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Lecture Notes Functional Analysis WS 2012/2013

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Lecture Notes Functional Analysis WS 2012/2013 Lecture Notes Functional Analysis WS 2012/2013 February 13, 2013 2 Contents I Normed vector spaces, Banach spaces and metric spaces5 1 Normed vector spaces and Banach spaces . .5 2 Basics of metric spaces . .9 3 Compactness in metric space . 19 4 The sequence spaces lp(N); 1 ≤ p ≤ 1 ............... 23 5 Hahn-Banach type theorems . 27 5.1 Some preparations . 27 5.2 The analytic form of Hahn-Banach: extension of linear functionals . 30 5.3 Geometric form of Hahn-Banach . 34 6 The Baire Category theorem and its applications . 39 6.1 The Baire Category theorem . 39 6.2 Application I: The set of discontinuities of a limit of con- tinuous functions . 41 6.3 Application II: Continuous but nowhere differentiable func- tions . 43 6.4 Application III: The uniform boundedness principle . 44 6.5 Application IV: The Open Mapping and the Closed Graph theorems . 46 7 Weak Topologies. Reflexive Spaces. Separable Spaces. Uniform Convexity . 49 7.1 The coarsest topology for which a collection of maps be- comes continuous . 49 7.2 The weakest topology σ(E; E∗)............... 51 7.3 Weak topology and convex sets . 55 7.4 The weak∗ topology σ(E∗;E)................ 57 7.5 Reflexive spaces . 60 7.6 Separable spaces . 64 7.7 Uniformly convex spaces . 68 8 Lp spaces . 71 8.1 Some results from integration everyone must know . 71 8.2 Definition and some properties of Lp spaces . 73 8.3 Reflexivity, Separability. The Dual of Lp .......... 77 9 Hilbert spaces . 87 9.1 Some elementary properties . 87 9.2 The dual space of a Hilbert space . 94 9.3 The Theorems of Stampacchia and Lax-Milgram . 96 3 4 CONTENTS Chapter I Normed vector spaces, Banach spaces and metric spaces 1 Normed vector spaces and Banach spaces In the following let X be a linear space (vector space) over the field F 2 fR; Cg. Definition 1.1. A seminorm on X is a map p : X ! R+ = [0; 1) s.t. (a) p(αx) = jαjp(x) 8α 2 F; 8x 2 X (homogeneity). (b) p(x + y) ≤ p(x) + p(y) 8x; y 2 X (triangle inequality). If, in addition, one has (c) p(x) = 0 ) x = 0 then p is called a norm. Usually one writes p(x) = kxk, p = k · k. The pair (X; k · k) is called a normed (vector) space. Remark 1.2. • If k · k is a seminorm on X then kxk − kyk ≤ kx − yk 8x; y 2 X (reverse triangle inequality): Proof. kxk = kx − y + yk ≤ kx − yk + kyk ) kxk − kyk ≤ kx − yk Now swap x&y: kxk − kyk ≤ ky − xk = k(−1)(x − y)k = kx − yk. Hence kxk − kyk = max(kxk − kyk; kyk − kxk) ≤ kx − yk: 5 6 CHAPTER I. NVS, BS AND MS •k ~0k = k0:~0k = j0j:k~0k = 0. Interpret kx − yk as distance between x and y. Definition 1.3. Let (xn)n2N = (xn)n be a sequence in a normed vector space X ((X; k · k)). Then (xn)n converges to a limit x 2 X if 8" > 09N" 2 N s.t. 8n ≥ N" it holds kxn − xk < " (or kxn − xk ≤ "). One writes xn ! x or limn!1 xn = x. (xn)n is a Cauchy sequence if 8" > 09N" 2 N s.t. 8n; m ≥ N" it holds kxn − xmk < " (or kxn − xmk ≤ "). (X; k · k) is complete if every Cauchy sequence converges. A complete normed space (X; k · k) is called a Banach space. Remark 1.4. Let X be a normed vector space, (xn)n a sequence in X. (a) If xn ! x in X, then (xn)n is a Cauchy sequence. " Proof. Given " > 09N" : 8n ≥ N" : kx − xnk < 2 . Hence for n; m ≥ N" we have kxn − xmk = kxn − x + x − xmk ≤ kxn − xk + kx − xmk < ": (b) Limits are unique! If xn ! x in X and xn ! y in X, then x = y. Proof. kx − yk = kx − xn + xn − yk ≤ kxn − xk + kxn − yk ! 0 + 0 = 0 as n ! 1: (c) If (xn)n converges or is Cauchy, then it is bounded, i.e. sup kxnk < 1: n2N Proof. Take " = 1. Then there exists N 2 N s.t. 8n; m ≥ N : kxn − xmk < 1. In particular, 8n ≥ N : kxn − xN k < 1: ) kxnk = kxn − xN + xN k ≤ kxn − xN k + kxN k < 1 + kxN k (8n 2 N) ) 8n 2 N : kxnk ≤ max(kx1k; kx2k;:::; kxN k; 1 + kxN k) < 1: 1. NORMED VECTOR SPACES AND BANACH SPACES 7 Let X be a normed vector space, S 6= ; a set. For functions f; g : S ! X, α; β 2 F define ( S ! X f + g : s 7! (f + g)(s) = f(s) + g(s) ( S ! X αf : s 7! (αf)(s) = αf(s) So the set of functions from S to X is a normed space itself! In case X = R, we write f ≥ α (or f > α) if f(s) ≥ α for all s 2 S (f(s) > α for all s 2 S). Similarly one defines α ≤ f ≤ β, f ≤ g, etc. Example 1.5. (a) X = Fd; d 2 N is a Banach space (or short B-space) with respect to (w.r.t) the norms n 1 X p p jxjp := jxjj ; 1 ≤ p < 1; j=1 jxj1 := max jxjj: j=1;:::;d d Here x = (x1; x2; : : : ; xd) 2 F . (b) Let Ω 6= ;, L1(Ω) = L1(Ω; R) = the set of all real-valued functions on Ω which are bounded, i.e. 1 f 2 L (Ω) then 9Mf < 1 : jf(!)j ≤ Mf for all ! 2 Ω: 1 1 Norm on L (Ω): for f 2 L (Ω) : kfk1 = sup!2Ω jf(!)j (check that this is a norm!). 1 Claim: (L (Ω); k · k1) is a Banach space. Proof. Normed vector space is clear. 1 Take (fn) a Cauchy sequence in L (Ω) w.r.t. k · k1. We have: 8" > 09N : 8n; m ≥ N : kfn − fmk1 < ". Fix ! 2 Ω, then fn(!) is a Cauchy sequence in R since jfn(!) − fm(!)j ≤ sup jfn(!) − fm(!)j = kfn − fmk1 < " 8n; m ≥ N: !2Ω Since R is complete, f(!) := limn!1 fn(!) exists (this f is the candidate for the limit). We have jf(!)j ≤ jf(!) − fn(!)j + jfn(!)j = lim jfm(!) − fn(!)j + jfn(!)j ≤ " + jfn(!)j m!1 | {z } <1 ) sup!2Ωjf(!)j ≤ 1; i.e., f 2 L1(Ω). Take " > 0. Then jfn(!) − f(!)j = lim jfn(!) − fm(!)j ≤ " if n ≥ N m!1 | {z } ≤" if n;m≥N 8 CHAPTER I. NVS, BS AND MS ) 8n ≥ N : kfn − fk1 ≤ "; i.e., fn ! f w.r.t. k · k1: 1 R (c) X = C([0; 1]); kfk1 := jf(t)jdt is a norm, C([0; 1]); k·k1 is not complete. 0 Proof. kfk1 ≥ 0 8f 2 C([0; 1]): 1 Z kf + gk1 = jf(t) + g(t)j dt ≤ kfk1 + kgk1 | {z } 0 ≤|f(t)j+jg(t)j 1 Z kαfk1 = jαf(t)jdt = jαjkfk1 0 So k · k is a seminorm. If f 6≡ 0 and f is continuous, we see that there exist an interval I ⊂ [0; 1], δ > 0 such that jf(t)j ≥ δ 8t 2 I. 1 Z Z ) kfk1 = jf(t)jdt ≥ jf(t)j dt ≥ δ:lehgth of I > 0: | {z } 0 I ≥δ So k · k1 is a seminorm. Now take a special sequence 8 1 1 >0; if 0 ≤ t ≤ 2 − n < n 1 1 1 fn(t) := nt − 2 + 1; if 2 − n < t < 2 (n ≥ 3) :> 1 1; if t ≥ 2 For m ≥ n ≥ 3: 1 n Z 1 kf − f k = jf (t) − f (t)jdt ≤ ! 0 as n ! 1; n m 1 n m n 1 1 2 − n so (fn) is a Cauchy sequence. 1 1 1 Assume that fn ! f 2 C([0; 1]). Fix α 2 0; 2 ; n : 2 − n ≥ α α α Z Z 0 ≤ jf(t)jdt = jfn(t) − f(t)jdt 0 0 1 Z ≤ jfn(t) − f(t)jdt = kfn − fk1 ! 0: 0 1 Hence f(t) = 0 for all 0 ≤ t ≤ α, all 0 ≤ α < 2 1 f(t) = 0 for all0 ≤ t < : 2 2. BASICS OF METRIC SPACES 9 On the other hand 1 1 Z Z 0 ≤ jf(t) − 1jdt = jf(t) − fn(t)jdt ≤ kf − fnk1 ! 0 as n ! 1 1 1 2 2 1 Since f is continuous on [0; 1], it follows that f(t) = 1 for 2 ≤ t ≤ 1. 1 So f cannot be continuous at t = 2 . A contradiction. 2 Basics of metric spaces Definition 2.1. Given a set M 6= ;, a metric (or distance) d on M is a function d : M × M ! R such that (a) d(x; y) ≥ 0 8x; y 2 M and d(x; y) = 0 () x = y. (b) d(x; y) = d(y; x) 8x; y 2 M (symmetry). (c) d(x; y) ≤ d(x; z) + d(z; y) 8x; y; z 2 M (triangle inequality). The pair (M; d) is called a metric space. We often simply write M if it is clear what d is. A sequence (xn)n in a metric space (M; d) converges to x 2 M if 8" > 09N" 2 N8n ≥ N" : d(x; xn) < " (or ≤ "). One writes lim xn = x or xn ! x. One always has d(x; z) − d(z; y) ≤ d(x; y) Hint for the proof: d(x; z) ≤ d(x; y) + d(y; z) and think and use symmetry. Example 2.2. • R with d(x; y) = jx − yj; • Any normed vector space (X; k · k) with d(x; y) = kx − yk; d 1 d d P 2 2 • Eucledian space R (or C ) with d2(x; y) = jxj −yjj or dp(x; y) = j=1 d 1 P p p jxj − yjj , or d1(x; y) = maxj=1;:::d jxj − yjj.
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