# Bounded operators on separable Hilbert spaces

Let $H$ be a separable Hilbert space. Show that every bounded operator from $H$ to itself can be approximated in the strong operator topology by a sequence of finite rank operators.

Im not sure what the strong operator topology implies (that the evaluation map for the operator is continuous?) Is it enough to show to use projection to some subspace of some orthogonal sequence?

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## 1 Answer

Strong operator topology is generated by the semi-norms $\rho_x(T):=\lVert Tx\rVert$.

What we have to show here is that if $T$ is linear and bounded, we can find $T_n$ of finite-rank such that $\rho_x(T-T_n)\to 0$ for all $x$. Let $\{e_n\}$ an Hilbert basis of $H$, and $P_N$ the projection over $\operatorname{Span}\{e_1,\dots,e_N\}$. Let $T_N:=TP_N$. It's a finite-ranked operator. As $P_Nx\to x$ as $N\to +\infty$ and $T$ is bounded, we have convergence in the strong operator topology.

This result helps us to see that strong operator topology is really different from the topology induced by operator norm. Indeed, in the later, the closure of finite ranked operators is the set of compact operators, no matter whether the Hilbert space is separable or not.

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thanks! I have not seen semi-norms before, how do you mean that they are different? – Johan Jan 3 '13 at 22:49
To say that $T_n\to T$ in norm is "much" stronger than saying that $T_n\to T$ in the strong operator topology. – Davide Giraudo Jan 3 '13 at 22:51