Subadditivity of the $n$th root of the volume of $r$-neighborhoods of a set

Question

Let $A$ be a closed subset of $\mathbb{R}^n$. For $r>0$, let $A_r$ be the $r$-neighborhood of $A$, namely the set $\{x:\operatorname{dist}(x,A)\le r\}$. Let $f(r) = \mu(A_r)^{1/n}$ where $\mu$ is the Lebesgue measure.

Is the function $f(r)/r$ decreasing?

Motivation

I previously asked about Concavity of the $n$th root of the volume of $r$-neighborhoods of a set where the answer turned out to be negative. But the counterexample by George Lowther still has $f(r)/r$ decreasing. Geometrically, this property means that the graph of $f$ lies above the line through $(0,0)$ and $(r,f(r))$. Although weaker than concavity, this is already a useful property to have: it implies the differential inequality $f'(r)\le f(r)/r$, as well as subadditivity $f(x+y)\le f(x)+f(y)$. The latter follows by comparing $f$ to the secant line through $(0,0)$ and $(x+y,f(x+y))$.

Among other things, the result stated above would simplify the proof here. As for ideas for its own proof, I hope that this time the Brunn-Minkowski inequality can be put to use.

Answer 1

Yes, this is true. $r^{-n}\mu(A_r)$ is decreasing in $r$.

By making limits, it is enough to prove this for $A$ finite. In that case, for each $a\in A$, let $S_r(a)$ be the set of points $p\in\mathbb{R}^n$ with $\lVert p-a\rVert=r$ and $\lVert p-b\rVert > r$ for all $b\in A\setminus\{a\}$. That is, $S_r(a)$ is the set of points on the boundary of $A_r$ which are closer to $a$ than the other elements of $A$. Then set $C_r(a)=\{(1-\lambda)a+\lambda p\colon\lambda\in[0,1],p\in S_r(a)\}$. The elements of $C_r(a)$ are within a distance $r$ of $a$, so $C_r(a)\subseteq A_r$. Setting $B_r=A_r\setminus\bigcup_{a\in A}C_r(a)$, $$ A_r=B_r\cup\bigcup_{a\in A}C_r(a). $$ In fact, this is a disjoint union. If $a\not=b$ and $C_r(a)\cap C_r(b)\not=\emptyset$ then, for some $p\in S_r(a)$, $q\in S_r(b)$, the line segments $ap$ and $bq$ intersect at a point $c$ giving the contradiction \begin{align} 2r & < \lVert p-b\rVert+\lVert q-a\rVert\\ &\le\lVert p-c\rVert+\lVert c-a\rVert+\lVert q-c\rVert+\lVert c-b\rVert\\ &=\lVert p-a\rVert+\lVert q-b\rVert=2r. \end{align} Next, if $t > 1$ then, $$ A_{tr}\subseteq B_r\cup\bigcup_{a\in A}\left(tC_r(a)+(1-t)a\right). $$ To see this, let $p\in A_{tr}$ and let $a\in A$ minimise $\lVert p-a\rVert\le tr$. If $\lVert p-a\rVert\le r$ then $p\in A_r$, which is contained in the right hand side of the above inclusion. If $\lVert p-a\rVert > r$, then let $c$ be the point on the line segment $ap$ at a distance $r$ from $a$. If any $b\in A\setminus\{a\}$ is within a distance $r$ of $c$, then $$ \lVert p - b\rVert < \lVert p - c\rVert + \lVert c - b\rVert\le\lVert p - c\rVert + \lVert c - a\rVert=\lVert p - a\rVert $$ contradicting the choice of $a$. So, $c\in S_r(a)$. Then, $t^{-1}p+(1-t^{-1})a$ is in the line segment $ac$, and is in $C_r(a)$. Hence, $p\in tC_r(a)+(1-t)a$. Therefore, $$ \mu(A_{tr})\le\mu(B_r)+\sum_{a\in A}t^n\mu(C_r(a)) \le t^n\mu(B_r)+\sum_{a\in A}t^n\mu(C_r(a)) =t^n\mu(A_r). $$ So, $t^{-n}\mu(A_{tr})\le\mu(A_r)$, showing that $r^{-n}\mu(A_r)$ is decreasing in $r$.