Let $ a,b \in \mathbb{R}$
Let $f(x) : [a,b] \rightarrow \mathbb{R} $ is a continuous function in $ [a ,b] $, differentiable and strictly convex in $ (a, b) $
and $g(x) : [a,b] \rightarrow \mathbb{R} $ is a continuous function in $ [a ,b] $, differentiable and strictly concave in $ (a, b) $
How can I prove the intersection of $ f $ and $ g $ can have a maximum of two roots $f(x)-g(x)=0 $ ?
$h(x)=f(x)-g(x)$ is a strictly convex function in $[a,b]$.
So we just have to prove that a convex function has at most $2$ roots in $[a,b]$.
If $h(x)$ has $3$ roots let's say, $p,q,r$ then there is $c \in (p,q)$ such that $h'(c)=0$ (by Rolle's Theorem) and there is $d \in (q,r)$ such that $h'(d)=0$.
But it means that there is $e \in (c,d)$ such that $(h')'(e)=0$, which is a contradiction because $h''(x)>0$.
If there were three distinct roots $x_1<x_2<x_3$ of $f(x)-g(x)=0$ then by choosing $\lambda_2$ so that $x_2=\lambda_2x_1+(1-\lambda_2)x_3$ we would have $$ (f-g)(x_2)=0=\lambda_2(f-g)(x_1)+(1-\lambda_2)(f-g)(x_3), $$ which would be in contradiction with the strict convexity of $f-g$.
If $f(x)$ is strictly convex and $g(x)$ is strictly concave, then $f(x) - g(x)$ is also strictly convex (the negative of a concave function is convex).
A strictly convex function has its derivative monotonically increasing and so can have at most two zeroes.
This comes about because a differentiable function with three zeroes on a closed interval $[a,b]$ must have at least two extrema on $(a,b)$ (a consequence of Rolle's Theorem, I think) but having two extrema violates monotonicity of the derivative.
