Proving T is self-adjoint iff $\left \langle Tx,y \right \rangle + \left \langle Ty,x \right \rangle \in \Bbb{R}$

Question

I'm struggling with proving $\left \langle Tx,y \right \rangle=\left \langle x,Ty \right \rangle \iff \left \langle Tx,y \right \rangle + \left \langle Ty,x \right \rangle \in \Bbb{R}$ for all $x,y \in H$ where $H$ is a Hilbert space over $\Bbb{C}$, and $T$ is a bounded linear operator from $H$ to $H$. I've proven $\left \langle Tx,y \right \rangle=\left \langle x,Ty \right \rangle \implies \left \langle Tx,y \right \rangle + \left \langle Ty,x \right \rangle \in \Bbb{R}$ as follows: $$\left \langle Tx,y \right \rangle=\left \langle x,Ty \right \rangle$$ $$\left \langle Tx,y \right \rangle=\overline{\left \langle Ty,x \right \rangle}$$ So $Im(\left \langle Tx,y \right \rangle)=-Im(\left \langle Ty,x \right \rangle)$, so the imaginary parts of $\left \langle Tx,y \right \rangle + \left \langle Ty,x \right \rangle$ cancel out and the result is a real number.

For the other direction $\left \langle Tx,y \right \rangle + \left \langle Ty,x \right \rangle \in \Bbb{R} \implies \left \langle Tx,y \right \rangle=\left \langle x,Ty \right \rangle$, I've gotten this far: $$\left \langle Tx,y \right \rangle + \left \langle Ty,x \right \rangle \in \Bbb{R}$$ $$\left \langle Tx,y \right \rangle + \overline{\left \langle x,Ty \right \rangle} \in \Bbb{R}$$ $$Im\left \langle Tx,y \right \rangle=-Im\overline{\left \langle x,Ty \right \rangle}$$ $$Im\left \langle Tx,y \right \rangle=Im{\left \langle x,Ty \right \rangle}$$ So I've proved $\left \langle Tx,y \right \rangle$ and $\left \langle x,Ty \right \rangle$ have the same imaginary parts but I'm stuck on proving they have the same real parts. I've been substituting the polarization identity and parallelogram law, equating the imaginary parts, etc, for a while, but haven't gotten anything useful. Any sort of push in the right direction would be hugely appreciated. Thank you :)

Answer 1

Hint: Your equality is true for all $y$. In particular, this is true if we replace $y$ by $iy$. Also $Im(iz)= Re(z)$ for all $z\in \mathbb{C}$.

Answer 2

For every complex numbers $u,v,$ we have $\bar v\in\Bbb R-v$ hence $$u+\bar v\in\Bbb R\iff u-v\in\Bbb R.$$ Therefore, $$\left \langle Tx,y \right \rangle + \overline{\left \langle x,Ty \right \rangle} \in \Bbb R\iff$$ $$\left \langle Tx,y \right \rangle-\left \langle x,Ty \right \rangle\in \Bbb R\iff$$ $$\left \langle\left(T-T^*\right)x,y \right \rangle\in \Bbb R.$$ For fixed vectors $x,y,$ the number $\left \langle\left(T-T^*\right)x,y \right \rangle$ can of course be real without being zero (e.g. $y=\left(T-T^*\right)x$). But $$\left[\forall x,y\in H\quad\left \langle\left(T-T^*\right)x,y \right \rangle\in \Bbb R\right]\implies T-T^*=0$$ because the range of a sesquilinear form is either $\Bbb C$ or $\{0\}.$

Answer 3

By the polarization identity $$\langle Tx,y\rangle = {1\over 4}\sum_{k=1}^4i^k\langle T(x+i^ky),x+i^ky\rangle $$two operators $S$ and $T$ are equal iff $\langle Sx,x\rangle =\langle Tx,x\rangle$ for any $x.$ Assume $\langle Tx,y\rangle +\langle Ty, x\rangle \in \mathbb{R}.$ Plugging in $y=x$ gives $\langle Tx,x\rangle\in \mathbb{R}.$ Therefore $$\langle T^*x,x\rangle =\langle x,Tx\rangle =\overline{\langle Tx,x\rangle}=\langle Tx,x\rangle$$ Hence $T^*=T.$ The converse implication is obvious.