Lecture Notes Functional Analysis (2014/15)

Lecture Notes Functional Analysis (2014/15)

Lecture Notes Functional Analysis (2014/15) Roland Schnaubelt These lecture notes are based on my course from winter semester 2014/15. Com- pared to the notes from three years ago, several details and very few subjects have been changed. I mostly kept the contents of the results discussed in the lectures, but the numbering has been shifted in some parts. Typically, the proofs and calculations in the notes are a bit shorter than those given in class. Moreover, the drawings and many additional, mostly oral remarks from the lectures are omitted here. On the other hand, in the notes I have added a few results (e.g., the Riesz{Thorin theorem) and a couple of proofs for peripheral statements not presented during the course. I am happy to thank Bernhard Konrad, J¨orgB¨auerle and Johannes Eilinghoff for their careful proof reading of my lecture notes from 2009 and 2011. Karlsruhe, December 30, 2019 Roland Schnaubelt Contents 1 Banach spaces3 1.1 Basic properties of Banach and metric spaces . .3 1.2 More examples of Banach spaces . 13 1.3 Compactness and separability . 20 2 Continuous linear operators 27 2.1 Basic properties and examples . 27 2.2 Standard constructions . 34 2.3 The interpolation theorem of Riesz and Thorin . 39 3 Two main theorems on linear operators 43 3.1 The principle of uniform boundedness . 43 3.2 The open mapping theorem and invertibility . 48 4 Duality 51 4.1 Hilbert spaces . 51 4.2 The duals of sequence and Lebesgue spaces . 54 4.3 The theorem of Hahn-Banach . 58 4.4 Reflexivity and weak convergence . 65 4.5 Adjoint operators . 70 5 The Fourier transform and generalized derivatives 75 5.1 The Fourier transform . 75 5.2 Sobolev spaces . 82 5.3 Tempered distributions . 88 Bibliography 92 2 Chapter 1 Banach spaces 1.1 Basic properties of Banach and metric spaces Throughout, X 6= f0g and Y 6= f0g are vector spaces over the field F 2 fR; Cg. Definition 1.1. A seminorm on X is a mapping p : X ! R+ satisfying a) p(αx) = jαj p(x) (homogeneity), b) p(x + y) ≤ p(x) + p(y) (triangle inquality) for all x; y 2 X and α 2 F. If in addition c) p(x) = 0 =) x = 0 (definiteness) holds for all x 2 X, then p is called a norm. One usually writes p(x) = kxk and p = k · k. The pair (X; k · k) (or just X) is called a normed vector space. In view of Example 1.4(a), we interpret kxk as the length of x and kx − yk as the distance between x and y. Definition 1.2. Let k·k be a seminorm on a vector space X. A sequence (xn)n2N = (xn) in X converges to a limit x 2 X if 8 " > 0 9 N" 2 N 8 n ≥ N" : kxn − xk ≤ ": We then write xn ! x as n ! 1 or x = limn!1 xn. Moreover, (xn) is a Cauchy sequence in X if 8 " > 0 9 N" 2 N 8 n; m ≥ N" : kxn − xmk ≤ ": A normed vector space (X; k · k) is called a Banach space if each Cauchy sequence in (X; k · k) converges in X. Then one also calls k · k or (X; k · k) complete. Remark 1.3. Let k · k be a seminorm on a vector space X and (xn) be a sequence in X. The following results are shown as in Analysis 2, see also Remark 1.7. a) jkxk − kykj ≤ kx − yk for all x; y 2 X. b) The vector 0 has the norm 0. c) If (xn) converges, then it is a Cauchy sequence. d) If (x ) converges or is Cauchy, then it is bounded, i.e., sup kx k < 1: n n2N n e) Limits are linear: If xn ! x and yn ! y in X as n ! 1 and α; β 2 F, then αxn + βyn converges to αx + βy in X. f) Limits are unique in the norm case: Let k · k be a norm. If xn ! x and xn ! y in X as n ! 1 for some x; y 2 X, then x = y. ♦ 3 4 Chapter 1. Banach spaces Let X be a vector space and M 6= ; be a set. For functions f; g : M ! X and numbers α; β 2 F one defines the functions f + g : M ! X;(f + g)(t) = f(t) + g(t) and αf : M ! X;(αf)(t) = αf(t): It is easily seen that the set ff : M ! Xg becomes a vector space if endowed with these operations. Function spaces are always equipped with this sum and scalar multiplication. Let X = F. Here one further puts fg : M ! F;(fg)(t) = f(t)g(t): Let α; β 2 R. We then write f ≥ α (f > α, respectively) if f(t) ≥ α (f(t) > α, respectively) for all t 2 M. Similarly one defines α ≤ f ≤ β, f ≤ g, and so on. Example 1.4. a) X = Fd is a Banach space with respect to the norms d 1 X p p jxjp = jxkj ; 1 ≤ p < 1; and jxj1 = maxfjxkj j k = 1; : : : ; dg k=1 d where x = (x1; : : : ; xd) 2 F , see Analysis 1+2 or the proof of Proposition 1.30. b) X = C([0; 1]) = ff : [0; 1] ! F j f is continuousg with the supremum norm kfk1 := sup jf(t)j (= max jf(t)j): t2[0;1] t2[0;1] is a Banach space. We always endow X with this norm, unless something else is specified. In fact, it is clear that X is a vector space and that kfk1 = 0 implies f = 0. We further compute kαfk1 = sup jαj jf(t)j = jαj sup jf(t)j = jαj kfk1; t2[0;1] t2[0;1] kf + gk1 = sup jf(t) + g(t)j ≤ sup (jf(t)j + jg(t)j) ≤ kfk1 + kgk1; t2[0;1] t2[0;1] for all f; g 2 X and α 2 F, so that X is a normed vector space. Let (fn) be a Cauchy sequence in X; i.e., for all " > 0 there is an index N" 2 N with jfn(t) − fm(t)j ≤ kfn − fmk1 ≤ " for all n; m ≥ N" and t 2 [0; 1]. We stress that N" does not depend on t. By this estimate, (fn(t))n2N is a Cauchy sequence in F. Since F is complete, there exists f(t) := limn!1 fn(t) in F for each t 2 [0; 1]. Let t 2 [0; 1] and " > 0 be given. Take N" from above and n ≥ N". We then deduce jf(t) − fn(t)j = lim jfm(t) − fn(t)j ≤ lim sup kfm − fnk1 ≤ ": m!1 m!1 Since this inequality holds for all t 2 [0; 1], it follows that kf − fnk1 ≤ " for all n ≥ N". Hence, fn converges uniformly (in t) to f, and so f is continuous by a convergence theorem from Analysis 1+2. Summing up, we have shown that fn converges in X to f 2 X, as required. R 1 c) We further introduce kfk1 = 0 jf(t)j dt for f 2 X = C([0; 1]). The number kf −gk1 corresponds to area between the graphs of f and g (if F = R), i.e., kf −gk1 is the distance in the mean; whereas kf − gk1 corresponds to the maximal vertical distance between the graphs. We claim that k · k1 is an incomplete norm on X. In fact, let α 2 F and f; g 2 X. Clearly, kfk1 ≥ 0. If f 6= 0, then there are 0 ≤ a < b ≤ 1 and δ > 0 such that jf(t)j ≥ δ for all t 2 [a; b], since f is continuous. It then follows Z b kfk1 ≥ jf(t)j dt ≥ (b − a)δ > 0: a 1.1. Basic properties of Banach and metric spaces 5 We also obtain Z 1 kαfk1 = jαj jf(t)j dt = jαj kfk1; 0 Z 1 Z 1 kf + gk1 = jf(t) + g(t)j dt ≤ (jf(t)j + jg(t)j) dt = kfk1 + kgk1: 0 0 As a result, k · k1 is a norm on X. Next, consider the functions given by 8 1 >1; 2 ≤ t ≤ 1; < n 1 1 1 fn(t) = nt − 2 + 1; 2 − n ≤ t ≤ 2 ; :> 1 1 0; 0 ≤ t ≤ 2 − n ; for n 2 N with n ≥ 3. For m ≥ n ≥ 3 we have 1 Z 2 1 kfn − fmk1 = jfn(t) − fm(t)j dt ≤ −! 0 1 1 n 2 − n as n ! 1, so that (fn) is a Cauchy sequence with respect to k · k1. We suppose 1 that (fn) would converge to some f 2 X with respect to k · k1. Fix any a 2 (0; 2 ) 1 1 and take n 2 N with 2 − n ≥ a. We then obtain that Z a Z a 0 ≤ jf(t)j dt = jf(t) − fn(t)j dt ≤ kfn − fk1 −! 0 0 0 R a as n ! 1, i.e., 0 jf(t)j dt = 0. The continuity of f implies as above that f = 0 on 1 [0; a] for every a < 2 , and hence f(1=2) = 0. On the other hand, Z 1 Z 1 0 ≤ jf(t) − 1j dt = jf(t) − fn(t)j dt ≤ kf − fnk1 −! 0 1 1 2 2 1 as n ! 1. As a consequence, f is equal to 1 on [ 2 ; 1], which is a contradiction. d) Let X = C(R) and a < b in R.

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