The isoperimetric problem for nonlocal perimeters

We consider a class of nonlocal generalized perimeters which includes fractional perimeters and Riesz type potentials. We prove a general isoperimetric inequality for such functionals, and we discuss some applications. In particular we prove existence of an isoperimetric profile, under suitable assumptions on the interaction kernel.


Introduction
In this paper we consider a family of geometric functionals, which in particular contains the fractional isotropic and anisotropic perimeter. More precisely, we define the following energy defined on measurable subsets E ⊂ R N : where the kernel K : R N → [0, +∞) satisfies the following assumptions: K(x) = K(−x) (2) min(|x|, 1) K(x) ∈ L 1 (R N ).
The functional (1) measures the interaction between points in E and in R N \ E, weighted by the kernel K.
Note that it is not restrictive to assume (2) since Per K (E) = Per K (E) for every E, where K(x) := (K(x)+K(−x))/2. Notice also that, if K ∈ L 1 (R N ), then for every E with |E| < ∞, we have (4) Per In the first part of the paper we deal with isoperimetric inequalities for such functionals. The main result is the following (see Corollary 3.4): if K(x) ≥ µχ Br (x) for some constants µ > 0 and r > 0, then for all measurable sets E there holds Per K (E) ≥ min(g(|E|), g(|R N \ E|)), where g(m) := Per K ⋆ (B m ), with K ⋆ the symmetric decreasing rearrangment of K, and B m the ball with volume m centered at 0. We discuss some property of the function g and we provide a Poincaré type inequality (see Proposition 4.1).
We recall that, in the case of fractional perimeters, sharp quantitative isoperimetric inequality, uniform with respect to the fractional exponent bounded away from 0, have been obtained in [9] (see also [11] for an anisotropic version), whereas Poincaré type inequalities have been discussed in [13].
An interesting related question is understanding which conditions on K imply the compact embedding of the functions with bounded energy J K into L p spaces, for some p ≥ 1.
In the second part of the paper, we consider the isoperimetric problem (5) min for a fixed volume m > 0.
In the case of the fractional perimeter, the existence of isoperimetric sets solving (5) has been studied in [2,9] (see also [4] where a bulk term is added to the energy), where it is shown that balls are the unique minimizers of the fractional perimeter among sets with the same volume. In the general case, the same result holds if the kernel K is a radially symmetric decreasing function, as a straightforward consequence of the Riesz rearrangement inequality [14]. So, we focus on the case in which K is not radially symmetric and decreasing. We provide an existence result of minimizers of the relaxed problem associated to (5) under the additional assumption that K ∈ L 1 (R N ) (see Theorem 5.6). The proof is based on a concentration compactness type argument. Finally, we show that if K has maximum at the origin (in an appropriate sense, see condition (38)), then every minimizer of the relaxed problem is actually the characteristic function of a compact set (see Theorem 5.7).
We are left with the open problem of extending the existence result to more general interaction kernels satisfying only (3).
Another interesting problem is to consider kernels which are just Radon measures on R N . In this case we don't expect in general compactness of minimizers.
Notation. We denote by B m (x) the ball centered at x with volume m, that is, the ball with radius r = m 1 N ω − 1 N N , and by B m for the ball centered at 0 with volume m. We also denote by B(x, r) the ball of center x and radius r.
For every measurable set E ⊆ R N , χ E denotes the characteristic function of E, that is the function which is 1 on E and 0 outside.
We recall that given a set E with |E| < ∞, its symmetric rearrangement E ⋆ is the ball B |E| that is the ball centered at 0 with volume |E|. Moreover the symmetric decreasing rearrangement of a nonnegative measurable function h with level sets of finite measure is defined as Note that if h is radially symmetric and decreasing, then , for all p ≥ 1.

Generalized fractional perimeters
In this section we discuss some properties of the K perimeters.
where C is a constant which depends on E.
Proposition 2.2. The following properties hold: Proof. We start by proving 1. The first equality is a direct consequence of the definition of Per K . In order to prove (6), we observe that . Therefore which gives (6). The proof of 2. is a consequence of Fatou lemma, observing that Per K (E) = R N R N |χ E (x)− χ E (y)|K(x − y)dxdy.

Examples. A first class of examples is given by the kernels K(x) which satisfies
for some s ∈ (0, 1) and 0 < λ ≤ Λ. This class includes the fractional perimeters, and its inhomogeneous and anisotropic versions.
The fractional perimeter, which has been introduced in [13] and further developed in [2], is defined as for s ∈ (0, 1). It is also possible to substitute the kernel 1 |x−y| N+s with more general heterogeneous, isotropic kernels of the type where a : R N → (0, +∞) is a measurable function such that 0 < λ ≤ a(x) ≤ Λ. The anisoptropic fractional perimeters have been defined in [11] as follows: let B ⊆ R N be a convex set which is symmetric with respect to the origin and let | · | B the norm in R N with unitary ball B, then we define Another class of examples, relevant for this paper, is given by the kernels K(x) ∈ L 1 (R N ), for which the representation formula (4) holds.

Coarea formula.
We introduce the following functional on functions u ∈ L 1 loc (R N ): We provide a coarea formula, linking the functional Per K to J K .

Proposition 2.3 (Coarea formula). The following formula holds
Proof. First of all we observe that, for every measurable function u, Moreover for every s ∈ R Therefore we get, recalling (2), and using Tonelli theorem,

Isoperimetric inequality
In this section we prove an isoperimetric inequality for generalized nonlocal perimeters.
where K * is the symmetric decreasing rearrangement of K.
In particular, if K is radially symmetric and decreasing, then

Moreover, equality holds if and only if E is a translated of B |E| .
Proof. First of all we consider the case in which K ∈ L 1 (R N ). Note that (χ E ) ⋆ = χ B |E| . By Riesz rearrangement inequality [14], we get that So, recalling (4) we get the conclusion. Finally, if K = K ⋆ , we have that equality in the Riesz rearrangement inequality holds if and only if χ E is equal, up to translation, to its symmetric-decreasing rearrangement, therefore if and only if E is equal, up to translation, to B |E| .
So as ε → 0, also K ⋆ ε → K ⋆ monotonically increasing. Therefore by the monotone convergence theorem if E is a measurable set we get that . By the previous argument we get Per Kε (E) ≥ Per K ⋆ ε (B |E| ). So, we conclude sending ε → 0.
For every m ≥ 0, we define g(m) := Per K ⋆ (B m ) where we recall that B m is the ball centered at 0 with volume m. We also set g(+∞) = +∞.
We provide some estimates on the function g.
Proof. For x ∈ R N and m > 0 we define We have which gives (11) and (12), sending m → 0.
Proposition 3.3. Assume that the kernel K satisfies the following condition: Proof. Let {Q i } i∈N be a partition of R N made of cubes of sidelength r/ √ n, where r is as in (13). Note that for all x, y ∈ Q i we have K(x − y) ≥ µ. Assume by contradiction that |E| = |R N \ E| = ∞. Then three possible cases may verify: This implies, recalling that Per Then the argument is the same as in Case 1, substituting E with R N \ E.
Case 3: assume that here exist two subsequences Q in , Q jn such that lim n |Q in ∩ E| = lim n |Q jn \ E| = |Q i |. Therefore for δ > 0 there exists i 0 such that for all j n , i n ≥ i 0 , we get giving a contradiction to Per K (E) < ∞.
As a consequence, neither of the three cases can arise, which implies that either |E| or |R N \ E| are finite.
From Propositions 3.1 and 3.3 we immediately get the following result: Corollary 3.4. Assume that K satisfies condition (13). Then, for all measurable sets E there holds

Poincaré inequality
Proposition 4.1. Assume that the kernel K satisfies (13) and that there exist k ≥ 1 and a constant C depending on k, N such that Then, for all u ∈ L 1 Proof. The argument is similar to the one in [1,Theorem 3.47]. First of all we observe that, since J K (u) < ∞, by the coarea formula (10) the set S of s ∈ R such that Per K ({x |u(x) > s}) < ∞ is dense in R. So by Proposition 3.3 for every s ∈ S, either |{u(x) > s}| < ∞ or |{u(x) ≤ s}| < ∞. Note that if s > m(u), then there exists t ∈ (m(u), s) such that |{u(x) > t}| < ∞ and then |{u(x) > s}| < ∞. Analogously, if s < m(u), then |{u(x) ≤ t}| < ∞. Moreover m(u) ∈ R. Indeed, if by contradiction this were not true, and e.g. m(u) = −∞ (the other case being similar), we would get that |{u(x) > t}| < ∞ for every t ∈ R. By the coarea formula −n −n−1 so there exist r > 0 and t n ∈ [−n − 1, −n] such that Per K ({u(x) > t n } ≤ r. By the isoperimetric inequality |{u(x) > t n }| ≤ (Cr) k , but this is contradiction with the fact that {u(x) > t n } → R N as n → +∞. Finally, if u ∈ L 1 (R N ), m(u) = 0. Indeed, by Chebychev inequality for every s > 0, we have that We denote by u + = max(u − m(u), 0) the positive part of u − m(u). Then, by definition of m(u), we get that |{x |u + (x) > s}| < ∞ for all s > 0.

Existence of an isoperimetric profile
In this section we show the existence of an isoperimetric profile, that is, a solution to Problem (5), under suitable assumptions on the kernel K. First of all we will assume throughout this section that K ≡ 0 and We observe that, if K = K ⋆ , then by Proposition 3.1 we know that the ball of volume m is the unique minimizer of (5), up to translations.
In order to get existence of minimizers, we first consider a relaxed version of the perimeter functional, obtained by extending it to general densities functions. More precisely, we define the new energy as follows: given f : Note that P K (χ E ) = Per K (E) and the constraint 0 ≤ f ≤ 1 is inherited by the original problem, naturally arising from the relaxation procedure. Note that the previous energy can be written as The relaxed version of the isoperimetric problem (5) can be restated as follows. Given m ≥ 0, we consider where the set of admissible functions is defined as Notice also that lim inf where the liminf is taken over all sequences E n with |E n | = m, such that χ En * ⇀ f weakly * in L ∞ .
Due to (21), the minimization problem (22) is equivalent to the maximization problem We now show monotonicity and subadditivity of the energy in (23) with respect to m. (23) is attained for all volumes m i 's, then m i = 0 for all i's except one.

Moreover, if the equality holds in the above inequality and if the supremum in
We havef ∈ A m 1 . Moreover, since K ≥ 0, ii) We consider the case l = 2, as the case l > 2 can be treated analogously. Let f i ∈ A m i , and let ε > 0 and R > 0 such that .
from which we conclude by the arbitrariness of ε.
Note that, since K ≡ 0, we can always choose x R such that so that the last inequality in (24) is in fact a strict inequality.

The potential function.
Given a function f ∈ A m , we can define the potential of f as In the following we give some properties of the potential V .
This implies that V is continuous. ii) Let ε > 0 and let A ε := {f > ε}. We have For R > 0 we have K(y)dy.
iii) Since K ≡ 0, there exist δ > 0 and a bounded set A with |A| = k > 0 such that 0 ∈ A and K(z) ≥ δ for a.e. z ∈ A. Let x 0 ∈ R N be a Lebesgue point of f such that f (x 0 ) > 0. Let y 0 ∈ A be a point of density 1 in A, such that x 0 + y 0 is a density point of f and we compute If, on the other hand, f (x 0 + y 0 ) = 1 for all points y 0 which are points of density 1 of A and density points for f (x 0 + ·), then we consider y 1 ∈ A be a point of density 1 in A, such that x 0 + y 0 + y 1 is a density point of f and we compute Repeating this argument, we construct a sequence y n ∈ A of points of density 1 such that x 0 + y 0 + · · · + y n is a density point of f , and such that V (x 0 + y 0 + · · · + y n ) > 0. Note that for some n ≥ 0, we get that f (x 0 + y 0 + · · · + y n ) < 1. If it were not the case, we would get which is impossible.

First and second variation.
We now compute the first and second variation of the energy in (22).
dx, and such that ψ ≡ 0 a.e. in S and φ ≡ 0 a.e. in N , the following holds ii) There exists a constant c > 0 such that Proof. i) We argue as in [6,Lemma 1.2]. First observe that for every λ, Moreover, for all λ ∈ [0, 1] and a.e. x ∈ S ∪ N , we get that By minimality of f we get So, sending λ → 0 and recalling that K is symmetric (2), we get So, sending ε → 0 and recalling that R N (φ(x) − ψ(x))dx = 0, we conclude ii) Choosing ψ, φ in (29) such that ψ = 0 = φ on S ∪ N , we can exchange the role of ψ and φ, and obtain that in R N \ (N ∪ S), V has to be constant. So, there exists c > 0 (by Choosing φ in (29) such that φ = 0 a.e. in S ∪ N , we get, since R N (ψ(x) − φ(x))dx = 0, for all ψ ∈ L 1 (R N , [0, 1]) such that ψ = 0 a.e. in S and R N (ψ(x) − φ(x))dx = 0. This implies that V ≤ c in N . With an analogous argument, exchanging the role of ψ and φ, we get V (x) ≥ c in S.
As immediate consequence of the first variation (30) and of the properties of the potential V we obtain that every minimizer of (22) has compact support. Proposition 5.4. Every minimizer f ∈ A m of (22) has compact support.
Proof. By Proposition 5.2 lim |x|→+∞ V (x) = 0. Hence we can find R > 0 such that 0 ≤ V (x) < c for |x| > R, where c > 0 is the constant appearing in (30). By (30) this implies immediately that the support of f is contained in B R (0).
We now consider the second variation of the functional.
Lemma 5.5. Let f ∈ A m be a minimizer of (22) and let S, N be as in Lemma 5.3. Then, for every ξ ∈ L 1 (R N , [−1, 1]) with R N ξ(x)dx = 0, and such that ξ ≡ 0 a.e. in N ∪ S, it holds Proof. We argue as in [6,Lemma 1.5]. Reasoning as in the proof of Lemma 5.3, item i), we can assume that there exists a sequence ξ ε → ξ in So for λ sufficiently small f + λξ ε ∈ A m and by minimality we get Recalling (30) and the fact that R N ξ ε (x)dx = 0, we conclude the desired inequality by letting ε → 0.

Existence of minimizers.
Theorem 5.6. For every m > 0 there exists at least one f ∈ A m which solves the minimization problem (22).
Proof. The proof is similar to that in [6, Theorem 1.9] (see also [4]), and is based on a concentration compactness argument. Let f n ∈ A m be a minimizing sequence, and recall that the energy can be written as We consider a partition of R N in disjoint cubes: let Q = [0, 1] n and let Q z := z + Q for z ∈ Z N . We fix ε > 0 small and divide the cubes in two subsets: For z ∈ I ε,n and w ∈ Z N , recalling Riesz rearrangement inequality, we have where the last inequality follows from the fact that z ∈ I ε,n , and the previous inequality from the fact that K ⋆ is symmetrically decreasing.
For R > 0, we compute By (32) the second addendum can be bounded as follows On the other hand the first addendum can be bounded as Collecting the two estimates, we get Since K ∈ L 1 , hence also K ⋆ ∈ L 1 , we can choose R = R(ε) in such a way that lim ε→0 R(ε) = +∞ and lim ε→0 R(ε) N Bε K ⋆ (y)dy = 0.
Observe that, due to the fact that R N f (x)dx = m, we have #J ε,n ≤ m/ε. Given w i , w j ∈ J ε,n , up to subsequence we get that |w i − w j | → c i,j ∈ N ∪ {+∞} as n → +∞. We consider the following sets, for l = 1, . . . , Note that by construction dist(Q l ε,n , Q k ε,n ) → +∞ if k = l as n → +∞. Moreover, always by construction, we get that where M ε does not depend on n.
Let f l,ε n := f n χ Q l ε,n and let x l,n such that f l,ε n (x l,n ) > 0. Up to subsequences we can assume that f l,ε n (· + x n,l ) * ⇀ f l,ε weakly * in L ∞ , as n → +∞. Observe that the support of f l,ε n (· + x n,l ) is contained in B(0, M ε ) for every n. Moreover, since the functional P K is continuous with respect to the tight convergence (see for instance [6]), we get that lim n R N R N f l,ε n (x + x n,l )f l,ε n (y + x n,l )K(x − y)dxdy f l,ε n (x + x n,l )f l,ε n (y + x n,l )K(x − y)dxdy − 2r(ε) = m K L 1 − Hε l=1 R N R N f l,ε (x)f l,ε (y)K(x − y)dxdy − 2r(ε).
We pass to a subsequence ε k → 0 such that ε k is decreasing. So H ε k → H ∈ (0, +∞]. Moreover, we can relabel the sequence in such a way that f l,ε k n and then also their limit f l,ε k are monotone in ε k . By monotone convergence f l,ε k → f l strongly in L 1 . Moreover if m l = R N f l (x)dx, then H l=1 m l =m ≤ m. Again by continuity of the functional with respect to the L 1 -convergence, from (37) and from Lemma 5.1 we get that Therefore the previous are all equalities, and f l is a minimizer of P K in A m l for all l's. In particular, recalling again Lemma 5.1, we get that H = 1, and f 1 is a minimizer of P K in A m .
We finally show that, under a further condition on K, the isoperimetric problem (5) admits a solution.
Then, for every m > 0 there exists a compact set E ⊆ R N such that |E| = m and E solves the isoperimetric problem (5).
Proof. By Theorem 5.6 there exists at least one f ∈ A m which solves the minimization problem (22). Moreover the support of f is compact due to Proposition 5.4.