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 $\sum_i p_i = 1$. Let $$ x = (x_1, \ldots, x_n) \in (0, \infty)^n $$ be a vector of positive real numbers.

In brief, my question is this:

Is the function $$ q \mapsto \biggl( \sum_{i = 1}^n p_i x_i^{q - 1} \biggr)^{1/(1 - q)} $$ ($q \geq 0$) necessarily convex?

Now let me give some context.

For each $t \in \mathbb{R}$, 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 $$ M_t(p, x) = \Bigl( \sum_{i = 1}^n p_i x_i^t \Bigr)^{1/t}. $$ 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 $$ t \mapsto 1/M_t(p, x) \qquad (t \geq -1) $$ for many different pairs $(p, x)$. The fact above tells us that the graph is always decreasing. But in every case I've seen, it's also appeared to be convex. Writing $q = t + 1$ ($q \geq 0$), the graph always looks something like this:

Graph of reciprocal power means http://www.maths.ed.ac.uk/%7Etl/images/profile.png

Notes:

1. If you can prove it when $p$ is the uniform distribution $(1/n, \ldots, 1/n)$, that's enough: there's a standard trick that will let us upgrade to an arbitrary $p$. (But my guess is that assuming $p = (1/n, \ldots, 1/n)$ is unlikely to be an enormous help.)

2. For $q \geq 0$, write $$ f(q) = 1/M_{q - 1}(p, x). $$ 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 $[0, \infty)$.