Functional Analysis Exercise Class

Transcription

1 Functional Analysis Exercise Class Week 2 November 6 November Deadline to hand in the homeworks: your exercise class on week 9 November 13 November Exercises (1) Let X be the following space of piecewise continuous functions from [0, 1] to K: X := { f K : f is piecewise continuous, right continuous Define f 1 := at every x [0, 1], and continuous at 1 }. f(t) dt, f X. a) Show that 1 is a norm on X. b) Show that C([0, 1], K) is not closed in X w.r.t. 1, and conclude that (C([0, 1], K) is not complete with the norm 1. (Hint: Construct a sequence of continuous functions in X that converges to a discontinuous function w.r.t. 1.) c) Recall the definition of the uniform norm f := sup t f(t). Show that Solution: 1, but 1 is not equivalent to. Contrast this with what you have learned about norms on finite-dimensional vector spaces. a) We have λf 1 = λf(t) dt = λ f(t) dt = λ f 1. 1

2 Since f(t) + g(t) f(t) + g(t), t [0, 1], we get f + g 1 = f(t) + g(t) dt f(t) dt + = f 1 + g 1, g(t) dt i.e., the triangle inequality holds. Obviously, f 1 0 for every f X. Finally, if f 0 then there exists an x 0 [0, 1) such that f(x 0 ) > 0. By piecewise continuity and right continuity, there exists a δ > 0 such that for all x [x 0, x 0 + δ], f(x) f(x 0 ) /2, so that f 1 f(t) dt δ f(x 0 ) /2 > 0. [x 0,x 0 +δ] b) Consider f X \ C([0, 1], K) given by { 0 0 x < 1/2 f(x) = 1 1/2 x 1. For every n N let f n be a continuons "smooth-over" of function f, 0, 0 x < 1/2 1/n n f n (x) = x + 1 n, 1 1 x n 2 n 1, 1/2 + 1/n x 1. Then f n C([0, 1], K), and f n f 1 = 1 0 f n (x) f(x) dx = 1 n 0 as n, i.e., (f n ) n N converges to f in the norm 1. Thus, (f n ) n N is a sequence in C([0, 1], K) whose limit is not in C([0, 1], K), and therefore C([0, 1], K) is not closed in (X, 1 ). Since every complete subspace of a metric space is closed, this shows that C([0, 1], K) is not complete w.r.t. the norm 1. c) Obviously, f(t) max x f(x) = f for every t [0, 1], and hence f 1 = f(t) dt f dt = f. Equivalence of 1 and would mean the existence of a positive constant c such that f c f 1 for all f X. In particular, this would imply that if a sequence (f n ) n N X converges to some f X w.r.t. 1 then it also converges to f w.r.t.. Consider, however, the function f and the 2

3 sequence (f n ) n N constructed in the previous point. Then f n f 1 0 as n +, but for all n N, f n f = sup f n (t) f(t) = 1/2, t and hence f n does not converge to f w.r.t.. This shows that there cannot exist a positive constant c such that f c f 1 for all f X, and hence 1 and are not equivalent. This is in contrast with the situation in finite-dimensional normed spaces, where any two norms are equivalent to each other. Alternative solution for the inequivalence: We know that C([0, 1], K) is complete w.r.t., and hence it is closed in X w.r.t.. In the previous point we have seen that it is not closed in X w.r.t. 1, and hence the two norms cannot be equivalent (because otherwise they would generate the same topology, and would have the same closed sets). (2) Consider the space of real-valued continuously differentiable functions C 1 ([0, 1], R) with the C 1 -norm f C 1 := f + f = sup f(x) + sup f (x). 0 x 1 0 x 1 Prove that this is indeed a norm, and with this norm C 1 ([0, 1], R) is a Banach space. Solution: It is obvious that C 1 is a seminorm, and since is strictly positive, and C 1, we see that C 1 is strictly positive, too. Now, suppose that (f n ) n N C 1 ([0, 1], R) is a Cauchy sequence. Then and hence f n f m C 1 0 as n, m, f n f m 0, f n f m 0, as n, m. ) That is, the sequences (f n ) n N and (f n) n N are Cauchy sequences in (C([0, 1], R),. Since this space is complete, both sequences have limits in this space, denote these limits by f and g, i.e. f n f 0, and f n g 0 as n. The proof will be complete if we can show that f C 1 ([0, 1], R) and lim n + f n f C 1 = 0. 3

4 Note that for every n N, Define f n (x) = f n (0) + f(x) := f(0) + x 0 x 0 f n(t) dt, x [0, 1]. g(t) dt, x [0, 1]. Then f n f = sup f n (x) f(x) x f n (0) f(0) + sup x x 0 f n(t) g(t) dt f n f + f n g n + 0, i.e., f n converges to f in the uniform norm. Since the limit of a sequence in a normed space is unique, we obtain that f = f. By construction, f is differentiable, and hence we see that f is differentiable. Finally, f n f C 1 = f n f + f n f = f n f + f n g 0 as n +, by the definition of f and g. (3) a) Let (M, d) be a metric space and let K M be compact. Show that every sequence in K has at least one accumulation point in K. b) Recall the definition of the 2-norm x 2 := ( + x(i) 2) 1/2 on K N, and the definition Solution: l 2 (N, K) := { x K N : x 2 < }. Show that in the space (l 2 (N, K), 2 ), the closed unit ball B 1 (0) = {x l 2 (C) : x 2 1} is not compact. Thus, closed and bounded sets need not be compact in an infinite-dimensional space. Contrast this with what you have learned about finite-dimensional normed spaces. (Hint: Consider the sequence e 1 = (1, 0, 0, 0,...), e 2 = (0, 1, 0, 0,...), e 3 = (0, 0, 1, 0,...),....) 4

5 a) Assume on the contrary that there is a sequence (x n ) n N in K that has no accumulation point. Then, for every y K, there is an ɛ y > 0 such that B ɛy (y) contains only finitely many points of (x n ) n N. Since K = y K B ɛ y (y), the B ɛy (y) cover K. Since K is compact, there is a finite subcover. But since every B ɛy (y) contains only finitely many points of (x n ) n N, we conclude that the sequence (x n ) n N is finite, a contradiction. b) The sequence (e n ) n N given in the Hint is in B 1 (0). We find that e n e m 2 = 2 for all n, m N with n m, and hence (e n ) n N has no accumulation point. By the previous point, this shows that B 1 (0) is not compact. Hence, B 1 (0) is a closed and bounded set that is not compact. This is in contrast with the case of finite-dimensional normed spaces where a set is compact if and only if it is closed and bounded. (4) a) Show that if W is a proper subspace in a normed space (i.e., it is not the whole space) then its interior is empty. b) Show that the algebraic dimension of any Banach space is either finite or uncountably infinite, i.e., there exists no countably infinite-dimensional Banach space. (Hint: Use the previous point and Baire s category theorem.) c) Let X be the set of polynomial functions on [0, 1]. Show that there exists no norm which would make X a Banach space. Solution: a) Let X be a normed space and W be a proper subspace. Let w W ; we have to show that for every ε > 0, B ε (y) is not contained in W. Since W is not the whole space, there exists some x X \ W. For any given ε > 0, let x ε := v + ε x. Then x 2 x ε is also outside of W ; indeed, if it was an element of W then so would 2 x (x ε ε v) = x be, a contradiction. On the other hand, x ε v = ε/2, i.e., x ε B ε (v). Hence, B ε (v) is not contained in W. b) Assume that X is a Banach space with an algebraic basis {e n } n N. For every N N, let W N be the subspace spanned by the first N basis vectors. Then W N is finite-dimensional, and hence closed, as it was shown in the lecture. By definition, any x X can be expanded in this basis, i.e., there exist n 1 <... < n r and c 1,..., c r K such that x = r c ie ni. Thus, x W nr. This shows that X = N N W N. However, by the previous point, each W N is closed and nowhere dense, and therefore we obtained a contradiction with Baire s category theorem. 5

6 c) It is easy to see that (f n (x) := x n ) n N forms a basis in the set of polynomial functions, and therefore this space is countably infinite dimensional. By the previous point, it cannot be a Banach space with any norm. 6

7 Homework with solutions (1) Recall that C 0 (R) := {f : R R f continuous, lim f(x) = lim f(x) = 0} x x denotes the space of real-valued continuous functions on R that vanish at infinity, and C c (R) := {f : R R f continuous, a > 0 s.t. f(x) = 0 if x > a} denotes the space of real-valued continuous, compactly supported functions on R. a) Prove that the closure of the space C c (R) with respect to the uniform norm f = sup x R f(x) is the space C 0 (R). b) Is the space C c (R) with the uniform norm complete? If not, give a concrete completion of it. Solution: a) It is clear that C c (R) C 0 (R), so what we have to show is that any ε- ball around a function f C 0 (R) contains a function from C c (R). By the definition of C 0 (R), there exists a t ε > 0 such that f(t) < ε/3 for all t R \ [ t ε, t ε ]. Define f(t), t t ε, (1 p)f(t ε ), t = t ε + p, 0 < p 1, f ε (t) := (1 p)f( t ε ), t = t ε p, 0 < p 1, 0, otherwise. Then it is easy to see that f C c (R), and and hence sup f(t) f ε (t) = 0, t [ t ε,t ε] sup f(t) f ε (t) f( t ε ) + sup f(t) 2ε/3, t [ t ε 1, t ε] t [ t ε 1, t ε] sup f(t) f ε (t) f(t ε ) + sup f(t) 2ε/3, t [t ε,t ε+1] t [t ε,t ε+1] sup f(t) f ε (t) = sup f(t) ε/3, t R\[ t ε,t ε] t R\[ t ε,t ε] f f ε = sup f(t) f ε (t) < ε. t R 7

8 b) The previous part shows that C c (R) is not closed. Since every complete subspace of a metric space is closed (see lecture), C c (R) is not complete. On the other hand, the previous point also shows that C 0 (R) with the identical embedding is a completion of C c (R). (2) For a finite n N, consider K n with the p-norms ( n ) 1/p x p := x(i) p, 1 p < +, x := max x(i). 1 i n For every 1 p q +, find constants c p,q, d p,q > 0 such that c p,q x q x p d p,q x q, x K n. Solution: By Exercise 6, week 2, we have q p, p q, and hence we can choose c p,q := 1. Note that this c p,q is optimal, as the inequality is saturated for any of the canonical basis vectors e i. When p = q, we can also choose d p,q := 1, and hence for the rest we assume p < q. When q < +, we can find an 1 < r < + such that 1 q/p + 1 r = 1; explicitly, r = q q p. Now we can apply Hölder s inequality with q/p and r in the following form: n x(i) p = n 1 x(i) p = 1 x p 1 1 r x p q/p Taking the p-th root, we get x p n q p qp x q. ( n ) 1/r ( n ) p/q = 1 r ( x(i) p ) q/p = n 1/r x p q. (0.1) Taking the limit q + (cf. sheet 2, Exercise 6/b), we get x p n 1 p x. Hence, we can choose d p,q := n q p qp for q < + and d p,q := n 1 p for p = +. This d p,q is actually optimal. Indeed, by taking the constant 1 vector x(i) = 1, i [n], we have x p = n 1/p = n q p qp n 1/q = n q p qp x q, 8

9 and similarly, x p = n 1 p = n 1 p x. Scores: finding c p,q : 2 points, finding d pq for q < + : 5 points, finding d p, : 1 point. Remarks: 1. One has to consider the q = + case separately, since it is otherwise difficult to make sense of the expression ( x(i) p ) q/p in (0.1). 2. There is a somewhat easier derivation by comparing p to, that goes like this: ( n ) 1/p x = max x(i) x 1 i n p = x(i) p (n x p )1/p = n 1 p x. This gives c p,+ = 1, d p,+ = n 1 p, and for q < + we get x p n 1 p x n 1 p x q, x q n 1 q x n 1 q x p, giving c p,q = n 1 q, dp,q = n 1 p. Note, however, that these constants are not optimal for q < +. 9