A condition on function $f(x+a)+f(x+b)=\frac{1}{2}f(2x)$ implies periodicity I stumbled upon the following question (I will provide more information about its source at the end):
Let $a$ and $b$ be distinct real numbers. Suppose $f:\mathcal{R}\to\mathcal{R}$ is continuous such that $f(x) = o(x^2)$ as $x\to\pm\infty$, and 
$$f(x+a)+f(x+b)=\frac{1}{2}f(2x),\ \forall x.$$
Then, $f$ must be periodic.
Here are my thoughts on this problem: if we could take the Fourier transform of the above equation, we get
$$\left(e^{iak}+e^{ibk}\right)\hat f(k) = \frac{1}{4}\hat f(k/2)$$
$$\Rightarrow |\hat f(k)| = \frac{|\hat f(k/2)|}{8|\cos\left(k\cdot(a-b)/2\right)|}.$$
Applying the above recursively, we see $\hat f(k) = 0$ for all $k$ except when $\cos\left(k\cdot (a-b)/2^n\right) = 0$, some $n$. This looks like the spectrum of a periodic function with period $2|a-b|$. This makes sense but it is far from an actual proof. Moreover, the Fourier approach could be the wrong way to go about this problem. So, any ideas?
About the source of this problem, it was published in Nieuw Archief voor Wiskunde 23(1975) p.176 (problem number 409) and I believe there is a solution published in the same journal issue 24(1976) p.101. Unfortunately, I don't have access to this journal (I found the problem on another site first and then in a book which referenced its proper source). If anyone has access to this journal and is willing to post the solution, I'd be very thankful.
 A: Replacing $x$ by $x+a$ and defining $f(x+2a)=g(x),$ we have $$g(x)+g(x+b-a)=\frac12\,g(2x).$$ Replacing $x$ by $2(b-a)x$ and defining $g(2(b-a)x)=h(x),$ we arrive at $$h(x)+h\left(x+\frac12\right)=\frac12\,h(2x).$$ Replacing $x$ by $x+\frac14$ and adding the result to the original equation, we get $$h(x)+h\left(x+\frac14\right)+h\left(x+\frac12\right)+h\left(x+\frac34\right)=\frac12\left[h(2x)+h\left(2x+\frac12\right)\right]=\frac14\,h(4x).$$ Iterating that process (i.e. replacing $x$ by $x+\frac1{2^n}$ and adding to the previous equation), we have
$$\sum^{2^n-1}_{k=0}h\left(x+\frac{k}{2^n}\right)=\frac1{2^n}\,h\left(2^n x\right),$$ i.e.
$$\frac1{2^n}\,\sum^{2^n-1}_{k=0}h\left(x+\frac{k}{2^n}\right)=\frac1{2^{2n}}\,h\left(2^n x\right).$$ Under our assumptions, $h$ is integrable, and $h(x)=o(x^2),$
so by letting $n\to\infty,$ we see that $$\int^1_0 h(x+t)\,dt=\int^{x+1}_x h(t)\,dt=0.$$ Since $h$ is continuous, we can differentiate, so $$h(x+1)-h(x)=0.$$ Thus, the original function $f$ must have period $2(b-a).$ 
