Let $f\left(p\right)$ and $g\left(p\right)$ be monotonically increasing convex functions on the interval $\left[0, 1\right]$.
We assume $f\left(0\right)=g\left(0\right)=0$, and $f\left(1\right)=g\left(1\right)=1$. In addition, $f\left(p\right) > g\left(p\right)$, $\forall p \in \left(0,1\right)$.
I am pretty sure that the distance function $D\left(p\right) = f\left(p\right) - g\left(p\right)$ must have only one maximum in $\left(0,1\right)$. It's easy to show existence, but I can't prove uniqueness.
Any help will be appreciated. Thanks!
There is no uniqueness - the distance function can have several maxima. The reason is that the difference between two increasing convex functions does not need to be increasing or convex at all.
I constructed the following counterexample using Mathematica:
$$f(x)=\frac13(x+1)^2 +\frac{1}{48}\left(x-\frac{1}{2}\right)^2-\frac{65}{192} \\ g(x) = \frac13(x+1)^2 +\frac13\left(x-\frac{1}{2}\right)^4 - \frac{17}{48}$$
These two functions are monotonously increasing and convex on the interval $[0,1]$ with $f(0)=g(0)=0$, $f(1)=g(1)=1$ and $f\geq g$, but the distance function defined by $D(x) = f(x)-g(x)$ has two maxima, one at $x=\frac12-\frac{\sqrt 2}{8}$ and one at $x=\frac12+\frac{\sqrt 2}{8}$.
I recommend using software to check that this indeed a valid counterexample. Heuristically, you can convince yourself that $D$ goes like $y^2-y^4$ (here $y=4x-2$) which has maxima at $y=\pm \frac1{\sqrt{2}}$.