that $\hat f$ is of the form $${\widehat}{f}(x) = \big(s_3(x^2) + x^2 s_4(x^2)\big)  e^{-\pi x^2}.$$

Let $\mathcal T$ be the operator that maps $x^{2k}$ to the function $\frac{k!}{\pi^k} L_k^{-1/2}(\pi x^2)$, where $L_k$ is the Laguerre polynomial of degree $k$ with parameter $-1/2$. Then, for $p$ an even polynomial, we have that $(\mathcal Tp)(x) e^{-\pi x^2}$ is the Fourier transform of $p(x) e^{-\pi x^2}$. We can now describe the functions of the form that are non-positive on $[R,\infty)$ and have nonnegative Fourier transform by positive semidefinite matrices $X_1,\ldots,X_4$ of size $d+1$ whose entries satisfy the linear relations coming from the identity $I(X_1,\ldots,X_4) = 0$, where $$I(X_1,\ldots,X_4) = \mathcal T\big(-s_1(x^2) - (x^2-R^2) s_2(x^2)\big) - \big(s_3(x^2) + x^2 s_4(x^2)\big).$$ Here $\mathcal T(-s_1(x^2) - (x^2-R^2) s_2(x^2))$ is a polynomial whose coefficients are linear combinations in the entries of $X_1$ and $X_2$, and the same for $s_3(x^2) + x^2 s_4(x^2)$ with $X_3$ and $X_4$. The linear constraints on the entries of $X_1,\ldots,X_4$ are then obtained by expressing $I(X_1,\ldots,X_4)$ in some polynomial basis and setting the coefficients to zero.

The conditions $f(0) = 1$ and $f(R) = 0$ are linear in the entries of $X_1$ and $X_2$, and the condition ${\widehat}f(0)=1$ is a linear condition on 