Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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

share|cite|improve this answer
  • 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 $v\in H$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.