Let $E$ Banach space, $\dim E = \infty$ and $T: E \longrightarrow E$ a linear isometry. Show that $T$ is not compact.

Question

Let $E$ Banach space, $\dim E = \infty$ and $T: E \longrightarrow E$ a linear isometry. Show that $T$ is not compact.

Comments: I'm trying to solve it like this:

Suppose T is compact, then $\overline{T(B_E)}$ is compact. I can not use the fact that $T$ is linear isometry to prove that $B_E$ is compact and to obtain a contradiction.

Answer 1

Since $T$ is an isometry, the range $F$ of $T$ is closed, hence a Banach space. Restricting the codomain to $F$, we obtain a bounded surjection, hence an open map. But as $T$ is injective, $F$ is infinite-dimensional, hence it's closed unit ball is not compact. But $T(B_E)=B_F$, so $T(B_E)$ is not compact, and therefore $T$ is not compact.

Answer 2

Let $\{x_n\}$ be a sequence in the closed unit ball. Then $\{Tx_n\}$ has a convergent subsequence. The corresponding subsequence of $\{x_n\}$ is Cauchy, hence convergent. This makes the unit ball compact and the space finite dimensional.