TITLE: Is image of ball of finite rank linear operator compact?
QUESTION [6 upvotes]: Let $X$ be a complex Banach space and $A:X\to \mathbb{C}^{n}$ be a continuous linear map. If $B_{X} = \{x\in X\,:\, ||x||\leq 1\}$ is a closed unit ball in $X$, is it true that $A(B_{X})$ is compact in $\mathbb{C}^{n}$? 
Since every finite rank operator is compact operator, we know that the closure $\overline{A(B_{X})}$ is compact, but I can't verify that the image $A(B_{X})$ itself is compact, although it seems true.

REPLY [5 votes]: Let $X=c_0$, the space of complex sequences that converge to 0 with the sup norm. Define $f:X \to \mathbb C$ by $f(c_n)=\sum \frac {c_n} {n^{2}}$ Then the image of the unit ball of $c_0$ is not closed because $|\sum \frac {c_n} {n^{2}}| \leq \sum \frac 1 {n^{2}}$ for all $(c_n)$ in this ball and $\sup \{ |f(c_n)|:||(c_n)|| \leq 1\} = \sum \frac 1 {n^{2}}$ but the value $\sum \frac 1 {n^{2}}$ is not attained (since we cannot have $|c_n| =1$ for all $n$).