I am wondering how to show that:  if $A$ is a non-negative operator, then $A$ is self-adjoint.    Def. 1. $A$ is non-negative if $\langle Ax,x \rangle \geq 0$ for $\forall x\in H$, where $H$ is a Hilbert space.  Def. 2. $A$ is self-adjoint if $A = A^*$.
Let $(\mathcal{H}, (\cdot, \cdot))$ be a complex Hilbert space, and $A :  \mathcal{H} \to \mathcal{H}$ a positive, bounded operator ($A$ being positive means $(Ax,x) \ge 0$ for all $x \in \mathcal{H}$). Prove that $A$ is self-adjoint. That is, prove that $(Ax,y) = (x, Ay)$ for all $x,y \in \mathcal{H}$. Here's what I have so far. Because $A$ is positive we have $\mathbb{R} \ni (Ax,x) = \overline{(x,Ax)} = (x,Ax)$, all $x \in \mathcal{H}$. Next, I have seen some hints that tell me to apply the polarization identity: $$(x,y) = \frac{1}{4}((\lVert x+y \rVert^2 + \lVert x-y \rVert^2) - i(\lVert x + iy \rVert^2 - \lVert x - iy \rVert^2)),$$ where of course the norm is defined by $\lVert \cdot \rVert^2 = (\cdot, \cdot)$. So my guess is that I need to start with the expressions: $$(Ax,y) = \frac{1}{4}((\lVert Ax+y \rVert^2 + \lVert Ax-y \rVert^2) - i(\lVert Ax + iy \rVert^2 - \lVert Ax - iy \rVert^2)),$$ $$(x,Ay) = \frac{1}{4}((\lVert x+Ay \rVert^2 + \lVert x-Ay \rVert^2) - i(\lVert x + iAy \rVert^2 - \lVert x - iAy \rVert^2)),$$ and somehow show they are equal. But here is where I have gotten stuck. Hints or solutions are greatly appreciated.
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