$A = \{ \sum_{k=1}^{\infty} a_ke_k :|a_k|\leq c_k \} \subset \mathcal H$ is compact, where $\sum c_k^2<\infty$

Question

The following is the exercise #24 in chapter 4 of Stein's Real Analysis.

Let $\{e_k\}$ be an orthonormal set in a Hilbert space $\mathcal H$. If $\{c_k\}$ is a sequence of positive real numbers such that $\sum c_k^2<\infty$, then the set $$A = \left\{ \sum_{k=1}^{\infty} a_ke_k :|a_k|\leq c_k \right\}$$ is compact in $H$.

I refered to this page and below is my trial.

Let $S_0 = \{(a_1,a_2,a_3,...):a_k \in \mathbb C, \sum_{k=1}^{\infty} a_k e_k \in A \}$, $f_n = \sum_{k=1}^{\infty} a_{n,k}e_k \in A$, and $A_1 = \{a_{n,1} \}_n$. $|a_{n,1}| \le c_1 \forall n$, so $A_1$ is a bounded sequence in $\mathbb C$. Then, there is a subsequence $\{a_{n_j,1} \}_j$ such that $\lim_{j\to \infty}a_{n_j,1} = b_1\quad (|b_1| \le |c_1|)$.

Let $S_1 = \{(a_1,a_2,a_3,...):a_1 \in A_1, S_1 \subset S_0 \}$, and $A_2 = \{a_{n,2} \}_n$....

Continuing this process inducively, $S_m$ can be obtained $(m \in \mathbb N)$ so that $S_{m+1} \subset S_m$ and $\lim_{j\to \infty}a_{n_j,k} = b_k \quad (|b_k| \le |c_k|), k = 1,...,m$. $\{a_{n_j,k}\}_j$ is a sequence of kth element of $S_m$.

Let $f_{n_j} = \sum_{k=1}^{\infty} a_{n_j,k}e_k$, and $f = \sum_{k=1}^{\infty} b_k e_k$.

Then, $\Vert f_{n_j} - f \Vert = \Vert \sum_k (a_{n_j,k}-b_k)e_k \Vert \le \sum_k \Vert (a_{n_j,k}-b_k)e_k \Vert = \sum_k |a_{n_j,k}-b_k|$

After then, I wanted to show that $\lim_{j\to\infty} \Vert f_{n_j} - f \Vert = 0$, but I found it difficult to show $$\lim_{j\to\infty}\sum_{k=1}^{\infty}|a_{n_j,k}-b_k| = 0$$

Any comments about my trial, whether my approach is correct or not, some errors if exists, how to finish this, or other better ideas, would be appreciated. Thank you.

Answer 1

The set $A$ can be described in a more convenient way as $$A=\left \{\sum_{k=1}^\infty x_kc_ke_k\,:\,|x_k|\le 1\right \}$$ The rest of the proof is standard but, in my opinion, slightly more transparent due to the modified description of $A.$

Let $u_n:=\sum_{k=1}^\infty x_k^{(n)}c_ke_k$ be a sequence in $A.$ Then $\{x^{(n)}_k\}_{n=1}^\infty$ form a family of sequences indexed by $k,$ all bounded by $1.$ By applying the diagonal method we get a subsequence $n_j$ such that $x^{(n_j)}_k$ is convergent for every $k.$ WLOG we may assume that $x^{(n)}_k$ is convergent for every $k.$ Let $x_k=\lim_n x^{(n)}_k.$ Then $|x_k|\le 1.$ Fix $\varepsilon>0.$ There exists $K$ such that $\sum_{k=K+1}^\infty c_k^2<\varepsilon ^2.$ Hence $$\left \|\sum_{k=1}^\infty x_k^{(n)}c_ke_k-\sum_{k=1}^\infty x_kc_ke_k\right \|\\ \le \sum_{k=1}^K |x_k^{(n)}-x_k|c_k+\left \|\sum_{k=K+1}^\infty x_k^{(n)}c_ke_k\right \|+\left \|\sum_{k=K+1}^\infty x_kc_ke_k\right \|\\ = \sum_{k=1}^K |x_k^{(n)}-x_k|c_k+\left (\sum_{k=K+1}^\infty |x_k^{(n)}|^2c_k^2\right )^{1/2}+\left (\sum_{k=K+1}^\infty |x_k|^2c_k^2\right)^{1/2} \\ \le \sum_{k=1}^K |x_k^{(n)}-x_k|c_k+2\varepsilon$$ Now the first finite sum is less than $\varepsilon$ if $n$ is large enough.