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Functional Analysis TMA401/MMA400 at Chalmers University of Technology and University of Gothenburg

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Functional Analysis TMA401/MMA400 at Chalmers University of Technology and University of Gothenburg Functional Analysis TMA401/MMA400 at Chalmers University of Technology and University of Gothenburg A Companion to the textbook Peter Kumlin 2019 ii iii Preface During the last years I have given a course in Functional Analysis for the master programs at Chalmers University of Technology and University of Gothenburg based on the textbook Hilbert Spaces with Applications, 3rd ed. by L.Debnath and P.Mikusinski. Pre- viously the book Applied Functional Analysis by D.H.Griffel was used and there is a vast literature on the subject. One nice recent book is Functional Analysis - Entering Hilbert Space by Vagn Lundsgaard Hansen. It should be noted that a (the) graduate course in functional analysis given at the department is based on Real Analysis: Modern Techniques and their Applications by G.B.Folland. The overlap between the two courses is substantial. These notes serve as a complement to the textbook [2]. The first chapter is an introduction to the course. Here we start out with a differential equation, reformulate it as an integral equation and prove the existence/uniqueness of solutions from scratch. The purpose is to motivate the different notions that will be discussed during the course. The second chapter treats fixed point theory more ambitiously than that in [2]. Although the results hold for complete metric spaces we formulate the theorems for Banach spaces. Chapter three gives a short survey on Lebesgue integration, based on measure theory in the spirit of Follands book. The fourth chapter is devoted to spectral theory and finally chapter five treats boundary value problems in one dimension. A description, a course diary, for the course given the first study period during fall 2018 makes up chapter six and the final chapter is a collection of exercises. A reference list is found at the end of the document. Exercises and Theorems in this text refer to exercises and theorems in this document. These notes coincides with the material [K] on the course homepage. Gothenburg May 2019 Peter Kumlin iv Contents 1 Introduction 1 2 Fixed point theory 11 2.1 Introduction . 11 2.2 Banach’s fixed point theorem . 13 2.3 Brouwer and Schauder fixed point theorems . 20 2.4 Continuity and applications . 24 2.5 Some more fixed point theorems . 29 3 Lp-spaces 33 3.1 Introduction . 33 3.2 Lebesgue measure on Rn ............................ 36 3.3 Lebesgue measurable functions . 38 3.4 Integrals and convergence theorems . 40 3.5 Lp–spaces, Hölder’s and Young’s inequalities . 44 4 Spectral theory 47 4.1 Introduction . 47 5 Ordinary differential equations 61 5.1 Introduction . 61 5.2 Existence of Green’s functions . 61 v vi CONTENTS 5.3 Spectral theory for ordinary differential equations . 68 6 What happened the year 2018/2019? A course diary 73 7 Exercises 89 7.1 Vector spaces . 89 7.2 Normed spaces . 93 7.3 Banach spaces . 98 7.4 Fixed point techniques . 112 7.5 Hilbert spaces . 124 7.6 Linear operators on Hilbert spaces . 131 7.7 Ordinary differential equations . 150 7.8 Calculus of variation . 158 Chapter 1 Introduction Functional analysis is one of the major fields of mathematics from the twentieth century. Early pioneers were Fredholm (a swede) and Volterra (an italian). Important mathemati- cian in the development were Hilbert (a german) and Riesz (a hungarian). And most importantly Stefan Banach (a pole) whose book Théorie des opérations linéares from 1932 is a cornerstone. Many more names could and should be mentioned but the ones given above will appear in the course. As a motivating example we discuss a problem from differential equations. Looking at the problem in retrospect we can easily identify many of the different notions that appear in course and also get a sense for the corresponding definitions as being "natural". So now let us roll up our sleeves and start with the calculations. Consider the differential equation f 00(x) + f(x) = g(x) (1.1) in the interval 0 ≤ x ≤ 1 with the solution satisfying the boundary conditions f(0) = 1; f 0(0) = 0: The problem itself is of no special interest and will only serve as a testing ground for our ideas. Here we first think of g(x) as a given continuous function on the interval x 2 [0; 1]. If g = 0 we know from our first calculus class that f(x) = A cos x + B sin x is a solution to the differential equation, where A; B are arbitrary constants. To treat the case with an arbitrary function g(x) we apply the method of variations of constants. If you are not familiar with the method it does not matter since this is the only time when we will use it. Set f(x) = A(x) cos x + B(x) sin x (1.2) 1 2 CHAPTER 1. INTRODUCTION and differentiate. We get f 0(x) = A0(x) cos x + B0(x) sin x − A(x) sin x + B(x) cos x: Assume (and this is part of the method) that A0(x) cos x + B0(x) sin x = 0; x 2 [0; 1]: Differentiate once more. This gives f 00(x) = −A(x) cos x − B(x) sin x − A0(x) sin x + B0(x) cos x: Hence (1:2) satisfies (1:1) if −A0(x) sin x + B0(x) cos x = g(x); x 2 [0; 1]: We now solve 8 0 0 < A (x) cos x + B (x) sin x = 0 : −A0(x) sin x + B0(x) cos x = g(x) This together with the boundary conditions give us 8 0 > A (x) = −g(x) sin x > > > 0 <> B (x) = g(x) cos x > > A(0) = 1 (= f(0)) > > :> B(0) = 0 (= f 0(0)) We conclude that 8 A(x) = A(0) + R x A0(t) dt = 1 + R x(−g(t) sin t) dt < 0 0 : R x 0 R x B(x) = B(0) + 0 B (t) dt = 0 g(t) cos t dt which finally implies that Z x f(x) = cos x + sin(x − t)g(t) dt: (1.3) 0 You can easily check that this function f(x) satisfies the differential equation and the imposed boundary conditions. (1:3) is a reformulation of the differential equation with the boundary conditions. To push things further consider the case with g(x) = k(x)f(x); x 2 [0; 1]. Here k is assumed to be a known continuous function on [0; 1]. The solution formula above implies Z x f(x) = cos x + sin(x − t)k(t)f(t) dt: (1.4) 0 3 (1:4) is a reformulation of the differential equation with the boundary conditions, i.e. the boundary value problem, as an integral equation. Note that now the function f appears on both sides. Here comes a main idea. Pick any f0(x) 2 C([0; 1]). C([0; 1]) denotes the set of all continuous functions on [0; 1]. Set 8 R x f1(x) = cos x + sin(x − t)k(t)f0(t) dt > 0 > < R x f2(x) = cos x + 0 sin(x − t)k(t)f1(t) dt > > :> ::: i.e. Z x fn(x) = cos x + sin(x − t)k(t)fn−1(t) dt n = 1; 2; 3;::: 0 Observe that all fn: s are continuous functions on I. To simplify notations set u(x) = cos x and Z x Kv(x) = sin(x − t)k(t)v(t) dt; v 2 C([0; 1]); x 2 [0; 1]: 0 Then equation (1:4) takes the form f = u + Kf: 1 Consider the sequence (fn)n=0 where fn = u + Kfn−1; n = 1; 2; 3;::: Hope: fn "tends to" a continuous function f and Kfn "tends to" Kf as n ! 1. Here we have to make "tends to", which we denote by !, precise. The wishful thinking is illustrated by the diagram fn = u + Kfn−1 ## (1.5) f = u + Kf The limit function f will be a solution to our problem. To proceed we recall some basic facts from first year calculus courses. 1 Definition 1.0.1. We say that a sequence (vn)n=1 of continuous functions on I = [0; 1] converges uniformly on I if max jvn(x) − vm(x)j ! 0; as n; m ! 1 (1.6) x2I i.e. if for all > 0 there exists N such that max jvn(x) − vm(x)j < , for all n; m ≥ N: x2I 4 CHAPTER 1. INTRODUCTION 1 Lemma 1.0.1. Suppose that (vn)n=1 converges uniformly on I. Then there exists a con- tinuous function v on I such that max jvn(x) − v(x)j ! 0; as n; m ! 1: (1.7) x2I Moreover we set khk = maxx2I jh(x)j for every h 2 C(I). Then (1:6) and (1:7) can be written as kvn − vmk ! 0; as n; m ! 1 and kv − vnk ! 0; as n ! 1 Back to our problem above: The question is now whether kfn − fmk ! 0; as n; m ! 1 or not? Does this depend on the choice of f0? To settle that question note that with notations from above K(v + w) = Kv + Kw for all v; w 2 C(I); where v + w is the (continuous!) function that is defined by (v + w)(x) = v(x) + w(x) for x 2 I (and also K(v + w) continuous). Moreover we set Knv = K(Kn−1v); for all v 2 C(I) and n = 2; 3; 4;::: Then we get f1 = u + Kf0 2 f2 = u + Kf1 = u + K(u + Kf0) = u + Ku + K f0 ::: 2 n−1 n fn = u + Ku + K u + ::: + K u + K f0 Assume n > m.
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