Application of closed graph theorem: proving continuity of $F : H \to H$ satisfying $\langle{Fx, y\rangle} = \langle{x, Fy\rangle}$

Question

Let $H$ be a hilbert space and let $F: H \to H$ be any map. Prove by using the Closed Graph Theorem that if $\langle{Fx, y\rangle} = \langle{x, Fy\rangle}$ for all $x, y \in H$, then $F$ is continuous and linear.

Answer 1

First, to show that $F$ is linear, $$ \langle F(\alpha x + \beta y),z\rangle=\langle \alpha x+\beta y,Fz\rangle \\ = \alpha \langle x,Fz\rangle+\beta\langle y,Fz\rangle \\ = \alpha\langle Fx,z\rangle+\beta\langle Fy,z\rangle \\ = \langle \alpha Fx + \beta Fy,z\rangle. $$ Because this holds for all $z$, then $F(\alpha x+\beta y)=\alpha Fx+\beta Fy$. So $F$ is linear.

To prove that $F$ is closed, suppose that $\{x_n\}$ converges to $x$ and $\{Fx_n\}$ converges to $y$. Then we must show that $Fx=y$. To do this, note that the following holds for all $z$: $$ \langle Fx,z\rangle = \langle x,Fz\rangle=\lim_n \langle x_n,Fz\rangle=\lim_n\langle Fx_n,z\rangle = \langle y,z\rangle. $$ Therefore $Fx=y$. So $F$ is closed. It follows that $F$ is continuous by the Closed Graph Theorem.