I am trying to find a closed form expression $f$ such that
$$f(g(x+1) - g(x)) + f(g(x) - g(x-1)) = f(g(x))$$
For all functions $g$
I have concluded that for polynomials
$$2^{n+1}f(0) = f(a_0 + a_1x + a_2x^2 ... a_nx^n)$$
But I am not sure how to extend this to rational functions etc... Any suggestions? My derivation of the aforementioned identity for polynomials is below:
My Derivations:
As an example if we set $g$ to $x$
$$f(x+1 - x) + f(x - x + 1) = f(x) \rightarrow 2f(1) = f(x)$$
Naturally we can attempt to pull a little more data out of this so we try $g = cx$ which yields:
$$ 2f(c) = f(cx)$$
Now I can attempt functions for the form $g = cx + d$ which yields
$$ 2f(c) = f(cx + d) $$
Therefore it may be true that $ f(g(x) + e) = f(g(x)) \forall e \in C $
We can now attempt to build up this functional for the quadratic functions $g = ax^2 + bx + c$ by noting
$$a(x+1)^2 + b(x+1) + c - ax^2 - bx - c = a(2x + 1) + b$$
and therefore
$$f(2ax + a + b) + f(2ax - a + b) = f(ax^2 + bx + c)$$
Thus:
$$2f(2a) + 2f(2a) = f(ax^2 + bx + c) \rightarrow 4f(2a) = f(ax^2 + bx + c)$$
From here it is easy to see that one can recursively take any polynomial $g = a_n x^n + ... + a_0$ and evaluate
$$f(a_n x^n + .... + a_0)$$
as simply $$2^n f(n! a_n)$$ and furthermore by observing that for constants
$$f(0) + f(0) = f(c) \rightarrow f(c) = 2f(0)$$
We can show now
$$f(a_0 + a_1x + ... + a_nx^n) = 2^{n+1}f(0)$$
For all $x \in \mathbb{R}$ we can find such a function $g$, that:
(i)$g(x)=x$
(ii)$g(x+1)=g(x-1)=0$
if we substitute it to equation $f(g(x+1) - g(x)) + f(g(x) - g(x-1)) = f(g(x))$ you have:
$$f(-x)+f(x)=f(x)$$
So for each $x \in \mathbb{R}$ you have:
$f(-x)=0$
So $f(x)=0$ for $x \in \mathbb{R}$