Let $X,Y,Z$ Banach spaces, $\text{dom}(S)\subset Y$, let $T:X\rightarrow Y$ be linear and continuous and let $S:\text{dom}(S)\rightarrow Z$ be linear and closed. Show that the composition $ST$ is also closed.
I think the open mapping theorem might be applicable, but I don't know weather $\text{dom}(S)\cap\text{Im}(T)$ is closed.
it is very straightforward. Let $(x_n) \in \mathrm{dom}(ST) = \{x \in X: Tx \in \mathrm{dom}(S)\}$, with $x_n \to x$ and $STx_n \to z$. Then, as $T$ is continuous, $Tx_n \to Tx$. Now $Tx_n \in \mathrm{dom}(S)$, and hence by closedness of $S$ and $S(Tx_n) \to z$ we have $Tx \in \mathrm{dom}(S)$ and $STx = z$. Hence $x \in \mathrm{dom}(ST)$ and $STx = z$.
So, $ST$ is closed.
By definition, a map as $S\circ T:T^{-1}\operatorname{dom}(S)\to Z$ is closed iff its graph $\Gamma(S\circ T)$ is closed in $X\times Z.$
Let be $\{x_n\}$ a sequence in $\operatorname{dom}(S\circ T)$ such that $x_n\to x$ and $S(Tx_n)\to y.$
Being $S$ closed by hypothesis, its graph $\Gamma(S)$ is closed in $Y\times Z$ therefore $Tx\in\operatorname{dom}(S)$ and $y=S(T(x)).$
So we get that $(x,y)=\lim_{n\to\infty}(x_n,(S\circ T)(x_n))\in\Gamma(S\circ T).$
By the arbitrariness of $(x_n)$ this means properly that $\Gamma(S\circ T)$ is closed in $X\times Z.$
