Two-dimensional Wick limits and a constructive interaction interface

In the formal power-series ring $\mathbb{R}\mathrm{⟦}s,t\mathrm{⟧}$, set $M(s,t)={\sum }_{a,b\ge 0}\mathbb{E}[X^a{Y}^b]s^a{t}^b\mathrm{/}(a!b!)$. By [F1], a pairing with $j$ $X$-self-pairs, $k$ $Y$-self-pairs, and $\mathrm{\ell }$ cross-pairs can occur only when $a=2j+\mathrm{\ell }$ and $b=2k+\mathrm{\ell }$. The number of such pairings is $a!b!\mathrm{/}(2^{j}j!2^{k}k!\mathrm{\ell }!)$: choose the $\mathrm{\ell }$ cross-indices on each side, match them in $\mathrm{\ell }!$ ways, then pair the remaining indices on each side. Thus coefficient-by-coefficient $M(s,t)=\exp ({\sigma }^2{s}^2\mathrm{/}2)\exp (cst)\exp ({\tau }^2{t}^2\mathrm{/}2)$. This is an identity of formal series, so no moment-generating-function convergence is assumed.

Multiplying the series in step 1.1 by $\exp (-{\sigma }^2{s}^2\mathrm{/}2)\exp (-{\tau }^2{t}^2\mathrm{/}2)$ leaves $\exp (cst)={\sum }_{r\ge 0}{c}^r{s}^r{t}^r\mathrm{/}r!$. By [F2], the coefficient of $s^n{t}^m$ on the multiplied left side is $\mathbb{E}[:X^n{:}_{{\sigma }^2}:Y^m{:}_{{\tau }^2}]\mathrm{/}(n!m!)$. The right side has zero coefficient when $n\mathrm{\ne }m$ and coefficient $c^n\mathrm{/}n!$ when $n=m$, giving the claimed formula. For $n=m=0$ the constant coefficient is $1$, including when $c=0$.

For $a>0$ and $T>0$, substitute $u=-at$ by [F6]. The function $u\mapsto \exp (u)$ is continuous and has derivative $\exp (u)$ by [F7], and this derivative is Riemann integrable on $[-aT,0]$ by [F12]; hence [F10] gives ${\int }_0^T\exp (-at)dt=a^{-1}(1-\exp (-aT))$. By [F9] with power $1$ and scale $a$, $T\mathrm{/}\exp (aT)\to 0$; for $T\ge 1$, [F8] gives $0<\exp (-aT)\le T\mathrm{/}\exp (aT)$, so $\exp (-aT)\to 0$. Thus ${\int }_0^{\mathrm{\infty }}\exp (-at)dt=1\mathrm{/}a$ as a nonnegative improper Riemann integral, and as a Lebesgue integral by [F11]. Next, [F2] and [F14] with $s=t\mathrm{/}\pi$ and $\xi =-z\mathrm{/}(2\pi )$ give $(2\pi {)}^{-2}\int e^{ip\cdot z}{e}^{-t\mathrm{\mid }p{\mathrm{\mid }}^2}{d}^{2}p=(2\pi {)}^{-2}(\pi \mathrm{/}t)e^{-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}(4t)}=(4\pi t{)}^{-1}{e}^{-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}(4t)}=:h_t(z)$. Its integral is one: [F5] in each coordinate and $x=2\sqrt{t}u$ by [F6] give $\int h_t(z)d^{2}z=1$. For every $p$, the Laplace identity gives $(\mathrm{\mid }p{\mathrm{\mid }}^2+m^2{)}^{-1}={\int }_0^{\mathrm{\infty }}{e}^{-t(\mathrm{\mid }p{\mathrm{\mid }}^2+m^2)}dt$. If $\psi \in \mathcal{S}({\mathbb{R}}^2)$, the displayed inverse-kernel formula and Fubini give $\int h_t(z)\psi (z)d^{2}z=(2\pi {)}^{-2}\int e^{-t\mathrm{\mid }p{\mathrm{\mid }}^2}\widehat{\psi }(-p)d^{2}p$; the exchanges are absolute because $e^{-t\mathrm{\mid }p{\mathrm{\mid }}^2}\in L^1$ and $\psi \in L^1$. Therefore ${\int }_0^{\mathrm{\infty }}{e}^{-m^{2}t}\int h_t(z)\mathrm{\mid }\psi (z)\mathrm{\mid }{d}^{2}zdt\le \mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}{m}^2$, while the Fourier-side integral is absolutely convergent since [F13] makes $\widehat{\psi }$ Schwartz and $m>0$. Integrating the Fourier identity over $t$ and applying Fubini [F3] in $(t,z)$ and $(t,p)$ gives the heat-kernel superposition's pairing $(2\pi {)}^{-2}\int \widehat{\psi }(-p)(\mathrm{\mid }p{\mathrm{\mid }}^2+m^2{)}^{-1}{d}^{2}p$; reflection $p\mapsto -p$ preserves Lebesgue measure by [F15] and yields the covariance pairing. For real Schwartz $f,g$, set $\mathrm{\Psi }(z)=\int \overline{f(y+z)}g(y)d^{2}y$. Differentiation under this integral is justified locally in $z$ by bounded derivatives of $f$ and the integrability of $g$. For multiindices $\alpha ,\beta$, write $x=y+z$ and expand $z^{\alpha }=(x-y{)}^{\alpha }$; each term of $z^{\alpha }{\mathrm{\partial }}^{\beta }\mathrm{\Psi }(z)$ is bounded uniformly in $z$ by a finite product of seminorms ${\sup }_x\mathrm{\mid }{x}^{\gamma }{\mathrm{\partial }}^{\beta }f(x)\mathrm{\mid }$ and integrals $\int \mathrm{\mid }{y}^{\alpha -\gamma }g(y)\mathrm{\mid }{d}^{2}y$, which are finite because $f,g$ are Schwartz. Thus $\mathrm{\Psi }\in \mathcal{S}({\mathbb{R}}^2)$. Absolute Fubini and the substitution $x=y+z$ give $\widehat{\mathrm{\Psi }}(-p)=\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int e^{ip\cdot (x-y)}\overline{f(x)}g(y)d^{2}x{d}^{2}y=\overline{\widehat{f}(p)}\widehat{g}(p)$; absolute integrability follows from $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1<\mathrm{\infty }$. Applying the just-proved kernel pairing to this Schwartz test $\mathrm{\Psi }$ recovers exactly the covariance $C_m(f,g)$ in [F1]. Thus the heat-kernel superposition is the a.e. spatial density of the covariance distribution, and its value for each $z\mathrm{\ne }0$ is the defining integral for $G_m(z)$.

For $0<r=\mathrm{\mid }z\mathrm{\mid }\le 1$, the part of the defining $t$-integral over $t\ge 1$ is bounded by ${\int }_1^{\mathrm{\infty }}{e}^{-m^{2}t}dt<\mathrm{\infty }$ by step 1.1. On $0<t<1$, drop $e^{-m^{2}t}\le 1$ and substitute $u=r^2\mathrm{/}(4t)$ by [F6] to bound the remaining integral by ${\int }_{r^2\mathrm{/}4}^{\mathrm{\infty }}{e}^{-u}du\mathrm{/}u\le \log (4\mathrm{/}{r}^2)+{\int }_1^{\mathrm{\infty }}{e}^{-u}du\mathrm{/}u$; the last tail is finite because $1\mathrm{/}u\le 1$ and step 1.1 gives ${\int }_1^{\mathrm{\infty }}{e}^{-u}du<\mathrm{\infty }$. This gives $\mathrm{\mid }{G}_m(z)\mathrm{\mid }\le K_m(1+\mathrm{\mid }\log r\mathrm{\mid })$. For $r\ge 1$, substitute $t=(r\mathrm{/}(2m))e^v$ in [F6]; then $m^{2}t+r^2\mathrm{/}(4t)=mr\cosh v$ and $dt\mathrm{/}t=dv$. Define $\cosh v=(\exp (v)+\exp (-v))\mathrm{/}2$. The exponential power series gives $\cosh v={\sum }_{j\ge 0}{v}^{2j}\mathrm{/}(2j)!\ge 1+v^2\mathrm{/}2$. [F5] and a further scaling substitution from [F6] give $G_m(z)\le (4\pi {)}^{-1}{e}^{-mr}{\int }_{\mathbb{R}}{e}^{-mr{v}^2\mathrm{/}2}dv\le K_m{e}^{-mr}\mathrm{/}\sqrt{mr}$. Together with the almost-everywhere density from step 1.1, this proves that $G_m$ is a locally integrable representative of the covariance distribution.

For every finite $p\ge 1$ and $0<s\le 1$, the same annuli scaled to $s$ show ${\int }_{\mathrm{\mid }z\mathrm{\mid }\le s}(1+\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }{)}^p{d}^{2}z\le K_p{s}^2(1+\mathrm{\mid }\log s\mathrm{\mid }{)}^p$: on $s{2}^{-j-1}<\mathrm{\mid }z\mathrm{\mid }\le s{2}^{-j}$ the area is at most $\pi s^2{2}^{-2j}$ and the logarithmic factor is at most $(1+\mathrm{\mid }\log s\mathrm{\mid }+(j+1)\log 2{)}^p$. Put $A=1+\mathrm{\mid }\log s\mathrm{\mid }\ge 1$; then this factor is at most $A^p(1+(j+1)\log 2{)}^p$. The ratio of successive terms $4^{-j}(1+(j+1)\log 2{)}^p$ tends to $1\mathrm{/}4$, so after an index depending only on $p$ it is at most $1\mathrm{/}2$; the remaining series is bounded by its finite initial sum plus a geometric tail. This proves the bound with $K_p$ independent of $s$, hence $G_m^p\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. The integral formula is continuous for $z\mathrm{\ne }0$: for any sequence $z_k\to z\mathrm{\ne }0$, choose $c>0$ so $\mathrm{\mid }{z}_k{\mathrm{\mid }}^2\mathrm{/}4\ge c$ for all sufficiently large $k$. On $0<t\le 1$, substitution $u=c\mathrm{/}t$ by [F6] gives ${\int }_0^1{t}^{-1}{e}^{-c\mathrm{/}t}dt={\int }_c^{\mathrm{\infty }}{u}^{-1}{e}^{-u}du<\mathrm{\infty }$; the finite interval up to $1$ is bounded and the tail is at most ${\int }_1^{\mathrm{\infty }}{e}^{-u}du$, finite by step 1.1. On $t\ge 1$, $t^{-1}{e}^{-m^{2}t}\le e^{-m^{2}t}$, integrable by step 1.1. Thus [F4] applies and gives sequential continuity.

Let $R_{\rho }>0$ satisfy $\mathrm{supp}\rho \subseteq B(0,R_{\rho })$. For $f_{ϵ,x}(u)={\rho }_{ϵ}(x-u)$ and $f_{\delta ,y}(v)={\rho }_{\delta }(y-v)$, the Gaussian covariance [F1] and the kernel representation in steps 1.1–2.1 give $\mathbb{E}[{\mathrm{\Phi }}_{ϵ}(x){\mathrm{\Phi }}_{\delta }(y)]=\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int f_{ϵ,x}(u)f_{\delta ,y}(v)G_m(u-v)d^{2}u{d}^{2}v$. This integral is absolute because the tests are bounded and compactly supported and $G_m\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. Put $z=x-y$, $a=x-u$, and $b=y-v$; [F15] preserves Lebesgue measure and yields $\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int {\rho }_{ϵ}(a){\rho }_{\delta }(b)G_m(z-a+b)d^{2}a{d}^{2}b$. With $w=a-b$, Fubini [F3] gives this as $\int G_m(z-w){\kappa }_{ϵ,\delta }(w)d^{2}w=C_{ϵ,\delta }(z)$. The nonnegative correlation kernel ${\kappa }_{ϵ,\delta }$ has integral one, support in a disk of radius $2R_{\rho }q$ for $q=\max (ϵ,\delta )$, and $\mathrm{\parallel }{\kappa }_{ϵ,\delta }{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}{q}^{-2}$. Its support shrinks to zero, so continuity of $G_m$ away from zero gives $C_{ϵ,\delta }(z)\to G_m(z)$ for each $z\mathrm{\ne }0$. To get the uniform bound, fix bounded $\mathrm{\mid }z\mathrm{\mid }\le R$. If $\mathrm{\mid }z\mathrm{\mid }$ exceeds twice the support radius, then $\mathrm{\mid }z-w\mathrm{\mid }\ge \mathrm{\mid }z\mathrm{\mid }\mathrm{/}2$ throughout the convolution and step 2.1 bounds the integrand by $K_{m,R}(1+\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid })$. Otherwise $\mathrm{\mid }z\mathrm{\mid }\le 4R_{\rho }q$ and $\mathrm{\mid }z-w\mathrm{\mid }\le 6R_{\rho }q$; the scaled dyadic-annulus estimate in step 3.1 gives ${\int }_{\mathrm{\mid }u\mathrm{\mid }\le 6R_{\rho }q}\mathrm{\mid }{G}_m(u)\mathrm{\mid }{d}^{2}u\le K_m{q}^2(1+\mathrm{\mid }\log q\mathrm{\mid })$. Multiplying by $\mathrm{\parallel }\kappa {\mathrm{\parallel }}_{\mathrm{\infty }}$ bounds the convolution by $K_m(1+\mathrm{\mid }\log q\mathrm{\mid })\le K_{m,R}(1+\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid })$ because $q\ge \mathrm{\mid }z\mathrm{\mid }\mathrm{/}(4R_{\rho })$. Choose ${ϵ}_{m,R}$ small enough that $6R_{\rho }q\le 1$. This proves the uniform estimate for all $0<ϵ,\delta \le {ϵ}_{m,R}$.

For fixed $ϵ,\delta >0$, the smoothed fields at $x,y$ are jointly centered Gaussian by [F1], and their cross-covariance is $C_{ϵ,\delta }(x-y)$ by [F2]. Apply [F3] to obtain $\mathbb{E}[\overline{:\text{\hspace{0.17em}⁣}{\mathrm{\Phi }}_{ϵ}(x{)}^n\text{\hspace{0.17em}⁣}:}:\text{\hspace{0.17em}⁣}{\mathrm{\Phi }}_{\delta }(y{)}^n\text{\hspace{0.17em}⁣}:]=n!C_{ϵ,\delta }(x-y{)}^n$. For fixed cutoffs, each variance is finite and independent of $x$ by translation invariance; Cauchy–Schwarz [F7] therefore bounds the absolute expected product by a finite constant on the bounded support of $f(x)g(y)$. Fubini [F4] applies and yields $\mathbb{E}[\overline{W_{ϵ,n}(f)}{W}_{\delta ,n}(g)]=n!\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int \overline{f(x)}g(y)C_{ϵ,\delta }(x-y{)}^n{d}^{2}x{d}^{2}y$. For measurability of the samplewise smearing, on a fixed bounded box containing the support replace the continuous function $x\mapsto :\text{\hspace{0.17em}⁣}{\mathrm{\Phi }}_{ϵ}(x{)}^n\text{\hspace{0.17em}⁣}:$ on each fine grid cell by its value at one fixed point of that cell. Its integral against $f$ is a finite sum of measurable evaluations with deterministic coefficients ${\int }_{\text{cell}}f$; uniform continuity on the box makes the sums converge to the samplewise integral. The same approximations make the field jointly measurable in sample and position, so Fubini applies to measurable functions.

Let $K_f,K_g$ be bounded supports of $f,g$. By [F2], $C_{ϵ,\delta }(z{)}^n\to G_m(z{)}^n$ for every $z\mathrm{\ne }0$ and, on the bounded difference set $K_f-K_g$, its absolute value is bounded by $K(1+\mathrm{\mid }\log \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }{)}^n$ for all sufficiently small cutoffs. This majorant is locally integrable: on the annulus $2^{-j-1}<\mathrm{\mid }z\mathrm{\mid }\le 2^{-j}$ its integral is at most $\pi 2^{-2j}(1+(j+1)\log 2{)}^n$, and the resulting series converges by the ratio test. Consequently $(x,y)\mapsto \mathrm{\mid }f(x)g(y)\mathrm{\mid }(1+\mathrm{\mid }\log \mathrm{\mid }x-y\mathrm{\mid }\mathrm{\mid }{)}^n$ is integrable on $K_f\times K_g$: for fixed $x$, the translated bounded set $x-K_g$ lies in one fixed disk on which the logarithmic power is integrable, uniformly in $x$, and $K_f$ has finite area. For every sequence $({ϵ}_k,{\delta }_k)\to (0,0)$, apply [F5] on this product to get $\mathbb{E}[\overline{W_{{ϵ}_k,n}(f)}{W}_{{\delta }_k,n}(g)]\to n!\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int \overline{f(x)}g(y)G_m(x-y{)}^n{d}^{2}x{d}^{2}y$. Sequential characterization of limits in ${\mathbb{R}}^2$ gives the same joint limit as $ϵ,\delta \downarrow 0$.

Take $g=f$. The right side of step 2.1 is a fixed finite real number, being the limit of the nonnegative diagonal covariances. Expanding the squared norm gives $\mathrm{\parallel }{W}_{ϵ,n}(f)-W_{\delta ,n}(f){\mathrm{\parallel }}_2^2=\mathbb{E}\mathrm{\mid }{W}_{ϵ,n}(f){\mathrm{\mid }}^2+\mathbb{E}\mathrm{\mid }{W}_{\delta ,n}(f){\mathrm{\mid }}^2-2\mathrm{Re}\mathbb{E}[\overline{W_{ϵ,n}(f)}{W}_{\delta ,n}(f)]$, which tends to zero by step 2.1. Thus the cutoff family is Cauchy in $L^2({\mu }_C)$, and [F6] supplies its limit $W_n(f)$. Applying the mixed covariance convergence in step 2.1 and [F7] to pass both factors to their $L^2$ limits proves the displayed covariance formula for $W_n(f),W_n(g)$. For real $f$, each cutoff smearing is real; continuity of the imaginary-part map in $L^2$ gives $\mathrm{\parallel }\mathrm{Im}{W}_n(f){\mathrm{\parallel }}_2\le \mathrm{\parallel }{W}_n(f)-W_{ϵ,n}(f){\mathrm{\parallel }}_2\to 0$, so the limit class is real.