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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:

$$(x,y) = \frac{1}{4}((||x+y||^2 + ||x-y||^2) - i(||x + iy||^2 - ||x - iy||^2)),$$

where of course the norm is defined by $|| \cdot||^2 = (\cdot, \cdot)$. So my guess is that I need to start with the expressions:

$$(Ax,y) = \frac{1}{4}((||Ax+y||^2 + ||Ax-y||^2) - i(||Ax + iy||^2 - ||Ax - iy||^2)),$$

$$(x,Ay) = \frac{1}{4}((||x+Ay||^2 + ||x-Ay||^2) - i(||x + iAy||^2 - ||x - iAy||^2)),$$

and somehow show they are equal. But here is where I have gotten stuck.

Hints or solutions are greatly appreciated.

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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.

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  • $\begingroup$ Consider the operator on 2D Hilbert space given by the matrix $A = \left[\begin{matrix}1&1\\-1&1\end{matrix}\right]$. Isn't this positive, but not self-adjoint? What am I missing? $\endgroup$ – Stephen Montgomery-Smith Nov 10 '13 at 22:14
  • $\begingroup$ $(Ax,x) = 2(1+i)$ isn't real for $x = (1,i)$. $\endgroup$ – Daniel Fischer Nov 10 '13 at 22:18
  • $\begingroup$ Yes, I just realized that 10 seconds before your post! $\endgroup$ – Stephen Montgomery-Smith Nov 10 '13 at 22:19
  • $\begingroup$ Thank you for the answer. Should we have a minus sign in front of the term $(A(x-y), x-y)$? $\endgroup$ – JZS Nov 10 '13 at 23:04
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    $\begingroup$ Can you give a hint of how this form of the polarization identity is derived? This is not the usual form that I have seen, which is just $(x,y)=\frac{1}{4}\sum_{k=0}^3 i^k \| x + i^k y\|$ $\endgroup$ – Squirtle Dec 10 '18 at 0:05
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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 $\langle Tx,x \rangle \in \mathbb{R}$.

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  • $\begingroup$ @jtms88 ,check my proof please.thank you $\endgroup$ – Haha Nov 24 '13 at 12:53
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    $\begingroup$ It isn't complete since you cannot conclude that $T=T^*$ from $\langle Tx,x\rangle=\langle T^*x,x\rangle$. You still need to use an argument similar to the one given by Daniel Fischer to deduce it. $\endgroup$ – Lolman Jul 24 '18 at 16:30

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