Positive spectrum gives bounded future-tube continuation

If $p\in {\bar{V}}_{+}$ and $b\in {\bar{V}}_{+}$, then $p\cdot b\ge p^0(b^0-\mathrm{\mid }\mathbf{b}\mathrm{\mid })\ge 0$, because $p^0\ge \mathrm{\mid }\mathbf{p}\mathrm{\mid }$ and $\mathrm{\mid }\mathbf{p}\cdot \mathbf{b}\mathrm{\mid }\le \mathrm{\mid }\mathbf{p}\mathrm{\mid }\mathrm{\mid }\mathbf{b}\mathrm{\mid }$. Thus for $z=a+ib$ in the closed tube, $\mathrm{\mid }{e}^{ip\cdot z}\mathrm{\mid }=e^{-p\cdot b}\le 1$.

Let $K\subset V_{+}$ be nonempty and compact and set ${\delta }_K:={\min }_{b\in K}(b^0-\mathrm{\mid }\mathbf{b}\mathrm{\mid })>0$. If $q\in {\bar{V}}_{+}$ and $\mathrm{\mid }q{\mathrm{\mid }}_{\mathrm{E}}=1$, then $q^0\ge 1\mathrm{/}\sqrt{2}$, since $(q^0{)}^2+\mathrm{\mid }\mathbf{q}{\mathrm{\mid }}^2=1$ and $q^0\ge \mathrm{\mid }\mathbf{q}\mathrm{\mid }$. Hence $q\cdot b\ge q^0(b^0-\mathrm{\mid }\mathbf{b}\mathrm{\mid })\ge {\delta }_K\mathrm{/}\sqrt{2}$ for every $b\in K$. By homogeneity, $p\cdot b\ge c_K\mathrm{\mid }p{\mathrm{\mid }}_{\mathrm{E}}$ for all $p\in {\bar{V}}_{+}$ and $b\in K$, with $c_K:={\delta }_K\mathrm{/}\sqrt{2}>0$.

To use a bounded Borel multiplier on all of ${\mathbb{R}}^4$, set $g_z(p):={\mathbf{1}}_{{\bar{V}}_{+}}(p)e^{ip\cdot z}$. By step 1.1, $\mathrm{\parallel }{g}_z{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$. Applying [F2] gives $U(z):={\mathrm{\Phi }}_E(g_z)$ and $\mathrm{\parallel }U(z)\mathrm{\parallel }\le 1$. Since $E({\mathbb{R}}^4\setminus {\bar{V}}_{+})=0$, this is the spectral integral of $e^{ip\cdot z}$ on the spectral support. For $b=0$, [F2] gives $\mathrm{\parallel }{\mathrm{\Phi }}_E(g_a-e^{ip\cdot a})\xi {\mathrm{\parallel }}^2=0$ for every $\xi$, because the multiplier vanishes on the support of each $E_{\xi }$. Thus ${\mathrm{\Phi }}_E(g_a)$ is the given $U(a)$.

Fix $z=a+ib\in {\mathcal{T}}_{+}$. By step 1.2 applied to the singleton $\{b\}$, $p\cdot b\ge c\mathrm{\mid }p{\mathrm{\mid }}_{\mathrm{E}}$ on the $E$-support for some $c>0$. For sufficiently small $h\in {\mathbb{C}}^4$ with $z+h\in {\mathcal{T}}_{+}$, define the bounded multiplier $g_h(p):={\mathbf{1}}_{{\bar{V}}_{+}}(p)i(p\cdot h)e^{ip\cdot z}$. It satisfies $\mathrm{\parallel }{g}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }h\mathrm{\mid }{\sup }_{r\ge 0}r{e}^{-cr}$ and is complex-linear in $h$. The exponential power-series bound $\mathrm{\mid }{e}^w-1-w\mathrm{\mid }\le \frac{1}{2}\mathrm{\mid }w{\mathrm{\mid }}^2{e}^{\mathrm{\mid }w\mathrm{\mid }}$ then gives $\mathrm{\parallel }U(z+h)-U(z)-{\mathrm{\Phi }}_E(g_h)\mathrm{\parallel }\le \frac{1}{2}\mathrm{\mid }h{\mathrm{\mid }}^2{\sup }_{r\ge 0}{r}^2{e}^{-(c-\mathrm{\mid }h\mathrm{\mid })r}=O(\mathrm{\mid }h{\mathrm{\mid }}^2),$ because the pairing identity and uniqueness in [F2] identify the difference operator with the PVM integral of the bounded multiplier ${\mathbf{1}}_{{\bar{V}}_{+}}(p)e^{ip\cdot z}(e^{ip\cdot h}-1-ip\cdot h)$, and [F2] supplies the norm bound. Here $\mathrm{\mid }p\cdot h\mathrm{\mid }\le \mathrm{\mid }p{\mathrm{\mid }}_{\mathrm{E}}\mathrm{\mid }h\mathrm{\mid }$. Applying the continuous linear functional $T\mapsto (T\psi ,\varphi )$ proves that $z\mapsto (U(z)\psi ,\varphi )$ has a complex-linear total derivative. Therefore this scalar matrix coefficient is holomorphic in the sense of [F3].

For $R>0$ put $E_R:=E(\{p:\mathrm{\mid }p{\mathrm{\mid }}_{\mathrm{E}}\le R\})$. Strong countable additivity and $E({\mathbb{R}}^4)=I$ imply $E_R\xi \to \xi$ for every $\xi \in \mathcal{H}$ as $R\to \mathrm{\infty }$, by writing the increasing balls as disjoint annuli. By the pairing identity and uniqueness in [F2], $U(z)-U(z_0)={\mathrm{\Phi }}_E(g_z-g_{z_0})$. The scalar measure $E_{E_R\xi }(S):=(E(S)E_R\xi ,E_R\xi )$ is supported in the compact ball $\{\mathrm{\mid }p{\mathrm{\mid }}_{\mathrm{E}}\le R\}$ by the PVM product law, so the quadratic identity in [F2] gives $\mathrm{\parallel }(U(z)-U(z_0))E_R\xi {\mathrm{\parallel }}^2=\int \mathrm{\mid }{g}_z-g_{z_0}{\mathrm{\mid }}^{2}d{E}_{E_R\xi }\le {\sup }_{\begin{array}{c}p\in {\bar{V}}_{+}\\ \mathrm{\mid }p{\mathrm{\mid }}_{\mathrm{E}}\le R\end{array}}\mathrm{\mid }{e}^{ip\cdot z}-e^{ip\cdot z_0}{\mathrm{\mid }}^2\mathrm{\parallel }{E}_R\xi {\mathrm{\parallel }}^2\mathrm{.}$ The supremum tends to zero as $z\to z_0$ in the closed tube by uniform continuity on the compact spectral ball. The remaining vector part has norm at most $2\mathrm{\parallel }\xi -E_R\xi \mathrm{\parallel }$ by step 2.1, so first taking large $R$ and then $z\to z_0$ proves strong continuity on the closed tube.

For $\mathbf{z}=(z_1,\dots ,z_n)$ in ${\mathcal{T}}_{+}^n$, apply step 2.2 to each factor: $U(z_j+h_j)=U(z_j)+{\mathrm{\Phi }}_E(g_{z_j,h_j})+R_j(h_j)$, with $\mathrm{\parallel }{R}_j(h_j)\mathrm{\parallel }=O(\mathrm{\mid }{h}_j{\mathrm{\mid }}^2)$ and $\mathrm{\parallel }{\mathrm{\Phi }}_E(g_{z_j,h_j})\mathrm{\parallel }=O(\mathrm{\mid }{h}_j\mathrm{\mid })$. Expanding the finite product defining $F(\mathbf{z}+\mathbf{h})$, the terms linear in $\mathbf{h}$ are the sum of the products with exactly one factor replaced by ${\mathrm{\Phi }}_E(g_{z_j,h_j})$; this is a complex-linear function of $\mathbf{h}$. Every remaining term contains either one $R_j(h_j)$ or at least two linear increments, so its norm is $O(\mathrm{\parallel }\mathbf{h}{\mathrm{\parallel }}^2)$ because the other factors $U(z_j)$ and $B_j$ are bounded. Pairing with $\psi ,\varphi$ preserves this remainder estimate. Thus $F$ is complex differentiable on the full product domain, hence holomorphic by [F3].