Let $U$ and $V$ be normed spaces. Show that a linear operator $T:V \to U$ is compact if and only if the image of any bounded set is relatively compact.

(Here $T$ is compact if the image of any bounded sequence has a convergent subsequence, and a relatively compact set is one whose closure is sequentially compact, which means every sequence in the closure has a convergent subsequence.)

The backward direction is straightforward: take any bounded sequence $v_1,v_2,\ldots$ in $V$. Then $W=\{v_1,v_2,\ldots\}$ is a bounded set, so $\overline{T(W)}$ is sequentially compact and it contains the sequence $T(v_1),T(v_2),\ldots$, so this sequence has a convergent subsequence. Therefore $T$ is compact.

I don't know how to do the forward direction though. One would consider a sequence in $\overline{T(W)}$ (where $W$ is a bounded set) and somehow use compactness of $T$ to exhibit a convergent subsequence. But these elements aren't even necessarily in $T(W)$; they're just limit points. I thought about taking sequences in $T(W)$ that converge to each element but couldn't get this to work. Help would be appreciated.

There's also this lemma that if $X,Y$ are normed spaces, $Y$ Banach, then a linear operator $A:X \to Y$ is compact if and only if the image of the unit sphere is compact. I can't prove the non-trivial direction of this. But I assume this isn't even useful here because $U$ is not assumed to be Banach.


Given $T$ compact and $W \subset V$ bounded, let $\{u_n \} \subset \overline{TW}$ be some sequence. The idea is to approximate each $u_n$ by elements in $TW,$ and extract a diagonal subsequence using compactness.

By definition, for each $n$ there is $\{v_{n,m}\} \subset V$ we can find $\lVert Tv_{n,m} - u_n\rVert \leq 2^{-m}$ for all $m,$ so in particular $Tv_{n,m} \rightarrow u_n$ as $m \rightarrow \infty.$ Then set $\tilde v_n = v_{n,n},$ so $\{\tilde v_n\}$ is a bounded sequence in $V$ and so $\{T\tilde v_n\}$ has a convergent subsequence $T\tilde v_{n_k} \rightarrow u \in \overline{TW}$ by compactness of $T.$ We then claim that $u_{n_k} \rightarrow u$ also; for this let $\varepsilon > 0.$ Then there is $N$ such that $n_k \geq N$ implies $\lVert Tv_{n_k,n_k} - u \rVert \leq \varepsilon /2.$ By increasing $N$ so $2^{-N} \leq \varepsilon/2$ also, we get $n_k \geq N$ implies $\lVert u_{n_k} - Tv_{n_k,n_k} \rVert \leq 2^{-n_k} \leq \varepsilon/2$ and hence $\lVert u_{n_k} - u \rVert \leq \varepsilon.$

  • $\begingroup$ Yes, with what I said a diagonal sequence would be the natural follow-up. But I have a few questions about this solution. I suppose you mean $T(v_{n,m}) \to u_n$. Why is $(Tv_{n,n})$ a bounded sequence? I see that $(T(v_{n,m}))$ in $m$ is a bounded sequence because it converges. Not sure about the diagonal sequence though. Also I don't understand how $||u_{n_k}−Tv_{n_k,n_k}||\leq\frac{\epsilon}{2}$ holds. I may be wrong, but it feels like a diagonal sequence isn't sufficient. $\endgroup$ – AlephNull Dec 24 '18 at 15:09
  • $\begingroup$ Maybe something like $\max(T(v_{1,n}),T(v_{2,n}),\ldots,T(v_{n,n}))$. Not sure though. $\endgroup$ – AlephNull Dec 24 '18 at 15:16
  • 1
    $\begingroup$ @AlephNull You're right, what I posted originally was incomplete. I've edited my proof to account for the issues you pointed out. The extra ingredient is that I can approximate each $u_n$ uniformly, because I have complete freedom in how I choose by $v_{n,m}$'s. $\endgroup$ – ktoi Dec 24 '18 at 15:19
  • $\begingroup$ Ah, that's a very nice modification, I'll make sure I remember that. I just have one more question, why is $\{\tilde{v}_n\}$ a bounded sequence? $\endgroup$ – AlephNull Dec 24 '18 at 16:09
  • 1
    $\begingroup$ @AlephNull That's because each $v_{n,m}$ lies in $W,$ which is bounded in $V.$ $\endgroup$ – ktoi Dec 24 '18 at 16:22

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.