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).

  • $\begingroup$ $h(x)=x^3$ is $C^\infty$ $\endgroup$ – Justin Benfield Mar 12 '16 at 16:14
  • $\begingroup$ For polynomials you can look at the multiplicity of the root. As long as that is even you will have the desired behavior. $\endgroup$ – Justin Benfield Mar 12 '16 at 16:16
  • $\begingroup$ 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 $x_1, x_2:=-x_1$ $\endgroup$ – HaroldF Mar 12 '16 at 16:39
  • $\begingroup$ 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) $\endgroup$ – Justin Benfield Mar 12 '16 at 16:42
  • $\begingroup$ 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?) $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.