Generalization of Cauchy-Schwarz to positive operators

Question

The problem I am given is:

Let $T$ be a positive operator. Prove that for all $x,y$ we have

$$|\langle Tx,y\rangle| \le \langle Tx,x\rangle^{\frac{1}{2}} \langle Ty,y\rangle^{\frac{1}{2}}.$$

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Now, it seems to me like, by positivity of $T$, we have some $A = T^{\frac{1}{2}}\ge 0$, thus $A=A^*$ and

$$\langle Tx,y \rangle = \langle A^2x,y \rangle = \langle Ax,Ay\rangle.$$

Then we notice $\langle A\cdot,A\cdot\rangle$ defines an inner product, and Cauchy-Schwarz gives us (is) the desired inequality? Am I missing something or is it this simple?

Answer 1

Well, you do not quite have that $\langle A \cdot, A \cdot \rangle$ is a inner product. $T=0$ would be a counterexample.

The Cauchy-Schwarz inequality for $\langle\cdot,\cdot\rangle$ suffices.

$$|\langle Tx,y\rangle |=|\langle Ax,Ay \rangle |\leq \|Ax\| \cdot \|Ay\| = \sqrt{\langle Ax, Ax\rangle \langle Ay, Ay\rangle } = \sqrt{\langle A^2 x, x\rangle \langle A^2y, y\rangle }= \langle Tx,x\rangle^{\frac 12} \langle Ty,y\rangle^{\frac 12} $$

Answer 2

Your reasoning is correct if 'positive' is interpreted as 'positive definite'. In fact you can directly verify that $(x,y)\mapsto\langle Tx,y\rangle$ is an inner product based on the fact that positive operators are hermitean.