Let $A\subset \mathbb{R}^n$ be a (measurable) bounded set, and consider the following optimization problem: minimize $P(X)$, the perimeter of a set $X$, where $X$ ranges over all Caccioppoli subsets of $\mathbb{R}^n$ such that $X\supset A$.

If $A$ is convex, then it's "obvious" that the minimizer is $A$ itself (up to a set of measure zero, of course). Is this true and where can I find a proof? Actually I only consider the case where $A$ is a ball.

Note that even in $\mathbb{R}^2$ it's not true that the minimizer is the convex hull of $A$, which can be seen by taking $A$ to be two balls which are placed at a far enough distance.


1 Answer 1


For any sensible definition of perimeter, an adaptation of following argument proves the claim. Consider the nearest-point projection map $p_A:\mathbb R^n\to A$ (that is, for every $x\in\mathbb R^n$, let $p_A(x)$ be the point of $A$ nearest to $x$). It is easy to see that $p_A$ is Lipschitz-1, i.e., $|p_A(x)-p_A(y)|\le |x-y|$ for all $x,y\in\mathbb R^n$. Such a map should decrease perimeters (or your definition of perimeter is flawed), hence $P(X)\ge P(A)$.

With Caccioppoli definition (which I just learned from Wikipedia), you need to translate this argument to the respective language. The resulting proof is e.g. the following. We may assume that $\partial A$ is smooth. Then consider the following smooth unit vector field $\phi$ in $\mathbb R^n\setminus A$: $$ \phi(x) =\operatorname{grad}(\operatorname{dist}(\cdot,A))(x) . $$ It is easy to see that $\operatorname{div}\phi\ge 0$ on $\mathbb R^n\setminus A$. Extend $\phi$ to $A$ so that it remains smooth and $|\phi(x)|\le 1$ for all $x\in\mathbb R^n$. Then $$ P(X) \ge \int_X \operatorname{div}\phi = area(\partial A) + \int_{X\setminus A}\operatorname{div}\phi \ge area(\partial A) $$ where the first inequality follows from the definition of the perimeter, the equality from Green's formula, and the second inequality from the fact that $\operatorname{div}\phi\ge 0$ on $\mathbb R^n\setminus A$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.