a trace class operator problem

Question

Could someone help me with this

Prove that If $A$ and $B$ are positive trace class operators on a Hilbert space, then so is $A^zB^{(1-z)}$ for a complex number $z$ such that $0 <Re(z)< 1$.

An operator $A$ is called trace class if $\sum_{n=1}^{\infty}\langle |A|e_n,e_n\rangle < \infty $

|A| is positive square root of $ A^*A$

Answer 1

Write $z=a+ib$. Then $A^zB^{1-z}=A^aA^{ib}B^{1-a}B^{-ib}=A^{ib}A^aB^{1-a}B^{-ib}$. Since the trace-class operators form an ideal, it is enough to show that $A^aB^{1-a}$ is trace-class. Even more, it is enough--by the polar decomposition--to show that $|A^aB^{1-a}|$ is trace class.

(julien has suggested a much shorter and efficient way to the answer; so I'm following his suggestion here, and leaving the rest of my original answer towards the end)

Now we can use Hölder's inequality $\|ST\|_1\leq\|S\|_p\,\|T\|_q$, which holds independently of whether $ST$ is trace-class or not. So, with $p=1/a$, $q=1/(1-a)$, $$ \|A^aB^{1-a}\|_1\leq\|A^a\|_{1/a}\,\|B^{1-a}\|_{1/(1-a)}=\|A\|_1^a\,\|B\|_1^{1-a}<\infty. $$ Thus, $|A^aB^{1-a}|$ is trace-class.

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(here is the rest of my original answer)

To this goal, we can use Young's inequality for compact operators (proven in a paper by Erlijman-Farenick-Zeng, from around 2001): there exists a partial isometry $U$ with initial space $(\ker|A^aB^{1-a}|)^\perp$ such that $$\tag{1} U|A^aB^{1-a}|U^*\leq \frac1p\,|A^a|^p+\frac1q\,|B^{1-a}|^q=aA+(1-a)B $$ (where we are using $p=1/a$, $q=1/(1-a)$; this is where we use the requirement $0<a<1$). So $U|A^aB^{1-a}|U^*$ is trace class. As $|A^aB^{1-a}|U^*U=|A^aB^{1-a}|$, we get $$ \mbox{Tr}(|A^aB^{1-a}|)=\mbox{Tr}(|A^aB^{1-a}|U^*U)=\mbox{Tr}(U|A^aB^{1-a}|U^*)<\infty. $$ This shows that $|A^aB^{1-a}|$ is trace class, since it is positive.

Answer 2

The trace-class operators are the particular case $p=1$ of the so-called Schatten class $S_p(H)$ which is defined, for $1\leq p<\infty$, by $$ S_p(H)=\{S\in B(H)\,;\,\|S\|_p=(\mbox{tr}\,|S|^p)^\frac{1}{p}<\infty\}. $$ that is $|S|^p$ is trace-class. For $p=2$, these are known as the Hilbert-Schmidt operators. For $1\leq p\leq q$, we have $S_1(H)\subseteq S_p(H)\subseteq S_q(H) \subseteq K(H)$, where $K(H)$ is the ideal of compact operators. A key fact in this theory is that the noncommutative version of Hölder's inequality holds, including $p$ or $q$ infinite, with $S_\infty(H)=B(H)$. For every $p,q\geq 1$ conjugate exponents, i.e. $\frac{1}{p}+\frac{1}{q}=1$:

$$ \forall S\in S_p(H), T\in S_q(H)\qquad \|ST\|_1\leq \|S\|_p\|T\|_q. $$ This says in particular that if $S$ is in $S_p(H)$ and $T$ is in $S_q(H)$, then $ST$ is trace-class. That's the generalization of the well-know case of Hilbert-Schmidt operators $S_2(H)S_2(H)\subseteq S_1(H)$, which is actually an equality, and the ideal property $B(H)S_1(H)B(H)\subseteq S_1(H)$.

Application: let $z=x+iy$ with $0<x<1$, and let $A,B$ be positive trace-class operators, i.e. in $S_1(H)$. Then $|A^z|=A^x$ and $|B^{1-z}|=B^{1-x}$, whence $$ \left(\|A^z\|_\frac{1}{x}\right)^\frac{1}{x}=\mbox{tr}\left(|A^z|^\frac{1}{x}\right)=\mbox{tr} A=\|A\|_1<\infty$$ and $$ \left(\|B^{1-z}\|_\frac{1}{1-x}\right)^\frac{1}{1-x}=\mbox{tr}\left(|B^{1-z}|^\frac{1}{1-x}\right)=\mbox{tr} B=\|B\|_1<\infty. $$ Therefore $$ \mbox{tr}\big|A^zB^{1-z} \big|=\|A^zB^{1-z}\|_1\leq \|A^z\|_\frac{1}{x}\|B^{1-z}\|_\frac{1}{1-x}=\|A\|_1^x\|B\|_1^{1-x}=(\mbox{tr} A)^x(\mbox{tr} B)^{1-x}<\infty. $$ In particular, $A^zB^{1-z}$ is trace-class.