27
$\begingroup$

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?

$\endgroup$

1 Answer 1

27
$\begingroup$

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 $0$.

$\endgroup$
12
  • $\begingroup$ 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 OT
    Nov 28, 2014 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 OT
    Nov 28, 2014 at 15:36
  • $\begingroup$ @SmileySam There was indeed an inaccuracy, thanks for pointed this out. I've edited. $\endgroup$ Nov 28, 2014 at 18:32
  • $\begingroup$ Is it correct now? "You can extract a convergent subsequence" is different to "Every subsequence is convergent$. $\endgroup$
    – Sam OT
    Nov 28, 2014 at 18:37
  • $\begingroup$ Yes it is; you extract a subsequence from a subsequence. $\endgroup$ Nov 28, 2014 at 21:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .