0
$\begingroup$

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.

Thanks in advance for help!

$\endgroup$
  • 1
    $\begingroup$ Hint: a closed subset of a compact set is compact. $\endgroup$ – Robert Israel Feb 15 at 15:22
  • $\begingroup$ 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 $\endgroup$ – reflexive Feb 15 at 15:24
2
$\begingroup$

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: $$\overline{Tc(A)}=\operatorname{cl}\{Ty:y\in cA\}=\operatorname{cl}\{Tcx:x\in A\}=\operatorname{cl}\{cTx:x\in A\}=\overline{cT(A)}.$$

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – reflexive Feb 15 at 15:28
  • $\begingroup$ @Coherent yes I see I made some typos. Should be ok now $\endgroup$ – K.Power Feb 15 at 15:36
  • $\begingroup$ closure of T (closure of B(o,R)) = R(closure of T(closure of B)), for this part can you just be a bit more explicit! $\endgroup$ – reflexive Feb 15 at 15:39
  • $\begingroup$ @Coherent see my edit $\endgroup$ – K.Power Feb 15 at 15:51

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.