Is there a lower bound to density at boundary points of a convex set?

Question

Let $X \subset \mathbb R^d$ be convex and compact. For each $x \in X$ define

$$D(x) = \lim_{r \to 0^+}\frac{\mu(X \cap B(x,r))}{\mu(B(x,r))}$$

where $B(r,d)$ is the ball with centre $x$ and radius $r$ and $\mu$ is the Lebesgue measure. The density measures what proportion of the ball is contained in $X$ as $r$ becomes very small.

For example if $X$ is a polygon then $D(x) = 1$ at interior points; and $D(x) = 1/2$ at every point on an edge but not a vertex; while for $x$ a vertex the density $D(x)$ is the angle at that vertex. Thus for polytopes at least

$$\min\{D(x): x \in X\} = \min\{D(v): v \in X \text{ is a vertex}\}>0.$$

For smooth bodies I would imagine $D(x) = 1/2$ at every boundary point, since the boundary is locally approximated by a hyperplane. Hence we have $\min\{D(x): x \in X\} =1/2$

For more general maybe-not-smooth bodies, is is known that $\min\{D(x): x \in X\} >0$?

Answer 1

I'll make a suggestion (maybe I'm wrong).

Let's fix $a$ - an interior point of $X$ and $\varepsilon > 0$ s.t. $B(a,\varepsilon) \subset X$. Then for arbitrary $x \in X$ you have that $x + t(a + z - x) \in X$ for $||z|| < \varepsilon$ and $t \in [0,1]$. So, for arbitrary $t < \frac{r}{b + \varepsilon}$ (where $b$ is the diameter of $X$) we have: $B(x + t(a - x), t\varepsilon) \subset X \bigcap B(x, r)$. And therefore $\frac{\mu(X \bigcap B(x, r))}{\mu(B(x, r))} \ge \frac{t^d \varepsilon^d}{r^d} \ge \frac{\varepsilon^d}{(b + \varepsilon)^d}$. Last expression is constant.

I think there can be more accurate estimate that uses measure of all cone $x + t(a + z - x)$ for $t \in [0,1]$ and $||z|| < \varepsilon$.

Answer 2

If $X$ is not of full rank (meaning $\mathrm{aff}(X)\ne\mathbb R^d$), the density will be zero at every point. If $X$ is of full rank, then it's well-known that it has an interior point $p$, and for any $x\in X$ we can choose $r>0$ such that $p\in B(x, r)$. Since an open set has positive measure, we have $$\frac{\mu(X\cap B(x,r))}{\mu(B(x,r))}>0.$$ Note that as $r\to0^+$, this value will only get bigger. For let $s<r$, and let $H:\mathbb R^d\to \mathbb R^d$ be a homothety centered at $x$ with ratio $s/r$. By convexity, $H(X\cap B(x,r)) \subseteq X\cap B(x,s)$, meaning that $$\frac{\mu(X\cap B(x,r))}{\mu(B(x,r))} = \frac{\mu(H(X\cap B(x,r)))}{\mu(H(B(x,r)))} \le \frac{\mu(X\cap B(x,s))}{\mu(B(x,s))}.$$ This shows that any point in a convex set of full rank has positive density.

Now suppose $X$ is compact. We can choose our initial value of $r$ as a continuous function of $x$, e.g. $\|x-p\|+1$. Then our initial bound for the density at a point will also be a continuous function, and thus attains a minimum. This serves as a positive lower bound for the density at any point, showing that $\inf\{D(x):x\in X\}>0$.