Suppose I want to find all the continuous solutions to the functional equation $f(x)+f(x+1)=f(2x+1),\tag{E1}$where $f$ is a continuous and $2$-periodic function defined on the dyadic rationals.
I know that if I have any continuous function $g$ satisfying $g(x)+g(x+1) = 0,\tag{E2}$ then I can build a continuous solution for the original equation as the uniformly convergent series $f(x) = g(x) - g(2x)/2 - g(4x)/4 - \cdots$
Indeed, we get $f(x)+f(x+1)-f(2x+1) = g(x)+g(x+1)-g(2x)-g(2x+1) = 0$.
Conversely, from a solution $f$ for $(\rm E1)$ I can build the corresponding solution $g$ for $(\rm E2)$, as the simply convergent series $g(x) = f(x) + f(2x)/2 + f(4x)/2 + \cdots - f(1)/2.$ For every dyadic number $x$, eventually $2^nx$ is an even integer and then $f(2^n x) = f(0) = 0$, so this is why the series converges.
And again, we get $\begin{align*} g(x)+g(x+1) &= f(x)+f(x+1)+f(2x)+f(4x)+\cdots -f(1) \\ & = f(2x)+f(2x+1)+f(4x)+\cdots-f(1) \\ & = \cdots = f(1)-f(1) \\ & = 0. \end{align*}$
But this time, because $g$ is not uniformly convergent, there is no guarantee that $g$ will be continuous.
These two operations are inverse of each other, so when looking at non-continuous solutions, there is a one-to-one correspondence between solutions of the two equations. But one direction preserves continuity, while the other doesn't.
Thus, my question is:
Are there continuous solutions for $(\rm E1)$ corresponding to non continuous solutions for $(\rm E2)$ ?
I don't see a priori why there wouldn't be, but I can't grasp a way to build one either. I also know that Fourier theory can be used to answer this kind of problems, but I don't know if it gives stronger results than what I have just exposed.