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)?


  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.