# 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.

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Try to calculate the distance between the elements $\{e_n\}$ and after calculate $A(e_n-e_m)$ – matgaio May 22 '12 at 22:24
What characterizations of compact operators do you know? (Hint: Maybe think about the weak topology on $H$.) – Scott LaLonde May 23 '12 at 1:50

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 $\lim_n\|Ae_n\|=0$.
• 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 $v\in H$.