Functional equation in one variable: $f(x)=\frac{1}{4}f(2x)+\frac{1}{4}f\left(\frac{x}{3}\right)+\frac{1}{2}f\left(\frac{x}{5}\right)$ Find all continuous functions $f$ from $\mathbb R$ to $\mathbb R$, such that
$$f(x)=\frac{1}{4}f(2x)+\frac{1}{4}f\left(\frac{x}{3}\right)+\frac{1}{2}f\left(\frac{x}{5}\right)$$
holds for all $x$.
It is easy to see that there is a solution of the form $f(x)=a\cdot x^t+b$ for a suitable $t$ (there are two possible values of $t$ that work, $t\approx1.921475$ and of course $t=0$). Are there any other solutions? What if we drop the continuity condition?
 A: The functional equation can be rewirtten as 
$$f(x)=4 f(\tfrac x2) -f(\tfrac x6)-2f(\tfrac x{10})$$
or as
$$f(x)=2f(5x)-\frac12f(10x)-\frac12f(\tfrac 53x) $$
Assume we have a function $f_I\colon I\to\Bbb R$ on some interval $I=[a,b]$ with $a>0$ and $b\ge 10a$ such that the functional equation holds for all $x$ such that $x,2x,\frac x3,\frac x5\in I$. Then let $b'=2b$ and $a'=\frac35a$ and define 
$$f_{I'}(x)=\begin{cases}f_I(x)&a\le x\le b\\
4 f_I(\tfrac x2) -f_I(\tfrac x6)-2f_I(\tfrac x{10})&b<x\le b'\\
2f(5x)-\frac12f(10x)-\frac12f(\tfrac 53x)&a'\le x<a
\end{cases} $$
Then the functional equation holds for $f_{I'}$ for all $x$ such that $x,2x,\frac x3,\frac x5\in I'$. Also, if $f_I$ was continuos, then so is $f_{I'}$.
With this in mind, we can start with $I=[1,10]$, where the only condition imposed by the functional equation is that $f(10)=4f(5)-f(\frac53)-2f(1)$. By repeated extension of the domain as above, we ultimately arrive at a function $f\colon(0,\infty)\to \Bbb R$ where the functional equation holds for all $x>0$ (and where $f$ is continuous provided we started with a continuous function).
We can do the same process for negative $x$ by starting with $[-10,-1]$, say.
Together, this gives us a (continuous if desired) function $f\colon \Bbb R\setminus\{0\}\to\Bbb R$.
Note that the functional equation imposes no condition at all about $f(0)$. The only problem we may run into there is continuity. Unfortunately, for arbitrary starting functions defined on $[1,10]$, and $[-10,-1]$, we cannot expect $\lim_{x\to0} f(x)$ to exist and hence it may not be possible to achieve continuity throughout.
