 Hilbert-Schmidt operators form a Hilbert space

On a Hilbert space $X$ with a countable Hilbertian basis $(e_k)$, the Hilbert-Schmidt operators form a Hilbert space under $⟨S,T⟩={\sum }_k⟨S{e}_k,T{e}_k⟩$, using the same linearity convention as the inner product on $X$. The Cauchy-Schwarz inequality proves convergence of this sum. The finite-sum triangle inequality and Cauchy-Schwarz inequality give $\Vert T\Vert \le \Vert T{\Vert }_2$. Thus a Cauchy sequence $(T_n)$ in Hilbert-Schmidt norm converges in operator norm to a bounded linear operator $T$. For $n,m\ge N$ with $\Vert T_n-T_m{\Vert }_2\le \epsilon$, taking $m\to \infty$ in each finite column sum gives ${\sum }_{k=1}^K\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$ for every $K$. Taking the supremum over $K$ proves $T_n-T\in {\mathcal{S}}_2(X)$ and $\Vert T_n-T{\Vert }_2\le \epsilon$, so $T\in {\mathcal{S}}_2(X)$ and completeness follows.