not hard but

If you know the solution, then this is not the place to post the problem. Try "Algebra Proposed and Own Problems" instead. If you don't know the solution (but, still, you find it "not hard but cool...") I'll post it.

Because $f$ is positive and increasing, it's easy to see that $\forall x,u>0 : uf(x) \leq \int _x^{x+u} f(t)dt=\int _{ax}^{a(x+u)} f(t)dt+\int _{bx}^{b(x+u)} f(t)dt $ $\leq auf(a(x+u)) + buf(b(x+u))$ so $f(x) \leq af(a(x+u)) + bf(b(x+u))$ . I take $u = \frac{1-a-b}{a+b}x \Longrightarrow (a(x+u) \leq x)\& ( b(x+u) \leq x) $ $\Longrightarrow (f(a(x+u)) \leq f(x))\& ( f(b(x+u)) \leq f(x))$ , so $f(x) \leq (a+b)f(x)$ But $0 < a+b < 1$ and $ 0 \leq f(x)$, so we have to have $\forall x>0, f(x)=0$ and $0 \leq f(0) \leq f(1)=0 \Longrightarrow f(0)=0$ q.e.d.= c.q.f.d.

enescu wrote: If you know the solution, then this is not the place to post the problem. Try "Algebra Proposed and Own Problems" instead. If you don't know the solution (but, still, you find it "not hard but cool...") I'll post it.