Show $ \langle Tx,x \rangle \in \Bbb R\ \forall x \in H\ \implies T$ is self-adjoint

Question

Show that a linear operator $T: H \to H$ is self adjoint if and only if $\langle Tx, x \rangle \in \Bbb R\ \forall x \in H$. You may use that the equality that for all $x,y \in H$

$4\langle T(x),y \rangle = \langle T(x+y),x+y \rangle - \langle T(x-y),x-y \rangle + i\langle T(x+iy), x+iy \rangle -i\langle T(x-iy),x-iy \rangle$

without proof.

I can show the forward implication quite easily:

If $T$ is self adjoint, then $T = T^*$ so $ \langle Tx,x \rangle = \langle x,Tx \rangle = \overline{\langle Tx,x \rangle}$

Only real values are equal to their conjugate so this show $ \langle Tx,x \rangle \in \mathbb R$ for all $x \in H$.

However, I am having trouble showing the other direction holds. I've tried using the parallelogram identity and then equating real parts but to little avail. I am wondering how to go about this.

Answer 1

Let $T$ be a linear bounded operator such that $\left<Tx, x \right> \in \mathbb{R}$ for all $x \in H$. Since $\left<Tx,x\right> \in \mathbb{R}$ then $$\left<Tx, x \right>=\overline{\left<Tx, x \right>}= \left<x, Tx\right> \ \ (\star)$$ for all $x \in H$.

Now take $x$, $y \in H$, by your polarisation identity (in the one which you wrote there is a mistake in the 3rd term it should be $i\left<T(x+iy), x+iy\right>$) and ($\star$) we obtain

$$\begin{align}\left<Tx, y\right> = \frac{1}{4} \left(\left<T(x+y), x+y \right> - \left<T(x-y), x-y \right> +i\left<T(x+iy), x+iy \right> -i\left< T(x-iy), x-iy\right> \right)= \frac{1}{4}\left(\left<x+y, T(x+y) \right> - \left<x-y, T(x-y) \right> +i\left<x+iy, T(x+iy) \right> -i\left< x-iy, T(x-iy)\right> \right)=\left<x, Ty\right> . \ _\square\end{align}$$

Answer 2

If $\langle Tx,x\rangle\in\mathbb{R}$ for all $x\in H$, it follows that $$\langle Tx,x\rangle=\overline{\langle Tx,x\rangle}=\langle x,Tx\rangle$$ for all $x\in H$. Now we can use the polarization identity you were given to extend this result. We have \begin{align*} 4\langle Tx,y \rangle &= \langle T(x+y),x+y \rangle - \langle T(x-y),x-y \rangle \\&\qquad + i\langle T(x+iy), x+iy \rangle -i\langle T(x-iy),x-iy \rangle\\ &= \langle x+y,T(x+y) \rangle - \langle x-y,T(x-y) \rangle\\&\qquad + i\langle x+iy, T(x+iy) \rangle -i\langle x-iy,T(x-iy) \rangle\\ &=4\langle x,Ty \rangle \end{align*} so $\langle Tx,y \rangle =\langle x,Ty \rangle$ for all $x,y\in H$ and therefore $T$ is self-adjoint.