General solution to the functional equation $ \frac{f(x + (1-2x)) - f(x)}{1-2x} = f(x)$

Question

I was wondering how to derive the solution to

$$ \frac{f\big(x + (1-2x)\big) - f(x)}{1-2x} = f(x)$$

Which can be simplified to

$$\frac{f(1-x) - f(x)}{1-2x} = f(x)$$

One idea is as follows. Consider the sequence $f_i(x)$ such that

$$f_0(x) = 1$$

and

$$ \frac{f_n(1-x)-f_n(x)}{1-2x} = f_{n-1}(x)$$

Then the function

$$ g(x) = \sum_{i=0}^{\infty}\left[ f_i(x) \right] $$

Has the property that

$$ \frac{g(1-x)-g(x)}{1-2x} = \frac{\sum_{i=0}^{\infty}\left[ f_i(1-x) \right]- \sum_{i=0}^{\infty}\left[ f_i(x) \right]}{1 - 2x} = \sum_{i=0}^{\infty}\frac{f_i(1-x)- f_i(x)}{1-2x} \\= 0 + \sum_{i=1}^{\infty} \left[ f_{i-1}(x) \right] = \sum_{i=0}^{\infty}\left[ f_i(x) \right] = g(x)$$

For a glimpse of a pattern:

$$f_0(x) = 1$$ $$f_1(x) = x, \frac{1}{2}x(3x-1)$$

I'm having trouble deriving further terms.

So basically I'm wondering

1. Whats the pattern to the terms $f_i(x)$ so I can construct $g(x)$ through series
2. Once I have that pattern does $g(x)$ have a closed form? If so, what is the closed form?
3. Is there a more analytic way of directly deriving $g(x)$? What would the technique be?

Answer 1

Notice that:

$\begin{align} f(1-x)=&\dfrac{f(1-(1-x))-f(1-x)}{1-2(1-x)}\\ =&\dfrac{f(x)-f(1-x)}{1-2+2x}\\ =&\dfrac{f(1-x)-f(x)}{1-2x}=f(x) \end{align}$

Which would show that the only solution to that equation would be $f\equiv0$.