Question

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.

Answer 1

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

Answer 2

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

Answer 3

Assume \(f(x) \in \mathbb{R}[x].\)

Let \(f(x)=a_0+a_1x+\dots+a_kx^k.\) \(f(0)=0\) implies \(a_0=0.\)

\[
\begin{align}
f(f(x))&=a_1(a_1x+\dots+a_kx^k)+\dots+a_k(a_1x+\dots+a_kx^k)^k\\
&=a_1(a_1x+\dots+a_kx^k)+a_2(a_1x+\dots+a_kx^k)^2\\
&+\dots+a_k(a_1x+\dots+a_kx^k)^k\\
&=2(a_1x+\dots+a_kx^k)\\
&\Rightarrow a_2,a_3,\dots,a_k=0\\
&\Rightarrow a_1^2x=2a_1x\\
&\Rightarrow a_1=2\\
\therefore f(x)&=2x
\end{align}
\]