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

In a forthcoming paper on nodal domains of Gaussian random functions, we (I and Misha Sodin) have a statement that is, roughly speaking, the following: if bounded nodal domains are possible at all, they have certain positive density. This sounds great until one asks a naive question "When are they possible at all?". Stripped of all irrelevant high tech terminology, this boils down to the following:

Let $K$ be an origin symmetric compact set in $\mathbb R^n$ having no isolated points and not contained in a hyperplane. Can one always construct a real-valued trigonometric polynomial $f(x)=\sum_{y\in K}\;c_y\; e^{i\,y\cdot x}$ (where all but finitely many $c_y$ vanish and $c_{-y}=\bar c_y$) such that the set $f\ge 0$ has at least one bounded connected component? If not, how to describe $K$ for which it is possible?

share|improve this question
    
Could you clarify what is meant by $e^{i(y,x)}$? –  Thierry Zell Aug 21 '11 at 17:07
    
presumably exp[ i * <y,x> ] where <y,x> is the standard Euclidean inner product? –  Anthony Quas Aug 21 '11 at 21:10
    
He uses cdot a few times in this one: arxiv.org/abs/1003.4237 but typically uses langle y,x rangle for, say, Fourier transform. –  Will Jagy Aug 21 '11 at 22:17
1  
Yes, it is $i$ (square root of $-1$) times the usual scalar product of $x$ and $y$. This time I was just a bit lazy to type langle and rangle and thought that the (,) notation would do. Well, a lazy person does everything twice, indeed :). Today I also noticed that just the hyperplane condition isn't quite enough but I'll be almost equally happy with "not contained in finitely many hyperplanes" or "has a point no neighborhood of which is contained in a hyperplane" as a condition. –  fedja Aug 22 '11 at 19:49
    
@PietroMajer Could you, please, finish the sentence? :-) –  fedja Oct 16 '13 at 13:20
show 1 more comment

1 Answer

Let's denote $F_K$ the family of real-valued trigonometric polynomials corresponding to $K$, and assume that $K$ has a point in the interior of its convex envelope. Then, there is a function $f$ in $F_K$ for which $\{f\ge 0\}$ has a bounded component.

To show this we can freely apply a linear transformation to $K$, for $F_{LK}=\{f\circ L^T\, :\, f\in F_K \}$. In particular we can assume that $K$ includes the standard basis $ \{ e_1,\dots, e_n\}$, and there is in $K$ one more point $y $ with $y_j\ge0$ and $\|y\|_1:=\sum_{j=1}^n y_j <1$. Consider a trigonometric polynomial $$f(x)= \sum_{j=1}^n \lambda_j \cos(x_j) -\cos(y\cdot x)\, .$$ It belongs to $F_K$ and has a second-order expansion at $0$ $$f(x)= \sum_{j=1}^n\lambda_j - 1 - \frac{1}{2}\sum_{j=1}^n \lambda_j x_j^2 + \frac{1}{2}(y\cdot x)^2+o(\|x\|^2)$$ $$\le \Big(\sum_{j=1}^n\lambda_j - 1\Big) -\frac{1}{2}\sum_{j=1}^n (\lambda_j -\|y\|_1y_j)x_j^2 +o(\|x\|^2) $$ because by Cauchy-Schwarz, $(y\cdot x)^2 = \big( \sum_{j=1}^n y_j ^{1/2} y_j ^{1/2} x_j\big)^2\le \|y\|_1\sum_{j=1}^n y_j x_j^2 $.

We can now take e.g. $\lambda_j= \|y\|_1y_j +\frac{1}{ n}(1-\|y\|_1^2+\epsilon)$ with $\epsilon>0$ so that $f(0)=\epsilon$ and $f(x)\le\epsilon-\frac{1}{2n}(1-\|y\|_1^2)\|x\|^2+o(\|x\|^2)$ (unif. on $\epsilon$). So for $\epsilon$ small enough $f(x)<0$ on the boundary of a ball around $0$, meaning that the connected component of $0$ in $\{f\ge0\}$ is contained in the ball.

share|improve this answer
    
this condition can be weakened a little more if we use more general $f$ in the above argument. For instance, a sufficient condition, always assuming wlog $\pm e_j\in K$ is: $$\mathrm{int} (\mathrm{co}(K))\cap S^{-1}\mathrm{co}(S(K))\neq\emptyset$$ where $S$ is the map $(x_1,\dots,x_n)\mapsto(x_1^2,\dots,x_n^2)$. However sated this way the latter is not invariant by linear transformations . –  Pietro Majer Oct 20 '13 at 12:44
add comment

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.