Finite Euclidean quartic interacting measure

A polynomial $U$ is continuous and hence Borel measurable. Since $U\ge 0$ and $\lambda \ge 0$, [F4] gives $f(q):=\exp (-\lambda U(q))$ Borel with $0<f(q)\le 1$. Thus $Z=\int fd{\gamma }_C$ is finite, and [F2] defines $\nu =fd{\gamma }_C$; [F3] gives its measure property. Its total mass is $\nu ({\mathbb{R}}^N)=Z$ by definition.

Under the displayed side condition $0<Z<\mathrm{\infty }$, the scaled measure $\mu =Z^{-1}\nu$ is a Borel measure and has total mass $Z^{-1}\nu ({\mathbb{R}}^N)=1$, so it is a probability measure by [F3] and step 1.1. For a polynomial $P$ integrable under $\mu$, $⟨P{⟩}_{C,U,\lambda }$ is then its well-defined integral by the notation in the Definition.

Let $Z\sim N(0,1)$. For an integer $k\ge 0$, choose an integer $r$ with $2r\ge k$. On $\mathrm{\mid }x\mathrm{\mid }\ge 1$, [F7] gives $e^{x^2\mathrm{/}4}\ge (x^2\mathrm{/}4{)}^r\mathrm{/}r!$, so $\mathrm{\mid }x{\mathrm{\mid }}^k{e}^{-x^2\mathrm{/}2}\le 4^{r}r!e^{-x^2\mathrm{/}4}$. The right side is integrable by [F6], and the integrand is bounded on $[-1,1]$. Since [F2] represents any scalar $N(0,v)$ as $\sqrt{v}Z$ with $\sqrt{v}$ supplied by [F8], every scalar normal has finite absolute moments of every integer order, including the degenerate $v=0$ case.

For fixed $k\ge 1$, define $T_k(u_1,\dots ,u_k):=\mathbb{E}{\prod }_{j=1}^k(u_j^{\mathsf{T}}Q)$. Each projection has a finite $k$th absolute moment by step 1.1 and [F1]. Applying [F5] successively shows the product is integrable. Linearity of each scalar projection and of the integral makes $T_k$ a symmetric $k$-linear form. The law exists under the stated Choice hypothesis [A1]. Since every polynomial is a finite linear combination of coordinate monomials, the triangle inequality also gives every polynomial a finite absolute moment.

On the diagonal, $T_k(u,\dots ,u)=\mathbb{E}[(u^{\mathsf{T}}Q{)}^k]$. Put $v=u^{\mathsf{T}}Cu\ge 0$. By [F1], $u^{\mathsf{T}}Q\sim N(0,v)$, so [F3] gives its characteristic function $e^{-v{t}^2\mathrm{/}2}$. At zero, [F4] gives ${\phi }^{(k)}(0)=i^k\mathbb{E}[(u^{\mathsf{T}}Q{)}^k]$; extracting the $t^k$ coefficient of $e^{-v{t}^2\mathrm{/}2}$ and dividing by $i^k$ yields zero for odd $k$, and for $k=2m$ gives $T_{2m}(u,\dots ,u)=\frac{(2m)!}{2^{m}m!}{v}^m=(2m-1)!!(u^{\mathsf{T}}Cu{)}^m\mathrm{.}$ The identity also holds for $v=0$, when the projected law is the point mass at zero.

For even $k=2m$, define $W_{2m}(u_1,\dots ,u_{2m})$ by the pairing sum in the Statement; for odd $k$, define $W_k=0$. This is a symmetric $k$-linear form. On the diagonal, every pairing contributes $(u^{\mathsf{T}}Cu{)}^m$, and there are $(2m-1)!!$ pairings: pair index $1$ with one of $2m-1$ indices and repeat on the remaining $2m-2$. Hence $W_k(u,\dots ,u)=T_k(u,\dots ,u)$ by step 3.1, also for odd $k$.

For $k\ge 1$, two symmetric $k$-linear forms agreeing on every diagonal agree everywhere. Indeed, for their difference $A$, expand ${\sum }_{\epsilon \in \{-1,1{\}}^k}\left({\prod }_{j=1}^k{\epsilon }_j\right)A\text{\hspace{0.17em}⁣}({\sum }_i{\epsilon }_i{u}_i,\dots ,{\sum }_i{\epsilon }_i{u}_i)\mathrm{.}$ In the multilinear expansion, the sum over signs cancels every term unless each label $1,\dots ,k$ occurs an odd number of times among the $k$ slots. Thus every surviving term uses each label once; symmetry makes all $k!$ such terms equal to $A(u_1,\dots ,u_k)$, and the sign sum is $2^{k}k!A(u_1,\dots ,u_k)$. Since the diagonal of $A=T_k-W_k$ vanishes by step 4.1, this sum is zero and $T_k=W_k$. For $k=0$, the empty product has expectation one and $\mathrm{Pair}(0)$ contains just the empty pairing, whose empty product is one, so the formula also holds.

Since $U(q)\ge 0$ and $\lambda \ge 0$, [F5] gives $0<\exp (-\lambda U(q))\le 1$ for every $q$, so $Z(\lambda )\le 1$. For each positive integer $n$, let $E_n=\{q:\exp (-\lambda U(q))\ge 1\mathrm{/}n\}$. These are Borel and their union is ${\mathbb{R}}^N$, because the integrand is strictly positive at every $q$. Since ${\gamma }_C$ is a probability measure, not all $E_n$ can have measure zero: otherwise countable subadditivity would give ${\gamma }_C({\mathbb{R}}^N)=0$. Thus some $E_n$ has positive measure, and $Z(\lambda )\ge {\gamma }_C(E_n)\mathrm{/}n>0$.

Fix $r\ge 0$. For $s=0,1,\dots ,r$, [F6] and [F7] give the pointwise derivative $(-1{)}^{s}P(q)U(q{)}^s\exp (-\lambda U(q))$ and its next derivative $(-1{)}^{s+1}P(q)U(q{)}^{s+1}\exp (-\lambda U(q))$. If ${\lambda }_0>0$ and $\mathrm{\mid }h\mathrm{\mid }<{\lambda }_0\mathrm{/}2$, [F8] bounds the difference quotient of the $s$th derivative by $\mathrm{\mid }P(q)\mathrm{\mid }U(q{)}^{s+1}\exp (-{\lambda }_{0}U(q)\mathrm{/}2)$, since every intermediate parameter is at least ${\lambda }_0\mathrm{/}2$ and [F5] makes the exponential increasing. Put $x=U(q)$. If $x=0$ the scalar factor is zero; if $x>0$, put $y={\lambda }_{0}x\mathrm{/}4$. The power-series estimate in [F5] at $2y$ gives $\exp (2y)\ge (2y{)}^{s+1}\mathrm{/}(s+1)!$, so its reciprocal gives $x^{s+1}\exp (-{\lambda }_{0}x\mathrm{/}2)\le (2\mathrm{/}{\lambda }_0{)}^{s+1}(s+1)!\mathrm{.}$ Thus these difference quotients are bounded by a finite constant times $\mathrm{\mid }P(q)\mathrm{\mid }$, integrable by [F2]. [F3] permits differentiation under the integral inductively through order $r$, giving $N_P^{(r)}(\lambda )=(-1{)}^r\int P{U}^r\exp (-\lambda U)d{\gamma }_C$ for $\lambda >0$. Taking $P=1$ gives the formula for $Z$.

The nonnegative Borel density $\exp (-\lambda U)\mathrm{/}Z(\lambda )$ has integral one by the definition of $Z$ and step 1.1. By [F4], it defines a Borel probability measure ${\mu }_{\lambda }$. For any polynomial $P$, [F2] gives $\int \mathrm{\mid }P\mathrm{\mid }d{\gamma }_C<\mathrm{\infty }$; since [F5] gives $\exp (-\lambda U)\le 1$ and step 1.1 gives $Z(\lambda )>0$, $\int \mathrm{\mid }P\mathrm{\mid }d{\mu }_{\lambda }\le Z(\lambda {)}^{-1}\int \mathrm{\mid }P\mathrm{\mid }d{\gamma }_C<\mathrm{\infty }\mathrm{.}$

At $\lambda =0$, use right limits and right difference quotients. For $r=0$, dominated convergence [F3] with dominator $\mathrm{\mid }P\mathrm{\mid }$ and [F5] gives $N_P(\lambda )\to N_P(0)$ as $\lambda \downarrow 0$; taking $P=1$ gives the same for $Z$. Now fix $r\ge 1$. For each $s=0,1,\dots ,r-1$ and $h>0$, [F8] bounds the difference quotient of $(-1{)}^{s}P{U}^s\exp (-\lambda U)$ at zero by $\mathrm{\mid }P\mathrm{\mid }{U}^{s+1}$, since [F5] gives $0<\exp (-cU)\le 1$ for $c\ge 0$. This bound is integrable by [F2]. Dominated convergence [F3] inductively gives $N_{P,+}^{(r)}(0)=(-1{)}^r\int P{U}^{r}d{\gamma }_C$; applying [F3] to the formula in step 1.2 as $\lambda \downarrow 0$ shows these derivatives converge continuously to their endpoint values. Taking $P=1$ gives the same assertions for $Z$.

Multiply the formal series for $e^{t{Q}_j}$ and $e^{-C_{jj}{t}^2\mathrm{/}2}$. The coefficient of $t^n$ is the finite sum over $0\le s\le \lfloor n\mathrm{/}2\rfloor$ of $Q_j^{n-2s}(-C_{jj}\mathrm{/}2{)}^s\mathrm{/}((n-2s)!s!)$. By the defining identity, multiplying this coefficient by $n!$ gives the displayed formula for $:Q_j^n{:}_C$. This is formal coefficient algebra and requires no convergence assertion.

For $n=4$, the terms $s=0,1,2$ are respectively $Q_j^4$, $-6C_{jj}{Q}_j^2$, and $3C_{jj}^2$, which gives the stated quartic Wick polynomial. The formula also covers $C_{jj}=0$.

By [F5], $f(x)=\exp (-x)$ is $C^{M+1}$ and $\mathrm{\mid }{f}^{(M+1)}(\xi )\mathrm{\mid }=\exp (-\xi )$. Apply [F4] at $x=\lambda U(q)\ge 0$; [F6] gives $\exp (-\xi )\le 1$. Thus pointwise $\mid \exp (-\lambda U(q))-{\sum }_{j=0}^M\frac{(-\lambda U(q){)}^j}{j!}\mid \le \frac{{\lambda }^{M+1}U(q{)}^{M+1}}{(M+1)!}\mathrm{.}$ The case $\lambda U(q)=0$ has zero remainder.

Let $Z_M(\lambda )={\sum }_{j=0}^M{z}_j{\lambda }^j$ and $A_M(\lambda )={\sum }_{j=0}^M{a}_j(P){\lambda }^j$. By the recurrence defining $b_j(P)$, every coefficient through degree $M$ in $A_M(\lambda )-({\sum }_{j=0}^M{b}_j(P){\lambda }^j)Z_M(\lambda )$ is zero, because $z_0=1$. The remaining polynomial terms have degree at least $M+1$, so their absolute value is bounded by a finite constant times ${\lambda }^{M+1}$ on $0\le \lambda \le 1$.

Integrate the inequality in step 1.1 against ${\gamma }_C$ to obtain the stated partition-function expansion and bound. Multiply the same pointwise inequality by $\mathrm{\mid }P(q)\mathrm{\mid }$ and integrate to obtain the numerator expansion and bound. Every right-hand expectation is finite by [F2].

On $0\le \lambda \le 1$, the polynomial $B_M(\lambda )={\sum }_{j=0}^M{b}_j(P){\lambda }^j$ is bounded. Using the remainder bounds in step 2.1 and the polynomial identity in step 1.2 therefore gives $\mathrm{\mid }{N}_P(\lambda )-B_M(\lambda )Z(\lambda )\mathrm{\mid }\le K_M{\lambda }^{M+1}$ for some finite $K_M$. As $\lambda \downarrow 0$, $\exp (-\lambda U(q))\to 1$ pointwise and $0<\exp (-\lambda U(q))\le 1$ by [F6], so [F3] gives $Z(\lambda )\to Z(0)=1$. Choose $\delta >0$ such that $Z(\lambda )\ge 1\mathrm{/}2$ for $0\le \lambda <\delta$. Dividing by $Z(\lambda )$ yields $\mathrm{\mid }⟨P{⟩}_{\lambda }-B_M(\lambda )\mathrm{\mid }\le 2K_M{\lambda }^{M+1}$ there. Finally, [F2] writes every coefficient moment as a finite covariance-pairing sum; no limit in $M$ was taken.