finite rank approximation on separable Hilbert spaces

Theorem Let $\mathscr{H}$ be a separable Hilbert space and let $T\in L(\mathscr{H})$. Then $T$ is a compact operator iff there is a sequence $\{F_n\}$ of finite rank operators with $\parallel T-F_n\parallel \to 0$.

Proof.

$(\Rightarrow )$: Assume $T$ is compact on $\mathscr{H}$ and $\{e_n\}$ is an orthonormal basis of $\mathscr{H}$. Define:

$P_{n}f$$={\displaystyle \sum _{k=0}^n}⟨f,e_k⟩e_k$

It is clear that the $P_n$ have finite rank and that we have $\parallel P_{n}f\parallel \le \parallel f\parallel$ for all $n\in \mathbb{N}$, $f\in \mathscr{H}$.

Let $ℬ$ be the unit ball in $\mathscr{H}$. We have that $P_n\to I$ pointwise. Since the $P_n$ are contractive they are equicontinuous, hence $P_n$ converges uniformly to $I$ on compact sets, and in particular on $\overline{T(ℬ)}$, which is compact by assumption.Therefore $P_{n}T\to T$ uniformly on $ℬ$, hence $\parallel P_{n}T-T\parallel \to 0$.Since $P_{n}T$ is bounded and of finite rank the first direction follows.

$(\Leftarrow )$: Now let $\{F_n\}$ be a sequence of bounded operators of finite rank with $\parallel T-F_n\parallel \to 0$.We have to show that $T(ℬ)$ is relatively compact in $\mathscr{H}$. This is equivalent to $T(ℬ)$ being totally bounded in $\mathscr{H}$.So we are left to show that for all $ϵ>0$ there is an $ϵ$-net $x_1,\mathrm{\cdots },x_n\in \mathscr{H}$ so that:

$T(ℬ)$$\subseteq {\displaystyle \bigcup _{k=1}^n}{B}_{ϵ}(x_k)$

So choose $ϵ>0$ and $n\in \mathbb{N}$ fixed so that:

Choose $x_1,\mathrm{\cdots },x_m\in \mathscr{H}$ with:

$F_n(ℬ)$$\subseteq {\displaystyle \bigcup _{k=1}^m}{B}_{\frac{ϵ}{2}}(x_k)$

Hence (by the triangle inequality):

$T(ℬ)$$\subseteq {\displaystyle \bigcup _{k=1}^m}{B}_{ϵ}(x_k)$

and we are done.∎