For $0 < x < \infty$, let $\phi(x)$ be positive and continuously twice differentiable satisfying $\phi(x+1)=\phi(x)$ and $\phi(\frac{x}{2})\phi(\frac{x+1}{2}) =d\phi(x)$ with $d$ being a constant.
I have to show that $\phi$ is constant, but I don't know how to start the proof here. Any hints?
On
This question is somehow interesting, as for $\phi$ is not a positive function, you can always find a solution of form :
$\phi(x)= ce^{ i2\pi x}$
To solve it in case, I'll assume $d=1$ and $\phi$ is also continuous at $0$( Note that this function is 1-periodic)
Define $h$ as $h(x)=ln(\phi(x))$
So we have easily that:
$h(x)= h(\frac{x}{2})+h( \frac{x+1}{2})$
Which leads to, by induction,:
$ h(x) = \left( \sum_{k=0}^{2^n-1} h \left( \frac{x}{2^n}+\frac{k}{2^n}\right)\right)$
So,
$ h'(x)=\frac{1}{2^n}\left( \sum_{k=0}^{2^n-1} h' \left( \frac{x}{2^n}+\frac{k}{2^n}\right)\right) \longrightarrow h(1)-h(0)$ when $n \rightarrow \infty$ (Because h is continuous)
So $h'(x)=0$
QED.
Iterating the second condition one you get \begin{align*} \phi\left(\frac x2\right)\phi\left(\frac{x+1}2\right) &\phi\left(\frac{x+2}2\right)\phi\left(\frac{x+3}2\right) \cdots\phi\left(\frac{x+2n}2\right)\phi\left(\frac{x+2n+1}2\right)\\ &=d^n\phi(x)\phi(x-2)\cdots\phi(x-2n)\\ &=d^n\left(\phi(x)\right)^n \end{align*} where the last equality follows from the first condition on $\phi$.
Can you conclude from here?