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Let $A$ and $B$ be positive-definite matrices such that $A \le B.$

By matrix monotonicity of the root, this also implies that $A^{\alpha} \le B^{\alpha}$ for $\alpha \in [0,1].$

I am now curious under what assumptions we have

$$ A^{\alpha} B A^{\alpha} \le B^{1+2\alpha}$$ in the sense of operators?

Edit: Notice that it does not seem to follow from Furuta's inequality, since we are changing the operator $A$.

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No, this is not true even if the matrices are $2 \times 2$ and $\alpha = 1/2$. For a concrete counter-example, consider $$A = \begin{bmatrix}1 & -\sqrt{3} \\ -\sqrt{3} & 3\end{bmatrix}, B = \begin{bmatrix}4 & -3 \\ -3 & 4\end{bmatrix}.$$ Here, $A$ is only positive semidefinite, but just add some small $\varepsilon > 0$ to the diagonal if you really need a positive definite counterexample.

It is straightforward to check that $A \leq B$, but $B^2 - A^{1/2}BA^{1/2}$ has minimal eigenvalue somewhere around $-2.391$, so $A^{1/2}BA^{1/2} \not\leq B^2$.

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This is a simple case of Furuta's inequality, see e.g. Theorem 1.1 of J IANGTAO Y UAN, FURUTA INEQUALITY AND q –HYPONORMAL OPERATORS Operators and Matrices Volume 4, Number 3 (2010), 405–415.

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    $\begingroup$ is it really? In that inequality it seems you only change the matrix in the center not the ones outside? $\endgroup$
    – António Borges Santos
    Commented Mar 18 at 0:02
  • $\begingroup$ @anto/nio,, I am not sure what the problem is, Theorem 1.1 uses A>=B>= 0 but you can just use inverses for the problem stated by the OP. $\endgroup$
    – Derek
    Commented Mar 18 at 1:05
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    $\begingroup$ The Theorem only implies that $A^{\alpha}A A^{\alpha} \le A^{\alpha}B A^{\alpha}$ and $B^{\alpha}A B^{\alpha} \le B^{\alpha}B B^{\alpha}$ but not the inequality claimed in the question. $\endgroup$
    – António Borges Santos
    Commented Mar 18 at 1:48

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