Let $C\subset \mathbb{R}^n$ be a compact, convex set. In any convex analysis course, it would be a standard homework exercise to prove that the functions $f(x)=\max_{y\in C} \|x-y\|$ and $f(x)=\min_{y\in C} \|x-y\|$ are convex (although the proof of the latter is a little harder than that of the former, for all proofs I've seen). I'm interested in a generalization of these functions: say that $C$ has volume $1$, define $a\in(0,1)$, and define $$f_a(x)=\min\{r:\mathrm{vol}(B_r(x)\cap C)=a \}$$ where $B_r(x)$ is the ball of radius $r$ about $x$. The two functions I described above would correspond to the limiting cases where $a=1$ and $a\to0$ respectively. (I'm not sure if such functions have a name, like "partial distance", "threshold distance", or something to that effect)

I was able to convince myself that (for fixed $a$), the function $f_a(x)$ is convex, although I had to appeal to Minkowski sums and treat $f_a(x)$ as a jointly convex function $f_a(x,S)$ over the set of all measurable subsets of $C$. This seems like a really complicated way to prove this fact; is there something more direct that I was missing?


1 Answer 1


Lemma. Let $B_r(x), B_s(y)$ be two balls in $\mathbb{R}^n$, $A \subseteq B_r(x)$, $B \subseteq B_s(y)$ convex sets such that $\mathrm{vol}(A), \mathrm{vol}(B) \geq a$, and $C = \mathrm{Conv}(A,B)$. Then, for every $t \in [0,1]$, $$ \mathrm{vol}(C \cap B_{(1-t)r+ts}((1-t)x+ty)) \geq a $$

Proof. Fix $t$. From the triangle inequality, it is easy to see that the set $$ (1-t)A + tB = \{ (1-t)z+tw : z\in A, w \in B\} $$ is contained in $B_{(1-t)r+ts}((1-t)x+ty)$. Using that, for $s \geq 0$ and $E \subseteq \mathbb{R}^n$, $$ \mathrm{vol}(sE) = s^n \mathrm{vol}(E) $$ and the Brunn–Minkowski inequality (see here), one gets $$ \mathrm{vol}((1-t)A + tB)^{1/n} \geq (1-t) \mathrm{vol}(A)^{1/n} + t \mathrm{vol}(B)^{1/n} \geq (1-t)a^{1/n} + ta^{1/n} = a^{1/n}, $$ hence $$ \mathrm{vol}(C \cap B_{(1-t)r+ts}((1-t)x+ty)) \geq \mathrm{vol}((1-t)A + tB) \geq a. $$

Now it is easy to prove convexity by taking $(x,r), (y,s)$ in the epigraph of your function $f_a$ and using the Lemma with $A = C \cap B_r(x)$, $B = C \cap B_s(y)$ to prove that $((1-t)x+ty, (1-t)r+ts)$ belongs to the epigraph.


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