Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ satisfying $$f(x)+f(\frac{x}{2})= \frac{x}{2}$$ $\forall x \in \mathbb{R}^+$.
My Attempt :
$-\frac{x}{3} + f(x) = \frac{x}{6} - f(\frac{x}{2})$
Let $g(x) = f(x) - \frac{x}{3}$
so $g(x)=-g(\frac{x}{2})\;$ $\forall x \in \mathbb{R}^+$
then $g(x)=g(\frac{x}{4})$
Please suggest how to proceed.
Let $x=2^z$. The equation is
$$f(2^z)+f(2^{z-1})=2^{z-1},$$ or
$$g(z)+g(z-1)=2^{z-1},$$ which is an ordinary linear recurrence.
The solution of the homogeneous equation $g(z)+g(z-1)=0$ is, by induction,
$$g(z)=(-1)^{\lfloor z\rfloor} g(\{z\})$$ where $\{z\}$ denotes the fractional part.
A particular solution of the non-homogeneous equation is
$$\frac23 e^{z-1}$$ (by undeterminate coefficients on $g(z)=ce^{z-1}$).
Finally,
$$f(x)=(-1)^{\lfloor\log_2x\rfloor}g(\{\log_2x\})+\frac x3$$ for some function $g$ defined on $[0,1)$.
If you want $f$ continuous, then $g$ must be continuous and $f(1^-)=f(1^+)$, or
$$(-1)^{-1}g(1^-)=(-1)^0g(0^+).$$