Any linear operator $T$ can be realised as the strong limit of compact operators

I'm able to show that the strong limit of compact operators need not be compact. In Stein and Shakarchi however, Question 21.(b) of Chapter 4 reads as follows:

Show that for any bounded operator $T$ there is a sequence $\{ T_n \}$ of bounded operators of finite rank so that $T_n \to T$ strongly as $n \to \infty$.

Clearly since $T_n$ has finite rank, it is compact. And note that strong convergence means that for all $f \in \mathcal{H}$, $T_nf \to Tf$. I'm unsure of how to go about this though.

• Is the Hilbert space assumed to be separable? – Aweygan Feb 15 '17 at 0:41
• @Aweygan Stein and Shakarchi define a Hilbert to be such, so yes. – user412674 Feb 15 '17 at 1:02

Since the Hilbert space $\mathcal H$ is separable, it has a countable orthonormal basis $\{e_k\}_{k=1}^\infty$. Now let $P_n$ be the projection onto the first $n$ coordinates, i.e. $$P_n\left(\sum_{k=1}^\infty\alpha_ke_k\right)=\sum_{k=1}^n\alpha_ke_k.$$ Now given an operator $T\in B(\mathcal H),$ put $T_n=P_nT$ for each $n\in\mathbb N$. Then each $T_n$ is finite-rank, and for any $f\in\mathcal H$, writing $Tf=\sum_{k=1}^\infty\alpha_ke_k$ we have $$\|(T_n-T)f\|=\left(\sum_{k=n+1}^\infty|\alpha_k|^2\right)^\frac{1}{2}\to 0$$ as $n\to\infty$. Since $f\in\mathcal H$ was arbitrary, we know $\{T_n\}$ converges strongly to $T$, and since $T\in B(\mathcal H)$ was arbitrary, the result is proven.