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A nonlocal approximation of the Gaussian perimeter: Gamma convergence and Isoperimetric properties

We study a non local approximation of the Gaussian perimeter, proving the Gamma convergence to the local one. Surprisingly, in contrast with the local setting, the halfspace turns out to be a volume constrained stationary point if and only if the boundary hyperplane passes through the origin. In particular, this implies that Ehrhard symmetrization can in general increase the considered non local Gaussian perimeter.


Introduction
The Gaussian isoperimetric inequality says that the halfspace has the smallest Gaussian perimeter among all sets with prescribed Gaussian measure, [4]. In the Euclidean setting, an increasing interest has been devoted to the study of non local approximations of the perimeter and their isoperimetric shapes, since the pioneering work of Caffarelli, Roquejoffre and Savin, [6].
The aim of this paper is to provide an analogous non local approximation of the Gaussian perimeter, showing the Gamma convergence to the local one. Moreover, we study the isoperimetric properties of this non local functional and observe that, in contrast with the local setting, an halfspace is a volume constrained critical point if and only if it has Gaussian measure 1 2 . In particular, we deduce that Ehrhard symmetrization can in general increase the considered non local Gaussian perimeter.
We remark that the non local approximation of the Gaussian perimeter we study is different from the one recently proposed in [17]. The non local functional we introduce has the advantage of having a more explicit formulation, while has the drawback that the isoperimetric shapes and the Ehrhard symmetrization are not preserved.

The Gamma-convergence
In this section we extend [1,Theorem 3] to the Gaussian case. Namely, we show: Theorem 1 (Convergence of local minimizers) Assume that s i ↑ 1, E i are local minimizers of J γ s i (·, Ω), and χ E i → χ E in L 1 loc (R n ). Then E is a local minimizer of P γ (·, Ω) and The proof of Theorem 1 is almost identical to the Euclidean one for [1, Theorem 3]. We limit our study to the parts which differ from it. In particular we will prove the following two propositions. Let ω k denote the volume of the unit ball in R k for k ≥ 1, and set ω 0 := 1.
w.r.t. the L 1 loc convergence of the corresponding characteristic functions in R n , i.e.

Proposition 3 For every measurable set
w.r.t. the L 1 loc convergence of the corresponding characteristic functions in R n . Inequality (3) means that for every measurable set E and sequence In these two propositions lurk the main differences between the Gaussian case and the Euclidean case. Once we have proved Proposition 2 and Proposition 3, we are done: the proof of Theorem 1 is completely identical to the proof of [1,Theorem 3], with the only forethought of adding a γ-superscript in every considered functional, and remembering the simple inequality γ(x, y) ≤ 1.
We will use the following notation: we write x ∈ R n as (x ′ , x n ) with x ′ ∈ R n−1 and x n ∈ R; we denote by H the halfspace {x : x n ≤ 0} and by Q = (−1/2, 1/2) n the canonical unit cube; we denote by B r (x) the ball of radius r centered at x and, unless otherwise specified, B r := B r (0); for every h ∈ R n and function u defined on U ⊂ R n we set τ h u(x) := u(x + h) for all x ∈ U − h. For the definition and basic properties of the perimeter P (E, Ω) in the sense of De Giorgi, we refer to [2,13].

Proof of proposition 2
We denote by C the family of all n-cubes in R n We can assume that the left-hand side of (4) is finite, otherwise the inequality is trivial. We choose an arbitrary Ω ′ ⋐ Ω, and find a positive constant c 0 = c 0 (Ω ′ ) so that c 0 ≤ γ(x, y), ∀ x, y ∈ Ω ′ . Then we easily obtain the inequality By [1, Theorem 1] and the arbitrariness of Ω ′ , we conclude that E has locally finite perimeter. We shall denote by µ its perimeter measure, i.e. µ(A) = |Dχ E |(A) for any Borel set A ⊂ Ω, and we shall use the following property of sets of finite perimeter: for µ-a.e. x 0 ∈ Ω there exists R x 0 ∈ SO(n) such that (E − x 0 )/r locally converge in measure to R x 0 H as r → 0. In addition, Indeed this property holds for every x 0 ∈ FE, see [2, Theorem 3.59(b)]. Now, given a cube C ∈ C contained in Ω, we set Moreover, we define C r (x 0 ) := x 0 + rR x 0 Q, where R x 0 is as in (5), and the measure for every E Borel set.
We claim that for µ-a.e. x 0 ∈ R n it holds If the claim is true, then we observe that for all ε > 0 the family is a fine covering of µ-almost all of Ω. By a suitable variant of Vitali's theorem (see [16]), we can extract a countable subfamily of disjoint cubes Since ε > 0 is arbitrary, we get the Γ − lim inf estimate. We now prove the inequality in (6) at any point x 0 such that (E − x 0 )/r converges locally in measure as r → 0 to R x 0 H and (5) holds. Because of (5) and the continuity of the exponential, we know that lim Thus we just need to show the inequality Since from now on x 0 is fixed, we assume R x 0 = I, so that the limit hyperplane is H and the cubes C r (x 0 ) are the standard ones x 0 + rQ. Let us choose a sequence r k → 0 such that For k > 0 we can choose i(k) large enough that the following conditions hold: We observe that, although J 1, γ s does not enjoy the nice scaling properties of J 1 s , it still satisfies the equality where we have set In particular, for r sufficiently small, and thus for every r k with k sufficiently big, the following inequality holds: Then we infer it has been proved in [1,Lemmata 7,11,12] that Γ n = ω n−1 . Hence we conclude the claimed inequality (7).

Proof of proposition 3
As in [1], it is enough to prove the Γ − lim sup inequality for the collection B of polyhedra Π of finite perimeter which satisfy P (Π, ∂Ω) = 0. B is dense in energy, i.e. such that for every set E of finite perimeter there exists E k ∈ B with χ E k → χ E in L 1 loc (R n ) as k → ∞ and lim sup k P γ (E k , Ω) = P γ (E, Ω). We recall that a polyhedron Π is in the class B if and only if where we have set We are going to prove that for a polyhedron Π ⊂ R n there holds where Γ * n := lim sup Again, as in [1,Lemmata 7,11,12] we have the equality Γ * n = ω n−1 . We shall divide the proof into two main steps.
Step 1. We first estimate J 1, γ s (Π, Ω). For a fixed ε > 0 set We can find N ε disjoint cubes Q ε i ⊂ Ω, 1 ≤ i ≤ N ε , of side length ε satisfying the following properties: (i) ifQ ε i denotes the dilation of Q ε i by a factor (1 + ε), then each cubeQ ε i intersects exactly one face Σ of ∂Π, its barycenter belongs to Σ and each of its sides is either parallel or orthogonal to Σ; Property (ii), combined with the continuity of the exponential and the property of measures, easily implies where we have set by x ε i the center of the cubes Q ǫ i . For x ∈ R n set We consider several cases.

hence (since all cubes are contained in Ω)
As for I 2 s (x) observe that if x ∈ Q ε i and |x − y| ≤ ε 2 , then y ∈Q ε i , whereQ ε i is the cube obtained by dilating Q ε i by a factor 1 + ε (hence the side length ofQ ε i is ε + ε 2 ). Then where in the last identity we used the scaling property Keeping ε > 0 fixed, letting s go to 1 and putting (13) Step 2. It now remains to estimate J 2,γ s . Let us start by considering the term In this case, using the same argument of case 1 for y ∈ Π c ∩ (Ω c \ Ω + δ ), we have An obvious similar estimate can be obtained by swapping Π and Π c , finally yielding

Using inequality (19) applied with the open set Ω
Since δ > 0 is arbitrary, letting δ go to zero, we conclude the proof of the Proposition.

First and second variation
In this section we calculate the first and second variation of J γ s (E). A similar analysis has been done in [12] in order to prove the local minimality of the ball for a functional involving nonlocal terms.
First, we fix some notation. Given a vector field X ∈ C 2 c (R n , R n ), the associated flow is defined as the solution of the Cauchy problem In the following, we shall always write Φ t to denote the map Φ(·, t). Note that for any given X there exists δ > 0 such that, for t ∈ [−δ, δ], the map Φ t is a diffeomorphism coinciding with the identity map outside a compact set. If E ⊂ R n is measurable, we set E t := Φ t (E). Denoting by JΦ t the n-dimensional Jacobian of Φ t , the first and second derivatives of JΦ t are given by Finally, given a sufficiently smooth bounded open set E ⊂ R n and a vector field X, we recall that the first variation of J γ s (E) along the vector field X is defined by where Φ t is the flow associated with X. The second variation of J γ s (E) along the vector field X is defined by If X is a vector field such that X := φν E on ∂E, where ν E denotes the exterior normal to E, using the area formula and the divergence theorem, the first variation of the Gaussian volume can be computed as If E is a set of class C 2 , given a smooth function φ : ∂E → R, it can be extended in a neighborhood U of ∂E so that The second variation of the Gaussian volume along the vector field X such that X = φν E on ∂E and φ satisfies (23), can be calculated using the divergence theorem and reads as Thus, we say that a vector field preserves the Gaussian volume of E if it satisfies We note that without this assumptions the expression of the second variation of the Gaussian perimeter even in the local framework is quite complicated, see [3,Eq. (17)].
Theorem 4 Let E be an open set of class C 2 and X ∈ C 2 c (R n , R n ) a vector field such that X = φν E on ∂E. Then the first variation of J γ s (E) along a vector field X is given by while the second variation reads as Moreover, if X is volume preserving, then Proof. Let us call J δ s the integral associated to the regularized kernel, namely By the definition of Φ t , the implicit function theorem gives the existence of ε > 0 such that the map Φ t is a diffeomorphism for all t ∈ [−δ, δ]. Using the area formula, we compute We use the first equation in (21) to compute the first variation of J δ (|x| 2 +|y| 2 ) )X(x) + D y (e − 1 4 (|x| 2 +|y| 2 ) )X(y) dxdy Now we compute the second variation of J δ s . E c E D x K δ (x − y)e − 1 4 (|x| 2 +|y| 2 ) , X(x) (div x X(x) + div y X(y))dxdy + 2 E c E D y K δ (x − y)e − 1 4 (|x| 2 +|y| 2 ) , X(y) (div x X(x) + div y X(y))dxdy + E c E K δ (x − y)e − 1 4 (|x| 2 +|y| 2 ) div[X(x) div(X(x))] + div[X(y) div(X(y))] dxdy + 2 E c E K δ (x − y)e − 1 4 (|x| 2 +|y| 2 ) div x X(x) div y X(y)dxdy We now use the divergence theorem and exploit the symmetry of K δ in order to simplify the above expression as follows div y X(y) div y K δ (x − y)e − 1 4 (|x| 2 +|y| 2 ) X(y) dxdy + E c E div x X(x) div y K δ (x − y)e − 1 4 (|x| 2 +|y| 2 ) X(y) dxdy + E c E div y X(y) div x K δ (x − y)e − 1 4 (|x| 2 +|y| 2 ) X(x) dxdy = I 1 + I 2 + I 3 + I 4 . (28) Using Fubini and the divergence theorems we have We remark that I 1 (resp. I 3 ) has the same expression of I 2 (resp. I 4 ) exchanging x and y. Using this observation and the symmetry of K δ , we compute where we used that X = φν E and then D τ f, X = 0 for every f ∈ C 1 (∂E). Regarding the second addend of the above expression, using again Fubini's theorem and the fact that D x K δ = −D y K δ , we get E ∂ ∂ν(x) K δ (x − y)e − 1 4 (|x| 2 +|y| 2 ) dy = −