If $T: X\to Y$ is a bijective closed linear operator, show that $T^{-1}$ is bounded.

Question

Let $X$ and $Y$ be normed spaces and $X$ compact. If $T: X\to Y$ is a bijective closed linear operator, show that $T^{-1}$ is bounded.

I don't know where to start here. Any help would be appreciated.

Answer 1

Select a compact set $C \subset X$ (which is closed and bounded). Therefore the preimage of $T^{-1}$ is also compact because $(T^{-1})^{-1}(C) = T(C) \subset Y$. Why? Because the preimage of $T$, $T^{-1}T(C) = C$ since $T$ is bijective, so the set $(T^{-1})^{-1}(C)$ is also compact (which is closed and bounded) meaning $T^{-1}$ is continuous, so $T^{-1}$ must be bounded.

I think there is a better argument where you can use the definition of being a closed linear operator, get continuity out of that, and then apply compactness to this.

Answer 2

To show that $T^{-1}$ is bounded it is enough to show that $T^{-1}$ is continuous.

Now $T^{-1}:Y\to X$

Select a closed set $C\subseteq X$ then since $C$ is closed hence it is compact .

Since $T$ is a closed linear operator so image of a compact set under $T$ is closed.

Hence $(T^{-1})^{-1}(C)=T(C)$ is closed .