$T^2=T$ implies $\dim(T(E))<\infty$

Question

Let $T:E\to E$ be a compact operator verifying $T^2=T$. I have to prove that $\dim(T(E))<\infty$.

I have proved that $T(E)$ is closed, but I don't know how to continue. Thank you!

Answer 1

Let $T(x_n)$ be a bounded sequence in $T(E)$. Then $T(x_n)=T(T(x_n))$ has a convergent subequence. Hence every sequnce in the closed unit ball of $T(E)$ has a convergent subsequence. This implies that $T(E)$ is finite dimensional.

Answer 2

First note that $T$ acts as identity on $T(E)$. Indeed, for $x \in E$ we have $T(Tx) = T^2x = Tx$.

Assume that $T(E)$ is infinite-dimensional. Using Riesz lemma we can construct a sequence $(y_n)_n$ in the unit ball of $T(E)$ such that $\|y_n - y_m\| \ge \frac12$ for $n \ne m$.

We get that $(y_n)_n$ is a sequence in the unit ball of $E$ such that $(Ty)_n = (y_n)_n$ has no convergent subsequence. Hence $T$ is not compact. Contradiction.