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.
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If $X$ is not reflexive, then a counterexample can be formed using James' theorem. Since $X$ is not reflexive, there must exist a functional $f \in X^*$ such that $f$ fails to achieve its maximum on the closed unit ball. Take such a functional, and $f(B_X)$ becomes the open ball of $\mathbb{C}$.
Theo Bendit
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1Thanks! So is it true for reflexive spaces? – Seewoo Lee Jun 01 '18 at 07:39
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Yep, it's true in reflexive spaces. You can reduce back to the functional case (i.e. $n = 1$) by considering $f_1, \ldots, f_n \in X^*$ such that $A(x) = (f_1(x), \ldots, f_n(x))$. If $A(x_k) \to v$, then $f_i(x_k) \to v_i$ as $k \to \infty$. Using sequential weak compactness of $B_X$ yields the result. – Theo Bendit Jun 01 '18 at 09:42
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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 $n$).
Kavi Rama Murthy
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