Causal support of the scalar commutator distribution

Assume Countable Choice as required by [F4]. For a spacetime test $\phi \in \mathcal{S}({\mathbb{R}}^4)$ set $q_{\phi }(t,\mathbf{k})={\int }_{{\mathbb{R}}^3}{e}^{i\mathbf{k}\cdot \mathbf{x}}\phi (t,\mathbf{x})d^3\mathbf{x}$. Integration by parts in the spatial variables gives $(1+\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2{)}^2{q}_{\phi }(t,\mathbf{k})=\int e^{i\mathbf{k}\cdot \mathbf{x}}(1-{\mathrm{\Delta }}_{\mathbf{x}}{)}^2\phi (t,\mathbf{x})d^3\mathbf{x}\mathrm{.}$ Let $P(\phi )$ be the maximum, over spatial multi-indices $\mathrm{\mid }\beta \mathrm{\mid }\le 4$, of ${\sup }_{t,\mathbf{x}}(1+\mathrm{\mid }t\mathrm{\mid }{)}^4(1+\mathrm{\mid }\mathbf{x}{\mathrm{\mid }}^2{)}^2\mathrm{\mid }{\mathrm{\partial }}_{\mathbf{x}}^{\beta }\phi (t,\mathbf{x})\mathrm{\mid }$. By [F9] this is bounded by a finite sum of the defining Schwartz seminorms. Since ${\int }_{{\mathbb{R}}^3}(1+\mathrm{\mid }\mathbf{x}{\mathrm{\mid }}^2{)}^{-2}{d}^3\mathbf{x}<\mathrm{\infty }$, $\mathrm{\mid }{q}_{\phi }(t,\mathbf{k})\mathrm{\mid }\le CP(\phi )(1+\mathrm{\mid }t\mathrm{\mid }{)}^{-4}(1+\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2{)}^{-2}\mathrm{.}$ For every $\mu \ge 0$, $\mathrm{\mid }\sin ({\omega }_{\mu }t)\mathrm{/}{\omega }_{\mu }\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }$ off the massless apex. Consequently the joint pairing $⟨{\mathrm{\Delta }}_{\mu },\phi ⟩=(2\pi {)}^{-3}{\int }_{\mathbb{R}}{\int }_{{\mathbb{R}}^3}\frac{\sin ({\omega }_{\mu }(\mathbf{k})t)}{{\omega }_{\mu }(\mathbf{k})}{q}_{\phi }(t,\mathbf{k})d^3\mathbf{k}dt$ is absolutely convergent and has modulus at most $C^{\mathrm{\prime }}P(\phi )$, because $\int \mathrm{\mid }t\mathrm{\mid }(1+\mathrm{\mid }t\mathrm{\mid }{)}^{-4}dt$ and $\int (1+\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2{)}^{-2}{d}^3\mathbf{k}$ are finite. It is complex-linear and continuous on $\mathcal{S}({\mathbb{R}}^4)$, so it defines the claimed spacetime tempered distribution [F5]. The estimate also supplies the required polynomial growth in time and proves that this pairing is the integral of its fixed-time spatial pairings. The apex value can be assigned arbitrarily without changing any integral; the separate $1\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ shell factor is locally integrable in three dimensions. For the fixed-time family, derivatives of $\sin ({\omega }_{\mu }t)\mathrm{/}{\omega }_{\mu }$ have at most polynomial growth in $\mathbf{k}$, uniformly on bounded time intervals. Taking enough spatial Schwartz weights therefore permits all time derivatives under its spatial pairing. It follows both for this family and, by integration by parts in $t$, for the joint distribution that $({\mathrm{\partial }}_t^2-{\mathrm{\Delta }}_{\mathbf{x}}+{\mu }^2){\mathrm{\Delta }}_{\mu }=0$, ${\mathrm{\Delta }}_{\mu }(0,\cdot )=0$, and ${\mathrm{\partial }}_t{\mathrm{\Delta }}_{\mu }(0,\cdot )={\delta }_0$; the last identity is Fourier inversion [F4].

Choose a nonnegative smooth spatial mollifier ${\rho }_{ϵ}$ of integral one supported in $\mathrm{\mid }\mathbf{x}\mathrm{\mid }<ϵ$, using [F7], and let $u_{ϵ}(t,\cdot )={\mathrm{\Delta }}_{\mu }(t,\cdot )\ast {\rho }_{ϵ}$. Its Fourier multiplier is ${\widehat{\rho }}_{ϵ}(\mathbf{k})\sin ({\omega }_{\mu }t)\mathrm{/}{\omega }_{\mu }$; this is Schwartz in $\mathbf{k}$ for every fixed $t$, including $\mu =0$ because $\sin (\mathrm{\mid }\mathbf{k}\mathrm{\mid }t)\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ has a power series in $\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2$. Thus $u_{ϵ}$ and its derivatives are Schwartz and have finite energy by [F8]. By 1.1 it is a smooth solution of the Klein–Gordon equation with data $u_{ϵ}(0)=0$, ${\mathrm{\partial }}_t{u}_{ϵ}(0)={\rho }_{ϵ}$. Its initial energy is finite. Choose $R>ϵ$ and for $t\ge 0$ set $E_R(t)=\frac{1}{2}{\int }_{\mathrm{\mid }\mathbf{x}\mathrm{\mid }>R+t}(\mathrm{\mid }{\mathrm{\partial }}_t{u}_{ϵ}{\mathrm{\mid }}^2+\mathrm{\mid }\mathrm{\nabla }{u}_{ϵ}{\mathrm{\mid }}^2+{\mu }^2\mathrm{\mid }{u}_{ϵ}{\mathrm{\mid }}^2)d^3\mathbf{x}\mathrm{.}$ The equation and integration by parts on the exterior region give $E_R^{\mathrm{\prime }}(t)=-\frac{1}{2}{\int }_{\mathrm{\mid }\mathbf{x}\mathrm{\mid }=R+t}(\mathrm{\mid }{\mathrm{\partial }}_t{u}_{ϵ}+{\mathrm{\partial }}_r{u}_{ϵ}{\mathrm{\mid }}^2+\mathrm{\mid }{\mathrm{\nabla }}_T{u}_{ϵ}{\mathrm{\mid }}^2+{\mu }^2\mathrm{\mid }{u}_{ϵ}{\mathrm{\mid }}^2)dS\le 0.$ For a complex solution apply this identity to real and imaginary parts. Cutoff at a large radius and use finite energy to pass to the exterior integral. Since $E_R(0)=0$ and $E_R(t)\ge 0$, it vanishes for all $t\ge 0$. For $\mu >0$ this forces $u_{ϵ}=0$ outside $R+t$; for $\mu =0$ it forces its spatial gradient and time derivative to vanish there, so it is constant on the connected exterior and the inverse Fourier transform tends to zero at spatial infinity, hence the constant is zero. Reversing time gives support inside $\mathrm{\mid }\mathbf{x}\mathrm{\mid }\le R+\mathrm{\mid }t\mathrm{\mid }$ for both signs of $t$.

Let $ϵ\downarrow 0$ and choose $R=2ϵ$. For every compact smooth test $\phi (t,\mathbf{x})$, its spatial convolution with the reflected mollifier converges with all derivatives uniformly on one compact set to $\phi$; this follows by writing the difference as the average of $\phi (t,\mathbf{x}-\mathbf{y})-\phi (t,\mathbf{x})$ for $\mathrm{\mid }\mathbf{y}\mathrm{\mid }<ϵ$ and applying the mean-value bound to each derivative. Continuity of the tempered distribution [F5] therefore gives ${\mathrm{\Delta }}_{\mu }{\ast }_{\mathbf{x}}{\rho }_{ϵ}\to {\mathrm{\Delta }}_{\mu }$ distributionally. If a compact test support lies in the spacelike open set $\mathrm{\mid }\mathbf{x}\mathrm{\mid }>\mathrm{\mid }t\mathrm{\mid }$, its positive distance from the closed cone on that compact set gives a margin $\delta >0$; choosing $3ϵ<\delta$ makes its pairing with $u_{ϵ}$ vanish by 2.1. Passing to the distributional limit proves the support inclusion.

Under [A1], [F3] supplies separability of $H_{\mu }$ for every $\mu \ge 0$, so the CCR lemma applies to the shell Fock space. For real $f,g$, expand the commutator using [F1]. The two like-type commutators vanish; [F3] leaves $⟨K_{\mu }\bar{f},K_{\mu }g{⟩}_2-⟨K_{\mu }\bar{g},K_{\mu }f{⟩}_2\mathrm{.}$ Define the compact smooth difference test $h(z)=\int f(z+y)g(y)d^{4}y$. Thus the double smearing in the Statement means $⟨{\mathrm{\Delta }}_{\mu },h⟩$, which is well-defined by 1.1. Substitute the shell integrals [F2], use the second-variable-linear pairing, and combine the two exponentials. The Fourier transform of $h$ is the product of the corresponding transforms of $f$ and $g$, by ordinary Fubini for compact tests. The resulting shell integrals are absolutely convergent: the transforms decay faster than any power and the massless $1\mathrm{/}\mathrm{\mid }\mathbf{k}\mathrm{\mid }$ factor is locally integrable. This justifies the comparison with the joint pairing in 1.1, without interchanging an unsmeared oscillatory integral. The result is $-i\int f(x)g(y){\mathrm{\Delta }}_{\mu }(x-y)d^{4}x{d}^{4}y$; the sign follows from $e^{-ip\circ (x-y)}-e^{ip\circ (x-y)}=-2i\sin (p\circ (x-y))$ and the definition of ${\mathrm{\Delta }}_{\mu }$. If every pair $(x,y)$ in the two supports is spacelike separated, then $x-y$ lies outside the support in 3.1, so the distributional pairing vanishes.