# $T :N \to N_1$ is compact iff closure of $T(S)$ is compact $\forall S$ bounded subset of $N$

In a normed linear space N, let $$B=\{x \in N : |x| <1\}$$ . Then define a linear map $$T :N \to N_1$$ ,( $$N_1$$ some normed linear space ) to be compact if closure of $$T(\bar{B})$$ is compact in $$N_1$$ . With this definition I was trying to prove :

$$T :N \to N_1$$ is compact iff closure of $$T(S)$$ is compact $$\forall S$$ bounded subset of $$N$$

Assuming, closure of $$T(S)$$ is compact $$\forall S$$ bounded subset of $$N$$, it follows that $$T$$ is compact by putting, $$S=\bar{B}$$ .

For the converse, I have tried : $$S$$ bounded $$\implies \exists R > 0$$ s.t. $$S \subset B(0,R) \subset \text{ closure of } B(0,R) \implies T(S) \subset T(\text{ closure of } B(0,R)) \implies \text{ closure of } T(S) \subset \text{ closure of } T(\text{ closure of } B(0,R))$$

So all I need to show is that $$\text{ closure of } T(\text{ closure of } B(0,R))$$ is compact. If $$\phi$$ be the scaling function that is, $$x \mapsto Rx$$ , then $$\phi(\bar{B}) = T(\text{ closure of } B(0,R))$$.

I am trying to use that fact that $$\phi$$ is a homeomorphism, but not quite sure how to finish the proof.

• Hint: a closed subset of a compact set is compact. – Robert Israel Feb 15 at 15:22
• Yeah I know that, and using it I have reduced that closure of 𝑇( closure of 𝐵(0,𝑅)) is all I need to prove – reflexive Feb 15 at 15:24

You are almost there: We know that $$\overline{T(S)}\subset \overline{T(\overline{B(0,R)})}$$, and also that $$\overline{B(0,R)}=R\overline{B(0,1)}$$. Thus $$\overline{T(\overline{B(0,R)})}=R\overline{T(\overline{B(0,1)})}$$ by the linearity of $$T$$, but as you have stated scalar multiplication is a homeomorphism, which preserves compactness, and by assumption $$\overline{T(\overline{B(0,1)})}$$ is compact. Thus $$\overline{T(S)}$$ is a closed subset of a compact set.
Crucially here we used the fact that for any $$c\in \mathbb C$$ we have, for any nonempty $$A\subset X$$, $$\overline{Tc(A)}=c\overline{T(A)}$$. This follows from the fact that scalar multiplication is a homeomorphism and from definition basically: $$\overline{Tc(A)}=\operatorname{cl}\{Ty:y\in cA\}=\operatorname{cl}\{Tcx:x\in A\}=\operatorname{cl}\{cTx:x\in A\}=\overline{cT(A)}.$$
• Note that closure of 𝑇(𝑆)⊂ closure of 𝑇( closure of 𝐵(0,𝑅)) not that closure of 𝑇(𝑆)⊂ closure of 𝐵(0,𝑅)). Moreover, since $T$ may be between two different n.l.s and thus your statement doesn't make any sense! – reflexive Feb 15 at 15:28