$Tx =x \iff \langle\,Tx,x\rangle = \|x\|^2$ in a Hilbert space

Question

$H$ is a Hilbert space, $T : H \to H$ is continous and linear with $\|T\| \leq 1$

prove $\forall x \in H, \,Tx =x \iff \langle\,Tx,x\rangle = \|x\|^2$

direct implication is trivial.

furthest I could reach concerning the reciprocal implication is this :

for $x = 0$ it's trivial, for $x \neq 0$ using the fact that $\|T\| = \sup_{x \neq 0}\frac{\|Tx\|}{\|x\|} \leq 1$

$\begin{align} \|(Tx-x)\|^2 & = \|Tx\|^2+\|x\|^2- \langle\,Tx,x\rangle- \langle\,x,Tx\rangle \\ & =\|x\|^2 (\frac{\|Tx\|}{\|x\|})^2 - \langle\,x,Tx\rangle \leq \|x\|^2 - \langle\,x,Tx\rangle = 0 \end{align}$

therefore $Tx =x,\,\forall x\in H$ however I'm aware that I assumed the field is the set of real numbers.

how do you do it in the complex case ?

Answer 1

In the same way: since $\Vert x \Vert^2 \in \mathbb{R}$, \begin{align} \Vert Tx-x \Vert ^2 &= \Vert Tx \Vert^2- 2\Re \langle Tx,x \rangle +\Vert x \Vert^2 = \Vert Tx \Vert^2- 2\Re \Vert x \Vert^2 +\Vert x \Vert^2\\ &= \Vert Tx \Vert^2 - \Vert x \Vert^2 \leq \Vert T \Vert^2 \Vert x \Vert^2 - \Vert x \Vert^2 \leq \Vert x \Vert^2 - \Vert x \Vert^2 = 0. \end{align}

Answer 2

Suppose $x\ne 0$. By the Cauchy-Schwarz inequality, $$ |\langle Tx,x\rangle| \le \|Tx\|\|x\| \le \|T\|\|x\|^2=\|x\|^2. $$ If you have $\langle Tx,x\rangle = \|x\|^2$, then you have equality in the Cauchy-Schwarz inequality, which forces $Tx=\alpha x$ for some scalar $\alpha$. That scalar must be $\alpha=1$ because $\alpha \|x\|^2 = \langle Tx,x\rangle = \|x\|^2$.