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Suppose that $A, B$ are Hermitian positive definite matrices of the same order and $0\le p\le 1$. Using a standard approach in matrix analysis, one can show that $\|A^{1-p}B^p\|\ge \|A\sharp_p B\|$, where $A\sharp_p B:=A^{1/2}(A^{-1/2}BA^{-1/2})^pA^{1/2}$ which is sometimes called the weighted geometric mean and has a geometric interpretation. The matrix norm here is spectral norm (i.e., largest singular value).

I tried to play with $\|A^{1-p}B^p\|\ge \|A\sharp_p B\|$ a little bit, a question suddenly occured to me: Is it true $$\|AB\|\ge \|(A\sharp_p B)(A\sharp_{1-p}B)\|?$$

I ran some simulations yet no counterexample showed up... the standard approach I know seems not work, so I am looking for some new ingredients.

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    $\begingroup$ what is the sketch proof for the first inequality ? does it use the spectral theorem ? the triangle, Cauchy-Schwartz or Hölder inequality ? $\endgroup$
    – reuns
    Commented Mar 6, 2016 at 8:29
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    $\begingroup$ Side question: can anything be said about $\|A^{1-p}B^pA^pB^{1-p}\|$ vs. $\|A B \|$? $\endgroup$
    – Wolfgang
    Commented Mar 6, 2016 at 15:20
  • $\begingroup$ @Wolfgang I think the Frobenius norm version of this is known to hold. $\endgroup$
    – Suvrit
    Commented Mar 6, 2016 at 19:43
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    $\begingroup$ @Wolfgang: This question was studied by Drury in his paper "OPERATOR NORMS OF WORDS FORMED FROM POSITIVE-DEFINITE MATRICES, Electronic Journal of Linear Algebra, Volume 18, pp. 13-20, January 2009" $\endgroup$
    – M. Lin
    Commented Mar 7, 2016 at 0:46
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    $\begingroup$ Fumio Hiai communicated to the MO proposer that the inequality is true for $1/4\le p\le 3/4$. $\endgroup$
    – M. Lin
    Commented May 3, 2016 at 2:10

2 Answers 2

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I'll write a proof for the (easy) case $p=1/2$. The general case seems more tricky.

For $p=1/2$, the task is to prove $\| (A\sharp B)^2\| \le \|AB\|$, or equivalently that $\|A\sharp B\|^2 \le \|AB\|$. But we know that $\|A\sharp B\| \le \|A^{1/2}B^{1/2}\|$, so if we show that $\|A^{1/2}B^{1/2}\|^2 \le \|AB\|$ we'll be done.

The latter inequality follows from the $k=1,r=2$ case of the well-known log-majorization: \begin{equation*} \prod_{i=1}^k \lambda^r(A^{1/2}BA^{1/2}) \le \prod_{i=1}^k \lambda(A^{r/2}B^rA^{r/2}),\qquad k=1,\ldots,n, r \ge 1, \end{equation*} for $n\times n$ positive definite matrices $A$ and $B$.

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    $\begingroup$ Thanks. I knew this case, which is an evidence in my mind before i ran simulations for the general 0<p<1/2. $\|A^{1/2}B^{1/2}\|^2$ is equal to the largest eigenvalue of $AB$, which is no larger than $\|AB\|$. $\endgroup$
    – M. Lin
    Commented Mar 7, 2016 at 0:51
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    $\begingroup$ Hmm, you knew that my question is equivalent to a log-majorization version. $\endgroup$
    – M. Lin
    Commented Mar 7, 2016 at 0:54
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    $\begingroup$ @M.Lin :-) it looked very much like a log-majorization. I'm wondering though how to "unwind" this in a manner related to Wolfgang's comment (and your reply to it).... $\endgroup$
    – Suvrit
    Commented Mar 7, 2016 at 2:22
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Her is an elementary proof of the case $p=\frac12$. As noted by Suvrit, it is enough to prove $\|A^{1/2}B^{1/2}\|^2 \le \|AB\|$, or equivalently $\|HK\|^2\le\|H^2K^2\|$. This is true because of $$\|HK\|^2=\rho((HK)^*HK)=\rho(KH^2K)=\rho(H^2K^2)\le\|H^2K^2\|.$$ Hereabove, $\rho$ is th spectral radius and the last inequality is true for every subordinated norm.

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    $\begingroup$ Thanks for your attention, but this has appeared in my comment to Suvrit's answer. $\endgroup$
    – M. Lin
    Commented Mar 11, 2016 at 12:29
  • $\begingroup$ Oups! I apologize $\endgroup$
    – Denis Serre
    Commented Mar 11, 2016 at 12:53

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