Let $$ p = (p_1, \ldots, p_n) $$ be a finite probability distribution, which for convenience I'll assume to have no zeroes: thus, $p_i > 0$ for all $i$ and $\sum_i p_i = 1$.
Is the function $$ q \mapsto \biggl( \sum_{i = 1}^n p_i^q \biggr)^{1/(1 - q)} $$ ($q \geq 0$) necessarily convex?
Now let me give some context.
For each $t \in \mathbb{R}$ and $x = (x_1, \ldots, x_n) \in (0, \infty)^n$, we can form the power mean of $x_1, \ldots, x_n$, weighted by $p_1, \ldots, p_n$, of order $t$. When $t \neq 0$, this is defined by $$ M_t(p, x) = \biggl( \sum_{i = 1}^n p_i x_i^t \biggr)^{1/t}. $$ We define $M_0(p, x) = \lim_{t \to 0} M_t(p, x)$, which works out to be $$ M_0(p, x) = \prod_{i = 1}^n x_i^{p_i}. $$
It's a well-known classical fact that $M_t(p, x)$ is increasing in $t$, for fixed $p$ and $x$. (I mean "increasing" non-strictly; e.g. it's constant in $t$ if $x_1 = \cdots = x_n$.) For instance, the fact that $M_0(p, x) \leq M_1(p, x)$ is the famous theorem that the geometric mean is less than or equal to the arithmetic mean. My question is, in some sense, at one level higher.
For reasons that probably aren't relevant here, I've seen plots of the function $$ t \mapsto 1/M_t(p, p) \qquad (t \geq -1) $$ for many different distributions $p$. The fact above tells us that the graph is always decreasing. But in every case I've seen, it has also looked as if it's convex. Write $q = t + 1$ ($q \geq 0$), so that $$ 1/M_t(p, p) = \begin{cases} \Bigl( \sum p_i^q \Bigr)^{1/(1 - q)} &\text{if } q \neq 1 \\ \prod p_i^{-p_i} &\text{if } q = 1. \end{cases} $$ Then in every case I've seen, the graph has looked something like this:
(source: ed.ac.uk)
Notes:
For $q \geq 0$, write $f(q) = \bigl( \sum p_i^q \bigr)^{1/(1 - q)}$. We know that $f \geq 0$ and $f' \leq 0$. I'm asking whether $f'' \geq 0$. If that's true, a natural conjecture is that $(-1)^k f^{(k)} \geq 0$ for all $k$: that is, $f$ is completely monotone. A theorem of Bernstein states that $f$ is completely monotone if and only if it's the Laplace transform of some finite measure on $[0, \infty)$.
For an arbitrary $x \in (0, \infty)^n$, it's not necessarily true that $t \mapsto 1/M_t(p, x)$ is convex in the region $t \geq -1$. There are counterexamples.
The quantity $\bigl( \sum p_i^q \bigr)^{1/(1 - q)}$ is the exponential of the Rényi entropy of $p$ of order $q$. That's why I've given this an "information theory" tag.
