If $\operatorname{ran}A$ is closed, is then $\operatorname{ran}A^2$ closed?

Question

Let $X$ be a Banach space or Hilbert space and $A : X\to X$ is bounded operator such that If $\operatorname{ran}A$ is closed. Is then also $\operatorname{ran}A^2$ closed?

I think not. Can anyone think of a counterexample?

Answer 1

An easy example: Let $X=E\oplus E$ with $E$ an infinite-dimensional Banach space, $T\colon E\to E$ be compact but not finite-rank. Then $$ A=\begin{pmatrix} T & 1_E\\ 0 & 0 \end{pmatrix}\colon X\to X $$ has $\operatorname{ran}A=E\oplus 0$ is closed, but $\operatorname{ran}A^2=\operatorname{ran}T\oplus 0$ is not closed.