# About closed graph of an unbounded operator

I am working on problems related to the closed graph of an unbounded operator. There is a proposition:

Let $$X,Y$$ be Banach spaces and let $$A:\mathrm{dom}(A)\to Y$$ be linear and defined on a linear subspace $$\mathrm{dom}(A)\subset X$$. Prove that the graph of $$A$$ is a closed subspace of $$X\times Y$$ if and only if $$\mathrm{dom}(A)$$ is Banach with respect to the graph norm.

I finished one direction. Suppose $$\mathrm{graph}(A)$$ is closed. We take any Cauchy sequence $$x_n$$ in $$\mathrm{dom}(A)$$, and since the norm is graph norm, we know $$x_n$$ and $$Ax_n$$ will both be Cauchy. Then we have a Cauchy sequence $$(x_n,Ax_n)$$ in the graph, so the pair converges to a certain $$(x_0,y_0)$$ since the graph is closed. Therefore $$x_0\in\mathrm{dom}(A)$$, which means that $$\mathrm{dom}(A)$$ is Banach.

However I encountered some trouble on the other direction. Suppose $$\operatorname{dom}(A)$$ is Banach with respect to the graph norm. If we take a Cauchy sequence $$(x_n,Ax_n)$$ in $$\mathrm{graph}(A)$$, since $$X,Y$$ are both Banach, it converges to a pair $$(x_0,y_0)\in X\times Y$$. Then we know $$x_n$$ converges to $$x_0$$ in the graph norm and so $$x_0\in\mathrm{dom}(A)$$, but this only tells $$(x_0,Ax_0)\in X\times Y$$. We still don't know whether $$Ax_0=y_0$$.

Let $$(x_n, Ax_n)$$ be a Cauchy sequence in $$\operatorname{graph}(A)$$. Then, by definition of the graph norm, $$(x_n)_n$$ is a Cauchy sequence in $$\operatorname{dom}(A)$$.
Since $$\operatorname{dom}(A)$$ is a Banach space w.r.t. the graph norm, $$(x_n)_n$$ converges to some $$x \in \operatorname{dom}(A)$$ w.r.t. the graph norm. This precisely means $$(x_n, Ax_n) \to (x, Ax)$$ in $$X \times Y$$. Hence, the sequence $$(x_n, Ax_n)$$ converges in $$\operatorname{graph}(A)$$ so $$\operatorname{graph}(A)$$ is a Banach space. In particular, it is a closed subspace of $$X \times Y$$.
• I think I just get stuck at "this precisely means". I kind of get it that since we already know that $x$ is in $\mathrm{dom}(A)$, $(x,Ax)$ must be in the graph of $A$, but somehow I still feel bad about it. It sounds stupid but if $x_n$ converges to $x$, why must $Ax_n$ converge to $Ax$? $A$ is not necessarily bounded. – Apocalypse Dec 1 '18 at 20:57
• @Apocalypse You seem to be confused by the convergence in the graph norm. The graph norm is for example defined as $\|x\|_A = \|x\|_X + \|Ax\|_Y$ for $x \in \operatorname{dom}(A)$. If $x_n \to x$ w.r.t. the graph norm, this means that $$\|x-x_n\|_X + \|Ax - Ax_n\|_Y = \|x-x_n\|_{A} \to 0$$ so in particular $x_n \to x$ in $X$ and $Ax_n \to Ax$ in $Y$. – mechanodroid Dec 1 '18 at 21:06
• @Apocalypse Indeed, if only $x_n \to x$ in $X$ (meaning $\|x_n - x\|_X \to 0$), we cannot conclude that $Ax_n \to Ax$ in $Y$ even if $x \in \operatorname{dom}(A)$ precisely because $A$ is unbounded. – mechanodroid Dec 1 '18 at 21:08
• Ohhhh! Indeed! I forgot that the second part $\|Ax-Ax_n\|$ in graph norm already requires $Ax_n\to Ax$. Thank you so much! – Apocalypse Dec 1 '18 at 21:15
Suppose $$\mathcal{D}(A)$$ is a Banach space in the graph norm of $$A$$. To show that $$A$$ is closed, suppose that $$x_n \rightarrow x$$ and $$Ax_n \rightarrow y$$, where $$\{ x_n\}\subset\mathcal{D}(A)$$. We want to show that $$x\in\mathcal{D}(A)$$ and $$Ax=y$$. Under these assumptions, $$\{ x_n \}$$ is a Cauchy sequence in the graph norm of $$A$$, which means that $$\{ x_n \}$$ converges in the graph norm to some $$x'\in\mathcal{D}(A)$$. So $$x_n\rightarrow x'$$ in $$X$$ and $$Ax_n\rightarrow Ax'$$ in $$Y$$. It follows that $$x=x'$$ and $$y=Ax'$$, which proves that $$A$$ is closed.