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The well-known Powers-Stormer inequality says the following: for positive semidefinite operators $A, B$, we have that $\mathrm{Tr}((A - B)(A - B)) \leq \| A^2 - B^2 \|_1$, where $\| \cdot \|_1$ indicates the Schatten-1 norm (also known as the trace norm). Does anyone know if the following extension might be true, or if there's an obvious counter-example?

I'm wondering if there is some constant $C$ such that for positive semidefinite operators (say finite dimensional) $A, B$, we have $$ \mathrm{Tr}((A - B)X^2 (A - B)) \leq C \| XA^2 X - X B^2 X \|_1 $$ where $X = \sqrt{A^2 + B^2}$ (the square root is uniquely defined because $A^2 + B^2$ is positive semidefinite).

Note that if $X$ is an arbitrary positive operator, then this is false. However, if $X$ is related to $A$ and $B$ in this nice way, could this be true (or is anything "like" this that is true)?

Thanks!

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  • $\begingroup$ What does $\dagger$ mean? $\endgroup$ – Fedor Petrov Aug 12 '15 at 18:38
  • $\begingroup$ He means $*$. I already edited the question but it hasn't been reviewed yet. $\endgroup$ – Chris Ramsey Aug 12 '15 at 18:51
  • $\begingroup$ But aren't all operators self-adjoint? $\endgroup$ – Fedor Petrov Aug 12 '15 at 19:29
  • $\begingroup$ @FedorPetrov Bingo! $\endgroup$ – Chris Ramsey Aug 12 '15 at 19:32
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Here is a partial answer:

If $A$ and $B$ commute then they also commute with $X$ and $X^{1/2}$. Hence, using the Powers-Stormer inequality you get \begin{align*}{\rm Tr}((A-B)X^2(A-B)) &= {\rm Tr}((X^{1/2}AX^{1/2} - X^{1/2}BX^{1/2})^2) \\ &\leq \|(X^{1/2}AX^{1/2})^2 - (X^{1/2}BX^{1/2})^2\|_1 \\ &= \|XA^2X - XB^2X\|_1 \end{align*} The general case eludes me and may be false.

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