Convexity of the exponential of the negative Renyi entropy For $r\ge -1$, the exponential of the negative Renyi entropy is defined as
$$M(p):=\Big(\sum_i p_i^{1+r}\Big)^{\frac1r},$$
for a probability measure as tuples $p:=(p_i)_i$
I would like to prove the convexity of $M(\cdot)$, or
$$M(ap+bq)\le aM(p)+bM(q),$$
$\forall\,a+b=1 \wedge a,b\ge0$, and two probability measures $p$ and $q$ with the same cardinalities.
For $r>0$, I can show the convexity
via the Minkowski inequality for $\big(\sum_i x_i^{1+r}\big)^{\frac1{1+r}}$ then the convexity of $f(x):=x^{1+\frac1r}$.
But how would one show the convexity for $-1<r<0$? The above technique does not work since the inequality signs from the two steps point in the opposite directions.
 A: This is actually easier than I thought. I came to the same idea slightly before @IosifPinelis suggested it.

Consider $x>0$, $h\in \mathbf R^n$ and
$$S(p+th):=M(p+th)^r=\sum_i(p_i+th_i)^{1+r}.$$
$$\frac d{dt}S(p+th)^{\frac1r}=\Big(1+\frac1r\Big)S^{\frac1r-1}\sum_i(p_i+th_i)^rh_i,$$
\begin{align}
\frac{d^2}{dt^2}M(p+th)&=\frac{d^2}{dt^2}S(p+th)^{\frac1r} \\
&=\Big(1+\frac1r\Big)S^{\frac1r-2}\Big(\frac{1-r^2}{r}\big(\sum_i(p_i+th_i)^rh_i\big)^2+rS\sum_i(p_i+th_i)^{r-1}h_i^2\Big) \\
&=(1+r)S^{\frac1r-2}\Big(\sum_ix_i^{r+1}\sum_ix_i^{r-1}h_i^2-\frac{r^2-1}{r^2}\big(\sum_ix_i^rh_i\big)^2\Big),
\end{align}
where $x_i:=p_i+th_i$. Now set $t=0$.

*

*For $r\in[-1,1]$, the last expression is obviously positive.


*For $r>1$, aside from the method I stated in my question, we can prove it via the Cauchy-Schwarz inequality as follows.
$$\Big(\sum_ix_i^{\frac {r+1}2}\big(x_i^{\frac{r-1}2}h_i\big)\Big)^2\le \sum_ix_i^{r+1}\sum_ix_i^{r-1}h_i^2.$$
Together with $\frac{r^2-1}{r^2}<1$,
$$\frac{d^2}{dt^2}M(p+th)\big|_{t=0}>0.$$
A: Let $h(t) = t^{1/r}$ and $g(x) = \sum_{i=1}^n x_i^{1+r}$, $0< r < -1$.
We have that $h'(t) = \frac{t^{1/r-1}}{r} \leq 0$ for all $0 \leq t$, so $h$ is nonincreasing on $\mathbb{R}_+$, and $h''(t) = (1/r-1)\frac{t^{1/r-2}}{r} \geq 0$ for all $0\leq t$, so $h$ is convex.
Similarly, $u(t) = t^{1+r}$ is concave on $0 \leq t$, since $u''(t) = r(1+r)t^{r-1} \leq 0$, which implies that $g(x)$ is concave (sum of component-wise concave functions).
Now, $g(\theta x + (1-\theta)y)\geq \theta g(x) + (1-\theta)g(y),$ so $h(g(\theta x + (1-\theta)y)) \leq h(\theta g(x) + (1-\theta)g(y)) \leq \theta h(g(x)) + (1-\theta)h(g(y))$.
Therefore $f=h\circ g$ is convex, as desired.
