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.

  • 1
    $\begingroup$ While you have proven it already there is an even easier way. Do note that $h(x) = x^{1/r}$ is convex and nonincreasing for $-1 < r < 0$ on $\mathbb{R}_+$, and $g(x) = \sum_{i=1}^n x_i^{1+r}$ is concave on $\mathbb{R}^n_+$ (very simple 2nd derivative tests). Therefore, $f = h\circ g$ must be convex, see page 84 in Vandenberghe and Boyd's "Convex Optimization". $\endgroup$
    – V.S.e.H.
    Jan 25 at 0:12
  • $\begingroup$ @V.S.e.H.: Right and thank you. But... darn! I was seeking a composition just like that. I should have thought about it. Why don't you write it up as an answer for posterity and get some credit? :-) $\endgroup$
    – Hans
    Jan 25 at 8:35
  • $\begingroup$ Yessir, as you wish :) $\endgroup$
    – V.S.e.H.
    Jan 25 at 9:31

2 Answers 2


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 $t=0$.

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

  2. 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 $\frac{r^2-1}{r^2}<1$, $$\frac{d^2}{dt^2}M(p+th)\big|_{t=0}>0.$$

  • $\begingroup$ (+1) Congratulations. $\endgroup$
    – River Li
    Jan 25 at 0:46

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.


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