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 $.

Since $\|y\|_1 < 1$, we can choose the $\lambda_j$'s so that $f(0) > 0$ and $f(x) < 0$ on the boundary of a ball around $0$, meaning that the connected component of $0$ in $\{f\ge0\}$ is contained there.

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.

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 ints 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 $.

Since $\|y\|_1 < 1$, we can choose the $\lambda_j$'s so that $f(0) > 0$ and $f(x) < 0$ on the boundary of a ball around $0$, meaning that the connected component of $0$ in $\{f\ge0\}$ is contained there.

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 $.

Since $\|y\|_1 < 1$, we can choose the $\lambda_j$'s so that $f(0) > 0$ and $f(x) < 0$ on the boundary of a ball around $0$, meaning that the connected component of $0$ in $\{f\ge0\}$ is contained there.