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Let $A, B$ and $C$ be Banach spaces, $T: \operatorname{dom}(T)\rightarrow B$ be a closed linear operator with $\operatorname{dom}(T) \subset A$ and let $S \in \operatorname{BL}(B,C)$ be invertible.
Can you now show that $ST$ is closed?

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The map $(a,b) \mapsto (a,Sb)$ is a homeomorphism $A \times B \to A \times C$ and it maps the graph of $T$ to the graph of $ST$. Therefore $T$ is closed if and only if $ST$ is closed.

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Yes, I believe so (if by invertible, you mean boundedly invertible, and by $BL$ you mean bounded linear operators).

First, recall the sequential characterization of closure: $R$ is a closed linear operator $A \to C$ if and only if, whenever $(\phi_n) \in \text{dom} (R)$ satisfies $$\phi_n \to \phi \, \text{(in $A$)}, \quad R\phi_n \to \psi \, \text{(in $C$)},$$ that $\phi \in \text{dom} (R)$ and $R \phi = \psi$.

If this is unfamiliar, see http://www2.math.ou.edu/~cremling/teaching/lecturenotes/fa-new/ln11.pdf (at the top). [The context here is Hilbert spaces, but graph-closures ought to be the same for Banach spaces.]

To test the case $R = ST$, note that if $ST \phi_n \to \psi$, then since $S$ is boundedly invertible, we note that $T\phi_n \to S^{-1}\psi$: indeed,

$$ \begin{align} \Vert T \phi_n - S^{-1} \psi \Vert_B & = \Vert S^{-1} (ST \phi_n -\psi) \Vert_B\\ & \leq \Vert S^{-1} \Vert_{C \to B} \cdot \Vert ST \phi_n - \psi \Vert_C \, , \end{align} $$ and since $ST \phi_n \to \psi$, $\displaystyle \lim_{n \to \infty} \Vert ST \phi_n - \psi \Vert_C = 0$, so by nonnegativity of norms and the Squeeze Theorem, $\displaystyle \lim_{n \to \infty} \Vert T \phi_n - S^{-1} \psi \Vert_B =0$, or $T\phi_n \to S^{-1} \psi$.

Yet you said that $T$ was closed. So by the above characterization, $\phi \in \text{dom}(T)$ and $T\phi = S^{-1} \psi$. Yet $S$ is bounded, so $\phi \in \text{dom}(ST)$, and by applying $S$ to both sides of $T\phi = S^{-1}\psi$, $ST \phi = \psi$. Case closed.

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  • $\begingroup$ +1: This surely is a much more helpful answer than mine. $\endgroup$ – user160629 Jun 30 '14 at 17:32
  • $\begingroup$ On the contrary, @user160629 , I up voted yours. Very short, actually uses the topology! $\endgroup$ – Charles Baker Jun 30 '14 at 22:30

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