Positive vacuum correlation for spacelike-separated massless tests

Choose $\delta >0$. By [F4], choose a nonnegative nonzero $\chi \in C_c^{\mathrm{\infty }}((-\delta ,\delta ))$. Choose two nonnegative nonzero $u,v\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ supported in closed balls of radii $r_1,r_2$ whose center separation $D$ satisfies $D-r_1-r_2>2\delta$. Such bumps follow by translating and rescaling [F4]. Set $f(t,\mathbf{x})=\chi (t)u(\mathbf{x})$ and $g(t,\mathbf{x})=\chi (t)v(\mathbf{x})$. For every point in the two supports, $\mathrm{\mid }t-s\mathrm{\mid }<2\delta$ and $\mathrm{\mid }\mathbf{x}-\mathbf{y}\mathrm{\mid }\ge D-r_1-r_2>2\delta$; therefore all support differences are spacelike. For example, $\delta =1$, $r_1=r_2=1$ and $D=5$ give a spatial gap of $3>2$, so these strict inequalities are attainable.

Insert a radial regulator $e^{-ϵ\mathrm{\mid }\mathbf{k}\mathrm{\mid }}$, $ϵ>0$, in the massless two-point kernel. Angular integration using [F3] gives for $r=\mathrm{\mid }\mathbf{x}-\mathbf{y}\mathrm{\mid }>0$ and $\tau =t-s$ $W_{ϵ}(\tau ,r)=\frac{1}{4{\pi }^{2}r}{\int }_0^{\mathrm{\infty }}{e}^{-(ϵ+i\tau )q}\sin (qr)dq=\frac{1}{4{\pi }^2(r^2+(ϵ+i\tau {)}^2)}\mathrm{.}$ The integral identity follows by integrating the two exponentials in the sine and using $\mathrm{Re}(ϵ+i\tau )>0$. On the compact support product in 1.1, $r\ge d:=D-r_1-r_2>2\delta >\mathrm{\mid }\tau \mathrm{\mid }$. Thus the denominator is bounded away from zero uniformly as $ϵ\downarrow 0$, and $W_{ϵ}$ converges uniformly there to $[4{\pi }^2(r^2-{\tau }^2){]}^{-1}$, which is strictly positive.

The regulated shell integrals converge to $⟨K_{0}f,K_{0}g{⟩}_2$ by dominated convergence [F5]: the absolute value of the unregulated product is integrable by Cauchy–Schwarz [F7], because $K_{0}f,K_{0}g\in H_0$. To identify the regulated shell integral with the spacetime pairing of $W_{ϵ}$, Fubini [F6] applies because its absolute integral is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1{\int }_{{\mathbb{R}}^3}{e}^{-ϵ\mathrm{\mid }\mathbf{k}\mathrm{\mid }}{d}^3\mathbf{k}\mathrm{/}[(2\pi {)}^{3}2\mathrm{\mid }\mathbf{k}\mathrm{\mid }]<\mathrm{\infty }$. The formula in 2.1 and uniform convergence on the compact support product give $⟨K_{0}f,K_{0}g{⟩}_2=\int f(t,\mathbf{x})g(s,\mathbf{y})\frac{d^{4}x{d}^{4}y}{4{\pi }^2(\mathrm{\mid }\mathbf{x}-\mathbf{y}{\mathrm{\mid }}^2-(s-t{)}^2)}>0,$ because both test functions are nonnegative and nonzero and the denominator is positive on the support product. More explicitly, $⟨K_{0}f,K_{0}g{⟩}_2\ge (\int f)(\int g)\mathrm{/}[4{\pi }^2(D+r_1+r_2{)}^2]>0$, since $\mathrm{\mid }\mathbf{x}-\mathbf{y}{\mathrm{\mid }}^2-(t-s{)}^2\le (D+r_1+r_2{)}^2$ there. The zero one-point values follow from [F1]; commutation follows from [F2] and separation in step 1.1.