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Let $P_1,P_2$ denote stochastic transition matrices on a countable set $I$. Consider $P_1,P_2$ as operators on $\ell^2(I)$ given by multiplication.

Question Under which conditions can we show that for $t\in (0,1)$, $$\rho(tP_1+(1-t)P_2)\le t\rho(P_1)+(1-t)\rho(P_2),$$ where $\rho(\cdot )$ denotes the spectral radius? (Of course, it would be nice to consider this problem also for more general classes of Markov operators acting on some function space.)

My hope is that the strong assumption that both matrices are stochastic, is enough to prove the inequality.

Example My favorite example is the case where $I$ is the free group $G$ of rank $d\ge2$ and $P_1,P_2$ are given by $P_i(g,g'):=\mu_i( g^{-1}g' )$, where $\mu_i$ is some measure on a set of generators of $G$. In this case, the spectral radius of $P_i$ is strictly less than one, since the group is non-amenable (H. Kesten 1959). I don't know if the above inequality holds in this setting. (Note that non-zero constant functions are not in $\ell^2(G)$.)

An additional assumption I'm fine with the additional assumption that $P_1$ and $P_2$ have zeros at the same places. So, regarding the previous example, the measures $\mu_i$ have the same support. A further simplification would be to assume that one of measures is equidistributed on its support.

Related results If $P_1,P_2$ commute, then the inequality holds. This is too restrictive.

If $P_1,P_2$ are selfadjoint, then the spectral radius can be replaced by the norm, and the inequality holds. I'm not interested in this case. I don't want to assume that $P_i$ is reversible with respect to some measure.

Finite matrices with real eigenvalues In the following paper, Lax proved that the spectral radius on the set of $n\times n$ matrices with real eigenvalues is convex.

Kingman proved that spectral radius is log-convex on the space of finite dimensional non-negative matrices.

Cohen proved that the spectral radius is a convex function of the diagonal elements on the space of finite dimensional non-negative matrices. There are generalisations for multiplication operators by Kato.

On amenability If the group is $G$ is amenable in the previous example, then for a large class of transition operators, the spectral radius is equal to one (e.g. uniform irreducibility and invariant measure bounded away from zero and infinity). Hence, under these conditions, all the spectral radii in the above inequality are equal to one. So, equality holds.

Example As an example, let $G$=$\mathbb{Z}$ and consider $P_1,P_2$ corresponding to the random walks, which go one step to the left resp. right with probabilities $p_i$ resp. $q_i$, $p_i+q_i=1$, $i=1,2$.

References LAX, P. D. Differential equations, difference equations and matrix theory. Comm. Pure Appl. Math. 11 (1958), 175-194.

Kingman, J.F.C.: A convexity property of positive matrices. Quart. J. Math. Oxford (2) 12, 283-284 (1961)

Cohen, J.E.: Convexity of the dominant eigenvalue of an essentially nonnegative matrix. Proc. Amer. Math. Soc. 81, 657-658 (1981)

Kato: Superconvexity of the Spectral Radius, and Convexity of the Spectral Bound and the Type, Mathematische Zeitschrift, (1982)

Using commutativity There are several papers of M.Zima about related properties for positive operators on partially ordered Banach spaces (see below). In there, some commutativity is required to prove the inequality in question. I don't want to assume commutativity.

M.Zima, A theorem on the spectral radius of the sum of two operators and its applications, Bull. Austral. Math. Soc. 48 (1993), 427{434. MR 94j:47006

M.Zima, On the local spectral radius in partially ordered Banach spaces, Czechoslovak Math. J. 49 (1999), 835{841. MR 2001m:47011

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For a lot of results of this sort (though probably not exactly what you are asking for), check out:

Convexity and log convexity for the spectral radius Roger D. Nussbaum Linear Algebra and its Applications Volume 73, January 1986, Pages 59–122

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Thanks for the reference. But, so far, I didn't find an answer to my question in it. It seems that the paper is more about eigenvalues. I don't see how to apply it to the l^2 setting. –  Mika Jun 27 '12 at 3:42
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