I am trying to solve this functional equation:
$$g(x)+g(1-x)=1$$
Over the reals. Unfortunately, since it's symmetric, usual tricks like trying to manipulate the expression in one of the functional summands doesn't yield anything useful. It "somewhat" resembles the Cauchy's equation but I failed to find a way to properly link them together (probably for a good reason – see below).
On the assumption that g(x) is a linear function we can find it rather easy:
$$Ax+B+A(1-x)+B=1
\\
\rightarrow A = 1-2B
\\
\rightarrow g(x) = (1-2B)x+B$$
and.. it's alright, but the problem is that I want to find all functions $g(x)$ if it's possible. And I was able to prove that $g(x)$ isn't limited to just linear functions. For example, if we set $g(x) = sin^2(Ax+B)$ then by choosing $A=\frac{\pi}{2}+2B$ we get:
$$g(x)+g(1-x)=sin^2(\frac{\pi x}{2}-2Bx+B)+sin^2(\frac{\pi}{2}-\frac{\pi x}{2}+2Bx-B)=
\\
sin^2(\frac{\pi x}{2}-2Bx+B)+sin^2(\frac{\pi}{2}-(\frac{\pi x}{2}-2Bx+B)) =
\\
sin^2(\frac{\pi x}{2}-2Bx+B)+cos^2(\frac{\pi x}{2}-2Bx+B) = 1$$
Thus $g(x) = sin^2((\frac{\pi}{2}-2B)x+B)$ also satisfies the equation but it's non-linear. It's obviously the same with cosine.
This begs the question – is it even possible to find all such functions $g(x)$? And if yes, how?
Best Answer
Let $g(x) = f(x-1/2)+1/2$ (a graph of $g$ is a graph of $f$ shifted up and right by $1/2$), then
$$ g(x)+g(1-x)=1,\\ f(x-1/2)+1/2 +f(1-x-1/2)+1/2 = 1,\\ f(x-1/2) = −f(1/2-x). $$
if $z=x-1/2$, we get $f(z)=-f(-z)$ a definition of an odd function. Thus $g(x)$ is any odd function shifted up and left by 1/2.