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?
How to prove that an operator is compact?
1 Answers
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 $0$.
-
0Sorry 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!) – 2014-11-28
-
0At 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!) – 2014-11-28
-
0@SmileySam There was indeed an inaccuracy, thanks for pointed this out. I've edited. – 2014-11-28
-
0Is it correct now? "You can extract a convergent subsequence" is different to "Every subsequence is convergent$. – 2014-11-28
-
0Yes it is; you extract a subsequence from a subsequence. – 2014-11-28
-
0Surely though you can always change the sequence to have every term 2 mod 4 as $10$ and 0 mod 4 as $20$. This only changes your subsequence, but doesn't all tend to zero? What I'm meaning is, you "miss out a lot of entries" when taking the first subsequence, then no matter how many subs you take of the sub, you'll never "get them back". – 2014-11-28
-
0What I meant is that given a subsequence of $(\lambda_j)$, you can extract from this subsequence a further subsequence which converges to $0$. This is equivalent to the fact that the whole sequence converges to $0$, is it? – 2014-11-28
-
0Ah ok, I didn't actually know that that was a thing, so thanks! So it's true that if you have a sequence $(\lambda_n)_n$ such that given any subsequence $(\lambda_{n(j)})_j$ there exists a further subsequence $(\lambda_{n(j(k))})_k$ converging to $0$, then $(\lambda_n)_n$ converges to $0$? Yes, I can well believe that! Thanks for your answer to this and another related question. Most helpful for me today! – 2014-11-28
-
0Yes it's true. It has been discussed in this site, maybe you will be faster than me to find a link. – 2014-11-28
-
0Yeah, that's fine. I believe it - I'll either look it up tomorrow or work it out myself. Thanks. :) – 2014-11-28
-
0Hello, I have this confusion... in an infinite dimensional hilbert space we can not express an element as a (possible infinite) sum of basis vectors correct? So then how come you are expressinng v as infinite sum of basis vectors? Please help clear my confusion. thanks – 2017-04-04