I'm reading a proof where they say that given
$$\phi(x) = \Gamma(x)\Gamma(1-x)\sin \pi x$$ and $$g(x) = [\log \phi(x)]''$$
then, since $g$ is periodic with period 1, it satisfies the functional equation $$\frac{1}{4} \left(g\left(\frac{x}{2}\right) + g\left(\frac{x+1}{2}\right)\right) = g(x).$$
I haven't been able to prove this even after expanding out $g(x)$. Is there a quick proof that $g$ satisfies this functional equation?
I figured it out myself. The only way seems to be by brute force.
$$g(x)=\frac{1}{\phi\left(\frac{x}{2}\right)^2 \phi\left(\frac{x+1}{2}\right)^2}\left[\phi\left(\frac{x}{2}\right) \phi\left(\frac{x+1}{2}\right) \left(\frac{1}{4} \phi\left(\frac{x+1}{2}\right) \phi''\left(\frac{x}{2}\right)+\frac{1}{4} \phi\left(\frac{x}{2}\right) \phi''\left(\frac{x+1}{2}\right)+\frac{1}{2} \phi'\left(\frac{x}{2}\right) \phi'\left(\frac{x+1}{2}\right)\right)-\left(\frac{1}{2} \phi\left(\frac{x+1}{2}\right) \phi'\left(\frac{x}{2}\right)+\frac{1}{2} \phi\left(\frac{x}{2}\right) \phi'\left(\frac{x+1}{2}\right)\right)^2\right]$$
And this simplifies to
$$g(x)=\frac{1}{4} \left(\frac{f''\left(\frac{x}{2}\right)}{f\left(\frac{x}{2}\right)}-\frac{f'\left( \frac{x}{2}\right)^2}{f\left(\frac{x}{2}\right)^2} + \frac{f\left(\frac{x+1}{2}\right) f''\left(\frac{x+1}{2}\right)-f'\left(\frac{x+1}{2}\right)^2}{f\left(\frac{x+1}{2} \right)^2}\right) = \frac{1}{4}\left[g\left( \frac{x}{2}\right)+g\left( \frac{x+1}{2}\right)\right]. $$
Would be interested in knowing if there were a faster and simpler way though...
