Let $A$ and $B$ be two non-empty compact subsets of $\mathbb{R}^d$.

Brunn-Minkowski inequality gives $(m(A+B))^{1/d} \geq (m(A))^{1/d} + (m(B))^{1/d}$.

But how to prove the following?

$(m(A+B))^{1/d} = (m(A))^{1/d} + (m(B))^{1/d} \implies$

$A$ and $B$ are convex and $\exists$ $\delta>0$, $h\in \mathbb{R}^d$ s.t. $A = \delta B + h$

(It's Problem $8$ in E.M.Stein Real Analysis Chapter $1$, Page $48$, though this problem doen't have much to do with real analysis.)

Edit:

As you might see, proving the inequality is easy, but find the necessary condition which makes the inequlity an equality is really hard. I searched for a long time about this, but most books (like Halmos's Measure Theory, Federer's Geometric Measure Theory etc) just give references instead of a proof and the proofs I found can date back to 1930s and are rather cmplicated.

Since E.M.Stein Real Analysis is a widely used book for analysis cource, I think if someone finished all problems in this book might have a better and modified proof. Thanks in advance.

You might also be interested in the following two papers by Fields medalist A. Figalli:

The stability result, proven with optimal transport techniques, implies in particular the equality condition.

Some proofs I found, which are really complicated since they contain a lot of definitions and lemmas new to others. Moreover, it's impossible to answer this question only in a few words, so I just list some references here for those who really need.

$(1)$ Rolf Schneider, Convex Bodies: the Brunn-Minkowski Theory. Section $6.1$, the Brunn-Minkowski theorem.

$\quad$Since this book mainly concentrates on convex bodies, it only proves for convex bodies when the inequality becomes equality without showing why it's not ture for non-convex sets.

$(2)$ IMRE Z. RUZSA, The Brunn–Minkowski Inequality and Nonconvex Sets.

$\quad$This article improves the inequlity for convex sets in $\mathbb R^n$.

$(3)$ DANIEL A. KLAIN, ON THE EQUALITY CONDITIONS OF THE BRUNN-MINKOWSKI THEOREM.

$\quad$This article describes a new proof of the equality condition for convex sets in $\mathbb R^n$.

To be continued...

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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