Compact linear operators composition on Banach space implies that one of them is compact?

Hi i just have some question regarding this problem , I cannot find a counter example and neither a proof:

Let $X$ be a Banach space of infinite dimension, and $S,T\in B(X)$ (the set of bounded linear operator from $X$ to $X$) is it true that:

if $ST$ is compact then $S$ or $T$ is compact?

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There exist Banach spaces where the answer is yes. gowers.wordpress.com/2009/02/07/… – Jonas Meyer Nov 28 '12 at 2:40

No. Let $X=\ell^2(\mathbb N)$. Define operators $S,T$ by $$Se_k=\begin{cases}e_k,&\text{ if }k\in2\mathbb N\\ 0,&\text{ if }k\not\in2\mathbb N \end{cases}$$ (where $e_1,e_2,\ldots$ is the canonical basis) and $T=I-S$. In matrix form, $$S=\begin{bmatrix}1\\&0\\& & 1\\ & & & 0 \\ & & & & & \ddots\end{bmatrix},\ \ \ T=\begin{bmatrix}0\\&1\\& & 0\\ & & & 1 \\ & & & & & \ddots\end{bmatrix}$$ Then $ST=0$, but neither is compact (both are projections with infinite-dimensional range).

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Even if $T^2=0$, $T$ need not be compact. E.g. modifying Martin's example to $Te_k=\begin{cases}e_{k-1},&\text{ if }k\in2\mathbb N\\ 0,&\text{ if }k\not\in2\mathbb N \end{cases}$ satisfies this. Operators $T$ such that there exists a nonzero polynomial $p$ with $p(T)$ compact are called polynomially compact. – Jonas Meyer Nov 28 '12 at 2:32
Good point!${\ }$ – Martin Argerami Nov 28 '12 at 4:32