Problem Set 6: Solutions Math 201A: Fall a n x n,

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1 Problem Set 6: Solutions Math 201A: Fall 2016 Problem 1. Is (x n ) n=0 a Schauder basis of C([0, 1])? No. If f(x) = a n x n, n=0 where the series converges uniformly on [0, 1], then f has a power series expansion at x = 0 with radius of convergence at least equal to 1, and it follows from the theory of power series that f is infinitely differentiable in [0, 1). For example, f(x) = x has no such expansion. Remark. The density of polynomials in C([0, 1]) is not sufficient to imply that (x n ) is a Schauder basis because one has to use polynomials with different coefficients to obtain better uniform approximations of a non-analytic function. 1

2 Problem 2. Let A : R n R m be a linear map with matrix (a ij ), so y = Ax if y i = n j=1 a ijx j. Equip R n and R m with the one-norms x 1 = n j=1 x j, y 1 = m y i. Show that the corresponding operator norm of A is given by the maximum absolute column sum A 1 = max a ij. We have which shows that y i n a ij x j j=1 ( n m ) a ij x j j=1 m n max a ij x j, j=1 m A 1 max a ij Conversely, for some 1 j 0 n, we have m a ij0 = max a ij. Define x = (x 1,..., x n ) by x j = 1 if j = j 0 0 if j j 0 Then x 1 = 1 and Ax 1 = m a ij 0, which shows that m A 1 max a ij. 2

3 Problem 3. Suppose that (X, ) be a Banach space and M X is a closed linear subspace. Let X/M = x + M : x X denote the the quotient space of X by M, where x + M = y X : y = x + m for some m M, and define x + M = inf m M x + m. Show that : X/M R defines a norm on X/M and X/M is a Banach space with respect to this norm. It is straightforward to check that X/M is a vector space with respect to the natural operations (x + M) + (y + M) = (x + y) + M, λ(x + M) = λx + M. In particular, these operations are well-defined since, for example, if x+m = x +M and y +M = y +M, then (x+y)+m = (x +y )+M. The zero-vector in X/M is the subspace M. We show that is a norm on X/M. (i) Clearly, x + M 0. If x + M = 0, then we can choose m k M such that x m k 0 as k, and then x = lim m k M since M is closed, so x + M = M. (ii) If λ = 0, then λ(x + M) = M; and if λ R \ 0, then, writing m = λm, we have λ(x + M) = inf λx + m = λ inf x + m M m M m = λ x + M. (iii) Let x, y X. Given ɛ > 0, there exists m 1, m 2 M such that Then x + m 1 x + M + ɛ 2, y + m 2 y + M + ɛ 2. (x + M) + (y + M) = inf x + y + m m M x + m 1 + y + m 2 Since ɛ > 0 is arbitrary, x + M + y + M + ɛ. (x + M) + (y + M) x + M + y + M. 3

4 Let (x k + M) be a sequence in X/M such that x k + M <. We will show that (x k + M) converges in X/M. Then Lemma 1, proved below, implies that X/M is complete. For each k N, there exists m k M such that y k = x k + m k satisfies y k x k + M k. Then y k <, so Lemma 1 implies that n y k z as n for some z X. Moreover, x k + M = y k + M. Since 0 M, we have x + M x for every x X. It follows that ( n n ) (x k + M) (z + M) = y k z + M n y k z, which shows that n (x k + M) (z + M) in X/M as n. Lemma 1. A normed linear space (X, ) is complete if and only if every absolutely convergent series converges. That is, if (x k ) is a sequence in X and x k <, then x k converges (in norm) in X. Proof. Suppose that X is complete. If x k converges, then the sequence (s n ) of partial sums s n = n x k is Cauchy in R. Since n n x k x k, k=m+1 k=m+1 the sequence (S n ) of partial sums S n = n x k is Cauchy in X, so x k converges. Conversely, suppose that every absolutely convergent series converges. Let (x n ) be a Cauchy sequence in X. Then (x n ) converges if some subsequence converges. After extracting a subsequence, still denoted by (x n ), we may assume that x k+1 x k < 1/2 k for every k N. Let y k = x k+1 x k. Then y k < and x n = n 1 y k + x 1, so (x n ) converges, meaning that X is complete. 4

5 Problem 4. (a) Suppose that M is a proper closed subspace of a normed linear space X and 0 < r < 1. Show that there exists x X such that x = 1 and d(x, M) = inf m M x m > r. (b) Show that the closed unit ball B = x X : x 1 of an infinitedimensional Banach space X is not compact. (a) Choose x 1 X \ M. Then a = d(x 1, M) > 0 since M is closed. Given any ɛ > 0, there exists m 1 M such that a x 1 m 1 < a+ɛ. Let x = x 1 m 1 x 1 m 1. Then x = 1 and, for each m M, x m = x 1 m x 1 m 1, m = m 1 + x 1 m 1 m. Since m runs over M as m runs over M, we have inf x m = 1 m M x 1 m 1 inf x m 1 m > M which proves the result with 0 < r = a/(a + ɛ) < 1. a a + ɛ, (b) Let X be an infinite-dimensional space with closed unit ball B. We recursively define a set x n : n N B with x m x n > 1/2 for all m n as follows. Pick x 1 B. Given E = x 1,..., x n, let M be the linear span of E. Then the finite-dimensional space M is a proper, closed subspace of X, so by (a) there exists x n+1 B with d(x n+1, M) > 1/2. Any ball of radius 1/4 contains at most one of the points x n, so B does not have a finite (1/4)-net, meaning that B is not totally bounded, and therefore not compact. Remark. The result in (a) is called the Riesz Lemma. If M is a closed proper subspace of a Banach space X, there may not exist a vector x X with x = 1 and d(x, M) = 1. If X is a Hilbert space, however, we can always find such a vector by taking x M. 5

6 Problem 5. Let c be the Banach space of convergent real sequences and let c 0 c be the Banach space of real sequences that converge to 0, both equipped with the sup-norm (x n ) = sup x n. (a) Show that c and c 0 are homeomorphic. Hint. Define Φ : c c 0 by Φ(x 1, x 2, x 3, x 4,... ) = (a, x 1 a, x 2 a, x 3 a,... ), a = lim n x n. (b) For any x c 0 with x = 1, show that there exist y, z c 0 such that y z, y = z = 1, and x = 1 2 (y + z). Deduce that c 0 and c are not isometrically isomorphic. (a) Clearly, Φ is a one-to-one, onto linear map. To show that Φ is a homeomorphism, we prove that Φ and Φ 1 are bounded. If x = (x n ) c, then Φx = max a, sup x n a a + sup x n 2 x, so Φ is bounded. In fact, Φ = 2 since, for example, For y = (y n ) c 0, we have Hence, Φ( 1, 1, 1, 1,... ) = (1, 2, 0, 0,... ). Φ 1 (y 1, y 2, y 3, y 4,... ) = (y 1 + y 2, y 1 + y 3, y 1 + y 4,... ). Φ 1 y = sup y 1 + y n+1 y 1 + sup y n 2 y, so Φ 1 is bounded. In fact, Φ 1 = 2 since, for example, Φ 1 (1, 1, 0, 0, 0,... ) = (2, 1, 1, 1, 1,... ). (b) Suppose that x = (x n ) c 0 and x = 1. Then, since x n 0 as n, we have x j = 1 for some j N and x k 1/2 for some k N. Define y = (y n ) and z = (z n ) by y n = x k + 1/2 if n = k, x n if n k, 6 z n = x k 1/2 if n = k, x n if n k.

7 It follows that: y z since y k z k ; y = z = 1 since y j = z j = 1 and y n, z n 1; and x = (y + z)/2. On the other hand, consider x = (1, 1, 1, 1,... ) c with x = 1. If x = (y + z)/2, then y n + z n = 2 for every n N, and y n > 1 or z n > 1 unless y n = z n = 1. Hence, there do not exist distinct y, z c such that y = z = 1 and x = (y + z)/2. The existence of such a y, z depends only on the geometry of the unit ball, and is preserved under an isometric isomorphism (a linear map that preserves norms). Hence, c 0 and c cannot be isometrically isomorphic. Remark. The result in (a) should be surprising from a finite-dimensional perspective: dimension is a topological invariant, so a finite-dimensional vector space cannot be homeomorphic to a proper subspace. If K X is a convex subset of a linear space X, then a point x K is said to be an extreme point of K if x = ty + (1 t)z for some y, z K and 0 < t < 1 implies that y = x, z = x. That is, x is not an interior point of any line segment in K. Part (b) shows that the closed unit ball in c 0 has no extreme points, but the closed unit ball in c has extreme points. Roughly speaking, the unit ball in c 0 resembles an infinite-dimensional cube with no corners, while the unit ball in c has lots of corners. It follows that the spaces cannot be isometrically isomorphic. As another application, the Krein-Milman theorem implies that every compact, convex set K in a locally convex topological vector space has extreme points (in fact, K is the closed convex hull of its extreme points). It follows from the Banach-Alaoglu theorem that the closed unit ball in the dual of a Banach space is compact and closed in the weak* topology, so it has extreme points. Consequently, if the closed unit ball of a Banach space has no extreme points, then the space cannot be isometrically isomorphic to the dual of a Banach space. For example, c 0 is not the dual of any Banach space. 7