$\{ x \in H: x=\sum_{k=1}^{\infty}c_{k}u_{k}$, $|c_{k}| \leq \frac{1}{k}\}$ is compact

Question

Let $H$ be a complex inner product space that is also a complete metric space with respect to the distance induced by the inner product. Assume $\{u_{k}\}_{k=1}^{\infty}$ be an orthonormal set in $H$ and $Q=\{ x \in H: x=\sum_{k=1}^{\infty}c_{k}u_{k}$, $|c_{k}| \leq \frac{1}{k}\}$. Show that $Q$ is a compact set.

I am aware of the condition $\sum c_k^2<\infty$ then $A=\{\sum_{k=1}^{\infty} a_ke_k :|a_k|\leq c_k \}$ is compact, but in this case I want to explicitly show $Q$ is compact by proving that every sequence in $Q$ has convergent subsequence.I just can't get started, anyone willing to lend a helpful hint? Cheers!

PS Any reference to Rudin - if you happen to have a copy nearby - would be appreciated. DS

Answer 1

Hint: there is a subsequence all of whose coordinates converge. Show that the sequence converges to the vector whose coordinates are these limits.

Details:

Let $x_n$ be a sequence in $Q$ and let $x_n^k = \langle x_n, u_k \rangle$ for $k = 1, 2,...$. Then by assumption $|x_n^k| \le \frac{1}{k}$ so that there is a subsequence of $\{x_n\}$ for which $x_n^1$ converges. This subsequence has a subsequence for which $x_n^2$ converges and so on. Using a diagonal argument, there is a subsequence of $x_n$ for which $x_n^k$ converges for all $k = 1,2,...$. We denote this subsequence again by $\{x_n\}$. Let $x \in H$ be defined by $$\langle x, u_k \rangle = \lim_{n \to \infty} x_n^k := x^k.$$ Then $x \in Q$ and we have $$\|x - x_n\|^2 = \sum_{k=1}^\infty |x^k - x_n^k|^2.$$ The summands tend to $0$ as $n \to \infty$ and are bounded by the summable sequence $\left(\frac{2}{k}\right)^2$. Thus the dominated convergence theorem implies that $x_n \to x$ in $H$.