Boundary regularity of an isotropically censored nonlocal operator

In a bounded domain, we consider a variable range nonlocal operator, which is maximally isotropic in the sense that its radius of interaction equals the distance to the boundary. We establish \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^{1,\alpha }$$\end{document}C1,α boundary regularity and existence results for the Dirichlet problem.


General setting and the operator
Let n ≥ 1 and ⊂ R n be a bounded, connected domain of class C 1,1 .Let d : → [0, +∞) be the distance to the boundary, When no confusion arises, we simply write d = d .It is convenient to smooth out the distance function by taking δ ∈ C 1,1 ( ) such that δ = d when d < d 0 , for a small d 0 > 0. A similar notation is employed for other domains.
(1.1)This is an isotropic regional fractional Laplacian, and the factor d(x) 2s−2 is inserted to ensure that L u(x) converges as x → ∂ to a nontrivial limit, namely − u(x).See Lemma A.1.Probabilistically speaking, the operator L generates a Lévy type process where a particle at x ∈ jumps randomly and isotropically in the largest possible ball B d(x) (x) contained inside .
We list some characteristic properties and consequences.
• L enjoys a mid-range maximum principle Proposition 2.2, whose strength lies between the local one for − and the global one for (− ) s : no absolute minima exist in the region where L u ≥ 0, provided that u is non-negative in the domain of interaction outside of that region.As a result, local barriers suffice to control boundary behaviors.• The domain of interaction depends on the point of evaluation.Thus, extra effort is needed in the construction of barriers in Sect.3, even following the established idea [11].• L is of order 2s but scales quadratically and satisfies the classical Hopf lemma, see Lemma A.4 and Lemma 4.6.• L is not variational.To see this, take = (0, 1).Upon integrating by parts, one immediate sees "hidden boundary terms" at x = 1/2.In higher dimensions the "hidden boundary" can be thought of as points having at least two projections to the boundary.Unfortunately, all these points contribute in such a different way that the resulting expression would not be manageable.Consequently, no weak formulations due to integration by parts can be expected.Existence is to be established in the viscosity sense, in Sect.8.
Generic and universal constants are denoted by C, c.They depend only on n, s and .
Our main results are the following.

Higher regularity up to the boundary remains open. There are two difficulties:
• The condition α < 1 is crucially used in the proof of Proposition 6.1, where the quadratic growth (linear in x times linear in x n ) contradicts the control |x| 1+α .• A complete study of the action of L R n + on monomials is missing (Lemma 4.3).
Remark 1. 3 Crucial to construction of the barrier (Proposition 3.1) is the uniform exterior ball condition.This is satisfied by bounded domains of class C 1,1 , as well as their blow-ups around a boundary point.

Main ideas
Let us explain the heuristics of the proof.By definition, This is implied (see Lemma C.1) by the expansion i.e. u/d is C α up to 0 ∈ ∂ .By building suitable barriers in Sect.3, we will be able to obtain global Hölder regularity (Proposition 5.4).This allows the use of a blow-up argument to reduce the problem to a half plane.It then suffices to prove a Liouville theorem (Proposition 6.1) for solutions to the homogeneous equation in the half space, namely under the growth u(x) = O(|x| 1+α ): the only solution is u = cx n .Since the homogeneous equation is preserved under scaling and tangential differentiations, solutions can be shown to be one dimensional (1D) (e.g.Lemma 6.4).Hence, we need to show that solutions u(x , x n ), which is independent of x , to which grow no faster than x 1+α n must be linear (Lemma 6.2).This is in turn implied by the boundary regularity in the half-line (Proposition 4.1), namely u(x n ) x n ∈ C α up to 0.
To show this, one simply proves a boundary Harnack inequality (Lemma 4.7) and an improvement of oscillation (Lemma 4.8).

Related works
The boundary Harnack inequality for a nonlocal elliptic operator in non-divergence form is proved by Ros-Oton-Serra in [12,Theorem 1.2].A similar type of nonlocal operator with fixed horizon (range of interaction) at every point has been considered by Bellido and Ortega [6].

Generalizations
We expect that the techniques introduced in this paper should be able to prove similar results in the parabolic setting, and when L is replaced by an analogous integro-differential operator of order 2s with any homogeneous kernel.

Preliminary results
Clearly L satisfies the global maximum principle.
Then either u ≡ 0 in or u > 0 in .
Proof At any interior minimum x 0 , L u(x 0 ) ≤ 0, with strict inequality unless u ≡ u(x 0 ) in B d(x 0 ) (x 0 ).But the latter ball contains a sequence converging to ∂ where u ≥ 0.
A more careful examination yields the following version of maximum principle.It is especially useful to study the blow-up equation when the domain becomes unbounded. (2.2) Thus equality holds and u ≡ u(x 0 ) = 0 in G.
Now we state the interior estimates which follows from the corresponding result for the restricted fractional Laplacian [9,10].We denote (2.3) Lemma 2.4 (Interior estimates) Suppose U ⊂ R n is not necessarily bounded, and Moreover, if f ∈ C β (B 1 ) for β ∈ (0, 1) and β + 2s is not an integer, then there exists where Remark 2. 5 We notice that, because the operator L U degenerates away from the boundary, so does the estimate (through the term Proof Let us extend u by zero outside U .In this proof we write d = d U .At each interior point x ∈ U , one can rewrite the equation in terms of the restricted fractional Laplacian, namely where | being the normalization constant for (− ) s .We first prove the C 2s (or C 1− ) regularity.Note that for x ∈ B 1 , we have d(x) ≥ 3 and Suppose first s = 1 2 .By [10, Theorem 1.1(a)], When s = 1 2 , using again [10, Theorem 1.1(a)], we replace accordingly the C 2s norm by C 1− norm for any ∈ (0, 1), with the constant depending also on .Now we prove the higher regularity.Up to multiplicative constants, we decompose g = g 1 + g 2 + g 3 where For x, y ∈ B 1 , we express For the integral in the first line, note that 3 ≤ d(x) ≤ |z|.By mean value theorem, there exists x * ∈ B 1 such that For the second line we note that there is a nontrivial contribution only in the symmetric difference B d(x) (x) c B d(y) (y) c , which lies in an annulus of width at most of order |x − y|.
3 The barriers

Super-solution near the boundary
We construct a barrier in the spirit of [10].Since the domain of interaction varies from point to point, we must compute at all points.
The idea is to consider powers of the distance function to a ball which touches the domain from the outside, since we want the super-solution to be strictly positive except at the contact point, however near the boundary.Proposition 3.1 (Super-solution) Let U ⊂ R n be a possibly unbounded domain of class C 1,1 .Suppose x 0 ∈ ∂U can be touched by an exterior ball of radius b > 0.Then, there exists a constant r 0 > 0 and a function ϕ Here ν(x 0 ) is the (outer) normal at x 0 ∈ ∂ , and r 0 and C depend only on n, s and b.
Upon a translation and a rotation, the exterior ball condition states that B b (−be n ) touches ∂U at x 0 = 0 ∈ ∂U from the outside.
The super-solution will be built from for p = 1 and p ∼ 2 − .As in [10], at each point x ∈ U we compare d B b (−be n ) (y) p with the one-dimensional function where are the hyperplanes which are orthogonal to and contain respectively −be n + bv and −be n .First we consider the planar barrier.This uses the fact that −L U behaves like the Laplacian near ∂U .(Notice that this computation is valid for d p T x only at the point x, although it is all we will need.)Lemma 3.2 For p > 0 and x ∈ U , we have Here the constant in O(| p − 2|) depends only on n and s.
Changing variable to y = d T x (x)z (recall that d T x (x) > 0 for x = 0 ∈ ∂U ) and choosing another coordinate system for z such that v is the direction of the last coordinate axis, we have (here ψ is defined in (B.1)) (3.1) as desired.
Next we compare d p B b (−be n ) and d p T x pointwise.For each fixed x ∈ U , hence v ∈ S n−1 , we denote the projection of a vector y ∈ R n onto P x by Proof For z ∈ U , we express to see it is non-negative.We make the following observations: • The sum of radii of the interior disjoint balls B d U (x) (x) and B b (−be n ) is at most the distance between the centers, giving d U (x) + b ≤ |x + be n |.This implies and Here C depends only on n and s.
Proof We split We are ready to prove Proposition 3.1.

Proof of Proposition 3.1 Let
Then On the other hand, we verify that Therefore, ϕ (0) = r −1 0 ϕ is the desired super-solution.

Super-solution for a bounded domain
A concave paraboloid serves as a simple global super-solution.Choose a coordinate system such that 0 ∈ .Let M = diam so that ⊂ B M .Consider the positive, strictly concave function When = B M , this is known as the torsion function.
Lemma 3.5 There holds Proof For any x ∈ and y ∈ B d(x) , the parallelogram law implies |y| 2 |y| n+2s dy = 1.Remark 3.6 By requiring that be compactly contained in B M , one obtains a strict supersolution.However, we will not need this.

Boundary Harnack inequality in 1D
We are interested in one-dimensional (1D) Dirichlet problems on the half space R n + .Note that for sin n−2 θ dθ dr.
Throughout this section we assume that u is 1D (i.e.u depends only on x n ).Then It is convenient to write x ∈ R + in place of x n .In this section we will prove the following There exists α * ∈ (0, 2 s ∧ 1) and C > 0 such that Here the C and α * depend only on n and s.Remark 4.2 Note that the function x is a model solution to (4.1), and in fact the unique solution on R + up to a constant multiple, as we will show in Lemma 6.2.In other words, any two solutions are comparable up to the boundary in a Hölder continuous way.

Preliminaries
The scaling property Lemma A.4 allows us to compute the action of L R + on monomials.
However, it is not clear from this expression if a( p) is monotone or signed for large p.
We will use the following version of strong maximum principle for functions with nonnegative data in the adjacent interval of the same length.

Boundary Harnack inequality
First of all we show Proposition 4.1 for α = 0, using interior Harnack inequality and comparison arguments.
Proof By replacing u by u/u(1) if necessary, we may assume that u(1) = 1.By Lemma 2.3, there exists C > 0 universal such that Then there exists C > 0 universal such that x .

Boundary Hölder regularity
Then there exists a universal constant c ∈ (0, 1) such that for any k ≥ 1, Proof By replacing u by u/u(1) if necessary, we may assume that u(1) = 1.By Lemma 4.7, we can take m 1 = C −1 3 and M 1 = C 3 .We assume in the following that u(x)/x is not a constant; otherwise we can trivially take By Lemma A.4, both the functions harmonic and non-negative on (0, 2).As they solve (4.2), the strong maximum principle Lemma 4.5, they are strictly positive on (0, 1).This means that Similarly, the functions x .
Rescaling and multiplying throughout by the normalizing factor 4 k+1 , we have This means that Adding up these two inequalities, Thus Now a standard iteration yields the Hölder continuity of the quotient.

Pointwise boundary Harnack inequality
Using the global maximum principle, we obtain a direct pointwise bound which is good for controlling the interior behavior.
We can now control a solution by the distance function.

Lemma 5.2 (Global boundary Harnack principle)
Then there exists a universal constant C such that

H. Chan
Proof Since is a bounded domain of class C 1,1 , there exists b > 0 such that an exterior tangent ball of radius b exists at each point x 0 ∈ ∂ .By Proposition 3.1, in a suitable coordinate system there exists ϕ (x 0 ) such that Since ϕ (x 0 ) grows linearly away from the boundary, we have The interior estimate simply follows from Lemma 5.1.
We present a local analogue in a half ball B + r = B r ∩ {x n > 0}, where r > 0.

Lemma 5.3 (Local boundary Harnack principle
Here C depends only on n and s.
The result follows by combining it with the trivial estimate

Hölder regularity up to boundary
As above, we give a global and a local result.While practically having an order of 2s in the interior, the operator satisfies the classical Hopf boundary lemma.Thus the minimum of the two yields the combined regularity.This effect is analogously seen with the spectral fractional Laplacian.
We need to show that Using the interior estimates Lemma 2.4, we have In view of (5.1) we observe that Then for any ∈ (0, 1), there exists a constant C = C(n, s, for s ∈ (0, 1 2 ).

Liouville-type results
In this section we classify solutions to homoegenous Dirichlet problems in a half space with controlled growth.We write Proposition 6.1 (Liouville-type result) Let v be a solution to which satisfies the growth condition for some α ∈ (0, α * ) with α * ∈ (0, 2s ∧ 1) given in Proposition 4.1.Then v is a 1D and linear, i.e.

Proof of the higher regularity
Proof of Theorem 1.1 In view of the interior estimates in Lemma 2.4, we just need to prove the following expansion: for some α = α(n, s) ∈ (0, 1) so small that Proposition 6.1 holds, for any z ∈ ∂ , there exists Q z ∈ R, r > 0 such that for any x ∈ ∩ B r (z), Indeed, as in the proof of Proposition 4.1, one can interpolate the (degenerate) interior estimate with (7.1) to obtain the full C 1,α ( ) regularity (for some α ∈ (0, α)).Suppose on the contrary that there exists z ∈ ∂ such that (7.1) does not hold for any

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We split the proof by contradiction into a number of steps.
Step 1: Choosing one Q for each r.
For each r > 0 small, we choose a Q(r ) that minimizers u − Qd L 2 (B r (z)) , i.e.
Step 2: The blow-up sequence and growth bound.Now we define the monotone quantity From lim r 0 θ(r ) = ∞, there is a sequence r m → 0 such that and, from the choice of Q(r m ), We claim the following growth control Indeed, the arguments in Step 1 (replacing θ by θ(r ), and the interval (0, 1] by [r , 1]) shows that Also since θ is non-increasing, Then (here we implicitly extend suitable functions by 0 outside ) This proves (7.5).
Step 3: Equation for the blow-up sequence.
Let m = (r m ) −1 ( − z), which converges to a halfspace {x • e > 0} as m → +∞, for e = −ν(z), the inward normal at z ∈ ∂ .By the properties in Lemma A.3 and Lemma A.4, the functions v m satisfy since L u and L d = L δ are bounded in view of Lemma A.2. Now, by Proposition 5.4, v m C 2β ( m ) ≤ C for some β > 0. So Arzelà-Ascoli Theorem asserts a subsequence of v m uniformly converging on compact sets in {x • e > 0} to some function v ∈ C β ({x • e > 0}), β ∈ (0, 1).Passing to the limit in (7.5) and (7.6) yields and Step 4: Classification of the limit, and the contradiction.By Proposition 6.1, v(x) = c 0 (x • e) for some constant c 0 ∈ R. Using the fact that we pass to the limit in (7.4) (upon dividing by r m ) to see that But this implies c 0 = 0 and hence v = 0, contradicting (7.3) in the limit m → +∞.Therefore (7.1) holds and the proof is complete.

Existence of viscosity solution
Consider the Dirichlet problem We will establish the existence of a continuous viscosity solution using Perron's method, carefully exploiting the mid-range maximum principle that L satisfies.Throughout the section we assume f ∈ C α ( ) and g ∈ C(∂ ), for some α > 0.
Our goal is to prove the following.The notion of viscosity solutions has been successfully used in nonlocal equations, see for example [1,7,11,13].For the proof, we extend the clean arguments described in [4] using the barrier constructed in Sect.3.

Definition 8.2 (Semi-continuous functions) We denote by
the space of upper (resp.lower) semi-continuous functions in .For u ∈ L ∞ ( ) we define the USC (resp.LSC) envelope as We also need a localized definition based on Proposition 2.

Supremum of sub-solutions
Define the family of admissible sub-solutions as The pointwise supremum of all sub-solutions in A is defined as We will prove that u is a viscosity solution by showing that u * = g on ∂ so that u * = u, and then verify that u * is a super-solution so that, by comparison, u * = u.
Proof The proof is similar to [4, Proof of Theorem 4.17, Step 1], but a mid-range comparison is to be employed.Indeed, for each x 0 ∈ ∂ , let r 0 be as in Proposition 3.1 and define the barrier where k ε , depending not only on ε but also on g, and sup u * , is chosen such that By Corollary 8.6, In particular, since ϕ (x 0 ) ∈ C( ), there exists δ(ε) > 0 such that This implies lim x k →x 0 u * (x k ) = g(x 0 ) and hence u * | ∂ = g ∈ C(∂ ).
Then, the maximum w = u ∨ v is also a sub-solution in .
Proof Suppose x ∈ N x ⊂ and φ ∈ C 2 (N x ) ∩ L 1 ( ) is such that w(x) = φ(x) and w ≤ φ in .We want to show that L φ(x) ≤ f (x).If w(x) = u(x), then since u ≤ w ≤ φ, the result follows from the fact that u is a viscosity sub-solution.If w(x) = v(x) = u(x), then x ∈ G and (using v ≤ w ≤ φ) the pointwise computation also gives L φ(x) ≤ L v ≤ f , as desired.
Lemma 8.12 The LSC envelope of u defined by (8.2) is a super-solution in the interior, i.e.
L u * ≥ f in .
Proof of Proposition 8.7 By Lemmas 8.9, 8.10 and 8.13, u ∈ C( ) is both a sub-and supersolution, and u = g on ∂ .

Regularity
Since suffices to obtain qualitative interior regularity, we compare to the (restricted) fractional Laplacian in R n as in Lemma 2.4, and invoke the ccorresponding regualrity results in [3,Chapter 3].

2 . 8 . 3 (
Definition Viscosity solutions) Let G be an non-empty, open set in .The domain of interaction of G is G * = y∈G B d (y) (y).

Lemma 8 . 10
It holds that u = u * ∈ U SC( ).Proof By definition, u ≤ u * .By Lemmas 8.8 and 8.9, u * ∈ A, so u * ≤ u.Lemma 8.11 Let G be an open subset of .Suppose u ∈ U SC( ) satisfies L u ≤ f in , in the viscosity sense, and v ∈ C 2 (G) ∩ L ∞ ( ) satisfies pointwise