Let us consider two functions $f(x)$ and $g(x)$ defined over the unit interval. These functions are characterised by the following properties:
- They are differentiable and convex.
- $f(0), g(0) < 0$, while $f(1), g(1) > 0$
- $f(x) \geq g(x)$
Properties 1. and 2. guarantee the existence and uniqueness of the roots of $f$ and $g$ (i.e. the solutions to $f(x) = 0$ and $g(x) = 0$).
I was wondering: can I say that the roots of $f$ and $g$, say $x_0$ and $x_1$, satisfy the inequality $x_0 \geq x_1$? This looks intuitive to me but I could not come up with a proof.
If $x_0$ is the root of $f$ and $x_1$ is the root of $g$, then I think you mean to say that $x_0 \leq x_1$. This can be argued by contradiction, as it leads to a violation of convexity for $f$. Argument below:
Suppose that $x_1 < x_0$. Based on (3) above $f(x_1) \geq g(x_1) = 0$.
Now, we have that $f(0) < 0$, $f(x_1) \geq 0$, and $f(x_0) = 0$. If one were to draw this scenario (keeping in mind that $x_1 < x_0$), it's clear this violates convexity of $f$.
Or, if you prefer: convexity dictates that $f(x_1)$ is bounded above by $$\frac{x_1}{x_0}f(x_0) + \left(1-\frac{x_1}{x_0}\right)f(0) = \left(1-\frac{x_1}{x_0}\right)f(0) < 0,$$ giving us that $f(x_1) < 0$, a contradiction.
I don't think differentiability from (1) was needed here at all, so that seems like a superfluous assumption.