# Ehrhard's inequality for convex sets *Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).* This lesson deduces Theorem A of [Gaussian heat semigroups and a gradient bound](gaussian-heat-semigroups-and-a-gradient-bound.md) from Theorem 2.1 of [A quantity monotone along the heat flow](a-quantity-monotone-along-the-heat-flow.md). Three limits are needed: the smoothing time \(\varepsilon\) tends to \(0\), which turns the hypothesis of Theorem 2.1 into a pointwise inequality between uniformly continuous functions (Proposition 3.1); Lipschitz functions approximate the indicator functions of convex sets (Proposition 3.3); and the correlation \(\rho\) tends to \(1\), so that \(\sigma\to1\). Corollary 3.4 records the concavity along translates in the form used in the course [The Gaussian propeller bound](course:the-gaussian-propeller-bound/gaussian-first-moments-of-partitions). The notation \(\lambda\), \(\mu=1-\lambda\), \(\rho\), \(\sigma\), \(\Xi\) and \(\mathcal B_N\) is that of the previous lesson. The measure-theoretic results are cited by number from Fremlin's *Measure Theory* in the core course [Measure and Integration](https://kokunoyumeto.github.io/program-matematika-indonesia/en/#course-D10): B. Levi's theorem on monotone convergence (123A), dominated convergence (123C) and the linear change of variables (263A). ## 1. A pointwise hypothesis **Proposition 3.1.** Let \(0<\lambda<1\), \(-1<\rho<1\), \(N>0\), \(\varepsilon_3>0\), and let \(f,g,h:\mathbf R^n\to[\Phi(-N),\Phi(N)]\) be uniformly continuous with, for all \(x,y\in\mathbf R^n\), \[ \Xi\bigl(f(x),g(y),h(\lambda x+\mu y)\bigr)\le-\varepsilon_3 . \] Then \(\Xi\bigl(\int f\,d\gamma_n,\int g\,d\gamma_n,\int h(\sigma w)\,d\gamma_n(w)\bigr)<0\). *Proof.* For a bounded uniformly continuous \(k\) with modulus of continuity \(\omega_k\), and \(t,R_1>0\), \[ \begin{aligned} &|Q_tk(x)-k(x)|\le\omega_k(\sqrt t\,R_1)\\ &\qquad+2\sup|k|\;\gamma_n(\{|w|>R_1\}), \end{aligned} \] by splitting the integral of \(|k(x+\sqrt t\,w)-k(x)|\) at \(|w|=R_1\). Choosing \(R_1\) large and then \(t\) small shows that \(Q_tk\to k\) uniformly as \(t\to0\). \(\Xi\) is Lipschitz on \(\mathcal B_N\) (its partial derivatives are bounded by \(1/\varphi(N)\) there). So for \(\varepsilon>0\) small enough, the functions \(F_{i,0}\) of Theorem 2.1, namely \(Q_\varepsilon f(x)\), \(Q_\varepsilon g(y)\) and \(Q_{\sigma^2\varepsilon}h(\lambda x+\mu y)\), satisfy \(\Xi(F_0(x,y))\le-\varepsilon_3/2\) for all \((x,y)\). Theorem 2.1 with \(\varepsilon_1=\varepsilon_3/2\) gives the conclusion. \(\square\) ## 2. Convex sets **Lemma 3.2.** Let \(S\subseteq\mathbf R^n\) be convex. If \(S\) has an interior point \(x_0\), then: 1. for \(00\), and \(y=\eta_t(y')\) with \(t=1/(1+\tau)\), so \(y\) is interior to \(S\) by (1). For \(R>0\) let \(S_R=\overline S\cap\overline{B(x_0,R)}\), a compact convex set with \(x_0\) in its interior. By (1) applied to \(S_R\), \(\eta_t(S_R)\) lies in the interior of \(S_R\), so the boundary of \(S_R\) lies in \(S_R\setminus\eta_t(S_R)\), whose Lebesgue measure is \((1-t^n)\) times that of \(S_R\) (263A). Letting \(t\to1\), the boundary of \(S_R\) is null. A boundary point \(y\) of \(S\) with \(|y-x_0|0\) and \(N\) with \[ \begin{aligned} \lambda N&\ge\mu m+\sigma N'+1,\\ \mu N&\ge\lambda m+\sigma N'+1, \end{aligned} \] and put \(T=(m+1)/\sigma\). For \(k\ge1\) let \[ \begin{aligned} a_k(x)&=\min\bigl(1,k\operatorname{dist}(x,\mathbf R^n\setminus\operatorname{int}K)\bigr),\\ b_k(y)&=\min\bigl(1,k\operatorname{dist}(y,\mathbf R^n\setminus\operatorname{int}L)\bigr),\\ c_k(z)&=\max\bigl(0,1-k\operatorname{dist}(z,M)\bigr). \end{aligned} \] These are Lipschitz with values in \([0,1]\); \(a_k=0\) outside \(\operatorname{int}K\), \(b_k=0\) outside \(\operatorname{int}L\), \(c_k=1\) on \(M\); as \(k\to\infty\), \(a_k\uparrow\mathbf 1_{\operatorname{int}K}\), \(b_k\uparrow\mathbf 1_{\operatorname{int}L}\) and \(c_k\downarrow\mathbf 1_M\) pointwise. Put \[ \begin{aligned} \alpha_k&=-N+(N+m)a_k,\\ \beta_k&=-N+(N+m)b_k,\\ \chi_k&=-N'+(N'+T)c_k, \end{aligned} \] and \(f=\Phi\circ\alpha_k\), \(g=\Phi\circ\beta_k\), \(h=\Phi\circ\chi_k\). They are uniformly continuous with values in \([\Phi(-N_1),\Phi(N_1)]\), \(N_1=\max(N,N',T)\). For all \(x,y\), \[ \begin{aligned} &\lambda\alpha_k(x)+\mu\beta_k(y)\\ &\quad-\sigma\chi_k(\lambda x+\mu y)\le-1 . \end{aligned} \] Indeed, if \(a_k(x)>0\) and \(b_k(y)>0\), then \(x\in K\), \(y\in L\), \(\lambda x+\mu y\in M\) and \(\chi_k=T\), so the left side is at most \(\lambda m+\mu m-\sigma T=-1\); if \(a_k(x)=0\), it is at most \(-\lambda N+\mu m+\sigma N'\le-1\); and likewise if \(b_k(y)=0\). Put \(p_k=\int\Phi\circ\alpha_k\,d\gamma_n\), \(p'_k=\int\Phi\circ\beta_k\,d\gamma_n\) and \(q_k=\int\Phi(\chi_k(\sigma w))\,d\gamma_n(w)\). Proposition 3.1 with \(\varepsilon_3=1\) gives \[ \lambda\Phi^{-1}(p_k)+\mu\Phi^{-1}(p'_k)<\sigma\Phi^{-1}(q_k). \] Let \(k\to\infty\), and write \(\kappa_K=\gamma_n(K)\), \(\kappa_L=\gamma_n(L)\) and \(s=\gamma_n(\sigma^{-1}M)\). By B. Levi's theorem (123A) and Lemma 3.2, \(p_k\to\bar p\) and \(p'_k\to\bar p'\), and by dominated convergence (123C), \(q_k\to\bar q\), where \[ \begin{aligned} \bar p&=\Phi(-N)(1-\kappa_K)+\Phi(m)\kappa_K,\\ \bar p'&=\Phi(-N)(1-\kappa_L)+\Phi(m)\kappa_L,\\ \bar q&=\Phi(-N')(1-s)+\Phi(T)s. \end{aligned} \] These limits lie in \((0,1)\), where \(\Phi^{-1}\) is continuous, so \[ \lambda\Phi^{-1}(\bar p)+\mu\Phi^{-1}(\bar p')\le\sigma\Phi^{-1}(\bar q). \] Let \(N\to\infty\), then \(N'\to\infty\) (with \(N\) still satisfying the two conditions), then \(m\to\infty\); the limits \(\Phi(m)\kappa_K\to\kappa_K\), \(\Phi(m)\kappa_L\to\kappa_L\) and \(\Phi(T)s\to s\) stay in \((0,1)\). So \[ \lambda\Phi^{-1}(\kappa_K)+\mu\Phi^{-1}(\kappa_L)\le\sigma\Phi^{-1}(s). \] Finally let \(\rho\to1\), so that \(\sigma\to1\) (Exercise 2.1 of the previous lesson). For \(w\) in the interior of \(M\), \(\sigma w\in M\) for \(\sigma\) close to \(1\); for \(w\notin M\), \(\sigma w\notin M\) for \(\sigma\) close to \(1\), as \(M\) is closed. So \(\mathbf 1_M(\sigma w)\to\mathbf 1_M(w)\) off the boundary of \(M\), which is null by Lemma 3.2, and \(\gamma_n(\sigma^{-1}M)\to\gamma_n(M)\) by dominated convergence. \(\square\) ## 3. Proof of Theorem A Let \(K,L\) be closed convex sets with \(\gamma_n(K),\gamma_n(L)\in(0,1)\), and \(0\le\lambda\le1\). For \(\lambda=0\) or \(1\) the two sides agree. Let \(0<\lambda<1\). The set \(M=\lambda K+\mu L\) is convex, hence measurable by Lemma 3.2 (it need not be closed, Exercise 3.3). If \(\gamma_n(M)=1\), the left side is \(+\infty\). By Lemma 3.2, \(K\) and \(L\) have interior points, so for large \(R\) the compact convex sets \(K_R=K\cap\overline{B(0,R)}\) and \(L_R=L\cap\overline{B(0,R)}\) have interior points; \(\gamma_n(K_R)\uparrow\gamma_n(K)\) and \(\gamma_n(L_R)\uparrow\gamma_n(L)\) as \(R\to\infty\) by monotone convergence. Since \(\lambda K_R+\mu L_R\subseteq M\), Proposition 3.3 gives \[ \begin{aligned} &\Phi^{-1}(\gamma_n(M))\\ &\quad\ge\Phi^{-1}\bigl(\gamma_n(\lambda K_R+\mu L_R)\bigr)\\ &\quad\ge\lambda\Phi^{-1}(\gamma_n(K_R))+\mu\Phi^{-1}(\gamma_n(L_R)), \end{aligned} \] and the right side tends to \(\lambda\Phi^{-1}(\gamma_n(K))+\mu\Phi^{-1}(\gamma_n(L))\). \(\square\) **Corollary 3.4** (concavity along translates). Let \(K\subseteq\mathbf R^n\) be closed and convex, \(v\in\mathbf R^n\), and suppose \(p(u)=\gamma_n(K+uv)\) lies in \((0,1)\) for every \(u\in\mathbf R\). Then \(u\mapsto\Phi^{-1}(p(u))\) is concave on \(\mathbf R\). *Proof.* Let \(u_1,u_2\in\mathbf R\), \(0\le\lambda\le1\) and \(u=\lambda u_1+\mu u_2\). Since \(\lambda K+\mu K=K\) for the convex set \(K\), \[ \lambda(K+u_1v)+\mu(K+u_2v)=K+uv . \] Theorem A for the closed convex sets \(K+u_1v\) and \(K+u_2v\) gives \[ \begin{aligned} \Phi^{-1}(p(u))&\ge\lambda\Phi^{-1}(p(u_1))\\ &\quad+\mu\Phi^{-1}(p(u_2)). \end{aligned} \] \(\square\) ## 4. Exercises **3.1** (easy). Show that Theorem A is an equality for \(K=L\), and for parallel half-spaces \(K=\{\langle e,x\rangle\le a\}\), \(L=\{\langle e,x\rangle\le b\}\) with \(|e|=1\). **3.2** (medium). Show that \(\log\Phi\) is concave, using \(x\Phi(x)+\varphi(x)=\int_{-\infty}^x\Phi(s)\,ds>0\), and deduce from Theorem A that \(\gamma_n(\lambda K+\mu L)\ge\gamma_n(K)^\lambda\gamma_n(L)^\mu\). **3.3** (medium). Show that \(K=\{(x_1,x_2):x_1>0,\ x_2\ge1/x_1\}\) and \(L=\{(x_1,0):x_1\le0\}\) are closed convex subsets of \(\mathbf R^2\) whose sum \(K+L\) is the open half-plane \(\{x_2>0\}\). **3.4** (easy). Deduce from Theorem A in dimension one that \((a,b)\mapsto\Phi^{-1}(\Phi(b)-\Phi(a))\) is concave on \(\{(a,b):a0\) and tends to \(0\) as \(x\to-\infty\), so it equals \(\int_{-\infty}^x\Phi\) and is positive. With \(a=\Phi^{-1}(\gamma_n(K))\) and \(b=\Phi^{-1}(\gamma_n(L))\), Theorem A gives \(\gamma_n(\lambda K+\mu L)\ge\Phi(\lambda a+\mu b)\), and this is at least \(\Phi(a)^\lambda\Phi(b)^\mu\) by the concavity of \(\log\Phi\). **3.3.** \(K\) is the region above the graph of the convex function \(1/x_1\) on \((0,\infty)\); it is convex, and closed because a limit of points of \(K\) has \(x_1>0\) (as \(x_2\ge1/x_1\) stays bounded along a convergent sequence). \(L\) is a closed half-line. A point of \(K+L\) is \((x_1+t,x_2)\) with \(t\le0\) and \(x_2\ge1/x_1>0\). Conversely, for \(c\in\mathbf R\) and \(d>0\), choose \(x_1\ge\max(c,1/d)\) with \(x_1>0\); then \((c,d)=(x_1,d)+(c-x_1,0)\) with \((x_1,d)\in K\) and \((c-x_1,0)\in L\). **3.4.** For \(a_i