An exponential polynomial with at least one bounded positivity component

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?

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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
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

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.
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