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The triangular function: as fedja remarks in his comment, when $p(x)$ is supported in $[0,1]$ and symmetric around 1/2, there is a solution from $\infty$ and a solution from $-\infty$. The condition that they coincide is that $1-g^{-1}(1-x)$ also satisfies (1) so $g^{-1}(x) = 1-g(1-x)$ and therefore $$\epsilon p(x) = g(x) + g(1-x) -1 (2)$$. Using this equation find $g(1/2) = (1 + \epsilon p(1/2))/2$. This incidentally gives us $$ g^{-1}((1 + \epsilon p(1/2))/2)$$ and using equation (1) find $g(g(1/2)$. Iterate, finding the value of g at all the points $g(1/2), g(g(1/2),...$ and iterating backwards, which you can using $g^{-1}$ in place of g and making the appropriate changes, find g at all values $g^n(1/2)$. Then, I claim, $g$ and $g^{-1}$ are linear in between. $$$$ Interpolate g linearly on $(1/2, g(1/2)$. p(x) is linear on this interval, so this wants to force $g^{-1}$ to be linear on the same interval. However, there is already a definition of $g^{-1}$, and I need to show that they are the same. To solve the equation $g^{-1}(x) = y$ for $x \in (1/2, g(1/2)$ we need to know for what values of $y$ is $g(y) \in (1/2, g(1/2)$, and that is the interval $(g^{-1}(1/2), 1/2)$ but we have already defined $g^{-1} $ to be linear on this interval. As the correct relations hold at the endpoints, and all functions are linear in between, this definition of $g$ works. $$$$ There is a lacuna I don't know how to deal with, this all seems good provided that $g^n(1/2) \rightarrow 1$ as $n \rightarrow \infty$. This can't always be true, but I don't know when what I have sketched above fails.