Theorem 9.3.4 (Hormander): Let $X_0,X_1,X_2$ be Banach spaces and let $T_1\colon X_0\to X_1$, $T_2\colon X_0\to X_2$ be linear maps such that $\text{Dom}T_1\subseteq\text{Dom}T_2$, $T_1$ is a closed graph operator and $T_2$ admits closure. Then, there exists $C>0$ such that
$||T_2x||\leq C(||(x,T_1x)||)$.
Proof. Since $T_1$ is a closed graph operator, the set
$$
E:=\{(x,T_1x):x\in\text{Dom}T_1\}\subseteq X_0\times X_1
$$
is a closed subspace of $X_0\times X_1$. Consider the operator $V\colon E\to X_2$ given by
$V(x,T_1x):=T_2x$. We claim that $V$ is itself a closed graph operator. Indeed, first consider the set
$$
\Gamma(V)=\{\left((x,T_1x),V(x,T_1x)=T_2x\right): (x,T_1x)\in E\}.
$$
Let $(x_n,T_1x_n)\to (x,y)$ and $T_2x_n\to z$. Our goal is to show that $(x,y)\in E$ and $z=T_2x$. Since $T_1$ is a closed graph operator, $x\in\text{Dom}T_1\subseteq\text{Dom}T_2$ and $y=T_1x$; hence, we have that $(x,y)\in E$. $\color{red}{\text{Since } x\in\text{Dom}T_2 \text{ and since } T_2 \text{ admits closure we have } z=T_2x}$.
Why is the text in red true (I know why $x\in\text{Dom}T_2$)?
Best Answer
If $S$ is the closure of $T_2$ then it is clear that $z=Sx$ (since $Sx_n \to z$ and $x_n \to x$). Once you know that $x \in \mathrm{Dom}(T_2)$ you can concludethat $Sx=T_2x$ so $z=T_2x$.