Let $X$ and $Y$ be Banach spaces and let $A:dom(A)\to Y$ be a linear operator, defined on a linear subspace $dom(A)\subset X $. Prof that the graph of $A$ is a closed subspace of $X\times Y$ if and only if $dom(A)$ is a Banach space w.r.t. the graph norm.
The graph norm of A on the vector space $dom(A)$ is the norm function $dom(A)\to [0,\infty):x\mapsto\Vert x\Vert_A$ defined by: \begin{equation} \Vert x \Vert_A:=\Vert x \Vert_X+\Vert Ax\Vert_Y \end{equation} for $x\in dom(A)$.
Sketch of my solution: First assume that $graph(A)\subset X\times Y$ closed. So let $(x_n,y_n)\in graph(A)$ be a sequence with $\lim_{n\to \infty}(x_n,y_n)=(x,y)$ and let $(x_n)_{n\in \mathbb{N}}$ be a cauchy sequence in $dom(A)$ w.r.t. the graph norm. That is for every $\varepsilon>0$ it exists a integer $N$ s.t. \begin{equation} \Vert x_n-x_m \Vert_A=\Vert x_n-x_m\Vert_X+\Vert Ax_n -Ax_m\Vert_Y<\varepsilon \end{equation} We also have that, since $X$ and $Y$ are both Banach space and by assumption the graph(A) is closed, so $X\times Y$ is a Banach space too: Fix $\varepsilon > 0 $ s.t. \begin{equation} \Vert (x_n,y_n)-(x,y) \Vert=\Vert x_n-x\Vert_X+\Vert y_n-y\Vert_Y<\varepsilon \end{equation} where $\Vert y_n-y\Vert_Y=\Vert A(x_n-x)\Vert_Y$. And the last line is exactly the deffinition of the Graph-norm.
I'm really not sure about my solution, could someone tell me if I'm on the right track or not? Thank you.