Logarithmic singularity and decay of the massive two-dimensional covariance

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 $0<ϵ,\delta \le {ϵ}_{m,R}$.