# A continuous composition of linear operators, when one is injective and continuous, implies the other to also be continuous.

Let $X, Y$ and $Z$ be Banach spaces, and $M: X \rightarrow Y$ linear. Let also $S: Y\rightarrow Z$ be linear, injective and continuous, and the composition of operators $SM: X \rightarrow Z$ continuous. I need to show that $M$ is also continuous.

I know that if $M, S$ are continuous, then the composition is continuous, and the Operator norm holds, $\left\lVert SM\right\rVert \leq \left\lVert S\right\rVert \left\lVert M\right\rVert,$ where the norms are taken in the corresponding operator spaces. The reversal must not be true. I know that i need to show that $M$ is bounded.

Can somebody provide a proposal of proof ?

$S$ gives an isomorphism $Y \cong \text{Im}\ S$. Since $S$ is continuous, its inverse $S^{-1}$ is continuous. We have that $M = S^{-1}SM$ is continuous as a composition of continuous operators.
• Why should $S^{-1}$ be continuous (or $im(S)$ closed)? – daw Jan 26 '18 at 7:24
Assume that $(x_{n})\subseteq X$ and $(Mx_{n})\subseteq Y$ are such that $x_{n}\rightarrow x$ and $Mx_{n}\rightarrow y$, then the continuity of $S$ implies that $SMx_{n}\rightarrow Sy$. The continuity of $SM$ implies that $SMx_{n}\rightarrow SMx$. So $SMx=Sy$, the injectivity of $S$ implies that $Mx=y$.
By Closed Graph Theorem, $M$ is bounded.
• Many thanks. Is it not necessary to suppose that $(x_n) \in K$, a vector subspace of $X,$ and then to prove that $x \in K$ ? – user249018 Jan 26 '18 at 8:55