# Inversion of mean value theorem

Given two functions $f,g: A \subset \mathbb{R} \rightarrow \mathbb{R}$ , $f,g \in C^1(A)$, I want to study the number of points of intersection of these two functions.

So I can take $h(x) := f(x)-g(x)$, $h \in C^1(A)$, and thus I want to study the roots of that function $h$. If I have two zeros, this means that, for mean value theorem, the derivative of $h$ is zero in some point between the roots.

My question is, there are conditions which allow us to say that given a root for the derivative $\alpha | h'(\alpha)=0$ there exist $\{x_1, x_2\} | h(x_1)=h(x_2)$?

Obviously there are counter examples to this ($h(x)=x^3$), but I intuitively suspect that if $h \in C^2, \alpha$ not a saddle point, then the result follow (intuitively because I'm thinking at the intermediate value theorem).

• $h(x)=x^3$ is $C^\infty$ – Justin Benfield Mar 12 '16 at 16:14
• For polynomials you can look at the multiplicity of the root. As long as that is even you will have the desired behavior. – Justin Benfield Mar 12 '16 at 16:16
• Yes, $h$ is $C^\infty$, but $0$ is a saddle point for $h$. The fact that if the polynomial is even then we have the result, holds for every even function because i can take $x_1, x_2:=-x_1$ – HaroldF Mar 12 '16 at 16:39
• You might be able to figure it out if you think about what happens between zeros of $h$ provided both $f$ and $g$ are continuous (They can be $C^0$ even) – Justin Benfield Mar 12 '16 at 16:42
• What relationship do you want between the root $\alpha$ of $h'$ and the numbers $x_1$ and $x_2$? Are you simply asking: if $h'$ has a root that's not at a saddle point, then it's not injective? (And how exactly are you defining saddle point?) – Greg Martin Mar 15 '16 at 7:40

A sufficient condition such that $h'(\alpha)=0$ implies $h(x_1)=h(x_2)$ for some $x_1 \neq x_2$ is:

$h''$ exists on $A$ (not necessarily continuous) and $h''(\alpha)\neq 0$.

Proof: It suffices to show that $h$ has a local extremum in $\alpha$. Suppose $h''(\alpha) < 0$.

Claim: There is an $x_1 < \alpha$ such that $h'(x)>0$ for all $x\in (x_1,\alpha)$.

Otherwise we can find a sequence $x_n \to \alpha,\,x_n < \alpha$ with $h'(x_n) \le 0$. Hence $$d_n := \frac{h'(\alpha)-h'(x_n)}{\alpha-x_n}=\frac{-h'(x_n)}{\alpha-x_n}\ge 0$$ and $h''(\alpha) =\lim_{n\to \infty}d_n \ge 0$. Contradiction!

Now, by mean value theorem, $h(x) < h(\alpha)$ for all $x \in (x_1,\alpha)$. Similarly one shows there is $x_2 > \alpha$ such that $h'(x) < 0$ for all $x \in (\alpha,x_2)$ which yields $h(x) < h(\alpha)$ for all $x\in (\alpha,x_2)$. Thus $h$ has a local maximum in $\alpha$.

If $h''(\alpha)>0$, then $h$ has a local minimum in $\alpha$. The proof is analogous. q.e.d.