dom(A) is a Banach space w.r.t. the Graph-norm 
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.
 A: The following are standard definitions and results ...


*

*The product topology on $X \oplus Y$ is induced by the norm $||(x, y)||_{X \oplus Y} = ||x||_X + ||y||_Y$
And, if $X, Y$ are Banach so is $X \oplus Y$

*$G(A)$, the graph of an operator $A: dom(A) \subset X \to Y$ is $\{(x, A(x)): x \in dom(A)\} \subset X \oplus Y$.
To say $G(A)$ is closed means it is closed in the product topology.

*When $dom(A)$ is a linear subspace of $X$ then $G(A)$ is a linear subspace of $X \oplus Y$

*The A-norm, $||x||_A = ||x||_X + ||A(x)||_Y$ is a norm on $dom(A)$

*In a Banach space, closure of a subspace is equivalent to completeness.


So, consider the function $\phi:dom(A) \to G(A): \phi(x) = (x, A(x))$
It's clear that $\phi$ is a linear bijection.
And $||x||_A  = ||x||_X + ||A(x)||_Y = ||\phi(x)||_{X \oplus Y}$ so $\phi$ and its inverse are continuous (given any $\epsilon$ there is $\delta = \epsilon$ etc).  
Then $\phi$ is a homeomorphism and therefore preserves completeness (among other things)
I.e. with the topologies induced by the A-norm on $dom(A)$ and the product norm on $X \oplus Y$ then
$dom(A)$ is Banach $\Leftrightarrow$ $dom(A)$ is complete $\Leftrightarrow G(A)$ is complete $\Leftrightarrow  G(A)$ is closed in $X \oplus Y$. 
A: let $(x_n)$ be a cauchy sequence in dom(A) with respect to the graph norm. That is  for every $\varepsilon >0 $ exists an integer N s.t.
$$\| x_n -x_m \|_A= \|x_n - x_m\|_X + \|Ax_n -Ax_m \|_Y  \leq \varepsilon \quad n,m > N $$
Since X and Y are Banach spaces, we have Cauchy sequences $(x_n)$ and $(Ax_n)$  converge to x and y respectively.
There is no way to be sure that $y= Ax$. It's possible that the sequence $y_n $ converges to $y=Ax'$, but $(x_n , Ax_n)$ converges to (x , y) and graph(A) is closed so we have y=Ax.
By definition of convergent sequences $(x_n \rightarrow x)$ and $(Ax_n \rightarrow Ax)$,we have, for every $\varepsilon/2$ exists an integer $N=\max\{N_{x_n} , N_{Ax_n}\}$ s.t.
$$\| x_n - x \|_A = \|x_n - x \|_X + \| Ax_n - Ax \|_Y \leq \varepsilon/2 +\varepsilon/2 = \varepsilon \quad  \forall n > N$$
hence every Cauchy sequence in dom (A) is convergent.
(Your answer is correct, just needs rearrangment)
A: I don't follow your solution: you write "let $(x_n, y_n)$...", and then "let $(x_n)$...", which already makes it unclear what is going on. One can't "let" a thing be something twice. 
I would prove that $\operatorname{dom}(A)$ equipped with the graph norm is isomorphic to the graph of $A$, using the isomorphism $x\mapsto (x,Ax)$. So one of these is complete if and only if the other is. 
