Closure of operator defined by extensions I am study (first time) about closure of operator and when I was reading about it, I thought in question below. I don't know if this is true or false, I think is true, but I don't know how prove it or find a counterexample.
Please somebody help me. Thanks in advance.

Let $X,Y$ Banach and $T:\mathcal{D}(T)\subseteq X\rightarrow Y$ a linear operator. Define the operator $\overline{T}:\mathcal{D}(\overline{T})\subseteq X\rightarrow Y$ such that


*

*$\overline{T}$ is linear closed operator

*$\overline{T}$ is an extension of $T$, i.e, $\mathcal{D}(T)\subseteq\mathcal{D}(\overline{T})$ and $T(x) = \overline{T}(x)$ for all $x\in\mathcal{D}(T)$.

*If $\hat{T}:\mathcal{D}(\hat{T})\subseteq X\rightarrow Y$ is any operator with properties 1. and 2., then $\hat{T}$ is an extension of $\overline{T}$.


Show that $\mathcal{G}(\overline{T}) = \overline{\mathcal{G}(T)}$. Where $\mathcal{G}(A)$ is the graph of operator $A$.

 A: We will distinguish two cases: Either $\def\cl{\mathop{\mathrm{cl}}\nolimits}\def\Gr{\mathcal G}$ $\cl_{X \times Y} \Gr(T)$ is the graph of an operator or not ;-). In the first case we will show that the operator having this graph has the properties 1.-3., in the second case we will show that no such operator exists, as in this case, $T$ does not have any closed extension.
So suppose that $G:=\cl_{X\times Y} \Gr(T)$ is the graph of an operator $\bar T \colon X \to Y$. Being the closure of a subspace, $G$ is a subspace, hence $\bar T$ is linear. To show that $\bar T$ is closed, suppose $x_n \to x$, $\bar Tx_n \to y$. Then $(x_n, \bar Tx_n) \to (x,y)$. As $G$ is closed, we have $(x,y) \in G$, hence $y = \bar Tx$ and $\bar T$ is closed. Obviously $\bar T$ is an extension of $T$. Now suppose $S \supseteq T$ is a closed extension. Let $(x,y) \in G$, then there is a sequence $(x_n) \in {}^{\mathbb N}X$ such that $(x_n, Tx_n) \to (x,y)$. So $x_n \to x$, $Sx_n = Tx_n \to y$. As $S$ is closed, we must have $Sx = y$, that is $(x,y) \in \Gr(S)$. So $S \supseteq \bar T$.
If $G$ (as above) is not a graph, there is some $x \in X$, and $y\ne y'\in Y$ such that $(x,y), (x,y') \in G$. As $G$ is a subspace, we have $(0, y) \in G$ for some $y \ne 0$, that is for some $x_n \to 0$, $Tx_n \to y$. Then $T$ has no closed extension: Suppose $S \supseteq T$ were closed. Then, as $(x_n, Sx_n) \to (0, y)$, we have $S0 = y$, but, as $T$ is linear $S0 = T0 = 0$. Contradiction.
