1 Introduction

As usual, we write \({\mathbb {R}}^n\) to represent the n-dimensional Euclidean space, and we denote by \(\mathrm {e}_i\) the i-th canonical unit vector. For \(i=1, \dots , n\), we represent by \(H_i=\bigl \{x=(x_1,\dots ,x_n)\in {\mathbb {R}}^n: \, x_i= 0\bigr \}\) the i-th coordinate hyperplane. The n-dimensional volume of a measurable set \(M\subset {\mathbb {R}}^n\), i.e., its n-dimensional Lebesgue measure, is denoted by \(\mathrm {vol}(M)\) (when integrating, as usual, \(\mathrm {d}x\) will stand for \(\mathrm {d}\mathrm {vol}(x)\)). We write \(M(t)=\{x\in {\mathbb {R}}^{n-1}: (x,t)\in M\}\) for the \((n-1)\)-dimensional section at height \(t\in {\mathbb {R}}\) (in the direction of \(\mathrm {e}_n\)), whereas the orthogonal projection of M onto an i-dimensional linear subspace H is denoted by M|H. Moreover, \(H^{\bot }\) represents the orthogonal complement of H and, for any \(x\in M|H_i\), we set \(M_i(x)=\{t\in {\mathbb {R}}: \, x+ t\mathrm {e}_i \in M\}\) to denote the one-dimensional section of M through the point x in the direction of \(\mathrm {e}_i\). Finally, given \(r>0\), rM stands for the set \(\{rm:\, m\in M\}\).

The Minkowski sum of two non-empty sets \(A, B\subset {\mathbb {R}}^n\) is the classical vector addition of them: \(A+B=\{a+b:\, a\in A, \, b\in B\}\). It is natural to wonder about the possibility of bounding the volume of the Minkowski sum of two sets in terms of their volumes; this is the statement of the Brunn–Minkowski inequality (for extensive and beautiful surveys on this inequality we refer the reader to [1, 7]). One form of it asserts that if \(\lambda \in (0,1)\) and A and B are non-empty measurable subsets of \({\mathbb {R}}^n\) such that \((1-\lambda )A+\lambda B\) is also measurable then

$$\begin{aligned} \mathrm {vol}\bigl ((1-\lambda )A+\lambda B\bigr )^{1/n}\ge (1-\lambda )\mathrm {vol}(A)^{1/n}+\lambda \mathrm {vol}(B)^{1/n}. \end{aligned}$$
(1.1)

The Brunn–Minkowski inequality was generalized to different types of measures, including the cases of log-concave measures [10, 15] and of p-concave measures (see e.g. [3, 4]). It is interesting to note that it was proved by Borell [2, 3] that such generalizations would require a p-concavity assumption on the density of the underlying measure (see (2.1) below for the precise definition). As a consequence of this approach (see also [21]), when dealing with arbitrary measurable sets and a Radon measure on \({\mathbb {R}}^n\), the (1/n)-form of the Brunn–Minkowski inequality (1.1) is only true, in general, for the volume (up to a constant). However, when considering some special families of sets (e.g. that of unconditional sets), the (1/n)-Brunn–Minkowski inequality holds for some types of measures, such as the standard Gaussian measure, which is given by

$$\begin{aligned} \mathrm {d}\gamma _n(x)=\frac{1}{(2\pi )^{n/2}}e^{\frac{-\left| {x}\right| ^2}{2}}\mathrm {d}x \end{aligned}$$

(see e.g. [8, 11, 12, 14, 16]). Furthermore, for the family of C-coconvex sets (complements of closed convex sets, of positive and finite volume, within a pointed closed convex cone with non-empty interior C), a “complemented” version of the Brunn–Minkowski inequality (1.1) holds for the volume (see [9, 19]), namely

$$\begin{aligned} \mathrm {vol}\bigl (C{\setminus }((1-\lambda )K+\lambda L)\bigr )^{1/n}\le (1-\lambda )\mathrm {vol}(C{\setminus } K)^{1/n}+\lambda \mathrm {vol}(C{\setminus } L)^{1/n} \end{aligned}$$

for all \(\lambda \in (0,1)\). And again, this (complemented) Brunn–Minkowski inequality can be also generalized for certain general measures (see [13]).

To complete the picture, one may ask about possible p-convexity conditions on the density of the underlying measure. Among others, what can be said about the measure \(\nu _n\) on \({\mathbb {R}}^n\) given by

$$\begin{aligned} \mathrm {d}\nu _n(x)=e^{\left| {x}\right| ^2}\mathrm {d}x, \end{aligned}$$

whose density is log-convex? In [13], when dealing with measures involving certain log-convex functions as part of their densities, the authors showed another type of complemented Brunn–Minkowski inequality. Nevertheless, not much more seems to be known regarding Brunn–Minkowski inequalities for log-convex densities or, more generally, quasi-convex densities (see (2.2) below for the precise definition).

To this regard, and inspired by the above-mentioned (complemented) Brunn–Minkowski inequalities, it is natural to wonder whether one may find certain classes of sets for which a measure on \({\mathbb {R}}^n\) of the kind of \(\nu _n\) satisfies the (1/n)-form of the Brunn–Minkowski inequality. Here we give a positive answer to this question, by showing that it is enough to consider congruous sets (see Definition 2.1): a family that contains, among others, the complements of unconditional sets within a centered box (cf. Example 2.1). This is the content of the following result, in the more general setting of product measures with quasi-convex densities (with minimum at the origin).

Theorem 1.1

Let \(\mu =\mu _1\otimes \dots \otimes \mu _n\) be a product measure on \({\mathbb {R}}^n\) such that \(\mu _i\) is the measure given by \(\mathrm {d}\mu _i(x)=\phi _i(x)\,\mathrm {d}x\), where \(\phi _i:{\mathbb {R}}\longrightarrow [0,\infty )\) is quasi-convex with \(\phi _i(0)=\min _{x\in {\mathbb {R}}}\phi _i(x)\), for all \(i=1,\dots ,n\).

Let \(\lambda \in (0,1)\) and let \(A, B\subset {\mathbb {R}}^n\) be non-empty measurable congruous sets such that \((1-\lambda )A+\lambda B\) is also measurable. Then

$$\begin{aligned} \mu \bigl ((1-\lambda )A+\lambda B\bigr )^{1/n}\ge (1-\lambda )\mu (A)^{1/n}+\lambda \mu (B)^{1/n}. \end{aligned}$$
(1.2)

Section 2 is mainly devoted to showing this result. Finally, in Sect. 3, we derive an isoperimetric type inequality as a consequence of (1.2).

2 Proof of the main result

2.1 Background

We recall that a function \(\phi :{\mathbb {R}}^n\longrightarrow [0,\infty )\) is p-concave, for \(p\in {\mathbb {R}}\cup \{\pm \infty \}\), if

$$\begin{aligned} \phi \bigl ((1-\lambda )x+\lambda y\bigr )\ge M_p\bigl (\phi (x),\phi (y),\lambda \bigr ) \end{aligned}$$
(2.1)

for all \(x,y\in {\mathbb {R}}^n\) such that \(\phi (x)\phi (y)>0\) and any \(\lambda \in (0,1)\). Here \(M_p\) denotes the p-mean of two non-negative numbers a, b:

$$\begin{aligned} M_p(a,b,\lambda )=\left\{ \begin{array}{ll} \bigl ((1-\lambda )a^p+\lambda b^p\bigr )^{1/p}, &{} \text { if }p\ne 0,\pm \infty ,\\ a^{1-\lambda }b^\lambda &{} \text { if }p=0,\\ \max \{a,b\} &{} \text { if }p=\infty ,\\ \min \{a,b\} &{} \text { if }p=-\infty . \end{array}\right. \end{aligned}$$

A 0-concave function is usually called log-concave whereas a \((-\infty )\)-concave function is called quasi-concave. Quasi-concavity is equivalent to the fact that the superlevel sets \(\{x\in {\mathbb {R}}^n: \phi (x)\ge t\}\) are convex for all \(t\in [0,1]\).

On the other side of the coin, one is led to p-convex functions, where \(p\in {\mathbb {R}}\cup \{\pm \infty \}\), i.e., those functions satisfying

$$\begin{aligned} \phi \bigl ((1-\lambda )x+\lambda y\bigr )\le M_p\bigl (\phi (x),\phi (y),\lambda \bigr ) \end{aligned}$$
(2.2)

for all \(x,y\in {\mathbb {R}}^n\) and all \(\lambda \in (0,1)\). Again, 0-convex functions are referred to as log-convex whereas \(\infty \)-convex functions are called quasi-convex.

Now we define a new class of (pairs of) sets that will play a relevant role throughout this paper.

Definition 2.1

Let \(A, B\subset {\mathbb {R}}^n\) be non-empty bounded sets. For \(n=1\), we say that A and B are congruous if one of the following assertions holds.

  1. (i)

    \(A\cap (-\infty ,0), B\cap (-\infty ,0)=\emptyset \) and \(\max (A)=\max (B)\).

  2. (ii)

    \(A\cap (0,\infty ), B\cap (0,\infty )=\emptyset \) and \(\min (A)=\min (B)\).

  3. (iii)

    \(A\cap (0,\infty ), B\cap (0,\infty ), A\cap (-\infty ,0), B\cap (-\infty ,0)\ne \emptyset \), \(\min (A)=\min (B)\) and \(\max (A)=\max (B)\).

For \(n\ge 2\), we say that A and B are congruous if, for any \(i=1, \dots , n\), the sets \(A_i(x)\) and \(B_i(y)\) are congruous for all \(x \in A|H_i\) and all \(y \in B|H_i\).

s

Fig. 1
figure 1

The congruous sets A (in gray) and B (the box), with the sections \(A_2(x)\), \(A_1(y)\) for given \(x\in A|H_2\), \(y\in A|H_1\)

Full size image

We notice that the fact that, for any \(i=1, \dots , n\), the sets \(A_i(x)\) and \(B_i(y)\) are congruous (for all \(x \in A|H_i\) and all \(y \in B|H_i\)) does not mean that the same condition in Definition 2.1 holds for all i (see Fig. 1; there \(A_2(x), B_2(x')\) satisfy condition (iii) of Definition 2.1, for all \(x\in A|H_2\) and all \(x'\in B|H_2\), whereas \(A_1(y), B_1(y')\) fulfil condition (i), for any \(y\in A|H_1\) and any \(y'\in B|H_1\)).

Unconditional convex sets are of particular interest in convexity, also regarding Brunn–Minkowski type inequalities (see e.g. [11, 18]). A subset \(A\subset {\mathbb {R}}^n\) is said to be unconditional (not necessarily convex) if for every \((x_1,\dots ,x_n)\in A\) and every \((\epsilon _1,\dots ,\epsilon _n)\in [-1,1]^n\) one has \((\epsilon _1x_1,\dots ,\epsilon _nx_n)\in A\). As announced before, the family of congruous sets contains certain complements of unconditional sets:

Example 2.1

Let \(P=\prod _{i=1}^n[-\alpha _i,\alpha _i]\), \(\alpha _i>0\) for \(i=1,\dots ,n\), be a centered orthogonal compact box and let \(A, B\subset P\) be non-empty compact sets such that \(P{\setminus } A, P{\setminus } B\) are unconditional. Then A and B are congruous. Indeed, from the unconditionality of \(P{\setminus } A\) and \(P{\setminus } B\), we have that \(\max \bigl (A_i(x)\bigr )=\max \bigl (B_i(y)\bigr )=\alpha _i\) and \(\min \bigl (A_i(x)\bigr )=\min \bigl (B_i(y)\bigr )=-\alpha _i\), for all \(x \in A|H_i\) and all \(y \in B|H_i\); thus \(A_i(x)\) and \(B_i(y)\) are congruous for any \(i=1, \dots , n\) since they satisfy condition (iii) in Definition 2.1 (see Fig. 2).

Fig. 2
figure 2

A set A (in gray) contained in a centered box P such that \(P{\setminus } A\) is unconditional

Full size image

The following result is well-known in the literature (see e.g. the one-dimensional case of [6, Theorem 4.1] and the references therein. Regarding its statement, and following the notation used in [6], we notice that for a quasi-concave function \(\phi :{\mathbb {R}}\longrightarrow [0,\infty )\) we have \((1-\lambda )\phi \chi _{_A}\star _{-\infty }\lambda \phi \chi _{_B}=\phi \chi _{_{(1-\lambda )A+\lambda B}}\), where \(\chi _{_M}\) denotes the characteristic function of the set \(M\subset {\mathbb {R}}\)).

Lemma 2.1

Let \(\mu \) be the measure on \({\mathbb {R}}\) given by \(\mathrm {d}\mu (x)=\phi (x)\mathrm {d}x\), where \(\phi :{\mathbb {R}}\longrightarrow [0,\infty )\) is quasi-concave with \(\phi (0)=\max _{x\in {\mathbb {R}}}\phi (x)\). Let \(\lambda \in (0,1)\) and let \(A, B\subset {\mathbb {R}}\) be measurable sets with \(0\in A\cap B\). Then

$$\begin{aligned} \mu (C)\ge (1-\lambda )\mu (A)+\lambda \mu (B) \end{aligned}$$

for any measurable set C such that \(C\supset (1-\lambda )A+\lambda B\).

As a consequence of such a Brunn–Minkowski inequality for quasi-concave densities on \({\mathbb {R}}\), we will obtain the one-dimensional Brunn–Minkowski inequality for measures associated to quasi-convex functions when working with congruous sets. This is the content of Lemma 2.2.

2.2 Proof

We start this subsection by showing the one-dimensional case of our main result, Theorem 1.1.

Lemma 2.2

Let \(\mu \) be the measure on \({\mathbb {R}}\) given by \(\mathrm {d}\mu (x)=\phi (x)\mathrm {d}x\), where \(\phi :{\mathbb {R}}\longrightarrow [0,\infty )\) is quasi-convex with \(\phi (0)=\min _{x\in {\mathbb {R}}}\phi (x)\). Let \(\lambda \in (0,1)\) and let \(A, B\subset {\mathbb {R}}\) be non-empty measurable congruous sets. Then

$$\begin{aligned} \mu (C)\ge (1-\lambda )\mu (A)+\lambda \mu (B) \end{aligned}$$

for any non-empty measurable set C such that \(C\supset (1-\lambda )A+\lambda B\).

Proof

Let A and B satisfy condition (iii) in Definition 2.1. Assuming that the result is true if either (i) or (ii) (of Definition 2.1) holds, it is enough to consider \(A^+, A^-, B^+, B^-, C^+,C^-\) where, for any \(M\subset {\mathbb {R}}\), the sets \(M^+\) and \(M^-\) stand for \(M^+=M\cap (0,\infty )\) and \(M^-=M\cap (-\infty ,0)\). Indeed, applying the result to the sets \(A^+, B^+, C^+\) and \(A^-, B^-, C^-\), respectively, we have

$$\begin{aligned} \begin{aligned} (1-\lambda )\mu (A)+\lambda \mu (B)&=(1-\lambda )\mu (A^+)+\lambda \mu (B^+)+(1-\lambda )\mu (A^-)+\lambda \mu (B^-)\\&\le \mu (C^+)+\mu (C^-)= \mu (C). \end{aligned} \end{aligned}$$

Moreover, we note that the function \({\bar{\phi }}:{\mathbb {R}}\longrightarrow [0,\infty )\) given by \({\bar{\phi }}(x)=\phi (-x)\) is quasi-convex (and, clearly, \({\bar{\phi }}(0)=\min _{x\in {\mathbb {R}}}{\bar{\phi }}(x)\)). Thus, considering if necessary \({\bar{A}}=-A\), \({\bar{B}}=-B\), \({\bar{C}}=-C\), and the measure \({\bar{\mu }}\) with density \({\bar{\phi }}\), it is enough to prove the result for congruous sets satisfying (i). Now, the quasi-convexity of \(\phi \) implies that \(\phi (x)\le \max \{\phi (0),\phi (y)\}=\phi (y)\) for any \(0<x<y\). This shows that \(\phi \) is increasing on \((0,\infty )\) and then \(\phi \cdot \chi _{_{(0,\infty )}}\) is quasi-concave. Thus, setting \(x_0=\max (A)=\max (B)\), the result follows from applying Lemma 2.1 to the function \(\psi :{\mathbb {R}}\longrightarrow [0,\infty )\) given by \(\psi (x)=\phi (x+x_0)\cdot \chi _{_{(-\infty ,0]}}(x)\) and the sets \(A-x_0, B-x_0, C-x_0\). \(\square \)

As stated in Theorem 1.1, the above result extends to dimension n. The approach we follow here is based on the underlying idea of [16, Theorem 1.3], and it goes back to some classical proofs of functional versions of the Brunn–Minkowski inequality (such as the Prékopa-Leindler inequality) and other related results.

Proof of Theorem 1.1

For the sake of brevity we write \(C=(1-\lambda )A+\lambda B\) and, given \(t_1, t_2\in {\mathbb {R}}\), \(t_\lambda =(1-\lambda )t_1+\lambda t_2\). We also set \({\bar{\mu }}=\mu _1\otimes \mu _2\otimes \dots \otimes \mu _{n-1}\) (i.e., \(\mu ={\bar{\mu }}\otimes \mu _n\)).

Since \(\mu \) is inner regular (i.e., \(\mu (A)=\sup \{\mu (K):\,K\subset A, K \text { compact}\}\) for any measurable set A), we may assume, without loss of generality, that A and B are compact. Indeed, given sequences of compact sets \((K_n)_{n\in {\mathbb {N}}}\), \((L_n)_{n\in {\mathbb {N}}}\) that approximate from inside the congruous sets A and B, respectively, one may clearly consider certain sequences of congruous compact sets \((K_n^\prime )_{n\in {\mathbb {N}}}\), \((L_n^\prime )_{n\in {\mathbb {N}}}\) such that \(\mu (K_n^\prime )=\mu (K_n)\) and \(\mu (L_n^\prime )=\mu (L_n)\), for all \(n\in {\mathbb {N}}\). In fact, it is enough to add to \(K_n\) and \(L_n\), respectively, the projections \(\bigl (A|H_i\bigr )\) and \(\bigl (B|H_i\bigr )\), located at the appropriate height(s) in the direction of \(\mathrm {e}_i\), for \(i=1,\dots ,n\).

Moreover, we observe that we may assume that \(\mu (A)\mu (B)>0\). Indeed, the case in which one of the sets, say B, has measure zero whereas the other one, A, has positive measure can be obtained (cf. [16, Proposition 2.7]) by applying the positive measures case to A and the following set: let P be an orthogonal compact box congruous with B (and so, with A) and let \(C_m\) be a decreasing sequence of (unions of) boxes, which are congruous with B, that shrinks (as \(m\rightarrow \infty \)) to the subset of vertices of P that belong to B; then we take \(B_m=B\cup C_m\), which is also congruous with A for all \(m\in {\mathbb {N}}\). We note that this congruence ensures that the points in the limit case belong to B, and hence \(\bigcap _{m\in {\mathbb {N}}}\bigl ((1-\lambda )A+\lambda B_m\bigr )=(1-\lambda )A+\lambda B\). Taking into account that

$$\begin{aligned} \mu \left( \bigcap _{m\in {\mathbb {N}}}\bigl ((1-\lambda )A+\lambda B_m\bigr )\right) =\lim _m\mu \bigl ((1-\lambda )A+\lambda B_m\bigr ), \end{aligned}$$

we get (1.2).

We then show the result by (finite) induction on the dimension n. The case \(n=1\) is just Lemma 2.2. So, we suppose that \(n\ge 2\) and that the inequality is true for dimension \(n-1\). The sets \(A(t_1), B(t_2)\), for \(t_1, t_2\in {\mathbb {R}}\) such that \(t_1\mathrm {e}_{n}\in A|H_{n}^\bot \), \(t_2\mathrm {e}_n\in B|H_{n}^\bot \), are clearly congruous and thus, applying the induction hypothesis (i.e., (1.2) in \({\mathbb {R}}^{n-1}\) for \({\bar{\mu }}\)) together with the fact that \(C(t_\lambda )\supset (1-\lambda )A(t_1)+\lambda B(t_2)\), we have

$$\begin{aligned} {\bar{\mu }}\bigl (C(t_\lambda )\bigr ) \ge \left( (1-\lambda ){\bar{\mu }}\bigl (A(t_1)\bigr )^{1/(n-1)} +\lambda {\bar{\mu }}\bigl (B(t_2)\bigr )^{1/(n-1)}\right) ^{n-1}. \end{aligned}$$
(2.3)

Now, we take the non-negative functions \(f,g,h:{\mathbb {R}}\longrightarrow [0,\infty )\) given by

$$\begin{aligned} f(t)=\frac{{\bar{\mu }}(A(t))}{\left| {{\bar{\mu }}(A(\cdot ))}\right| _\infty }, \ g(t)=\frac{{\bar{\mu }}(B(t))}{\left| {{\bar{\mu }}(B(\cdot ))}\right| _\infty }, \ h(t)=\frac{{\bar{\mu }}(C(t))}{c}, \end{aligned}$$

where

$$\begin{aligned} c=\left( (1-\lambda )\left| {{\bar{\mu }}(A(\cdot ))}\right| _\infty ^{1/(n-1)}+ \lambda \left| {{\bar{\mu }}(B(\cdot ))}\right| _\infty ^{1/(n-1)}\right) ^{n-1}. \end{aligned}$$

We notice that the above functions are well-defined: denominators are positive since \(\mu (A)\mu (B)>0\), and they are finite because \(A|H_{n-1}\) and \(B|H_{n-1}\) are compact sets and \({\bar{\mu }}\) is locally finite. Furthermore, \(\sup _{t\in {\mathbb {R}}}f(t)=\sup _{t\in {\mathbb {R}}}g(t)=1\).

Using (2.3), and setting \(\theta =\displaystyle \frac{\lambda \left| {{\bar{\mu }}(B(\cdot ))}\right| _\infty ^{1/(n-1)}}{c^{1/(n-1)}}\in (0,1)\), we get

$$\begin{aligned} \begin{aligned} {\bar{\mu }}\bigl (C(t_\lambda )\bigr )&\ge \left( (1-\lambda ){\bar{\mu }}\bigl (A(t_1)\bigr )^{1/(n-1)}+\lambda {\bar{\mu }}\bigl (B(t_2)\bigr )^{1/(n-1)}\right) ^{n-1}\\&=c\left( (1-\theta )f(t_1)^{1/(n-1)}+\theta g(t_2)^{1/(n-1)}\right) ^{n-1}\\&\ge c\,\min \bigl \{f(t_1),g(t_2)\bigr \}. \end{aligned} \end{aligned}$$

This shows that \(h((1-\lambda )t_1+\lambda t_2)\ge \min \{f(t_1),g(t_2)\}\) for any \(t_1, t_2\in {\mathbb {R}}\), which clearly implies that

$$\begin{aligned} \{t\in {\mathbb {R}}: h(t)\ge s\}\supset (1-\lambda )\{t\in {\mathbb {R}}: f(t)\ge s\}+\lambda \{t\in {\mathbb {R}}: g(t)\ge s\} \end{aligned}$$
(2.4)

for all \(s\in [0,1)\). Moreover, since \(A_n(x)\) and \(B_n(y)\) are congruous for all \(x \in A|H_n\) and all \(y \in B|H_n\) then the superlevel sets \(\{t\in {\mathbb {R}}:f(t)\ge s\}\) and \(\{t\in {\mathbb {R}}: g(t)\ge s\}\) are also congruous for any \(s\in [0,1)\). Indeed, assuming without loss of generality that \(A_n(x), B_n(y)\) satisfy condition (i) of Definition 2.1, for all \(x \in A|H_n\) and all \(y \in B|H_n\), then there exists \(s_0>0\) such that \(\bigl (A|H_n\bigr )+s_0\mathrm {e}_n\subset A\), \(\bigl (B|H_n\bigr )+s_0\mathrm {e}_n\subset B\) and \(A, B \subset [0, s_0\mathrm {e}_n]+H_n\). Hence, both f and g attain their maximum at \(s_0\) and vanish on \((-\infty ,0)\cup (s_0,\infty )\), which implies that their superlevel sets satisfy condition (i) of Definition 2.1 and thus are congruous.

Therefore, we may apply Lemma 2.2 to get

$$\begin{aligned} \mu _{n}\bigl (\{t\in {\mathbb {R}}: h(t)\ge s\}\bigr )\ge (1-\lambda )\mu _{n}\bigl (\{t\in {\mathbb {R}}: f(t)\ge s\}\bigr )+\lambda \mu _{n}\bigl (\{t\in {\mathbb {R}}: g(t)\ge s\}\bigr ) \end{aligned}$$

for any \(s\in [0,1)\). This, together with Fubini’s theorem and the Cavalieri Principle

$$\begin{aligned} \int _{{\mathbb {R}}} \psi (x)\,\mathrm {d}\mu _n(x) = \int _0^{\left| {\psi }\right| _\infty }\mu _n\bigl (\{t\in {\mathbb {R}}: \psi (t)\ge s\}\bigr )\, \mathrm {d}s \end{aligned}$$

for \(\psi = f, g, h\), jointly with the fact that \(\left| {h}\right| _\infty \ge 1=\left| {f}\right| _\infty =\left| {g}\right| _\infty \) (cf. (2.4)), allows us to obtain

$$\begin{aligned} \begin{aligned} \mu \bigl ((1-\lambda )A+\lambda B\bigr )&=c\int _{{\mathbb {R}}} h(x)\,\mathrm {d}\mu _n(x)\\&\ge c\left( (1-\lambda )\int _{{\mathbb {R}}} f(x)\,\mathrm {d}\mu _n(x) + \lambda \int _{{\mathbb {R}}} g(x)\,\mathrm {d}\mu _n(x)\right) \\&=c\left( (1-\lambda )\frac{\mu (A)}{\left| {{\bar{\mu }}(A(\cdot ))}\right| _\infty }+ \lambda \frac{\mu (B)}{\left| {{\bar{\mu }}(B(\cdot ))}\right| _\infty }\right) . \end{aligned} \end{aligned}$$

And then, applying the (reverse) Hölder inequality (see e.g. [5, Theorem 1, page 178]),

$$\begin{aligned} a_1b_1 + a_2b_2 \ge \left( a_1^{p} + a_2^{p}\right) ^{1/p} \left( b_1^q + b_2^q \right) ^{1/q}, \end{aligned}$$

with parameters \(p=1/n\) and \(q=-1/(n-1)\), and taking \(a_1 = (1-\lambda )^{1/p} \mu (A)\), \(a_2 = \lambda ^{1/p} \mu (B)\), \(b_1 = (1-\lambda )^{1/q}\left| {{\bar{\mu }}(A(\cdot ))}\right| _\infty ^{-1}\) and \(b_2 = \lambda ^{1/q}\left| {{\bar{\mu }}(B(\cdot ))}\right| _\infty ^{-1}\), we conclude that

$$\begin{aligned} \mu \bigl ((1-\lambda )A+\lambda B\bigr )\ge \left( (1-\lambda )\mu (A)^{1/n}+\lambda \mu (B)^{1/n}\right) ^{n}, \end{aligned}$$

as desired. \(\square \)

3 A remark on an isoperimetric inequality

Given a set \(M\subset {\mathbb {R}}^n\), let \(\mathop {\mathrm {pos}}\nolimits M\) and \({{\,\mathrm{int}\,}}M\) denote, respectively, the positive hull and interior of M. Moreover, let \(\varepsilon _1,\dots ,\varepsilon _{2^n}\) denote the elements of \(\{-1,1\}^n\). Then, setting \(\varepsilon _j=(\varepsilon _1^j,\dots ,\varepsilon _n^j)\) for any \(j=1,\dots ,2^n\), we write

$$\begin{aligned} O_j=\mathop {\mathrm {pos}}\nolimits \{\varepsilon _1^j\mathrm {e}_1,\dots ,\varepsilon _n^j\mathrm {e}_n\} \end{aligned}$$

for the corresponding orthant of \({\mathbb {R}}^n\).

Along this section, we deal with certain sets contained in an orthogonal compact box (which, for the sake of simplicity, will be assumed to be centered): fixing a box \(P=\prod _{i=1}^n[-\alpha _i,\alpha _i]\), with \(\alpha _i>0\) for all i, we consider unions of orthants of unconditional compact convex sets ‘embedded’ in the corners of P. More precisely, such a set A satisfies that, for all \(j=1,\dots ,2^n\),

$$\begin{aligned} A\cap O_j=x_j+(K_j\cap (-O_j)) \end{aligned}$$
(3.1)

for some unconditional compact convex set \(K_j\subset {{\,\mathrm{int}\,}}P\) (cf. Fig. 3), where \(x_j=(\varepsilon _1^j\alpha _1,\dots ,\varepsilon _n^j\alpha _n)\) is the corresponding vertex of P. In the following, for the sake of brevity, we will write \(A_j=A\cap O_j\).

Fig. 3
figure 3

Union of orthants \(A_j\) of unconditional compact convex sets (left) and the corresponding orthants of balls \(x_j+r_j(B_n\cap (-O_j))\) of the same measure (right)

Full size image

As in the Euclidean setting, we will obtain an isoperimetric type inequality as a consequence of (1.2). To this aim, we introduce some notation. Let

$$\begin{aligned} \mathrm {W}^{\mu }_1(A;B)=\frac{1}{n}\liminf _{t\rightarrow 0^+}\frac{\mu (A+tB)-\mu (A)}{t} \end{aligned}$$

be the first quermassintegral of A with respect to the set B associated to the measure \(\mu \). Here we assume that A and B are measurable sets such that \(A+tB\) is also measurable for all \(t\ge 0\).

In a similar way, and denoting by \(B_n\) the n-dimensional Euclidean (closed) unit ball, we may define

$$\begin{aligned} \mu ^+(A)=\liminf _{t\rightarrow 0^+}\frac{\mu (A+tB_n)-\mu (A)}{t}, \end{aligned}$$

the surface area measure associated to \(\mu \), i.e., its (lower) Minkowski content. Clearly, \(\mu ^+(A)=n\mathrm {W}^{\mu }_1(A;B_n)\). The relative Minkowski content of a set \(A\subset {\mathbb {R}}^n\) with respect to a second set \(\Omega \subset {\mathbb {R}}^n\) is defined by

$$\begin{aligned} \mu ^+(A,\Omega )=\liminf _{t\rightarrow 0^+}\frac{\mu \bigl ((A+tB_n)\cap \Omega \bigr )-\mu (A\cap \Omega )}{t}. \end{aligned}$$

Moreover, given \(x\in {\mathbb {R}}^n\), we set

$$\begin{aligned} M^\mu (x,A)=n\mu (x+A)-\frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!-}\Bigg |_{t=1}\,\mu (x+tA), \end{aligned}$$

provided that \(((x,A),\mu )\) is so that the above (left) derivative exists. When dealing with a set \(A\subset {\mathbb {R}}^n\) satisfying (3.1) for all \(j=1,\dots ,2^n\), we also write \(M^\mu (A)=\sum _{j=1}^{2^n} M_j^\mu (A_j)\), where \(M_j^\mu (A_j)=M^\mu (x_j,K_j\cap (-O_j))\). We notice that, from the convexity of \(K_j\cap (-O_j)\) and using Theorem 1.1, the function \(t\mapsto \mu (x_j+t(K_j\cap (-O_j)))^{1/n}\) is (increasing and) concave on (0, 1] for any product measure \(\mu \) in the conditions of the latter result. This implies that the left derivative of \(\mu (x_j+t(K_j\cap (-O_j)))\) at \(t=1\) (possibly infinite) exists (cf. [17, Theorem 23.1]) and hence, for all \(j=1,\dots ,2^n\), \(M_j^\mu (A_j)\) (and so \(M^\mu (A)\)) is well-defined. Clearly, \(M^{\mathrm {vol}}(A)=0\) for such a set A and thus this functional does not appear in the classical isoperimetric inequality. For more information about this functional, we refer the reader to [11, 16] and the references therein.

Now we show an isoperimetric type inequality for unions of orthants of unconditional compact convex sets embedded in the corners of a fixed orthogonal box, in the setting of product measures with quasi-convex densities. This a straightforward consequence of the following result for (such) a sole orthant.

Theorem 3.1

Let \(\mu =\mu _1\otimes \dots \otimes \mu _n\) be a product measure on \({\mathbb {R}}^n\) such that \(\mu _i\) is the measure given by \(\mathrm {d}\mu _i(x)=\phi _i(x)\,\mathrm {d}x\), where \(\phi _i:{\mathbb {R}}\longrightarrow [0,\infty )\) is quasi-convex with \(\phi _i(0)=\min _{x\in {\mathbb {R}}}\phi _i(x)\), for all \(i=1,\dots ,n\).

Let \(P=\prod _{i=1}^n[-\alpha _i,\alpha _i]\), with \(\alpha _i>0\) for all i and let \(K\subset {{\,\mathrm{int}\,}}P\) be a non-empty unconditional compact convex set. Let \(A=x_1+(K\cap (-O_1))\), where \(x_1=(\alpha _1,\dots ,\alpha _n)\) and \(O_1=\mathop {\mathrm {pos}}\nolimits \{\mathrm {e}_1,\dots ,\mathrm {e}_n\}\). Then, for any \(r>0\) such that \(rB_n\subset {{\,\mathrm{int}\,}}P\),

$$\begin{aligned} r\mu ^+(A,P)+ M^\mu (x_1,K_1\cap (-O_1))\ge n\mu (A)^{1-1/n}\mu \bigl (x_1+(rB_n\cap (-O_1))\bigr )^{1/n}, \end{aligned}$$

with equality if \(A=x_1 + (rB_n\cap (-O_1))\).

Following the same argument for any orthant \(A_j\) of a non-empty set \(A\subset P\) satisfying (3.1) for all \(j=1,\dots ,2^n\), we get that, for any \(r_1,\dots ,r_{2^n}>0\) such that \(r_jB_n\subset {{\,\mathrm{int}\,}}P\) for all j, we have

$$\begin{aligned} \sum _{j=1}^{2^n}\left( r_j\mu ^+(A_j,P)+ M_j^\mu (A_j)\right) \ge n\sum _{j=1}^{2^n}\mu (A_j)^{1-1/n}\mu \bigl (x_j+(r_jB_n\cap (-O_j))\bigr )^{1/n}, \end{aligned}$$

with equality if \(A_j=x_j + (r_jB_n\cap (-O_j))\) for all \(j=1,\dots ,2^n\).

The particular case \(r_1=\dots =r_{2^n}(=:r)\) of the latter inequality shows that

$$\begin{aligned} r\mu ^+(A,P)+ M^\mu (A)\ge n\sum _{j=1}^{2^n}\mu (A_j)^{1-1/n}\mu \bigl (x_j+(rB_n\cap (-O_j))\bigr )^{1/n}. \end{aligned}$$

In other words: among all unions A of orthants of unconditional compact convex sets embedded in the corners of a fixed centered orthogonal box P (i.e., satisfying (3.1) for all \(j=1,\dots ,2^n\)) with predetermined measure\(\mu (A_j)=\mu (x_j+(rB_n\cap (-O_j))\), (union of orthants embedded in the corners of P of) Euclidean balls\(rB_n\)minimize the functional\(r\mu ^+(A,P)+ M^\mu (A)\).

The main idea of the proof we present here goes back to the classical proof of the Minkowski first inequality that can be found in [20, Theorem 7.2.1]. We refer also the reader to [16, Sect. 4] and the references therein.

Proof

We consider \(L=rB_n\) and we denote by \(B=x_1+L^-\), where \(L^-=L\cap (-O_1)\). In the same way, we will write \(K^-=K\cap (-O_1)\).

Notice that, for any \(\epsilon >0\) such that \(K^- + \epsilon L^-\subset P\), we have that \(x_1+K^- + t_1L^-\) and \(x_1+K^- + t_2L^-\) are congruous for all \(t_1,t_2\in [0,\epsilon ]\) (since each one-dimensional section of them in the direction of \(\mathrm {e}_i\), \(i=1,\dots ,n\), satisfies condition (i) in Definition 2.1, with maximum equal to \(\alpha _i\)). Then, from the convexity of \(L^-\) (and \(K^-\)) and using Theorem 1.1, the function \(t\mapsto \mu (A+tL^-)^{1/n}\) is concave on \([0,\epsilon ]\). This implies that the right derivative of \(\mu (A+tL^-)\) at \(t=0\) (possibly infinite) exists (cf. [17, Theorem 23.1]). Similarly, the left derivative of \(\mu (x_1+tK^-)\) at \(t=1\) exists.

Now, we consider the function \(f:[0,1]\longrightarrow {\mathbb {R}}_{\ge 0}\) given by

$$\begin{aligned} f(t)=\mu \bigl ((1-t)A+tB\bigr )^{1/n}-\bigl ((1-t)\mu (A)^{1/n}+t\mu (B)^{1/n}\bigr ). \end{aligned}$$

By Theorem 1.1 (and from the convexity of both \(K^-\) and \(L^-\)) f is concave (we notice that the fact of being an unconditional set is closed under convex combinations) and, moreover, \(f(0)=f(1)=0\). Thus, the right derivative of f at \(t=0\) exists and furthermore

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!+}\Bigg |_{t=0}\,f(t)\ge 0 \end{aligned}$$
(3.2)

with equality if and only if \(f(t)=0\) for all \(t\in [0,1]\), i.e., if and only if (1.2) holds with equality for all \(t\in [0,1]\).

Now, since

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!+}\Bigg |_{t=0}\,f(t)=\frac{1}{n}\mu (A)^{(1/n)-1}\frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!+}\Bigg |_{t=0}\,\mu \bigl ((1-t)A+tB\bigr )+\mu (A)^{1/n}-\mu (B)^{1/n}, \end{aligned}$$

we just must compute the right derivative at 0 of \(\mu \bigl ((1-t)A+tB\bigr )\). Writing \(g(r,s)=\mu \bigl (x_1+r(K^- +sL^-)\bigr )\), we have

$$\begin{aligned} \begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!+}\Bigg |_{t=0}\,\mu \bigl ((1-t)A+tB\bigr )&=\frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!+}\Bigg |_{t=0}\, g\left( 1-t,\frac{t}{1-t}\right) \\&=-\frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!-}\Bigg |_{t=1}\,\mu (x_1+tK^-)+\frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!+}\Bigg |_{t=0}\,\mu (A+tL^-)\\&=M^\mu (x_1,K^-)-n\mu (A)+ n\mathrm {W}^{\mu }_1(A;L^-), \end{aligned} \end{aligned}$$

and thus

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}^{\!\!\!+}\Bigg |_{t=0}\,f(t)=\frac{1}{n}\mu (A)^{(1/n)-1}\bigl (M^\mu (x_1,K^-)+ n\mathrm {W}^{\mu }_1(A;L^-)\bigr )-\mu (B)^{1/n}. \end{aligned}$$

Hence, the latter identity, together with (3.2), gives

$$\begin{aligned} \mathrm {W}^{\mu }_1(A;L^-)+\frac{1}{n}M^\mu (x_1,K^-)\ge \mu (A)^{1-1/n}\mu (B)^{1/n}, \end{aligned}$$

with equality if \(A=B\).

Finally, from the unconditionality of \(K^-\) we clearly have that \(\bigl ((A+tL)\cap P\bigr )=A+tL^-\), which yields \(n\mathrm {W}^{\mu }_1(A;L^-)=r\mu ^+(A,P)\). Then, we have

$$\begin{aligned} r\mu ^+(A,P)+ M^\mu (x_1,K_1\cap (-O_1))\ge n\mu (A)^{1-1/n}\mu \bigl (x_1+(rB_n)^-\bigr )^{1/n}, \end{aligned}$$

with equality if \(A=x_1 + (rB_n\cap (-O_1))\). This concludes the proof. \(\square \)