I wish to show the following theorem:
Let $T:H\to H$ be a bounded linear operator on a complex Hilbert space $H$. Then if $\left\langle Tx,x\right\rangle \in\mathbb{R}$ for all $x\in H$, then $T$ is self-adjoint.
My proof is as follows:
If $\left\langle Tx,x\right\rangle \in\mathbb{R}$ for each $x\in H$ then for each $x \in \mathbb{R}$: $$ \left\langle Tx,x\right\rangle =\overline{\left\langle Tx,x\right\rangle }=\overline{\overline{\left\langle x,Tx\right\rangle }}=\left\langle x,Tx\right\rangle $$ so that $T$ is self-adjoint by definition. (For the first equality, I have used the fact that $x \in \mathbb{R} \Leftrightarrow \bar{x} = x$, for the second, I have used the general property in a Hilbert space that $<x,y> = \overline{<y,x>}$ and for the third that for all $x \in \mathbb{C}: \overline{\bar{x}} = x$.)
I'm satisfied with this proof, but the one I have in my textbook is more complicated, showing that $0 = <(T - T^{*})x, x>$ and a Lemma that says that if $Q : X \to X$ is a bounded linear operator on a complex inner product space $X$ and $ <Qx, x> = 0 \forall x \in X$ then $Q = 0$. Am I missing something subtle?