (f, g ) = ( σ(g), σ (f)) .

It is easily seen that H∗ (with the above scalar product) is a Hilbert space. If we apply Riesz’ representation theorem for H∗, we obtain another isometry from τ : H∗∗ H∗, and with a similar reasoning as above we get that → H∗∗ is also a Hilbert space. Now, let φ H∗∗ . For all f H∗ we have ∈ ∈ φ, f = ( f, τ (φ)) = ( σ τ(φ), σ (f)) = f, σ τ(φ) , h i ◦ h ◦ i i. e. φ = J(σ τ(φ)) J(H), hence H∗∗ = J(H). ◦ ∈ 14.2. DUAL OF A HILBERT SPACE AND REFLEXIVITY OF HILBERT SPACES 101

Remark 14.2.1 . As a consequence of the above results, a sequence xn in a Hilbert space H converges weakly to x H if and only if ∈ (x , y ) (x, y ) , n H → H for all y H. ∈ Corollary 14.2.3. Let H be a Hilbert space. If M H is a closed linear subspace with M ⊥ = 0 , then M = H. ⊂ { } If S H is an arbitrary subset, then S⊥⊥ = [S], where [S] is the linear subspace generated by S (i. e. the set of all ﬁnite⊂ linear combinations of elements in S).

Proof. To prove the ﬁrst statement, if M = H then there exists z H M, which implies 0 = Q(z) M ⊥, which is a contradiction. 6 ∈ \ 6 ∈ For the second statement, S⊥⊥ is clearly a closed linear space (since S⊥⊥ = ( S⊥)⊥, see exercise 14.1.3). Moreover, S⊥⊥ S (easy to check!), and therefore S⊥⊥ [S]. But the opposite inclusion is also satisﬁed, since ⊃ ⊃ for a given z S⊥⊥ [S], let z be the orthogonal projection of z onto [S], then z z S⊥, and therefore ∈ \ 0 − 0 ∈ z z S⊥ S⊥⊥ = 0 . The proof is complete. − 0 ∈ ∩ { } We conclude with the following important application of the Riesz’ representation theorem, namely the Lax- Milgram theorem. We start by proving the following lemma. Lemma 14.2.4. Let T : X Y be a bounded linear map between Banach spaces X, Y . The following statements are equivalent: →

(a) There is a constant c > 0 such that

c x T x for all x X, k k ≤ k k ∈ (b) T has closed range, and the only solution of the equation T x = 0 is x = 0 .

Proof. First, suppose that T satisﬁes (a). Then T x = 0 implies that x = 0, so x = 0. To show that (T ) is closed, suppose that ( y ) is a convergent sequence in (T ), with y ky k Y . Then there is a sequence ( xR) in X n R n → ∈ n such that T x n = yn. The sequence ( xn) is Cauchy. since ( yn) is Cauchy and 1 1 x x T (x x ) = y y . k n − mk ≤ c k n − m k c k n − mk Hence, since X is complete, we have x x for some x X. Since T is bounded, we have n → ∈

T x = lim T x n = lim yn = y, n + n + → ∞ → ∞ so y (T ) and (T ) is closed. ∈ R R Conversely, suppose that T satisﬁes (b) Since (T ) is closed, it is a Banach space. Since T : X Y is one-to- one, the operator T : X (T ) is a one-to-one,R onto map between Banach spaces. The open mapping→ theorem 1 → R 10.3.1 implies that T − : (T ) X is bounded, and hence there is a constant C > 0 such that R → 1 T − y C y for all y (T ). k k ≤ k k ∈ R Setting y = T x , we see that c x T x for all x X, where c = 1 /C . k k ≤ k k ∈ Theorem 14.2.5 (Lax-Milgram Theorem) . Let B : H H R be a bilinear map for which there exist α, β > 0 such that × →

i) B[u, v ] α u v for any u, v H, | | ≤ k kk k ∈ ii) β u 2 B[u, u ] for any u H. k k ≤ ∈ Finally, let f : H R be a bounded linear functional on H. Then, there exists a unique u H such that → ∈ B[u, v ] = < f, v > for any v H. ∈ 102 CHAPTER 14. HILBERT SPACES

Proof. Since B is a bilinear form, for any ﬁxed u H the mapping ∈ v B[u, v ] → is a bounded linear functional on H. Hence, by Theorem 14.2.1 there exists a unique w H such that ∈ B[u, v ] = ( w, v ) for any v H. ∈ Let Au = w. We claim that A : H H is a bounded linear operator. Indeed, the linearity is trivial, while concerning the boundedness one has that→ for any u H ∈ Au 2 = ( Au, Au ) = B[u, Au ] α u Au . k k ≤ k kk k Morover, we note that for any u H ∈ β u 2 B[u, u ] = ( Au, u ) Au u , k k ≤ ≤ k kk k then, β u Au . (14.2) k k ≤ k k By Lemma 14.2.4 it follows that A is one-to-one and the range of A, (A), is closed in H. The next step is to prove that R (A) = H. R To prove this we argue per absurdum . Then, since (A) is closed, by Theorem 14.1.2, there would exist a non-zero R w H such that w (A)⊥. But this is a contradiction because ∈ ∈ R β w 2 B[w, w ] = ( Aw, w ) = 0 . k k ≤ We are ready to conclude. Let f H∗ , by Theorem 14.2.1, there exists w H such that < f, v > = ( w, v ) for any v H. Since (A) is onto, there∈ exists u H such that Au = w. Then, for∈ any v H ∈ R ∈ ∈ B[u, v ] = ( Au, v ) = ( w, v ) = .

Finally, let u, u˜ H such that B[u, v ] = < f,v > and B[˜u, v ] = < f,v > with v H. Then, since B is a bilinear map ∈ ∈ B[u u,˜ v ] = 0 . − By choosing v = u u˜ we have that − β u u˜ 2 B[u u,˜ v ] = 0 k − k ≤ − Then, u =u ˜ and the uniqueness is proved.

14.3 Hilbert sums. Orthonormal bases

Deﬁnition 14.3.1. Let H be a Hilbert space. A set S H is called an orthonormal set if x = 1 for all x S ⊆ k k ∈ and if ( x, y ) = 0 for all x, y S with x = y. Let x H. The numbers (x, y ) y S are called Fourier coeﬃcients of ∈ 6 ∈ { } ∈ x with respect to S. An orthonormal set S is called complete if S⊥ = 0 , or equivalently if [S] = H). { } Remark 14.3.1 (Finite orthonormal sets) . Let S = x ,...,x be an orthonormal set in H. This implies { 1 N } (xi, x j) = δij , where δ is the Kronocker delta symbol, i. e. δ = 1 if i = j, δ = 0 if i = j. Let M = [ S], M is a closed linear ij ij ij 6 subspace of H. Hence, H = M M ⊥. In this case, we claim that the orthogonal projection on M is ⊕ N

P (x) = (x, x k)xk. kX=1 Indeed, let v = N (x, x )x . Clearly v M. Moreover, for all h = 1 , . . . , N , k=1 k k ∈ P N N (x v, x ) = ( x, x ) (x, x )( x , x ) = ( x, x ) (x, x )δ = 0 , − h h − k k h h − k hk i=1 i=1 X X 14.3. HILBERT SUMS. ORTHONORMAL BASES 103

which implies that x v M ⊥. Therefore, − ∈ x = v + ( x v) − is the unique decomposition of x into v M and x v M ⊥, or equivalently, v = P x and x v = Qx with P ∈ − ∈ − orthogonal projection on M and Q orthogonal projection on M ⊥.

Proposition 14.3.1 (Bessel’s inequality) . Let xk k N be a countable orthonormal set, or orthonormal sequence. Then, for all x H, the following Bessel’s inequality{ } ∈holds ∈ + ∞ x 2 (x, x ) 2. k k ≥ | k | kX=1 Proof. Let N N, and consider S = x ,...,x . From the previous remark, v = N (x, x )x = P (x) is the ∈ N { 1 N } k=1 k k orthogonal projection of x onto [ SN ], and x = v + ( x v) is the unique decomposition of x in H = [ SN ] [SN ]⊥. Therefore, by Pythagoras’ theorem we have − P ⊕

N N N x 2 = v 2 + x v 2 = ( (x, x )x , (x, x )x ) + x v 2 = (x, x ) 2 + x v 2. k k k k k − k k k k k k − k | k | k − k kX=1 kX=1 kX=1 Therefore, N (x, x ) 2 x 2, for all N N. | k | ≤ k k ∈ kX=1 By sending N + we get the assertion. → ∞

Proposition 14.3.2. Let xk k N be an orthonormal sequence, and let λk k N be a sequence in C. The inﬁnite sum λ x converges if and{ } only∈ if λ 2 < + . In this case, the sum{ }x∈= λ x does not depend on the k k k k | k| ∞ k k k order of the xk’s, and we have P P P x 2 = λ 2. k k | k| k N X∈ Proof. Similarly to the proof of Bessel’s inequality, for all m, n N, m > n , we have ∈ m 2 m 2 λkxk = λk . | | k=n+1 k=n+1 X X 2 Since H is complete, with the Cauchy criterion we get that k λx k converges if and only if k λk does. Moreover, if we choose n = 0 and let m + we get | | → ∞ P P 2 2 λkxk = λk . | | k N k N X∈ X∈

Finally, let k λk′ xk′ be a rearrangement of the above infinite sum. A well known result on real (or complex) infinite series states that if an infinite series converges absolutely, then every rearrangement of the series converges absolutely andP the sum does not depend on the order. This implies that

2 2 λ = λ′ . | k| | k| Xk Xk

Now, let x = λ x and z = λ′ x′ . From above we know x = z . Now, for all n N we set k k k k k k k k k k ∈ P P n n sn = λkxk, t n = λk′ xk′ . kX=1 kX=1 + 2 We claim that ( sn, t n) k=1∞ λk as n + . To see this, we observe that the scalar product ( sn, t n) equals the 2 → | | → ∞ sum of the λk corresponding to elements xk of the sequence that appear in both ﬁnite sums sn and tn. Clearly, | | P 104 CHAPTER 14. HILBERT SPACES the sequence increases with n (there are more and more nonnegative terms added to the previous values of the + 2 sequence!), and one can prove that the limit cannot be less than ∞ λ . Now, the claim implies k=1 | k| + P ∞ (x, z ) = λ 2 = (z, x ) = ( z, x ). | k| kX=1 Hence, x z 2 = x 2 + z 2 (x, z ) (z, x ) = 0 . k − k k k k k − −

Theorem 14.3.3. Let S H be an orthonormal set, and let M = [S], x H. Then, ⊆ ∈ (i) The set S = y S : ( x, y ) = 0 x { ∈ 6 } is at most countable.

(ii) The inﬁnite sum P (x) = (x, y )y

y Sx X∈ converges independently of the order of the terms in the sum, and

P (x) 2 = (x, y ) 2. k k | | y Sx X∈

(iii) P (x) is the orthogonal projection of x onto M.

Proof. Proof of (i). Let ǫ > 0, and let S = y S : (x, y ) ǫ . x,ǫ { ∈ | | ≥ } Now, we claim that the set Sx,ǫ is finite. Assume by contradiction that it is not finite. Now, since every infinite set 1 contains an infinite countable subset , let S˜x,ǫ be such set. Due to Bessel’s inequality,

x 2 (x, y ) 2 ǫ2♯(S˜ ), k k ≥ | | ≥ x,ǫ y S˜x,ǫ ∈X and this implies that the set S˜x,ǫ is ﬁnite, which is a contradiction. Hence, Sx,ǫ is ﬁnite. Now, clearly

+ ∞ Sx = Sx, 1/n , n=1 [ which means that Sx is the countable union of finite sets, i. e. Sx is countable at most. Proof of (ii). Let us write Sx as an ordered sequence, i. e. Sx = yk k N. Bessel’s inequality implies { } ∈ (x, y ) 2 x 2, | k | ≤ k k k 1 X≥ therefore, Proposition 14.3.2 implies that the series k 1(x, y k)yk = P (x) converges and the sum is independent of the order of the terms in the series. Moreover, ≥ P P (x) 2 = (x, y ) 2. k k | k | k 1 X≥ 1 Let X be an infinite set. We define the sequence xn ∈ X as follows. By the axiom of choice, choose x1 to be an element in X. Since X \ x1 is non empty, choose x2 ∈ X \ x1. Inductively, assume that x1,...,x n are well defined. Since X is infinite, choose xn+1 ∈ X \ { x1,...,x n}. The resulting sequence is made up by distinct points. 14.3. HILBERT SUMS. ORTHONORMAL BASES 105

N Proof of (iii). It is clear that P (x) M. Indeed, for each N N, k=1 (x, y k)yk [S], and therefore N ∈ ∈ ∈ P (x) = lim N + (x, y k)yk [S] = M. In order to show that P (x) is the orthogonal projection of x onto M, → ∞ k=1 ∈ P we need to prove that v = x P (x) M ⊥. Once again, let P − ∈ N SN = (x, y k)yk. kX=1 For all k N, we have ≤ N N N (x S , y ) = ( x (x, y )y , y ) = ( x, y ) (x, y )( y , y ) = ( x, y ) (x, y )δ = 0 . − N k − h h k k − h h k k − h hk hX=1 hX=1 hX=1 Let N + to obtain (by continuity of the scalar product) → ∞ (x P (x), y ) = 0 for all k 1. − k ≥

Hence, x P (x) is orthogonal to S, hence x P (x) M ⊥. − − ∈ Deﬁnition 14.3.2. An orthonormal base of the Hilbert space H is a orthonormal set S H with the property ⊂ that [S] = H, or equivalently that S⊥ = 0 . { } Remark 14.3.2 . From Theorem 14.3.3, since x = P (x) for all x H, we have ∈ x = (x, y )y, x 2 = (x, y ) 2, k k | | y Sx y Sx X∈ X∈ where S = y S : ( x, y ) = 0 , which we know to be at most countable. x { ∈ 6 } Theorem 14.3.4. Every Hilbert space has an orthonormal base.

Proof. Let be the set of all orthonormal sets in H, ordered with the relation S1 S2 if and only if S1 S2. As explained inM the second example 1.1.2, this set is inductive, and therefore Zorn’s lemma≺ implies that there⊆ exists a maximal element, that we call S. Assume by contradiction that S⊥ = 0 . Let x = 0, x S⊥. Take S = S x . 6 { } 6 ∈ 1 ∪ { } Clearly S1 = S and S S1, which contradicts the fact that S is maximal. Hence, S⊥ = 0 , which proves the assertion. 6 ≺ { }

Theorem 14.3.5. A Hilbert space H has a countable orthonormal base if and only if H is separable.

Proof. Assume first that H has a countable base S = yk k N. Hence, [ S] is dense in H. Now, set { } ∈ N D = x = α y : y S, α = q + ir , q , r Q, N N . { k k k ∈ k k k k k ∈ ∈ } kX=1 Clearly, D is countable and D = [S] = H. Therefore H is separable. Vice versa, assume H is separable. Let xk k N be a dense sequence in H. We define the orthonormal sequence { } ∈ xk y N as follows. Let x the first nonzero vector of the sequence x , and let y = 1 . Set X = Cy . k k k1 k k 1 xk 1 1 { } ∈ { } k 1 k Inductively, for a given n N and given y ,...,y orthonormal, set X = [ y ,...,y ], i. e. X is the linear ∈ 1 n n 1 n n subspace generated by y1,...,y n. Call Pn the orthogonal projection onto Xn. Let xnk+1 be the first vector of the sequence which does not belong to X . Decompose x in X (X )⊥, i. e. n nk+1 n ⊕ n

x = P (x ) +y ˜ , for some y (X )⊥. nk+1 n nk+1 n+1 n+1 ∈ n Set y = yñ+1 . Clearly, y is an orthonormal base. Moreover, [ y ,...,y ] [x ,...,x ] for all N N, n+1 yñ+1 n n 1 N 1 N k k { } ⊇ ∈ hence ( yk k N) is dense in H as desired. { } ∈ The process above, by which we define an orthonormal basis y such that [ y ,...,y ] = [ x ,...,x ] for a { n} 1 n 1 n given sequence xn of independent vectors, is called Gram-Schmidt orthogonalisation . 106 CHAPTER 14. HILBERT SPACES

Exercise 14.3.1. We recall that the space ℓ2 of squared convergent sequences can be also deﬁned for sequences in 2 C. If x = xn n N ℓ , then ∈ { } ∈ + 2 ∞ 2 x 2 = x . k kℓ | n| n=1 X Such space is also a Hilbert space in C with the scalar product

+ ∞ (x, y ) 2 = x y , x = x , y = y C. ℓ n n { n}n { n}n ⊂ n=1 X Corollary 14.3.6. Let H be a separable Hilbert space. Then H is isometric to ℓ2.

Proof. Due to Theorems 14.3.3 and 14.3.5, H has a countable base e , and the map { n}n H x x˜ = x˜ ℓ2, x˜ = ( x, e ) ∋ 7→ { n}n ∈ n n H 2 + 2 is invertible (exercise) and an isometry, since x = ∞ x˜ . k kH n=1 | n| P We conclude this section with an example that shows how Hilbert spaces arose naturally from problems in classical analysis. If f(x) is an integrable function on [0 , 2π], we can deﬁne the numbers

2π 1 inx c = e− f(x)dx. n (2 π)1/2 Z0 The formal series + ∞ 1/2 inx cn(2 π)− e n= X−∞ is called the Fourier series of f. The classical problem is: for which f and in what sense does the Fourier series of f converge to f? This problem which originated from Fourier in 1811 has had a rich and eventful history. It has given rise to an entire branch of modern mathematics (abstract harmonic analysis). Furthermore, some of the nicest results on the classical case have just been proven in the second half of the 20th century. As an example of a classical result, we state the following Theorem without proof.

Theorem 14.3.7. Suppose that f(x) is periodic of period 2π and is continuously diﬀerentiable. Then the functions

M 1/2 inx cn(2 π)− e M −X converge uniformly to f(x) as M + → ∞ This theorem gives suﬃcient conditions for the Fourier series of a function to converge uniformly. But ﬁnding the exact class of functions whose Fourier series converge uniformly or converge pointwise has proved to be a hard problem. We can, however, get a nice answer to this question if we change our notion of ‘convergence’ and this 1/2 inx + is just where Hilbert spaces come in. The collection of functions (2 π)− e n=∞ is clearly an orthonormal set (exercise!) in L2([0 , 2π]). If we knew that it was an orthonormal{ base, then} Remark−∞ 14.3.2 would allow us to conclude that for all functions f L2([0 , 2π]), ∈ M 1/2 inx f(x) = lim cn(2 π)− e M + → ∞ M −X 2 1/2 inx + where convergence means convergence in the L norm. In fact, (2 π)− e n=∞ is a base. We will give a proof that relies on the classical result stated above in Theorem 14.3.7.{ } −∞

2 M 1/2 inx 2 Theorem 14.3.8. If f L ([0 , 2π]) , then M cn(2 π)− e converges to f in the L norm as M + . ∈ − → ∞ P 14.3. HILBERT SUMS. ORTHONORMAL BASES 107

2 Proof. We need to prove that the periodic, continuously diﬀerentiable functions C1([0 , 2π]) are dense in L ([0 , 2π]). In Theorem 13.1.2 we proved that continuous compactly supported functions are dense in L2. But a continuous function can be approximated in the L2 norm by a C1 function by rounding oﬀ the corners in a smooth way (the details are omitted). We should convince ourselves easily that this can be done so that the resulting function is arbitrarily close to the continuous function in the L2 norm. Hence, C1([0 , 2π]) = L2([0 , 2π]). To show that the orthonormal set we are considering is a base we need only to show that ( einx , g ) = 0 for all n implies g = 0. Suppose f C1([0 , 2π]), then by Theorem 14.3.7 ∈ M 1/2 inx c (2 π)− e f, n → M −X uniformly and thus in the L2 norm (the domain is bounded!) also. Therefore, by continuity of the scalar product in L2 we get M 1/2 inx (f, g ) = lim ( cn(2 π)− e , g ) = 0 M + → ∞ M −X due to ( einx , g ) = 0 for all n. This implies that g is orthogonal to al f in the dense set C1([0 , 2π]), which implies g = 0, which proves the assertion. 108 CHAPTER 14. HILBERT SPACES Chapter 15

Exercises

Exercise 15.1. Assume that 1 s t r and θ [0 , 1] are real numbers such that ≤ ≤ ≤ ≤ ∞ ∈ 1 θ 1 θ = + − . t s r Let u Ls(Rn) Lr(RN ). Show that u Lt(RN ) and ∈ ∩ ∈ θ 1 θ u u u − . k kt ≤ k ksk kr

Exercise 15.2. Let E be a Banach space let xn n be a sequence such that xn ⇀ x in the weak topology σ(E, E ∗). Set { } 1 n y = x . n n i i=1 X Show that yn ⇀ x in the weak topology σ(E, E ∗).

Exercise 15.3. Let L1(0 , 1) be a bounded set. Prove that is relatively compact in L1(0 , 1) equipped with the 1 F ⊂ F weak topology σ(L (0 , 1) , L ∞(0 , 1) if and only if is equi-integrable. F

Exercise 15.4. Construct a sequence f in L1(0 , 1) , f 0, such that: { n}n n ≥ 1. f f weakly in L1 n → 2. f f , k nk1 → k k1 1 3. fn does not converge strongly to f in L .

1 Exercise 15.5. Let 1 p and let f be a periodic function of period T in Lloc (R). Let un(x) = f(nx ) for x (0 , 1) . Deﬁne ≤ ≤ ∞ ∈ 1 T f¯ = f(x) dx. T Z0 Show that

p 1. un converges weakly to f¯ in L (0 , 1) ,

2. un converges weakly star to f¯ in L∞(0 , 1) .

Exercise 15.6. Let 1 < p < and f be the sequence deﬁned for x (0 , 1) given by ∞ { n}n ∈ 1 nx f (x) = n p e− . for any n N. n ∈ 109 110 CHAPTER 15. EXERCISES

1. f 0 almost everywhere. n → 2. f is bounded in Lp((0 , 1)) . { n}n 3. f does not converge strongly to 0 in Lp((0 , 1)) . { n}n 4. f converges weakly to 0 in Lp((0 , 1)) . { n}n 5. f converges strongly to 0 in Lq((0 , 1)) for 1 q < p . { n}n ≤

Exercise 15.7. Let f be the sequence deﬁned for x (0 , 1) given by { n}n ∈ nx f (x) = ne − . for any n N. n ∈ Prove that: 1. f 0 almost everywhere. n → 2. f is bounded in L1((0 , 1)) . { n}n 3. f does not converge strongly to 0 in L1((0 , 1)) . { n}n 4. f does not converge weakly to 0 in L1((0 , 1)) . { n}n

Exercise 15.8. Let A be a subset of a Hilbert space H. Prove that A⊥ = A⊥.

Exercise 15.9. Let x be a sequence in a Hilbert space H which converges weakly to x H. Assume that { n}n ∈ lim xn = x . n + → ∞ k k k k Prove that x x strongly in H. n →

Exercise 15.10. Let f, g H, H a Hilbert space. Assume that equality holds in Cauchy-Schwartz inequality for f and g, i. e ∈ (f, g ) = f g . k kk k Prove that f = cg for some scalar c C. Hint: assume ﬁrst that f = g = 1 and use Pythagoras’ theorem. ∈ k k k k

Exercise 15.11. Let η : [ a, b ] R be a continuous function such that η(t) > 0 for all t [a, b ]. For two given functions f, g : [ a, b ] C deﬁne→ the product ∈ → b (f, g )η := f(x)g(x)η(x)dx. Za Prove that ( , ) is a scalar product, and prove that the resulting normed space is a Hilbert space. · · η

Exercise 15.12. Let H = L2(R), and set M = f H : f(t) = f( t) almost everywhere in R . { ∈ − } 1. Show that M is a closed subspace. 2. Express the orthogonal projection operator P : H M explicitly. → 3. Find M ⊥.

Exercise 15.13. Prove that the space of polynomials on [0 , 2π] is a dense subspace of L2([0 , 2π]) . Hint: use the fact that 1 einx is an othonormal base for L2([0 , 2π]), and approximate the elements of the base { √2π }n by a polynomial using some classical results in basic analysis. 111

Exercise 15.14. * Deﬁne the Legendre polynomials Pn by 1 dn P (x) = (x2 1) n. n 2nn! dx n − 1. Compute the ﬁrst few Legendre polynomials.

2. Show that the Legendre polynomials are orthogonal in L2([ 1, 1]) , and that they are obtained by Gram-Schmidt orthogonalisation of the monomials. − 3. Show that 1 2 2 Pn(x) dx = . 1 2n + 1 Z− 4. Prove that the set 2n + 1 Pn, n N (r 2 ∈ ) is an orthonormal basis for L2([ 1, 1]) −

Exercise 15.15. Let xn n N and yn n N be two sequences in the Hilbert space H. Suppose that xn 1 and y 1. Moreover, suppose{ } ∈ that { } ∈ k k ≤ k nk ≤ lim (xn, y n) = 1 . n →∞ Show that lim xn yn = 0 . n →∞ k − k

Exercise 15.16. Assume that en n is an orthonormal basis of H. Let an n be a bounded sequence in R and set 1 n { } { } un = n i=1 aiei.

1. CheckP that en converges weakly to 0 in H. 2. Prove that u 0. | n| → 3. Show that √nu n converges weakly to 0 in H.

Exercise 15.17. Let λn n be a sequence of positive numbers such that λn when n . Let V be the space of sequences u of{ real} numbers such that → ∞ → ∞ { n}n

∞ λ u < . n| n| ∞ i=1 X The space V is equipped with the scalar product

∞ (( u, v )) = λnunvn. i=1 X Show that V is an Hilbert space and that V l2 with compact injection. ⊂ 112 CHAPTER 15. EXERCISES

Exercise 15.18. Let be the set of all trigonometric polynomials in C(R), i.e the set of all 2π-periodic functions f : R R in C(R) Pwhich are ﬁnite linear combinations of trigonometric functions of the form einx , for n = ,...,→+ . Given u C(R) a 2π-periodic function, consider the truncated Fourier series −∞ ∞ ∈ 2π 1 ikt ikx U (x) = u(t)e− dt e . n 2π k n Z0 |X|≤ (i) Find a suitable continuous periodic function D : R R such that n → 2π U (x) = D (x t)u(t)dt. n n − Z0 (ii) Set 1 n V (x) = U . n n + 1 l Xl=0 Find a suitable periodic function S : R R such that n → 2π V (x) = S (x t)u(t)dt. n n − Z0

(iii) Prove that Sn satisﬁes

(a) 2π S (x t)dt = 1 for all x R. 0 n − ∈ (b) S 0 everywhere. R n ≥ (c) max x [δ, 2 δ] Sn(x) 0 as n + . ∈ − → → ∞ (iv) Use the above steps to prove that is dense in the set of periodic continuous functions on R with respect to the uniform norm. P

2 Exercise 15.19. Let (xn) be the sequence in ℓ deﬁned by

1 if n = k xn,k = (0 otherwise .

2 1. Prove that xn converges to zero weakly in ℓ .

2. Is (xn) converging to zero strongly?

Exercise 15.20. For any n N consider the sequences u (x) = sin( nx ) e v (x) = cos( nx ). Show that ∈ n n 1. u v is uniformly bounded in Lp((0 , π )) for any p [1 , ]. { n · n}n ∈ ∞ 2. Find the weak limit of the sequence u v in Lp((0 , π ) for p (1 , ) and the weak* limit of u v in { n · n}n ∈ ∞ { n · n}n L∞((0 , π )) . 3. Is u v weakly convergent in L1((0 , π )) ? { n · n}n

Exercise 15.21. Consider the sequence let u (x) = arctan( xn) for x [0 , ). Show that n ∈ ∞ 1. u is converging almost everywhere in [0 , ) and determine the limit. n ∞ 2. u does not converge strongly in L∞((0 , )) . n ∞ 3. u does not converge strongly in L∞((1 , )) . n ∞ 113

4. Let ε > 0 ﬁxed. Is u convergent strongly in L∞((1 + ε, )) ? n ∞ 5. Is u weakly star convergent in L∞((0 , )) ? n ∞

Exercise 15.22. Let n n n 2 sin(2 x) if x [0 , 2− π] fn(x) = ∈ n 0 if x [2 − π, π ]. ( ∈ Show that

1. fn converges to zero almost everywhere and in measure.

1 2. fn is uniformly bounded in L ((0 , 2π)) .

1 3. fn does not converge strongly in L ((0 , 2π)) .

1 4. Is fn weakly convergent in L ((0 , 2π)) ?

Exercise 15.23. Let α if x [0 , 1/2] f˜(x) = ∈ β if x (1 /2, 1] . ( ∈

Let f be the 1-periodic extension of f˜ to the whole space R and fn(x) = f(x). Show that 1. f is weakly convergent in Lp(0 , 1) for any p [1 , ) and ﬁnd the weak limit. n ∈ ∞ 2. fn is weakly star convergent in L∞(0 , 1) and ﬁnd the weak star limit. 114 CHAPTER 15. EXERCISES Part V

Operator theory

115

Chapter 16

Introductory concepts

Throughout this part we shall work with Banach (or Hilbert) spaces on R.

16.1 Basic concepts and examples

Let E, F be Banach spaces. We recall that a linear operator T : E F is bounded (or continuous) if →

T (E,F ) = sup T (x) < + . k kL x E , x 1 k k ∞ ∈ k k≤ We denote by (E, F ) the space of bounded linear operators from E into F , which is a normed space equipped L with (E,F ). If F is a Banach space, so is (E, F ). k · k L L For a given T (E, F ), we set ∈ L Ker( T ) = x E : T (x) = 0 , { ∈ } i. e. the null set , or kernel of T . We denote by R F the set T ⊂ (T ) = T (E), R called the range of T (or the image of T ). We use the notation (E) for (E, E ). The identity operator I (E) is the operator I(x) = x for all x E. L L ∈ L ∈ Exercise 16.1.1. Let T (E). Prove that ∈ L 2 2 T (E) T (E). k kL ≤ k kL

Theorem 16.1.1 (Neumann’s lemma) . Let E be a normed space and let T (E) with T (E) < 1. Then, (1 T ) (E) is invertible, and ∈ L k kL − ∈ L + 1 ∞ n (I T )− = T . − n=0 X

Proof. Let q = T (E) < 1. From a simple geometric series we get k kL + 1 ∞ n (1 q)− = q . − n=0 X Now, let N N. We have ∈ N (I T ) T n = I T + T T 2 + T 2 . . . T N+1 = I T N+1 . − − − − − − n=0 X We clearly have N+1 N+1 N+1 T (E) T (E) = q , k kL ≤ k kL 117 118 CHAPTER 16. INTRODUCTORY CONCEPTS and the right hand side goes to zero as N + . This implies that → ∞ N n lim (I T ) T I (E) = 0 , N + k − − kL → ∞ n=0 X and hence the operator + ∞ (I T ) T n − n=0 X is well deﬁned and continuous, and it is equal to the identity operator. This means that

+ ∞ T n n=0 X is the inverse of ( I T ) as expected. − The algebra of bounded linear operators on a normed space E is the linear space (E) equipped with the product L ST = S T. ◦ We now introduce the basic concepts of the spectral theory of linear operators on normed spaces. Definition 16.1.1. Let E be a normed space, and let A (E) be a continuous linear operator on E. The resolvent set of A is the set ρ(A) R of all complex numbers ∈λ L such that λI A is invertible and has a continuous inverse. The set σ(A) = R ρ(A⊂) is called the spectrum of A. An element −λ σ(A) is called eigenvalue for A if there exists v E, v = 0, such\ that A(v) = λv (i. e. λI A is not injective). In∈ this case v is called an eigenvector corresponding∈ to eigenvalue6 λ. The eigenspace of λ is− the subspace Ker( λI A). The dimension of the closed − subspace Ker( λI A) is called the multiplicity of λ. The set of all eigenvalues is denoted by σp(A), and is called point spectrum of −A. It is useful to keep in mind that, due to the open mapping theorem, in order to have λ ρ(A) it suffices to show that λI A is bijective. ∈ The− concepts in definition 16.1.1 are very important, as they allow (in some cases) to deepen the understanding of a given linear operator. For instance, in finite dimension it is well known that a linear operator on a vector space ES can be represented by a squared matrix. If the matrix is fully diagonalisable in R, then there exists a change of coordinates on E after which the operator can be written as a diagonal matrix. In such a new coordinate system, the operator can be then simply seen as the combination of the following operations: projecting a vector on all the eigenspaces, multiplying each projection by the corresponding eigenvalue, and finally taking the sum of the resulting vectors.

Remark 16.1.1 . It is clear that σp(A) σ(A). In general, this inclusion can be strict: there might be some λ R such that ⊂ ∈ Ker( A λI) = 0 , − { } and (A λI) = E. R − 6 We also point out that, in general, the point spectrum of a linear operator may be empty if we work on R. If we work on C the situation is totally diﬀerent, as every operator on ﬁnite dimensional spaces has at least some eigenvector. Proposition 16.1.2. The spectrum σ(A) of a bounded linear operator A on E is compact and

σ(A) [ A , A ]. ⊂ −k k k k 1 Proof. Let λ R be such that λ > A . Clearly this implies λ = 0. Now, λI A = λ(I λ− A), and 1 ∈1 | | k k 6 − − λ− A = λ − A < 1. Therefore, Neumann’s lemma in Theorem 16.1.1 implies that λI A is invertible. This kshows thatk |σ(|A)k k[ A , A ]. − We now prove⊂ that−k ρ(kAk) isk open. Let λ ρ(A). Given λ R (close to λ ) and f E, we try to solve 0 ∈ ∈ 0 ∈ A(u) λu = f. − 16.1. BASIC CONCEPTS AND EXAMPLES 119

The latter equation can be written as A(u) λ u = f + ( λ λ )u, − 0 − 0 i. e. 1 u = ( A λ I)− [f + ( λ λ)u]. (16.1) − 0 0 − We are therefore trying to ﬁnd a ﬁxed point for the map

1 E u [u] := ( A λ I)− [f + ( λ λ)u] E. ∋ 7→ A − 0 0 − ∈ Assuming we are working on the metric space u C for some C > 0 to be determined, we compute {k k ≤ } 1 1 [u] = (A λ I)− [f + ( λ λ)u] (A λ I)− [ f + λ λ C]. kA k k − 0 0 − k ≤ k − 0 k k k | − 0| Now it is easily seen that if we require 1 λ λ , (16.2) | − 0| ≤ 2 (A λ I) 1 k − 0 − k we can ﬁnd a suitable C > 0 such that

1 (A λ I)− [ f + λ λ C] C. k − 0 k k k | − 0| ≤

For such choice of C, the map maps the complete metric space BC = u C into itself. Moreover, let u , u B , and let us compute A {k k ≤ } 1 2 ∈ C 1 [u ] [u ] (A λ I)− (λ λ ) u u , kA 1 − A 2 k ≤ k − 0 − 0 kk 1 − 2k which shows that is a contraction provided λ λ0 satisﬁes (16.2). Hence, Banach ﬁxed point theorem implies A 1 − 1 the assertion, since every λ (λ0 I −1 , λ 0 + I −1 ) is an element of ρ(A), which implies that ρ(A) ∈ − 2 (A λ0 ) 2 (A λ0 ) is open, and hence σ(A) is closed, andk thus− compact.k k − k

2 2 Example 16.1.1 (Bounded operators on ℓ ). Consider the Hilbert space H = ℓ . Let T (H). Given x = xn n ℓ2, we denote ∈ L { } ∈ T (x) = (T (x)) ℓ . { n}n ∈ 2 The set e ℓ2 with ( e ) = δ is an orthonormal basis for ℓ2. Given x = (x, e )e , we have { n}n ⊂ n j nj n n n P T (x) = T (x, e )e = (x, e )T (e ) . n  j j j j n j j X n X   We set the infinite matrix + MT = ( T (ej)n)j,n∞=1 . Then, if we think of x and T (x) as infinite vertical vectors, we have { n}n { }n T (x) = M x, T · where above should be seen as a standard matrix product. · The case of MT being zero everywhere except in an N N upper left squared sub-matrix is, of course, a special case of interest. We therefore advise the reader to review× all the material on spectral theory in finite dimensions (squared matrices, eigenvectors, eigenvalues, eigenspaces, diagonalisation of a squared matrix, etc.). Another special case is the following one. Let α be a sequence of real numbers. For a given x ℓ2, set n ∈

T (x)n = αnxn.

In this case, the inﬁnite matrix MT is diagonal . A sharp requirement on the sequence αn such that T is a bounded 2 operator on ℓ is that α ℓ∞. Indeed, H¨older’s inequality implies { n}n ∈ 2 2 2 2 2 T (x) ℓ = T (x)n = αnxn sup αn x ℓ2 . k k | | | | ≤ n | |k k n n X X 120 CHAPTER 16. INTRODUCTORY CONCEPTS

Now, let nk be a sequence in N such that

lim αnk = α := sup αn . k + | | ∞ | | → ∞ n Set x˜ ℓ2 as { k}k ⊂ x˜k,j = δj,n k . We have

T (˜xk)j = δj,n k αnk . Hence, T (˜x) 2 = α , k kkℓ | nk | which shows that T (ℓ2) = α . In this case, the calculus of the spectrum of T is extremely easy: σ(T ) = σp(T ), k kL ∞ and all the elements of the sequence αn are eigenvalues.

16.2 Adjoint operators

Deﬁnition 16.2.1. Let E and F be normed spaces, and let T (E, F ). The adjoint T ∗ of T is an operator from ∈ L F ∗ into E∗ deﬁned by

T ∗[g], x E∗ E = g, T (x) F ∗ F , for all g F ∗ and for all x E. h i × h i × ∈ ∈

Remark 16.2.1 . The above deﬁnition is well posed. That is, given g F ∗, T ∗[g] is an element of E∗. The linearity is trivial. The continuity follows from ∈

T ∗[g], x = g, T (x) g F ∗ T (x) F g F ∗ T (E,F ) x E, |h i| |h i| ≤ k k k k ≤ k k k kL k k in particular T ∗[g] E∗ g F ∗ T (E,F ). k k ≤ k k k kL Theorem 16.2.1. The mapping (E, F ) T T ∗ (F ∗, E ∗) L ∋ 7→ ∈ L is an isometric isomorphism.

Proof. The fact that the map is linear is trivial. It remains to show that

T ∗ (F ∗,E ∗) = T (E,F ). k kL k kL

The above remark already shows that T ∗ (F ∗,E ∗) T (E,F ). To prove the opposite inequality, Corollary 9.1.5 implies that k kL ≤ k kL T (x) F = sup g, T (x) , k k g F ∗, g 1 |h i| ∈ k k≤ which implies

T ∗ (F ∗,E ∗) = sup T ∗[g] E∗ = sup sup T ∗[g], x k kL g F ∗, g 1 k k g F ∗, g 1 x E, x 1 |h i| ∈ k k≤ ∈ k k≤ ∈ k k≤ = sup sup g, T (x) = sup sup g, T (x) = sup T (x) = T . g F ∗, g 1 x E, x 1 |h i| x E, x 1 g F ∗, g 1 |h i| x E, x 1 k k k k ∈ k k≤ ∈ k k≤ ∈ k k≤ ∈ k k≤ ∈ k k≤

Proposition 16.2.2. Let E, F, G be normed spaces, and let T (E, F ), S (F, G ). Then, (S T )∗ = T ∗ S∗. ∈ L ∈ L ◦ ◦ Moreover, the adjoint of the identity I : E E, I (E), is the identity I : E∗ E∗. → ∈ L →

Proof. Let h G∗. We have ( S T )∗[h] E∗. Now, for x E we have ∈ ◦ ∈ ∈

(S T )∗[h], x = h, (S T )( x) = h, S (T (x)) = S∗[h], T (x) = T ∗ S∗[h], x . h ◦ i h ◦ i h i h i h ◦ i The statement on the identity is a simple exercise. 16.2. ADJOINT OPERATORS 121

Exercise 16.2.1. Let E, F be normed spaces and let T (E, F ). Then, prove that the second adjoint T ∗∗ : ∈ L E∗∗ F ∗∗ is an extension of T (by identifying J(E) with E in E∗∗ ). Moreover, prove that if E is reﬂexive, then → T ∗∗ = T . Proposition 16.2.3. Let E be a Banach space and let F be a normed linear space. A linear operator T (E, F ) 1 1 ∈ L has a bounded inverse T − (with domain F ) if and only if T ∗ has a bounded inverse (T ∗)− (with domain E∗). In 1 1 that case, (T − )∗ = ( T ∗)− .

1 1 1 Proof. Suppose T − exists. By Proposition 16.2.2, I∗ = ( T T − )∗ = ( T − )∗ T ∗, where I∗ is the identity operator 1 ◦ ◦ 1 1 on Y ∗. Similarly, T ∗ (T − )∗ is the identity operator on E∗. It follows that ( T ∗)− exists and is equal to ( T − )∗. ◦ 1 1 Suppose conversely that ( T ∗)− is a bounded operator in (E∗, F ∗). Then, from the previous step, ( T ∗∗ )− exists L and is in (F ∗∗ , E ∗∗ ). From the exercise 16.2.1, T ∗∗ is an extension of T , hence T is one-to-one. If we can prove L 1 1 that T (E) = F then the proof is complete, since, on T (E), T − = ( T ∗∗ )− is a bounded operator. Now, since T ∗∗ has a bounded inverse and since J(E) is closed in E∗∗ , the set T ∗∗ (J(E)) is closed. Hence, by deﬁnition of T ∗∗ , we easily see that T (E) is closed. Suppose now that T (E) = F . From corollary 9.2.6, we get that there exists 6 g F ∗ such that g = 0 and g, T (x) = 0 for all x E, which implies T ∗[g] = 0, and this is impossible since T ∗ is one-to-one.∈ 6 h i ∈ 122 CHAPTER 16. INTRODUCTORY CONCEPTS Chapter 17

Compact operators

17.1 Deﬁnitions and basic properties

In this chapter, E and F denote two Banach spaces, unless otherwise speciﬁed. The set BE E denotes the closed set ⊂ B = x E , x 1 . E { ∈ k k ≤ } Deﬁnition 17.1.1. A bounded linear operator T (E, F ) is said to be compact (or completely continuous) if ∈ L T (BE) has compact closure in F (in the strong topology). We denote by (E, F ) (E, F ) the set of all compact operators from E into F . We use the notation (E) = (E, E ). K ⊂ L K K Exercise 17.1.1. Prove that T : E F belongs to (E, F ) if and only if T maps bounded sets of E into pre- compact sets (i. e. sets with compact→ closure) of F . K

Exercise 17.1.2. Prove that the set (E, F ) is a closed linear subspace of (E, F ) (in the topology induced by the K L norm (E,F ). Solution.k · k L The fact that (E, F ) is a linear subspace is straightforward. Let T be a sequence of compact K { n}n opeartors, and T a bounded operator such that Tn T (E,F ) 0. We claim that T is compact. Since F is complete, k − kL → it suffices to show that T (BE) F is totally bounded. Let ǫ > 0. Fix n N such that Tn T (E,F ) < ǫ/ 2. Since ⊂ ∈ k − kL Tn(BE) has compact closure, there is a finite covering of Tn(BE) by balls of radius ǫ/ 2, say Tn(BE) i I B(yi, ǫ/ 2) with I finite. It follows that ⊂ ∈ S T (B ) B(y , ǫ ). E ⊂ i i I [∈ We recall that the rank of a linear operator T : E F is the dimension of the linear subspace T (E) F , i. e. the dimension of the image (or range ) of T . → ⊂

Definition 17.1.2. An operator T (E, F ) is said to be of finite rank if T (E) is finite dimensional. ∈ L Exercise 17.1.3. Any finite rank operator is a compact operator. Hint: use Riesz’ lemma for normed spaces.

Proposition 17.1.1. Let T be a sequence of ﬁnite rank operators, and let T (E, F ) be such that T n ∈ L k n − T (E,F ) 0. Then, T (E, F ). kL → ∈ K Proof. It is an immediate consequence of the closedness of (E, F ). K The converse of the previous proposition is valid if F a Hilbert space.

Proposition 17.1.2. Let F be a Hilbert space. Let T (E, F ). Then, for every ǫ > 0 there exists a ﬁnite rank operator T (E, F ) such that T T < 2ǫ. ∈ K ǫ ∈ K k ǫ − k

Proof. Since T is compact, there exists a ﬁnite covering of T (BE) with balls of radius ǫ, say

T (B ) B (x ), E ⊂ ǫ i i[=I 123 124 CHAPTER 17. COMPACT OPERATORS with I ﬁnite. Let G be the vector space generated by the x ’s with i I, and set T = P T , i. e. T is the i ∈ ǫ G ǫ orthogonal projection of T onto G. Clearly, Tǫ is a ﬁnite rank operator. Now, for every x BE there exists i0 I such that ∈ ∈ T (x) x < ǫ. k − i0 k Thus, P (T (x)) P (x ) < ǫ, k G − G i0 k that is P (T (x)) x < ǫ. k G − i0 k Combining the two above inequalities we get

P (T (x)) T (x) < 2ǫ, for all x B , k G − k ∈ E which implies the assertion.

17.2 The Riesz-Fredholm theory

The next proposition collects some useful properties of compact operators. Proposition 17.2.1. (i) A compact operator T (E, F ) maps weakly convergent sequences in E into strongly convergent sequences in F . ∈ K

(ii) If A (E, F ) and B (F, G ), or if A (E, F ) and B (F, G ), then BA (E, G ). ∈ K ∈ L ∈ L ∈ K ∈ K (iii) (F. Riesz) Let A (E), with E Banach space, and let λ R 0 . Then, (λI A) is closed, and if λI A is one-to-one then∈ K (λI A) = E. ∈ \ { } R − − R − (iv) If A (E, F ), then A∗ (F ∗, E ∗). Vice versa, if A∗ (F ∗, E ∗) and F is a Banach space, then A ∈(E, K F ). ∈ K ∈ K ∈ K Proof. Proof of (i). Let x E be a sequence which converges weakly to x E. We claim that T (x ) F n ∈ ∈ n ∈ converges weakly to T (x) in F . To see this, let g F ∗, and compute ∈ g, T (x ) = g T, x h n i h ◦ ni and the last term above converges to g T, x since g T E∗ and from the definition of weak convergence h ◦ i ◦ ∈ in E. Now, we claim that the convergence of T (xn) to T (x) is strong. Indeed, assume by contradiction that T (x ) T (x) ǫ > 0 for some subsequence x of x and for some ǫ > 0. Since x is weakly convergent, k nk − kF ≥ 0 nk n 0 n then xn is bounded due to Proposition 11.2.3. Hence, by definition of compact operator, T (xnk ) is pre-compact in F , and that means that there exists a further subsequence of T (x ) that converges strongly to some y F . nk ∈ By assumption on the subsequence xnk , we get y T (x) ǫ0. On the other hand, the whole sequence T (xn) converges weakly to T (x), and since strong convergencek − impliesk ≥ weak convergence we get y = T (x), which is a contradiction. Proof of (ii). In the first case, A maps bounded sets into pre-compact ones, and B maps compact sets into compact sets ( B is continuous). Hence, BA maps bounded sets into pre-compact sets. In the second case, A maps bounded sets into bounded sets, and B maps bounded sets into pre-compact sets. Hence, BA maps bounded sets into pre-compact ones. Proof of (iii). Let us first prove that (λI A) is closed. Let yn n (λI A) such that yn y0 E strongly. We know that R − { } ⊂ R − → ∈ y = λx A(x ), n n − n for some sequence x E. We need to prove that there exists some x E such that { n}n ⊂ 0 ∈ y = λx A(x ). 0 0 − 0

Let us assume ﬁrst that xn is a bounded sequence. In this case, y + A(x ) x = n n , n λ 17.2. THE RIESZ-FREDHOLM THEORY 125 and since A(x ) is pre-compact, there exists a subsequence x of x such that A(x ) converges. Therefore, { n }n nk n nk y + A(x ) x = nk nk nk λ converges to some x0, which implies A(xnk ) converges to A(x0), and hence

y + A(x ) x = 0 0 , 0 λ as desired. Let us now assume that x is not bounded. Let M = Ker( λI A). Let n −

αn = dist( xn, M ).

There exists a sequence w M such that n ∈ 1 α x w α 1 + . n ≤ k n − nk ≤ n n Since w Ker( λI A), we have n ∈ − (λI A)( x w ) = y . − n − n n If x w is bounded, then we are in the previous case and the assertion is proven. We shall now prove that n − n xn wn is always bounded. To see this, we prove that αn is bounded in R. Assume by contradiction that αn is not− bounded. By extracting a subsequence, we can assume that α + . Set n → ∞ x w z = n − n . n x w k n − nk

Clearly, zn = 1. Now, k k y (λI A)z = n , − n x w k n − nk and since x w α + and y y , we have k n − nk ≥ n → ∞ n → 0 (λI A)z 0. − n →

Since A(zn) is a compact set, we can extract a subsequence znk such that A(znk ), converges, and the above formula{ implies} z z for some z , and the continuity of A implies z = A(z ). Therefore nk → 0 0 0 0 (λI A)z = 0 , − 0 i. e z0 M. But then ∈ 1 z z = [x (w + z x w )] , n − 0 x w n − n 0k n − nk k n − nk with w + z x w M, which implies n 0k n − nk ∈ x (w + z x w ) α , k n − n 0k n − nk k ≥ n and hence αn zn z0 1 1, k − k ≥ αn 1 + n → and this is a contradiction because znk z0. → For the ﬁnal statement in (iii), assume Aλ = λI A is one-to-one. The set (Aλ) is a closed subspace of E. − R 1 Hence, A : E (A ) is a bijection between two Banach spaces. By the open mapping theorem 10.3.1, A− λ → R λ λ is a bounded operator. If, by contradiction, we have E1 := (Aλ) = E, then let E2 := Aλ(E1). We claim that E cannot be equal to E . Indeed, let x E E . Then, AR(x) 6E with x E . Assuming that A (x) E 2 1 ∈ \ 1 λ ∈ 1 6∈ 1 λ ∈ 2 would imply that Aλ(x) = Aλ(Aλ(y)) for some y E, and since Aλ is injective this would imply x = Aλ(y), i. e. x E , a contradiction. Hence, A (x) E E∈ . Now, E is clearly a closed subspace of E. We inductively ∈ 1 λ ∈ 1 \ 2 2 set En+1 = Aλ(En), and we have just proved that En+1 ( En. By Riesz’s lemma 7.1.2, there exists a sequence 126 CHAPTER 17. COMPACT OPERATORS

xn n E with xn En 1 En for all n N, xn = 1 for all n E, and xn xn 1 1/2. Hence, for n > m , { } ⊂ ∈ − \ ∈ k k ∈ k − − k ≥ we have by deﬁnition of Aλ

1 1 (A(x ) A(x )) = x x + (A (x ) A (x )) , λ m − n m − n λ λ m − λ n and since the term multiplied by 1 /λ above belongs to Em+1 , then the whole squared bracket is an element of Em+1 (n > m ), and that implies λ A(x ) A(x ) | |, k m − n k ≥ 2 but this is a contradiction with the fact that A is a compact operator (the sequence xn is bounded, and therefore the sequence A(xn) has a convergent subsequence, which contradicts the above inequality for all n > m ). Proof of (iv). Assume A (E, F ). We claim that the space Z = A(E) is separable. To see this, since A is ∈ K compact the set A(B1(0)) is compact, and thus separable (it can be covered with ﬁnitely many balls with radius 1 /k for all k N, and the centers of those balls are a countable dense set). Then, A(E) = n 1 nA(B1(0)), which proves ∈ ≥ that A(E) is separable, as it is a countable union of separable sets. Since Z is separable,S Banach-Alaoglu-Bourbaki theorem implies that every bounded sequence in Z∗ admits a convergent subsequence in the weak ∗ topology. We need to prove that A∗(B (0)) is pre-compact, where B (0) F ∗ is the open unit ball in F ∗. Let f be a sequence 1 1 ⊂ n in A∗(B (0)), we need to show that f has a convergent subsequence. We know that f = A∗(g ) = g A, with 1 n n n n ◦ g F ∗ and g ∗ 1. We can restrict g to Z F and consider g Z∗, still with g ∗ 1. Since Z is n ∈ k nkF ≤ n ⊂ n ∈ k nkZ ≤ separable, we can extract a subsequence g which converges weakly to some g Z∗. Now, let f = g A. Let us nk ∗ ∈ ◦ consider fnk = gnk A. By deﬁnition of norm of a functional, there exists a sequence xnk E with xnk 1 such that ◦ ∈ k k ≤ 1 (f f)( x ) f f . k nk − nk k ≥ 2k nk − k

Since xnk is bounded, then A(xnk ) has a convergent subsequence, that we rename A(xnk ) for notation convenience.

Let y0 be its limit. Then, g(A(xnk )) g(y0). Moreover, since A(xnk ) y0 strongly, and gn converges to g weakly star, from Exercise 11.3.1 we have g →(A(x )) g(y ). But then → nk nk → 0 1 f f (f f)( x ) = g (A(x )) g(A(x )) g (A(x )) g(y ) + g(A(x )) g(y ) , 2k nk − k ≤ k nk − nk k k nk nk − nk k ≤ k nk nk − 0 k k nk − 0 k and the above shows that the ﬁrst term above goes to zero, whereas the second term goes to zero since g is continuous. Hence, f f strongly, which proves the assertion. nk → Vice versa, if A∗ (F ∗, E ∗) and F is a Banach space, then A∗∗ (E∗∗ , F ∗∗ ) because of the previous step. ∈ K ∈ K Now let B E be bounded. Then, A∗∗ (J(B)) is pre-compact, because J(B) is bounded and A∗∗ is a compact ⊂ operator. Now, we claim that A∗∗ (J(B)) = J(A(B)). To see this, let η A∗∗ (J(B)). By deﬁnition of adjoint ∈ operator, for all g F ∗ we have ∈

η, g F ∗∗ F ∗ = A∗∗ (J(x)) , g F ∗∗ F ∗ = J(x), A ∗(g) E∗∗ E∗ = J(x), g A E∗∗ E∗ , h i × h i × h i × h ◦ i × and by deﬁnition of J, we have

η, g F ∗∗ F ∗ = J(x), g A E∗∗ E∗ = g A, x E∗ E = g, A (x) F ∗ F = J(A(x)) , g F ∗∗ F ∗ , h i × h ◦ i × h ◦ i × h i × h i × i. e. η = J(A(x)) J(A(B)). The same combination of identities shows that J(A(B)) = A∗∗ (J(B)). Now, this implies that J(A(B))∈ is pre-compact, and since J is an isometry, then A(B) is pre-compact. Hence, A (E, F ). ∈ K Remark 17.2.1 . Let E be a normed space, and let M E be a subspace of ﬁnite dimension. Then E can be written as direct sum E = M N with N being another closed⊂ subspace of M, i. e. there exists a closed subspace N of E such that M N = 0 and such that every x E can be uniquely written as x = y + z with y M and z N. To see this, let m∩ , . . . ,{ m L} be a base for M, so that∈ every y M can be uniquely written as y = ∈λ m + . . . +∈ λ m . 1 k ∈ 1 1 k k With such notation, set ϕj : M R, ϕj(y) = λj. By the Hahn-Banach theorem 9.1.1, for each j = 1 ,...,k there → k exists a linear and continuous functional ϕj on E that extends ϕj . Set N = j=1 Kerϕ ˜ j. Then N is a closed subspace, and it satisﬁes the requested properties. T Notation 17.2.1 . Let M be a linear subspacee of a normed space E. We set

M ⊥ = f E∗ : f, x = 0 for all x M E∗. { ∈ h i ∈ } ⊂ 17.2. THE RIESZ-FREDHOLM THEORY 127

If N E∗ is a linear subspace, we set ⊂ N ⊥ = x E : f, x = 0 for all f N E. { ∈ h i ∈ } ⊂

M ⊥ and N ⊥ are closed subspaces (easy exercise). Note that, by deﬁnition, N ⊥ is a subset of E rather than E∗∗ . We say that M ⊥ (resp. N ⊥) is the space orthogonal to M (resp. N).

Exercise 17.2.1. Let M be as above. Prove that (M ⊥)⊥ M. ⊃ In what follows we shall consider I A instead of λI A. In fact all properties of I A are easily extended to λI A for λ C 0 . − − − − ∈ \ { } Theorem 17.2.2 (Fredholm) . Let E be a Banach space and A (E). Then, ∈ K (a) Ker( I A) is ﬁnite dimensional. − (b) (I A) = Ker(( I A)∗)⊥ = Ker( I A∗)⊥. R − − − (c) Ker( I A) = 0 if and only if (I A) = E. − { } R − (d) dim(Ker( I A)) = dim(Ker( I A∗)) . − − Proof. Proof of (a). Let N = Ker( I A). Then x N if and only if x = A(x). Hence, given BN the closed unit ball of N, we have − ∈ BN = A(BN ), which implies that BN is compact in the strong topology. Theorem 7.1.3 implies that N has ﬁnite dimension. Proof of (b). Let x (I A). This means that x = y A(y) for some y E. Now, an element z E∗ belongs ∈ R − − ∈ ∈ to Ker(( I A)∗)⊥ if and only if f, z = 0 for all f Ker(( I A)∗). The latter means ( I A)∗(f) = f A∗(f) = 0. Now, for such− f, let x as above,h andi compute ∈ − − −

f, x = f, y A(y) = (I A∗)f, y = 0 , h i h − i h − i and this implies x Ker(( I A)∗)⊥. The opposite inclusion follows as an easy exercise. Proof of (c). ∈The fact− that Ker( I A) = 0 implies (I A) = E is a consequence of Proposition 17.2.1 − { } R − (iii). Assume now that (I A) = E. (b) above implies Ker( I A∗) = (I A)⊥ = 0 . But A∗ is a compact R − − R − { } operator on E∗, hence Proposition 17.2.1 (iii) implies (I A∗∗ ) = E∗∗ . But since A∗∗ is an extension of A, this implies (together with (b)) R − Ker( I A) = (I A∗)⊥ = 0 . − R − { } Proof of (d). Since both dimensions are finite in view of (a), set d = dim(Ker( I A)) and d∗ = dim(Ker( I A∗)). − − We will first prove that d∗ d. Suppose not, that d < d ∗. Since d is finite, due to Remark 17.2.1 there exists a closed subspace M E such≤ that E = Ker( I A) M. Thus, there exists a continuous projection P from E onto Ker( I A). ⊂ − − L On the other hand, (I A) = Ker( I A∗)⊥ has finite codimension d∗, i. e. there exists a d∗-dimensional space Z E such that E = R(I −A) Z. To− see this, we recall that ⊂ R −

(I A) = Ker(L I A∗)⊥ = x E : f, x = 0 , for all f Ker( I A∗) . R − − { ∈ h i ∈ − } ∗ d Now, let f ,...,f ∗ be a basis of Ker( I A∗). Consider the map Φ : E R deﬁned by 1 d − → Φ( x) = ( f , x ,..., f ∗ , x ). h 1 i h d i ∗ Clearly, Φ is surjective. Otherwise, there would be a nonzero vector α = ( α ,...,α ∗ ) Rd orthogonal to Φ( E), 1 d ∈ i. e. ∗ d α Φ( x) = α f , x = 0 , for all x E, · jh j i ∈ j=1 X which is a contradiction with f ,...,f ∗ being a basis. Consequently, there exist e ,...,e ∗ E such that 1 d 1 d ∈

f , e = δ , for all i, j = 1 ,...,d ∗. h i j i ij

It is easy to check that the ej’s are linearly independent and that they generate the desired space Z. Since d < d ∗, there is a linear map Λ : Ker( I A) Z that is injective and not surjective. − → 128 CHAPTER 17. COMPACT OPERATORS

Set S = A + Λ P . Since Λ P has ﬁnite rank, then S is a compact operator. We claim that Ker( I S) = 0 . Indeed, if ◦ ◦ − { } 0 = u S(u) = ( u A(u)) (Λ( P (u))) , − − − since u A(u) = (I A) and Λ( P (u)) Z, then we have − R − ∈ u A(u) = Λ( P (u)) = 0 , − i. e. u Ker( I A), which implies P (u) = u and therefore Λ( u) = 0, and thus u = 0 because Λ is injective. Applying∈ (c)− to the operator S we get (I S) = E. This is a contradiction. Indeed, let f F such that f (Λ), which exists since Λ is not surjective.R − This means that the equation u S(u) = f has no∈ solution (easy exercise).6∈ R Hence, (I S) cannot be equal to the whole space E. − R − We have proved that d∗ d. Applying this fact to A∗. we obtain ≤

dim(Ker( I A∗∗ )) dim(Ker( I A∗)) dim(Ker( I A)) . − ≤ − ≤ −

But due to Exercise 16.2.1 we know that I A∗∗ is an extension of I A), which implies − −

Ker( I A∗∗ ) Ker( I A), − ⊃ − hence d = d∗. The above theorem is the main tool to solve the class of functional equation

u A(u) = f (17.1) − on a given Banach space E. Given a compact linear operator A on E, the Fredholm alternative principle says that

either for every f E the equation (17.1) has a unique solution u E, • ∈ ∈ or the homogeneous equation u A(u) = 0 admits n linearly independent solutions, and in this case the • inhomogeneous equation (17.1) is− solvable if and only if f satisﬁes n orthogonality conditions, i. e.

f Ker( I A∗)⊥. ∈ − This is proven as follows: if (17.1) has a solution for every f, then I A is surjective. But then the solution is unique, because given two solutions u , u E we have − 1 2 ∈ (u u ) A(u u ) = u A(u ) (u A(u )) = f f = 0 , 1 − 2 − 1 − 2 1 − 1 − 2 − 2 − and (c) implies that u1 = u2. If, otherwise, there is some f E for which (17.1) has no solution, it means that I A is no longer surjective, and therefore ∈ − the kernel of I A is no longer the null space. However, (a) says it will be finite dimensional, say dim(Ker( I • A)) = n, and therefore− the homogeneous equation u A(u) = 0 has n independent solutions. Moreover, − − (17.1) is solvable if and only if f (I A), and (b) implies this is possible if and only if f is orthogonal to • ∈ R − Ker( I A∗). Now, (d) establishes that the latter condition is equivalent to exactly n equations. − Remark 17.2.2 . The property (c) in Theorem 17.2.2 is familiar in finite dimensional spaces. If dim( E) < + , a linear operator from E into itself is injective if and only if it is surjective. However, in infinite dimensional spaces∞ a bounded operator may be injective without being surjective and conversely. Therefore (c) is a remarkable property of operators of the form I A with A (E). − ∈ K 17.3 The spectrum of a compact operator

We devote this section to properties of compact operators related with the concepts in deﬁnition 16.1.1. We start with the following lemma which shows that every nonzero element of the spectrum of a compact operator is an eigenvalue. Throughout the whole section, E will be a Banach space. Lemma 17.3.1. Let A (E) and let λ σ(A) 0 . Then, λ σ (A), i. e. λ is an eigenvalue of A. ∈ K ∈ \ { } ∈ p 17.3. THE SPECTRUM OF A COMPACT OPERATOR 129

Proof. Suppose λ = 0 is not an eigenvalue. Then λI A is injective. From Proposition 17.2.1 we know that λI A is surjective, and from6 the open mapping theorem we− know that λI A is invertible, hence λ ρ(A), and λ cannot− be in the spectrum. − ∈

We continue with the following lemma, which basically states that all the nonzero points in the spectrum of a compact operators are isolated.

Lemma 17.3.2. Let A (E) and let λ be a sequence of distinct real numbers such that λ λ and ∈ K n n → λ σ(A) 0 , for all n N. n ∈ \ { } ∈ Then λ = 0 .

Proof. From Lemma 17.3.1 we know that the λn’s are eigenvalues. Let en = 0 such that ( A λnI)en = 0. Let E E be the linear subspace generated by e ,...,e . Clearly E E .6 We claim that E−= E for all n. n ⊂ 1 n n ⊂ n+1 n 6 n+1 To see that, it suﬃces to show that the vectors en are all linearly independent. We prove that by induction on n. n Assume this holds up to n and suppose en+1 = i=1 αiei. Then

P n n A(en+1 ) = αiA(ei) = αiλiei. i=1 i=1 X X On the other hand, n A(en+1 ) = λn+1 en+1 = λn+1 αiei. i=1 X The two above identities imply αi(λi λn+1 ) = 0 for all i = 1 ,...,n since e1,...,e n are linearly independent. Since the eigenvalues are all distinct, we have− α = 0 for all i = 1 ,...,n , a contradiction with λ = 0. i n+1 6 Applying Riesz’s lemma 7.1.2, we construct a sequence u E such that u E for all n N, u = 1, and n ∈ n ∈ n ∈ k nk dist( un, E n 1) 1/2 for all n 2. For 2 m < n we have − ≥ ≥ ≤

Em 1 Em En 1 En. − ⊂ ⊂ − ⊂

On the other hand, ( A λnI)En En 1. Indeed, let y (A λnI)En, i. e. y = ( A λnI)( gn 1 + αe n) for some − ⊂ − ∈ − − − gn 1 En 1. We have − ∈ − y = ( A λnI)( gn 1) + α(A λnI)( en) = ( A λnI)( gn 1), − − − − − and the last term above is an element of En 1 (because A(gn 1) is a linear combination of vectors in En 1). Therefore, we can write − − −

A(un) A(um) (A(un) λnun) (A(um) λmum) = − − + un um dist( un, E n 1) 1/2. λ − λ λ − λ − ≥ − ≥ n m n m

Now, assume by contradiction that λ λ = 0. Then, A(u ) has a convergent subsequence because A is a n → 6 { n }n compact operator. Therefore, A(un) has a convergent subsequence too, but that contradicts the above inequality. λn

Theorem 17.3.3 (Schauder) . Let A (E) with dim( E) = + . Then, we have: ∈ K ∞ (a) 0 σ(A). ∈ (b) σ(A) 0 = σ (A) 0 . \ { } p \ { } (c) One of the following cases holds:

– σ(A) = 0 , { } – σ(A) 0 is a ﬁnite set, \ { } – σ(A) 0 is a sequence converging to zero. \ { } 130 CHAPTER 17. COMPACT OPERATORS

Proof. Proof of (a). Suppose that 0 σ(A). Then, 0 ρ(A), which means that A is invertible with a bounded 1 6∈ ∈ inverse. Therefore, I = A A− , i. e. I is a compact operator, i. e. the unit ball BE is pre-compact, i. e. the dimension of E is finite, a contradiction.◦ Proof of (b). This is proven in Lemma 17.3.1. Proof of (c). For every integer n 1, the set ≥ σ(A) λ R : λ 1/n ∩ { ∈ | | ≥ } is either empty of finite (if it had infinitely many distinct points we would have had a subsequence convergent to some λ = 0, which contradicts Lemma 17.3.2). Hence, if σ(A) 0 has infinitely many points, we may order them as a sequence6 tending to zero. \ { } Chapter 18

Self-adjoint operators on Hilbert spaces

In what follows we assume that E = H is a Hilbert space and T (H). Identifying H∗ with H, we may view the ∈ L adjoint operator T ∗ as a bounded operator from H into itself.

Definition 18.0.1. A bounded operator T (H) is said to be self-adjoint if T ∗ = T , i. e. ∈ L (T (u), v ) = ( u, T (v)) , for all u, v H. ∈ Proposition 18.0.1. Let T (H) be a self-adjoint operator. Set ∈ L m = inf (T (u), u ), and M = sup (T (u), u ). u H, u =1 u H, u =1 ∈ k k ∈ k k Then, σ(T ) [m, M ], m, M σ(T ). Moreover, T = max m , M . ⊂ ∈ k k {| | | |} Proof. Let λ > M ; then since ( T u, u ) M u 2 for any u H we have that ≤ k k ∈ (λu T u, u ) = λ u 2 (T u, u ) (λ M) u 2. − k k − ≥ − k k Then, by using Theorem 14.2.5 the operator λI T is a bijection from H into itself and then λ ρ(T ). By the same argument is follows that if λ < m then λ −ρ(T ). Then, σ(T ) [m, M ]. We prove that m,∈ M σ(T ). We prove only that M σ(T ) since the proof that m∈ σ(T ) is similar. First,⊂ we note that a(u, v ) := ( Mu∈ T u, v ) is symmetric since T is∈ self-adjoint and satisfies a(u,∈ u ) 0 for any u H. Then, by Exercise 14.1.1, a(u,− v ) satisfies the Cauchy-Schwartz inequality and then ≥ ∈

1 1 (Mu T u, v ) (Mu T u, u ) 2 (Mv T v, v ) 2 for any u, v H | − | ≤ − − ∈ and then 1 Mu T u C(Mu T u, u ) 2 for any u H. k − k ≤ − ∈ Finally, by the very deﬁnition of M it follows that there exists un n H with un = 1 and ( T u n, u n) M as n . Then, since Mu T u 0 it follows M σ(T ), otherwise{ } ⊂MI T wouldk k be a bijection and→ then → ∞ k n − nk → ∈ − 1 u = ( MI T )− (MI T )u 0 n − − n → which is a contraction to the fact that un = 1. Finally we prove that T = µ = maxk km , M . For any u, v H it holds that k k {| | | |} ∈ (T (u + v), u + v) = ( T u, u ) + ( T v, v ) + 2( T u, v ), (T (u v), u v) = ( T u, u ) + ( T v, v ) 2( T u, v ). − − − By subtracting the above two equalities and using the deﬁnition of m and M we have 4 (T u, v ) µ( u + v 2 + u v 2) µ( u 2 + v 2). | | ≤ k k k − k ≤ k k k k By replacing v with αv with α = u / v we get k k k k (T u, v ) µ u v for any u, v H, | | ≤ k kk k ∈ then T µ. On the other hand since (T u, u ) T u 2 it is clear that m T and M T and then µ < kT k. ≤ | | ≤ k kk k | | ≤ k k | | ≤ k k k k

131 132 CHAPTER 18. SELF-ADJOINT OPERATORS ON HILBERT SPACES

Corollary 18.0.2. Let T (H) be a self-adjoint operator such that σ(T ) = 0 . Then, T 0. ∈ L { } ≡ Proof. Straightforward consequence of the previous proposition. Our last statement is a fundamental result in operator theory. It states that every compact self-adjoint operator on a Hilbert space may be diagonalised in some suitable basis. Theorem 18.0.3. Let T (H), H separable Hilbert space, T self-adjoint. Then, there exists an orthonormal basis composed by eigenvector∈ K of T .

Proof. Let λn n be the sequence of all distinct non-zero eigenvalues of T . We set { } λ = 0 , E = Ker (T ), E = Ker (T λ I). 0 0 n − n and we recall that by using Theorem 17.2.2 and 17.3.3 we have

0 dim E 0 < dim E < . ≤ 0 ≤ ∞ n ∞ It follows that the sets E are mutually orthogonal. Indeed, if u E and v E with m = n, then { n}n ∈ n ∈ m 6

T u = λnu, Tv = λmv and then (T u, v ) = λn(u, v ) = ( u, T v ) = λm(u, v ) and then ( u, v ) = 0. Let F be the vector space spanned by the space E . We want to prove that F is dense in H. We start { n}n by noticing that T (F ) F and T (F ⊥) F ⊥. Indeed, the ﬁrst inclusion is trivial and the second one follows by ⊂ ⊂ considering u F ⊥ and using that T is self-adjoint we have ∈ (T u, v ) = ( u, T v ) = 0 for any v F, ∈ then, T u F ⊥. Let T0 be the restriction of T to F ⊥. T0 is a self-adjoint operator compact on F ⊥ and σ(T0) = 0 . Indeed, suppose∈ by contradiction σ(T ) = 0 . Then, there is λ = 0 in σ(T ) and since T is compact λ is{ an} 0 6 { } 6 0 0 eigenvalue. Then, there exists u F ⊥ such that ∈

T0u = λu. (18.1)

But since T0 is the restriction of T to F ⊥ (18.1) implies that λ is an eigenvalue of T , namely λ = λn for some n N. Therefore u is in E F and then in F ⊥ F . Then, u = 0 which is a contradiction. In particular, by ∈ n ⊂ ∩ applying Corollary 18.0.2 it follows that F ⊥ Ker T and then, since Ker T F , F ⊥ F and then F ⊥ = 0 which implies that F is dense in H. ⊂ ⊂ ⊂ { } In order to construct a base for H we argue as following. Since H is separable and E0 is closed it follows that E0 is a separable Hilbert space and then Theorem 14.3.5 ensures the existence of a countable orthonormal base for E0. To this orthonormal base we add all the orthonormal basis of En for n N whose existence is trivial since they are ﬁnite dimensional. Of course the resulting orthonormal system is a base∈ of eigenvector for H.

As a consequence of the above theorem, let us denote by λn n the sequence of eigenvalues of A, with each eigenvalue repeated by its multiplicity. Let x H be a the{ basis} of eigenvectors as in the theorem. Then, we { n}n ⊂ write H = Ker( A) Ker( A)⊥. Therefore, any x H can be uniquely decomposed as ∈ L x = x0 + (x, x k)xk, Xk with x Ker( A). Therefore, 0 ∈ A(x) = λk(x, x k)xk, Xk i. e. the operator A is entirely described in terms of the basis x and the corresponding eigenvalues λ . { k} k Chapter 19

Exercises

Exercise 19.1. Let E be a Banach space and T (E) be an invertible operator. ∈ L 1. 0 / σ(T ), ∈ 1 2. λ σ(T ) if and only if 1/λ σT − , ∈ ∈ 1 3. If T = T − = 1 then λ = 1 for any λ σ(T ). k k k k | | ∈

Exercise 19.2. Let k : [0 , 1] [0 , 1] R be a continuos function. Consider the operator × → K : C([0 , 1]) C([0 , 1]) , → 1 (Kf )( x) = k(x, y )f(y) dy. Z0 Show that 1. T is continuous, 2. T is compact.

Exercise 19.3. Let k : [0 , 1] [0 , 1] R be a function in L2([0 , 1] [0 , 1]) . Consider the operator × → × K : L2(0 , 1) L2(0 , 1) , → 1 (Kf )( x) = k(x, y )f(y) dy. Z0 Show that 1. T is continuous, 2. T is compact.

Exercise 19.4. Let U Rm compact. Let K : C(U) C(U) be the operator ∈ → (Kf )( x) = k(x, y )f(y) dy. ZU Suppose k is deﬁned and continuos for all x, y U, x = y, and there exist positive constant M and α (0 , m ] such that ∈ 6 ∈ α m K(x, y ) M x y − x, y U, x = y. | | ≤ | − | ∈ 6 Show that K is compact.

133 134 CHAPTER 19. EXERCISES

Deﬁnition 19.0.1. Let H be a separable Hilbert space and T be a bounded operator from H into itself. We say 2 that T is an Hilbert-Schmidt operator if there exists an othornormal basis e such that ∞ T e < . { n}n i=1 k nk ∞ P Exercise 19.5. Prove the following statement.

1. T is an Hilbert-Schmidt operator if and only if T ∗ is an Hilbert-Schmidt operator. 2. If T is an Hilbert-Schmidt operator, then T is compact.

Exercise 19.6. Let K : L2(0 , 1) L2(0 , 1) be the operator → 1 (Kf )( x) = k(x, y )f(y) dy Z0 with k L2((0 , 1) (0 , 1)) . Show that ∈ × 1. K is an Hilbert-Schmidt operator. 2. Any Hilbert-Schmidt operator on L2((0 , 1)) has the form of K

n Exercise 19.7. Let H be an Hilbert space and e n N be a orthonormal set in H. Let λn n N. Consider the operator { } ∈ { } ⊂

∞ n n T x = λn(x, e )e . n=1 X Show that

1. T (H) if and only if λ l∞, ∈ L { n}n ∈ 2. T is compact if and only if λ 0 as n , n → → ∞ 3. T is an Hilbert-Schmidt operator if and only if λ l2. { n}n ∈

Exercise 19.8. Discuss the existence of solution(s) to the integral equation in L2(0 , 2π) given by

2π u(x) = f(x) λ sin( x + y)u(y) dx, − Z0 for the two cases

1 2 a) λ = √π , f(x) = x ,

1 b) λ = π , f(x) = sin(3 x).

Exercise 19.9. Consider the operator

T : L2( π, π ) L2( π, π ), − π→ − (T f )( x) = sin( x + y)f(y) dy. π Z− Show that 1. T is bounded operator, 2. T is a compact operator, 3. determine the dimension of the range of T . 135

Exercise 19.10. Consider the operator T : L2(0 , 1) L2(0 , 1) , →x (T f )( x) = f(y) dy. Z0 Show that 1. show that T is continuous, 2. Determine whether T is compact or not,

3. Find σ(T ) and σp(T ), 4. Compute the norm of T .

Exercise 19.11. Consider the operator

T : l2 l2, → (T x )n = xn 1, (T x )1 = 0 . − Show that 1. T is a continuous operator, 2. 0 σ(T ) σ (T ), ∈ \ p 3. σ (T ) = , p ∅ 4. λ σ(T ) if and only if λ 1. ∈ | | ≤

Exercise 19.12. Consider the operator T : C([0 , 1]) C([0 , 1]) , → (T f )( x) = xf (x). Show that 1. T is a continuous operator, 2. σ (T ) = , p ∅ 3. σ(T ) = [0 , 1] , 4. If λ [0 , 1] then T λI is not dense in C([0 , 1]) . ∈ −

Exercise 19.13. Consider the operator T : L2(R) L2(R), → (T f )( x) = f(x + 1) .

Show that T is continuous but not compact. Find σ(T ) and σp(T ).

Exercise 19.14. Consider the operator T : L2(R) L2(R), → f(x) (T f )( x) = . 1 + x | | Show that T is continuous but not compact. Find σ(T ) and σp(T ). 136 CHAPTER 19. EXERCISES

Exercise 19.15. Let a (0 , 1) . Consider the operator ∈ T : L2(R) L2(R), → (Taf)( x) = f(ax ).

Show that

1. T is continuous but not compact,

2. ﬁnd a formula for the adjoint Ta∗,

3. compute the norm of Ta∗ and Ta.

4. ﬁnd σ(T ) and σp(T ).

Exercise 19.16. Consider the operator

T : C([ 1, 1]) C([ 1, 1]) , − → − (T f )( x) = f( x). − Show that

1. 1 and 1 are eigenvalues of T , − 2. σ(T ) = 1, 1 , {− } 3. T is not compact.

Exercise 19.17. Consider the operator T : l2 l2, → x (T x ) = n . n n Then,

1. show that T is continuous,

2. Determine whether T is compact or not,

3. Find σ(T ) and σp(T ).

Exercise 19.18. Consider the operator

T : l2 l2, → xn for n odd, (T x )n = (0 for n even.

Then,

1. show that T is continuous,

2. Determine whether T is compact or not,

3. Find σ(T ) and σp(T ). 137

Exercise 19.19. Consider the operator

T : L2( π, π ) L2( π, π ), − → − π ∞ (T f )( x) = sin( nx ) sin( ny )f(y) dy. n=1 π X Z− Then, 1. show that T is continuous, 2. Determine whether T is compact or not,

3. Find σ(T ) and σp(T ).

Exercise 19.20. Consider the operator T : C([0 , 1]) C([0 , 1]) , →x (T f )( x) = yf (y) dy. Z0 Then, 1. show that T is continuous, 2. Determine whether T is compact or not,

3. Find σ(T ) and σp(T ).

Exercise 19.21. Consider the operator T : C([0 , 1]) C([0 , 1]) , → x f(y) (T f )( x) = dy. √y Z0 Then, 1. show that T is continuous, 2. Determine whether T is compact or not,

3. Find σ(T ) and σp(T ).

Exercise 19.22. Consider the operator

T : L1(R) L1(R), → x y (T f )( x) = e−| − |f(y) dy. R Z Then, 1. show that T is continuous, 2. Determine whether T is compact or not,

3. Find σ(T ) and σp(T ). Exercise 19.23. Deﬁne the left-shift operator on ℓp, p [1 , + ), ∈ ∞ (T a) = a , n 1, a = ( a ) . ℓ n n+1 ≥ n n Prove that T (ℓp) and compute T . • ℓ ∈ L k ℓk 138 CHAPTER 19. EXERCISES

Prove that σ (T ) = ( 1, 1) and compute the corresponding eigenvalues. • p ℓ − Prove that σ(T ) = [ 1, 1] . • ℓ − Exercise 19.24. Deﬁne the right-shift operator on ℓp, p [1 , + ), ∈ ∞ 0 if n = 1 (Tra)n = , a = ( an)n. (an 1 if n 2, − ≥ Prove that T (ℓp) and compute T . • r ∈ L k rk Prove that σ(T ) = [ 1, 1] . Hint: try to solve λa T a = b for b = (1 , 0, 0, 0,... ). • r − − r Prove that σ (T ) = . • p r ∅ Exercise 19.25. Let K : L2([0 , 1]) L2([0 , 1]) be the integral operator deﬁned by → x Kf (x) = f(y)dy. Z0

(a) Find the adjoint operator K∗

2 (b) Find the operator norm K (Hint: use K∗K = K ). k k k k k k (c) Show that the spectral radius of K is equal to zero. (d) Show that 0 belongs to the continuum spectrum of K.

Exercise 19.26. Given a bounded linear operator A (X) on a Banach space X, the continuum spectrum of A is the set of λ R such that λI A is one-to-one but∈ L not onto and the range of λI A is dense. The residual spectrum is the of∈ elements in the spectrum− which are neither in the point spectrum nor− in the continuum spectrum. Now, let G be the multiplication operator on L2(R) deﬁned by

Gf (x) = g(x)f(x), where g is C1 and bounded. Prove that G is bounded and with spectrum

σ(G) = g(x) , x R . { ∈ } Can an operator of this form have eigenvalues? Does G have any element in the residual spectrum? Bibliography

[1] Haim Brezis. Functional Analysis, Sobolev Spaces, and Partial Differential Equations . Springer, 2011. [2] Nelson Dunford and Jacob T. Schwartz. Linear operators. Part I . Wiley Classics Library. John Wiley & Sons, Inc., New York, 1988. General theory, With the assistance of William G. Bade and Robert G. Bartle, Reprint of the 1958 original, A Wiley-Interscience Publication. [3] Henri Leon Lebesgue. Le¸cons sur l’intégration et la recherche des fonctions primitives professéesau Collège de France . Cambridge Library Collection. Cambridge University Press, Cambridge, 2009. Reprint of the 1904 original, URL: http://dx.doi.org/10.1017/CBO9780511701825.

[4] Terence Tao. An introduction to measure theory , volume 126 of Graduate Studies in Mathematics . Ameri- can Mathematical Society, Providence, RI, 2011. URL: http://terrytao.files.wordpress.com/2011/01/ measure-book1.pdf. [5] E. Zermelo. Beweis, daß jede Menge wohlgeordnet werden kann. Math. Ann. , 59(4):514–516, 1904.

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