2 Preliminaries

As usual, $S^{n-1}$ denotes the unit sphere in the Euclidean n-space $\mathbb {R}^n$ with Euclidean norm $\lVert {\cdot \rVert }$ . The term $ball$ will always mean an n-dimensional Euclidean ball, and the unit ball in $\mathbb {R}^n$ will be denoted by $B^n$ . $B(x,r)$ is the open ball with center x and radius r. If $x,y \in \mathbb {R}^n$ , we write $x\cdot y$ for the inner product. If $x \in \mathbb {R}^n \setminus \{o\}$ , then $x^{\perp }$ is the $(n-1)$ -dimensional subspace orthogonal to x. $\mathcal {G}(n,i)$ denotes the Grassmanian of the i-dimensional subspaces of $\mathbb {R}^n$ , $1\leq i \leq n-1$ , and if $H \in \mathcal {G}(n,i)$ , $H^{\perp }$ is the $(n-i)$ -dimensional subspace orthogonal to H. By subspace, we mean linear subspace. Given $x \in \mathbb {R}$ , $\lfloor x \rfloor $ is the floor function of x.

If X is a set, we denote by convX its convex envelope, and by $\partial X$ its boundary. If $H \in \mathcal {G}(n,i)$ , then $X|H$ is the (orthogonal) projection of X on H. If X and Y are sets in $\mathbb {R}^n$ and $t\geq 0$ , then $tX:=\{tx:x \in X\}$ and

$$ \begin{align*} X+Y:=\{x+y: x \in X, y \in Y \} \end{align*} $$

denotes the Minkowski sum of X and Y. For X measurable set, its volume will be $\lvert X\rvert $ .

When $H \in \mathcal {G}(n,i),$ we write $R_H$ for the reflection with respect to H, i.e., the map that takes $x\in \mathbb {R}^n$ to $2(x|H)-x$ , where $x|H$ is the projection of x onto H. If $R_H X=X$ , we say that X is H-symmetric.

We denote by $\mathcal {C}^n$ the class of nonempty compact subsets of $\mathbb {R}^n$ . $\mathcal {K}^n$ will be the class of nonempty compact convex subsets of $\mathbb {R}^n$ , and $\mathcal {K}_n^n$ is the class of convex bodies. A body is a compact subset of $\mathbb {R}^n$ equal to the closure of its interior. In the analogous way, we define $\mathcal {C}_n^n$ . If $K \in \mathcal {K}^n$ , then

$$ \begin{align*} h_K(x):=\sup\{x\cdot y: y \in K\}, \end{align*} $$

for $x \in \mathbb {R}^n$ , defines the support function $h_K$ of K. With the support function, we can define the mean width of a convex body K, which is

$$ \begin{align*} w(K):=\frac{1}{\lvert{\partial B^n}\rvert } \int_{S^{n-1}} (h_K(\nu)+h_K(-\nu)) d\nu, \end{align*} $$

where $\lvert {\partial B^n}\rvert $ is the $(n-1)$ -dimensional measure of the unit sphere in $\mathbb {R}^n$ . The aforementioned spaces $\mathcal {C}^n$ and $\mathcal {K}^n$ are metric spaces with the Hausdorff metric, which is given in general for two sets $A,B$ by

$$ \begin{align*} d_{\mathcal{H}}(A,B):=\sup\{e(A,B),e(B,A)\}, \end{align*} $$

where

$$ \begin{align*} e(A,B):=\sup_{x\in B}d(x,A) \end{align*} $$

is the excess of the set A from the set B, and $d(x,A)$ is the usual distance between a point and a set. The completeness of such metric spaces is a classic result [Reference Blaschke6], we will refer to it as the Blaschke selection theorem both in convex and compact context.

Another classical result we will refer to is the Brunn–Minkowski inequality. Given two measurable sets $A,B$ such that $A+B$ is measurable (the sum of measurable set is not always measurable), it states that

$$ \begin{align*} \lvert A+B \rvert^{1/n}\geq \lvert{A\rvert}^{1/n}+\lvert{B\rvert}^{1/n}, \end{align*} $$

where equality holds if and only if one of the following holds: either A is convex with nonempty interior and B is a homothetic copy of A (up to removing subsets of volume zero), or both A and B have null measure and lie on parallel hyperplanes. See Gardner [Reference Gardner10] for a precise and comprehensive survey on the general case of this inequality. For more background material in convex geometry see Gardner [Reference Gardner11], Gruber [Reference Gruber12], Lay [Reference Lay16] and Schneider [Reference Schneider20].

Given $C \in \mathcal {C}^n, H \in \mathcal {G}(n,i),1\leq i \leq n-1$ , we recall the definition of some symmetrizations.

• • Schwarz symmetrization:

  $$ \begin{align*} S_H C:=\bigcup_{x \in H} B(x,r_x), \end{align*} $$
  where $r_x$ is such that for the $(n-i)$ -dimensional measure of every section holds $\lvert {B(x,r_x)\rvert } = \lvert {C \cap (H^{\perp }+x)\rvert }$ , and $B(x,r_x) \subset H^{\perp } +x$ . If $\lvert {C \cap (H^{\perp }+x)\rvert }=0$ , then $r_x=0$ when $C \cap (H^{\perp }+x) \neq \emptyset $ , while when the section is empty, then the symmetrization keeps it empty. Observe that thanks to Fubini’s theorem, Schwarz symmetrization preserves the volume.

  For $i=n-1$ , we have Steiner symmetrization. See Gardner [Reference Gardner11], Chapter 1 and 2 for a comprehensive exposition on this symmetrization.

• • Minkowski symmetrization:

  $$ \begin{align*} M_H C:=\frac{1}{2}(C+R_H C). \end{align*} $$
  We will also consider the case $i=0$ , which is called the central Minkowski symmetrization
  $$ \begin{align*} M_o K=\frac{K-K}{2}. \end{align*} $$

  Notice that for support functions, we have the following:

  $$ \begin{align*} h_{K}+h_L=h_{K+L} \qquad \forall K,L \in \mathcal{K}^n, \end{align*} $$
  thus Minkowski symmetrization preserves mean width.

• • Fiber symmetrization:

  (2.1) $$ \begin{align} F_H C:= \bigcup_{x \in H} \left[\frac{1}{2}\left(C\cap(H^{\perp}+x)\right)+\frac{1}{2}\left(R_H C \cap (H^{\perp}+x)\right) \right]. \end{align} $$
  Observe that, defining $M_{H^{\perp },x}$ the central Minkowski symmetrization with respect to x in $H^{\perp }+x$ identified with $\mathbb {R}^{n-i}$ , we can write
  $$ \begin{align*}F_H K=\bigcup_{x \in H} M_{H^{\perp},x} (K \cap (H^{\perp}+x)).\end{align*} $$

• • Minkowski–Blaschke symmetrization: If K is a convex compact set, we define

  
  for every $u \in S^{n-1}$ . At the end of Section 3, we will see that we can extend this definition to any compact set using the support function of its convex envelope.

Consider a family of bodies $\mathcal {E}$ and a subspace $H \in \mathcal {G}(n,i)$ , then an i-symmetrization is a map

$$ \begin{align*} \diamondsuit_H:\mathcal{E} \to \mathcal{E}_H , \end{align*} $$

where $\mathcal {E}_H$ are the H-symmetric elements of $\mathcal {E}$ .

We now state some properties of i-symmetrizations. As we will see, the interaction between these properties and eventually their lack can be crucial in the study of convergence. Consider $K, L \in \mathcal {E}$ , H a subspace in $\mathbb {R}^n$ , and the map $\diamondsuit _H$ satisfies the properties of:

• Monotonicity if $K \subset L \Rightarrow \diamondsuit _H K \subset \diamondsuit _H L$ ;

• H-symmetric invariance if $R_H K=K \Rightarrow \diamondsuit _H K= K$ ;

• Invariance under translations orthogonal to H of H-symmetric sets if $R_H K=K, y \in H^{\perp } \Rightarrow \diamondsuit _H(K+y)=\diamondsuit _H K$ .

Notice that these properties hold for Steiner, Minkowski, and fiber symmetrizations, while the first and the third hold for Schwarz symmetrization. More examples of symmetrizations with peculiar behaviors can be found in Saroglou [Reference Saroglou19].

3 Klain’s theorem for Minkowski symmetrization of compact sets

Two of the main features of Steiner, Schwarz, Minkowski, and fiber symmetrizations are the idempotency and the invariance for H-symmetric bodies in the class of convex sets. These two properties no longer hold when we switch to the class of generic compact sets.

An immediate example regarding Minkowski symmetrization is the following. Consider in $\mathbb {R}^2$ the compact set $C=\{(-1,0),(1,0)\}$ . This set is obviously symmetric with respect to the vertical axis, which we can identify with a subspace H. Then, we have

$$ \begin{align*} M_H C=\{(-1,0),(0,0),(1,0)\}, \end{align*} $$

thus the invariance for symmetric sets no longer holds. If we apply again the same symmetrization,

$$ \begin{align*} M_H(M_H C)=\{(-1,0),(-1/2,0),(0,0),(1/2,0),(1,0)\}, \end{align*} $$

showing that the same happens to idempotency. In Figures 1 and 2, we see an example for the fiber symmetrization of a compact set in the plane.

Figure 1 Original body and first symmetrization.

Figure 2 Second and third symmetrization.

If we iterate this process for $C=\{(-1,0),(1,0)\}$ , we see that in this case there is no finite degree of idempotency, i.e., there does not exist an index $\ell \in \mathbb {N}$ such that

$$ \begin{align*} M_H^{\ell} C=M_H^{k+\ell} C, \end{align*} $$

for every $k \in \mathbb {N}$ , where in general

$$ \begin{align*} \underbrace{M_H \dots M_H}_{\ell\text{-times}}:=M_H^{\ell}. \end{align*} $$

Moreover, the iterated symmetrals converge to the set given by conv $(C)$ . This is the main idea behind the next result, after proving a technical lemma.

Proof The first statement follows from the explicit calculations

$$ \begin{align*} M_H(K+v)=\frac{K+v+R_H(K+v)}{2}=\frac{K+R_H(K)}{2}+ \end{align*} $$

$$ \begin{align*} \frac{v|H^{\perp}+v|H-v|H^{\perp}+v|H}{2}=M_H(K)+v|H, \end{align*} $$

where we used the linearity of the reflections and the decomposition $v=v|H+v|H^{\perp }$ .

For the second statement, by hypothesis, we have that $R_H K=K$ , i.e., $R_H(x) \in K$ for every $x \in K$ . Then, taking $x \in K$ , $(x+R_H(R_H(x)))/2=x \in M_H K$ , concluding the proof.

Consider now K such that $K=M_H K$ . Then, obviously K must be H-symmetric, and $K=R_H K$ . Then, for every $x,y \in K$ , we have that $(x+y)/2 \in K$ , thus for every point z in the segment $[x,y]$ , we can build a sequence by bisection such that it converges to z. K is compact, henceforth it contains z. The other implication is trivial.▪

If we consider the iterated symmetral

(3.1) $$ \begin{align} K_m:=M_H^m K=\underbrace{M_H\ldots M_H}_{m\text{-times}} K, \end{align} $$

we notice that the second statement implies that $K_m \subseteq K_{m+1}$ for every $m \in \mathbb {N}$ .

The next theorem presents the intuition behind the rest of the work. It will actually be an easy corollary of the results we will prove later, but we present it, because its proof is self-contained and it is useful to glimpse the underlying structure of Minkowski symmetrization.

Theorem 3.2 Let $K \in \mathcal {C}^n$ , $H \in \mathcal {G}(n,i), 1 \leq i \leq n-1$ . Then, the sequence $K_m$ in (3.1) converges in Hausdorff distance to the H-symmetric convex compact set

$$ \begin{align*}L=\text{conv}(M_H K).\end{align*} $$

Proof We observe preliminarily that for the properties of convex envelope and Minkowski sum, we have $K_m \subseteq L$ for every $m \in \mathbb {N}$ . Then, we only need to prove that for every $x \in L$ , we can find a sequence $x_m \in K_m$ such that $x_m \to x$ . We can represent K as $\bar {K}+v, v \in K$ , where $\bar {K}$ contains the origin. Because Minkowski symmetrization is invariant under H-orthogonal translations, we can take $v \in H$ .

For every m, we have $R_H K_m=K_m$ , and thus we can write

$$ \begin{align*} K_{m+1}=M_H K_m=\frac{K_m+K_m}{2}=\frac{\overbrace{{K_1+\cdots+K_1}}^{2^m-\text{times}}}{2^{m}}. \end{align*} $$

Considering the aforementioned representation of K, $R_H K=R_H \bar {K} + v$ , and we have

$$ \begin{align*} K_{m}=\bar{K}_m+v, \quad \text{ where } \bar{K}_m:=M_H^m \bar{K}, \end{align*} $$

and thus we can write every point $y \in K_m$ as $y=\bar {y}+v, \bar {y} \in \bar {K}_m$ .

Given $x\in L$ , thanks to the Carathèodory theorem, there exist $x_k \in K_1, \lambda _k \in (0,1), k=1,\ldots ,n+1$ such that $\sum _{k=1}^{n+1} \lambda _i=1$ and

$$ \begin{align*} x=\sum_{k=1}^{n+1} \lambda_k x_k=\sum_{k=1}^{n+1} \lambda_k \bar{x}_k + v, \end{align*} $$

where $x_k=\bar {x}_k+v, \bar {x}_k \in \bar {K}_1$ . For every $\lambda _k$ , we consider its binary representation

$$ \begin{align*} \lambda_k=\sum_{\ell=1}^{+\infty} \frac{a_{\ell,k}}{2^{\ell}}, \qquad a_{\ell,k} \in\{0,1\} \end{align*} $$

(we do not consider $\ell =0$ , because $\lambda _i < 1$ ), and its mth approximation given by the partial sum

$$ \begin{align*} \lambda_{m,k}:=\sum_{\ell=1}^{m} \frac{a_{\ell,k}}{2^{\ell}}=\frac{1}{2^m}\sum_{\ell=1}^m a_{\ell,k} 2^{m-\ell}. \end{align*} $$

We notice for later use that $\lvert {\lambda _k-\lambda _{m,k}\rvert }\leq 1/2^m$ .

Calling $q_s:=\lfloor 2^{s}/(n+1) \rfloor $ , we now build the sequence

$$ \begin{align*} x_s:=\sum_{k=1}^{n+1} \lambda_{q_s,k} \bar{x}_k+v = \frac{1}{2^{q_s}} \sum_{k=1}^{n+1} \left(\sum_{\ell=1}^{q_s} a_{\ell,k} 2^{q_s-\ell}\right)\bar{x}_k +v, \end{align*} $$

where the $2^{s}- q_s(n+1)$ spare terms in $\bar {K}_1$ can be taken as the origin in the sum representing $\bar {K_s}$ .

Then, we have that every $x_s$ belongs to $K_s$ , and

$$ \begin{align*} \lVert{x-x_s\rVert} &= \lVert{\bar{x}+v-(\bar{x}_s+v)\rVert} \\ &\leq \sum_{k=1}^{n+1} \lVert{\bar{x}_k\rVert} \lvert{\lambda_k-\lambda_{q_s,k}\rvert} \leq \frac{1}{2^{q_s}} \sum_{k=1}^{n+1} \lVert{\bar{x}_k\rVert}\leq (n+1) \frac{\max_{y \in K_1} \lVert{y-v\rVert}}{2^{q_s}}. \end{align*} $$

Clearly, $\lVert {x-x_s\rVert } \to 0$ as $s \to +\infty $ , which concludes our proof.▪

As an immediate consequence, we have the following result.

Corollary 3.3 Let $K \in \mathcal {C}^n$ , $H \in \mathcal {G}(n,i), 1 \leq i \leq n-1$ . Then, we have that the sequence

$$ \begin{align*} K_m:=F_H^m K=\underbrace{F_H\ldots F_H}_{\text{m-times}} K \end{align*} $$

converges in Hausdorff distance to the H-symmetric compact set

$$ \begin{align*} L=\bigcup_{x \in H} \text{conv}(F_H K \cap (x+H^{\perp})). \end{align*} $$

Proof Recalling the definition of fiber symmetrization,

$$ \begin{align*} F_H K=\bigcup_{x \in H} \frac{1}{2}((K\cap(x+H^{\perp}))+(R_H K \cap (x+H^{\perp})))=\bigcup_{x \in H} M_{H^{\perp},x}(K\cap (x+H^{\perp})). \end{align*} $$

The result is a straightforward application of Theorem 3.2 to the sections of K.▪

We now state the Shapley–Folkman–Starr theorem ([Reference Schneider20, Theorem 3.1.2], [Reference Starr21]) for using it in the next proof.

Theorem 3.4 (Shapley–Folkman–Starr)

Let $A_1,\ldots ,A_k \in \mathcal {C}^n$ . Then,

$$ \begin{align*} d_{\mathcal{H}}\left(\sum_{j=1}^k A_j,\text{conv}\left(\sum_{j=1}^k A_j \right) \right)\leq \sqrt{n} \max_{1\leq j \leq k} D(A_j), \end{align*} $$

where $D(\cdot )$ is the diameter function $D(K):=\sup \{\lVert {x-y\rVert }: x,y \in K\}$ .

In Theorem 3.2, we already saw how the convexification effect of Minkowski addition works when we iterate the same symmetrization. Now, with the inequality given by the Shapley–Folkman–Starr theorem, we generalize the former result proving Theorem 1.2.

Proof of Theorem 1.2 First notice that orthogonal transformations and Minkowski addition commute with convex envelope. Thus, for $C_m=\sum _{j=1}^m \mathbb {A}_j C/m$ , where $C \in \mathcal {C}^n$ and $\{\mathbb {A}_j\}$ is a sequence of isometries as in the hypothesis,

$$ \begin{align*} \text{conv}(C_m)= \text{conv} \left( \frac{1}{m}\sum_{j=1}^m \mathbb{A}_j C \right)= \frac{1}{m} \sum_{j=1}^m \mathbb{A}_j \text{conv} (C)= \frac{1}{m} \sum_{j=1}^m \mathbb{A}_j K= K_m. \end{align*} $$

We now apply the Shapley–Folkman–Starr theorem, obtaining

$$ \begin{align*} d_{\mathcal{H}} (C_m,K_m)=d_{\mathcal{H}}(C_m, \text{conv}C_m) \leq \frac{\sqrt{n}}{m}\max_{1\leq j \leq m} D(\mathbb{A}_j C)=\frac{\sqrt{n}}{m}\max_{1\leq j \leq m} D(C). \end{align*} $$

C is compact and thus bounded, hence $d_{{\mathcal {H}}}(C_m,K_m)\to 0$ , completing the proof. In fact, the compactness is not necessary, and boundedness would suffice, but this is beyond the interest of the present paper.▪

Corollary 3.5 Let K be a convex compact set, and let $\{H_m\}$ be a sequence of subspaces of $\mathbb {R}^n$ (not necessarily of the same dimension) such that the sequence of iterated symmetrals

$$ \begin{align*}K_m:=M_{H_m}\ldots M_{H_1} K\end{align*} $$

converges to a convex compact set L in the Hausdorff distance. Then, the same happens for every compact set $\tilde {K}$ such that $conv(\tilde {K})=K$ , and the sequence $\tilde {K}_m$ , defined as $\tilde {K}_m:= M_{H_m}\dots M_{H_1} \tilde {K}$ , converges to the same limit L.

Proof We will show that the theorem holds proving that

$$ \begin{align*} d_{\mathcal{H}}(\tilde{K}_m, K_m) \to 0, \end{align*} $$

for $m \to \infty $ .

We can write $K_m$ as the mean of Minkowski sum of composition of reflections of K. Indeed, we have

$$ \begin{align*} \begin{array}{c} K_1=\displaystyle\frac{K+R_{H_1} K}{2},\\ K_2=\displaystyle\frac{K+R_{H_1} K + R_{H_2}(K+R_{H_1} K)}{4}=\displaystyle\frac{K+R_{H_1}K+R_{H_2} K + R_{H_2} R_{H_1} K}{4},\\ \ldots \end{array} \end{align*} $$

and so on. The same obviously holds for $\tilde {K}_m$ . We call these compositions of reflections $\mathbb {A}_j, 1\leq j \leq 2^m$ , and defining $A_j:=\mathbb {A}_j \tilde {K}$ , we can write

$$ \begin{align*} \tilde{K}_m=\frac{1}{2^m}\sum_{j=1}^{2^m} \mathbb{A}_j \tilde{K}=\frac{1}{2^m}\sum_{j=1}^{2^m} A_j. \end{align*} $$

The proof follows applying Theorem 1.2.▪

We now have, as a consequence of Corollary 3.5, our generalization of Klain’s result.

Corollary 3.6 Let $K \in \mathcal {C}^n$ , $\mathcal {F}=\{Q_1,\ldots ,Q_s\} \subset \mathcal {G}(n,i), 1\leq i \leq n-1$ , $\{H_m\}$ a sequence of elements of $\mathcal {F}$ . Then, the sequence

$$ \begin{align*} K_m:=M_{H_m}\ldots M_{H_1}K \end{align*} $$

converges to a convex set L such that it is the limit of the same symmetrization process applied to $\bar {K}=conv(K)$ . Moreover, L is symmetric with respect to all the subspaces of $\mathcal {F}$ which appear infinitely often in $\{H_m\}$ .

We can use a similar method to generalize the following classical result from Hadwiger (see, for example, [Reference Schneider20, Theorem 3.3.5]).

Then, the next result is obtained combining Theorems 1.2 and 3.7.

Remark Corollary 3.5 gives us an answer regarding the possibility of extending the Minkowski–Blaschke symmetrization $\overline {M}_H$ to compact sets. This symmetrization, that we have defined in Section 2 for convex bodies, can be practically seen as the mean of rotations of a compact set $K \in \mathcal {K}^n$ by a subgroup of $SO(n)$ , and thus can be approximated by

$$ \begin{align*} \frac{1}{N}\sum_{k=1}^N\mathbb{A}_k K, \end{align*} $$

where $\{\mathbb {A}_k\}_{k=1}^N \subset \{\mathbb {A}_k\}_{k \in \mathbb {N}}$ is a suitable set of rotations which is dense in said subgroup.

Indeed, from the definition of $\overline {M}_H$ in terms of the support function, we have that the integral can be approximated by

$$ \begin{align*} \sum_{k=1}^N \frac{h_K(\mathbb{A}_k^{*} x)}{N}=\frac{1}{N}\sum_{k=1}^N h_{\mathbb{A}_k K} (x), \end{align*} $$

which corresponds naturally to the Minkowski sum written above.

Then again, following the proof of Corollary 3.5, we can write the symmetral as the limit of a mean of Minkowski sum of isometries of a fixed $K \in \mathcal {K}^n$ , and thus the Minkowski–Blaschke symmetrization actually gives the same result for every $C \in \mathcal {C}^n$ such that conv $(C)=K$ .

This shows that this symmetrization is sensible only to the extremal points of a set, thus it makes no difference in using it with compact sets or convex sets.

4 The case of convex outer boundary

One of the main properties of Minkowski symmetrization is that, as a consequence of the Brunn–Minkowski inequality, it increases the volume. Indeed, for every measurable set $K \subset \mathbb {R}^n$ such that $\lvert {K\rvert }> 0$ and $M_H K$ is measurable, we have

$$ \begin{align*} \lvert{M_H K\rvert}^{1/n}= \lvert{1/2(K+R_H K)\rvert}^{1/n}\geq \frac{1}{2}\lvert{K\rvert}^{1/n}+\frac{1}{2}\lvert{R_H K\rvert}^{1/n}= \lvert{K\rvert}^{1/n}, \end{align*} $$

where equality holds if and only if K and $R_H K$ are homothetic convex bodies from which sets of measure zero have been removed. See [Reference Gardner10] for a general proof of this result. We work only with compact sets, thus the equality condition is possible only if the two bodies are homothetic and convex. This happens if and only if $K=M_H K$ , thus we would like to state that the iteration of Minkowski symmetrization increases the volume until the sequence of symmetrals reaches $M_H \text {conv} (K)$ .

With Theorem 3.2, we proved that, regardless of the volume, the limit of $\tilde {K}_m$ is actually $M_H \text {conv}(K)$ , but now we raise one more question: Can we obtain this limit in a finite number of iterations? Under which hypothesis is this possible?

We start by giving an answer for compact sets of $\mathbb {R}$ . This case is more complicated than for similar objects in $\mathbb {R}^n, n \geq 2$ , as we will prove later.

Lemma 4.1 Let $K \in \mathbb {R}$ be a compact set such that $\text {conv}(K)=[a,b]$ with the following property:

$$ \begin{align*} \exists \varepsilon>0 \text{ s.t. } [a,a+\varepsilon] \subset K \text{ or } [b-\varepsilon,b]\subset K. \end{align*} $$

Then, there exists an index $\ell \in \mathbb {N}$ depending on $\varepsilon $ and $(b-a)$ such that

$$ \begin{align*} M_o^{\ell} K= M_o^{\ell+k} K, \end{align*} $$

for every $k\in \mathbb {N}$ .

Moreover, $\ell $ increases with $(b-a)$ and decreases if $\varepsilon $ increases.

Proof First consider the case $K \supseteq \{a\}\cup [b-\varepsilon ,b]$ . Then,

$$ \begin{align*} M_o K \supseteq M_o (\{a\}\cup[b-\varepsilon,b])\supset \left[\frac{a-b}{2},\frac{a-b}{2}+\frac{\varepsilon}{2}\right] \cup \left[\frac{b-a}{2}-\frac{\varepsilon}{2},\frac{b-a}{2}\right]. \end{align*} $$

Easy calculations show that the same happens when $K \supseteq [a,a+\varepsilon ]\cup \{b\}$ . Then, naming

$$ \begin{align*} M:=\frac{b-a}{2}, \quad m:=\frac{b-a}{2}-\frac{\varepsilon}{2}, \end{align*} $$

and we can work with a set containing a subset the form

$$ \begin{align*} [-M,-m] \cup [m,M]=:\tilde{K}, \end{align*} $$

where $M-m=\varepsilon /2$ .

If now we apply the symmetrization, we obtain

(4.1) $$ \begin{align} M_o K\supseteq [-M,-m]\cup \left[ \frac{m-M}{2},\frac{M-m}{2}\right]\cup [m,M]=M_o \tilde{K}. \end{align} $$

If $(M-m)/2\geq m$ , that is, $m \leq M/3$ , then $M_o K=\text {conv}(K)$ , and the result holds with $\ell =1$ .

In the general case, we can show by induction that the following inclusion holds:

$$ \begin{align*} M_o^{k+1} K \supseteq M_o^{k+1} \tilde{K} \supseteq \bigcup_{j=0}^{2^{k+1}} \left[ \frac{(2^{k+1}-j)m-jM}{2^{k+1}},\frac{(2^{k+1} -j)M-jm}{2^{k+1}} \right], \end{align*} $$

where the first inclusion is trivial thanks to the monotonicity of Minkowski symmetrization. In particular, we will show that

$$ \begin{align*} M_o^{k+1}\tilde{K}\supseteq M_o^{k}\tilde{K}\cup \bigcup_{j=1}^{2^{k}} \left[\frac{(2^{k+1}-2j+1)m-(2j-1)M}{2^{k+1}}, \frac{(2^{k+1}-2j+1)M-(2j-1)m}{2^{k+1}} \right], \end{align*} $$

which is the desired set. This inclusion is actually an equality, but proving this fact is beyond our goal here.

For $k=0$ , we have already seen in (4.1) that the inclusion holds. By inductive hypothesis, at the $(k+1)$ th step, the means of adjacent intervals of $M_o^k \tilde {K}$ are given by

$$ \begin{align*} &\displaystyle\frac{1}{2}\left[\displaystyle\frac{(2^k-(j+1))m-(j+1)M}{2^k},\displaystyle\frac{(2^k-(j+1))M-(j+1)m}{2^k}\right] \\ &\qquad +\displaystyle\frac{1}{2}\left[\displaystyle\frac{(2^k-j)m-jM}{2^k},\displaystyle\frac{(2^k -j)M-jm}{2^k} \right] \\ &\!\! = \left[\displaystyle\frac{(2^{k+1}-2(j+1)+1)m-(2(j+1)-1)M}{2^{k+1}},\displaystyle\frac{(2^{k+1}-2(j+1)+1)M-(2(j+1)-1)m}{2^{k+1}} \right], \end{align*} $$

for every $j=0,\ldots ,2^{k}-1$ , giving us the elements of the union with odd indices.

Observe that $M_o^{k}\tilde {K}$ is invariant under reflection. Thus, thanks to Lemma 3.1 and the monotonicity of Minkowski symmetrization, we have $M_o^{k}\tilde {K} \subseteq M_o^{k+1} K$ , concluding the induction.

Taking at the kth step two adjacent intervals, we have that they are connected if

$$ \begin{align*} \frac{(2^k-(j+1))M-(j+1)m}{2^k}\geq \frac{(2^k-j)m-jM}{2^k}. \end{align*} $$

It follows that the condition for filling the whole segment conv $(M_H^k K)$ is

$$ \begin{align*} \frac{m}{M}\leq\frac{2^k-1}{2^k+1}. \end{align*} $$

Observe that the dependence on the index j disappeared after calculations, confirming that this holds for every couple of adjacent intervals.

By hypothesis, $M-m=\varepsilon /2$ and $(2^k-1)/(2^k+1) \to 1 $ . We have

$$ \begin{align*} \frac{m}{M}=1+\frac{m-M}{M}=1-\frac{\varepsilon}{2M}, \end{align*} $$

then there exists $\ell \in \mathbb {N}$ such that

$$ \begin{align*} 1-\frac{\varepsilon}{2M} < \frac{2^{\ell}-1}{2^{\ell}+1}, \end{align*} $$

thus $M_o^{\ell } K= \text {conv}(K)$ for

$$ \begin{align*} \ell \geq \log_2 \left(\frac{4M}{\varepsilon}-1 \right). \end{align*} $$

This set is convex and o-symmetric, and thus is invariant under Minkowski symmetrization. The dependence from M and $\varepsilon $ is clear from the last inequality.▪

Notice that it is crucial that either a or b belongs to an interval with positive measure contained in K. Indeed, if the two extremes both were isolated points, there would occur a situation analogous to the example presented at the beginning of Section 3, thus there would be a part of the set which stabilizes itself only at the limit.

Remark With wider generality, the previous lemma holds for the means of Minkowski sums. Indeed, if $K \subset \mathbb {R}$ , for every $x \in \mathbb {R}$ holds

$$ \begin{align*} \frac{1}{m}\sum_{j=1}^m(K-x)=\frac{1}{m}\sum_{j=1}^m K -x, \end{align*} $$

and taking x as the average of the extreme points of K, we reduce ourselves to the same context of the lemma, which can be restated as follows.

Lemma 4.2 Let $K \in \mathbb {R}$ be a compact set such that $\text {conv}(K)=[a,b]$ with the following property:

$$ \begin{align*} \exists \varepsilon>0 \text{ s.t. } [a,a+\varepsilon] \cup [b-\varepsilon,b]\subset K. \end{align*} $$

Then, there exist an index $\ell \in \mathbb {N}$ depending on $\varepsilon $ and $(b-a)$ such that

$$ \begin{align*} \frac{1}{2^{\ell}}\sum_{j=1}^{2^{\ell}} K=\frac{1}{2^{\ell+k}}\sum_{j=1}^{2^{k+\ell}} K, \end{align*} $$

for every $k\in \mathbb {N}$ .

Moreover, $\ell $ increases with $(b-a)$ and decreases if $\varepsilon $ increases.

Proof First, we remind the reader that, as we have seen in Theorem 3.2, when we iterate $M_H$ , after the first symmetrization, we are just computing the mean

$$ \begin{align*} \frac{1}{2^{m-1}}\sum_{j=1}^{2^{m-1}} M_H K = M_H^{m} K. \end{align*} $$

Moreover, we observe that the only difference with the previous lemma is that we do not have the sum with the reflection, so we have to require in the hypothesis that both the end points of K belong to segments included in K.

Now, we can work with a set

$$ \begin{align*} \tilde{K}:=([-M,-m] \cup [m,M] )+ x \end{align*} $$

for a suitable $x \in \mathbb {R}$ , and the rest of the proof follows as the previous one.▪

A weaker property that these sets have is to contain the boundary of their convex envelope. When $n \geq 2$ , this is enough to prove the stronger and more general result in Corollary 4.4.

If A is a connected compact set, then we call the external connected component of $\mathbb {R}^n\setminus A$ the unbounded connected component of such a set. Then, we notice that, as in [Reference Fradelizi, Madiman, Marsiglietti and Zvavitch8], this result holds also for the boundary of the external connected component of $\mathbb {R}^n \setminus K$ and $\mathbb {R}^n \setminus L$ . Moreover, we point out that the hypothesis of Lemma 4.3 immediately rules out the case $n=1$ . This will indeed be an issue in Corollary 4.4 and Theorem 1.3.

Now, we can prove Theorem 1.4.

Proof of Theorem 1.4 Let $x \in K+L$ , then there exist $\kappa \in K, \ell \in L$ such that $x=\kappa +\ell $ . If we define $\tilde {K}:=K+x-\kappa , \tilde {L}:=-L+x+\ell $ , we have that $x \in \tilde {K} \cap \tilde {L}$ , hence $\tilde {K}$ and $\tilde {L}$ satisfy the hypothesis of Lemma 4.3. Thus, $\partial \tilde {K} \cap \partial \tilde {L} \neq \emptyset $ .

Let $z \in \partial \tilde {K} \cap \partial \tilde {L}$ , then

$$ \begin{align*} z-x+\kappa \in \partial K, \ell-z+x \in \partial L. \end{align*} $$

Now,

$$ \begin{align*} (z-x+\kappa)+(\ell-z+x)=\kappa+\ell, \end{align*} $$

proving our assertion.▪

Corollary 4.4 Let $K \in \mathcal {K}^n$ , and let H be a subspace of $\mathbb {R}^n$ . Then,

(4.2) $$ \begin{align} M_H K=M_H \partial K, \end{align} $$

In particular, if $C \in \mathcal {C}^n$ and $C \supseteq \partial \text {conv}(C)$ , then $M_H C$ is convex, and

$$ \begin{align*} M_H \text{conv}(C)=M_H C. \end{align*} $$

The same holds for fiber symmetrization if H is not a hyperplane.

Proof We first prove the result regarding Minkowski symmetrization. We apply Theorem 1.4 to $K/2$ and $R_H K/2$ . Indeed, observe that the two sets are convex, hence with connected boundary. Moreover, because they have the same volume, no translate of one set is strictly contained in the other set. Then, Theorem 1.4 yields $M_H K=M_H \partial K$ .

Consider now a set $C \in \mathcal {C}^n$ with $\partial \text {conv}(C) \subseteq C$ . From equation (4.2), $\partial \text {conv}(C) \subseteq \partial C$ , and the monotonicity of Minkowski symmetrization, we infer

$$ \begin{align*} M_H C \supset \frac{\partial C + \partial R_H C}{2} \supseteq \frac{\partial \text{conv}(C)+ \partial R_H \text{conv}(C)}{2} = M_H \text{conv}(C). \end{align*} $$

Since the reverse inclusion trivial, this concludes the proof in the case of Minkowski symmetrization.

Regarding fiber symmetrization, notice that if H was a hyperplane, then the sections are one-dimensional, and in Lemma 4.1, we proved that we need certain conditions on the boundary to obtain idempotency. In general, we know that fiber symmetrization preserves convexity, thus $F_H \text {conv} C$ is convex, and its boundary is given by the union of the boundaries of the sections by $H^{\perp } + x, x \in H$ . If H is not a hyperplane, these sections are obtained by the Minkowski symmetrization of convex sets of dimension greater or equal than 2, completing the proof.▪

The proof of Theorem 1.3 now follows immediately.

We conclude this section with another immediate application, a small addition to Klartag’s following result (cf. Theorem 1.1 in [Reference Klartag15]). The same generalization holds for similar results in [Reference Klartag14].

Theorem 4.5 Let $n \geq 2, 0 < \epsilon < 1/2$ , and let $K \subset \mathbb {R}^n $ be a compact set such that $K \supseteq \partial \text {conv} K$ . Then, there exist $cn \log 1/\epsilon $ Minkowski symmetrizations with respect to hyperplanes, that transform K into a body $\tilde {K}$ that satisfies

$$ \begin{align*} (1-\epsilon) w(K) B^n \subset \tilde{K} \subset (1+\epsilon) w(K) B^n, \end{align*} $$

where $c>0$ is some numerical constant.