Closed Graph Theorem: let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a linear map with closed graph. Then $T$ is continuous.
Definition: let $X$ and $Y$ be two normed spaces and let $T : X \to Y$ be a mapping. We say that $T$ is an isomorphism if it is continuous, opened and bijective.
Banach Isomorphism Theorem: let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a linear bijective continuous map. Then $T$ is isomorphism.
Is it possible to show the Banach Isomorphism theorem using Closed Graph theorem? it should be sufficient if we had the graph from $T^{- 1}$, $G(T^{-1}) = \{(x , y) \in X \times Y : x = T^{- 1}(y)\}$ is closed in $X \times Y$, but I don't think that it is as easy.
Best Answer
If $$G=\{(y,x) \in Y \times X: x=T^{-1}(y)\}$$ and we take a sequence $\{(y_n,x_n)\} \subset G$ such that is convergent to some point $(y,x) \in Y \times X$ then we have that $x_n=T^{-1}(y_n)$ and as $T$ is bijective, $y_n=T(x_n)$.
So, on the one hand $T(x_n) \rightarrow y$ but on the other hand, as $T$ is continuous, $T(x_n) \rightarrow T(x)$, so $T(x)=y$ and then $x=T^{-1}(y)$. We conclude then that $(y,x) \in G$ and $G$ is closed.