Functional Analysis – Lecture script Prof. D. Salamon February 12, 2007 Coordinating: Lukas Lewark Writing: Urs Fuchs, SaˇsaParad-, Andrin Schmidt Correcting: Philipp Arbenz, David Umbricht, Dominik Staub, Thomas Rast If you want to be informed in case of a new version or if you find any mistakes please write to [email protected]. Warning: We are sure there are lots of mistakes in these notes. Use at your own risk! Corrections would be appreciated and can be sent to [email protected]; please always state what version (look in the Id line below) you found the error in. For further information see: http://vmp.ethz.ch/wiki/index.php/Vorlesungsmitschriften $Id: fa.tex 1894 2007-02-12 15:05:18 charon$ i Contents 0 Introduction 1 1 Basic Notions 2 1.1 Finite dimensional vector spaces . 2 1.2 LinearOperators ........................... 5 1.3 Infinite dimensional vector spaces . 6 1.4 The Theorem of Arzela-Ascoli . 8 1.5 The Baire Category Theorem . 12 1.6 Dualspaces.............................. 15 1.7 Quotient spaces . 21 2 Functional Analysis 22 2.1 Basics ................................. 22 2.2 Product spaces . 25 2.3 Extension of bounded linear functionals . 29 2.4 Reflexive Banach Spaces . 34 3 The weak and weak* topologies 38 3.1 Theweaktopology .......................... 38 3.2 Theweak*topology ......................... 40 3.3 Ergodicmeasures........................... 50 4 Compact operators and Fredholm theory 56 4.1 Compactoperators.......................... 56 4.2 Fredholmoperators.......................... 60 5 Spectral Theory 65 5.1 Eigenvectors.............................. 65 5.2 Integrals................................ 66 5.3 Compact operators on Banach spaces . 73 5.4 Spectral Measure . 78 6 Unbounded operators 84 Index 89 ii 0 Introduction 25.10.2006 0 Introduction Remark: Functional Analysis can be viewed as a combination of linear alge- bra and topology: Linear Algebra Topology Functional Analysis vector spaces metric spaces normed vector spaces linear maps continuous maps continuous linear maps subspaces closed subsets closed subspaces The vector spaces concerned in Functional Analysis generally have infinite di- mension. 1 1 Basic Notions 25.10.2006 1 Basic Notions 1.1 Finite dimensional vector spaces Definition: A normed vector space is a pair (X, k · k) where X is a vector space (we consider only real spaces) and X → [0, ∞),x 7→ kxk is a norm, i.e. 1. kxk = 0 ⇐⇒ x = 0 2. kλxk = |λ| · kxk ∀x ∈ X,λ ∈ R 3. kx + yk ≤ kxk + kyk ∀x,y ∈ X Remark: A norm induces a metric on the vector space, by d(x,y) := kx − yk. Definition: A Banach space is a complete normed vector space (X, k.k), i.e. every Cauchy sequence in (X, d) converges. Definition: Two norms k · k1, k · k2 on a real vector space X are called equiv- alent, if 1 ∃c > 0 ∀x ∈ X : kxk ≤ kxk ≤ ckxk . c 1 2 1 Example: n n 1. X = R ,x = (x1,...,xn) ∈ R 1 n p p kxkp := |xi| , 1 < p < ∞ i=1 ! X kxk∞ := max{|xi| | 1 ≤ i ≤ n} 2. (M, A, µ) measure space p p L (µ) = {f : M → R | f measurable , |f| dµ < ∞}/∼ ZM where ∼ means equal almost everywhere. 1 p p kfkp := |f| dµ , 1 ≤ p < ∞ ZM p (L (µ), k · kp) is a Banach space, and if M = {1,...,n} we get Example 1. 3. Let M be a locally compact and hausdorff topologic space. Cc(M) := {f : M → R | f is continuous and has compact support} kfk∞ := sup {|f(m)|} m∈M Combine 2. and 3.: Let B ⊂ 2M be the Borel σ-Algebra and µ : B → [0, ∞] a Radon measure. Then one can define kfkp, kfk∞ for all f ∈ Cc(M). These two norms are not equivalent, because there are Cauchy sequences converging in k · k∞ which are not convergent in k · kp, e.g. 1 1 p x ∈ [0, n] fn : R → R, fn = n 0 otherwise ( 2 1 Basic Notions 25.10.2006 4. k n n k X = Cb (R ) = {f : R → R | f ∈ C and α n sup |∂ f(x)| < ∞∀α = (α1,...αn) ∈ N } x∈Rn α α kfkck := sup sup |∂ f(x)| = sup k∂ fk∞ |α|≤k x∈Rn |α|≤k with |α| := α1 + . + αn. k n The normed vector space (Cb (R ), k.kck ) is called Sobolev space. Lemma 1: Let X be a finite dimensional vector space. ⇒ Any two norms on X are equivalent. Proof: w.l.o.g. X = Rn Let e1,...,en ∈ X be the standard basis of X. Let Rn → R : x 7→ kxk be any norm and n c := ke k2 v i ui=1 uX t n ⇒ kxk = k xieik (1) i=1 nX ≤ kxieik (2) i=1 Xn = |xi|keik by Cauchy-Schwarz (3) i=1 X n n ≤ x2 ke k2 (4) v i v i ui=1 ui=1 uX uX = ctkxk2 t (5) That proves one half of the inequality. It follows that the function Rn → R : x 7→ kxk is continuous with respect to the Euclidian norm on Rn: |kxk − kyk| ≤ kx − yk ≤ ckx − yk2 n n The set S := {x ∈ R | kxk2 = 1} is compact with respect to the Euclidian norm. n ⇒ ∃x0 ∈ S∀x ∈ S : kxk ≥ kx0k =: δ > 0 x ⇒ ∀x ∈ Rn : ∈ Sn kxk2 and so x ≥ δ kxk 2 and therefore kxk ≥ δkxk2 Which is the other half of the inequality. 2 Lemma 2: Every finite dimensional vector space (X, k · k) is complete. n Proof: True for (R , k · k2). ⇒ true for Rn with any norm. ⇒ true for any finite dimensional vector space. 2 3 1 Basic Notions 25.10.2006 Lemma 3: Let (X, k · k) be any normed vector space and Y ⊂ X a finite dimensional linear subspace. ⇒ Y is a closed subset of X. Proof: Y is a finite dimensional normed vector space. Lemma⇒ 2 Y is complete. (yn)n∈N ⊂ Y, lim yn = y ∈ X n→∞ Y complete ⇒ y ∈ Y ⇒ Y is closed. 2 Theorem 1: Let (X, k.k) be a normed vector space and B := {x ∈ X | kxk ≤ 1} be the unit ball. Then dim(X) < ∞ ⇐⇒ B is compact. Proof of Theorem 1, “⇒”: Let e1,...,en be a basis of X and define n n T : R → X by T ξ := i=1 ξiei ⇒ The function Rn → R : ξ 7→ kT ξk is a norm on Rn P Lemma 1 n ⇒ ∃c > 0 ∀ξ ∈ R : max |ξi| ≤ ckT ξk i=1,...,n ν ν ν ν −1 ν Let (x )ν∈N ∈ B be any sequence and denote ξ = (ξ1 ,...,ξn) := T x ν ν ν ⇒ |ξi | ≤ ckT ξ k = ckx k ≤ c Heine-Borel ν νk ⇒ (ξ )ν∈N has a convergent subsequence (ξ )k∈R,ν1<ν2<... νk ⇒ (ξi )k∈N converges in R for i = 1,...,n νk vk νk ⇒ x = ξ1 e1 + . + ξn en converges; so B is sequentially compact. We use that on metric spaces sequential compactness and compactness defined by existence of finite subcoverings are equivalent; that will be proven in Theorem 2. 2 Lemma 4: 0 < δ < 1, (X, k · k) a normed vector space, Y ( X a closed subspace. ⇒ ∃x ∈ X so that kxk = 1, inf kx − yk > 1 − δ y∈Y Proof: Let x0 ∈ X \ Y . Denote d := inf kx0 − yk > 0 y∈Y d (d > 0 because Y is closed.) ∃y0 ∈ Y so that kx0 − y0k < 1−δ Let x := x0−y0 ⇒ kxk = 1 kx0−y0k x − y kx − yk = 0 0 − y = kx − y k 0 0 1 d kx0 − y0 − kx 0 − y0kyk ≥ > 1 − δ kx0 − y0k kx0 − y0k ∈Y | ≥d {z } 2 | {z } 4 1 Basic Notions 26.10.2006 Proof of Theorem 1, “⇐”: Suppose dim(X) = ∞ We construct a sequence x1,x2 . in B so that 1 kx − x k ≥ ∀i 6= j i j 2 Then (xi)i∈N has no convergent subsequence. 1 We construct by induction sets {x1,...,xn} ⊂ B so that kxi − xj k ≥ 2 ∀i 6= j n = 1: pick any vector x ∈ B. n ≥ 1: Suppose x1,...,xn have been constructed. Define n Y := span{x1,...,xn} = λixi λi ∈ R ( X ( ) i1 X ⇒ Y is closed. So by Lemma 4 ∃xn+1 ∈ X so that 1 kx k = 1, kx − yk ≥ ∀y ∈ Y n+1 n+1 2 1 ⇒ kx − x k ≥ ∀i = 1,...,n n+1 i 2 This completes the inductive construction of the sequence. 2 1.2 Linear Operators (X, k · kX ), (Y, k · kY ) normed vector spaces. Definition: A linear operator T : X → Y is called bounded if ∃c > 0∀x ∈ X : kTxkY ≤ ckxkX The number kT k := sup kT xkY is called the norm of T . x∈X,x6=0 kxkX Notation : L(X,Y ) := {T : X → Y | T is a bounded linear operator} is a normed vector space, and complete whenever Y is complete. (Analysis II) Lemma 5: T : X → Y linear operator. Equivalent are i. T is bounded ii. T is continuous iii. T is continuous at 0. Proof: i. ⇒ ii. kTx − TykY ≤ kT kkx − ykX ⇒ Lipschitz continuous ii. ⇒ iii. trivial iii. ⇒ i. ε = 1 ⇒ ∃δ > 0 ∀x ∈ X : kxkX ≤ δ ⇒ kTxkY ≤ 1 δx 0 6= x ∈ X ⇒ = δ kxk X X δTx ⇒ ≤ 1 kxk X Y 1 ⇒ kTx k ≤ kxk Y δ X c 2 |{z} 5 1 Basic Notions 26.10.2006 Lemma 6: Let X,Y be normed vector spaces of finite dimension ⇒ every linear operator T : X → Y is bounded. Proof: Choose a basis e1,...en ∈ X. a) n c1 := kT e1kY i=1 X b) The map n n R → R : (x1,...xn) → xiei i=1 X X n is a norm on R . By Lemma 1, ∃c2 > 0 so that n n max |xi| ≤ c2 xiei ∀(x1,...xn) ∈ R i=1,...n i=1 X X n a)&b) ⇒ ∀x = i=1 xiei ∈ X we have P n n kTxkY = xiT ei ≤ |xi| · kT eikY i=1 Y i=1 X X n ≤ (max |xi|) · kT eikY = ci max |xi| < c1c2kxkX i i=1 X 2 What for infinite dimensions? 1.3 Infinite dimensional vector spaces 1 Example 1: Let X = C ([0, 1]; X), kxkX := sup0≤t≤1 |f(t)|, Y = R and Tx :=x ˙(0).