Spring 2014 - Problem 6

Solution.

1. $\set{a_1}$ is linearly independent and orthonormal. We now throw out all $a_n$ such that $\set{a_1, a_n}$ is linearly independent. If all that remains is $\set{a_1}$, then $\mathcal{H}$ is finite dimensional by assumption. Otherwise, after renumbering, $\set{a_1, a_2}$ is linearly independent, and by Gram-Schmidt, we may assume without loss of generality that $\set{a_1, a_2}$ is orthonormal.

   Repeating this process, we obtain $\set{a_1, \ldots, a_N}$ an orthonormal set at the $N$-th stage. If this process terminates, then $\mathcal{H} \simeq \C^N$ and we are done. Now suppose that this does not terminate, so we get an orthonormal sequence $\set{a_n}_n$ which still spans $\mathcal{H}$ linearly.

   Consider $x = \sum_{n=1}^\infty \frac{a_n}{2^n}$ and notice that the partial sums are Cauchy:

   $$\norm{\sum_{n=N}^M \frac{a_n}{2^n}}^2 = \sum_{n=N}^M \frac{1}{4^n} \leq \frac{1}{4^N} \cdot \frac{1}{1 - \frac{1}{4}} = \frac{1}{3 \cdot 4^{N-1}} \xrightarrow{N\to\infty} 0,$$

   so $x \in \mathcal{H} \setminus \set{0}$. Hence, by assumption, we may express it as a finite linear combination $x = \sum_{n=1}^N c_n a_n$. But observe that

   $$c_k = \sum_{n=1}^N c_n \inner{a_n, a_k} = \inner{x, a_k} = \sum_{n=1}^\infty \frac{\inner{a_n, a_k}}{2^n} = \frac{1}{2^k},$$

   i.e., the coefficients of the finite linear combination matches up with those of the infinite one. This means that any finite linear combination cannot represent $x$, a contradiction, so $\mathcal{H}$ must have been finite dimensional to begin with.

2. Let $K$ be the closed convex hull in question. Since $K$ is closed and convex, there exists a unique $x \in K$ such that $\norm{x} = d\p{0, K}$. Since $x$ is a minimizer, for any $y \in K$, $\p{1 - \theta}y + \theta x \in K$ by convexity and so $\norm{\p{1 - \theta}y + \theta x}^2 \geq \norm{x}^2$. Expanding,

   $$\begin{aligned} &\p{1 - \theta}^2\norm{y}^2 + \theta^2 \norm{x}^2 + 2\Re\inner{\p{1 - \theta} y, \theta x} \geq \norm{x}^2 \\ \implies{}& 2\p{1 - \theta}\theta\Re\inner{y, x} \geq \p{1 - \theta^2} \norm{x}^2 - \p{1 - \theta}^2\norm{y}^2 \\ \implies{}& 2\theta\Re\inner{y, x} \geq \p{1 + \theta}\norm{x}^2 - \p{1 - \theta}\norm{y}^2. \end{aligned}$$

   Sending $\theta \to 1$, we see $\Re\inner{y, x} \geq \norm{x}^2$. But if we let $y = a_n$, then

   $$\norm{x}^2 \leq \Re\inner{a_n, x} \xrightarrow{n\to\infty} 0$$

   by assumption, so $x = 0$.