TITLE: Closure of operators
QUESTION [6 upvotes]: Let $X$ and $Y$ Banach. We say that the linear operator $A:\mathcal{D}(A)\subseteq X\rightarrow Y$ admits a closure if there's a linear operator $B:\mathcal{D}(B)\subseteq X\rightarrow Y$ such that $\mathcal{D}(A)\subseteq \mathcal{D}(B)$, $B|_{\mathcal{D}(A)} = A$
and $\mathcal{G}(B)= \overline{\mathcal{G}(A)}$, where $\mathcal{G}(Z)$ is the graph of $Z$.

I want to prove that $A$ admits a closure if and only if every sequence $\{x_n\}_{n\in\mathbb{N}}\subseteq \mathcal{D}(A)$ such that $(x_n,A(x_n))\xrightarrow{n\rightarrow\infty}(0,y)$, with $y\in Y$, satisfy that $y=0$.

I proved ($\Rightarrow$), but I have had problems in ($\Leftarrow$), because I don't know how to define the operator $B$. Please, somebody can help me?
Thanks in advance.

REPLY [5 votes]: The operator $B$ is defined as follows:

If $x \in \mathcal{D}(A)$, then $Bx := Ax$.
If $x \notin \mathcal{D}(A)$ happens to be the limit of a sequence $\{x_n\}_n \subset \mathcal{D}(A)$ such that $\{Ax_n\}_n$ converges in $Y$---note that such a sequence need not exist in general, in which case we can't define $B$ on $x$---then $Bx := \lim_{n\to\infty}Ax_n$.

The point, then, is to check that $B$ is actually well-defined in the second case. So, suppose that $x \notin \mathcal{D}(A)$ is the limit both of $\{x_n\}_n \subset \mathcal{D}(A)$ and of $\{x_n^\prime\}_n \subset \mathcal{D}(A)$, such that $\{Ax_n\}_n$ and $\{Ax_n^\prime\}_n$ converge in $Y$. Then $$\{(x_n-x_n^\prime,A(x_n-x_n^\prime))\}_n = \{(x_n-x_n^\prime,Ax_n - Ax_n^\prime)\}_n \to_{n\to\infty} \left(0,\lim_{n\to\infty}Ax_n - \lim_{n\to\infty}Ax_n^\prime\right);$$
what can you now conclude?