The definition of compact operator

Question

One of the definitions of compact operator is so: the closing of the image of unit Ball is compact. Can we take a sphere instead of a Ball?

Answer 1

Let $V$, $W$ be normed vector spaces. Consider the map $\Phi:\Bbb R\times V\to V$, $\Phi(t,x)=tx$, let $T:W\to V$ be a linear map, let $S$ be the unit sphere (in $W$), let $E$ be the closed unit ball (in $W$) and let $B$ be the open unit ball (in $W$). Then, $T(E)=\Phi\left([0,1]\times T(S)\right)$. If $\overline{T(S)}$ is compact, then so are $[0,1]\times \overline{T(S)}$ and $\Phi\left([0,1]\times\overline{ T(S)}\right)$. It follows that $\Phi\left([0,1]\times\overline{ T(S)}\right)$ is closed and it contains $\Phi\left([0,1]\times T(S)\right)=T(E)$. Therefore $T(E)$ is relatively compact.

On the other hand, if $T(B)$ is relatively compact, then so are $T\left(\frac12 S\right)$ and $T(S)=2T\left(\frac12 S\right)$.