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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 ?

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$S$ 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.

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  • $\begingroup$ Why should $S^{-1}$ be continuous (or $im(S)$ closed)? $\endgroup$ – daw Jan 26 '18 at 7:24
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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.

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  • $\begingroup$ Many thanks. Is it not necessary to suppose that $(x_n) \in K$, a vector subspace of $X,$ and then to prove that $x \in K$ ? $\endgroup$ – user249018 Jan 26 '18 at 8:55
  • $\begingroup$ There is no such requirement in Closed Graph Theorem. $\endgroup$ – user284331 Jan 26 '18 at 14:47

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