Skip to main content
MathOverflow

Return to Answer

fixed a typo in the main ineq.
Source Link
Suvrit
Suvrit
  • 28.6k
  • 7
  • 82
  • 150

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$$k=1,r=2$ case of the well-known log-majorization inequality: \begin{equation*} \prod_{i=1}^k \lambda^r(A^{1/2}BA^{1/2}) \le \prod_{i=1}^k \lambda(A^{r/2}BA^{r/2}),\qquad k=1,\ldots,n, r \ge 1, \end{equation*}\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 $B$.

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

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 $B$.

Source Link
Suvrit
Suvrit
  • 28.6k
  • 7
  • 82
  • 150

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