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.
 A: First of all if we plug in $x=1$ we get
$$f(1) = \frac 12 + \frac 14 f(1). $$
This implies
$$f(1) = \frac 23$$
so it appears that you cannot have $f(1)=1$.
A: $$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right) \tag 1$$
As already pointed out this implies $f(1)=\frac23$ which is contradictory to the specified condition $f(1)=1$.
Thus the answer is : There is no solution to the problem with the original wording.
Nevertheless it is interesting to look for the general solution of Eq.$(1)$ without the condition.
We will check the solution :
$$f(x)=\frac23+c\left(\frac{x}{x-1}\right)^2 \tag 2$$
$c=$any constant.
$$f\left( \frac{2x}{1+x} \right)=\frac23+c\left(\frac{ \frac{2x}{1+x}}{ \frac{2x}{1+x}-1}\right)^2 \tag 3$$
After simplification :
$$f\left( \frac{2x}{1+x} \right)= \frac23+c\left(\frac{2x}{x-1}\right)^2$$
We put $(3)$ in $(1)$ :
$$\frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right)=\frac{1}{2} + \frac{1}{4}\left( \frac23+c\left(\frac{2x}{x-1}\right)^2\right)$$
After simplification :
$$\frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right)=\frac23+c\left(\frac{x}{x-1}\right)^2$$
Comparing to Eq.$(2)$ we see that :
$$\frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right)=f(x)$$
This agrees with Eq.$(2)$. Thus the general solution of $(1)$ is :
$$f(x)=\frac23+c\left(\frac{x}{x-1}\right)^2 \qquad (x\neq 1)$$
The particular case $c=0$ corresponds to the trivial solution $f(x)=\frac23$.
This confirmes that the condition $f(1)=1$ must be rejected if we want Eq.$(1)$ have solutions.
A: $f(x)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2x}{1+x}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2\times\dfrac{2^x}{2^x-1}}{1+\dfrac{2^x}{2^x-1}}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{\dfrac{2^{x+1}}{2^x-1}}{\dfrac{2^{x+1}-1}{2^x-1}}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2^{x+1}}{2^{x+1}-1}\right)$
$f\left(\dfrac{2^{x+1}}{2^{x+1}-1}\right)-4f\left(\dfrac{2^x}{2^x-1}\right)=-2$
$f\left(\dfrac{2^x}{2^x-1}\right)=4^x\Theta(x)-2x+1$ , 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
