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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

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up vote 7 down vote accepted

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 $X_1,\dots, X_K$ and $p_1,\dots,p_K \in [1,\infty]$ with $\sum_i 1/p_i=1$, $$ |Tr(X_1\dots X_K)| \leq \prod \|X_i\|_{p_i}.$$ 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.

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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.

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