Convex bodies have more volume on the outside near the boundary

Question

I am looking for a reference for a result from convex geometry that I suspect has already been proven. The result seems geometrically obvious, but I couldn't find a similar result in Peter Gruber's book nor could I prove it succinctly with my limited knowledge of convex geometry!

For each $\epsilon > 0$ and $A \subseteq \mathbb{R}^d$ we define $$ B_\epsilon(A) = \{ x \in \mathbb{R}^d : \inf_{y \in A} |x-y| < \epsilon \},$$ where $|\cdot|$ is the usual Euclidean norm.

Let $P$ is a proper, compact convex body in $\mathbb{R}^d$ and let $m$ denote Lebesgue measure. Setting aside measurability concerns, we need the inequality $$ m(B_\epsilon(\partial P) \cap P ) \le m(B_\epsilon(\partial P) \cap P^C ) $$ for each $\epsilon > 0$. As the title suggests, this is just the statement that there is more volume close to the boundary of a proper, compact convex body on the outside than on the inside.

For our application it suffices to consider the case where $P$ is a polytope.

Answer 1

Here is a proof for the case of a convex polytope. (Edit: See Wlodek Kuperberg's answer for an elementary version of this argument stripped of all unnecessary notation and extended to the general case.)

If $P\subset \Bbb{R}^d$ is a compact convex polytope containing the origin $O$ in its interior then it can be broken up into cones over its facets and it suffices to verify the inequality for each of these (polyhedral) cones individually.

Let $F$ be a facet with area $A$. The hyperplane $H$ supporting $F$ separates $\Bbb{R}^d$ into two half-spaces, the inner one, $H^{-}$, containing $O$, and the outer one, $H^{+}$. The convex cone $C=\Bbb{R}^{+}F$ is separated into the "inner" and "outer" parts, $C^{-}=C\cap H^{-}$ and $C^{+}=C\cap H^{+}$. Then $B_{\epsilon}(F)\cap C^{-}$, the "inner" $\epsilon$-neighborhood of $F$, is contained in the inner cylinder inward-facing rectangular prism with height $\epsilon$ based on $F$ and $B_{\epsilon}(F)\cap {C}^{+}$, the "outer" $\epsilon$-neighborhood of $F$, contains the outer cylinder outward-facing rectangular prism with height $\epsilon$ based on $F$. Therefore,

(the inner volume) $\leq \epsilon A\leq $ (the outer volume).

Answer 2

I am sure it IS in Gruber's book - what you want is Steiner's formula on the volume of parallel bodies (or tubular neighborhoods, if you prefer) of a convex body. These have the form:

$$V_r = \sum_{i=0}^d Q_i r^i,$$ where the $Q_i$ are the Quermassintegrals, which are positive, $r$ is the distance along the outward normal and $d$ the dimension. $Q_1$ is the area of the boundary. Consider the epsilon-parallel body $B_\epsilon$ inside $B$ Then, the volume of the inside neighborhood is $$ V(\epsilon) = \sum_{i=1}^d Q_i(B_\epsilon) \epsilon^i,$$ while the volume of the outside $|epsilon$-neighborhood is $$V_{2\epsilon} - V_\epsilon.$$ Notice that all the terms in the difference are positive.

This has been generalized by Weyl to Riemannian manifolds, and there is a whole (quite good) book on the subject - Tubes, by Alfred Gray.

Answer 3

A modification of Victor Protsak's answer yields a short and almost elementary proof:

Since the set of convex $n$-dimensional convex polytopes is dense in the space of all $n$-dimensional convex bodies, it suffices to prove the inequality in question for convex polytopes, and this is quite obvious when you erect perpendicular prisms of height $\varepsilon$ based on the polytope's facets, one family outwards the polytope, the other one inwards. (There is no need to break up the polytope into cones.)

(Compare this with my answer to an earlier question, A convex curve inside the unit circle and see the drawing in it.)

Answer 4

I guess proper means nonempty interior, right? I claim that the inequality holds for every $\epsilon>0$ (not just for small values), without assuming $P$ to be a polytope. Assume first that $P$ has smooth boundary.

Look at the map $F:\partial P\times\mathbb R\to\mathbb R^d$, $F(x,t):=x+t\nu(x)$, $\nu(x)$ being the outward unit normal at $x$. Then $F(\partial P\times(-\epsilon,0])\supseteq B_\epsilon(\partial P)\cap P$ and, being $P$ convex, $F(\partial P\times (0,\epsilon))=B_\epsilon(\partial P)\cap P^c$ bijectively.

Given $x\in\partial P$, take a basis of tangent vectors $e_1,\dots,e_{d-1}$ diagonalizing the shape operator, i.e. $d\nu_x[e_j]=\lambda_je_j$. Since $\langle d\nu_x[e_j],e_j\rangle$ equals $-\langle\nu(x),\ddot\gamma(0)$ for any curve $\gamma:(-\delta,\delta)\to\partial P$ with $\dot\gamma(0)=e_j$, we get by convexity that $\lambda_j\ge 0$.

Hence, for any $t>0$, the Jacobian determinant of $F$ satisfies $$JF(x,-t)=\prod_{j=1}^{d-1}|1-\lambda_jt|\le\prod_{j=1}^{d-1}(1+\lambda_jt)=JF(x,t).$$ The area formula gives $$m(B_\epsilon(\partial P)\cap P)\le\int_{\partial S\times(-\epsilon,0)}JF\le \int_{\partial S\times(0,\epsilon)}JF=m(B_\epsilon(\partial P)\cap P^C).$$

You can easily remove the smoothness assumption: choose a nonnegative convex $f\in C^\infty(\mathbb R)$ with $f^{-1}(0)=(-\infty,0]$ and let $\psi(x):=\int_{S^{d-1}}f\circ d_v(x)\,dv$, where $d_v$ is the distance from the supporting half-space in direction $v$. To be more explicit, $f\circ d_v(x)$ equals $f(\inf_{y\in P}\langle v,y\rangle-\langle v,x\rangle)$.

$\psi$ is convex, smooth and vanishes on $P$ (elsewhere $\psi>0$). With Sard, take a smooth sublevel set $P_\lambda:=\{\psi\le\lambda\}$, with $\lambda>0$ arbitrarily small. Then $B_\epsilon(\partial P)\cap P\subseteq B_{\epsilon+\epsilon_\lambda}(\partial P_\lambda)\cap P_\lambda$ and $B_{\epsilon+\epsilon_\lambda}(\partial P_\lambda)\cap P_\lambda^c\subseteq B_{\epsilon+2\epsilon_\lambda}(\partial P)\cap P^c$, with $\epsilon_\lambda\to 0$ as $\lambda\to 0$ (you may take e.g. $\epsilon_\lambda$ to be the Hausdorff distance between $\partial P$ and $\partial P_\lambda$, which is actually the one between $P$ and its superset $P_\lambda$).