# A linear operator is compact if and only if the image of any bounded set is relatively compact

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 $$\lVert u_{n_k} - u \rVert \leq \varepsilon.$$
• 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. – AlephNull Dec 24 '18 at 15:09
• Maybe something like $\max(T(v_{1,n}),T(v_{2,n}),\ldots,T(v_{n,n}))$. Not sure though. – AlephNull Dec 24 '18 at 15:16
• @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. – ktoi Dec 24 '18 at 15:19
• 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? – AlephNull Dec 24 '18 at 16:09
• @AlephNull That's because each $v_{n,m}$ lies in $W,$ which is bounded in $V.$ – ktoi Dec 24 '18 at 16:22