Gaussian matching and a rigorous derivative-expansion remainder

By [F1]–[F2], ${\gamma }_{K^{-1}}$ has Lebesgue density $c_K\exp (-\frac{1}{2}{x}^{T}Kx)$, where $c_K=(2\pi {)}^{-d\mathrm{/}2}\det (K^{-1}{)}^{-1\mathrm{/}2}>0$. The expectation in the Statement is therefore the ratio of the displayed Lebesgue integrals, provided the numerator is finite.

Since $K$ is invertible, direct expansion gives $\frac{1}{2}{x}^{T}Kx-j^{T}x=\frac{1}{2}(x-K^{-1}j{)}^{T}K(x-K^{-1}j)-\frac{1}{2}{j}^T{K}^{-1}j\mathrm{.}$ The translation $x\mapsto x-K^{-1}j$ preserves Lebesgue measure by [F3]–[F5], so the numerator is the denominator multiplied by $\exp (\frac{1}{2}{j}^T{K}^{-1}j)$. The denominator is finite and positive because it is the normalizing integral of the probability density in [F2]; hence the numerator is also finite, and division proves both equalities. Expanding the same square proves the action identity.

The actual proof of [F1] uses Cauchy–Schwarz for its resolvent lower bound and closedness of $A$ for its resolvent graph argument. [F2] supplies the former. By [A1]–[A2] countable choice is available, so [F3] applies and self-adjointness gives $A=A^{\ast }$, hence $A$ is closed. This records the actual inputs of [F1]'s proof; the carrier's complete conditional proof was inspected.

Set $g_M(a)=(M^2+a{)}^{-1}$ and ${\mu }_u(S)=⟨E_A(S)u,u⟩$. By [F1], $g_M(A)$ is bounded with norm at most $M^{-2}$. Since $0\le a{g}_M(a)\le 1$, $\int a^2{g}_M(a{)}^{2}d{\mu }_u(a)\le {\mu }_u([0,\mathrm{\infty }))=\mathrm{\parallel }u{\mathrm{\parallel }}^2$ for every $u\in H$, so the domain formula in [F1] gives $g_M(A)u\in D(A)$. To see the left-inverse identity, let $x=g_M(A)u$ and $x_n=E_A([0,n])x$. The domain formula gives $x_n\to x$ and $A{x}_n\to Ax$ in $H$. The spectral representation in [F1] and the multiplicativity in [F5] applied to bounded truncations give $(M^{2}I+A)x_n={\int }_{[0,n]}(M^2+a)g_M(a)d{E}_A(a)u=E_A([0,n])u$; hence $(M^{2}I+A)g_M(A)u=u$. For the right-inverse identity, if $v\in D(A)$, set $v_n=E_A([0,n])v$. Then $v_n\to v$ and $(M^{2}I+A)v_n\to (M^{2}I+A)v$ in $H$. The same spectral representation and multiplicativity give $g_M(A)(M^{2}I+A)v_n={\int }_{[0,n]}{g}_M(a)(M^2+a)d{E}_A(a)v=v_n$, and boundedness of $g_M(A)$ gives $g_M(A)(M^{2}I+A)v=v$. Thus $K^{-1}=g_M(A)$ and $\mathrm{\parallel }{K}^{-1}\mathrm{\parallel }\le M^{-2}$. Finally, the positive scalar multiplier gives $⟨J,K^{-1}J⟩=\int g_M(a)d{\mu }_J(a)\ge 0$, so the quadratic form is real.

In a finite real regulator with symmetric positive-definite matrix $K$, [F4] gives the normalized source integral $\exp (\frac{1}{2}{j}^T{K}^{-1}j)$. Taking its negative logarithm using [F6] gives the effective-action shift $-\frac{1}{2}{j}^T{K}^{-1}j$. No determinant or Lebesgue integral beyond that finite regulator enters the definition.

For every $a\ge 0$, putting $x=a\mathrm{/}{M}^2$ and multiplying by $1+x$ gives $\frac{1}{M^2+a}=M^{-2}{\sum }_{r=0}^N(-a\mathrm{/}{M}^2{)}^r+M^{-2}\frac{(-a\mathrm{/}{M}^2{)}^{N+1}}{1+a\mathrm{/}{M}^2}\mathrm{.}$ This finite geometric identity also holds at $a=0$ and for $N=0$.

On $J=P_{E}J$, the spectral variable satisfies $0\le a\le E^2$. Therefore every polynomial term $A^{r}J$ is defined, and the remainder multiplier satisfies $\mid M^{-2}\frac{(-a\mathrm{/}{M}^2{)}^{N+1}}{1+a\mathrm{/}{M}^2}\mid \le M^{-2}(E^2\mathrm{/}{M}^2{)}^{N+1},$ because $1+a\mathrm{/}{M}^2\ge 1$. Extend this bounded multiplier by zero outside $[0,E^2]$ and apply [F1]. With [F2] this identifies the inverse kernel and proves the vector identity defining $R_{N}J$ and its norm bound.

By [F2], the action remainder equals $\frac{1}{2}\mathrm{\mid }⟨J,R_{N}J⟩\mathrm{\mid }$. Cauchy–Schwarz [F3] and step 2.1 give the claimed bound. The factor $(E^2\mathrm{/}{M}^2{)}^{N+1}$ decreases to zero only when $E<M$; for $E\ge M$ the same finite-order bound remains true but need not be small.

Write ${\mathcal{B}}_d$ for the Borel sigma-algebra of ${\mathbb{R}}^d$ and ${\lambda }_B$ for Lebesgue measure restricted to ${\mathcal{B}}_d$. The countable basis of Euclidean open rectangles gives $\mathcal{B}({\mathbb{R}}^d\times {\mathbb{R}}^d)={\mathcal{B}}_d\otimes {\mathcal{B}}_d$. Hence $(x,z)\mapsto f(x+z)$ is product-measurable because addition is continuous and $f$ is Borel. The measure space $({\mathbb{R}}^d,{\mathcal{B}}_d,{\lambda }_B)$ is sigma-finite because ${\mathbb{R}}^d={\bigcup }_{n\ge 1}[-n,n{]}^d$ and each cube has finite measure; $({\mathbb{R}}^d,{\mathcal{B}}_d,{\gamma }_C)$ is a probability space by [F1]. Applying [F2] to these two spaces makes $x\mapsto T_{C}f(x)$ measurable for ${\mathcal{B}}_d$, hence Borel, including when the integral is infinite.

For finite nonnegative Borel $V$, $g_x(z):=e^{-V(x+z)}$ is Borel and satisfies $0<g_x(z)\le 1$. Its integral against the probability ${\gamma }_C$ is at most one. It is strictly positive: if the integral were zero, each set $\{z:g_x(z)\ge 1\mathrm{/}n\}$ would have measure zero, while their countable union is all of ${\mathbb{R}}^d$, contradicting ${\gamma }_C({\mathbb{R}}^d)=1$. Thus $T_C(e^{-V})(x)\in (0,1]$.

By [F3], the logarithm is defined at these values. Its continuity follows directly from [F4]: for $y_0>0$ and $ϵ>0$, strict increase and continuity of $\exp$ give $\exp (\log y_0-ϵ)<y_0<\exp (\log y_0+ϵ)$; put $\delta =\min \{y_0-\exp (\log y_0-ϵ),\exp (\log y_0+ϵ)-y_0\}>0$. If $y>0$ and $\mathrm{\mid }y-y_0\mathrm{\mid }<\delta$, then $\exp (\log y_0-ϵ)<y<\exp (\log y_0+ϵ)$, and strict monotonicity gives $\mathrm{\mid }\log y-\log y_0\mathrm{\mid }<ϵ$. Hence $V_C$ is finite, nonnegative, and Borel. If $C=0$, then for each coordinate $Z_i$, $\mathbb{E}{Z}_i^2=C_{ii}=0$. For every $n\ge 1$, $0=\mathbb{E}{Z}_i^2\ge n^{-2}\mathbb{P}(\mathrm{\mid }{Z}_i\mathrm{\mid }\ge n^{-1})$, so $Z_i=0$ almost surely; the finite intersection gives $Z=0$ almost surely and ${\gamma }_0={\delta }_0$; hence $T_{0}f=f$ and $V_0=V$.

On $({\mathbb{R}}^d,{\gamma }_{C_1})\times ({\mathbb{R}}^d,{\gamma }_{C_2})$, the sum map $(y,z)\mapsto y+z$ has characteristic function $\int e^{iu\cdot (y+z)}d({\gamma }_{C_1}\times {\gamma }_{C_2})=\exp (-u^T{C}_{1}u\mathrm{/}2)\exp (-u^T{C}_{2}u\mathrm{/}2)=\exp [-u^T(C_1+C_2)u\mathrm{/}2]\mathrm{.}$ The product measure exists by [F3], the integrand is bounded and therefore integrable, and [F4] justifies the iterated integral/product factorization. Since $C_1+C_2$ is positive semidefinite, [F2] and [F5] identify the pushforward under addition as ${\gamma }_{C_1+C_2}$. This includes singular and zero covariances.

The function $(x,y,z)\mapsto f(x+y+z)$ is nonnegative and Borel. Apply [F6] to integrate first in $y$ and then in $z$; the resulting iterated integral equals the integral against the product law of $(y,z)$. By step 1.1 its pushforward under addition is ${\gamma }_{C_1+C_2}$. Therefore $\int (\int f(x+z+y)d{\gamma }_{C_1}(y))d{\gamma }_{C_2}(z)=\int f(x+w)d{\gamma }_{C_1+C_2}(w),$ including the value $+\mathrm{\infty }$. By [F1] this is exactly $T_{C_2}(T_{C_1}f)(x)=T_{C_1+C_2}f(x)$.

For finite nonnegative Borel $V$, [F1] gives $\exp [-V_{C_1}(x)]=T_{C_1}(e^{-V})(x)$ and the corresponding identity after applying $C_2$. Substituting $f=e^{-V}$ into step 2.1 proves the stated potential-weight identity. Since the normalized Gaussian integrals are expectations under probability laws, no determinant factor occurs. No differentiation or covariance-path limit was used.