# Why is the inverse of a bounded bijective operator continuous?

I'm facing some troubles with the following theorem,

Let $X,Y$ be Banach spaces and let $T \in B(X,Y)$. I want to show that if $T$ is bijective then its inverse is continuous. Now if $T$ is bijective than its surjectivity implies that it is open by the open mapping theorem. From here on I need do deduce that $T^{-1}$ is continuous but I don't see it really.

Can someone help me with this?

• The fact that a bijective function $f$ between topological spaces has continuous inverse if and only if it is open is an exercise of general topology. It really uses nothing but the definitions. – user228113 Mar 26 '17 at 9:38
For $y \in Y$, define $||y||_T = ||T^{-1}y||_X$. Then $||\cdot||_T$ is a norm on $Y$, and $I : (Y, ||\cdot||_T) \to (Y, ||\cdot||)$ is continuous (check this).
Check that $(Y, ||\cdot||_T)$ is a Banach space (take a sequence $y_n$ such that $\sum_{n=1}^\infty ||y_n||_T < \infty$, and show that $\sum y_n$ is an element of $Y$).
It is a simple exercise to see that when the inclusion is continuous, then the norms are equivalent. That is, there is a constant $c$ such that $||T^{-1}y||_X \leq c ||y||$, but this is what had to be proved. So, $T^{-1}$ is continuous.