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?
 A: 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.
