## anonymous 4 years ago Consider the continuous function f(x)R:R->R that satisfies f(0)=0 and f(f(x))=2f(x)∀x∈R Find f(x). Problem taken from AoPS.

1. Mr.Math

An obvious function that satisfies the given conditions is $$f(x)=2x.$$ But is this the only such function?! I doubt that.

2. anonymous

That's why there should be a formally written proof. :3

3. anonymous

Assume $$f(x) \in \mathbb{R}[x].$$ Let $$f(x)=a_0+a_1x+\dots+a_kx^k.$$ $$f(0)=0$$ implies $$a_0=0.$$ \begin{align} f(f(x))&=a_1(a_1x+\dots+a_kx^k)+\dots+a_k(a_1x+\dots+a_kx^k)^k\\ &=a_1(a_1x+\dots+a_kx^k)+a_2(a_1x+\dots+a_kx^k)^2\\ &+\dots+a_k(a_1x+\dots+a_kx^k)^k\\ &=2(a_1x+\dots+a_kx^k)\\ &\Rightarrow a_2,a_3,\dots,a_k=0\\ &\Rightarrow a_1^2x=2a_1x\\ &\Rightarrow a_1=2\\ \therefore f(x)&=2x \end{align}