$T$ is a compact operator but not a finite rank operator

Question

Prove that if $X$ is a Banach Space and if $T$ is a compact operator but is not a finite rank operator, then $0 \in \bar {T(S_X)}$(where $S_X$ is the unite sphere).

Please provide some solution (hints). Thanks for your time.

Answer 1

That's wrong. Consider $X = \ell^2$ and $Tx = (\frac{x_n}{n+1})_{n \ge 0}$. Then $T \in K(\ell^2)$ as limit of the finite-rank operators $$T_n x = \left(x_0,\frac{x_1}2, \ldots, \frac{x_n}{n+1}, 0, \ldots\right) $$ but $T$ is one-to-one. Hence $Tx \ne 0$ for all $x \in S_X$.

But, $0 \in \overline{T(S_X)}$ does always hold. Suppose not, let $T \in K(X)$ with $\alpha := \inf_{x \in S_X} \|Tx\| > 0$. Then $T$ is one-to-one and has closed image, as for $Tx_n \to y$ we have that $$ \|x_n - x_m\| \le \alpha^{-1}\|Tx_n - Tx_m\| \to 0 $$ and hence $x_n \to x$ (as $X$ is complete), so $y = Tx \in T(X)$. Therefore $T^{-1}\colon TX \to X$ is bounded by $\alpha^{-1}$ and hence $\mathrm{Id}_X = T^{-1}T \in L(X)$ is compact. Hence $X$ is finite-dimensional, which contradicts the fact the $T$ is not of finite rank.