A functional inequality 
$g:[0,1]\to[0,1]$ continuously differentiable and increasing such that for
   all integers $t>0$ and for all $r\in(0,1)$, $g(r^{t+1})>g(r)\cdot g(r^t)$. Does this imply
   that for all $r\in(0,1)$, and $m,n \in N$, $g(r^{m+n})>g(r^m)\cdot g(r^n)$?
  If not, what can be a counter example?

This is the last part of a larger question that I had asked before here: https://math.stackexchange.com/questions/810277/a-unsolved-puzzle-from-number-theory-functional-inequalities. Overcoming this would help me solve the whole question.
Edit: I missed this out, when I posted initially: $g(0)=0$ and $g(1)=1$
 A: $\let\eps\varepsilon$It is easier to consider a function $f(x)=-\log g(e^{-x})$. Then $f\colon (0,\infty)\to(0,\infty)$ is still continuously differentiable and increasing (the bordering conditions are almost not in the game), and it satisfies 
$$
  f((t+1)x)<f(tx)+f(x)  \qquad (*)
$$ for all $x\in(0,+\infty)$ and positive integer $t$ (I do not know why you set $t>1$...).
Now let us consruct such an $f$ violating the desired condition. Firstly, we set $f(x)=x+\eps$ for a small $\eps>0$, so that it satisfies $(*)$. Next, we perturb $f$ in a small neighborhood of $x=5$ so that now $f(5)$ is slighly greater than $f(2)+f(3)$, thus violating the desired condition. Surely, now it also violates $(*)$, but this happens only for $x$ being in a small neighborhood of $5/2$, or for $x$ being almost in $(0,5/3)$ (or, recursively from $5/2$, for $x$ around $5/4$ and smaller ones). Thus, now it is easy to perturb $f$ in a neighborhood of $5/2$, and then simply to increase it on, say, $(0,\,1.9)$ so that it becomes, say, $1.1x+\eps$ on $(0,\,1.7)$. The details can be simply recovered.
