If there exists a unique function $$f : R → R$$ such that $f$ is continuous at $x = 0$, and such that for all $x ∈ R$ such that $$f(x) + f(\frac{x}{2}) = x$$ Then Will it be continuous$?$
I tried many functions to fullfill tthe above criteria but unable to find any. Thanks in advance.
Hint:
$$f(x)+(-1)^{n-1}f\left(\frac{x}{2^{n}}\right)=\sum_{i=0}^{n-1} (-1)^i \cdot\left( f\left(\frac{x}{2^i}\right)+f\left(\frac{x}{2^{i+1}}\right) \right) = \sum_{i=0}^{n-1} (-1)^i\cdot \frac{x}{2^{i}}= \frac{1-2^{-n}}{3/2}x$$ $$\implies f(x) =\frac{2}{3}x -\frac{2^{-(n-1)}}{3}x - (-1)^{n-1}f\left(\frac{x}{2^{n}}\right)$$ And it suffices to use the continuity of the function $f$ at $x = 0$.