A unsolved puzzle from Number Theory/ Functional inequalities

Question

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 $p,q\in(0,1)$, $g(pq)\geq g(p)g(q)$.

Answer 1

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.