convex set in $R^n$ cannot have its barycenter at boundary?

Question

Suppose $p$ is a probability measure in $R^n$ which satisfies

1. $p$ is absolutely continuous with respect to Lebesgue measure.
2. $p(D)=1$ for a closed and bounded convex set $D\subset R^n$ and $p(A)>0$ for every open set $A\subset D$.

Let $S\subset D$ and $p(S)>\delta>0$. Also $S$ is the intersection of several half space $\{x\in R^3 | a^T x + b > 0\}$($S$ is convex). The mean with respect to $S$ is defined by $$ u(S)=\frac{1}{p(S)}\int_{S} x dp(S) $$

Can we show that $u(S)$ cannot be arbitrarily near to the boundary of $S$?

Motivation: I encountered this question when reading the first paper discussing k-means method. The paper can be found at projecteuclid and the pdf file is open accessible. On the fourth page when proving lemma 1, the author claims that "a convex set of $p$ measure at least $\delta>0$ cannot have its conditional mean arbitrarily near its boundary". But I cannot configure out why it is so.

Even if $p$ is uniform, how to show that the barycenter of $S$ cannot be arbitrarily near the boundary? Or the claim does not hold generally ?

Answer 1

Depending on how you "weigh" $p$, you can get arbitrarily close to the boundary. You cannot get on the boundary ( or outside)- you can see that by taking a supporting hyperplane and using that $p$ is $>0$ for open non-void subsets.

If $p$ is the uniform measure then the center of mass is an affine notion. That is, if a linear invertible map takes $K$ to $K'$ it will map $u_K$ to $u_{K'}$ (note invertible here, $K$,$K'$ need not be convex, just closed, bounded with non-void interior).

To see how far is $u_K$ from the boundary of $K$ there are several ways to measure that:

1. First, notice that hyperplanes through $u_K$ do not in general divide $K$ into $2$ pieces of equal mass. It's not the masses on each sides of the hyperplane that are balanced, by the moments of the masses. However, there are bounds for the rations of the pieces.

2. There are bounds for the ratios of the "radiuses" of chords through $u_K$.

Note that the above bounds are affinely invariant. In general, the worst case is if $K$ is a tetrahedron, while the best is the sphere ( and in some ways, centrally symmetric bodies).

You should do a Google search for the above notions, and perhas try to come up with other ways to measure the "distance" from a ball or a centrally symmetric body.

Added:

there are at least two results that I am aware of.

1.First, it's Winternitz theorem that says that any hyperplane through the center of mass of a convex body will divide it into whose volumes have ratios $$\frac{\operatorname{vol}K_1}{\operatorname{vol}K_1} \le \frac{1- \left(\frac{n}{n+1}\right)^n}{ \left(\frac{n}{n+1}\right)^n}= \left(\frac{n+1}{n}\right)^n-1$$

( the constant is calculated easily for hyperplanes through the center of mass of an $n$ dimensional simplex that are parallel to one of the $n-1$ faces). Note that it approaches increasingly $e-1$.

2.If $K$ convex body has width $d$ ( the closest distance two parallel supporting planes) then $u_K$ is at the distance at least $\frac{1}{n+1}$ from any supporting hyperplane.

Answer 2

Let $p$ be as given. If $S\subset D$ is a convex set $S\subset D$ with $p(S)>0$ and $s\in\partial S$, we will find a positive lower bound for $|u(S)-s|$ that depends only on (properties of $p$ and) $p(S)$.

By absolute continuity of $p$, there exists $V_0>0$ such that $p(E)<\frac 12p(S)$ holds for all $E$ with $\operatorname{vol}(E)<V_0$.

If $a\in\Bbb R^n$, $|a|=1$, and $b\in\Bbb R$ let $H_{a,b}=\{\,x\in\Bbb R^n\mid a^Tx+b>0\,\}$. We know that we can write $S$ as $$ S=\bigcap_{i\in I}H_{a_i,b_i}.$$ Clearly, $S\subset H_{a,b}$ implies $$ a^Tu(S)+b=\frac1{p(S)}\int_S(a^t x+b)\,\mathrm dp(S)> 0,$$ and we conclude $u(S)\in S$. A priori, this would still allow $u(S)$ to be arbitrarily close to $\partial S$.

For $\rho>0$, we define $$ S\ominus\rho:=\bigcap_{i\in I}H_{a_i,b_i-\rho}$$ as the set $S$ eroded by $\rho$. Let us write $$\partial_\rho S:=S\setminus(S\ominus\rho)$$ for the points near the boundary that got eroded away. Note that every point in $S\ominus \rho$ has distance $\ge \rho$ from every point in $\partial S$.

First one can show(!) that $\partial S$ is a $(n-1)$-dimensional hypersurface for which $$\operatorname{vol}_{n-1}(\partial S)\le \operatorname{vol}_{n-1}(\partial D)$$ holds. Moreover, $$\operatorname{vol}_{n}(\partial_\rho S)\le \rho\cdot \operatorname{vol}_{n-1}(\partial S).$$ For $\rho<\frac{V_0}{\operatorname{vol}_{n-1}(\partial D)}$, we can conclude that $p(\partial_\rho S)<\frac 12p(S)$ and so $$ p(S\ominus\rho)=p(S)-p(\partial_\rho S)>\frac 12p(S), \qquad \operatorname{vol}_n(S\ominus \rho)\ge V_0.$$ We have $$ u(S)=\frac{p(\partial_\rho S)\cdot u(\partial_\rho S) +p(S\ominus \rho)\cdot u(S\ominus \rho)}{p(S)}=t u(\partial_\rho S) + (1-t) u(S\ominus \rho)$$ with $0<t<\frac12$. Then if $|a|=1$ and $S\subseteq H_{a,b}$, we find $$ a^Tu(S)=t a^Tu(\partial_\rho S)+(1-t) a^Tu(S\ominus \rho)\ge t\cdot (-b) + (1-t)(-b+\rho)>-b+\frac12\rho.$$ We conclude that $$|u(S)-s|\ge \frac12\rho$$ for all $s\in\partial S$.