Prove that there exists a unique function $f: [0, \infty) \to [0, \infty)$ which satisfies the equation $f(f(x))=6x-f(x)$

Question

Prove that there exists a unique function $f: [0, \infty) \to [0, \infty)$ which satisfies the equation $f(f(x))=6x-f(x)$

$f(x)=2x$ seems to satisfy except $f(0)=0$ but $0$ is not in the codomain. Let $f(0)=a \implies f(a)=a$ which means there's a fixed point. Also, I can show that it's an injective function.

Answer 1

Let us fix $x$ and let $a_n= f^n(x)$, so $a_n$ is $n$-th iteration of $x$

Then $$ a_{n+1} = f^2(f^{n-1}(x)) = 6f^{n-1}(x) -f^n(x) = -a_n+6a_{n-1}$$

Solving this recursive equation we get: $$a_n = b(-3)^n+c2^n$$ for some real $b,c$. Now if $b \neq 0$ then for $n$ big enough we get $a_n<0$ which is not true since the domain of $f$ is $(0,\infty)$. So $b=0$ and thus $a_n =c2^n$ and thus $a_1=2c$ and $a_0 = c$ so $f(x) = 2x$.