About $ f(x^2) = f(x) + f(x/2) $

Question

I was thinking about the equation

$$ f(x^2) = f(x) + f(x/2) $$

This should be consistant for $x>1$. And probably for all reals. But I focus on $x>1$.

This equation implies that $f$ grows slower than a logarithm or any power of a logarithm. But still faster than a double logarithm.

[*] Maybe like $\exp( \sqrt \ln (\ln(x)) ) $ ??

It is a simple Equation, but I have not seen it before.

The estimate [*] comes from The similar equation

$$ g(n+1) = g(n) + g(n/2) $$

Or

$$ h’(x) = h(x/2) $$

Which I discussed here before and is strongly related to the so-called binary partitions function.

$h$ can be given by an infinite sum. Can $f$ be given by an infinite sum ?

See If $f(x) = \sum \limits_{n=0}^{\infty} \frac{x^n}{2^{n(n-1)/2} n!}$ then $f^{-1}(f(x)-f(x-1))-\frac{x}{2}$ is bounded

Answer 1

$f(x)=1+\log_2(x)$ solves the equation for $x > 0$:

$f(x^2)=1+\log_2(x^2)= 1+2\log_2(x)$ $f(x)+f(x/2)=1+\log_2(x) + 1+\log_2(x/2)=1+\log_2(x) + 1+\log_2(x)-1=1+2\log_2(x)$.