# 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.

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 $$f(1)=1$$.

$$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right) \tag 1$$ 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.

$$f(x)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2x}{1+x}\right)$$

$$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2\times\dfrac{2^x}{2^x-1}}{1+\dfrac{2^x}{2^x-1}}\right)$$

$$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{\dfrac{2^{x+1}}{2^x-1}}{\dfrac{2^{x+1}-1}{2^x-1}}\right)$$

$$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2^{x+1}}{2^{x+1}-1}\right)$$

$$f\left(\dfrac{2^{x+1}}{2^{x+1}-1}\right)-4f\left(\dfrac{2^x}{2^x-1}\right)=-2$$

$$f\left(\dfrac{2^x}{2^x-1}\right)=4^x\Theta(x)-2x+1$$ , 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