Solutions: Problem Set 1 Math 201B, Winter 2007

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1 Solutions: Problem Set 1 Math 21B, Winter 27 Problem 1. Suppose that X is a linear space with inner product (, ). If x n x and n as n, prove that (x n, n ) (x, ) as n. Using the Cauch-Schwarz and triangle inequalities, we have (x n, n ) (x, ) (x n x, n ) + (x, n ) x n x n + x n x n x ( n + ) + x n x n x n + x n x + x n. Since x n x, n as n, we see that (x n, n ) (x, ) as n.

2 Problem 2. (a) Consider the linear space C([, 1]) equipped with the L 1 - norm, f 1 = f(x) dx. Prove that there is no inner product (, ) on C([, 1]) such that f 1 = (f, f). (b) Suppose that X is a normed linear space (over C) whose norm satisfies the parallelogram law. Define (, ) : X X C b (x, ) = 1 4 { x + 2 x 2 i x + i 2 + i x i 2}. Prove that (, ) is an inner product on X such that x = (x, x). (a) Consider, for example, f(x) = 1 and g(x) = 2x. Then while Thus f 1 = 1, g 1 = 1, f g 1 = f + g 1 = 1 2x dx = 1 2, (1 + 2x) dx = 2. f g f + g 2 1 = 17 4, 2 ( ) f g 2 1 = 4, so the norm does not satisf the parallelogram law. Hence it is not obtained from an inner product. (b) Note that we ma write the expression for (, ) as (x, ) = i k x + i k 2. k=

3 It follows immediatel from the definition that (x, x) = x 2, (, x) = (x, ). So the main thing we need to prove is that (x, ) is linear in. Using the parallelogram law, we find for an x,, x X that x + + 2i k z 2 + x 2 = ( x + i k z ) + ( + i k z ) 2 + ( x + i k z ) ( + i k z) 2 = 2 x + i k z i k z 2. Multipling this equation b i k, summing the result over k 3, using the fact that 3 k= i k =, and using the definition of (, ), we get (x +, 2z) = 2(x, z) + 2(, z). Setting = in this equation, and using the fact immediate from the definition that (, z) =, we get (x, 2z) = 2(x, z). It then follows that Since (, x) = (x, ), we also get (x +, z) = (x, z) + (, z). (x, + z) = (x, ) + (x, z). Repeated application of this result implies that for an m N, (x, m) = m (x, ). It follows that for an m, n N, n (x, m ) n = (x, m) = m (x, ), so (x, m ) n = m (x, ). n

4 Finall, using the densit of the rationals in the reals, the continuit of the norm, and the immediate properties (x, ) = (x, ), (x, i) = i(x, ), we conclude that (x, λ) = λ (x, ) for all λ C This proves that (, ) defines an inner product. Remark. According to P. Lax in Functional Analsis, this observation is due to von Neumann.

5 Problem 3. Let M be a linear subspace of a Hilbert space H. Prove that M = M. If x M then x, = for all M, so x M. It follows that M M. Since M is an orthogonal complement, it is a closed linear subspace, so M M. If x H then, b the projection theorem, we ma write x = +z where M, z M. If x M, then x, z =, so, z + z, z =. Since M = M, we have, z =, so z, z =. Hence z = and x = M. It follows that M M. Combining these results, we get M = M. Remark. The same argument shows that if E H is an subset, then E = [E], where [E] is the closed linear span of E.

6 Problem 4. Consider C([, 1]) equipped with the sup-norm, and define the closed linear subspace { } M = g C([, 1]) g() =, g(x) dx =. Let f C([, 1]) \ M be the function f(x) = x. Prove that d(f, M) = inf g M f g = 1 2, but that the infimum is not attained for an g M. (Meaning that there is no closest element to f in M.) We consider real-valued functions for simplicit. We have for all x [, 1] that f(x) g(x) f g. Integrating this equation with respect to x, we get f(x) dx g(x) dx f g Hence, if f(x) = x and g M so g(x) dx =, then It follows that d(f, M) 1/ f g. For sufficientl small ɛ >, define the function { kx if x δ, g ɛ (x) = x 1/2 ɛ if δ < x 1, where we choose kδ = δ 1/2 ɛ to ensure the continuit of g ɛ at x = δ > and ( ) kδ 2 + ɛ = 1 ( ) ɛ

7 to ensure that the integral of g ɛ is zero. Explicitl, δ = 2ɛ (1/2 ɛ)2, k =. 1/2 + ɛ 2ɛ Then g ɛ M and f g ɛ = 1/2 + ɛ 1/2 as ɛ +. It follows that d(f, M) 1/2, so d(f, M) = 1/2. Suppose, for contradiction, that the infimum is attained, and g M is such that f g = 1/2. Then g(x) f(x) 1 2 for all x [, 1]. Thus, writing h(x) = g(x) x + 1 2, we see that h. Since g(x) dx =, we have h(x) dx =. However, since g() =, we have h() = 1/2. Since h is continuous there is an interval [, δ] of width δ > on which h 1/4. Hence, since h is nonnegative, h(x) dx δ h(x) dx δ 4 >. This contradiction proves that the infimum is not attained.

8 Problem 5. We denote the Hölder semi-norm with exponent 1/2 and the L 2 -norm of a function f : [, 1] R b [f] = ( f(x) f() 1 1/2 sup, f x [,1] x 1/2 2 = f(x) dx) 2. We denote the sup-norm of f b f. (a) If f is continuousl differentiable on [, 1], with derivative f, prove that (b) Given R >, let [f] f 2. F = {f : [, 1] R f is continuousl differentiable, f 2 R, f 2 R} Prove that F is a precompact subset of C([, 1]) equipped with the sup-norm. (a) B the fundamental theorem of calculus, f(x) f() = x f (t) dt. The Cauch-Schwarz inequalit implies that x x f (t) dt = 1 f (t) dt ( x ) 1/2 ( x ) 1/2 1 dt [f (t)] 2 dt ( x 1/2 [f (t)] 2 dt x 1/2 f 2. Hence for all x 1, we have f(x) f() x 1/2 f 2. Taking the supremum over x, we get [f] f 2. ) 1/2

9 (b) Given ɛ >, let δ = ɛ 2 /R 2. If f F and x < δ, then f(x) f() < ɛ, so the famil F is equicontinuous. It follows from (a) that if f F, then for all x, [, 1]. Hence, f(x) f() R f(x) f(x) f() + f() R + f(). Integrating this equation over [, 1] with respect to, we get for ever x [, 1] that f(x) R + f() d. B the Cauch-Schwarz inequalit if f F, then f() d = ( 1 f() d Thus, f(x) 2R, so f 2R and F is bounded. ) 1/2 ( 1/2 1 2 d f() d) 2 R. The Arzelà-Ascoli theorem implies that F is a precompact subset of C([, 1]).