I would like to try my luck here for the following question after failing to elicit an answer to [it on math.stackexchange.com](https://math.stackexchange.com/q/4363044/64809).




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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.



Convexity of the exponential of the negative Renyi entropy
Take any $r\in(-1,0)$, any vector $(p\_i)\_{i=1}^n$ with $p\_i>0$ for all $i$, and any vector $(h\_i)\_{i=1}^n\in\mathbb R^n$. For all real $t$ close enough to $0$, let
$$g(t):=M(p+th).$$
Then
$$g''(0)=(1+r)\Big(\sum\_{i=1}^n p\_i^{r-1}h\_i^2\,\sum\_{i=1}^n p\_i^{1+r}
+\frac{1-r^2}{r^2}\,\Big(\sum\_{i=1}^n p\_i^r h\_i\Big)^2\Big)
\Big(\sum\_{i=1}^n p\_i^{1+r}\Big)^{1/r-2},$$
which is manifestly $\ge0$.


It follows that $M$ is indeed convex (actually, convex on the entire positive orthant of $\mathbb R^n$).


