General Cauchy-Schwarz for adjoint positive operators

Question

I'm trying to prove the next inequality, like Cauchy-Schwarz standard inequality:

$$|\langle Tx,y\rangle |\leq\langle Tx,x\rangle ^{1/2}\langle Ty,y\rangle ^{1/2}\space\forall x,y\in\mathcal{H},$$ where $\mathcal{H}$ is a complex Hilbert space, $T$ bounded linear operator, $T\geq 0$ and $T=T^{*}.$

If we consider that, for $t\in\mathbb{R},$ $$0\leq\langle T(y-tx),y-tx\rangle =\langle Ty,y\rangle -2t\mathcal{Re}(\langle Ty,x\rangle )+t^{2}\langle Tx,x\rangle :=P(t),$$ then $P(t)$ is a polynomial of second grade, so its discrimante have to be $$\mathcal{Re}^{2}(\langle Ty,x\rangle )\leq\langle Tx,x\rangle \langle Ty,y\rangle ,$$ but I would like to conclude that $$|\langle Ty,x\rangle |\leq\langle Tx,x\rangle \langle Ty,y\rangle ,$$

How can we conclude the desire inequality? Is there something wrong?

Answer 1

For any $\epsilon > 0$, $$ [x,y]_{\epsilon} = \langle (T+\epsilon I)x, y\rangle $$ is an inner product. Therefore, the Cauchy-Schwarz inequality gives $$ |[x,y]_{\epsilon}| \le [x,x]_{\epsilon}^{1/2}[y,y]_{\epsilon}^{1/2}. $$ That is, the following holds for all $\epsilon > 0$ and $x,y\in H$ $$ |\langle (T+\epsilon I)x,y\rangle| \le \langle (T+\epsilon I)x,x\rangle^{1/2}\langle (T+\epsilon I)y,y\rangle^{1/2} $$ Now let $\epsilon\downarrow 0$ to obtain the desired result that $$ |\langle Tx,y\rangle| \le \langle Tx,x\rangle^{1/2}\langle Ty,y\rangle^{1/2}. $$

Answer 2

Replace $y$ in the inequality you have obtained by $ay$ where $a$ is a complex number. You get $\{Re (a\langle Ty,x \rangle \})^{2} \leq |a|^{2}\langle Tx, x\rangle \langle Ty, y\rangle $. Now put $\bar a=\frac {\langle Ty,x \rangle} {|\langle Ty,x \rangle|}$.