Inequality involving positive definite operator

[ismaili]

1.
Given that $$|\langle f|g\rangle|^2 \leqslant \langle f|f\rangle\langle g|g\rangle$$
prove that $$|\langle f|H|g\rangle|^2 \leqslant \langle f|H|f\rangle \langle g|H|g\rangle$$
where $$H$$ is a Hermitian and positive definite operator.




3. I tried to identify $$H|g\rangle$$ as a state and put it into the given inequality, but not so help.

Is there any ideas to prove this inequality? Thanks in advance.

 

[weejee]

How about trying sqrt(H)?

You will also need to think about the conditions H should meet for its square root to exist.

 

[ismaili]

How about trying sqrt(H)?

You will also need to think about the conditions H should meet for its square root to exist.

Thank you so much!
I got it, and in this way, Hermitian and positive definite properties are both used.
Thanks!

 

[Ben Niehoff]

The problem is kind of silly to begin with. If you can show that

$$\langle \cdot | H | \cdot \rangle$$

satisfies the defining properties of an inner product:

1. Positive definite,

2. Linear on second slot,

3. Conjugate-linear on first slot,

then the expression automatically obeys the Schwartz inequality, because the Schwartz inequality is true of inner products in general.

 

[weejee]

The problem is kind of silly to begin with. If you can show that

$$\langle \cdot | H | \cdot \rangle$$

satisfies the defining properties of an inner product:

1. Positive definite,

2. Linear on second slot,

3. Conjugate-linear on first slot,

then the expression automatically obeys the Schwartz inequality, because the Schwartz inequality is true of inner products in general.

Well, you are right. I keep forget that an inner product doesn't always have to be what we usually consider as 'the inner product'.