I am not experienced with functional equations, and I have been wondering about equations of the form
$$ f(x) = f(mx+c) $$
For positve $m,c$, where we further restrict the solutions to be continuous.
It is clear that if we have defined a function on the interval $$[0,c]$$
that satisfies this equation, we can expand it to a function on all positive real numbers, since
$$ g([0,c]) = [c,(m+1)c] $$
(Where $g(x) = mx + c$ )
So that a repeated application of $g$ to the existing domain allows us to expand the domain of $f$ arbitrarily.
Moreover, the graph of $f$ on this second interval will be a stretched version of the graph of $f$ on the first interval, so the function must have some sort of periodic form, where it repeats itself but more stretched in increasingly larger disjoint, covering intervals.
My questions then is, is it sufficient to define any continuous function on the interval $[0,c]$ which satisfies $f(0)= f(c)$? I suspect the answer is yes, since the boundary conditions ensure continuity, and no function can violate the condition on this interval, but I am unsure
My second question is, how does one extend this to the negative real numbers? You could rewrite the functional equation as
$$ f\left(\frac{x-c}{m}\right) = f(x) $$
And if we let $h(x) = \frac{x-c}{m}$ then
$$h([0,c]) = [-\frac{c}{m},0]$$
And
$$ h^{2}([0,c]) = [-\frac{c}{m^2}-\frac{c}{m},-\frac{c}{m}]$$
And it looks like repeatedly applying this again yields a disjoint cover of the negative real numbers, so it seems as if it is sufficient to define any continuous function on $$[0,c]$$ satisfying the boundary conditions, but I am not quite sure how to prove some of these statements, and am just a bit unsure about my reasoning in general, any help, including more elegant solutions or hints would be appreciated.