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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 $Ax_0=y_0$.

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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 $X \times Y$.

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  • $\begingroup$ 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. $\endgroup$ – Apocalypse Dec 1 '18 at 20:57
  • $\begingroup$ @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 $Y$. $\endgroup$ – mechanodroid Dec 1 '18 at 21:06
  • $\begingroup$ @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. $\endgroup$ – mechanodroid Dec 1 '18 at 21:08
  • $\begingroup$ Ohhhh! Indeed! I forgot that the second part $\|Ax-Ax_n\|$ in graph norm already requires $Ax_n\to Ax$. Thank you so much! $\endgroup$ – Apocalypse Dec 1 '18 at 21:15
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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.

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  • $\begingroup$ I know where I got myself confused! Thank you! $\endgroup$ – Apocalypse Dec 1 '18 at 21:24

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