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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.

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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 $\lVert u_{n_k} - u \rVert \leq \varepsilon.$

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @AlephNull That's because each $v_{n,m}$ lies in $W,$ which is bounded in $V.$ $\endgroup$ – ktoi Dec 24 '18 at 16:22

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