Intersection of strictly monotone, convex functions on interval with two given intersections

Question

Given two functions $f,g:[0,1]\rightarrow[0,1]$ that are smooth, strictly convex, strictly monotone s.t. $f(1)=g(1)=0$, $g(0)=f(0)=1$ and $f(x_0)>g(x_0)$ holds for some point $x_0$, does it follow that $f(x)\geq g(x)$?

Answer 1

No.

Let $f(x) = { x(x^2-1) \over 2} + x$, $g(x) = { (1-x)((1-x)^2-1) \over 2}$, both are clearly smooth.

$f''(x) = 3x$, $g''(x) = 3(1-x)$, hence strictly convex on $[0,1]$.

$f'(x) = { 3 x^2+1 \over 2}$, $g'(x) = { 3 (2-x) x \over 2}$, hence monotone on $[0,1]$.

Finally, $\delta(x) = f(x)-g(x) = x(x-{1 \over 2}) (x-1)$, hence $\delta({1 \over 4}) > 0$, $\delta(1-{1 \over 4}) < 0$.