Let $f : \mathbb R^+ \rightarrow \mathbb R^+$ satisfies $f(f(x)) + f(x) = 6x$. Show that $f(x) = 2x$ only.

Question

Let $f : \mathbb R^+ \rightarrow \mathbb R^+$ satisfies $$f(f(x)) + f(x) = 6x.$$

Show that $f(x) = 2x$ only solution.

I define $a_{n+1} = f(a_n)$ and $a_0 = x$

Next , I put $x = f^{n-1}(x)$ in the parent equation. This makes the recurrence relation $a_{n+2} + a_{n+1} = 6a_n$.

By using the characteristic root technique, we get $a_n = c_1(2^n) + c_2(-3)^n$. But I don’t know how to use this to make $f(x) = 2x$ only.

Thanks in advance!

Answer 1

You're most of the way there!

Hint: Show that if $c_2 \neq 0$ , then eventually some $a_n < 0$ ,contradicting $ a_n > 0$.