Spectral theory for compact operators on Banach spaces
|
|
- Letitia Ball
- 5 years ago
- Views:
Transcription
1 68
2 Chapter 9 Spectral theory for compact operators on Banach spaces Recall that a subset S of a metric space X is precompact if its closure is compact, or equivalently every sequence contains a Cauchy subsequence. Another characterization is that S is totally bounded, namely for any ɛ > 0 one could cover S by finitely many ɛ-balls. If X is a normed linear space we can add/multiply, and we have the following basic properties: (i) If S is precompact then so is αs for any α scalar. (ii) If S 1 and S 2 are precompact then S 1 +S 2 := {s 1 +s 2, s 1 S 1, s 2 S 2 } is precompact. (iii) If S is precompact then so is the convex hull of S. (iv) Let T : X Y where X and Y are say normed linear spaces. If S X is precompact then so is T S Y. 9.1 Compact operators Let T : X Y a bounded linear map between Banach spaces X and Y. Let B 1 be the unit ball in X. We say that T is pre-compact if T B 1 is a precompact subset of Y. Properties: (i) A finite rank operator (namely the range of the operator is finite dimensional) is compact. (ii) If T 1 : X Y and T 1 : X Y are compact operators then α 1 T 1 + α 2 T 2 is also compact for any scalar α 1, α 2. (iii) If T : X Y is compact and M : U X and N : Y V are bounded linear maps between Banach spaces then NT M : U V is compact. (iv) If T : X Y is compact then it maps a weakly convergent sequence in X to a strongly convergent sequence in Y. Such operator is called completely continuous, so we could say that compactness implies complete continuity. 1 1 If T : X X is completely continuous and X is reflexive then T is compact, this is left as an exercise. 69
3 70 CHAPTER 9. COMPACT OPERATORS To see property (iv), first note that T x n converges weakly to T x. To see this, let l X, then l(t x n ) = (T l)x n converges to T lx which is the same as l(t x). It then follows that any strongly convergent subsequence of T x n has to converge to T x. Now since x n converges weakly to x in X it follows that x n is a family of bounded operator on X that is uniformly bounded pointwise, consequently by the principle of uniform boundedness sup n x n <, thus (T x n ) is precompact, so any subsequent contains a Cauchy subsequence; thus by a routine argument it follows that T x n is convergent to T x. (v) If T n is a sequence of compact operators from X to Y and T n T 0 as n for some T : X Y bounded linear, then T is compact. To see property (v), note that for any ɛ > 0 one could choose N large such that T N T < ɛ/2. Now, we could use finitely many ɛ/2 balls to cover T N B 1. The ɛ balls with the same centers will then cover T B 1. (Recall that B 1 is the unit ball in X.) As a corollary, we have Corollary 3. If T : X Y is the limit of a sequence of finite rank operators in the norm topology then T is compact. The converse direction of this corollary is not true in general (for Banach spaces X and Y ), the first construction of counter examples is due to P. Enflo ( 73). However, if Y is a Hilbert space then the converse is true. Theorem 32. Any compact operator T : X Y where X is Banach and Y is Hilbert can be approximated by a sequence of finite rank operators. Proof. We sketch the main ideas of the proof. For every n the set T B 1 is covered by Mn k=1 B Y (y k, 1 n ). We may assume that M n is an increasing sequence. Let P n be the projection onto the span of y 1,..., y Mn, which is clearly a finite rank operator, thus P n T is also finite rank. It remains to show that T P n T = O( 1 n ). We observe that P ny y k y y k for every 1 k M n and every y Y, since projection are contractions. It follows that P n T y T y P n T y y k + T y y k 2 T y y k and clearly given any y B 1 there is a k such that T y y k 1/n, therefore P n T y T y 2 n for every y B 1, thus P n T T 2/n as desired. Theorem 33 (Schauder). Let T : X Y be a bounded linear operator between Banach spaces X and Y. Then T is compact if any only if T : Y X is compact. Proof. It suffices to show the forward direction, namely if T is compact then T is also compact. For the other direction, apply the forward direction it follows that T is compact from Y to X, and by restricting T to X we obtain T therefore T is also compact. Now, assume that T is compact. Given any sequence l n X with l n 1 we will show that (T l n ) has a Cauchy subsequence T l nj, in other words given any ɛ > 0 we have sup l nj (T x) l nk (T x) ɛ x 1
4 9.2. COMPACTNESS OF INTEGRAL OPERATORS 71 if j and k are large enough. Let B be the unit ball in X and let K = T B which is a compact subset of X. Then the above estimate is a consequence of sup y K l nj y l nk y ɛ. We may view l n as a sequence of continuous funcitons on K, which are uniformly bounded pointwise and equicontinuous on K: sup l n y y n sup l n y 1 l n y 2 y 1 y 2 n Thus by the Arzela Ascoli theorem the sequence l n has an uniformly convergent subsequence, as desired. 9.2 Compactness of integral operators We now discuss compactness of the integral operator T T f(y) = K(x, y)f(x)dµ(x), U y V where U and V are say compact metric spaces, viewing T as operator on different function spaces. Viewing T as a map from L 2 (U, dµ) to L 2 (V, dν) for some measure ν (note that both spaces are separable Hilbert spaces), we know that one sufficient condition that guarantee boundedness of T is U V K 2 dµdν <. It turns out that this would also imply compactness of T. To see this, for each x we expand K(x, y) into the (countable) orthogonal basis of L 2 (Y, dν), which we may denote by φ 1, φ 2,... K(x, y) = K j (x)φ j (y) j=1 note that for almost every x X the function K(x.y) is L 2 (Y, dν) integrable in y and so K(x, y) 2 dµ(x)dν(y) = j K j (x) 2 dµ(x) thus we may approximate T with the finite rank operator T n f(y) = K j (x)f(x)dµ(x)u j (y) j n X so T is compact.
5 72 CHAPTER 9. COMPACT OPERATORS 9.3 Spectral properties of compact operators Riesz s theorem One of the main results about compact operators is the following fact: if T : X X is a compact operator on a Banach space X and 1 T is injective, then 1 T is boundedly invertible. Note that it is possible for T to be large, so the basic theory about small perturbation of 1 does not applies here. The key to showing this fact is the following theorem of Riesz. Theorem 34 (Riesz). Let T : X X be a compact operator on a Banach space X. Then range of 1 T is a closed subspace of X and furthermore dim(ker(1 T )) and codim(1 T ) are finite and equal to each other. Part of the proof of this theorem is the following lemma. Lemma 7. Let T : X X be a compact operator on a Banach space X. Then (i) ker(1 T ) is finite dimensional. (ii) There exists some k 1 such that ker((1 T ) m ) = ker((1 T ) m+1 ) for every m k. (iii) range(1 T ) is a closed subspace of X. Proof of Lemma 7. (i) If y ker(1 T ) then T y = y. Assume towards a contradiction that ker(1 T ) is infinite dimensional. Then by another result of Riesz we could find an infinite sequence (y n ) in this kernel such that y n = 1 and y n y m 1/2 for all m n. This implies the set {T y n, n 1} is not precompact, contradiction. (ii) Note that ker([1 T ] k ) ker([1 T ] k+1 ) for all k. Now, if ker([1 T ] k ) = ker([1 T ] k+1 ) then ker([1 T ] m ) = ker([1 T ] m+1 ) for all m k. Assume towards a contradiction that we have the strict inclusion ker([1 T ] k ) ker([1 T ] k+1 ) for all k 1. Since ker([1 T ] k ) are closed, by Riesz lemma we may find x n ker([1 T ] n ) such that x n = 1 and dist(x n, ker([1 T ] n 1 )) 1/2. We will show that T x n T x m 1/2 for all m n, which will contradict compactness of T. Now, without loss of generality assume that m < n, then T x n T x m = x n (1 T )x n T x m and (1 T )x n ker([1 T ] n 1 ) and T x m ker([1 T ] n 1 ) (since T commutes with (1 T ) n 1 ). Therefore by choice of x n we obtain T x n T x m dist(x n, ker([1 T ] n 1 ) 1/2 (iii) Assume that y k = (1 T )x k is a sequence in range(1 T ) that converges to some y Y, we will show that for some x X we have y = T x. Certainly if x k has a convergent subsequence we could simply take x to be the corresponding limit. Now, x k = y k + T x k so it suffices to obtain convergence of some subsequence of T x k, and using compactness of T this would follow if x k is uniformly bounded. Unfortunately it is possible for x k to be unbounded, but we could correct this by modifying x k by an appropriate term in ker(1 T ) to make it
6 9.3. SPECTRAL PROPERTIES OF COMPACT OPERATORS 73 bounded. Note that this correction does not change y k and hence does not change the goal of this part. Let d k = dist(x k, ker(1 T )) By modifying x k by an amount inside ker(1 T ) we may assume that d k x k 2d k. Thus it suffices to show that sup d k < k Assume towards a contradiction that some subsequence of d k conveges to. Without loss of generality we may assume lim d k =. We then have y k d k = (1 T )( x k d k ) now y k /d k 0 and x k /d k is uniformly bounded, so using compactness of T it follows that some subsequence of T (x k /d k ) converges, which in turn implies that some subsequence of x k /d k converges to some x X. We obtain 0 = (1 T )x so x ker(1 T ), on the other hand it is clear that dist(x k /d k, ker(1 T )) 1, contradiction. We now prove Riesz s theorem using the lemma. It remains to show that ker(1 T ) has the same dimension as the codimension of 1 T. We first reduce the proof to the case when ker(1 T ) is trivial. Let k be the index given by part (ii) of the lemma and let Y = ker[(1 T ) k+1 ] a closed subspace of X. Since (1 T )Y ker[(1 T ) k ] = Y by choice of k, it follows that T Y Y, thus T induces an operator T Z on Z := X/Y, which is a Banach space with the induced norm. It follows immediately that T Z is compact on Z, and 1 T Z is also injective on Z. Now, the dimension of ker(1 T ) is the same as the dimension of the kernel of 1 T viewing as an operator on Y, which by standard linear algebra is the same as the codimension of the range of 1 T viewing as an operator on Y. Thus it remains to show that the codimension of 1 T on Z is 0. One could see that we have reduced the proof to the case of the compact operator T Z on the Banach space Z which is also injective. Now, if (1 T Z )Z is a strict subspace of Z it follows that (1 T Z ) n Z is a strictly decreasing sequence of closed subspaces of Z, and again we may find z n (1 T Z ) n Z such that z n = 1 and dist(z n, (1 T Z ) n+1 Z) 1/2. Then for n < m we have T Z z n T Z z m = z n (1 T Z )z n T Z z m dist(z n, (1 T Z ) n+1 Z) 1/2, violating the compactness of T Z. It follows that the codimension of 1 T Z is 0 as desired Spectral properties Theorem 35. Let T be a compact operator on a Banach space X. Then (i) its spectrum σ(t ) consists of at most countably many elements, all of them are eigenvalues with finite dimensional eiegenspace. (ii) 0 is the only possible accumulation point for the elements of σ(t ) if such an accumulation point exists, and 0 will belong to σ(t ) if the dimension of X is infinite. Note that if X is a (separable) Hilbert space then we could furthermore diagonalize T using the eigenvectors.
7 74 CHAPTER 9. COMPACT OPERATORS We now prove Theorem 35. First, given any λ σ(t ) such that λ 0 we show that it is an eigenvalue with finite dimensional eigenspace. Clearly 1 T is compact. Thus, if ker(1 1 T ) λ λ is trivial then by Riesz s theorem it is also onto, therefore by the open mapping theorem 1 1 T is boundedly invertible and therefore λ σ(t ). Thus ker(1 1 T ) is nontrivial and λ λ also finite dimensional by Riesz theorem, and so λ is an eigenvalue with finite dim eigen space. Now, we will show that 0 is the only possible accumulation point, which also implies that σ(t ) is at most countable. It suffices to show that given any ɛ > 0 there is some C = C(T, ɛ) finite such that at most C(T, ɛ) elements of σ(t ) would be outside [ ɛ, ɛ]. Let λ 1,..., λ m be a finite collection of distinct elements of σ(t ) with λ j > ɛ, it suffices to show that m < O T,ɛ (1). The idea is to let Y n be the eigenspace associated with λ n and observe that for any n it holds that Y n span{y m, m < n} is trivial. Let Y <n = span{y m, m < n} which is now a strictly nested sequence. Then by Riesz s lemma one could choose y n Y n such that y n = 1 and dist(y n, Y <n ) 1/2. We will show that T y n T y m ɛ/2 for any n m. Indeed, without loss of generality assume n > m, then using the fact that y n Y n and the fact that T leaves Y m invariant we have T y n T y m = T y m = λ n y n T y m λ n dist(y n, Y <n ) λ n /2 ɛ/2 Consequently, m is bounded above by the maximum number of points in T B where B is the unit ball inside X that are ɛ/2 apart. Since T B is precompact this is finite and depends only on T and ɛ (and certainly X), and independent of the sequence (λ k ).
Compact operators on Banach spaces
Compact operators on Banach spaces Jordan Bell jordan.bell@gmail.com Department of Mathematics, University of Toronto November 12, 2017 1 Introduction In this note I prove several things about compact
More information08a. Operators on Hilbert spaces. 1. Boundedness, continuity, operator norms
(February 24, 2017) 08a. Operators on Hilbert spaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/real/notes 2016-17/08a-ops
More informationFunctional Analysis. Franck Sueur Metric spaces Definitions Completeness Compactness Separability...
Functional Analysis Franck Sueur 2018-2019 Contents 1 Metric spaces 1 1.1 Definitions........................................ 1 1.2 Completeness...................................... 3 1.3 Compactness......................................
More informationMAT 578 FUNCTIONAL ANALYSIS EXERCISES
MAT 578 FUNCTIONAL ANALYSIS EXERCISES JOHN QUIGG Exercise 1. Prove that if A is bounded in a topological vector space, then for every neighborhood V of 0 there exists c > 0 such that tv A for all t > c.
More informationAnalysis Preliminary Exam Workshop: Hilbert Spaces
Analysis Preliminary Exam Workshop: Hilbert Spaces 1. Hilbert spaces A Hilbert space H is a complete real or complex inner product space. Consider complex Hilbert spaces for definiteness. If (, ) : H H
More informationOverview of normed linear spaces
20 Chapter 2 Overview of normed linear spaces Starting from this chapter, we begin examining linear spaces with at least one extra structure (topology or geometry). We assume linearity; this is a natural
More informationCOMPACT OPERATORS. 1. Definitions
COMPACT OPERATORS. Definitions S:defi An operator M : X Y, X, Y Banach, is compact if M(B X (0, )) is relatively compact, i.e. it has compact closure. We denote { E:kk (.) K(X, Y ) = M L(X, Y ), M compact
More informationCONTENTS. 4 Hausdorff Measure Introduction The Cantor Set Rectifiable Curves Cantor Set-Like Objects...
Contents 1 Functional Analysis 1 1.1 Hilbert Spaces................................... 1 1.1.1 Spectral Theorem............................. 4 1.2 Normed Vector Spaces.............................. 7 1.2.1
More informationMath 118B Solutions. Charles Martin. March 6, d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i=1
Math 8B Solutions Charles Martin March 6, Homework Problems. Let (X i, d i ), i n, be finitely many metric spaces. Construct a metric on the product space X = X X n. Proof. Denote points in X as x = (x,
More informationNotes on Banach Algebras and Functional Calculus
Notes on Banach Algebras and Functional Calculus April 23, 2014 1 The Gelfand-Naimark theorem (proved on Feb 7) Theorem 1. If A is a commutative C -algebra and M is the maximal ideal space, of A then the
More informationFredholm Theory. April 25, 2018
Fredholm Theory April 25, 208 Roughly speaking, Fredholm theory consists of the study of operators of the form I + A where A is compact. From this point on, we will also refer to I + A as Fredholm operators.
More informationSPECTRAL THEOREM FOR COMPACT SELF-ADJOINT OPERATORS
SPECTRAL THEOREM FOR COMPACT SELF-ADJOINT OPERATORS G. RAMESH Contents Introduction 1 1. Bounded Operators 1 1.3. Examples 3 2. Compact Operators 5 2.1. Properties 6 3. The Spectral Theorem 9 3.3. Self-adjoint
More informationYour first day at work MATH 806 (Fall 2015)
Your first day at work MATH 806 (Fall 2015) 1. Let X be a set (with no particular algebraic structure). A function d : X X R is called a metric on X (and then X is called a metric space) when d satisfies
More informationPart III. 10 Topological Space Basics. Topological Spaces
Part III 10 Topological Space Basics Topological Spaces Using the metric space results above as motivation we will axiomatize the notion of being an open set to more general settings. Definition 10.1.
More informationContinuity of convex functions in normed spaces
Continuity of convex functions in normed spaces In this chapter, we consider continuity properties of real-valued convex functions defined on open convex sets in normed spaces. Recall that every infinitedimensional
More informationSpaces of continuous functions
Chapter 2 Spaces of continuous functions 2.8 Baire s Category Theorem Recall that a subset A of a metric space (X, d) is dense if for all x X there is a sequence from A converging to x. An equivalent definition
More informationProblem Set 5. 2 n k. Then a nk (x) = 1+( 1)k
Problem Set 5 1. (Folland 2.43) For x [, 1), let 1 a n (x)2 n (a n (x) = or 1) be the base-2 expansion of x. (If x is a dyadic rational, choose the expansion such that a n (x) = for large n.) Then the
More informationProblem Set 2: Solutions Math 201A: Fall 2016
Problem Set 2: s Math 201A: Fall 2016 Problem 1. (a) Prove that a closed subset of a complete metric space is complete. (b) Prove that a closed subset of a compact metric space is compact. (c) Prove that
More informationOPERATOR THEORY ON HILBERT SPACE. Class notes. John Petrovic
OPERATOR THEORY ON HILBERT SPACE Class notes John Petrovic Contents Chapter 1. Hilbert space 1 1.1. Definition and Properties 1 1.2. Orthogonality 3 1.3. Subspaces 7 1.4. Weak topology 9 Chapter 2. Operators
More informationPROBLEMS. (b) (Polarization Identity) Show that in any inner product space
1 Professor Carl Cowen Math 54600 Fall 09 PROBLEMS 1. (Geometry in Inner Product Spaces) (a) (Parallelogram Law) Show that in any inner product space x + y 2 + x y 2 = 2( x 2 + y 2 ). (b) (Polarization
More informationREAL ANALYSIS II HOMEWORK 3. Conway, Page 49
REAL ANALYSIS II HOMEWORK 3 CİHAN BAHRAN Conway, Page 49 3. Let K and k be as in Proposition 4.7 and suppose that k(x, y) k(y, x). Show that K is self-adjoint and if {µ n } are the eigenvalues of K, each
More informationMA5206 Homework 4. Group 4. April 26, ϕ 1 = 1, ϕ n (x) = 1 n 2 ϕ 1(n 2 x). = 1 and h n C 0. For any ξ ( 1 n, 2 n 2 ), n 3, h n (t) ξ t dt
MA526 Homework 4 Group 4 April 26, 26 Qn 6.2 Show that H is not bounded as a map: L L. Deduce from this that H is not bounded as a map L L. Let {ϕ n } be an approximation of the identity s.t. ϕ C, sptϕ
More informationReal Analysis Notes. Thomas Goller
Real Analysis Notes Thomas Goller September 4, 2011 Contents 1 Abstract Measure Spaces 2 1.1 Basic Definitions........................... 2 1.2 Measurable Functions........................ 2 1.3 Integration..............................
More information7 Complete metric spaces and function spaces
7 Complete metric spaces and function spaces 7.1 Completeness Let (X, d) be a metric space. Definition 7.1. A sequence (x n ) n N in X is a Cauchy sequence if for any ɛ > 0, there is N N such that n, m
More informationPrinciples of Real Analysis I Fall VII. Sequences of Functions
21-355 Principles of Real Analysis I Fall 2004 VII. Sequences of Functions In Section II, we studied sequences of real numbers. It is very useful to consider extensions of this concept. More generally,
More informationShort note on compact operators - Monday 24 th March, Sylvester Eriksson-Bique
Short note on compact operators - Monday 24 th March, 2014 Sylvester Eriksson-Bique 1 Introduction In this note I will give a short outline about the structure theory of compact operators. I restrict attention
More information(convex combination!). Use convexity of f and multiply by the common denominator to get. Interchanging the role of x and y, we obtain that f is ( 2M ε
1. Continuity of convex functions in normed spaces In this chapter, we consider continuity properties of real-valued convex functions defined on open convex sets in normed spaces. Recall that every infinitedimensional
More informationCHAPTER V DUAL SPACES
CHAPTER V DUAL SPACES DEFINITION Let (X, T ) be a (real) locally convex topological vector space. By the dual space X, or (X, T ), of X we mean the set of all continuous linear functionals on X. By the
More informationLax Solution Part 4. October 27, 2016
Lax Solution Part 4 www.mathtuition88.com October 27, 2016 Textbook: Functional Analysis by Peter D. Lax Exercises: Ch 16: Q2 4. Ch 21: Q1, 2, 9, 10. Ch 28: 1, 5, 9, 10. 1 Chapter 16 Exercise 2 Let h =
More informationTOEPLITZ OPERATORS. Toeplitz studied infinite matrices with NW-SE diagonals constant. f e C :
TOEPLITZ OPERATORS EFTON PARK 1. Introduction to Toeplitz Operators Otto Toeplitz lived from 1881-1940 in Goettingen, and it was pretty rough there, so he eventually went to Palestine and eventually contracted
More information16 1 Basic Facts from Functional Analysis and Banach Lattices
16 1 Basic Facts from Functional Analysis and Banach Lattices 1.2.3 Banach Steinhaus Theorem Another fundamental theorem of functional analysis is the Banach Steinhaus theorem, or the Uniform Boundedness
More informationFUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITE-DIMENSIONAL SPACES. 1. Compact Sets
FUNCTIONAL ANALYSIS LECTURE NOTES: COMPACT SETS AND FINITE-DIMENSIONAL SPACES CHRISTOPHER HEIL 1. Compact Sets Definition 1.1 (Compact and Totally Bounded Sets). Let X be a metric space, and let E X be
More informationFUNCTIONAL ANALYSIS-NORMED SPACE
MAT641- MSC Mathematics, MNIT Jaipur FUNCTIONAL ANALYSIS-NORMED SPACE DR. RITU AGARWAL MALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR 1. Normed space Norm generalizes the concept of length in an arbitrary
More informationFunctional Analysis HW #1
Functional Analysis HW #1 Sangchul Lee October 9, 2015 1 Solutions Solution of #1.1. Suppose that X
More informationA Spectral Characterization of Closed Range Operators 1
A Spectral Characterization of Closed Range Operators 1 M.THAMBAN NAIR (IIT Madras) 1 Closed Range Operators Operator equations of the form T x = y, where T : X Y is a linear operator between normed linear
More informationOn compact operators
On compact operators Alen Aleanderian * Abstract In this basic note, we consider some basic properties of compact operators. We also consider the spectrum of compact operators on Hilbert spaces. A basic
More informationProblem Set 5: Solutions Math 201A: Fall 2016
Problem Set 5: s Math 21A: Fall 216 Problem 1. Define f : [1, ) [1, ) by f(x) = x + 1/x. Show that f(x) f(y) < x y for all x, y [1, ) with x y, but f has no fixed point. Why doesn t this example contradict
More informationCHAPTER VIII HILBERT SPACES
CHAPTER VIII HILBERT SPACES DEFINITION Let X and Y be two complex vector spaces. A map T : X Y is called a conjugate-linear transformation if it is a reallinear transformation from X into Y, and if T (λx)
More informationSOLUTIONS TO SOME PROBLEMS
23 FUNCTIONAL ANALYSIS Spring 23 SOLUTIONS TO SOME PROBLEMS Warning:These solutions may contain errors!! PREPARED BY SULEYMAN ULUSOY PROBLEM 1. Prove that a necessary and sufficient condition that the
More informationLecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University
Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University February 7, 2007 2 Contents 1 Metric Spaces 1 1.1 Basic definitions...........................
More informationSPECTRAL THEOREM FOR SYMMETRIC OPERATORS WITH COMPACT RESOLVENT
SPECTRAL THEOREM FOR SYMMETRIC OPERATORS WITH COMPACT RESOLVENT Abstract. These are the letcure notes prepared for the workshop on Functional Analysis and Operator Algebras to be held at NIT-Karnataka,
More informationAnalysis Comprehensive Exam Questions Fall 2008
Analysis Comprehensive xam Questions Fall 28. (a) Let R be measurable with finite Lebesgue measure. Suppose that {f n } n N is a bounded sequence in L 2 () and there exists a function f such that f n (x)
More informationReal Analysis, 2nd Edition, G.B.Folland Elements of Functional Analysis
Real Analysis, 2nd Edition, G.B.Folland Chapter 5 Elements of Functional Analysis Yung-Hsiang Huang 5.1 Normed Vector Spaces 1. Note for any x, y X and a, b K, x+y x + y and by ax b y x + b a x. 2. It
More informationFunctional Analysis, Math 7321 Lecture Notes from April 04, 2017 taken by Chandi Bhandari
Functional Analysis, Math 7321 Lecture Notes from April 0, 2017 taken by Chandi Bhandari Last time:we have completed direct sum decomposition with generalized eigen space. 2. Theorem. Let X be separable
More informationA Brief Introduction to Functional Analysis
A Brief Introduction to Functional Analysis Sungwook Lee Department of Mathematics University of Southern Mississippi sunglee@usm.edu July 5, 2007 Definition 1. An algebra A is a vector space over C with
More informationIntroduction to Bases in Banach Spaces
Introduction to Bases in Banach Spaces Matt Daws June 5, 2005 Abstract We introduce the notion of Schauder bases in Banach spaces, aiming to be able to give a statement of, and make sense of, the Gowers
More informationChapter 2 Linear Transformations
Chapter 2 Linear Transformations Linear Transformations Loosely speaking, a linear transformation is a function from one vector space to another that preserves the vector space operations. Let us be more
More informationFunctional Analysis HW #5
Functional Analysis HW #5 Sangchul Lee October 29, 2015 Contents 1 Solutions........................................ 1 1 Solutions Exercise 3.4. Show that C([0, 1]) is not a Hilbert space, that is, there
More informationSPECTRAL PROPERTIES OF THE LAPLACIAN ON BOUNDED DOMAINS
SPECTRAL PROPERTIES OF THE LAPLACIAN ON BOUNDED DOMAINS TSOGTGEREL GANTUMUR Abstract. After establishing discrete spectra for a large class of elliptic operators, we present some fundamental spectral properties
More informationExercise Solutions to Functional Analysis
Exercise Solutions to Functional Analysis Note: References refer to M. Schechter, Principles of Functional Analysis Exersize that. Let φ,..., φ n be an orthonormal set in a Hilbert space H. Show n f n
More informationFive Mini-Courses on Analysis
Christopher Heil Five Mini-Courses on Analysis Metrics, Norms, Inner Products, and Topology Lebesgue Measure and Integral Operator Theory and Functional Analysis Borel and Radon Measures Topological Vector
More informationAssignment-10. (Due 11/21) Solution: Any continuous function on a compact set is uniformly continuous.
Assignment-1 (Due 11/21) 1. Consider the sequence of functions f n (x) = x n on [, 1]. (a) Show that each function f n is uniformly continuous on [, 1]. Solution: Any continuous function on a compact set
More informationMathematical foundations - linear algebra
Mathematical foundations - linear algebra Andrea Passerini passerini@disi.unitn.it Machine Learning Vector space Definition (over reals) A set X is called a vector space over IR if addition and scalar
More informationEberlein-Šmulian theorem and some of its applications
Eberlein-Šmulian theorem and some of its applications Kristina Qarri Supervisors Trond Abrahamsen Associate professor, PhD University of Agder Norway Olav Nygaard Professor, PhD University of Agder Norway
More informationFunctional analysis II Prof. Dr. Thomas Sørensen winter term 2014/15
Functional analysis II Prof. Dr. Thomas Sørensen winter term 2014/15 Marcel Schaub February 4, 2015 1 Contents 0 Motivation and repetition 3 1 Spectral Theory for compact operators 10 1.1 Spectral Theorem
More informationE.7 Alaoglu s Theorem
E.7 Alaoglu s Theorem 359 E.7 Alaoglu s Theorem If X is a normed space, then the closed unit ball in X or X is compact if and only if X is finite-dimensional (Problem A.25). Even so, Alaoglu s Theorem
More informationProblem Set 6: Solutions Math 201A: Fall a n x n,
Problem Set 6: Solutions Math 201A: Fall 2016 Problem 1. Is (x n ) n=0 a Schauder basis of C([0, 1])? No. If f(x) = a n x n, n=0 where the series converges uniformly on [0, 1], then f has a power series
More informationThe Arzelà-Ascoli Theorem
John Nachbar Washington University March 27, 2016 The Arzelà-Ascoli Theorem The Arzelà-Ascoli Theorem gives sufficient conditions for compactness in certain function spaces. Among other things, it helps
More information5 Compact linear operators
5 Compact linear operators One of the most important results of Linear Algebra is that for every selfadjoint linear map A on a finite-dimensional space, there exists a basis consisting of eigenvectors.
More informationChapter 7: Bounded Operators in Hilbert Spaces
Chapter 7: Bounded Operators in Hilbert Spaces I-Liang Chern Department of Applied Mathematics National Chiao Tung University and Department of Mathematics National Taiwan University Fall, 2013 1 / 84
More informationTopological vectorspaces
(July 25, 2011) Topological vectorspaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ Natural non-fréchet spaces Topological vector spaces Quotients and linear maps More topological
More informationBiholomorphic functions on dual of Banach Space
Biholomorphic functions on dual of Banach Space Mary Lilian Lourenço University of São Paulo - Brazil Joint work with H. Carrión and P. Galindo Conference on Non Linear Functional Analysis. Workshop on
More informationReal Variables # 10 : Hilbert Spaces II
randon ehring Real Variables # 0 : Hilbert Spaces II Exercise 20 For any sequence {f n } in H with f n = for all n, there exists f H and a subsequence {f nk } such that for all g H, one has lim (f n k,
More informationI teach myself... Hilbert spaces
I teach myself... Hilbert spaces by F.J.Sayas, for MATH 806 November 4, 2015 This document will be growing with the semester. Every in red is for you to justify. Even if we start with the basic definition
More informationContinuous Functions on Metric Spaces
Continuous Functions on Metric Spaces Math 201A, Fall 2016 1 Continuous functions Definition 1. Let (X, d X ) and (Y, d Y ) be metric spaces. A function f : X Y is continuous at a X if for every ɛ > 0
More information1 Definition and Basic Properties of Compa Operator
1 Definition and Basic Properties of Compa Operator 1.1 Let X be a infinite dimensional Banach space. Show that if A C(X ), A does not have bounded inverse. Proof. Denote the unit ball of X by B and the
More informationNotes for Functional Analysis
Notes for Functional Analysis Wang Zuoqin (typed by Xiyu Zhai) September 29, 2015 1 Lecture 09 1.1 Equicontinuity First let s recall the conception of equicontinuity for family of functions that we learned
More informationMath 209B Homework 2
Math 29B Homework 2 Edward Burkard Note: All vector spaces are over the field F = R or C 4.6. Two Compactness Theorems. 4. Point Set Topology Exercise 6 The product of countably many sequentally compact
More informationC.6 Adjoints for Operators on Hilbert Spaces
C.6 Adjoints for Operators on Hilbert Spaces 317 Additional Problems C.11. Let E R be measurable. Given 1 p and a measurable weight function w: E (0, ), the weighted L p space L p s (R) consists of all
More informationDEFINABLE OPERATORS ON HILBERT SPACES
DEFINABLE OPERATORS ON HILBERT SPACES ISAAC GOLDBRING Abstract. Let H be an infinite-dimensional (real or complex) Hilbert space, viewed as a metric structure in its natural signature. We characterize
More information11. Spectral theory For operators on finite dimensional vectors spaces, we can often find a basis of eigenvectors (which we use to diagonalize the
11. Spectral theory For operators on finite dimensional vectors spaces, we can often find a basis of eigenvectors (which we use to diagonalize the matrix). If the operator is symmetric, this is always
More information10.1. The spectrum of an operator. Lemma If A < 1 then I A is invertible with bounded inverse
10. Spectral theory For operators on finite dimensional vectors spaces, we can often find a basis of eigenvectors (which we use to diagonalize the matrix). If the operator is symmetric, this is always
More information4 Hilbert spaces. The proof of the Hilbert basis theorem is not mathematics, it is theology. Camille Jordan
The proof of the Hilbert basis theorem is not mathematics, it is theology. Camille Jordan Wir müssen wissen, wir werden wissen. David Hilbert We now continue to study a special class of Banach spaces,
More information285K Homework #1. Sangchul Lee. April 28, 2017
285K Homework #1 Sangchul Lee April 28, 2017 Problem 1. Suppose that X is a Banach space with respect to two norms: 1 and 2. Prove that if there is c (0, such that x 1 c x 2 for each x X, then there is
More informationChapter 1. Introduction
Chapter 1 Introduction Functional analysis can be seen as a natural extension of the real analysis to more general spaces. As an example we can think at the Heine - Borel theorem (closed and bounded is
More informationFunctional Analysis. Martin Brokate. 1 Normed Spaces 2. 2 Hilbert Spaces The Principle of Uniform Boundedness 32
Functional Analysis Martin Brokate Contents 1 Normed Spaces 2 2 Hilbert Spaces 2 3 The Principle of Uniform Boundedness 32 4 Extension, Reflexivity, Separation 37 5 Compact subsets of C and L p 46 6 Weak
More informationN-WEAKLY SUPERCYCLIC MATRICES
N-WEAKLY SUPERCYCLIC MATRICES NATHAN S. FELDMAN Abstract. We define an operator to n-weakly hypercyclic if it has an orbit that has a dense projection onto every n-dimensional subspace. Similarly, an operator
More information(1) Consider the space S consisting of all continuous real-valued functions on the closed interval [0, 1]. For f, g S, define
Homework, Real Analysis I, Fall, 2010. (1) Consider the space S consisting of all continuous real-valued functions on the closed interval [0, 1]. For f, g S, define ρ(f, g) = 1 0 f(x) g(x) dx. Show that
More informationBanach Spaces V: A Closer Look at the w- and the w -Topologies
BS V c Gabriel Nagy Banach Spaces V: A Closer Look at the w- and the w -Topologies Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we discuss two important, but highly non-trivial,
More informationLocally convex spaces, the hyperplane separation theorem, and the Krein-Milman theorem
56 Chapter 7 Locally convex spaces, the hyperplane separation theorem, and the Krein-Milman theorem Recall that C(X) is not a normed linear space when X is not compact. On the other hand we could use semi
More informationMATH 113 SPRING 2015
MATH 113 SPRING 2015 DIARY Effective syllabus I. Metric spaces - 6 Lectures and 2 problem sessions I.1. Definitions and examples I.2. Metric topology I.3. Complete spaces I.4. The Ascoli-Arzelà Theorem
More information************************************* Partial Differential Equations II (Math 849, Spring 2019) Baisheng Yan
************************************* Partial Differential Equations II (Math 849, Spring 2019) by Baisheng Yan Department of Mathematics Michigan State University yan@math.msu.edu Contents Chapter 1.
More informationSpectral Theory, with an Introduction to Operator Means. William L. Green
Spectral Theory, with an Introduction to Operator Means William L. Green January 30, 2008 Contents Introduction............................... 1 Hilbert Space.............................. 4 Linear Maps
More informationHilbert spaces. 1. Cauchy-Schwarz-Bunyakowsky inequality
(October 29, 2016) Hilbert spaces Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/fun/notes 2016-17/03 hsp.pdf] Hilbert spaces are
More informationFUNCTIONAL ANALYSIS LECTURE NOTES: WEAK AND WEAK* CONVERGENCE
FUNCTIONAL ANALYSIS LECTURE NOTES: WEAK AND WEAK* CONVERGENCE CHRISTOPHER HEIL 1. Weak and Weak* Convergence of Vectors Definition 1.1. Let X be a normed linear space, and let x n, x X. a. We say that
More information1 Compact and Precompact Subsets of H
Compact Sets and Compact Operators by Francis J. Narcowich November, 2014 Throughout these notes, H denotes a separable Hilbert space. We will use the notation B(H) to denote the set of bounded linear
More informationEigenvalues and Eigenfunctions of the Laplacian
The Waterloo Mathematics Review 23 Eigenvalues and Eigenfunctions of the Laplacian Mihai Nica University of Waterloo mcnica@uwaterloo.ca Abstract: The problem of determining the eigenvalues and eigenvectors
More informationDiffun2, Fredholm Operators
Diffun2, Fredholm Operators Camilla Frantzen June 8, 2012 H 1 and H 2 denote Hilbert spaces in the following. Definition 1. A Fredholm operator is an operator T B(H 1, H 2 ) such that ker T and cokert
More informationNOTES ON REPRESENTATIONS OF COMPACT GROUPS
NOTES ON REPRESENTATIONS OF COMPACT ROUPS DRAAN MILIČIĆ 1. Haar measure on compact groups 1.1. Compact groups. Let be a group. We say that is a topological group if is equipped with hausdorff topology
More informationReal Analysis Chapter 4 Solutions Jonathan Conder
2. Let x, y X and suppose that x y. Then {x} c is open in the cofinite topology and contains y but not x. The cofinite topology on X is therefore T 1. Since X is infinite it contains two distinct points
More informationCombinatorics in Banach space theory Lecture 12
Combinatorics in Banach space theory Lecture The next lemma considerably strengthens the assertion of Lemma.6(b). Lemma.9. For every Banach space X and any n N, either all the numbers n b n (X), c n (X)
More informationLECTURE 7. k=1 (, v k)u k. Moreover r
LECTURE 7 Finite rank operators Definition. T is said to be of rank r (r < ) if dim T(H) = r. The class of operators of rank r is denoted by K r and K := r K r. Theorem 1. T K r iff T K r. Proof. Let T
More informationMath Solutions to homework 5
Math 75 - Solutions to homework 5 Cédric De Groote November 9, 207 Problem (7. in the book): Let {e n } be a complete orthonormal sequence in a Hilbert space H and let λ n C for n N. Show that there is
More informationHILBERT SPACES AND THE RADON-NIKODYM THEOREM. where the bar in the first equation denotes complex conjugation. In either case, for any x V define
HILBERT SPACES AND THE RADON-NIKODYM THEOREM STEVEN P. LALLEY 1. DEFINITIONS Definition 1. A real inner product space is a real vector space V together with a symmetric, bilinear, positive-definite mapping,
More informationSPECTRAL THEORY EVAN JENKINS
SPECTRAL THEORY EVAN JENKINS Abstract. These are notes from two lectures given in MATH 27200, Basic Functional Analysis, at the University of Chicago in March 2010. The proof of the spectral theorem for
More informationCHAPTER 3. Hilbert spaces
CHAPTER 3 Hilbert spaces There are really three types of Hilbert spaces (over C). The finite dimensional ones, essentially just C n, for different integer values of n, with which you are pretty familiar,
More informationMAT 544 Problem Set 2 Solutions
MAT 544 Problem Set 2 Solutions Problems. Problem 1 A metric space is separable if it contains a dense subset which is finite or countably infinite. Prove that every totally bounded metric space X is separable.
More informationLECTURE OCTOBER, 2016
18.155 LECTURE 11 18 OCTOBER, 2016 RICHARD MELROSE Abstract. Notes before and after lecture if you have questions, ask! Read: Notes Chapter 2. Unfortunately my proof of the Closed Graph Theorem in lecture
More informationRESTRICTED UNIFORM BOUNDEDNESS IN BANACH SPACES
RESTRICTED UNIFORM BOUNDEDNESS IN BANACH SPACES OLAV NYGAARD AND MÄRT PÕLDVERE Abstract. Precise conditions for a subset A of a Banach space X are known in order that pointwise bounded on A sequences of
More informationHilbert space methods for quantum mechanics. S. Richard
Hilbert space methods for quantum mechanics S. Richard Spring Semester 2016 2 Contents 1 Hilbert space and bounded linear operators 5 1.1 Hilbert space................................ 5 1.2 Vector-valued
More information