Bounding the trace of a matrix product by the operator norms; generalized Hölder inequality?

Question

$\DeclareMathOperator\Tr{Tr}$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.

Answer 1

The claim is true. More generally the Hölder inequality holds for the Schatten $p$-norms. The statement is here without proof 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.

Answer 2

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 $$\lVert X\rVert_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\lVert X\rVert_p = \lVert\sigma(X)\rVert_p$, where $\sigma(X)$ is the vector of singular values.

From von Neumann's trace inequality we know that $$|{\operatorname{trace}(XY)}| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have $$\langle \sigma(X), \sigma(Y)\rangle \le \lVert\sigma(X)\rVert_p\lVert\sigma(Y)\rVert_q,$$ where $1/p + 1/q = 1$.

On combining the above two, we immediately get the alleged Schatten-$p$ norm Hölder inequality.