# General Cauchy-Schwarz for adjoint positive operators

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?

• (This is Schwarz without a "t"...) – paul garrett Mar 15 '18 at 12:12

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 $$|\langle Tx,y\rangle| \le \langle Tx,x\rangle^{1/2}\langle Ty,y\rangle^{1/2}.$$
• Thanks to answer @DisintegratingByParts, but I don't get how does work Cauchy-Schwarz inequality in this case because $[x,y]_{\epsilon}\leq||(T+\epsilon I)x||||y||$ – Squird37 Mar 15 '18 at 16:45
• @Squird37 : The second displayed inequality is the Cauchy-Schwarz inequality for the inner product $[x,y]_{\epsilon}$. Write this out in terms of the definition above it, and then let $\epsilon\downarrow 0$ to get what you want. – Disintegrating By Parts Mar 15 '18 at 16:55
• I get it @DisintegratingByParts! I wasn't paying attention to $[x,y]_{\epsilon}$ is a inner product. Many thanks for the help; it's a way to prove it that I didn't imagine. – Squird37 Mar 15 '18 at 17:03
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 $\bar a=\frac {\langle Ty,x \rangle} {|\langle Ty,x \rangle|}$.