The problem I am given is:

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

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

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?

  • $\begingroup$ You can prove the given inequality for a pseudo inner product, which is what $\langle Tx,y \rangle$ is. And you don't need to be able to take the square root of $T$. Or you can use Cauchy-Schwarz for $\langle (T+\epsilon )x,y\rangle$ for $\epsilon > 0$, which is an inner product; then you can let $\epsilon\rightarrow 0$. $\endgroup$ – Disintegrating By Parts Nov 27 '15 at 20:41

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.

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

  • $\begingroup$ First, your last line has a typo, multiplication not addition. Second, you're right, my solution fails for T=0, I noticed this and was just going to do that separately: are there any other counterexamples, or would that be valid? $\endgroup$ – Ryan Nov 27 '15 at 18:45
  • $\begingroup$ @Ryan Wait: 0 is positive semidefinite, but not positive definite. A necessary and sufficient condition for an operator to be positive definite is that the sesquilinear form $\langle\langle x,y\rangle\rangle:=\langle Ax,y\rangle$ is an inner product. So no, 0 is not the unique counterexample: take $T$ to be your favourite $2\times 2$ matrix with one vanishing eigenvalue and by evaluating it at a nontrivial vector in its null space you will see that the positive definiteness fails. So yes, your reasoning is perfectly fine. $\endgroup$ – DeM Nov 27 '15 at 19:36
  • $\begingroup$ Actually I'm pretty sure we say positive to mean positive semi-definite in my class. $\endgroup$ – Ryan Nov 27 '15 at 19:40
  • $\begingroup$ Yes, this is just a matter of definition. We actually had that $A$ is said to be positive iff $\langle x,Ax\rangle \geq 0$ for all $x\in H$. In the other case, you're right that it definies an inner product. This is a statement which is easy to prove. My (obvious) calculation does not use any other statement, so I thought it might help. $\endgroup$ – Peter Nov 27 '15 at 20:10

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.


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