# Show that if $T: X \to Y$ is a bijective, closed operator, then $T^{-1}$ is bounded.

Suppose that $X$ and $Y$ are two normed spaces, where $X$ is complete. Show that if $T: X \to Y$ is a bijective, closed operator, then $T^{-1}$ is bounded.

My idea is to prove that $Y$ is a Banach space, and then I may use the Closed Graph theorem to conclude that $T$ is bounded, and then the Inverse Mapping theorem for boundedness of $T^{-1}$.

I have shown that $T$ maps compact subsets of $X$ to closed subsets of $Y$. And I (think) that I can shown that $T^{-1}$ is also a closed operator.

I think I am on the wrong footing here, so I would appreciate hints. This question is already on the site but is still open, and the given answer is, to my understanding, vague.

• A compact normed space? – Thomas Dec 2 '16 at 14:21
• (Did you intend to write locally compact?) – Thomas Dec 2 '16 at 14:22
• If $X$ is locally compact, it is finite dimensional. Any linear map whose domain is finite-dimensional is continuous. – ncmathsadist Dec 2 '16 at 14:24
• A bijective closed function has continuous inverse almost by definition. Indeed a function is continuous if and only if preimages of closed sets are closed. – Crostul Dec 2 '16 at 14:24
• What about $X$ being complete ? – daw Dec 2 '16 at 14:35

Try a concrete example where you know $T$ does not have a bounded inverse. For example, suppose $X=\ell^2$ and define $T : X\rightarrow X$ by $$Tx = \sum_{n=0}^{\infty}\frac{1}{2^n}\langle x,e_n\rangle e_n$$ Define $Y=TX$ with the norm of $\ell^2$. Then $T : X\rightarrow Y$ is a bijection, and it's bounded, which makes it closed. To see why, suppose $\{ x_n \}$ converges to some $x\in X$ and $\{ y_n =Tx_n \}$ converges to some $y \in Y$. Then, because $T$ is bounded, $y_n = Tx_n \rightarrow Tx$, giving $Tx=y$.
But the inverse $T^{-1}$ is not bounded because $T^{-1}e_n=2^ne_n$.