Deriving a logarithmic solution for a functional equation

Question

While studying a heuristic argument for the Collatz conjecture, I found the following functional equation $$f(n)=\frac{1}{2}\left(3+f\left(\frac{1}{2}n\right)+f\left(\frac{3}{2}n\right)\right)$$ $f(n)$ behaves logarithmically, and indeed it is possible to get a solution by substituting $f(n)=a\ln(n)+b$. But is it the only solution? If so, how would one prove that it must be a logarithmic function? If not, what are all the families of solutions (using common assumptions, such as continuity)?

Answer 1

No, it is not the only solution. If $g(x)$ is solves $$g(x)-2g(2x)+g(3x)=0\tag{*}$$ and $f(x)$ solves your equation, then $f(x)+g(x)$ also solves your equation. This works also in the other direction: if $f_1(x)$, $f_2(x)$ are two solutions of your equation, then $g(x)=f_1(x)-f_2(x)$ must satisfy $(*)$.

So you might want to focus on solving $(*)$ instead, but as an example notice that linear functions $g(x)=bx+c$ are some of its solutions. Combining with $f(x)=a\ln x$ you have found (where necessarily $a=3/(2\ln 2-\ln 3)$), we get another family of solutions is $f(x)=a\ln x+bx+c$ for any $b,c$ constants.

It is possible there are other solutions, but this is enough to disprove that logarithm is the only one.

Note: As noted in comments, there will be many solutions unless additional assumptions are imposed, such as continuity. As a simple discontinuous example consider $$ f(x)= \begin{cases} a \ln x & x=2^u3^v \text{ for some }u,v\in\mathbb{Z}\\ a \ln x +x& x=2^u3^v\pi \text{ for some }u,v\in\mathbb{Z}\\ a \ln x +x+1& \text{else}\\ \end{cases} $$ You can verify it satisfies the equation, more wild examples are possible of course.