intersecting a convex body with a small ball whose center is in its interior

Question

Let $C$ be a convex body in $\mathbb{R}^d$, i.e. $C$ is a compact, convex set with nonempty interior.

For any example I can think of, for each $C$ there exists positive constants $\lambda, \epsilon$, such that for every ball $B$ of radius $r$ with center in $C$ and $r < \epsilon$, at least a $\lambda$ fraction of the volume of $B$ is inside $C$.

That is, for every $x \in C$, $r < \epsilon$, if $B=B(x,r)$ then we have $$\frac{\mu( B \cap C)}{\mu(B)} \ge \lambda.$$

The quantity that is interesting to me is $\lambda$, and it seems to be essentially an invariant of the convex body. For example, if $C$ is itself a ball, then $\lambda$ can be chosen to be arbitrarily close to $1/2$. If $C = [0,1]^d$ is a cube, then $\lambda$ can be chosen arbitrarily close to $1 \big/ 2^d$. On the other hand, $\lambda$ can forced to be arbitrarily small by varying the convex body $C$, for example by taking triangles with an arbitrarily small angle, and placing $x$ at this vertex.

Is this true that there exists such a positive $\lambda(C)$ for each convex body $C$? If so, is there an associated invariant and does it have a name?

Answer 1

Just a thought. Welcome to fill in the details.

First let's fix the ball radius as $r$. Let $$ f(x,r)=\frac{\mu(B(x,r)\cap C)}{\mu(B(x,r))} $$

Lemma 1. (i) For each fixed $r$, $f(x,r)$ is continuous w.r.t. $x$. (ii) $f(x,r)>0$ for $x\in C$ and $r>0$.

This should be rather obvious but I am not sure how to give a formal proof.

Lemma 2. As $r$ decreases, $f(x,r)$ is non-decreasing for each $x\in C$.
Proof. Let $0<r'<r$. Then $$ f(x,r')=\frac{\mu(B(x,r')\cap C)}{\mu(B(x,r'))} $$ Let us define $C_{x,r,r'}$ as the convex body obtained by shrinking $C$ towards $x$ with a scaling factor of $r'/r$. Clearly, if we scale the ball and the convex body simultaneously, the ratio is invariant, i.e. $$ \frac{\mu(B(x,r')\cap C_{x,r,r'})}{\mu(B(x,r'))}=\frac{\mu(B(x,r)\cap C)}{\mu(B(x,r))}=f(x,r) $$ On the other hand, since $C$ is convex and $x\in C$, we have $C_{x,r,r'}\subset C$. Hence, $$ f(x,r')=\frac{\mu(B(x,r')\cap C)}{\mu(B(x,r'))}\geq\frac{\mu(B(x,r')\cap C_{x,r,r'})}{\mu(B(x,r'))}=f(x,r) $$

Now you can fix any $r_0>0$ as a start and obtain $F(x)=f(x,r_0)$ as a continuous and positive function on $C$ by Lemma 1. Since $C$ is compact, $F(x)$ assumes a positive minimum value, say $\lambda(C)=\inf_{x\in C}F(x)>0$. Now choose $r_0$ as your $\epsilon$. Then for any $x\in C$, $r<\epsilon=r_0$, we have from Lemma 2: $$ f(x,r)\geq f(x,r_0)\geq\lambda(C) $$

Note that $\lambda(C)$ depends on $r_0$ and is only one lower bound but not necessarily the greatest one. I conjecture if you let $r_0\to0$, then $\lambda(C)$ will approach or even become the greatest lower bound.