TITLE: bounding the trace of a matrix product by the operator norms; generalized Hölder inequality?
QUESTION [9 upvotes]: 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

REPLY [7 votes]: I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.
The Schatten-$p$ norm of a matrix $X$ is defined as
$$\|X\|_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$
where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.
From this definition, we see that $\|X\|_p = \|\sigma(X)\|_p$, where $\sigma(X)$ is the vector of singular values. 
From von Neumann's trace inequality we know that
$$|\mbox{trace}(XY)| \le \langle \sigma(X), \sigma(Y)\rangle,$$
while from the Hölder inequality for vectors, we have
$$\langle \sigma(X), \sigma(Y)\rangle \le \|\sigma(X)\|_p\|\sigma(Y)\|_q,$$
where $1/p + 1/q = 1$.
On combining the above two, we immediately get alleged Schatten-p norm Hölder inequality.