Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What are sufficient conditions on $f, g : \mathbb{R}^n \to \mathbb{R}$ such that $\{x : f(x) \geq t \} \cup \{ x : g(x) \geq u\}$ has piecewise-smooth boundary?

Some remarks:

  1. I don't mind if the conditions are stronger than necessary. In my application, $f$ and $g$ will be extremely nice functions anyway. My dream is just to have a simple-to-state condition backed up by a citation.

  2. Feel free to assume $f$ and $g$ are nonnegative, compactly supported, and $0 < t < \|f\|_\infty$, $0 < u < \|g\|_\infty$. I imagine you'll want them to be smooth, too :)

  3. To be honest, I'm naive enough that I don't even know sufficient conditions on $f$ such that $\{x : f(x) = t\}$ is a piecewise-smooth $(n-1)$-dimensional hypersurface. Maybe it's enough that $f$ is smooth and $\nabla f$ has only finitely many zeros?

  4. Perhaps it depends on the definition of "piecewise"? At first I thought that if $\{x : f(x) \geq t \}$ and $\{ x : g(x) \geq u\}$ both had piecewise-smooth boundary then their union would too. But now that looks to me like it could be wrong. E.g., for $n = 2$ the set of points $\{(x,y) : -1 \leq x \leq 1, y \leq \exp(-1/x^2) \sin(1/x)\}$ has piecewise smooth boundary. If we take its union with $\{(x,y) : -1 \leq x \leq 1, y \leq 0\}$, then the "top part" of the resulting set's boundary is the curve $\max(\exp(-1/x^2) \sin(1/x), 0)$. Is that a piecewise-smooth curve? Seems like you need to break it into a (countably) infinite number of pieces to get all pieces smooth. On the other hand, perhaps one could/should (nonstandardly?) define "piecewise-smooth" to allow for countably many pieces.

  5. Why do I even want the surface to be "piecewise-smooth"? Well, I want to apply a (higher-dimensional) version of the Cauchy-Crofton formula to it. In the textbooks I've looked at (e.g., Santalo) they usually assume that their surfaces are piecewise-$\mathcal{C}^1$. So really, I only need that, but I'm happy to require piecewise-smoothness if that makes things simpler. What I'd prefer not to have to do is to investigate what weaker conditions suffice for these Cauchy-Crofton-type formulas (e.g., is it okay for "piecewise" to allow for countably many pieces?).

Thanks very much, sorry for my naivete!

share|improve this question

2 Answers 2

up vote 3 down vote accepted

If you need a general regularity assumption on $f,g$ but no structural assumption on the shape of the level sets, I think the best one can do is to assume that $f,g$ are real analytic functions. (Of course this requires to drop the assumption of compact support). Then your set is subanalytic and in particular its boundary is made of a finite number of real analytic pieces. Anything less, e.g. $f,g$ in $C^\infty$, may produce a countable number of pieces plus a fat set where the two level sets touch each other (think Cantor). Can't you approximate your problem with real analytic functions?

share|improve this answer
    
Yes, that may do the trick. To be 100% honest, my functions $f$ and $g$ are defined on the torus $\mathbb{R}^n / \mathbb{Z}^n$, so I think they can even be genuinely analytic. I did not know the keyword "subanalytic" -- I think this should get me on my way. Thanks! –  Ryan O'Donnell Nov 9 '10 at 14:52

Assume that $\nabla f(x)$ does not vanish on $M:=\{ f=t \}$ and that $\nabla g(x)$ does not vanish on $N:=\{ g=u \}$. This implies that $M$ and $N$ are smooth $n-1$ dimensional submanifolds. Assume further that $\nabla f(x)$ and $\nabla g(x)$ are never linearly dependent at any point of $M\cap N,$ that is, $M$ and $N$ meet transversally; in particular the intersection is a smooth $n-2$ dimensional submanifold of both. So this makes the set $M\cup N$ a union of simple smooth pieces.

share|improve this answer
    
Thanks! Will this survive if $\nabla f$ and $\nabla g$ have, say, finitely many zeros, or are linearly dependent at finitely many points of $M \cap N$? –  Ryan O'Donnell Nov 9 '10 at 15:04
2  
But this means you assume a priori something on the geometry. Then it is simpler just to assume that the set is made of a finite number of pieces. At least that's how I understood the question... –  Piero D'Ancona Nov 9 '10 at 15:04
    
Piero, yes, how natural are those assumptions depends of course on the problem (if e.g. one can move a little $t$ and $u$, one can apply Sard'd theorem and make them regular levels). –  Pietro Majer Nov 9 '10 at 15:31
    
Ryan: in general, no: that is e.g the case of your example. In these cases, to understand the geometry of the level sets and their intersections, basically you need to look at further Taylor expansions of $f$ and $g$ till you find a non-degenerate situation. In this situation, having analytic functions would certainly help. For a general description you should check a book on Real Geometry (I don't have in mind one right now). –  Pietro Majer Nov 9 '10 at 15:39
    
Thank you both for the help! –  Ryan O'Donnell Nov 9 '10 at 23:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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