<Home>Qualifying Exams><Analysis>Spring 2010 - Problem 9

Spring 2010 - Problem 9

Hilbert spaces

Let

A={x∈ℓ2 | ∑n≥1n∣xn∣2≤1}.A = \set{x \in \ell^2 \st \sum_{n\geq1} n\abs{x_n}^2 \leq 1}.

  1. Show that AA is compact in the ℓ2\ell^2 topology.

  2. Show that the mapping from AA to R\R defined by

    x↦∫02π∣∑n≥1xneinθ∣dθ2πx \mapsto \int_0^{2\pi} \abs{\sum_{n\geq1} x_ne^{in\theta}} \frac{\diff\theta}{2\pi}

    achieves its maximum on AA.
Solution.

We will show that AA is complete and totally bounded.

Since ℓ2\ell^2 is complete, it suffices to show that AA is closed. Let {x(n)}n⊆A\set{x^{\p{n}}}_n \subseteq A be a sequence which converges to some x∈ℓ2x \in \ell^2. By Fatou's lemma with summation viewed as integration with respect to the counting measure,

∑n≥1n∣xn∣2≤lim inf⁡k→∞∑n≥1n∣xn(k)∣≤1,\sum_{n\geq1} n\abs{x_n}^2 \leq \liminf_{k\to\infty} \sum_{n\geq1} n\abs{x^{\p{k}}_n} \leq 1,

so x∈Ax \in A, which means AA is closed, hence complete.

To show that AA is totally bounded, let ε>0\epsilon > 0. For any x∈Ax \in A and N≥1N \geq 1,

∑n≥N∣xn∣2≤∑n≥NN∣xn∣2≤∑n≥Nn∣xn∣2≤1  ⟹  ∑n≥N∣xn∣2≤1N.\sum_{n\geq N} \abs{x_n}^2 \leq \sum_{n\geq N} N\abs{x_n}^2 \leq \sum_{n\geq N} n\abs{x_n}^2 \leq 1 \implies \sum_{n\geq N} \abs{x_n}^2 \leq \frac{1}{N}.

Pick NN large enough so that 1N<ε\frac{1}{N} < \epsilon. Observe that if 1≤n≤N1 \leq n \leq N, then n∣xn∣2≤1n\abs{x_n}^2 \leq 1, so xn∈[−1,1]x_n \in \br{-1, 1}. Since [−1,1]\br{-1, 1} is compact, there exist z1,…,zM∈[−1,1]z_1, \ldots, z_M \in \br{-1, 1} such that the B(zi,εN)B\p{z_i, \sqrt{\frac{\epsilon}{N}}} cover [−1,1]\br{-1, 1}. Hence, given any x∈Bx \in B, there exist zi1,…,ziNz_{i_1}, \ldots, z_{i_N} so that ∣zij−xj∣<εN\abs{z_{i_j} - x_j} < \sqrt{\frac{\epsilon}{N}}, and so

∑n=1N∣xn−zin∣2+∑n>N∣xn∣2≤∑n=1NεN+1N≤2ε.\sum_{n=1}^N \abs{x_n - z_{i_n}}^2 + \sum_{n>N} \abs{x_n}^2 \leq \sum_{n=1}^N \frac{\epsilon}{N} + \frac{1}{N} \leq 2\epsilon.

Thus, AA is totally bounded, hence compact.

  1. It is enough to show the mapping, which we call ff, is continuous on ℓ2\ell^2, since AA is compact. First notice that

    ⟨einθ,eimθ⟩=12π∫02πeinθeimθ‾ dθ={0if n≠m,1if n=m,\inner{e^{in\theta}, e^{im\theta}} = \frac{1}{2\pi} \int_0^{2\pi} e^{in\theta}\conj{e^{im\theta}} \,\diff\theta = \begin{cases} 0 & \text{if } n \neq m, \\ 1 & \text{if } n = m, \end{cases}

    i.e., {einθ}n\set{e^{in\theta}}_n is an orthonormal set on L2([0,2π])L^2\p{\br{0, 2\pi}} with respect to the normalized Lebesgue measure dθ2π\frac{\diff\theta}{2\pi}. Thus,

    ∥∑n≥1xneinθ∥L22=∑n≥1xn2.\norm{\sum_{n\geq1} x_ne^{in\theta}}_{L^2}^2 = \sum_{n\geq1} x_n^2.

    By Hölder's inequality,

    ∣f(x)−f(y)∣=12π∣∫02π∣∑n≥1xneinθ∣−∣∑n≥1yneinθ∣ dθ∣≤12π∫02π∣∑n≥1(xn−yn)einθ∣ dθ≤(∫02π∣∑n≥1(xn−yn)einθ∣2 dθ2π)1/2(∫02π dθ2π)1/2=∥∑n≥1(xn−yn)einθ∥L2=∑n≥1∣xn−yn∣2.\begin{aligned} \abs{f\p{x} - f\p{y}} &= \frac{1}{2\pi} \abs{\int_0^{2\pi} \abs{\sum_{n\geq1} x_ne^{in\theta}} - \abs{\sum_{n\geq1} y_ne^{in\theta}} \,\diff\theta} \\ &\leq \frac{1}{2\pi} \int_0^{2\pi} \abs{\sum_{n\geq1} \p{x_n - y_n}e^{in\theta}} \,\diff\theta \\ &\leq \p{\int_0^{2\pi} \abs{\sum_{n\geq1} \p{x_n - y_n}e^{in\theta}}^2 \,\frac{\diff\theta}{2\pi}}^{1/2} \p{\int_0^{2\pi} \,\frac{\diff\theta}{2\pi}}^{1/2} \\ &= \norm{\sum_{n\geq1} \p{x_n - y_n}e^{in\theta}}_{L^2} \\ &= \sqrt{\sum_{n\geq1} \abs{x_n - y_n}^2}. \end{aligned}

    We have shown that ff is 11-Lipschitz, hence continuous, which shows that ff attains its maximum on AA.