"Theoretical" exercise on convex function

Question

Good evening, i've been struggling with this exercise since i read it and i was unable to find the solution,maybe is very simple but i can't figured it out.

The Exercise was :

$f : \mathbb R \to \mathbb R $. convex function:

(a) what we can say about the solution $f(x)=0$?

(b) what if the function was strongly convex?

I tried to do some consideration with $f''(x)$ and $f'(x)$ but (a) seems to general and i really don't what to answer. Maybe that there at least two? But then i don't know how to prove it.

Thanks.

Answer 1

Set $S := \{x\in\mathbb{R}\mid f(x)=0\}$.

(a): $S$ is empty, contains one point, contains two points, or is an interval.
(All these cases can occur: $x^2+1$, $x^2$, $x^2-1$, $-1$.)

(b): In this case, we are down to: $S$ is empty, contains one point, contains two points.

Answer 2

Suppose $f$ is convex and let $S =\{x\in\mathbb{R}\mid f(x)=0\}$. As seen in the other answer, $S$ can be empty, and have cardinality $1$ or $2$. If $S$ has more than $2$ elements, it must be an interval. Indeed, if $a<b<c$ are in $S$, since $\tau_b:x\mapsto \frac{f(x)-f(b)}{x-b}$ is increasing and $\tau_b(a) = \tau_b(c)=0$, it must be that $\forall x\in (a,c)\setminus\{b\}, \tau_b(x) = 0$, thus $\forall x\in (a,c)\setminus\{b\}, f(x) = 0.$

If $f$ is strongly convex, $f$ cannot be $0$ on an interval: if there is such an interval $I$, $f$ is twice-differentiable over $I$, thus must satisfy $f''\geq m$ for some $m>0$ on that interval, a contradiction.