If $T^2$ is a compact operator on a Banach space then is $T$ also compact?

Question

If $T^2$ is a compact operator on a Banach space $X$ then is it true that $T$ is also compact operator on X?

Here $T:X\to X.$

First I was trying to prove(assuming it is true). Since $X$ is Banach the set of all compact operators is closed in $\beta(X;X)$ ($\beta(X;X)$ is the set of all bounded operators in X). So I was trying to find a sequence $\{T_n\}$ in $\beta_0(X;X)$ (the set of all compact operators on $X$) which would converge to $T$ . Then using the given condition that $T^2\in\beta_0(X;X)$, we could say that $T\in \beta(X;X)$.

So the main point is to find a suitable sequence $\{T_n\}$ from $\beta_0(X;X)$ that will converge to $T.$

Please someone help . Can we find such sequence or the statement is false?

Thank you..

Answer 1

It is not true. Take, for instance $T\colon\ell^\infty\longrightarrow\ell^\infty$ defined by$$T(a_1,a_2,a_3,a_4,a_5,a_6,\ldots)=(a_2,0,a_4,0,a_6,0,\ldots).$$Then $T$ is not compact (with respect to the $\|\cdot\|_\infty$ norm), but $T^2=0$, which, of course, is compact.