22
$\begingroup$

Consider $T\colon\ell^2\to\ell^2$ an operator such that $Te_k=\lambda_k e_k$ with $\lambda_k\to 0$ as $k \to \infty$ how to prove that it is compact?

$\endgroup$
21
$\begingroup$

Let $T_n$ such that $T_n(e_k)=\begin{cases}\lambda_ke_k&\mbox{ if }k\leq n\\\ 0&\mbox{ if }k>n \end{cases}$. Then $T_n$ is finite ranked hence compact and for $v\in\ell^2$, $v=(v_0,v_1,\ldots)$ $$\lVert (T-T_n)v\rVert^2=\sum_{k=0}^{+\infty}|\langle((T-T_n)v)_k\rangle|^2=\sum_{k\geq n+1}|(T-T_n)(v_k)|^2=\sum_{k\geq n+1}|\lambda_k|^2\cdot|v_k|^2\\\leq \sup_{k\geq n+1}|\lambda_k|^2\lVert v\rVert_{\ell^2}^2 $$ so $\lVert T-T_n\rVert\leq \sup_{k\geq n+1}|\lambda_k|$ and we conclude that $T_n\to T$ in norm. A norm-limit of compact operators is compact so $T$ is compact.

Conversely, if $T$ is compact, then you can extract from $\{Te_n\}$ a converging subsequence so you can extract from $\{\lambda_ne_n\}$ a converging subsequence, say $\{\lambda_{n_k}e_{n_k}\}$. Since $\lVert \lambda_{n_{k+1}}e_{n_k{+1}}-\lambda_{n_k}e_{n_k}\rVert^2=|\lambda_{n_{k+1}}|^2+|\lambda_{n_k}|^2\to 0$, we should have $\lambda_{n_k}\to 0$. We thus can prove that for each subsequence $\{\lambda_{n_j}\}$, we can extract that a further subsequence which converges to $0$, hence the whole sequence converges to $0$.

$\endgroup$
  • $\begingroup$ Sorry to resurrect an old post, but I was stuck on the same question so had a look on here, and I think I have found an error in your proof. (That said, I'm a third year maths student and your page says that you're doing a PhD - the odds aren't in my favour!) $\endgroup$ – Sam T Nov 28 '14 at 15:35
  • $\begingroup$ At the end, you say $$\lVert \lambda_{n+1}e_{n+1}-\lambda_ne_n\rVert^2=|\lambda_{n+1}|^2+|\lambda_n|^2\to 0$$ but actually you only know that a subsequence of $\{ \lambda_n e_n \}_n$ converges, so surely you only know your statement, not for $n$, but for some $n_j$ forming the convergent subsequence $\{ \lambda_{n_j} e_{n_j} \}_j$? This gives $\lambda_{n_j} \rightarrow_j 0$, which doesn't imply $\lambda_n \rightarrow_n 0$. (As I said, you're the one doing the PhD, so apologies if I am mistaken!) $\endgroup$ – Sam T Nov 28 '14 at 15:36
  • $\begingroup$ @SmileySam There was indeed an inaccuracy, thanks for pointed this out. I've edited. $\endgroup$ – Davide Giraudo Nov 28 '14 at 18:32
  • $\begingroup$ Is it correct now? "You can extract a convergent subsequence" is different to "Every subsequence is convergent$. $\endgroup$ – Sam T Nov 28 '14 at 18:37
  • $\begingroup$ Yes it is; you extract a subsequence from a subsequence. $\endgroup$ – Davide Giraudo Nov 28 '14 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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