Why is the Sine Kernel admissible as the kernel of a DPP? The Sine DPP is given by the kernel $$K(x,y)=\frac{\sin(\pi(x-y))}{\pi(x-y)}$$
and is a well-known example of a Kernel for a determinantal point process. However, this kernel is not always positive. Why is this kernel admissible as the kernel of a determinantal point process if it is not positive? Wouldn't this contradict the following theorem from this paper?

Theorem 3. Hermitian locally trace class operator $K$ on $L^{2}(E)$ defines a determinantal random point field if and only if $0 \leq K \leq 1$. If the corresponding random point field exists it is unique.

 A: This seems to be a notation confusion coming from the fact that Soshnikov uses the notation $K$ for a kernel and for the associated integral operator.
In his paper, Soshnikov states that the operator $K$ needs to satisfy $0\leq K\leq 1$. Note that the expression $0\leq K\leq 1$ basically means that the operator $K$ needs to have eigenvalues between $0$ and $1$. Here, $0$ corresponds to the null operator on $L^2(E)$, $1$ corresponds to the identity operator on $L^2(E)$ and $\leq$ is the Loewner partial order on the set of hermitian operator of $L^2(E)$ where $K\leq L$ means that the operator $L-K$ is positive semi-definite.
In your case, you considered the kernel $K(x,y)=\frac{\sin(y-x)}{y-x}$. The associated integral operator is
$$\mathcal{K}: f\mapsto \left(x\mapsto \int_{\mathbb{R}}K(x,y)f(y)\mathrm{d} y\right)$$
and Soshnikov's result states that if $\mathcal{K}$ has eigenvalues between $0$ and $1$ then it defined a determinantal random point field. One way to check that this condition is satisfied for a kernel $K$ of the form $K(x,y)=C(y-x)$ is to use Corollary 3.3. in https://arxiv.org/abs/1205.4818 that shows that $0\leq \mathcal{K}\leq 1$ if and only if the Fourier transform of $C$ has values between $0$ and $1$. This is the case for the sine kernel since $\hat C$ is an indicator function in this case.
