Show that a positive operator on a complex Hilbert space is self-adjoint

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.

Solution 1:

You should apply the polarization identity in the form

$$4(Ax,y) = (A(x+y),x+y) - (A(x-y),x-y) -i(A(x+iy),x+iy) + i(A(x-iy),x-iy).$$

Since you already know $(Az,z) = (z,Az)$ for all $z \in \mathcal{H}$, it is not difficult to deduce $A^\ast = A$ from that.

Solution 2:

Another solution is this one:

$T$ is self-adjoint iff $\langle Tx,x\rangle \in \mathbb{R}$ for every $x\in \mathcal{H}$.

Let $x\in \mathcal{H}$. Then $0=\langle Tx,x \rangle-\langle T^{*}x,x \rangle = \langle Tx,x \rangle-\overline{\langle Tx,x \rangle}=2i\operatorname{Im}\langle Tx,x\rangle \iff \langle Tx,x \rangle \in \mathbb{R}$. Thus $T=T^{*} \iff \langle Tx,x\rangle \in \mathbb{R}$ for every $x\in \mathcal{H}$.

Because $T$ is positive we have that $\langle Tx,x \rangle \in \mathbb{R}$.