If $A$ is an operator linear on a Hilbert space $H$ and there exists an operator $B$ on $H$ such that $\langle Ax,y\rangle=\langle x,By\rangle$ for all $x,y\in H$, show that $A$ is bounded and $B=A*$.
What I thought was to take advantage of the fact that H is a Banach space and A is a linear operator. So I wanted to prove that A is a closed operator and then use the closed graph theorem and conclude that A is continuous. But I haven't been able to do it.
On
Closed graph theorem can be applied as well. Assume $x_n\to x$ and $Ax_n\to z.$ Then for any $y$ we have $$\langle z,y\rangle =\lim_n\langle Ax_n,y\rangle =\lim_n\langle x_n,By\rangle=\langle x,By\rangle =\langle Ax,y\rangle$$ Hence $z=Ax.$
Remark It is worthwhile pointing out that for linear operator $A$ the graph is closed if $x_n\to 0$ and $Ax_n\to z$ imply $z=0.$ The above reasoning gets shorter: $$\langle z,z\rangle= \lim_n\langle Ax_n,z\rangle =\lim_n\langle x_n,Bz\rangle=0 $$ hence $z=0.$
Denote by $D$ the closed ball of radius $1$ in $H$, that is $$ D = \{x\in H \; | \; \|x\|\leq 1\}. $$ For each $x\in D$ let $\varphi_x:H\to \mathbb{R}$ be the linear functional $$ \varphi_x(y) = \langle Ax,y\rangle = \langle x,By\rangle, \quad y\in H. $$ Note that $$ |\varphi_x(y)| = |\langle Ax,y\rangle|\leq \|Ax\|\|y\| $$ so that $\varphi_x$ is continuous for all $x\in D$. On the other hand, for $x\in D$ and $y\in H$ $$ |\varphi_x(y)| = |\langle x,By\rangle|\leq \|x\|\|By\| \leq \|By\| $$ so that $\sup_{x\in D} |\varphi_x(y)|$ is finite for all $y\in H$. By the uniform boundedness principle (or Banach-Steinhaus theorem if you wish) we have that $$ M:=\sup_{x\in D}\|\varphi_x\| $$ is finite. Now, by Riesz theorem and (a corollary of) Hahn-Banach theorem, $$ \|Ax\| = \sup_{y\in D}|\langle Ax,y\rangle| = \sup_{y\in D}|\varphi_x(y)| = \|\varphi_x\|\leq M, $$ for all $x\in D$ so that $A$ is bounded. The fact that $B=A^*$ follows from the definition of Hilbert adjoint.