Is there a lower bound to density at boundary points of a convex set? 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 $\min\{D(x): x \in X\}  >0$?

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 $||z|| < \varepsilon$.
