Euclidean positivity, controlled limits and reconstruction boundaries

By continuity [F3], choose a basic neighbourhood $V_u=\{f:p_{{\alpha }_j{\beta }_j}(f)<{\epsilon }_j,1\le j\le k\}$ on which $\mathrm{\mid }u(f)\mathrm{\mid }<1$. Intersecting with $p_{00}(f)<1$ if necessary ensures $k\ge 1$. Set $M_u={\max }_{1\le j\le k}\{\mathrm{\mid }{\alpha }_j\mathrm{\mid },\mathrm{\mid }{\beta }_j\mathrm{\mid }\}$, $P_{M_u}(f)={\max }_{\mathrm{\mid }\alpha \mathrm{\mid },\mathrm{\mid }\beta \mathrm{\mid }\le M_u}{p}_{\alpha \beta }(f)$, and $c_u=2{\max }_j{\epsilon }_j^{-1}$. For $D(f)={\max }_j(p_{{\alpha }_j{\beta }_j}(f)\mathrm{/}{\epsilon }_j)>0$, the test $f\mathrm{/}(2D(f))$ lies in $V_u$, so $\mathrm{\mid }u(f)\mathrm{\mid }\le 2D(f)\le c_u{P}_{M_u}(f)$. If $D(f)=0$, every positive scalar multiple of $f$ lies in $V_u$, forcing $u(f)=0$ and the same bound. Applying this argument to $v$ gives finite $M_v,c_v$ with $\mathrm{\mid }v(g)\mathrm{\mid }\le c_v{P}_{M_v}(g)$.

For fixed $\mathrm{\Phi }\in \mathcal{S}({\mathbb{R}}^{p+q})$, the map $x\mapsto \mathrm{\Phi }(x,\cdot )$ is smooth with values in $\mathcal{S}({\mathbb{R}}^q)$: Taylor's formula in the $x$ coordinates, with its remainder differentiated in $y$, tends to zero in each $p_{\gamma \delta }$ on every compact $x$ set because every mixed Schwartz seminorm of $\mathrm{\Phi }$ is finite. Continuity of $v$ therefore makes $h(x):=v(\mathrm{\Phi }(x,\cdot ))$ smooth and gives ${\mathrm{\partial }}_x^{\beta }h(x)=v({\mathrm{\partial }}_x^{\beta }\mathrm{\Phi }(x,\cdot ))$. For every $\alpha ,\beta$, [F3] and step 1.1 yield ${\sup }_x\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}_x^{\beta }h(x)\mathrm{\mid }\le c_v{\max }_{\mathrm{\mid }\gamma \mathrm{\mid },\mathrm{\mid }\delta \mathrm{\mid }\le M_v}{\sup }_{x,y}\mathrm{\mid }{x}^{\alpha }{y}^{\gamma }{\mathrm{\partial }}_x^{\beta }{\mathrm{\partial }}_y^{\delta }\mathrm{\Phi }(x,y)\mathrm{\mid }<\mathrm{\infty }\mathrm{.}$ Thus $h\in \mathcal{S}({\mathbb{R}}^p)$.

Define $w(\mathrm{\Phi })=u(h)$. Applying the bound in step 1.1 to the finitely many $x$ seminorms with indices at most $M_u$, and then the displayed estimate in step 2.1, bounds $\mathrm{\mid }w(\mathrm{\Phi })\mathrm{\mid }$ by a finite constant times the maximum of finitely many product Schwartz seminorms of $\mathrm{\Phi }$. Hence $w$ is continuous on $\mathcal{S}({\mathbb{R}}^{p+q})$. For $\mathrm{\Phi }=\phi \otimes \psi$, the inner pairing is $\phi (x)v(\psi )$, giving the claimed product value; on compactly supported tests the formula is the same iterated pairing as [F6].

Finite sums of Schwartz product tests are dense in $\mathcal{S}({\mathbb{R}}^{p+q})$. Indeed, choose $\chi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^p)$ and $\eta \in C_c^{\mathrm{\infty }}({\mathbb{R}}^q)$ equal to one near the unit balls by [F4], and set ${\chi }_R(x)=\chi (x\mathrm{/}R)$ and ${\eta }_R(y)=\eta (y\mathrm{/}R)$. The product rule and rapid decay show ${\chi }_R(x){\eta }_R(y)\mathrm{\Phi }(x,y)\to \mathrm{\Phi }(x,y)$ in every product Schwartz seminorm: the undifferentiated tails lie outside balls of radius proportional to $R$, and each differentiated cutoff contributes a factor $R^{-j}$ on the same tails. For fixed $R$, [F5] approximates this compactly supported smooth function by finite sums of product tests with common compact support; on that support, uniform convergence of all derivatives implies convergence in every Schwartz seminorm. Thus any two continuous functionals with the same product values agree on a dense subspace and hence everywhere. The same density argument, iterated over finitely many factors, shows that different parenthesizations of the constructed tensor product agree; this also agrees with the compact-test associativity in [F6].

For each $n\ge 1$, define the product-test map $J_n:S_d^{{\otimes }_{\mathrm{a}\mathrm{l}\mathrm{g}}n}\to S_{d,n}$ by $J_n(f_1\otimes \cdots \otimes f_n)(x_1,\dots ,x_n)={\prod }_{i=1}^n{f}_i(x_i)$ and linear extension; the product of Schwartz functions is a Schwartz function. Set $J_0(c)=c$ and, for $a\in T_d$, define $Q_S(a):={\omega }_S(\mathrm{\Theta }aa)$. Fix a finite positive-time polynomial $a={\sum }_{j=0}^N{a}_j$, with each $a_j$ a finite sum of degree-$j$ product tests. Expanding concatenation and the moment functional gives the finite sum $Q_{S^{\lambda }}(a)={\sum }_{j,k=0}^N⟨S_{j+k}^{\lambda },J_{j+k}((\mathrm{\Theta }{a}_j)a_k)⟩\mathrm{.}$ Every test in this finite sum is fixed independently of $\lambda$, so [A2] gives $Q_{S^{\lambda }}(a)\to Q_S(a)$. By [A1], every value $Q_{S^{\lambda }}(a)$ is a nonnegative real number. Since $[0,\mathrm{\infty })$ is closed in $\mathbb{C}$, its limit $Q_S(a)$ is also a nonnegative real number. As $a$ was arbitrary, $S$ is reflection positive.

Fix one Euclidean isometry $\gamma (x)=Rx+b$ and one order-$n$ test $\mathrm{\Phi }$. The pullback $\mathrm{\Phi }\mapsto \mathrm{\Phi }\circ ({\gamma }^{-1}{)}^{\times n}$ is continuous on Schwartz space: the chain rule writes each derivative as a finite linear combination of derivatives of $\mathrm{\Phi }$ evaluated at the inverse affine image, and each polynomial weight in $x$ becomes a finite sum of polynomial weights in the new variable with fixed coefficients. Thus each transformed Schwartz seminorm is bounded by finitely many original seminorms. If every $S_n^{\lambda }$ is invariant under this fixed pullback, [A2] applied to both $\mathrm{\Phi }$ and its pullback passes the equality to $S_n$. A permutation of argument blocks is a coordinate permutation and preserves the same seminorms exactly, so the same limit argument passes permutation symmetry. Step 1.1 uses only a fixed finite number of orders and tests; it supplies neither a uniform-in-$n$ bound nor a limit as a translation parameter tends to infinity, so it proves no order-uniform growth or clustering claim.

The full Axiom of Choice assumed in [A1] implies Countable Choice: if $(X_n{)}_{n\in \mathbb{N}}$ is any sequence of nonempty sets, AC supplies a choice function $g$ on the family $\{X_n:n\in \mathbb{N}\}$, and $n\mapsto g(X_n)$ chooses from every $X_n$. Thus the actual Countable Choice hypotheses of [F6] and [F7] are satisfied before those suppliers are used. The full Fourier transform $\widehat{\mathrm{\Phi }}$ of $\mathrm{\Phi }\in \mathcal{S}(E_d\times E_d)$ is Schwartz by [F6]. For $\mathrm{\mid }p\mathrm{\mid }\ge 1$, some coordinate $p_j$ satisfies $\mathrm{\mid }{p}_j\mathrm{\mid }\ge \mathrm{\mid }p\mathrm{\mid }\mathrm{/}\sqrt{d}$; the monomial-weight seminorm $P_N$ includes the weight given by the $N$th power of that coordinate in the first Fourier block. For $\mathrm{\mid }p\mathrm{\mid }<1$, its zero-weight seminorm bounds the value. Thus, with the finite dimension/order constant $K_{d,N}:=2^N{d}^{N\mathrm{/}2}$, $\mathrm{\mid }\widehat{\mathrm{\Phi }}(-p,p)\mathrm{\mid }\le K_{d,N}{P}_N(\widehat{\mathrm{\Phi }})(1+\mathrm{\mid }p\mathrm{\mid }{)}^{-N}\mathrm{.}$ For an integer $N>d$, this proves absolute convergence and bounds the defining pairing by $(2\pi {)}^{-d}{m}^{-2}{K}_{d,N}{P}_N(\widehat{\mathrm{\Phi }}){\int }_{{\mathbb{R}}^d}(1+\mathrm{\mid }p\mathrm{\mid }{)}^{-N}dp$. The last integral is finite: dyadic annuli have volume $O(2^{jd})$ while the integrand is $O(2^{-jN})$. Fourier continuity in [F6] bounds $P_N(\widehat{\mathrm{\Phi }})$ by finitely many Schwartz seminorms of $\mathrm{\Phi }$, so $C_m^{(2)}$ is continuous and tempered. Under simultaneous translation by $a\in E_d$, the full Fourier transform acquires the factor $e^{-i(q_1+q_2)\cdot a}$; at $(q_1,q_2)=(-p,p)$ this factor is $1$, proving translation invariance. On product tests, $C_m(\bar{f},g)=(2\pi {)}^{-d}{\int }_{{\mathbb{R}}^d}\frac{\overline{\widehat{f}(p)}\widehat{g}(p)}{p_0^2+\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2+m^2}dp,$ which is nonnegative for $g=f$.

The zeroth term $S_0=1$ is a continuous functional on $\mathbb{C}$, and every odd-order term $S_n=0$ is the zero continuous functional. For even $n$, each pairing $\pi$ partitions the $n$ test-variable blocks into disjoint pairs. The corresponding tensor product of copies of $C_m^{(2)}$ is a tempered distribution by [F2]; the finite sum over $\pi$ is therefore in $S_{d,n}^{\mathrm{\prime }}$. Its values on elementary tests are the displayed Wick formula. The bilinear form $C_m$ is symmetric by changing $p$ to $-p$ in step 1.1, so the sum is invariant under every permutation of its argument blocks. For any finite list of real Schwartz tests, $G_{ij}=C_m(f_i,f_j)$ is symmetric and real: for real $f_i$, ${\widehat{f}}_i(-p)=\overline{{\widehat{f}}_i(p)}$, so the Fourier pairing is Hermitian, while symmetry makes it real. It is positive semidefinite because ${\sum }_{i,j}{\bar{z}}_i{G}_{ij}{z}_j=C_m(\overline{{\sum }_i{z}_i{f}_i},{\sum }_j{z}_j{f}_j)\ge 0$. By [A1] and [F4] the finite centered Gaussian law ${\gamma }_G$ exists, including when $G$ is singular, and [F5] shows its moments are exactly the displayed pairing values. Splitting complex tests into real and imaginary parts gives the same multilinear extension. The symmetry of the moments makes the algebraic moment functional commute on this finite family of generators, so every finite polynomial $a$ satisfies ${\omega }_S(a^{\ast }a)={\mathbb{E}}_{{\gamma }_G}\mathrm{\mid }a(Q){\mathrm{\mid }}^2\ge 0$ and ${\omega }_S(1)=S_0=1$.

By step 1.1, the Countable Choice hypotheses of [F6] and [F7] are satisfied. Put ${\omega }_{\mathbf{k}}=\sqrt{\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2+m^2}>0$. The function $K_{\omega }(u)=e^{-\omega \mathrm{\mid }u\mathrm{\mid }}\mathrm{/}(2\omega )$ is integrable. Splitting its Fourier transform into the two half-lines gives ${\int }_{\mathbb{R}}{e}^{-i{p}_{0}u}{K}_{\omega }(u)du=\frac{1}{2\omega }(\frac{1}{\omega +i{p}_0}+\frac{1}{\omega -i{p}_0})=\frac{1}{p_0^2+{\omega }^2}\mathrm{.}$ By injectivity of the Fourier automorphism on tempered distributions [F6], this identifies the inverse kernel of $(p_0^2+{\omega }^2{)}^{-1}$; the inverse normalization is $1\mathrm{/}(2\pi )$ under the stated convention [F7]. Since $f,g$ are compactly supported in $t>0$, reflecting the first time variable changes $t-s$ to $-(t+s)$; its absolute value is $t+s$. Integrating the spatial variables by their Fourier transforms yields the stated formula. The exchanges of integrals are valid: ${\int }_{\mathbb{R}}(p_0^2+{\omega }_{\mathbf{k}}^2{)}^{-1}d{p}_0=\pi \mathrm{/}{\omega }_{\mathbf{k}}\le \pi \mathrm{/}m$, while repeated spatial integration by parts gives rapid $\mathbf{k}$-decay of $\widehat{f}(t,\mathbf{k})$ and $\widehat{g}(s,\mathbf{k})$ uniformly on their compact time supports.

Define $h_f(\mathbf{k})={\int }_0^{\mathrm{\infty }}{e}^{-{\omega }_{\mathbf{k}}t}\widehat{f}(t,\mathbf{k})dt$ and use the measure $d{\nu }_m(\mathbf{k})=d\mathbf{k}\mathrm{/}((2\pi {)}^{d-1}2{\omega }_{\mathbf{k}})$, with the one-point convention from the Statement when $d=1$. Compact time support and the rapid spatial decay just proved give $h_f\in L^2({\nu }_m)$, and step 2.2 gives $C_m(\overline{f\circ \vartheta },g)=⟨h_f,h_g{⟩}_{L^2({\nu }_m)}$, where $⟨u,v{⟩}_{L^2({\nu }_m)}:=\int \overline{u(\mathbf{k})}v(\mathbf{k})d{\nu }_m(\mathbf{k})$ is linear in its second argument. To pass from this Gram form to the full Gaussian reflection form, fix any finite set of positive-time tests $f_1,\dots ,f_N$ used by $a$. Since all $S_n$ are symmetric, their moment functional on this finite set of field variables factors through its commutative polynomial algebra. Write $X(f_i)$ for the degree-one generator $f_i$ in $T_d$. The formal Wick exponential $W(z)=\exp \text{\hspace{0.17em}⁣}({\sum }_i{z}_{i}X(f_i)-\frac{1}{2}{\sum }_{i,j}{z}_i{z}_j{C}_m(f_i,f_j))$ has polynomial coefficients; no analytic convergence is asserted. Write $s_i^{\mathrm{\#}}$ for the formal conjugate of $s_i$: reflection conjugates coefficients, so $\mathrm{\Theta }W(s)$ uses $s_i^{\mathrm{\#}}$ and conjugates the coefficients of the quadratic counterterm. Put $G_{ij}=C_m(f_i,f_j)$ and $R_{ij}=C_m(\overline{f_i\circ \vartheta },f_j)$. Applying [F5] to the joint finite real list consisting of the real and imaginary parts of $f_i$ and $\overline{f_i\circ \vartheta }$ gives, by complex multilinearity, the coefficientwise formal identity. Each coefficient on the right is the sum over pairings of the corresponding finite test list: the quadratic exponential chooses unordered pairs, and its factorial denominator cancels their ordering multiplicity. Thus the identity follows coefficient by coefficient, with no convergence of an infinite series asserted. ${\omega }_S\text{\hspace{0.17em}⁣}(\exp \text{\hspace{0.17em}⁣}({\sum }_i{s}_i^{\mathrm{\#}}X(\overline{f_i\circ \vartheta }))\exp \text{\hspace{0.17em}⁣}({\sum }_j{z}_{j}X(f_j)))=\exp \text{\hspace{0.17em}⁣}(\frac{1}{2}{\sum }_{i,j}{s}_i^{\mathrm{\#}}\overline{G_{ij}}{s}_j^{\mathrm{\#}}+{\sum }_{i,j}{s}_i^{\mathrm{\#}}{R}_{ij}{z}_j+\frac{1}{2}{\sum }_{i,j}{z}_i{G}_{ij}{z}_j)\mathrm{.}$ The identity $C_m(\overline{f_i\circ \vartheta },\overline{f_j\circ \vartheta })=\overline{G_{ij}}$ for the reflected covariance follows from $\widehat{\overline{f\circ \vartheta }}(p_0,\mathbf{k})=\overline{\widehat{f}(p_0,-\mathbf{k})}$: substitute this in the Fourier formula for $C_m$ and change $\mathbf{k}$ to $-\mathbf{k}$, using the real multiplier $p_0^2+\mathrm{\mid }\mathbf{k}{\mathrm{\mid }}^2+m^2$. Multiplying by the two Wick counterterms therefore cancels the first and third quadratic sums and leaves ${\omega }_S(\mathrm{\Theta }W(s)W(z))=\exp \text{\hspace{0.17em}⁣}({\sum }_{i,j}{s}_i^{\mathrm{\#}}{C}_m(\overline{f_i\circ \vartheta },f_j)z_j),$ where $s$ and $z$ are independent formal variables and the symbol $s_i^{\mathrm{\#}}$ is conjugated when coefficients are evaluated. The coefficient of unequal degrees in $s^{\mathrm{\#}},z$ is zero. Differentiating $k$ times in each variable gives the cross-pairing matrix ${\sum }_{\sigma \in {\mathrm{\Sigma }}_k}{\prod }_{j=1}^k{R}_{i_j{\mathrm{\ell }}_{\sigma (j)}}$, where $R_{i\mathrm{\ell }}=⟨h_{f_i},h_{f_{\mathrm{\ell }}}⟩$. Write the finite Wick expansion of a polynomial as $a={\sum }_{k=0}^K{\sum }_{i_1,\dots ,i_k}{a}_{i_1\dots i_k}^{(k)}{\mathrm{\partial }}_{z_{i_1}}\cdots {\mathrm{\partial }}_{z_{i_k}}W(0)$, with symmetric coefficient tensors, and let $A_k=\sum a_{i_1\dots i_k}^{(k)}{e}_{i_1}\otimes \cdots \otimes e_{i_k}$. The triangular relation between ordinary monomials and Wick polynomials, whose leading terms agree and whose remaining terms have lower degree, proves that every finite polynomial has such a finite expansion. Let $H:{\mathbb{C}}^N\to L^2({\nu }_m)$ be the linear map $H{e}_i=h_{f_i}$, and let $\mathrm{Sym}$ be the average over the $k!$ permutations of tensor factors. Expanding the squared norm of $\mathrm{Sym}{H}^{\otimes k}{A}_k$ gives $1\mathrm{/}k!$ times the sum over those permutations, so the displayed cross-pairing matrix makes the degree-$k$ reflected quadratic form exactly $k!\mathrm{\parallel }\mathrm{Sym}{H}^{\otimes k}{A}_k{\mathrm{\parallel }}^2\ge 0$. There are no cross terms between different degrees. Therefore for every $a\in T_{d,+}$, ${\omega }_S(\mathrm{\Theta }aa)$ is a finite sum of nonnegative squared norms.