A functional equation (another) I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $x\in [1/2,1]$
$$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right).$$
I am actually not sure about existence of a solution.

$f(x)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2x}{1+x}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2\times\dfrac{2^x}{2^x-1}}{1+\dfrac{2^x}{2^x-1}}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{\dfrac{2^{x+1}}{2^x-1}}{\dfrac{2^{x+1}-1}{2^x-1}}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2^{x+1}}{2^{x+1}-1}\right)$
$f\left(\dfrac{2^{x+1}}{2^{x+1}-1}\right)-4f\left(\dfrac{2^x}{2^x-1}\right)=-2$
$f\left(\dfrac{2^x}{2^x-1}\right)=4^x\Theta(x)-2x+1$ , where $\Theta(x)$ is an arbitrary periodic function with unit period (according to http://eqworld.ipmnet.ru/en/solutions/fe/fe1102.pdf)
$f(x)=\dfrac{x^2}{(x-1)^2}\Theta\left(\log_2\dfrac{x}{x-1}\right)-2\log_2\dfrac{x}{x-1}+1$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
