Banach Gaussian priors, RKHS support, and concentration

Choose a countable dense sequence $(d_j{)}_{j\ge 1}$ in $B$. For each $n\ge 1$, assign $\omega$ to the first index $j$ for which $\mathrm{\parallel }f(\omega )-d_j\mathrm{\parallel }<2^{-n}$; such an index exists by density. The resulting sets $A_{n,j}$ form a measurable countable partition: each is the preimage of an open ball under $f$, with the finitely many earlier preimages removed.

Since the partial unions ${\bigcup }_{j\le N}{A}_{n,j}$ increase to $\mathrm{\Omega }$, [F3] gives an integer $N_n$ for which the omitted tail has measure at most $2^{-2n}$. Define $s_n=d_j$ on $A_{n,j}$ for $j\le N_n$, and $s_n=0$ on the omitted tail. Then $s_n$ is a finite-valued measurable simple function.

The tail events $T_n={\bigcup }_{j>N_n}{A}_{n,j}$ satisfy ${\sum }_{n}P(T_n)\le {\sum }_n{2}^{-2n}<\mathrm{\infty }$. By [F4], for every $N$, $P({\bigcup }_{n\ge N}{T}_n)\le {\sum }_{n\ge N}{2}^{-2n}$; continuity from above makes the probability of belonging to infinitely many $T_n$ equal to zero. Outside that null set, for all sufficiently large $n$ the point is not in $T_n$, and then $\mathrm{\parallel }{s}_n(\omega )-f(\omega )\mathrm{\parallel }<2^{-n}$. Thus $s_n\to f$ almost everywhere, which is strong measurability by [F2].

If $B=\{0\}$, take the constant zero sequence. Otherwise let $D$ be a countable dense subset of $B$. The normalized set $\{d\mathrm{/}\mathrm{\parallel }d\mathrm{\parallel }:d\in D,d\mathrm{\ne }0\}$ is countable and dense in the unit sphere: for $\mathrm{\parallel }x\mathrm{\parallel }=1$ and $d\to x$, eventually $d\mathrm{\ne }0$ and $\mathrm{\parallel }d\mathrm{/}\mathrm{\parallel }d\mathrm{\parallel }-x\mathrm{\parallel }\le 2\mathrm{\parallel }d-x\mathrm{\parallel }$. Enumerate it as $(x_j)$. By [F2] and [A1], choose ${\mathrm{\ell }}_j\in B^{\ast }$ with $\mathrm{\parallel }{\mathrm{\ell }}_j\mathrm{\parallel }=1$ and ${\mathrm{\ell }}_j(x_j)=1$ for each $j$.

For every $x\in B$ and every $j$, $\mathrm{\mid }{\mathrm{\ell }}_j(x)\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }$, so the supremum is at most $\mathrm{\parallel }x\mathrm{\parallel }$. If $x\mathrm{\ne }0$, choose unit vectors $x_{j_k}\to x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }$. Then ${\mathrm{\ell }}_{j_k}(x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel })\ge 1-\mathrm{\parallel }{x}_{j_k}-x\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }\to 1$, so ${\sup }_j\mathrm{\mid }{\mathrm{\ell }}_j(x)\mathrm{\mid }\ge \mathrm{\parallel }x\mathrm{\parallel }$. The zero vector gives equality directly.

Write $\phi (x)=(2\pi {)}^{-1\mathrm{/}2}{e}^{-x^2\mathrm{/}2}$. For $h\mathrm{\ne }0$, the density formula gives $[\mathrm{\Phi }(x+h)-\mathrm{\Phi }(x)]\mathrm{/}h=h^{-1}{\int }_x^{x+h}\phi (s)ds$, which tends to $\phi (x)$ by continuity; hence ${\mathrm{\Phi }}^{\mathrm{\prime }}(x)=\phi (x)>0$. The CDF lemma gives continuity, strict increase and limits $0,1$ at the two infinities. For any $u\in (0,1)$, choose $a<b$ with $\mathrm{\Phi }(a)<u<\mathrm{\Phi }(b)$ using those limits; [F8] applied to $\mathrm{\Phi }$ on $[a,b]$ gives an $x$ with $\mathrm{\Phi }(x)=u$, and strict increase makes it unique. Thus $\mathrm{\Phi }$ is a bijection onto $(0,1)$. Set $I(u)=\phi ({\mathrm{\Phi }}^{-1}(u))$ for $0<u<1$, and set $I(0)=I(1)=0$. The derivative-of-inverse rule [F7] therefore gives $({\mathrm{\Phi }}^{-1}{)}^{\mathrm{\prime }}(u)=1\mathrm{/}\phi ({\mathrm{\Phi }}^{-1}(u))=1\mathrm{/}I(u)$. From $\phi (x+h)=\phi (x)e^{-xh-h^2\mathrm{/}2}$ and $(\exp {)}^{\mathrm{\prime }}(0)=1$ in [F7], the difference quotient gives ${\phi }^{\mathrm{\prime }}(x)=-x\phi (x)$. Writing $x={\mathrm{\Phi }}^{-1}(u)$, the chain rule [F7] yields $I^{\mathrm{\prime }}(u)={\phi }^{\mathrm{\prime }}(x)\mathrm{/}\phi (x)=-x$ and then $I^{\mathrm{\prime }\mathrm{\prime }}(u)=-({\mathrm{\Phi }}^{-1}{)}^{\mathrm{\prime }}(u)=-1\mathrm{/}I(u)$. Thus $I{I}^{\mathrm{\prime }\mathrm{\prime }}=-1$. As $u\downarrow 0$ or $u\uparrow 1$, ${\mathrm{\Phi }}^{-1}(u)$ tends to the corresponding infinity and $\phi ({\mathrm{\Phi }}^{-1}(u))\to 0$, so $I$ is continuous on $[0,1]$; also $0<I(u)\le (2\pi {)}^{-1\mathrm{/}2}$ on $(0,1)$. First take a smooth $f$ valued in $[\eta ,1-\eta ]$, $0<\eta <1\mathrm{/}2$, with all derivatives bounded. Define $P_{t}f(x)=\int f(e^{-t}x+\sqrt{1-e^{-2t}}y){\gamma }_d(dy)$ and $u_t=P_{t}f$. Gaussian integration by parts in $y$ gives ${\mathrm{\partial }}_t{u}_t=L{u}_t$, $L=\mathrm{\Delta }-x\cdot \mathrm{\nabla }$, while $\mathrm{\nabla }{P}_{t}f=e^{-t}{P}_t\mathrm{\nabla }f$. Invariance $\int P_{t}gd{\gamma }_d=\int gd{\gamma }_d$ follows because $e^{-t}X+\sqrt{1-e^{-2t}}Y$ is standard Gaussian for independent standard $X,Y$. Differentiation is justified by [F2]; the integration-by-parts boundary terms vanish by bounded derivatives and Gaussian decay.

Let $D$ be a nonempty finite union of closed balls of one common positive radius. Its boundary is contained in finitely many spheres, which have Lebesgue and hence Gaussian measure zero. For $h>0$ put $f_h=(1-\mathrm{dist}(\cdot ,D)\mathrm{/}h{)}_{+}$. Distance is $1$-Lipschitz, since $\mathrm{dist}(x,D)\le \mathrm{\mid }x-y\mathrm{\mid }+\mathrm{dist}(y,D)$ and one may interchange $x,y$. Thus $0\le f_h\le 1$ and it is $1\mathrm{/}h$-Lipschitz. Outside finitely many inner and outer cutoff spheres and nearest-center bisector hyperplanes it is smooth, its gradient has norm at most $1\mathrm{/}h$, and its gradient vanishes outside $(D+h{B}_2^d)\setminus D$. These exceptional sets have Lebesgue measure zero by Fubini in suitable orthogonal coordinates.

Put $v=\mathrm{\nabla }{u}_t$, $H=D^2{u}_t$, $K=I(u_t{)}^2+\mathrm{\mid }v{\mathrm{\mid }}^2$ and $J(t)=\int \sqrt{K}d{\gamma }_d$. The identity $\mathrm{\nabla }Lu=L\mathrm{\nabla }u-\mathrm{\nabla }u$, Gaussian integration by parts, and $I{I}^{\mathrm{\prime }\mathrm{\prime }}=-1$ give $J^{\mathrm{\prime }}(t)=\int \frac{-\mathrm{\parallel }{I}^{\mathrm{\prime }}(u_t)v{v}^T-I(u_t)H{\mathrm{\parallel }}_F^2+\mathrm{\mid }Hv{\mathrm{\mid }}^2-\mathrm{\parallel }H{\mathrm{\parallel }}_F^2\mathrm{\mid }v{\mathrm{\mid }}^2}{K^{3\mathrm{/}2}}d{\gamma }_d\le 0.$ To verify the displayed computation, differentiating $J$ first gives $\int [I{I}^{\mathrm{\prime }}Lu+v\cdot \mathrm{\nabla }Lu]\mathrm{/}\sqrt{K}$. Integrating both $Lu$ and each $L{v}_i$ by parts leaves numerator $-I^{\mathrm{\prime }2}\mathrm{\mid }v{\mathrm{\mid }}^4+2I{I}^{\mathrm{\prime }}{v}^{T}Hv-I^2\mathrm{\parallel }H{\mathrm{\parallel }}_F^2-\mathrm{\parallel }H{\mathrm{\parallel }}_F^2\mathrm{\mid }v{\mathrm{\mid }}^2+\mathrm{\mid }Hv{\mathrm{\mid }}^2$, the displayed square expansion. Applying [F9] to row $i$ gives $({\sum }_j{H}_{ij}{v}_j{)}^2\le ({\sum }_j{H}_{ij}^2)({\sum }_j{v}_j^2)$; summing over the finitely many rows gives $\mathrm{\mid }Hv{\mathrm{\mid }}^2\le \mathrm{\parallel }H{\mathrm{\parallel }}_F^2\mathrm{\mid }v{\mathrm{\mid }}^2$. All operations on compact time intervals are dominated because $I(u_t)$ has a fixed positive lower bound, derivatives are bounded, and Gaussian polynomial moments are finite.

As $t\to \mathrm{\infty }$, $u_t(x)\to \int fd{\gamma }_d$ and $\mathrm{\mid }\mathrm{\nabla }{u}_t\mathrm{\mid }\le e^{-t}\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Dominated convergence and step 2.1 yield $I(\int fd{\gamma }_d)\le \int \sqrt{I(f{)}^2+\mathrm{\mid }\mathrm{\nabla }f{\mathrm{\mid }}^2}d{\gamma }_d\le \int I(f)d{\gamma }_d+\int \mathrm{\mid }\mathrm{\nabla }f\mathrm{\mid }d{\gamma }_d\mathrm{.}$ Replacing a smooth $[0,1]$-valued $f$ by $\eta +(1-2\eta )f$ and letting $\eta \downarrow 0$ extends this inequality to smooth $[0,1]$-valued functions with bounded derivatives.

Convolve $f_h$ with the nonnegative bumps ${\rho }_{\delta }$ from [F5]. The convolutions are smooth, valued in $[0,1]$, have bounded derivatives, and converge uniformly to $f_h$. On almost every coordinate line the distance formula is a finite piecewise smooth minimum of distances to centers, so $f_h$ is absolutely continuous there with derivative bounded by $1\mathrm{/}h$. One-dimensional integration by parts and Fubini show $\mathrm{\nabla }(f_h\ast {\rho }_{\delta })=(\mathrm{\nabla }{f}_h)\ast {\rho }_{\delta }$. At each point off the exceptional hypersurfaces the latter converges to $\mathrm{\nabla }{f}_h$, and its norm is at most $1\mathrm{/}h$. Dominated convergence against Gaussian probability therefore passes step 3.1 to $f_h$ and gives $I(\int f_{h}d{\gamma }_d)\le \int I(f_h)d{\gamma }_d+\frac{{\gamma }_d(D+h{B}_2^d)-{\gamma }_d(D)}{h}\mathrm{.}$ As $h\downarrow 0$, $f_h\to {\mathbf{1}}_D$ and $I(f_h)\to 0$, so $I({\gamma }_d(D))$ is at most the lower right derivative of the neighborhood mass.

Set $D_t=D+t{B}_2^d$ and $V(t)={\gamma }_d(D_t)$. On each bounded interval $[0,R]$, $V$ is Lipschitz: for $0\le t<t+h\le R$, the mass difference is bounded by the sum of masses of finitely many spherical shells. Gaussian density is bounded, and [F6] gives each shell volume as $v_d((a+t+h{)}^d-(a+t{)}^d)\le C_{R}h$, where $a>0$ is the common initial radius. Thus $\mathrm{\mid }V(t+h)-V(t)\mathrm{\mid }\le L_{R}h$ for a finite constant $L_R$. For any finite disjoint interval family, the sum of its endpoint increments is therefore at most $L_R$ times its total length; taking $\delta =\epsilon \mathrm{/}(L_R+1)$ proves absolute continuity directly from [F10]. By [F3], $V$ is differentiable almost everywhere. Each $D_t$ is again a finite union of equal-radius balls, so the lower-right derivative bound in step 4.1 implies $V^{\mathrm{\prime }}(t)\ge I(V(t))$ at each differentiability point. Since $D_0$ has interior and $D_R$ is bounded, $0<V(0)\le V(t)\le V(R)<1$. On this compact range, $I$ has a positive minimum, so the inverse derivative formula in step 1.1 and the mean value theorem [F10] show that ${\mathrm{\Phi }}^{-1}$ is Lipschitz there. Hence $G={\mathrm{\Phi }}^{-1}\circ V$ is Lipschitz on $[0,R]$; the same finite-total-length argument proves that $G$ is absolutely continuous. At every differentiability point of $V$, the chain rule [F7] gives $G^{\mathrm{\prime }}=V^{\mathrm{\prime }}\mathrm{/}I(V)\ge 1$. Applying the fundamental theorem [F3] to $G$ and taking $R=r$ (with $r=0$ immediate) yields $G(r)-G(0)\ge r$, or ${\gamma }_d(D+r{B}_2^d)\ge \mathrm{\Phi }({\mathrm{\Phi }}^{-1}({\gamma }_d(D))+r)\mathrm{.}$

For nonempty compact $K$ and $r>0$, choose $0<\delta <r$ and cover $K$ by finitely many closed $\delta$-balls centered in $K$; call their union $D_{\delta }$. Then $K\subseteq D_{\delta }\subseteq K+\delta B_2^d$ and $D_{\delta }+(r-\delta )B_2^d\subseteq K+r{B}_2^d$. Step 5.1 gives ${\gamma }_d(K+r{B}_2^d)\ge \mathrm{\Phi }({\mathrm{\Phi }}^{-1}({\gamma }_d(K))+r-\delta )$; let $\delta \downarrow 0$. For Borel $A$, inner regularity provides compact $K\subseteq A$ with mass arbitrarily close to ${\gamma }_d(A)$. Each $K+r{B}_2^d$ is compact and contained in $A+r{B}_2^d$, so taking the supremum gives the asserted inner-measure bound. If ${\gamma }_d(A)=0$ the right side is zero; if it is one, compact inner approximation gives lower bound one. At $r=0$, [F4] identifies the inner and ordinary measures of $A$.

For $t>0$, put $a=e^{-t}$, $b=\sqrt{1-e^{-2t}}$, and $P_{t}g(x)=\int g(ax+by)d{\gamma }_d(y)$. By [F1], $aX+bY$ is standard Gaussian for independent $X,Y\sim {\gamma }_d$; Fubini [F4] gives $\int P_{t}gd{\gamma }_d=\int gd{\gamma }_d$. Differentiation under the integral [F2] gives $\mathrm{\nabla }{P}_{t}g=a{P}_t(\mathrm{\nabla }g)$ and $\mathrm{\Delta }{P}_{t}g=a^2{P}_t(\mathrm{\Delta }g)$. It also gives ${\mathrm{\partial }}_t{P}_{t}g=-ax\cdot P_t(\mathrm{\nabla }g)+(a^2\mathrm{/}b)\int y\cdot \mathrm{\nabla }g(ax+by)d{\gamma }_d(y)$. For each coordinate, [F3] applied against the Gaussian density gives $\int y_i{\mathrm{\partial }}_{i}g(ax+by)d{\gamma }_d=b\int {\mathrm{\partial }}_{ii}g(ax+by)d{\gamma }_d$; the boundary terms vanish because ${\mathrm{\partial }}_{i}g$ is bounded and the Gaussian density tends to zero at both ends. Thus ${\mathrm{\partial }}_t{P}_{t}g=\mathrm{\Delta }{P}_{t}g-x\cdot \mathrm{\nabla }{P}_{t}g$.

Put $u_t=P_{t}g$. Differentiate the entropy using step 1.1 and [F2]: $\frac{d}{dt}{\mathrm{Ent}}_{{\gamma }_d}(u_t)=\int (1+\log u_t)(\mathrm{\Delta }{u}_t-x\cdot \mathrm{\nabla }{u}_t)d{\gamma }_d$. For each coordinate, [F3] applied to $v=1+\log u_t$ and ${\mathrm{\partial }}_i{u}_t$ against the Gaussian density [F1] gives $\int v({\mathrm{\partial }}_{ii}{u}_t-x_i{\mathrm{\partial }}_i{u}_t)d{\gamma }_d=-\int ({\mathrm{\partial }}_i{u}_t{)}^2\mathrm{/}{u}_{t}d{\gamma }_d$; the boundary term vanishes because $v$ and $\mathrm{\nabla }{u}_t$ are bounded and the density tends to zero. Summing coordinates yields $\frac{d}{dt}{\mathrm{Ent}}_{{\gamma }_d}(u_t)=-\int \frac{\mathrm{\mid }\mathrm{\nabla }{u}_t{\mathrm{\mid }}^2}{u_t}d{\gamma }_d$. The differentiated integrands are dominated on each compact time interval because $u_t$ stays between the positive bounds for $g$ and its first two derivatives are bounded. As $t\downarrow 0$, continuity of $g$ and bounded convergence [F4] give $P_{t}g(x)\to g(x)$ and ${\mathrm{Ent}}_{{\gamma }_d}(u_t)\to {\mathrm{Ent}}_{{\gamma }_d}(g)$. As $t\to \mathrm{\infty }$, $u_t(x)\to \int gd{\gamma }_d$, so [F4] gives entropy limit zero. Integrate the derivative on $[\delta ,T]$ by [F6], then let $\delta \downarrow 0$ and $T\uparrow \mathrm{\infty }$ using [F5] on the nonnegative dissipation. Therefore ${\mathrm{Ent}}_{{\gamma }_d}(g)={\int }_0^{\mathrm{\infty }}\int \mathrm{\mid }\mathrm{\nabla }{P}_{t}g{\mathrm{\mid }}^2\mathrm{/}(P_{t}g)d{\gamma }_{d}dt$.

For each $x,t$, apply [F7] to each coordinate after writing $\mathrm{\nabla }g=(\mathrm{\nabla }g\mathrm{/}\sqrt{g})\sqrt{g}$ inside the probability integral defining $P_t$. Summing the coordinate inequalities gives $\frac{\mathrm{\mid }{P}_t(\mathrm{\nabla }g)(x){\mathrm{\mid }}^2}{P_{t}g(x)}\le P_t\text{\hspace{0.17em}⁣}\left(\frac{\mathrm{\mid }\mathrm{\nabla }g{\mathrm{\mid }}^2}{g}\right)(x)\mathrm{.}$

By step 1.1, $\mathrm{\mid }\mathrm{\nabla }{P}_{t}g{\mathrm{\mid }}^2=e^{-2t}\mathrm{\mid }{P}_t(\mathrm{\nabla }g){\mathrm{\mid }}^2$; apply step 2.2, integrate using invariance of ${\gamma }_d$, and use ${\int }_0^{\mathrm{\infty }}{e}^{-2t}dt=1\mathrm{/}2$ in the entropy representation of step 2.1. This gives ${\mathrm{Ent}}_{{\gamma }_d}(g)\le \frac{1}{2}\int \frac{\mathrm{\mid }\mathrm{\nabla }g{\mathrm{\mid }}^2}{g}d{\gamma }_d\mathrm{.}$

For every $x\in B$ and $r>0$, the given norming identity gives the identity $B(x,r)={\bigcup }_{\begin{array}{c}q\in \mathbb{Q}\\ 0\le q<r\end{array}}{\bigcap }_{j\ge 1}\{y:\mathrm{\mid }{\mathrm{\ell }}_j(y-x)\mathrm{\mid }\le q\}\mathrm{.}$ If $\mathrm{\parallel }y-x\mathrm{\parallel }<r$, use [F2] to choose rational $q$ between that norm and $r$; conversely, membership in one intersection implies $\mathrm{\parallel }y-x\mathrm{\parallel }\le q<r$. Thus every open ball belongs to $\sigma ({\mathrm{\ell }}_j:j\ge 1)$. By [F3], fix a countable dense set $D\subseteq B$. The balls $B(d,s)$ with $d\in D$ and positive rational $s$ form a countable family by [F2]. For any open $U$ and $y\in U$, take $a>0$ with $B(y,a)\subseteq U$, choose $d\in D$ with $\mathrm{\parallel }d-y\mathrm{\parallel }<a\mathrm{/}4$, and choose rational $s$ with $a\mathrm{/}4<s<a\mathrm{/}2$ by [F2]. Then $y\in B(d,s)\subseteq B(y,a)\subseteq U$. Thus this family is a base, and every open set is the union of its members contained in that set, a countable union. All those balls are in $\sigma ({\mathrm{\ell }}_j)$ by the first part, so every open set is in that sigma-algebra. Conversely each continuous linear ${\mathrm{\ell }}_j$ is Borel measurable, giving the reverse inclusion; hence it is the Borel sigma-algebra.

The finite-coordinate cylinder sets form a pi-system generating $\sigma ({\mathrm{\ell }}_j:j\ge 1)$. The assumed equality of finite-dimensional pushforwards makes $\mu$ and $\nu$ agree on each such cylinder. The cylinder pi-system contains $B$, and both probabilities assign it mass one; the constant exhaustion by $B$ meets [F1], which gives equality on the generated Borel sigma-algebra.

First suppose $f$ is smooth and bounded with bounded derivatives. For $\lambda >0$, apply [F1] to $g=e^{\lambda f}$. With $\psi (\lambda )=\log E{e}^{\lambda (f(Z)-Ef(Z))}$, the entropy identity and $\mathrm{\mid }\mathrm{\nabla }f\mathrm{\mid }\le L$ give $\lambda {\psi }^{\mathrm{\prime }}(\lambda )-\psi (\lambda )\le {\lambda }^2{L}^2\mathrm{/}2$.

Assume $L>0$. The function $q(\lambda )=\psi (\lambda )\mathrm{/}\lambda$ extends continuously to $q(0)=0$ because $\psi (0)={\psi }^{\mathrm{\prime }}(0)=0$. Its derivative on $(0,\mathrm{\infty })$ is $q^{\mathrm{\prime }}(\lambda )=(\lambda {\psi }^{\mathrm{\prime }}(\lambda )-\psi (\lambda ))\mathrm{/}{\lambda }^2\le L^2\mathrm{/}2$ by step 1.1. The mean value theorem [F5] on $[0,\lambda ]$ gives $\psi (\lambda )\le {\lambda }^2{L}^2\mathrm{/}2$. Markov's inequality [F6] therefore gives $P(f(Z)-Ef(Z)\ge r)\le e^{-\lambda r+{\lambda }^2{L}^2\mathrm{/}2}$; choosing $\lambda =r\mathrm{/}{L}^2$ proves the displayed bound for this $f$.

Suppose $L>0$. For a general Lipschitz $f$, clip its values to $[-n,n]$ to obtain a bounded $L$-Lipschitz function $f_n$. For $s>0$ set $P_s{f}_n(x)=\int f_n(e^{-s}x+\sqrt{1-e^{-2s}}y)d{\gamma }_d(y)$. Equivalently, write this as an integral against the smooth Gaussian density ${\rho }_s(y-e^{-s}x)$. Every derivative of the kernel in $x$ is a polynomial times a Gaussian density; on compact $x$-sets it has an integrable Gaussian majorant, so [F4] permits differentiation under the integral to every order. The global $L^1$ norm of each kernel derivative is finite, hence the derivatives of $P_s{f}_n$ are bounded by $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}$ times finite constants. Using the same integration variable for two points shows $P_s{f}_n$ is $L$-Lipschitz, so step 2.1 applies. For each fixed $n$, $P_s{f}_n(x)\to f_n(x)$ as $s\downarrow 0$, and then $f_n(x)\to f(x)$ as $n\to \mathrm{\infty }$. Throughout, $\mathrm{\mid }{P}_s{f}_n(x)\mathrm{\mid }\le \mathrm{\mid }f(0)\mathrm{\mid }+L\mathrm{\mid }x\mathrm{\mid }+L\int \mathrm{\mid }y\mathrm{\mid }d{\gamma }_d(y)$. For each fixed $\lambda$, $\lambda L\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}4+(\lambda L{)}^2$ and the density in [F2] give an integrable Gaussian majorant for $e^{\lambda \mathrm{\mid }{P}_s{f}_n(x)\mathrm{\mid }}$. Dominated convergence [F3] passes the means and exponential moments first as $s\downarrow 0$ and then as $n\to \mathrm{\infty }$, proving the bound for $f$.

If $L=0$, then $f$ is constant on ${\mathbb{R}}^d$, so $P\{f(Z)\ge Ef(Z)+r\}=0$ for every $r>0$. This handles the zero-Lipschitz case and completes the theorem.

Choose countable norming families $({\mathrm{\ell }}_i)$ for $B_1$ and $(m_j)$ for $B_2$. For any finite subfamilies, the joint coordinate vector is centered multivariate normal. Its covariance matrix has zero off-diagonal block by the hypothesis, so the characteristic function in [F1] is the product of the two marginal characteristic functions. The product law has that same characteristic function, so uniqueness in [F1] proves the finite coordinate vectors are independent.

Finite-coordinate cylinder rectangles form a pi-system generating $\mathcal{B}(B_1)\otimes \mathcal{B}(B_2)$ by [F2]. The joint law of $(X,Y)$ and the product of its marginal laws agree on every such rectangle by step 1.1; [F3] supplies that product measure. The cylinder pi-system contains $B_1\times B_2$, and both laws have mass one, so the constant exhaustion meets [F4] and gives equality on the generated sigma-algebra. The rectangle criterion [F5] now gives independence of $X$ and $Y$.

On $(B\times B,\mathcal{B}(B)\otimes \mathcal{B}(B),\mu \otimes \mu )$ let $W(x,y)=x$ and $W^{\mathrm{\prime }}(x,y)=y$ be the coordinate copies. The product-measure rectangle formula makes them independent. A countable dense subset of $B$ and rational-radius balls give a countable base, and products of base sets give one for $B\times B$, so [F5] identifies the product sigma-algebra with $\mathcal{B}(B\times B)$. The coordinate projections are continuous, hence Borel; the maps $(x,y)\mapsto (x+y)\mathrm{/}\sqrt{2}$ and $(x,y)\mapsto (x-y)\mathrm{/}\sqrt{2}$ are continuous as well, so [F5] makes $U=(W+W^{\mathrm{\prime }})\mathrm{/}\sqrt{2}$ and $V=(W-W^{\mathrm{\prime }})\mathrm{/}\sqrt{2}$ Borel random elements. For any finite list of functionals used on either $U$ or $V$, the corresponding vectors from $W$ and $W^{\mathrm{\prime }}$ are independent centered Gaussian vectors, so [F6] makes their stack multivariate Gaussian. The finite coordinate vector of $(U,V)$ is a linear image of that stack, hence is also centered multivariate Gaussian.

Centering and independence give, for all $\mathrm{\ell },m\in B^{\ast }$, $E[\mathrm{\ell }(U)m(V)]=\frac{1}{2}(E[\mathrm{\ell }(W)m(W)]-E[\mathrm{\ell }(W)m(W^{\mathrm{\prime }})]+E[\mathrm{\ell }(W^{\mathrm{\prime }})m(W)]-E[\mathrm{\ell }(W^{\mathrm{\prime }})m(W^{\mathrm{\prime }})])=0$. Independence and centering give $E[\mathrm{\ell }(U)m(U)]=E[\mathrm{\ell }(W)m(W)]$ and $E[\mathrm{\ell }(V)m(V)]=E[\mathrm{\ell }(W)m(W)]$. Thus each finite coordinate vector of $U$ or $V$ is centered Gaussian with the same covariance as the corresponding vector of $W$, so [F2] gives matching finite laws.