How is the Hahn-Banach theorem is applied here?

Question

I am trying to understand how the Banach theorem is applied to get the following highlighted conclusions.

The only way I could think of is the following: Since the dual $X^*$ is a normed space, if $x^*_1,x^*_2 \in X^*$ are distinct functional, then by the Hahn-Banach theorem, there exists $x \in X^{**}=X$ (by reflexivity) such that $\langle x,x^*_1 \rangle \neq \langle x,x^*_2 \rangle$ which clearly is what is concluded here. Any help, please?

Answer 1

As the proof says, $(0,x_0)$ is not in the (closed) graph of $S$. So you want to apply Hahn-Banach on the space $X\times X$. It is easy to check that you can identify $(X\times X)^*$ with $X^*\times X^*$ via $$\tag1 \langle (x_1^*,x_2^*),(x_1,x_2)\rangle=\langle x_1^*,x_1\rangle+\langle x_2^*,x_2\rangle $$ and $\|(x_1^*,x_2^*)\|=\|x_1^*\|+\|x_2^*\|$.

So Hahn-Banach tells you that there exists $(x_1^*,x_2^*)\in (X\times X)^*$ such that $\langle(x_1^*,x_2^*),(0,x_0)\rangle\ne0$ and $(x_1^*,x_2^*)$ is $0$ on the graph of $S$, which means that $$ \langle x_1^*,0\rangle+\langle x_2^*,x_0\rangle=0 $$ and $$ \langle x_1^*,x\rangle+\langle x_2^*,Sx\rangle=0 $$ for all $x\in D(S)$. You can get the minus sign if replace the duality $(1)$ with
$$\tag2 \langle (x_1^*,x_2^*),(x_1,x_2)\rangle=\langle x_1^*,x_1\rangle-\langle x_2^*,x_2\rangle. $$

Answer 2

Consider the graph of $S,$ i.e. $\Gamma_S:=\{(x,Sx)\,:\, x\in X\}\subset X\times X.$ We have $(0,x_0)\notin \Gamma_S.$ Therefore there exists a linear functional $\varphi$ on $X\times X$ such that $\varphi(x,Sx)=0$ for any $x\in X$ and $\varphi(0,x_0)\neq 0.$ We have $(X\times X)^*=X^*\times X^*.$ Hence $\varphi=(x_1^*,x_2^*),$ and $\varphi(x,x')=x_1^*(x)+x_2^*(x').$ Thus $$0=\varphi(x,Sx)=x_1^*(x)+x_2^*(Sx),\quad 0\neq \varphi(0,x_0)=x_1^*(0)+x_2^*(x_0)$$