# Less complicated proof of this "obvious" fact about convexity

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?

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.