I'd like to find all the functions f: $[0,+\infty] \to \mathbb{R}$ wich satisfy the following requirements:
1) $f$ is continuous on $\mathbb{R+}$
2) $\forall x \ge 0, f(x)=f(e^x-1)$
I managed to prove that $\forall x \ge 0, f(x)=f(e^x-1)=f(\ln(1+x))$ but I don't even know whether it's actually useful or not.
Thanks for helping.
Let $g(x)=\ln(x+1)$. Fix $x\geq 0$, let $$a_n=g^{(n)}(x)=(g\circ g\circ\cdots\circ g)(x),$$ so $a_{n+1}=g(a_n)=\ln(1+a_n)$. Since $\ln(x+1)\leq x$, $a_{n+1}\leq a_n$. But $a_n\geq 0$, so $\lim_{n\to\infty} a_n=a$ exists and $a=\ln (1+a)$ thus $\lim_{n\to\infty}a_n=a=0$. By the continuity, we have $$f(x)=f(g^{(n)}(x))\to f(0).$$ Hence $f$ is constant.