If $f(2x)=f(x)+f(2), \forall x > 1$ prove that $f$ in increasing

Question

Let $f:(1, +\infty)\to(0,+\infty)$ and $f(2x)=f(x)+f(2), \forall x > 1$. Prove that f is increasing, if $f$ is a strictly monotone function.

I thought maybe I could use proof by contradiction, but I don't know how to do that. I also tried to prove it by showing that $\forall x_1,x_2 \in (1,+\infty)$ with $x_1<x_2 \Rightarrow f(x_1)<f(x_2)$ but again dead end. Any suggestions?

Answer 1

Since you know $f$ is strictly monotone and positive, note that $f(4) = 2f(2) > f(2)$ so $f$ must be strictly increasing.

Interestingly, $f(x) = a\ln(x)$ is an example for any $a \in \mathbb{R}$, and it is indeed increasing...

Answer 2

As

$$ f\left(2^{\log_2(2x)}\right)=f\left(2^{\log_2(x)}\right)+f(2) $$

calling $F(\cdot) = f\left(2^{(\cdot)}\right)$ and $z = \log_2 x$ we have the functional recurrence

$$ F(z+1)=F(z)+f(2) $$

with solution

$$ F(z) = f(2)z + \Phi(z) $$

with $\Phi(z)$ a generic periodic/constant function with period $1$ and going backwards

$$ f(x) = f(2)\log_2 x + \Phi(\log_2 x) $$

now to be in accordance with $f(4) = 2f(2)$ we need $\Phi(\log_2 x) = C_0 = 0$ hence

$$ f(x) = f(2)\log_2 x $$

which is strictly increasing.