I discovered something interesting, and I would like to know whether it is a known result or not. Say that a function $f: \Omega \subset \mathbb{R} \rightarrow \mathbb{R_+^*}$ is $\alpha$-concave if $f^\alpha$ is concave.

Let $f$ a $\alpha$-concave function, $g$ a $\beta$-concave function. Let $\gamma$ be the half of the harmonic mean of $\alpha$ and $\beta$, ie. \begin{equation*}\frac{1}{\gamma} = \frac{1}{\alpha} + \frac{1}{\beta}.\end{equation*} Then $fg$ is a $\gamma$-concave function.

This can be proved by computing the hessian of $fg$. Have you already seen this result in the literature?

  • $\begingroup$ A small remark: $\gamma$ is not the harmonic mean, but half of it. $\endgroup$ – Ivan Izmestiev Nov 24 '18 at 6:59
  • $\begingroup$ You're perfectly right, thanks! I've just corrected it. $\endgroup$ – LacXav Nov 26 '18 at 15:51

$\newcommand{\a}{\alpha}$ $\newcommand{\b}{\beta}$ $\newcommand{\g}{\gamma}$

We want to prove that $ (f(ax+by)g(ax+by))^\g \ge a(f(x)g(x))^\g + b (f(y)g(y))^\g$ for every $x, y$ and $a+b = 1$. Since we know that $f(ax+by)^\a \ge af(x)^\a + bf(y)^\a$ and $g(ax+by)^\b \ge ag(x)^\b + bg(y)^\b$ it is enough to prove that

$$(af(x)^\a + bf(y)^\a)^{\frac{\g}{\a}} (ag(x)^\b + bg(y)^\b)^{\frac{\g}{\b}}\ge af(x)^\g g(x)^\g + b f(y)^\g g(y)^\g.$$

Consider set $\{x, y\}$ as measurable space with $m(x) = a$, $m(y) = b$. Then our desired inequality (after rising to the power $\frac{1}{\g}$) is nothing but Holder inequality with parameters $\a, \b$.

Unfortunatly, I too do not have any reference for this cute result, but I'm sure it is known.


This is nice, and I do not know a reference for this.

The notion of $p$-concavity is mentioned, for example, in Section 9 of

Gardner, R. J., The Brunn-Minkowski inequality, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008.

Note that a natural interpretation of $0$-concavity is log-concavity. Your result generalizes the fact that the product of log-concave functions is log-concave.

Also your result is equivalent to the following.

Theorem. If functions $F$ and $G$ are concave, then so is $F^tG^{1-t}$ for all $t \in [0,1]$.

Indeed, substitute $f^\alpha = F$, $g^\beta = G$, $t = \frac{\beta}{\alpha+\beta}$.

The concavity of $F^tG^{1-t}$ can be proved as follows. For every $\lambda \in (0,1)$ and $\mu = 1-\lambda$ the concavity of $F$ and $G$ imply \begin{multline*} F^tG^{1-t}(\lambda x + \mu y) = F^t(\lambda x + \mu y) G^{1-t}(\lambda x + \mu y)\\ \ge (\lambda F(x) + \mu F(y))^t (\lambda G(x) + \mu G(y))^{1-t}\\ \ge (\lambda F(x))^t(\lambda G(x))^{1-t} + (\mu F(y))^t(\mu G(y))^{1-t}\\ = \lambda F^tG^{1-t}(x) + \mu F^tG^{1-t}(y) \end{multline*} (In the middle we have used the inequality $(1+u)^t(1+v)^{1-t} \ge 1+u^t v^{1-t}$, for which I have only a nasty proof.)

One can prove the theorem also by computing the second derivative of $F^tG^{1-t}$ (which is fun), but the above argument is more general because it does not require differentiability.

I think these results should be known, but have no reference.

EDIT: As Alexei Kulikov pointed out, one can prove the Theorem by first proving the special case $t=\frac12$ as a lemma (in this case the nasty inequality becomes $\sqrt{(1+u)(1+v)} \ge 1 + \sqrt{uv}$, which follows from Cauchy-Schwarz). From the special case one infers by induction all $t = p/2^n$ cases, which implies the general case by a limit argument.

EDIT: A reformulation of the theorem: the space of exp-concave functions is convex.

  • 1
    $\begingroup$ It seems to me that your inequality(and the initial problem as well) can be redused to the case $t=\frac{1}{2}$ via mimicking proof of the fact that mid-point concavity implies concavity for continuous functions. And for $t=\frac{1}{2}$ it is just C-S. $\endgroup$ – Aleksei Kulikov Nov 23 '18 at 23:33
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    $\begingroup$ And moreover it is actually Holder inequality on $2$-point measurable space with appropriate functions. $\endgroup$ – Aleksei Kulikov Nov 24 '18 at 0:37
  • $\begingroup$ @AlekseiKulikov: I will add the $t=1/2$ approach to my answer. What do you mean by Hoelder inequality on $2$-point measurable space? Could you write this up as an answer? $\endgroup$ – Ivan Izmestiev Nov 24 '18 at 6:52
  • $\begingroup$ Thanks for the reference, and having put in perspective. $\endgroup$ – LacXav Nov 26 '18 at 16:35

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