Continuous $f$ satisfying $f(2x)=f(x-1/4)+f(x+1/4)$ on $(-1/2,1/2)$ What are the continuous functions $f\colon (-\frac{1}{2},\frac{1}{2}) \to \mathbb{C}$ that satisfy the following functional equation, and how are they derived?
$$f(2x)=f(x-\frac{1}{4})+f(x+\frac{1}{4})\,\,\,\,\,\,\,\,\text{for }x \in (-\frac{1}{4},\frac{1}{4})$$
I think this could be interesting because in addition to the obvious solutions $f(x)=zx$, this is also satisfied by $f(x)=\ln (2\cos(\pi x))$.
 A: If $f$ is as in the problem statement, then, from the functional equation and continuity of $f$ at $x=0$, we see that 
$$\lim_{x \ \to  -\frac{1}{4}} f(2x) - f(x-\frac{1}{4}) = f(0).$$ 
Now, suppose $f_0$ is any continuous function on $(-\frac{1}{2},0]$ such that
\begin{align*} \lim_{x \ \to  -\frac{1}{4}} f_0(2x) - f_0(x-\frac{1}{4}) = f_0(0). && && (*) \end{align*}
A simple sufficient condition on $f_0$ implying the above condition is $\lim_{x \to - \frac{1}{2}} f_0(x)$ exists and $f_0(0) = 0$. 
I claim that any $f_0$ as above extends uniquely to an $f$ as in the problem statement. 
We partition $(-\frac{1}{2},\frac{1}{2})$ into subintervals so that $f$ can be defined on one interval at a time. Put
\begin{align*}
x_{-1}=-\frac{1}{2} && x_0 = 0 && x_1 = \frac{1}{4} && x_2 = \frac{3}{8} && x_3 = \frac{7}{16} && \ldots
\end{align*}
\begin{align*}
I_0 = (x_{-1},x_0] && I_1 = (x_0, x_1] && I_2 = (x_1,x_2]  && I_3 = (x_2,x_3] && \ldots 
\end{align*}
Generally:
\begin{align*} x_n = \frac{1}{2} - \frac{1}{2^{n+1}} && n \geq -1 \end{align*}
\begin{align*}I_n = (x_{n-1}, x_n ] && n \geq 0.  \end{align*}
Now, whenever $x \in I_n$ for $n \geq 1$, a quick calculation shows that $2x -1/2 \in I_{n-1}$ and $x - 1/2 \in I_0$. So, we get a function $f_n$ on $I_n$ for all $n \geq 0$ by the recursive definition
\begin{align*}
f_n(x) = f_{n-1}(2x -\frac{1}{2}) - f_0(x-\frac{1}{2}) && x \in I_n && n \geq 1.
\end{align*}
By construction, the function $f$ on $(-\frac{1}{2},\frac{1}{2})$ obtained by glueing all the $f_n$ together satisfies the functional equation, this recursive definition having been derived directly from the functional equation. Since $f_0$ is continuous, it is clear that each $f_n$ is continuous. It remains to see the $f_n$ agree at the endpoints so that $f$ is also continuous. We check inductively that 
\begin{align*} \lim_{x \to x_n} f_{n+1}(x) = f_n(x_n) && n \geq 0. && && (**)\end{align*}
When $n=0$, this says
$$ \lim_{x \to 0} f_0(2x - \frac{1}{2}) - f_0(x-\frac{1}{2}) = f_0(0)$$
which is just a restatement of $(*)$, so the induction starts. Suppose the claim holds when for some $n \geq 0$ look at
$$ f_{n+2}(x)  =  f_{n+1}(2x - \frac{1}{2}) - f_0(x - \frac{1}{2})$$
with $x \in I_{n+2}$ converging to $x_{n+1}$. Then $2x - \frac{1}{2}$ is converging to $x_n$ and $x - \frac{1}{2}$ is converging to $x_{n+1} - \frac{1}{2} \in I_0$. So, by induction and continuity of $f_0$, the whole expression above is converging to 
$$f_n(x_n) - f_0(x_{n+1} - \frac{1}{2}) = f_n(2x_{n+1} - \frac{1}{2}) - f_0(x_{n+1} - \frac{1}{2}) = f_{n+1}(x_{n+1})$$
and the claim follows by induction. 
We have obtained the continuous $f$ extending $f_0$ as claimed and by design it is the unique extension satisfing the functional equation.
A: example 
Using Mike's method, with $f_0(x) = x^2$

Here is an unbounded one, $f_0(x) = \log(4x+2)$
 
So now, what about differentiable, or real-analytic solutions?
Or even solutions continuous at both endpoints.
