What are the real analytic functions $f\colon (-\frac{1}{2},\frac{1}{2}) \to \mathbb{C}$ that satisfy the following functional equation, and how are they derived? $$f(2x)=f(x-\frac{1}{4})+f(x+\frac{1}{4})\quad\text{for }x \in (-\frac{1}{4},\frac{1}{4})$$
I think this could be interesting because in addition to the obvious solutions $f(x)=zx$, this is also satisfied by $f(x)=\ln (2\cos(\pi x))$.
Note: The question of which continuous functions satisfy this property has been answered (and for $f \in \mathcal{C}^n$ see the comments to the answers there); however, the restriction on functions in that case is rather weak, and it is not at all clear to me how to characterize the real analytic solutions of this fuctional equation.
case 1. $f'$ extends continuously to $[-1/2,1/2]$ (or, more generally, $f'$ is Riemann integrable on $[-1/2,1/2]$).
Take identity $f(2x) = f(x-1/4)+f(x+1/4)$, $-1/4<x<1/4$, differentiate it to get $2f'(2x) = f'(x-1/4)+f'(x-1/4)$, $-1/4 < x < 1/4$ or: $$ f'(x) = \frac{1}{2}\;f'\left(\frac{x}{2}-\frac{1}{4}\right) +\frac{1}{2}\;f'\left(\frac{x}{2}+\frac{1}{4}\right) ,\qquad \frac{-1}{2} < x < \frac{1}{2} . \tag{1} $$ Apply (1) to each of the terms on the right to get $$ f'(x) = \frac{1}{4}\;f'\left(\frac{x}{4}-\frac{3}{8}\right) +\frac{1}{4}\;f'\left(\frac{x}{4}-\frac{1}{8}\right) +\frac{1}{4}\;f'\left(\frac{x}{4}+\frac{1}{8}\right) +\frac{1}{4}\;f'\left(\frac{x}{4}+\frac{3}{8}\right) . $$ and by induction, for any $k$, $$ f'(x) = \frac{1}{2^k}\sum_{j=1}^{2^k} f'\left(\frac{x}{2^k}+\frac{-1-2^k+2j}{2^{k+1}}\right) \tag{2}$$ Now the right-hand side in (2) is a Riemann sum for the integral $\int_{-1/2}^{1/2} f'(t)\,dt$. Using a partition of $2^k$ equal size subintervals, we evaluate at one point in each of the subintervals.
Since we assumed $f'$ is Riemann integrable on $[-1/2,1/2]$, we may take the limit in (2) to conclude that $f'(x)$ is constant. Therefore $f(x) = ax+b$ for some constants $a$ and $b$, and plugging in to the functional equation we get $f(0)=0$, so we conclude $f(x) = ax$.
case 2. remains to be done. We still get Riemann sums (2), though. In case $\int_{-1/2}^{1/2} f'(t)\,dt = \infty$ we will get $f'(x) = \infty$, no good. So the interesting case will be where the improper Riemann integral $\int_{-1/2}^{1/2} f'(t) dt$ exists, but the Riemann sums do not necessarily converge to it. Perhaps it is enough for a "principal value" to exist of the form $$ \lim_{\delta \to 0^+} \int_{-1/2+\delta}^{1/2-\delta} f'(t)\,dt $$
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If $\varphi$ is any function on $(-1/2,1/2)$, then $$ \psi(x) = \lim_{k\to\infty}\frac{1}{2^k}\sum_{j=1}^{2^k} \varphi\left(\frac{x}{2^k}+\frac{-1-2^k+2j}{2^{k+1}}\right) \tag{2'} $$ (if it exists) satisfies the functional equation $$ \psi(x) = \frac{1}{2}\;\psi\left(\frac{x}{2}-\frac{1}{4}\right) +\frac{1}{2}\;\psi\left(\frac{x}{2}+\frac{1}{4}\right) \tag{1'}$$
So integrate $\psi$ to get a solution of the original functional equation.
Now in most cases, $(2')$ yields either a constant (case 1) or $\infty$. But carefully choosing $\varphi$ will give us something interesting.
Examples ... take $$ \varphi(x) = \frac{1}{x-1/2}+\frac{1}{x+1/2} $$
and apply $(2')$ ... the result is $\psi(x) = -\pi\tan(\pi x)$, and integrating this, we get Malper's original example $\log(2\cos(\pi x))$.
Trying to go to infinity faster or slower than $1/(x+1/2)$ at the endpoints didn't give me anything interesting: either a constant or $\pm \infty$. But I did come up with a convergent case (apparently) with oscillatory discontinuity at the endpoints. Start with $$ \varphi(x) = -\frac{1}{x+1/2}\sin \left( {\frac {2\pi \,\ln \left( x+1/2 \right) }{\ln \left( 2 \right) }} \right) -\frac{1}{x-1/2}\sin \left( { \frac {2\pi \,\ln \left( -x+1/2 \right) }{\ln \left( 2 \right) }} \right) $$
Using this in $(2')$, we get this solution of $(1')$:
This oscillatory $\psi$ works out to $$ \psi(x) = \text{Im} \sum_{n=0}^\infty \left[-\left(n+\frac{1}{2}+x\right)^{-1+ia} +\left(n+\frac{1}{2}-x\right)^{-1+ia}\right] , $$ where $a = 2\pi/\log 2$. Summed, $$ \psi(x) = \text{Im}\left( -\zeta\left(1-\frac{2\pi i}{\log 2},\frac{1}{2}+x\right) +\zeta\left(1-\frac{2\pi i}{\log 2},\frac{1}{2}-x\right)\right) $$ in terms of the Hurwitz Zeta Function $\zeta(s,z) = \sum_{n=0}^\infty (n+z)^{-s}$. Its integral
satisfies the original functional equation.