# Product of concave functions and harmonic mean

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?

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

• 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. – Aleksei Kulikov Nov 23 '18 at 23:33
• And moreover it is actually Holder inequality on $2$-point measurable space with appropriate functions. – Aleksei Kulikov Nov 24 '18 at 0:37
• @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? – Ivan Izmestiev Nov 24 '18 at 6:52
• Thanks for the reference, and having put in perspective. – LacXav Nov 26 '18 at 16:35