finding the value of $f(\frac{1}{7})$

Question

$f$ 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 $f( \frac{1}{7} )$ .

Answer 1

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 $a^2(1-a)\ne1$.

Answer 2

For each $n$ with $1\leq n\leq 6$ we have the equation $f(\frac{n}{7}) = (1-a)f(\frac{n-1}{7}) + af(\frac{n+1}{7})$. Since we know $f(0)$ and $f(1)$, this gives us a system of $6$ linear equations in $6$ unknowns.