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Measure Theorey and Functional Analysis MeasureTheoreyandFunctional Analysis DMTH505 MEASURE THEORY AND FUNCTIONAL ANALYSIS Copyright © 2012 Anuradha All rights reserved Produced & Printed by EXCEL BOOKS PRIVATE LIMITED A-45, Naraina, Phase-I, New Delhi-110028 for Lovely Professional University Phagwara SYLLABUS Measure Theory and Functional Analysis Objectives: This course is designed for the of analysis of various types of spaces like Banach Spaces, Hilbert Space, etc. and also Sr. No Description 1 Differentiation and Integration: Differentiation of monotone functions, Functions of bounded variation, 2 Differentiation of an integral, Absolute continuity 3 Spaces, Holder, Minkowski inequalities, Convergence and Completeness 4 Bounded linear functional on the Lp spaces, Measure spaces, Measurable Functions, Integration 5 General Convergence Theorems, Signed Measures, Radon-Nikodym theorem. 6 Banach spaces: Definition and some examples, Continuous linear transformations, The Hahn-Banach theorem 7 The natural imbedding of N in N**, The open mapping theorem, The closed graph theorem, 8 The conjugate of an operator, The uniform boundedness theorem, The uniform boundedness theorem, Hilbert spaces : The definition and some simple properties 9 Orthogonal complements, Orthonormal Sets, The conjugate space H*, The Adjoint of an Operator, Self Adjoint Operators 10 Normal and Unitary Operators, Projections, Finite dimensional spectral theory : the spectrum of an operator on a finite dimensional Hilbert space, the Spectral theorem CONTENTS Unit 1: Differentiation and Integration: Differentiation of Monotone Functions 1 Unit 2: Functions of Bounded Variation 11 Unit 3: Differentiation of an Integral 27 Unit 4: Absolute Continuity 36 Unit 5: Spaces, Hölder 50 Unit 6: Minkowski Inequalities 65 Unit 7: Convergence and Completeness 72 Unit 8: Bounded Linear Functional on the Lp-spaces 82 Unit 9: Measure Spaces 100 Unit 10: Measurable Functions 106 Unit 11: Integration 121 Unit 12: General Convergence Theorems 141 Unit 13: Signed Measures 158 Unit 14: Radon-Nikodym Theorem 166 Unit 15: Banach Space: Definition and Some Examples 174 Unit 16: Continuous Linear Transformations 183 Unit 17: The Hahn-Banach Theorem 201 Unit 18: The Natural Imbedding of N in N** 211 Unit 19: The Open Mapping Theorem 217 Unit 20: The Closed Graph Theorem 225 Unit 21: The Conjugate of an Operator 231 Unit 22: The Uniform Boundedness Theorem 238 Unit 23: Hilbert Spaces: The Definition and Some Simple Properties 244 Unit 24: Orthogonal Complements 253 Unit 25: Orthonormal Sets 262 Unit 26: The Conjugate Space H* 273 Unit 27: The Adjoint of an Operator 283 Unit 28: Self Adjoint Operators 294 Unit 29: Normal and Unitary Operators 305 Unit 30: Projections 316 Unit 31: Finite Dimensional Spectral Theory 329 Unit 1: Differentiation and Integration: Differentiation of Monotone Functions Unit 1: Differentiation and Integration: Differentiation Notes of Monotone Functions CONTENTS Objectives Introduction 1.1 Differentiation and Integration 1.1.1 Lipschitz Condition 1.1.2 Lebesgue Point of a Function 1.1.3 Covering in the Sense of Vitali 1.1.4 Four Dini's Derivatives 1.1.5 Lebesgue Differentiation Theorem 1.2 Summary 1.3 Keywords 1.4 Review Questions 1.5 Further Readings Objectives After studying this unit, you will be able to: Understand differentiation and integration Describe Lipschitz condition and Lebesgue point of a function State Vitali’s Lemma and understand its proof. Explain four Dini’s derivatives and its properties Describe Lebesgue differentiation theorem. Introduction Differentiation and integration are closely connected. The fundamental theorem of the integral calculus is that differentiation and integration are inverse processes. The general principle may be interpreted in two different ways: 1. If f is a Riemann integrable function over [a, b], then its indefinite integral i.e. x F : [a, b] R defined by F (x) = f (t) dt is continuous on [a, b]. Furthermore if f is a continuous at a point xo [a, b], then F is differentiable thereat and F (xo) = f (xo). 2. If f is Riemann integrable over [a, b] and if there is a differentiable function F on [a, b] such that F f(x) for x [a, b], then x f (t) dt = F (x) – F (a) [a x b]. a LOVELY PROFESSIONAL UNIVERSITY 1 Measure Theory and Functional Analysis Notes 1.1 Differentiation and Integration 1.1.1 Lipschitz Condition Definition: A function f defined on [a, b] is said to satisfy Lipschitz condition (or Lipschitzian function), if a constant M > 0 s.t. |f (x) – f (y)| M |x – y|, x, y [a, b]. 1.1.2 Lebesgue Point of a Function Definition: A point x is said to be a Lebesgue point of the function f (t), if 1 x n Lim |f(t) f(x)|dt 0. h 0 h x 1.1.3 Covering in the Sense of Vitali Definition: A set E is said to be covered in the sense of Vitali by a family of intervals (may be open, closed or half open), M in which none is a singleton set, if every point of the set E is contained in some small interval of M i.e., for each x E, and > 0, an interval I M s.t. x I and (I) . The family M is called the Vitali Cover of set E. Example: If E = {q : q is a rational number in the interval [a, b]}, then the family Iqi 1 1 where I q , q , i N is a vitali cover of [a, b]. qi i i Vitali's Lemma Let E be a set of finite outer measure and M be a family of intervals which cover E in the sense of Vitali; then for a given > 0, it is possible to find a finite family of disjoint intervals {Ik, k = 1, 2, … n} of M, such that n m * E I k < . k 1 Proof: Without any loss of generality, we assume that every interval of family M is a closed interval, because if not we replace each interval by its closure and observe that the set of end points of I1, I2, …… In has measure zero. [Due to this property some authors take family M of closed intervals in the definition of Vitali’s covering]. Suppose 0 is an open set containing E s.t. m* (0) < m* (E) + 1 < we assume that each interval in M is contained in 0, if this can be achieved by discarding the intervals of M extending beyond 0 and still the family M will cover the set E in the sense of Vitali. Now we shall use the induction method to determine the sequence <Ik : k = 1, 2, … n> of disjoint intervals of M as follows: 2 LOVELY PROFESSIONAL UNIVERSITY Unit 1: Differentiation and Integration: Differentiation of Monotone Functions Notes Let I1 be any interval in M and let 1 be the supremum (least upper bound of the lengths of the intervals in M disjoint from I1 (i.e., which do not have any point common with I1). Obviously 1 < as 1 m (0) < . 1 Now we choose an interval I from M, disjoint from I , such that (I ) . Let be the 2 1 22 1 2 supremums of lengths of all those intervals of M which do not have any point common with I 2 or I2 obviously 2 < . In general, suppose we have already chosen r intervals I 1, I2, … Ir (mutually disjoint). Let r be the supremums of the length of those intervals of M which do not have any point in common with I (i.e., which do not meet any of the intervals I , I , … I . Then m (0) < . i 1 2 r r i 1 r r Now if E is contained in I , then Lemma established. Suppose IE . Then we can find i i i 1 i 1 1 interval I s.t. (I ) which is disjoint from I , I , … I . r+1 r 12 r 1 2 r Thus at some finite iteration either the Lemma will be established or we shall get an infinite 1 sequence <I > of disjoint intervals of M s.t. (I ) and r < , n = 1, 2, 3 …. r r 12 r Note that < r > is a monotonically decreasing sequence of non-negative real numbers. Obviously, we have that I 0 ( ) m(0) hence for any arbitrary > 0, we can r r i 1 r 1 find an integer N s.t. 1 (Ir ) . r N 1 5 N I Let a set F = r . r 1 N The lemma will be established if we show that m* (F) < . For, let x F, then x I x is an r r 1 N element of E not belonging to the closed set I an interval I in M s.t. x I and (I) is so r r 1 N small that I does not meet the I , i.e. r r 1 I Ir = , r = 1, 2, … N. Therefore we shall have (I) N 2 (I n 1 ) as by the method of construction we take 1 I . n 12 N LOVELY PROFESSIONAL UNIVERSITY 3 Measure Theory and Functional Analysis Notes It also I IN+1 = , we should have I N 1 . Further if the interval I does not meet any of the intervals in the sequence <Ir>, we must have I r , r which is not true as r2 (I r 1 ) 0 as r . I must meet at least one of the intervals of the sequence <Ir>. Let p be the least integer s.t. I meets Ip. Then p > N and (I) p 1 2 (I p ). Further let x I as well x Ip, then the distance of x from the mid point of Ip is at most 1 1 5 (I) ( )2(I) (I) (I) 2p p 2 p 2 p Thus if Ip is an interval having the same mid point as Ip but length 5 times the length of Ip, i.e.
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