# 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: $$\Vert x \Vert_A:=\Vert x \Vert_X+\Vert Ax\Vert_Y$$ 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. $$\Vert x_n-x_m \Vert_A=\Vert x_n-x_m\Vert_X+\Vert Ax_n -Ax_m\Vert_Y<\varepsilon$$ 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. $$\Vert (x_n,y_n)-(x,y) \Vert=\Vert x_n-x\Vert_X+\Vert y_n-y\Vert_Y<\varepsilon$$ 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.

The following are standard definitions and results ...

1. 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 $X \oplus Y$
2. $G(A)$, 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.
3. When $dom(A)$ is a linear subspace of $X$ then $G(A)$ is a linear subspace of $X \oplus Y$
4. The A-norm, $||x||_A = ||x||_X + ||A(x)||_Y$ is a norm on $dom(A)$
5. 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 $X \oplus Y$.

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.

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.

• One-liners rarely make good answers. Please expand on your answer to show how it relates to the larger problem, showing that the graph of $A$ is closed if and only "$dom(A)$ is a Banach space w.r.t. the graph norm." – hardmath Aug 3 '17 at 13:26