The image of orthonormal basis under compact operator I need a help to prove that statement: if $\{e_n\}$ an orthonormal basis in Hilbert space $H$ and $A$ is a compact operator from $H$ to $H$, then $Ae_n\rightarrow 0$. Thx for any help. 
 A: Here's a different proof. 
Assume first that $A$ is finite-rank. Then $\text{Tr}(A^*A)<\infty$, and so
$$
0\leq\text{ Tr}(A^*A)=\sum_{n=1}^\infty\langle A^*Ae_n,e_n\rangle=\sum_{n=1}^\infty\|Ae_n\|^2<\infty,
$$
and so $\|Ae_n\|\to0$.
If $A$ is any compact operator, there exists a sequence of finite-rank operators $\{A_m\}$ with $\|A_m- A\|\to0.$ Then
$$
\|Ae_n\|\leq\|(A-A_m)e_n\|+\|A_me_n\|\leq\|A_m-A\|+\|A_me_n\|.
$$
So
$$
0\leq\limsup_n\|Ae_n\|\leq\|A_m-A\|+0=\|A_m-A\|.
$$
As $m$ was arbitrary, we conclude that $0\leq\limsup_n\|Ae_n\|=0$, and so $\lim_n\|Ae_n\|=0$.
A: I think the simplest proof goes as follows. Assume that $(Ae_n)$ does not converge to $0$ strongly. Then, by the compactness of $A$, there is a subsequence of $(Ae_n)$ that converges strongly to some nonzero vector $v\neq 0$. On the other hand, by Bessel's inequality, $(e_n)$ converges to $0$ weakly, hence $(Ae_n)$ also converges to $0$ weakly. As strong convergence implies weak convergence, we conclude that $v=0$, a contradiction.
A: *

*We show that each subsequence of $\{Ae_n\}$ has a further converging to $0$ subsequence. Let $\{Ae_{n_k}\}$ such a subsequence. By compactness of $A$, we can find a subsequence denoted $\{Ae_{\varphi(j)}\}$ or $\{f_j\}$ where $f_j=Ae_{\varphi(j)}$, which is convergent (for the norm) to, say $y\in H$. 

*We show that $f_j$ converges weakly to $0$. Indeed, fix an integer $k$. Then 
$$|\langle f_j,e_k\rangle|=\langle e_{\varphi(j)},A^*e_k\rangle\to 0,$$
using the fact that $e_{\varphi(j)}$ is weakly convergent to $0$ (this follows from the fact that $\{e_n\}$ is an orthonormal basis, hence you can approximate each vector by one of the vector space spanned by the $\{e_n\}$, denoted $V$).  Then we can show that for each $v\in V$, $\langle f_j,v\rangle \to 0$, and using the fact that $\{f_j\}$ is bounded and $V$ dense in $H$, we get this convergence for $v\in H$.  

