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!!
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.