Problem Set 7: Solutions Math 21A: Fall 216 Problem 1. (a) A subset A X of a metric space X is nowhere dense in X if Ā = i.e., the interior of the closure is empty. Show that A is nowhere dense if and only if Āc is an open dense subset of X. (b) Which of the following sets are nowhere dense in R: A = {1/n : n N}; B = Q (, 1); C = Cantor set? (c) Show that the Baire category theorem implies that a complete metric space is not a countable union of closed, nowhere dense sets. (d) Show that a Hamel basis of an infinite-dimensional Banach space is uncountable. (a) The set Āc is open since Ā is closed. For any set B X, we have B c = [ ] c {F : F B c closed} = {G : G B open} = (B ) c. Hence, setting B = Ā, we get that (Āc ) = (Ā ) c, so Āc is dense in X if and only if Ā =. (b) The set A is nowhere dense since Ā = {} A has no interior points. The set B is not nowhere dense since B = [, 1] and B = (, 1). The Cantor set C is nowhere dense, since C = C and C contains no intervals so C =. (c) This follows directly from de Morgan s law. If F α is closed and nowhere dense, then G α = Fα c is open and dense, so G α and Fα = ( G α ) c X. (d) Suppose for contradiction that a Banach space X has a countably infinite Hamel basis {e n : n N}. Let E n = e 1,..., e n denote the linear span of the first n basis vectors. Then E n = X. However, E n is closed, since it is a finite-dimensional subspace, and E n is nowhere dense, since a finite-dimensional space is not a neighborhood of any point in X. The Baire category theorem implies that E n X, which is a contradiction. 1
Problem 2. Let X = M N where M, N are closed subspaces of a Banach space X with M N = {} and M + N = X. Define the projection P : X X of X onto M along N by P x = y where x = y + z with y M and z N. Prove that P is bounded. Let (x n ) be a sequence in X such that x n x and P x n y. If x n = y n + z n where y n M and z n N, then y n + z n x and y n y, so z n x y. Since M is closed, y M, and since N is closed, z = x y N. It follows that x = y + z where y M and z N, so P x = y, meaning that P is closed. The closed graph theorem implies that P is bounded. 2
Problem 3. (a) Let K : X X be a bounded linear operator on a Banach space X with K < 1. Show that I K has a bounded inverse given by the uniformly convergent series (I K) 1 = I + K + K 2 + K 3 +.... Also show that (I K) 1 1/(1 K ). (b) Suppose that k : [, 1] [, 1] R is a continuous function with sup { k(x, y) : (x, y) [, 1] [, 1]} < 1. Use (a) to show that the Fredholm integral equation of the second kind u(x) k(x, y)u(y) dy = f(x) x 1, has a unique solution u C([, 1]) for every f C([, 1]). Express the solution u as a series, and write out explicitly an approximation of u(x) valid for small k(x, y), up to terms of cubic order in k. (c) Show that the result in (b) can also be obtained from a contraction mapping iteration u n+1 = T u n with u = f where T : C([, 1]) C([, 1]) is defined by (T u)(x) = k(x, y)u(y) dy + f(x). (a) The series K n converges absolutely, since K n K n 1 1 K, n= n= so it converges in B(X) since B(X) is a Banach space. For every N N, we have ( N ) ( N ) (I K) K n = K n (I K) = I K N+1. n= n= If x X, then K N+1 x as N, and ( ) ( ) (I K) K n x = K n (I K)x = x, so n= Kn = (I K) 1, and (1 K) 1 1/(1 K ). 3
(b) Define K : C([, 1]) C([, 1]) by Then K < 1 since Ku(x) = k(x, y)f(y) dy. Ku θ u, θ = sup k(x, y) < 1, x,y [,1] so by (a) the equation has a unique solution u = (I K) 1 f. The expansion u = f + Kf + K 2 f + K 3 f +... gives u(x) = f(x) + k 2 (x, y) = k 3 (x, y) = + k(x, y)f(y) dy + k(x, s)k(s, y) ds, k 3 (x, y)f(y) dy +..., k(x, s)k(s, t)k(t, y) dsdt. k 2 (x, y)f(y) dy (c) The map T u = f + Ku is a contraction on C([, 1]), since T u T v = K(u v) θ u v, so the contraction mapping theorem implies that there is a unique solution of u = f + Ku. It follows by induction that the contraction mapping iterates u N+1 = T u N, starting with with u = f, are given by u N = N n= Kn f. Remark. The series for (I K) 1 is called the Neumann series. It shows that the invertible operators form an open set in B(X). The Neumann series is also used in problems from wave propagation and quantum mechanics (where it is called the Born approximation) to describe the scattering of waves by nonuniformities or a potential. Successive terms in the series describe contributions to the solution from singly, doubly, triply,... scattered waves. 4
Problem 4. Define the multiplication operator Φ : C([, 1]) C([, 1]) associated with a function φ C([, 1]) by Φf = φ f. (a) Equip C([, 1]) with the sup-norm f = sup x [,1] f(x). Show that Φ = φ. If (Φ n ) is a sequence of multiplication operators, show that Φ n strongly in B (C([, 1]), ) if and only if Φ n uniformly. (b) Equip C([, 1]) with the one-norm f 1 = f(x) dx. Give an example of a sequence of functions (φ n ) in C([, 1]) with associated multi- plication operators (Φ n ) such that Φ n strongly but not uniformly in B (C([, 1]), 1 ). (a) We have so Φ φ. Φf = sup φ(x)f(x) φ f, x [,1] Let 1 C([, 1]) denote the function 1(x) = 1. Then Φ Φ1 1 = φ. Thus, Φ = φ. Furthermore, Φ = Φ1. If Φ n strongly, then Φ n 1, which implies that Φ n, so Φ n uniformly. (b) Choose ψ C([, )) such that ψ 1 and { 1 if x 1, ψ(x) = if 2 x <. Define φ n C([, 1]) by φ n (x) = ψ(nx). Then φ n (x) dx 2 n, φ 2 n(x) dx 1 n. 5
If f C([, 1]), then there exists M > such that f(x) M for all x [, 1]. Hence Φ n f 1 = so Φ n strongly as n. On the other hand, Φ n 1 Φ nφ n 1 φ n 1 = so Φ n uniformly as n. φ n (x) f(x) dx 2M n, φ2 n(x) dx φ n(x) dx 1 2, 6
Problem 5. Let K : X X be a compact linear operator on an infinitedimensional Banach space X. If K is one-to-one, prove that the range of K is not closed. Suppose that K : X X is a compact operator on a Banach space X and M = ran K is closed. Then M is a Banach space and K : X M is one-to-one and onto. The open mapping theorem implies that K is open, so K is a homeomorphism between X and M. Let B X be the closed unit ball in X. Then K : B K(B) is a homeomorphism, and K(B) is compact since K is compact and K(B) X is closed. It follows that B is compact, which implies that X is finite-dimensional. 7
Problem 6. Let 1 < p <. Define the Hölder conjugate p of p by 1 p + 1 p = 1. (a) Prove Young s inequality: If a, b, then ab ap p + bp p. (b) Let l p = {(x n ) : x n R and (x n ) p < }, where ( ) 1/p (x n ) p = x n p. Prove Hölder s inequality: If x = (x n ) l p and y = (y n ) l p, then x n y n x p y p. (c) Show that (l p ) = l p. (a) Young s inequality is trivial if ab =, so assume that a, b >. By Lemma 1, the function log x is concave in x >, since (log x) = 1/x 2 <. It follows that for a, b >, log a + log b = 1 p log ap + 1 ( 1 p log bp log p ap + 1 ) p bp. Taking the exponential of this inequality gives the result. (b) First, consider finite sequences x = (x 1,..., x n ), y = (y 1,..., y n ) in R n, and assume that x p = ( n x k p ) 1/p, y p = ( n y k p ) 1/p are nonzero, otherwise the result is trivial. For any α, β >, x k y k α β x k p pα p + y k p p β p, 8
so n x k y k β pα p 1 n x k p + α p β p 1 n y k p. Choosing α = x p, β = y p, we get Hölder s inequality: n n x k y k x k y k 1 p x p y p + 1 p x p y p = x p y p. The inequality for x l p, y l p n. then follows by taking the limit as (c) For y l p, define a linear functional F : l p R by F (x) = x n y n. Hölder s inequality implies that F is bounded and F y p. Conversely, define x = (x n ) by x n = (sgn y n )y p 1 n. Since ( p 1 = p 1 1 ) = p p p, we have x l p with x p = y p /p p, so F F (x) x p = which proves that F = y p. Suppose that F (l p ). Let y p y p /p p = y p, y n = F (e n ) where e n = (,...,, 1,... ) is the nth basis vector in l p. If x = (y p 1 1,..., y p 1 n,... ), x p = 9 ( n ) 1/p y k p,
Then x belongs to the subspace c of sequences that have finitely many nonzero terms, and by linearity n F (x) = y k p. It follows that ( n ) 1/p y k p = which shows that y l p. F (x) x p F, Since c is dense in l p and F is bounded, F : c R extends by continuity to a unique linear functional F : l p R given by F (x) = x n y n, which shows that (l p ) = l p. The same argument shows that (l 1 ) = l. The proof fails for (l ) because c is not dense in l, and l 1 is strictly contained in (l ). Lemma 1. If f : (a, b) R is twice differentiable and f, then f is convex. Proof. First, suppose that f(y) f(x) + f (x)(y x) for all x, y (a, b), meaning that the graph of f lies above every tangent line. Let x, y (a, b) and z = tx + (1 t)y for t 1. Then f(x) f(z) + f (z)(x z), and it follows that f(y) f(z) + f (z)(y z), tf(x) + (1 t)f(y) f(z) + f (z) [t(x z) + (1 t)(y z)] = f(z), so f is convex. Now suppose that f. If x, y (a, b), then by Taylor s theorem with Lagrange remainder, there exists c between x, y such that f(y) = f(x) + f (x)(y x) + 1 2 f (c)(y x) 2 f(x) + f (x)(y x), so the graph of f lies above its tangent lines, and f is convex. 1