# Difference of concave increasing and convex increasing function

Let $$f:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$$, and $$g:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$$ be two functions. Suppose that $$f(0) > g(0) = 0$$, and $$f$$ is strictly increasing and concave, while $$g$$ is strictly increasing and strictly convex (so $$f'(x)>0,\; f''(x) \leq 0,\; g'(x)>0$$, and $$g''(x) >0)$$. Suppose that for some $$x>0$$ we have that $$f(x)- g(x)>0$$. Is it true that then for every $$y \in [0,x]$$ it must be that $$f(y) - g(y) \geq 0$$? Seems to be correct, at least for some special cases but, but I'm unable to prove it for the above outlined general case.

$$h(x) = f(x) - g(x)$$ is strictly concave. If $$h(0)\ge 0$$ and $$h(x) \ge 0$$ for some $$x > 0$$ then $$h(y) > \frac{x-y}{x} \, h(0) + \frac yx \, h(x) \ge 0$$ for $$y \in (0, x)$$.
The monotony of $$f$$ and $$g$$ is not needed for this conclusion.