$\DeclareMathOperator\Tr{Tr}$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.



Bounding the trace of a matrix product by the operator norms; generalized Hölder inequality?
The claim is true. More generally the Hölder inequality holds for the Schatten $p$-norms. The statement is [here without proof](https://en.wikipedia.org/wiki/Schatten_class_operator) in Wikipedia, and by induction it implies that for matrices $X\_1,\dotsc, X\_K$ and $p\_1,\dotsc,p\_K \in [1,\infty]$ with $\sum\_i 1/p\_i=1$,
$$ \lvert\operatorname{Tr}(X\_1\dotsm X\_K)\rvert \leq \prod \lVert X\_i\rVert\_{p\_i}.$$
In your setting, take $K=2N$, $X\_{2k} = p^{a\_k}$, $X\_{2k-1} = A\_k$, $p\_{2k} = 1/a\_k$ and $p\_{2k-1} = \infty$.


I guess that all this is clearly explained (and with no mistake?) in Barry Simon's book [Trace ideals and their applications](https://doi.org/http://dx.doi.org/10.1090/surv/120).


