Let $A_i$ with $i=1,\dots,N$ and $p$ be real $M\times M$ matrices. Further, let $p$ be positive definite, i.e., $p\succ 0$, with $Tr(p)=1$. Let $0< a_i<1$ and $\sum_{i=1}^N a_i = 1$.

Claim: $Tr( A_1 p^{a_1} A_2 p^{a_2} \dots A_N p^{a_N} ) \leq \|A_1\| \|A_2\|\dots \|A_N\|$

Here, |X| denotes the operator norm of $X$ (= largest singular value).

Can you prove this claim (at least for symmetric matrices A_i)? It is trivial for N=1: $Tr(A p)=\sum_i A_{i,i} p_i\leq \|A\|\sum_i p_i = \|A\|$, where I used the representation of $A$ in the eigenbasis of $p$.

The claim even seems to hold if one uses N different matrices $p_i\succ 0$, $Tr(p_i)=1$ on the left-hand side of the claim.

thanks a lot, Tom

$\DeclareMathOperator\Tr{Tr}$Let $A_i$ with $i=1,\dotsc,N$ and $p$ be real $M\times M$ matrices. Further, let $p$ be positive definite, i.e., $p\succ 0$, with $\Tr(p)=1$. Let $0< a_i<1$ and $\sum_{i=1}^N a_i = 1$.

Claim: $$\Tr( A_1 p^{a_1} A_2 p^{a_2} \dotsm A_N p^{a_N} ) \leq \lVert A_1\rVert \lVert A_2\rVert\dotsm \lVert A_N\rVert.$$

Here, $\lVert X\rVert$ denotes the operator norm of $X$ (= largest singular value).

Can you prove this claim (at least for symmetric matrices $A_i$)? It is trivial for $N=1$: $\Tr(A p)=\sum_i A_{i,i} p_i\leq \lVert A\rVert\sum_i p_i = \lVert A\rVert$, where I used the representation of $A$ in the eigenbasis of $p$.

The claim even seems to hold if one uses N different matrices $p_i\succ 0$, $\Tr(p_i)=1$ on the left-hand side of the claim.