# A unsolved puzzle from Number Theory/ Functional inequalities

The function $g:[0,1]\to[0,1]$ is continuously differentiable and increasing. Also, $g(0)=0$ and $g(1)=1$. Continuity and differentiability of higher orders can be assumed if necessary. The proposition on hand is the following:

If for all integers $t>0$ and for all $r\in(0,1)$, $g(r^{t+1})>g(r)\cdot g(r^t)$, then for all $p,q\in(0,1)$, $g(pq)\geq g(p)g(q)$.

• It seems to me that you need to approximate $\ln p/\ln q$ well enough by a rational number $m/n$. Then use $r=q^{1/n}$, so $r^n=q$, and $r^m$ is close to $p$ as you want. You can easily prove by induction that the given condition implies $g(r^{m+n})>g(r^m)g(r^n)$. Letting the approximation improve in the limit you get at least $g(pq)\ge g(p)g(q)$. May be another idea is needed to get the strict inequality? Commented May 26, 2014 at 18:51
• Hello Jyrki, thanks for your reply. I like your idea of approximating lnp/lnq by a rational number m/n. But, I do not see how, induction would imply g(r^(m+n))>g(r^m)g(r^n).For example, g(r^(t+2))>g(r^(t+1))g(r)>g(r^t)[g(r)]^2, which does not help the induction hypothesis. Were you alluding to a different way of using induction? Commented May 26, 2014 at 19:32
• Ahh! May be I just made a mistake there :-) Commented May 26, 2014 at 19:46
• In your overflow posting you did not include $g(0)=0$ and $g(1)=1$. Is this maintained?
– user145413
Commented May 28, 2014 at 17:09
• It is easier to visualize if you write it instead $h(0)=1$, $\lim_{x\to\infty}h(x)=0$, and $$h((n+1)x)>h(x)+h(nx)$$ for all $x>0$. Your question is then does it imply $$h(x+y)>h(x)+h(y).$$
– user145413
Commented May 29, 2014 at 12:00

By setting $F(x)=-\log g(e^{-x})$ we get $F(0)=0,\lim_{x\to+\infty}F(x)=+\infty$ and: $$\forall x\in\mathbb{R}^+,\forall t\in\mathbb{N},\quad F((t+1)x)-F(tx)\leq F(x)-F(0),\tag{1}$$ that is a sort of concavity condition. $(1)$ can be re-written as: $$\forall x\in\mathbb{R}^+,\forall t\in\mathbb{N},\quad \int_{t}^{t+1} f(xt)\, dt \leq \int_{0}^{1}f(xt)\,dt.\tag{2}$$ It comes, almost by magic, that the function $$f(x)=\frac{1}{x+1}+\frac{2}{3}\exp\left(-3(x-9/2)^2\right)$$ satisfies $(1)$ and the boundary conditions, however: $$\int_{4}^{5}f(x)\,dx = 0.713993\ldots > 0.693147\ldots = \int_{0}^{1}f(x)\,dx,$$ so we just found a $C^\infty(\mathbb{R}^+)$ counter-example.
• I had a look at @doetoe's solution in math.stackexchange.com/questions/844013/… and try to found a regular function with the same characteristic; so I took $f(x)=\frac{1}{x+1}$ that suits the inequality nicely and introduced a gaussian perturbation by trial and error. Commented Jul 17, 2014 at 18:02
• I checked the inequality by the graphical way with the aid of Mathematica. It happens that since the perturbation is concentrated between $4$ and $5$, only the small values of $t$ (the ones below $6$) are really critical, for the remaining ones the separation between $F((t+1)x)-F(tx)$ and $F(x)$ is quite neat. Commented Jul 17, 2014 at 18:07