Find all continuous functions $f:\mathbb{R}^{+}\to \mathbb{R}$ such that $f(x)-2f(2x)+f(3x)=0$ When thinking about this problem, the following subproblem came up:

Find all continuous functions $f:\mathbb{R}^{+}\to \mathbb{R}$ such that
$$f(x)-2f(2x)+f(3x)=0$$
for all $x \in \mathbb{R}^{+}$.

I suspect that only such solutions are linear functions $f(x)=ax+b$. If we try higher polynomials, we reach a contradictions as comparing the coefficients we get $1-2^{n+1}+3^n=0$ which has solution in non-negative integers only for $n=0,1$. It's also clear if $f_1(x),f_2(x)$ solve the problem, then $f_1(x)+f_2(x)$ and $cf_1(x)$ solve the problem as well.
I've added the continuity requirement since otherwise the equation relates only rational multiples of $2^m3^n$ for $m,n\in \mathbb{Z}$ and this leads to infinitely many discontinuous solutions. I've made the domain $\mathbb{R}^{+}$ since the equation relates positive values to themselves, similarly negative values as well as zero is by itself.
As for an attempt to solve this version of the problem, I've gone the path to find at least a solution to $x=2^m3^n$ for $m,n\in \mathbb{Z}$ and then we could use that such $x$'s are dense in $\mathbb{R}^{+}$. Then I wrote $f(2^m3^n)=g(m,n)$ and obtained $g(m,n)-2g(m+1,n)+g(m,n+1)=0$, but since it goes in both directions it required way too many initial conditions so it looked like a rabbit hole.
Edit: Thinking a bit more about the approach with solution on $\{ 2^m3^n \}$, there are in fact other than linear solutions just on this subset, so this approach does not seem to work.  One example is
$$
f(x)=f(2^m3^n)=m+2n=v_2(x)+2v_3(x).
$$
where $v_p(x)$ is a $p$-adic valuation of $x$. Clearly $m+2n\not\equiv a(2^m3^n)+b$ is not linear in $x$.
Then for $x=2^m3^n$ the functional equation gives
\begin{align}
f(x)-2f(2x)+f(3x)&=f(2^m3^n)-2f(2^{m+1}3^n)+f(2^m3^{n+1})\\
&=(m+2n)-2(m+1+2n)+(m+2(n+1))\\
&=0.
\end{align}
And there seem to be even more crazy examples, since the solutions of the corresponding recurrence $g(m,n)-2g(m+1,n)+g(m,n+1)=0$ actually require infinitely many starting points... Still this doesn't mean there are non-linear solutions when extending to $\mathbb{R}^{+}$ where we also require continuity, but I don't know how to prove/disprove that...
 A: Given any two starting values $f(1)$ and $f(2)$, you've probably already shown that $f(x) = (f(2)-f(1))(x-1)+f(1)$ for any real number of the form $2^m3^n$ with $m,n\in\Bbb Z$. But such real numbers are dense in $\Bbb R^+$, and so there can be at most one continuous function on $\Bbb R^+$ that takes those values. Therefore linear functions are indeed the only continuous solutions.
A: Suppose we consider functions $f: \mathbb{R^+}\to\mathbb{C}$. These can easily be converted back to real solutions by taking the real or imaginary part. Look for solutions of the form $f(x)=x^a$ for complex $a$.
This gives $x^a-2\cdot(2x)^a+(3x)^a = x^a(1-2\cdot2^a+3^a) = 0$
So any solution to $1-2\cdot2^a+3^a = 0$ will give a solution. $0$ and $1$ are clearly solutions to this, but Wolfram Alpha can find two complex solutions https://www.wolframalpha.com/input?i=1-2*2%5Ex%2B3%5Ex+%3D+0+complex+solutions, and I'm pretty sure there are infinitely many (Picard's theorem says there are infinitely many solutions for all but two values on the RHS).
$$x^{a+bi}=e^{a\ln x+bi\ln x}=x^a(\cos(b\ln x)+i\sin(b\ln x))$$
So there should be infinitely many solutions of the form $x^a\cos(b\ln x)$ and $x^a\sin(b\ln x)$. If the function was required to be $\mathbb{R}_{\ge0}\to\mathbb{R}$ these solutions would disappear when $a\le 0$.
