Trace cauchy schwarz inequality

Question

This was an interesting result I found while reading Holevo's Quantum Information theory book. They call it the non-commutative operator Cauchy Schwarz.

Let $S$ be a state i.e. a non negative operator with unit trace on a Hilbert space $\mathcal{H}$ in $\mathbb{C}$. Then for arbitrary operators $X$ and $Y$ on $\mathcal{H}$, \begin{equation} |Tr(SX^*Y)|^2 \leq Tr(SX^*X)Tr(SY^*Y)\label{Thi} \end{equation}

I was able to show this by mirroring the usual Cauchy Schwarz inequality (I assumed RHS is finite). Then a friend suggested that it might follow easily if we show that $Tr(SX^*Y)$ is an inner product on the operator space. I was able to show that it is a semi inner product. To prove it is an inner product, I needed to show $$Tr(SX^*X) = 0 \Rightarrow X=0$$

I wasn't successful with this one. I was able to show it if $S$ is a strictly positive operator (at least I think I did) but not for non negative. My guess is that in general, $Tr(SX^*Y)$ is a seminorm but not necessarily a norm.

But I was unable to furnish an $X$ such that $X\ne 0$ but $Tr(SX^*X) = 0$. I seek such an example. Kindly give me some ideas. We could think in terms of matrices here.

Update: Upon further inspection, I realized that my "proof that mirrors cauchy schwarz" was in fact flawed. I assumed implicitly that both terms in RHS were $>0$. In one of the steps, I divide by the square root of RHS, which would be illegal if it were $0$. This wasn't a problem in usual Cauchy schwarz because we were given that LHS is square of mod of an inner product. This changes my perception and I now believe that the above is an inner product but I need to show the final step.

Update: I'll show my work so far. Let $\langle X,Y\rangle_T \triangleq Tr(SX^*Y)$. Then

1. Linearity in second argument (In Quantum theory, it is the second argument) and Conjugation: Easy to show.
2. Non-Negative for equal arguments: We have $$Tr(SX^*X) = Tr(XSX^*)$$ Since $S \geq 0$, $XSX^* \geq 0$ and hence $Tr(XSX^*) \geq 0$. Also if $X=0$, then $\langle X,X\rangle_T = 0$.
3. Need to show if $Tr(SX^*X) = 0$ then $X=0$. If $S$ is strictly positive definite, then for any orthonormal basis $e_i$

$$ 0 = \langle X,X\rangle_T = Tr(\sqrt{S}X^*X\sqrt{S}) = \sum_{i=1}^d \langle e_i,\sqrt{S}X^*X\sqrt{S}e_i\rangle \\ = \sum_{i=1}^d \langle X\sqrt{S}e_i,X\sqrt{S}e_i\rangle = \sum_{i=1}^d \|X\sqrt{S}e_i\|^2$$

Hence $\|X\sqrt{S}e_i\| =0 $ for every $i$ and this further implies $X\sqrt{S} = 0$. Since $\sqrt{S}$ is invertible owing to strict positivity, we get $X=0$.This completes the proof when $S$ is strictly positive definite. For non-negative definite, I don't know how to tackle it as $S$ may have a non-trivial kernel/nullspace.

Answer 1

Looks like I figured it out. I think one doesn't require 3) in my second update to prove the result. If $S$ has a nontrivial nullspace, it isn't true. So we tackle that case by showing that if either $Tr(SX^*X)=0$ or $Tr(SY^*Y)=0$ then $Tr(SX^*Y)=0$.

Suppose $Tr(SX^*X)=0$, then given a basis $e_i$, we have $$0=Tr(SX^*X) = Tr(XSX^*)=\sum_{i=1}^d \langle e_i, XSX^*e_i \rangle =\sum_{i=1}^d \langle \sqrt{S}X^*e_i, \sqrt{S}X^*e_i \rangle \\= \sum_{i=1}^d\| \sqrt{S}X^*e_i\|^2$$ The last step implies $\|\sqrt{S}X^*e_i \|=0$ and hence $\sqrt{S}X^*e_i=0$ for all $i$. Hence $\sqrt{S}X^*=0$. Now $$Tr(SX^*Y) = Tr(Y\sqrt{S}\sqrt{S}X^*) = 0$$ The other case is similar. For the case where $Tr(SX^*X)\ne 0,Tr(SY^*Y)\neq 0$ but finite, we can mirror the Cauchy Schwarz proof.