The image of orthonormal basis under compact operator

Question

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.

Answer 1

• 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 $v\in H$.

Answer 2

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 $\lim_n\|Ae_n\|=0$.