Let $T:E\to F$ and $S:F\to G$ be bounded linear maps with closed range between the Banach spaces $E, F$ and $G$. Does $ST:E\to G$ also have closed range? I can't find any theorem that makes the proof obvious so, this being functional analysis, I'm assuming it is false. However, the counterexample is still elusive.
Can someone give a hint (or full answer) of some easy counterexample?
Thanks
Let $E = F = G = \ell^2$.
It is straightforward to see that the map $U: \ell^2 \to \ell^2$ given by $(Ux)_n = \frac{x_n}{n}$ does not have closed range. To get the desired example we will exhibit a decomposition of $U$ into two maps with closed range.
Let $Tx = (x_1, 0, x_2, 0, x_3, 0, \dots)$ and let $(Sx)_n = \frac{x_{2n-1}}{n} + x_{2n}$. It is clear that $T$ has closed range since convergence in $\ell^2$ implies coordinatewise convergence. The range of $S$ is all of $\ell^2$ since if $y \in \ell^2$ then we can define $x \in \ell^2$ by $x_{2n} = y_n$ and $x_{2n-1} = 0$ so that $Sx = y$. In particular, $S$ and $T$ both have closed range but $ST = U$ as desired.