Suppose $H$ is a separable Hilbert space, $A$ is a Hilbert Schmidt operator on $H$, and $P_n$ is an increasing sequence of finite rank orthogonal projections of $H$ (so $P_nx\rightarrow x$ for all $x\in H$). Then we certainly have that, for $A_n=P_nAP_n$, $A_n\rightarrow A$ in the Hilbert-Schmidt norm; see for example this very complete post on Approximating a Hilbert-Schmidt operator. My follow-up question: If the singular values of an operator are always ordered from largest to smallest, including multiplicities, is it (trivially?) true that the $k^\mathrm{th}$ singular value of the the sequence of operators $\{A_n\}_{n\ge k}$ converges to the $k^\mathrm{th}$ singular value of $A$?
Thanks!
Yes. By the max-min principle for singular values, $$\sigma_k(A)=\sup_{\dim S=k}\min \{\|Ax\|/\|x\|:x\in S\} \tag1$$ It follows that $$ \begin{split}\sigma_k(A_n) &=\sup_{\dim S=k}\min \{\|P_nAP_nx\|/\|x\|:x\in S\} \\ &\le \sup_{\dim S=k}\min \{\|Ay\|/\|y\|:x\in P_n S\} \\ &\le \sigma_k(A) \end{split} \tag2$$ In the reverse direction, pick a subspace $S$ such that $\dim S=k$ and $\min \{\|Ax\|/\|x\|:x\in S\} > \sigma_k(A)-\epsilon$. Show* that the restriction of $A-A_n$ to the unit sphere of $S$ tends to zero uniformly. Conclude that $\limsup_{n\to\infty} \sigma_k(A_n) \ge \sigma_k(A)$, finishing the proof.
(*) One way to do this is to pick a finite $\epsilon$-net $E$ in the unit sphere of $S$ and find $N$ such that $\|Ax-A_n x\|<\epsilon $ for all $x\in E$ and $n\ge N$. Then use the uniform boundedness boundedness of $\|A_n\|$.