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The Question: Let $X$ and $Y$ be Banach spaces and $T: X \rightarrow Y$ an injective bounded linear operator. Show that if $R(T)$ is closed in Y, then $T^{-1} : R(T) \rightarrow X $ is bounded.

My attempt: So I was going to show that $T^{-1} $ is a continuous function. Therefore leading to showing $T^{-1} $ is bounded. Since $T$ is a bounded linear operator, $T$ is continuous. Is this the way to go?

Thank you very much!!

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You're right on track! Note that if we take a closed subset of a complete space, we obtain a complete space again. So, $R(T)$ is another Banach space. Therefore, we see that $T$ is a bijective bounded linear operator on its range. The main trick is to apply the open mapping theorem, which says that for surjective bounded linear operators in $\mathcal{L}(X, Y)$, with $X$, $Y$ Banach, open sets map to open sets.

To show that $T^{-1}$ is continuous, we can show that the pre-image of open sets are again open. So, $(T^{-1})^{-1} (U) = T(U) = O$, with $U$ open. Now, $O$ is also open by the open mapping theorem, so we see that $T^{-1}$ is indeed continuous.

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  • $\begingroup$ Okay! Now If T is an open map, does it imply that T is continuous? $\endgroup$ – Overachiever Nov 14 at 18:29
  • $\begingroup$ In general, no. There is no reason to think that the inverse image of open set should also be open again if you have an open map for a linear operator. I can't think of a counter-example right now, but I will come back to you when I know one. $\endgroup$ – JHoogendijk Nov 14 at 18:41

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