<s>[INST] finding the value of $f(\frac{1}{7})$ $f$ is a function mapping positive reals between $0$ and $1$ to reals. Let $f$ be given by, $f( \frac{x+y}{2} ) = (1-a)f(x)+af(y)$ where $y > x$ and $a$ being a constant. Also,$f(0) = 0$ and $f(1) = 1$. Find $f( \frac{1}{7} )$.
I have tried some things but it is not working out. How to find the $f( \frac{1}{7} )$ . 
 [/INST] Here are the calculations:
$$f(\frac{1}{7})=f(\frac{0+\frac{2}{7}}{2})=(1-a)f(0)+af(\frac{2}{7})=af(\frac{2}{7})$$
$$f(\frac{2}{7})=f(\frac{0+\frac{4}{7}}{2})=(1-a)f(0)+af(\frac{4}{7})=af(\frac{4}{7})$$
$$f(\frac{4}{7})=f(\frac{\frac{1}{7}+1}{2})=(1-a)f(\frac{1}{7})+af(1)$$
So we have
$$f(\frac{1}{7})=a^2f(\frac{4}{7})=a^2[(1-a)f(\frac{1}{7})+af(1)]$$
then
$$f(\frac{1}{7})=\frac{a^3f(1)}{1-a^2(1-a)}=\frac{a^3}{1-a^2(1-a)}$$
where $a$ satisfies $a^2(1-a)\ne1$.
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