3
$\begingroup$

I ran into the following "simple" question and I am wondering whether there are any references, which might help me. I am coming from statistics, so I am not so aware which branch of math could have already dealt with this question.

Let $S\subset\mathbb{R}^D$ be compact and assume that $f,g: S\rightarrow \mathbb{R}$ are $C^2$ Morse functions (so basically requiring finitely many critical values). Further assume that $f+g$ has only finitely many isolated critical points. Denote with $C_f$ and $C_g$ the number of critical points of $f$ and $g$.

Is it possible to find a bound on $C_{f+g}$ depending only on $C_f, C_g$ and $D$ and maybe $S$?

For polynomials this is easy and simply Bezout's Theorem, yet I need something like it for more general functions.

$\endgroup$
1
  • 1
    $\begingroup$ Suppose that $f$,$g$ are also stable, meaning that different critical points have different critical values. This is a generic situation, i.e., probability 1 event. Consider the function $h_t=f+tg$. For $t$ very small $h_t$ has the same number of critical points as $f$ and for $t$ large it has the same number of critical points as $g$. $\endgroup$ Commented Aug 28, 2020 at 10:41

1 Answer 1

5
$\begingroup$

Is the following example helpful? This is inspired by the fact that no bound on the "degree" of the two functions $f$ and $g$ is assumed. As far as I understand Bezout's theorem, this would make a bound difficult even for polynomial functions.

Here let $A,B > 1$ be large and $0 < \epsilon < 1$ be small. Let $D = 1$, and $S = [0,1]$ be the closed unit interval. Define $f: x \mapsto Ax + \epsilon \cos(Bx)$ and $g: x \mapsto Ax$. Then $f'(x) = A - B\epsilon \sin(Bx)$ and $g'(x) = A$, so that neither function has critical points in $[0,1]$ provided just that $A > B \epsilon$. However their difference $f-g$ has derivative $(f-g)'(x) = - \epsilon \sin(Bx)$. By taking suitably large $A,B$ this function has arbitrarily many critical points in $[0,1]$.

Perhaps it is worth pointing that really only $B$ needs to be large in the argument above. In particular, by first picking $B$ large and then a small $\epsilon > 0$ in terms of it, one is free to pick a small $A$ as well. Therefore the functions $f,g$ can be taken to have arbitrarily small first and second derivatives.

$\endgroup$
1
  • $\begingroup$ Hey, thanks. That is a nice, simple example. Still trying to bound moments of the critical value of a "simple" random field obtained through basis functions, yet I might need to use Kac-Rice-formulas to do so. $\endgroup$
    – BrainSlap
    Commented Sep 1, 2020 at 6:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .