Let $T:H \to H$ be a self-adjoint operator on a Hilbert space H. Suppose there is a sequence $(T_n)$ of finite rank operators that converge strongly (but not necessarily in norm) to $T$. Must $T$ be compact?
Strong convergence: $\|T_n(x) – T(x)\| \to 0$ for each $x \in H$.
Best Answer
No. The strong closure of the finite-rank operators is all of $B(H)$. This follows easily from von Neumann's Double Commutant Theorem.
If you want a concrete counterexample the canonical way is the one mentioned by Kavi Rama Murthy, that the identitiy is a strong limit of finite-rank projections. That fact can also be used to show that the finite-rank operators are dense in $B(H)$. Namely, let $\{P_n\}$ be a sequence of finite-rank projections such that $P_n\to I$ strongly. Then, for any $T\in B(H)$ we have $$ Tx=\lim_nP_nTx, $$ so that $T=\lim P_nT$, where $P_nT$ is finite-rank for all $n$.