Let $H$ be a separable Hilbert space. and we define a compact operator: $A:H\rightarrow H.$
Let $P_n$ be the projection on the basis $(e_n)_n$: $P_nx=\sum_{0\leq k\leq n}(x,e_k).e_k$
I want to prove that: $$||P_nA-A||\rightarrow 0. (n\rightarrow \infty)$$
well, we have : fo all $x\in B_H(0,1)$:$||P_nAx-Ax||=\sum_{k>n}(Ax,e_k).e_k$
I did not know how to use the fact that $A$ is compact
Best Answer
$\|P_nAx-Ax\|^{2}=\sum_{k>n} | \langle Ax, e_k \rangle|^{2}$. Since $A$ is compact it is enough to show that $\sum_{k>n} | \langle u, e_k \rangle|^{2} \to 0$ uniformly for $u$ in a compact set. This is an immediate consequence of Dini's Theorem.
The compact set in this proof is the closure of $\{Ax: \|x\| \leq 1\}$.