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Let $p:\mathbb R^n \rightarrow \mathbb R$ be a polynomial of degree at most $d$. Restrict $p$ to the unit cube $Q=[0,1]^n\subset\mathbb R^n$. We assume that $p$ has mean value zero on the unit cube $Q$: $$\int_Q p(x) dx = 0.$$

For $\alpha>0$ consider the sublevel sets of $P$,

$$E_\alpha= \{x\in Q: |p(x)|\leq \alpha\}$$

There are several known estimates for the Lebesgue measure of this set which in some sense or another are uniform over some classes of polynomials. For example, we have that

$$|E_\alpha| \lesssim \min(pd,n) \frac{ \alpha^{1/d} }{ \|p\|_{ L^p(Q) }^{1/d} } $$

This particular estimate is due to Carbery and Wright and can be found here.

I'm interested in studying the (induced Lebesgue) measure of the boundary of this set

$$|\partial E_\alpha|=|\{x\in Q: |P(x)|=\alpha\}| $$

Consider first the easy case of dimension $n=1$. Then the set $E_\alpha$ is a finite union of closed intervals and the question is trivial. It is obvious that in this case there are at most $O(d)$ intervals so the $0$-dimensional measure of the boundary is $O(d)$.

Now in many variables things will be much more complicated. For example can we say that the set $E_\alpha$ has $O(d)$ connected components? Is there an estimate for the measure of the boundary $\partial E_\alpha $ in terms of $\alpha$, $d$ and $n$, assuming (say) that $\|p\| _ {L^1(Q)}=1$ ?

This question comes up naturally if one tries to study an oscillatory integral with phase $p$ and apply integration by parts (i.e Gauss theorem) imitating the one dimensional method of proving the van der Corput lemma (for example).

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  • $\begingroup$ The link to the estimate of Carbery and Wright appears to be broken. Could you perhaps replace it with citation data (which would presumably be more durable)? $\endgroup$
    – Ian Morris
    Commented Aug 5, 2014 at 23:26

3 Answers 3

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There is a trivial estimate for the measure of the boundary based on the observation that $|p|^2$ is still a polynomial, so the corresponding surface intersects any line at most $2d$ times. Averaging over directions, we get $O(d\sqrt[] n)$ regardless of $\alpha$. Now, the question is what exactly you want: a dimension-independent bound, a bound that is small for large $\alpha$, or anything else. It may really help to try it from the other end: figure out what exactly you need and we'll try to figure out whether it is true or not. Otherwise, you may get plenty of answers with trivial and not so trivial estimates for everything, none of which will fit your real needs.

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  • $\begingroup$ Thank you for the answer. I am a bit confused. So you consider all the lines that meet the set $E_ \alpha$. If you average over all the lines that meet the set you will get the surface measure of the boundary (which is what we want). So we need to compute this in another way as well. I'm missing this other way. To answer your question now, I need an estimate of the measure of the boundary that stays bounded in $n$ when $$d<<n$$. So we can suppose that $d$ is fixed and $n\rightarrow \infty$. I'm not sure if this is even possible of course. You should think that $\alpha \sim d$ $\endgroup$
    – ioannis.parissis
    Commented Nov 9, 2009 at 0:45
  • $\begingroup$ This other way is just noting that coordinate lines (they are enough to average over) are restricted to the cube (in other words, the area of the projection of your set to each coordinate hyperplane is bounded by $1$). As to the bound independent of $n$ in the cube, it holds when $d=1$ (sections of the cube by hyperplanes have area at most $\sqrt[]2$) but it'll take me some time (possibly, infinite) to figure out if it holds for other $d$ as well. That the cube is unit is crucial here, of course. Seems like a really nice question (unless there is some trivial counterexample)! $\endgroup$
    – fedja
    Commented Nov 9, 2009 at 1:30
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    $\begingroup$ A small update: apparently, even the question about dimension-free estimates for the Gaussian perimeters of algebraic surfaces of fixed degree is open (your question is not easier because we can easily simulate independent Gaussians by long sums of coordinates). Even for $d=2$, the best that is known is that the Gaussian perimeters of all balls are uniformly bounded and the Gaussian perimeters of all origin-symmetric quadratic surfaces are uniformly bounded. Fascinating! It might make a nice polymath project... $\endgroup$
    – fedja
    Commented Nov 9, 2009 at 4:01
  • $\begingroup$ Let me rephrase your argument the way i understand it better. By crofton's formula for reasonable surfaced $S\subset \mathbb R ^n$,$ |S|=\int_{\mathcal L_S}n _{S}(l) d\nu(l)$ where $\mathcal {L_S}$ is the space of lines that intersect $S$, $\nu$ is the motion invariant measure on the set of lines and $n_S(l)$ is the number of intersections of $l$ with $S$. $\endgroup$
    – ioannis.parissis
    Commented Nov 9, 2009 at 15:46
  • $\begingroup$ In our case this number is at most $2d$,and we can make the integral bigger by considering all lines that intersect the unit cube. We end up with $|S|\lesssim d n$, $n$ coming from the measure of lines that intersect $Q$. We have a $\sqrt{n}$ discrepancy here! $\endgroup$
    – ioannis.parissis
    Commented Nov 9, 2009 at 15:46
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I don't know how to estimate the measure of the level sets, but I can answer the question about the number of connected components to an extent. Let $P_f(x) = (x-1)(x-2)\cdots(x-f)$, and define $p(x_1, \ldots, x_n) = P_f(x_1)^2 + \cdots + P_f(x_n)^2$. Then $p$ has degree $2f$, and the level sets $p = \varepsilon$ for small $\varepsilon$ have $f^n = (d/2)^n$ connected components (each $x_i$ must be close to one of the roots of $P_f$).

Theorem 11.5.3 of Bochnak, Coste, & Roy, Real Algebraic Geometry, says the sum of the Betti numbers, hence the number of connected components, of an algebraic set in $R^n$ defined by degree ≤d polynomials is at most $d(2d-1)^{n-1}$. So for fixed n, the maximum number of connected components of the level sets of a degree d polynomial in n variables is $\Theta(d^n)$.

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  • $\begingroup$ Thank you for the answer. I forgot to add hypothesis that the polynomial $p$ is assumed to have mean value zero on the unit cube. The second part of your answer still applies of course. I will edit my question to add this hypothesis. $\endgroup$
    – ioannis.parissis
    Commented Nov 8, 2009 at 21:30
  • $\begingroup$ I think the example can be adapted; take (p-ɛ)*L where L is some linear function whose zero set does not meet the grid; you should be able to find some hypercube on which the mean value is zero, and rescale the variables to make it the unit hypercube. Then consider the zero set of the resulting polynomial. $\endgroup$
    – Reid Barton
    Commented Nov 8, 2009 at 21:39
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An encouraging update: Daniel Kane posted his proof for the Gaussian case on arXiv yesterday. The proof is both very simple and brilliant. I won't be surprised if his technique can be modified to cover the cube case.

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