Suppose $H$ is a Hilbert space and $A \in B(H)$ satisfies $\langle Ax, y\rangle = \langle x, Ay\rangle , \, \forall x, y \in H$. (Notice that this means $A = A^*$) Prove:
If $\DeclareMathOperator*{\range}{range} \range(A)$ is dense in $H$, then for any fixed $y \in \range{(A)}$, the equation $Ax = y$ has a unique solution.
Let $A(H)$ denote the range of $A$ and $N(A)$ the kernel of $A$.
We have
$$ H=\overline{A(H)}=N(A^*)^{\perp}=N(A)^{\perp}.$$
This gives $N(A)=\{0\}$ and the assertion follows.
I don't know if this is correct:
First of all, we can assert the existence of such $x$. Now, suppose $\exists a,b \in H$, s.t. $Aa = y_0 = Ab$ for fixed $y_0 \in \DeclareMathOperator*{\range}{range}\range(A)$. Then $ 0 = y_0 - y_0 = Ax_1 - Ax_2 = A(a - b )\implies a-b\in \DeclareMathOperator*{\null}{null} \null(A)$. Now $\forall y \in H$, since $\range(A)$ is dense, we know $\exists \{x_n\}\in H,$ s.t. $Ax_n\xrightarrow{n \to \infty} y$. Thus, $\langle a - b, y\rangle = \langle a- b, \lim_\limits{n \to \infty} Ax_n\rangle = \lim_\limits{n \to \infty} \langle a - b, Ax_n\rangle = \lim_\limits{n \to \infty}\langle A(a-b), x_n\rangle = \lim_\limits{n \to \infty}\langle 0, x_n \rangle = 0$. This implies $a - b =0$ and thus, $a = b$.