Regions when a concave function is smaller than another concave function

Question

Let $f_1,f_2:[0,1]\mapsto\mathbb{R}$ be two bounded and concave functions. Assume $f_1(0)<f_2(0)$ and $f_1(1)<f_2(1)$. I want to investigate the set $\mathcal{X}\triangleq\{x\in[0,1]: f_1(x)>f_2(x)\}$. Can I say there exists a finite number $M$ such that $\mathcal{X}$ is of the form $\bigcup_{i=1}^M[a_i, b_i]$ where $0<a_1<b_1<a_2<b_2<\cdots<a_M<b_M<1$? If so, how can I show it? Otherwise, can you give a counterexample? Thanks!

Answer 1

Of course not. Let me construct an example on some other interval than $[0, 1]$, you can easily adjust it by the linear change of variables.

Let $f_1(x) = -10^9 x^2$, $f_2(x) = f_1(x) + e^{\frac{-1}{x^2}}\sin(\frac{1}{x})$. Both of these functions are $C^\infty(\mathbb{R})$ and are concave on $[-1, 1]$, say, since $10^9$ is a pretty big number.

Now, their difference is $e^{\frac{-1}{x^2}}\sin(\frac{1}{x})$, which oscilates infinitely often around $0$. In particular, it is negative on an infinite number of disjoint intervals. Now, pick two small numbers $-1 < -\varepsilon_1 < 0 < \varepsilon_2 < 1$ such that $\sin(\frac{1}{\varepsilon_1}) > 0 > \sin(\frac{1}{\varepsilon_2})$, then on $[-\varepsilon_1, \varepsilon_2]$ we will meet all our conditions.