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.