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