How to find number of distinct solutions of equation $f(x) + f(x/2) =x$.

Question

How to find number of distinct solutions of equation $f(x) + f(x/2) =x$.

by replacing x by x/2? and so on, at end i m left with f(0).

Answer 1

Let $x=2^t$ so that $f(2^t)+f(2^{t-1})=2^t$ or $g(t)+g(t-1)=2^t$. We now have an ordinary linear recurrence. The homogeneous equation has the general solution

$$g(t)=(-1)^{\lfloor t\rfloor}\phi(\{t\})$$ where $\phi$ is an arbitrary function defined over $[0,1)$. This is because the sign alternates in successive unit intervals.

The non-homogeneous equation has the particular solution

$$g(t)=\frac232^t.$$

Finally the general solution in $x$ is

$$f(x)=(-1)^{\lfloor\log_2 x\rfloor}\phi(2^{\{\log_2 x\}})+\frac23x.$$

Below, an example smooth solution with $\phi(t)=\sin(\pi t)$ (giving $\sin(\log_2 x)+\frac{2x}3$).

Note that there is an uncountable infinity of solutions.

Answer 2

$$f\left(x\right)+f\left(\frac x2\right)=x,f\left(\frac x2\right)+f\left(\frac x4\right)=\frac x2\implies f\left(x\right)-f\left(\frac x4\right)=\frac x2$$

$$f\left(x\right)-f\left(\frac x4\right)=\frac x2,f\left(\frac x4\right)+f\left(\frac x8\right)=\frac x4\implies f\left(x\right)+f\left(\frac x8\right)=\frac{3x}4$$ $$\cdots$$ $$ f\left(x\right)+f\left(\frac x{2^{2n}}\right)=\left(\frac{2}3-\frac 1{2^{2n}}\right)x.$$

Hence you can be tempted to "take the limit"

$$f(x)+f(0)=\frac{2x}3$$

and conclude $$f(x)=\frac{2x}3.$$

But this doesn't work if $\lim_{x\to0}f(x)$ does not exist.

---

Disclaimer: the right term in the RHS, third equation, is not exact, but this doesn't change the conclusion. (It gives a possible solution, but not all of them.)