# Trying to prove the inverse operator theorem using the closed graph theorem?

I am trying to show the inverse operator theorem follows from the closed graph theorem. I am trying to show it directly, rather than using that the closed graph theorem leads to the open mapping theorem, and then showing the inverse is bounded based on that.

Let $A:X\to Y$ be a bounded, linear, bijective operator between the Banach spaces $X$ and $Y$. Suppose $A$ is closed.

Let $y\in Y$ and let $(y_n)$ be a sequence in $Y$ such that $y_n \to y$. As $A$ is bijective we have that there exists $x$ and a sequence $(x_n)$ in $X$ such that $x_n \to x$, $Ax_n = y_n \to y = Ax$.

Then \begin{align} ||A^{-1}|| & = \sup_{||y||_Y = 1} ||A^{-1}y||_X \\ & = \sup_{||Ax||_y = 1} ||A^{-1}Ax||_X \\ & = \sup_{||Ax||_y = 1} ||x||_X \\ & = \lim_{n\to\infty} \sup_{||Ax_n||_y = 1} ||x_n||_X \\ & = ??? \end{align}

I don't see how to get a bound on $||A^{-1}||$, or where the fact that $A$ is closed gets used? It seems that $A$ being bijective provides the closed property, which is that for a sequence $(x_n)$ in $X$ such that $x_n \to x\in X$ and $Ax_n \to y \in Y$, then $y = \lim_{n\to \infty} Ax_n = Ax$. If its bijective surely this holds automatically..

So how does the CGT get used, and how can I show $A^{-1}$ is bounded?

Since we aim to apply the closed graph theorem, there is no need to get a bound on $\|A^{-1}\|$. We just apply the theorem.
Suppose that $A:X\to Y$ is a bounded linear bijection between Banach spaces. Due to the closed graph theorem, it suffices to prove that the graph $\Gamma(A^{-1})$ of $A^{-1}$ is a closed subset of $Y\times X$. To this end, suppose $(y_n)$ is a sequence in $Y$ such that $y_n\to y$ in $Y$ and $A^{-1}y_n\to x$ in $X$ for some $y\in Y$ and $x\in X$. We need to show that $A^{-1}y=x$. The continuity of $A$ yields that $y_n=AA^{-1}y_n\to Ax$, and thus by uniqueness of limits, $y=Ax$. Therefore $A^{-1}y=x$. This completes the proof that $\Gamma(A^{-1})$ is closed, and consequently that $A^{-1}$ is bounded.
• Did we ever use that $A$ is closed in this argument? Thanks! – csiqrucalree Nov 13 '20 at 4:35