self-adjoint operator is bounded in Hilbert space

Question

$X$ is a Hilbert space. $A$ is a linear and defined everywhere on $H$. $A$ satisfies $$<x,Ay>=<Ax,y>$$ for all $x,y\in H$, then $A$ is bounded. $<,>$ denote the inner product

Answer 1

There is another way to prove this: For fixed $y \in H$ let $T_y: H \rightarrow \mathbb{R}$ be the operator $$ T_y(x) := \langle Ax, y \rangle_H. $$ Because of symmetry and Cauchy-Schwarz, it is continuous ($x \in H$ arbitrary): $$ T_y(x) = \langle x, Ay \rangle_H \leq \lVert x \rVert_H \lVert Ay \rVert_H $$ Now let $$ \Gamma := \lbrace T_y \in L(H, \mathbb{R}): \lVert y \rVert_H = 1 \rbrace $$ be a family of linear operators. Then we have for fixed $x \in H$ by Cauchy-Schwarz: $$ \sup_{T_y \in \Gamma} \lvert T_y(x) \rvert = \sup_{\lVert y \rVert_H = 1} \lvert \langle Ax, y \rangle \rvert \leq \sup_{\lVert y \rVert_H = 1} \lVert Ax \rVert_H \lVert y \rVert_H = \lVert Ax \rVert_H \leq \infty $$

Now the Banach-Steinhaus theorem tells us: $$ \sup_{T_y \in \Gamma} \lVert T_y \rVert_{L(H, \mathbb{R})} = \sup_{\lVert y \rVert_H = 1} \sup_{\lVert x \rVert_H = 1} \lvert \langle Ax, y \rangle_H \rvert < \infty $$

Hence: $$ \sup_{\lVert x \rVert_H = 1} \lVert Ax \rVert_H = \sup_{\lVert x \rVert_H = 1} \frac{\lvert \langle Ax, Ax\rangle_H \rvert}{\lVert Ax \rVert_H} = \sup_{\lVert x \rVert_H = 1} \left \lvert \left \langle Ax, \frac{Ax}{\lVert Ax \rVert_H} \right \rangle_H \right\rvert \leq \sup_{\lVert y \rVert_H = 1} \sup_{\lVert x \rVert_H = 1} \lvert \langle Ax, y \rangle_H \rvert < \infty $$

So $A$ is bounded.