I am dealing with the following functional equation:
$$ f(x)=\frac 1T \sum_{t=1}^Tf(a+tx)$$ where $a \in \mathbb{R}$ is a parameter. Is there a way to find a possible $f$ that satisfies the above equation?
EDIT: $f$ should be a cumulative distribution function, in particular it must be non-decreasing and right-continuous. I need a function that works for each $a$ and $f$ can depend on $T$.
Let's assume $\lim_{y\to 0} f(y)$ exists: $$ f(x)=\frac{f(a+x)+\dots+f(a+nx)}{n}=\frac{f(a+x)+\dots+f(a+nx)+f(a+(n+1)x)}{n+1} $$ $$ (n+1)(f(a+x)+\dots+f(a+nx))=n(f(a+x)+\dots+f(a+nx)+f(a+(n+1)x)) $$
$$ f(x)=\frac{f(a+x)+\dots+f(a+nx)}{n}=f(a+(n+1)x) $$ We get that $y=a+(n+1)x \implies x=\frac{y-a}{n+1}$ $$ f(x)=f(\frac{x-a}{n+1}) $$ Then by letting $n\to \infty$ we have $$ f(x)=\lim_{y\to 0} f(y) $$ So the function is constant.