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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.

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

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    $\begingroup$ It's worth noting that this argument only requires the identity operator can be approximated by a sequence of finite rank operators in the strong topology. $\endgroup$ – Aweygan Feb 15 '17 at 3:14

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