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 from Theorem 2.1 of A quantity monotone along the heat flow. 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.

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: 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 \(0<t<1\), the map \(\eta_t(y)=x_0+t(y-x_0)\) sends \(\overline S\) into the interior of \(S\);
  2. \(\overline S\) and \(S\) have the same interior, and the boundary of \(S\) is Lebesgue-null.

If \(S\) has no interior point, it lies in an affine hyperplane and is Lebesgue-null. In both cases \(S\) is Lebesgue measurable, and if \(S\) has an interior point, \(\gamma_n(S)=\gamma_n(\operatorname{int}S)=\gamma_n(\overline S)\).

Proof. (1) Let the open ball \(B(x_0,r_0)\) lie in \(S\), let \(y\in\overline S\) and \(z=\eta_t(y)\), and let \(|v|<(1-t)r_0\). Choose \(y'\in S\) so close to \(y\) that \(w=(v+t(y-y'))/(1-t)\) has \(|w|<r_0\). Then \(z+v=ty'+(1-t)(x_0+w)\) is a convex combination of points of \(S\), so the ball of radius \((1-t)r_0\) about \(z\) lies in \(S\).

(2) If \(y\ne x_0\) is an interior point of \(\overline S\), then \(y'=y+\tau(y-x_0)\in\overline S\) for some \(\tau>0\), 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 \(|y-x_0|<R\) lies in \(S_R\) and is not interior to \(\overline S\), hence not interior to \(S_R\); so the boundary of \(S\) is null.

If \(S\) contained \(n+1\) affinely independent points, it would contain the simplex they span, which has interior points; so a convex set without interior points lies in an affine hyperplane, which is null. Finally, \(S\) differs from its interior by a subset of its boundary. \(\square\)

Proposition 3.3. Let \(K,L\subseteq\mathbf R^n\) be compact convex sets with interior points and \(0<\lambda<1\). Then \[ \begin{aligned} &\Phi^{-1}\bigl(\gamma_n(\lambda K+\mu L)\bigr)\\ &\quad\ge\lambda\Phi^{-1}(\gamma_n(K))+\mu\Phi^{-1}(\gamma_n(L)). \end{aligned} \]

Proof. \(M=\lambda K+\mu L\) is compact and convex, as the image of \(K\times L\) under \((x,y)\mapsto\lambda x+\mu y\), and has interior points; all of \(\gamma_n(K)\), \(\gamma_n(L)\), \(\gamma_n(M)\) lie in \((0,1)\). Fix \(\rho\in(0,1)\), and numbers \(m,N'>0\) 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 \(\{(a,b):a<b\}\).

5. Solutions

3.1. For \(K=L\), \(\lambda K+\mu K=K\) and both sides equal \(\Phi^{-1}(\gamma_n(K))\). For the half-spaces, \(\gamma_n(K)=\Phi(a)\), \(\gamma_n(L)=\Phi(b)\), and \(\lambda K+\mu L=\{\langle e,x\rangle\le\lambda a+\mu b\}\) has measure \(\Phi(\lambda a+\mu b)\).

3.2. Since \(\varphi'(x)=-x\varphi(x)\), \[ (\log\Phi)''=\frac{\varphi'\Phi-\varphi^2}{\Phi^2}=-\frac{\varphi\,(x\Phi+\varphi)}{\Phi^2}. \] The function \(x\Phi(x)+\varphi(x)\) has derivative \(\Phi(x)>0\) 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 \(a_i<b_i\), the intervals \(K=[a_1,b_1]\) and \(L=[a_2,b_2]\) have \(\lambda K+\mu L=[\lambda a_1+\mu a_2,\lambda b_1+\mu b_2]\), and \(\gamma_1([a,b])=\Phi(b)-\Phi(a)\in(0,1)\); Theorem A is the concavity inequality for \(\psi(a,b)=\Phi^{-1}(\Phi(b)-\Phi(a))\) at the two points.

References