Let $(e_k)_{k=1}^{\infty}$ denote an orthonormal set in a Hilbert space $H$. If $(c_k)_{k=1}^{\infty}$ 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 $H$.
Some hint? Thanks in advance.
$\prod_k [-c_k, c_k]$ is compact under product topology; the map $f:\prod_k [-c_k, c_k] \to A$ defined by $f(a_k)=\sum_k a_ke_k$ is continuous and onto. Hence $A$ is compact. Hints for proving that $f$ is continuous: The maps $f_N(a_k)=\sum_{k=1}^{N} a_ke_k$ are obviously continuous. Prove that these converge to $f$ uniformly. ( given $\epsilon >0$ there exists $N$ such that $\sum_{n>N} c_k^{2} <\epsilon$, etc ).
Second approach: take a sequence $\sum a_k^{(j)} e_k$ in $A$. For fixed $k$ the sequence $(a_k^{(j)})_j$ is bounded and hence it has a convergent subsequence. Use a diagonal argument to get one subsequence along which all the coordinates converge. If $a_k=\lim_{j\to \infty} a_k^{(j)}$ prove that $\sum a_ke_k$ is in $A$ and $\sum a_k^{(j)} e_k \to \sum a_ke_k$. This proves that $A$ is sequentially compact, hence compact.
