Finite-dimensional Gaussian isoperimetric expansion

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 $A$.