Compact projection on Banach space has finite rank

Question

Let $E$ be a Banach space. Show that every compact projection has finite rank.

I have no idea where to start.

Answer 1

Hint: If $P$ is a projection the image of the closed ball is the closed ball in the space you are projecting onto.

Answer 2

Assume the projection $\pi$ has infinite rank, and let $X:=\pi(E)$. Let $x_n$ be an orthonormal base of $X$. Clearly $x_n$ belong to the unit ball of $E$, but contains no Cauchy sequence.

Answer 3

Let $B\subset P(X)$ be a closed and limited subset in $P(X)$, if we have that B is compact, then $P(X)$ is finite-dimensional. We have $P^2=P$, then $P(B)=B$ and it implies that $\overline{P(B)}$ is compact, since $P$ is compact and $B$ is limited. Hence B is closed, then $\overline{P(B)}=\overline{B}=B$ and $B$ is compact, so $P(X)$ has finite rank.

Answer 4

If some Fredholm theory is assumed this is immediate. If $u: E \to E$ is a compact operator then $u-\lambda$ is Fredholm for any $\lambda \in \mathbb{C} \setminus \{0\}$ (here we only need that $\ker(u-\lambda)$ is finite dimensional). Now let $p: E \to E$ be any compact idempotent, we have $p(E) = \ker(1 - p)$ and we are done.