The trace fractional Laplacian and the mid-range fractional Laplacian

In this paper we introduce two new fractional versions of the Laplacian. The first one is based on the classical formula that writes the usual Laplacian as the sum of the eigenvalues of the Hessian. The second one comes from looking at the classical fractional Laplacian as the mean value (in the sphere) of the 1-dimensional fractional Laplacians in lines with directions in the sphere. To obtain this second new fractional operator we just replace the mean value by the mid-range of 1-dimensional fractional Laplacians with directions in the sphere. For these two new fractional operators we prove a comparison principle for viscosity sub and supersolutions and then we obtain existence and uniqueness for the Dirichlet problem. We also show that solutions are $C^\gamma$ smooth up to the boundary when the exterior datum is also H\"older continuous. Finally, we prove that for the first operator we recover the classical Laplacian in the limit as $s\nearrow 1$.

This may be the best known and most famous second order differential operator.Written as in (1.1) it is an operator in divergence form.This allows to use techniques form calculus of variations a framework in which solutions are understood in a weak sense integrating against test functions (typically solutions are functions in the Sobolev space H 1 ).When one introduces coefficients in this context a natural operator to look at is (1.2) Lu(x) = div(A(x)∇u(x)), with a given matrix (that is usually assumed to be symmetric) with spatial dependence, A(x), see for instance [1,29].
A different way of writing the Laplacian is as Here λ 1 (D 2 u) ≤ λ 2 (D 2 u) ≤ ... ≤ λ N (D 2 u) stands for the eigenvalues of the Hessian, D 2 u = (∂ 2 ij u) ij .This way of writing the Laplacian is not in divergence form but as an operator for which solutions are understood in viscosity sense [21] (here solutions are just continuous functions and the operator is applied to smooth test functions that touches the solution from above or below).Introducing coefficients thinking in this way one finds (1.4) F (D 2 u)(x) = tr(A(x)D 2 u(x)) with an x−dependent matrix A(x), see [16].
For the classical Laplacian both (1.1) and (1.3) are equivalent ways of writing the same operator.For the Dirichlet problem for ∆u = 0 the notions of weak and viscosity solutions coincide (and in fact the Dirichlet problem has a unique classical solution), see [25] and [30] (the equivalence between weak and viscosity solutions include quasi-linear equations, [26], and some non-local equations, [6,15]).However, when one introduces coefficients, the operators (1.2) and (1.4) are not equivalent (in fact, the notion of weak solution using Sobolev spaces is not appropriate to deal with (1.4)).
In recent years an operator that has become quite popular is the well-known fractional Laplacian defined as (1.5) (−∆) s u(x) = c(s) p.v.
Here s ∈ (0, 1), p.v refers to the principal value of the integral and c(s) is a constant that depends also on the dimension N and goes to zero as (1 − s) as s → 1 (we will make explicit and use the constant only in dimension 1).For several different ways of writing the fractional Laplacian we refer to [27].The operator in (1.5) is also well suited for variational methods (and typically solutions are functions in the fractional Sobolev space H s ).In this context one can introduce spatial dependence in the operator using a general symmetric kernel k(x, y) (this symmetry assumption allows to integrate by parts and use variational techniques) and consider Here a natural assumption is to ask that the kernel k is comparable to the one of the fractional Laplacian, in the sense that, for two positive constants c 1 , c 2 , we have c 1 |x−y| −N −2s ≤ k(x, y) ≤ c 2 |x−y| −N −2s .One possible choice of the kernel is given by k(x, y) = |A(x − y)| −N −2s , obtaining an operator that is similar to (1.2).When one wants to look for non-divergence form nonlocal operators one can consider nonsymmetric kernels k(x, y).These kind of operators have been intensively studied recently, we refer to [17,18,19,20,23,24,28,31] and references therein.
However, up to now, there is no clear analogous to the classical way of understanding the Laplacian as the sum of the eigenvalues of the Hessian, as in (1.3).Our main goal in the paper is to introduce a new nonlocal operator that is a natural analogous to this way of looking at the classical Laplacian.To this end we first recall that, from the classical Courant-Hilbert formulas for the eigenvalues of a symmetric N × N matrix, we have (1.6) Here the maximum is taken among all possible subspaces S of R N of dimension N −i+1 and the minimum among unitary vectors in S. For the Dirichlet problem for the equation λ i (D 2 u(x)) = 0 we refer to [14].Related operators are the truncated Laplacians studied in [8,9,10].Notice that (1.6) can be written as version of the eigenvalues (that we will call fractional eigenvalues) is given by that is, we are computing the same max-min procedure as before, but now we are taking the one-dimensional fractional derivative of order 2s.The Dirichlet problem for the first fractional eigenvalue is related to fractional convex envelopes, see [22].Operators that are fractional analogous to truncated laplacians are studied in [7,11,12,22].Notice that due to their definition, the fractional eigenvalues are ordered Λ s Now, let us introduce the operator that we call the trace fractional Laplacian, Here Λ s i u is given by (1.7).Notice that (1.8) is not in divergence form and therefore we will use viscosity theory to study this operator.We remark that this fractional version of the classical Laplacian is not equivalent to the usual fractional Laplacian given by (1.5).This is a striking difference between the fractional setting and the classical local context (the variational fractional Laplacian does not coincide with the trace fractional Laplacian).
Our main goal in this paper is to show that the Dirichlet problem for the trace fractional Laplacian is well posed in the framework of viscosity solutions.Given a bounded domain Ω and an exterior datum g we will deal with (1.9) Since we want a continuous up to the boundary solution (some of our arguments requiere this property) we will assume that s ∈ (1/2, 1).Now, let us introduce a second fractional operator that we will call the mid-range fractional Laplacian.To this end, notice that the classical fractional Laplacian, given by (1.5), of a smooth function that decays at infinity can be written as Therefore, the fractional Laplacian can be written (up to a negative constant) as the mean value (on the sphere) of the function Θ s φ : That is, to obtain the fractional Laplacian one computes the mean value in the directions of the one-dimensional fractional derivatives of order 2s.With this idea in mind let us introduce a different fractional operator.Instead of the mean value we just take the mid-range (the measure of central tendency that is given by the average of the lowest and highest values in a set of data) of the function Θ s φ in S N −1 and we obtain The mid-range and the trace fractional Laplacians are closely related.They are part of a large family of operators (those given in terms of combinations of fractional eigenvalues, we will add more comments on this in the final section of this paper).Moreover, in dimension two, that is, for N = 2 in our notation, the mid-range and the trace fractional Laplacians coincide up to a constant In addition, remark that these two new fractional operators that we introduced here, (−∆) s tr and (−∆) s mid , are 1-homogeneous (it holds that (−∆) s tr (ku , and are invariant under rotations, as the usual Laplacians (both local and fractional) are.
For the mid-range fractional Laplacian we also study the Dirichlet problem, that in this case reads as Here we dropped the constant 1/2 in front of the fractional eigenvalues to simplify the notation.
Our first result says that when the domain is smooth and the exterior datum g is continuous and bounded, there is a unique viscosity solution for (1.9) or for (1.10).
These operators share many properties with the corresponding local Laplacian and the classical fractional Laplacian (as the validity of a comparison principle and a strong maximum principle and the continuous dependence of the solutions on the exterior data).
Theorem 1.2.Under the hypothesis of Theorem 1.1, a comparison principle holds, let u 1 and u 2 denote the solutions to (1.9) (or to (1.10)) with exterior data g 1 and g 2 respectively, then and, as a consequence, the solution depends continuously on the exterior data; it holds that Moreover, a strong maximum principle holds both for (1.9) and for (1.10).If there exists x 0 ∈ Ω such that then, the solution and the exterior datum are constant, u ≡ g ≡ cte.
A striking difference with the classical Laplacian (and also with the classical fractional Laplacian) is that these operators are nonlinear.
Theorem 1.3.The problems (1.9) and (1.10) are nonlinear problems.There exists data g 1 , g 2 such that but the corresponding solutions verify That the operators are nonlinear is due to the fact that there are maxima and minima involved in the definition of ∆ s tr and ∆ s mid .In fact, what is really surprising is that the usual local Laplacian, written as is a linear operator.
In addition, we study the limit as s ր 1 and prove that solutions to (1.9) converge uniformly to the unique solution to the Dirichlet problem for the classical local Laplacian (1.11) ∆u(x) = 0 x ∈ Ω, x ∈ ∂Ω.
Here we will use the explicit constant that appears front of the integral in (1.7).We just remark that any c(s) such that c(s) ∼ (1 − s) will also work when taking this limit, but the explicit form of (1.12) is the one that corresponds to the 1-dimensional fractional Laplacian.
Theorem 1.4.Let Ω be a C 2 bounded domain and fix g ∈ C(R N \ Ω) and bounded.Let u s denote the unique solution to to the problem for the trace fractional Laplacian, (1.9).Then, it holds that lim sր1 u s = u in C(Ω).The limit u is given by the unique solution to (1.11).
Moreover, when u s is the unique solution to the problem for the mid-range fractional Laplacian, (1.10), it holds that lim sր1 u s = u in C(Ω) where the limit u is given by the unique solution to the local problem x ∈ ∂Ω.
Remark that the first limit result also holds for the usual fractional Laplacian (1.5) with the appropriate N dimensional constant c(s).
In these limits, as s ր 1, the limit is unique and is characterized as the solution to (1.11) or to (1.13).Hence, we have convergence of the whole family u s as s ր 1 (not only along subsequences).
Concerning regularity of solutions we quote the recent paper [13] where the authors prove a Hölder regularity result for solutions to Λ s 1 u(x) + Λ s N u(x) = f (x) in Ω with homogeneous Dirichlet boundary conditions, u(x) = 0 in R N \ Ω and s close to 1. Regularity issues for these operators is a delicate issue since they are very degenerate.Now, to end the introduction, let us comment briefly on the ideas and methods used in the proofs.
In most of our arguments, to simplify the notation and clarify the ideas used in the proofs we will only include the details for the mid-range fractional Laplacian, (−∆) s mid (u)(x), that involve the smallest and the largest fractional eigenvalues.Since for this operator the main difficulties arise, we will just briefly comment on how to extend the results for the fractional trace Laplacian, (−∆) s tr (u)(x), in which the intermediate eigenvalues appear.Recall again that for N = 2 the mid-range fractional Laplacian and the trace fractional Laplacian coincide.In addition, we will drop the notation p.v. in front of the integrals (that need to be understood in the principal value sense when appropriate) and, when we prove results for a fixed s, we will also drop the constant c(s).
Since our problem is fully nonlinear we use the concept of viscosity solutions in a nonlocal framework.We will use ideas from [2,3,4].First, we prove a comparison principle for sub and supersolutions to our problems (1.9) and (1.10).The proof follows ideas from [4] and [22].Notice that in [22] it is assumed that the domain is strictly convex.This condition is not needed here since the maximum and the minimum among directions are involved in our operator and we can choose any direction either in the max or in the min to obtain a super or a subsolution.Once we have a comparison principle existence and uniqueness of solutions are an easy consequence of Perron's method.
To show that the operator is nonlinear we find a domain in two dimensions, a smooth function u inside Ω and an exterior datum g such that the lines corresponding to directions associated with the maximum and the infimum of the 1−dimensional fractional derivatives of u inside Ω do not intersect a region outside Ω.Then, we show that the exterior datum g can be slightly perturbed in that region keeping the same u as the solution inside Ω.In this way we find two different exterior data (that, in addition, are ordered) with the same solution inside the domain and we conclude the nonlinearity of our problem from the strong maximum principle.
To recover the usual Laplacian in the limit as s ր 1 we just have to observe that with the choice of the constant c(s) given in (1.12) it holds that the 1-dimensional fractional Laplacian converges to the usual second derivative as s ր 1 (for c(s) ∼ (1 − s) we just obtain a multiple of the Laplacian in the limit).Then, the proof follows just taking care of the max and min involved in the fractional eigenvalues using viscosity tricks.The uniform convergence in Theorem 1.4 will follow from the fact that we show that the upper and lower half-relaxed limits of {u s } s coincide.
Finally, let us point out again that to compute the 1-dimensional fractional Laplacian in directions z on some particular test functions we need to restrict ourselves to consider only the case s > 1/2.We believe that without this condition solutions may noy be continuous up to the boundary of the domain.
The paper is organized as follows: In Section 2 we prove the comparison principle for viscosity sub and supersolutions to our problems and then we obtain existence and uniqueness for the Dirichlet problems.In Section 3 we show that the operators are nonlinear.In Section 4 we deal with the limit as s ր 1.Finally, in Section 5 we will comment on possible extensions of our results and describe how to introduce coefficients in fractional trace operators.

Comparison principle. Existence and uniqueness of solutions
The main result in this section is to prove a comparison principle for the problem (2.1) To this end, we borrow ideas from [4] (see also [22]).
Since in this section s is fixed we drop the constant c(s) that plays no role in our arguments.Also, as mentioned in the introduction, to simplify the notation in the proofs we will only analyze in detail the problem and at the end of the section comment on how to obtain the results for (2.1).In fact, we can consider any combination with nonnegative coefficients of fractional eigenvalues as long as Λ s 1 u and Λ s N u appear.
2.1.Basic notations and definition of solution.We use the notion of viscosity solution from [4], which is the nonlocal extension of the classical theory, see [21].
To state the precise notion of solution, we need the following: Given g : R N \Ω → R, for a function u : Ω → R, we define the upper g-extension of u as In the analogous way we define u g , the lower g-extension of u, replacing max by min.
Now we introduce the definition of the upper and lower semicontinuous envelope, that we will denote by ũ and u ˜respectively of u, that are given by ũ(x) := inf r>0 sup u(y) : y ∈ B(y, r) An important fact, that can be easily verified, is that for any continuous function g : R N \ Ω → R and any upper semicontinuous function u : Ω → R, it holds that Here 1 A denotes the indicator function of a set A in R N .
We now introduce a useful notation, for δ > 0 we write and then define Now, with these notations at hand, we can introduce our notion of viscosity solution to (2.2) testing with N −dimensional functions as usual.

Definition 2.1. A bounded upper semicontinuous function
In an analogous way, we define viscosity supersolutions (reversing the inequalities and replacing u g by u g ) and viscosity solutions (asking that u is both a supersolution and a subsolution) to (2.2).
Remark 2.1.For the definition of a viscosity solution to the Dirichlet problem for the trace fractional Laplacian, (2.1), we just have to take in the previous definition.

Comparison principle. Now, our goal is to prove a comparison principle between sub and supersolutions to (2.2).
In order to prove the comparison principle, we first show that sub and supersolutions behave well on the boundary of the domain.
Let u, v : R N → R be viscosity sub and supersolution of (2.2) in Ω, in the sense of Definition 2.1, respectively.Then, (i) u ≤ g on ∂Ω; (ii) v ≥ g on ∂Ω.
Proof.We begin by proving (i).Suppose by contradiction that there is Hence, we have that u g (x 0 ) = u(x 0 ).Since g is continuous, there exists We may with no loss of generality assume that R 0 < max{ x − y : x, y ∈ Ω}.
We now introduce two auxiliary functions: and D 2 a is bounded; for instance we can just take • b : R → R, a smooth bounded and increasing function which is concave in (0, +∞), and such that b(0 Next, we use these two functions to define for any ε > 0 the penalized test function here d is a smooth extension of the signed distance to the boundary, ∂Ω.It is at this point where we use the that is upper semicontinuous for any ε small.Then, for any ε small, Ψ ε attains a global maximum at a point x ε .Therefore, we have and hence, From here, we get that In particular, x ε ∈ B(x 0 , 2R 0 ) for any ε small enough.Now, using again the properties of a and b we get Therefore x ε ∈ Ω.Notice that we used the function a to penalize that x ε is far from x 0 and the function b with the signed distance to ∂Ω to obtain that x ε is not in Since a is non-negative, by (2.3), we have Hence, due to the fact that u is upper semicontinuous, we obtain Thus, we have that a ) for any ε small enough.Now, using that u is a viscosity subsolution of (2.2) in Ω in the sense of Definition 2.1, we have that Here we need to introduce changes in the arguments used in [5].The key point is that the directions z that are associated to the max and min that appear in the equation may be different (and we will take advantage of this fact).
We have To bound the infimum we choose Now, for the supremum we argue as follows: the main idea is that what we obtained with our choice in the infimum is negative enough to absorb the supremum and reach a contradiction at the end (the contradiction arrives from the fact that the sum of the infimum and the supremum is greater or equal to zero, see (2.4), but the infimum is negative enough in order to obtain that the sum is also negative).
Given η ε > 0, we choose a direction, z ε , such that ).We can prove both integrals are of order ε −2s in a similar way as the integrals in the infimum: we estimate I 1 z,ε (ω ε , x ε ) using the Taylor expansion of ω ε and we estimate I 2 z,ε (u, x ε ) simply by using the boundedness of u g .Thus, Adding the bounds for the infimum and the supremum and using (2.5) we obtain and we reach a contradiction taking η ε small enough.
Since we are dealing with subsolutions, in the associated inequality we can choose any direction to obtain a bound for the infimum, but we have to take care of the supremum.When one deals with supersolutions the situation is exactly the opposite.Notice that in our problem (2.2) both the infimum and the supremum appear.
Hence, as we are dealing with subsolutions, for the supremum, given η ε , we have a direction z ε as before.On the other hand, for the infimum we are free to choose the direction.We choose z ε that points inwards the domain (for example the inner unit normal to ∂Ω at the point x ε will do the job).We also choose a distance δ ε such that δε ε → 0. Now, we use exactly the same computations as in the previous case.For the infimum we use δ ε and z ε and for the supremum ε and z ε .As in the previous case, we reach a contradiction since the infimum is negative enough and we have a control of the supremum in such a way that the sum is still negative.
In the case of supersolutions the arguments works in an analogous way since we can interchange the roles of the infimum and the supremum in the previous computations (notice that for supersolutions we are reversing the inequalities).
Remark 2.2.To deal with the problem involving the trace fractional Laplacian ∆ s tr , we just observe that, for the terms that involve intermediate fractional eigenvalues, with i = 2, ..., N − 1, we can choose a direction that almost reaches this quantity.
In fact, when we deal with subsolutions we use that, since Λ s 1 u involves only the infimum, we have freedom to choose the direction.For the terms that involve the other eigenvalues we have that, given η ε > 0, we can choose a direction, z ε , such that and then, by the same computations that we made before (we just have to add a finite number of terms involving η ε ) we reach a contradiction.
Notice that, when dealing with supersolutions we use the freedom in the choice of the direction in the term that comes from Λ s N u (this is the term that involves a supremum) and bound all the other terms (that involve inf/sup).
Remark 2.3.In [5] is is used that the domain is strictly convex.Here we do not need this condition, since we have both the infimum and the supremum among directions in our operator and hence we can choose the direction in one of the terms (the one with the infimum or the one with the supremum) when dealing with sub and super solutions.Now, we are ready to state and prove the comparison principle for sub and super solutions to our problem.In this proof we again follow ideas from [5] but we have to introduce a different function Ψ ε (see below) and the integrals that appear are also split in a different way.Again, here we are not assuming that the domain is strictly convex (as was needed for the arguments in [5]).
Theorem 2.2.Assume that g ∈ C(R N \ Ω) is bounded and that Ω is a bounded C 2 −domain.Let u, v : R N → R be a viscosity sub and supersolution to (2.2) in Ω, in the sense of Definition 2.1, then As usual, we argue by contradiction, that is, we assume that M > 0. Since u g and v g are upper and lower semicontinuous functions, For any ε > 0, we define Observe that Moreover, M ε1 ≤ M ε2 for all ε 1 ≤ ε 2 .Then, there exists the limit lim On the other hand, since u g and −v g are upper semicontinuous functions, for any ε, Ψ ε is an upper semicontinuous function.Thus, there is ( Observe that Since B R is compact, extracting a subsequence if necessary, we can assume that Thus M = u g (x) − v g (x).This limit point x cannot be outside Ω, since M > 0, and by Theorem 2.1, it cannot be on ∂Ω.Consequently, x ∈ Ω and we may assume (without loss of generality) that provided ε is small enough.
On the other hand, by (2.6), for any w ∈ R N such that we have ε are test functions for u and v at x ε and y ε , respectively.Then, we have that for all δ ∈ (0, d ε ).
At this point we have to choose two sequences of directions, one to approximate the supremum and another one for the infimum.As before, for the subsolution we are free to choose directions in the infimum; while for the supersolution we can choose a direction close to the supremum.
By the definition of E δ , for each h > 0 there exists and thus, we get (2.9) Now, our goal is to obtain upper estimates for the differences 2 .In this way, when we substract the second expression to the first one in (2.9), we get a contradiction, provided we controlled the differences properly.
We can assume that To estimate the first difference, let us write where and We first observe that there is a positive constant C independent of δ, ε and h such that For the estimate of a.e. as ε, h → 0. Hence, Then, we divide the integrals I 2 and J 2 further Now, we observe that the terms in which we have the difference of characteristic functions go to zero, using the boundedness of u g , v g and dominated convergence theorem.The difference of the terms that have a segment in common is negative thanks to (2.8).
Collecting all these bounds we get lim sup δ,ε,h→0 Finally, we observe that we have a.e. as ε, h → 0, and hence, arguing similarly to the previous case and using that g is a bounded continuous function, we obtain lim ε,h→0 The negative term, −M , comes from the fact that Therefore, letting first δ → 0, then ε → 0, and h → 0, we get For the supremum part the limit can be bounded by exactly the same quantity (with a possible different limit direction z 0 ), The fact that we get a negative bound came from having and not from the fact that we analyze the infimum.Thus, substracting both expressions of (2.9) and taking limits we obtain we get and we end up with the desired contradiction.
Remark 2.4.This proof can be extended to deal with the problem involving the trace fractional laplacian (−∆) s tr .As before we just observe that, for the terms that involve the intermediate fractional eigenvalues, with i = 2, ..., N − 1, we can choose a subspace and then a direction that almost reach the associated quantity With the same computations one can obtain a comparison principle (and then existence and uniqueness of solutions, see below) for C 2 domains and continuous and bounded exterior data, g, for problems of the form (2.10) as long as a 1 > 0, a N > 0 and a i ≥ 0, i = 2, ..., N − 1 with g continuous and bounded and Ω a C 2 bounded domain.
If only one of the maximum or minimum fractional eigenvalues (Λ s 1 u or Λ s N u) is involved in the operator, then the proof still works (as in [5]) with the extra assumption that the domain is strictly convex.This assumption (strict convexity) ensures that, close to the boundary, the line in any direction reaches the boundary close to the nearest point on the boundary.We refer to Section 5 for extra comments on extensions of our results.

2.3.
Existence and uniqueness of a solution.This part is standard in the viscosity theory once one has at hand a comparison principle, but we include the details here for completeness.Now our goal is to show existence and uniqueness of a solution to (2.2) (the equation involves only Λ s 1 u and Λ s N u) and (2.1) (the equation is given by the trace fractional Laplacian, that is, the sum of the fractional eigenvalues).The same proof works for any of the operators that involve fractional eigenvalues as long as we have a comparsion principle, see Remark 2.4.
The proof of existence is by using Perron's method and uniqueness is immediate from the comparison principle.
Theorem 2.3.Assume that g ∈ C(R N \ Ω) is bounded and Ω is a bounded C 2 −domain.Then, there is a unique viscosity solution u to (2.2) or to (2.1) in Ω, in the sense of Definition 2.1.This unique solution is continuous in Ω and the datum g is taken with continuity, that is, u| ∂Ω = g| ∂Ω .
Proof.Again we use ideas from [4] to obtain the existence of a viscosity solution to our Dirichlet problem.
Existence of u, v : R N → R that are a viscosity subsolution and a viscosity supersolution in Ω, in the sense of Definition 2.1, follows easily taking large constants (here we are using that g is bounded).
We take a one-parameter family of continuous functions Then for all k ∈ N we consider the obstacle problem (2.11) for x ∈ R N , which is degenerate elliptic, that is, it satisfies the general assumption (E) of [3].It has ± g ∞ as viscosity super and subsolution.Recall that these viscosity supersolution and subsolution do not depend on the L ∞ bounds of ψ k ± .Then, in view of the general Perron's method given in [3] for problems in R N , since condition (E) holds, we conclude the existence of a continuous bounded viscosity solution u k to (2.11) for each k and, in addition, this family of solutions is equal to g in R N \ Ω for all k.Moreover, we have that The following corollary is immediate and we just estate it here since it is a part of Theorem 1.2.
Corollary 2.1.The comparison principle implies that if g 1 ≥ g 2 are boundary conditions of solutions u 1 and u 2 respectively, we have that u 1 ≥ u 2 .
Proof.This follows from considering u 2 a supersolution for the boundary condition g 1 or considering u 2 a subsolution for the boundary condition g 2 .
Also as an immediate corollary of the comparison principle we obtain continuous dependence of the solution with respect to the exterior data, this gives another part of Theorem 1.2.
Corollary 2.2.The comparison principle implies that the solution depends continuously on the exterior data; it holds that Proof.Just observe that is a supersolution to the problem with exterior datum g 1 , hence, by comparison we get In a similar way, we get that since the left hand side is a subsolution to the problem with exterior datum g 1 .Now, our goal is to show that (2.2) has a strong maximum principle, a viscosity subsolution to the problem can not attain the maximum inside Ω unless it is constant.
Theorem 2.4 (Strong maximum principle).Let u be a subsolution to problem (2.2) such that u g attains a maximum at some point x ∈ Ω.Then u is constant.
Proof.We argue by contradiction.Let us assume u is not constant.That means there is some x 0 in Ω such that u(x) > u(x 0 ).Since x ∈ Ω, we can take 0 < δ < |x − x 0 |/2 and φ(x) = u(x) for x ∈ B δ (x).We extend this φ to all R N in a smooth way.Since u is a subsolution, inf For the infimum, we can choose a direction freely, so let us choose z 0 = x 0 − x.For the supremum, fix z k such that sup This way we get Both expressions lack the integral where the test function appears due to the fact that the test function is constant.What we have to estimate is For every z we have that I(z) ≤ 0 thanks to the fact that the difference inside the integral is nonpositive.But for z 0 , we get an estrict inequality.The reason behind this is that upper continuity grants the existence of a ball centered at x 0 of radius sufficiently small so that for every y in that ball, u(y) < u(x).This yields that for a certain interval centered around t = 1, the difference u(x + tz 0 ) − u(x) < 0, so we can conclude that I(z 0 ) < 0. Hence, Taking the limit k → +∞ we get 0 < 0, arriving at the desired contradiction.
With the same idea one can show that a supersolution to the problem (2.2) that attains a minimum at some point x ∈ Ω must be constant.
Remark 2.5.In our proof of the strong maximum principle we need to choose directions.When we deal with subsolutions we can choose the direction that is involved in the infimum and hence any non-constant solution to an equation given in terms of a sum of fractional eigenvalues can not attain an interior maximum provided Λ s 1 u appears in the operator.Analogously, when Λ s N u appears in the operator non-constant solutions can not have interior minima.
Corollary 2.3.For an exterior datum g 0 we have that the corresponding solution to (2.2) is strictly positive in Ω, u(x) > 0, x ∈ Ω.
Finally, we show a strong comparison principle, provided that the exterior data verify that g 1 ≥ g 2 and there exists a point x in every line that passes trough Ω such that g 1 (x) > g 2 (x).Theorem 2.5 (Strong comparison principle).Assume g 1 , g 2 ∈ C(R N \ Ω) with g 1 ≥ g 2 and in every line that passes trough Ω there exists a point x ∈ R N \ Ω such that g 1 (x) > g 2 (x).Let u 1 , u 2 be solutions of our problem with boundary conditions g 1 and g 2 respectively.Then, u 1 > u 2 in Ω.
Proof.We already know, thanks to the comparison principle, that u 1 ≥ u 2 .Arguing by contradiction, assume that there exists and define Next, we proceed to construct an auxiliary function that will yield a test function for u 1 from below and a test function for u 2 from above like the one used in the proof of Theorem 2.2.We aim to use the fact that u 1 is a viscosity supersolution and u 2 a viscosity subsolution. Let It is simple to observe, following similar steps to the ones in the proof of Theorem 2.2, that S ε2 ≤ S ε1 for ε 1 ≤ ε 2 and S ε ≤ 0. Using the continuity of ψ ε we know that there exists (x ε , y ε ) Taking a subsequence, we may assume x ε and y ε converge to x and y, and we know that x = y thanks to lim Hence, we get This implies x ε and y ε converge to a minimum point of u 1 − u 2 , and since this point belongs to Ω −η , it follows that it is inside Ω.With no loss of generality we will assume x = x 0 .
Thanks to the previous arguments, we have found test functions for u 1 and u 2 at the points x ε and y ε respectively.We now use the fact that u 1 is a viscosity supersolution and u 2 a subsolution so that Now we use the same strategy as in Theorem 2.2 to get rid of the infimum and the supremum of the expressions.For each h > 0 there exists z Then, Here the contradiction will follow from the fact that, lim By hypothesis, for some t 0 , (x 0 + t 0 z 0 ) ∈ Ω we get Now, using the continuity of the exterior data g 1 and g 2 , the strict inequality holds on some interval I. Doing exactly the same computations on the other set of directions, after taking limits, we find the desired contradiction Therefore, we conclude that u 1 > u 2 in Ω.
Remark 2.6.As a simple example where Theorem 2.5 can be applied, one can take functions g 1 , g 2 that verify g 1 ≥ g 2 with g 1 (x) > g 2 (x) for x inside an annulus r < |x| < R and a domain Ω ⊂ B r (0).
Remark 2.7.To obtain that u 1 > u 2 we need to assume that in every line that passes trough Ω there exists a point x ∈ R N \ Ω such that g 1 (x) > g 2 (x).This condition is necessary, in fact, in the next section we will construct data with g 1 g 2 in R N \ Ω and such that the corresponding solutions coincide, u 1 ≡ u 2 in Ω.

The trace fractional Laplacian is nonlinear
Our goal in this section is to show that the trace fractional Laplacian and the mid-range fractional Laplacian are nonlinear operators.Since in R 2 both operators coincide up to a constant we perform the argument for N = 2.
Theorem 3.1.The problems (1.9) and (1.10) are nonlinear problems.There exist data g 1 , g 2 such that Proof.We will argue in R 2 and denote a point as (x, y).To begin, we consider a(x) a viscosity solution in R to . This viscosity solution is going to be a strong solution in (−1, 1), and in fact we can assume it is at least C 2 , so we can drop the principal value.
Define in R 2 the function u(x, y) = a(x) − a(y).
Observe that, for a unitary vector v = (v x , v y ), we have We can easily check that the supremum and the infimum of this expression are achieved at v = (1, 0) and v = (0, 1) respectively.Thus, this particular u(x, y) satisfies our equation in Ω = (−1, 1) × (−1, 1).We have, This problem has the unique solution that we already constructed, w(x, y) = u(x, y) (uniqueness comes from the comparison principle).Now, if we adequately add a little perturbation far from the origin supported close to a point (x, x) on the diagonal to obtain a new exterior datum, we will get that for this different exterior datum we get the same solution.
Let f (x, y) be a radially non-increasing nonnegative and nontrivial cut-off function such that f (x, y) = 1 for (x, y) ∈ B r (x, ŷ) and f (x, y) = 0 for (x, y) ∈ R 2 \ B r (x, x) for some far away point (x, x) in a diagonal and some small r.Now, let us take g(x, y) = u(x, y) + εf (x, y) as exterior datum.Then, using calculations similar to those performed before we get that R g(x + tv x , y + tv y ) − u(x, y) The ε > 0, which we can assume to be small, is the effect of the perturbation f .Both the supremum and infimum are still going to be achieved for v = (1, 0) and v = (0, 1).This is quite immediate for the infimum, since ε is positive.The supremum does not change because, if |v x | ∼ |v y |, then in a diagonal direction the above expression will be something similar to 2ε, which can be chosen smaller than 1 and thus, z = (1, 0) achieves a greater value.Therefore, even after altering the exterior datum by adding a nonegative and nontrivial perturbation, u(x, y) is still a solution inside B 1 (0).Now, let us substract the two functions in order to obtain a nonnegative and nontrivial exterior datum (the perturbation).If our operator was linear, from our previous results we obtain that the solution with this exterior datum is strictly positive inside B 1 (0) (we use the strong maximum principle, Corollary 2.3).This proves the nonlinearity of the problem, since the difference of two solutions that coincide in B 1 (0) is not the solution for the difference of the exterior data.

Limit as s ր 1
In this section, for a fixed C 2 domain Ω and a fixed continuous and bounded exterior datum g we study the limit as s ր 1 of the solutions u s (we make explicit the dependence of the solution in s along this section).
Recall that the fractional eigenvalues are given by Here we will use the explicit constant given by however, as we have mentioned in the introduction, any c(s) with c(s) ∼ (1 − s) will give the same limit.
Next, our goal is to show that u s , the unique solution to converges uniformly as s ր 1 to the unique solution to x ∈ ∂Ω.
In addition, with the same arguments, we obtain that when u s is the unique solution to in C(Ω) with the limit u is given by the unique solution to the local problem (4.5) x ∈ ∂Ω.
First, we show that the half-relaxed upper limit of the u s is a subsolution to the limit problem.Then u is a viscosity subsolution to the Dirichlet problem for the classical local Laplacian (4.4).
Proof.First, we notice that u s are uniformly bounded in s.This fact can be easily obtained using the comparison principle and that w = g ∞ is a supersolution to (4.3).Therefore, the half-relaxed upper limit, u(x) := lim sup is well-defined and bounded.By definition, u is an upper semicontinuous function.
Again, to simplify the notation, we will prove the result in R 2 , that is, we take N = 2, since in this case we have only two eigenvalues (one is given by an infimum among directions and the other by a supremum).At the end of the proof we will add a few lines on how to treat the general case.
Choose x ∈ Ω and φ a test function such that u − φ attains a maximum at x.We can assume that (φ − u)(x) = 0 and that (ū − φ)(x) > 0 for x = x.Since u s is upper semicontinuous by definition, ψ s = u s − φ reaches a maximum point at x s ∈ Ω.By the definition of half-relaxed limit, we have \ {x} , ∀k ∈ N Choose some k.For this k, we can find s k , y k such that By extracting a subsequence, we can assume x s k → x 0 for some x 0 ∈ Ω.Then, Since the only maximum point of ψ was x, we deduce x 0 = x, and thus lim that is, the sequence of maximum points converge to the maximum of the halfrelaxed limit.Now, we can use φ as a test function at x s k for the subsolutions u s k .Let us choose z k , z k unitary vectors such that inf for some 0 < δ < d (x,∂Ω) 2 < d (x s , ∂Ω), a condition that we can assume with no loss of generality.Here E s z,δ (•, φ, x) is given as in Section 2 (without multiplying by c(s) the corresponding integrals).Notice that here we write E s z,δ (•, φ, x) to make explicit that the operator depends on s.Now, as before, we write From this previous estimate it is clear we must take limits first in s, and then in δ.
For the first integral, using a second order Taylor expansion of φ(x x k + tz k,ε ) around t = 0, we get Hence, using the precise expression for c(s) (that implies c(s) ∼ (1 − s) as s ր 1), taking a subsequence such that z k → ẑ for some ẑ, and using continuity of D 2 φ(x)z, z in both x and z, we obtain lim Therefore, collecting the previous results, we obtain that for any convergent sequence z k → ẑ, it holds that lim We would like to check that We can get this last inequality taking limits in In fact, if we take the limit as s ր 1 (k → ∞) and as δ ց 0 we have lim Then, from (4.6), we obtain as we wanted to show.
For the supremum the proof is exactly the same, and hence we conclude that As u s is a subsolution, we have 0 ≤ inf and then in the limit we get that showing that u is a viscosity subsolution.
Moreover, from Theorem 2.1 we obtain that the half-relaxed limit satisfies u g ≤ g on ∂Ω (we have a uniform barrier at every boundary point that implies that u s ≤ g on ∂Ω).
For the general case of N eigenvalues, we just observe similar arguments to the previous ones, imply that the limit lim from where the result follows as before.
Remark 4.1.Analogously, we can obtain that the half-relaxed lower limit of supersolutions u s , given by, u(x) := lim inf s→1 − y→x u s (y), is a supersolution to the limit problem (4.4).Now, we are ready to prove the convergence result.It also holds that the solutions to the mid-range fractional Laplacian converge to the solution to the limit problem (4.5).
Proof.By the previous result, we know half-relaxed limits of viscosity subsolutions and supersolutions converge respectively to subsolutions and supersolutions of the equivalent Laplacian problem.Using the definition of the half-relaxed limit and comparison, we conclude that the half-relaxed limits coincide lim inf and is the solution to (4.4) (since it is both a sub and a supersolution).Hence, we have that the solutions u s converge to the solution of the limit problem in the following sense, lim s→1 − y→x u s (y) = u(x).
We only have to check that the convergence is uniform.If the convergence was not uniform, for any sequence s n ր 1 there exists some ε 0 > 0 and a corresponding sequence {x sn } ⊂ Ω such that |u sn (x sn ) − u(x sn )| ≥ ε 0 But in a compact set Ω we can assume x sn converges to some x 0 ∈ Ω after taking a subsequence.Due to its definition, lim n u sn (x sn ) = u(x 0 ).We arrive at a contradiction and the convergence had to be uniform.

Possible extensions
In this last section we briefly comment on possible extensions of our results.

5.1.
A nontrivial righthand side.One can obtain similar existence, uniqueness and comparison results for as long as f is continuous in Ω.For Hölder regularity results for the solutions when g = 0 and s is close to 1 we refer to [13].Here, to use our previous arguments, we need that a i are continuous in Ω and nonnegative with a 1 and a N strictly positive.
Associated with this idea of introducing coefficients in our model problem we can define fractional Pucci operators (described in terms of the fractional eigenvalues) considering, for two real constants 0 < θ ≤ Θ, Λ s i u(x) + θ These operators are extremal operators in the class of fractional trace Laplacians with coefficients between θ and Θ.
Existence, uniqueness and a comparison principle for the Dirichlet problem for the operators P + θ,Θ (u) and P − θ,Θ (u) in C 2 domains with continuous and bounded exterior data can be proved as in Section 2. Notice that here we are computing the supremum (or the infimum) of fractional Laplacians of the function u restricted to subspaces of dimension j and hence the singularity of the kernel is of the form |y| −j−2s .
Then, with these operators at hand one can define a different version of the fractional Laplacian, Remark that the same procedure with the maximum and minimum of the local usual Laplacian acting on subspaces gives the Laplacian in the whole space.In fact, we have that is, the supremum over subspaces S of dimension j os the Laplacians of u restricted to S is given by the sum of the j largest eigenvalues of D 2 u(x) and similarly the infimum among subspaces of dimension i is the sum of the smallest eigenvalues of D 2 u(x).Therefore, for any pair (j, i) such that j + i = N we have Here we are computing the supremum among subspaces S of dimension j of the infimum of fractional Laplacians (of dimension i) of u restricted to subspaces T included in S. Notice that the fractional eigenvalues Λ j (u)(x) that we used here to define the trace fractional Laplacian are given by With these operators W + j,i (u) we can obtain a fractional version of the Laplacian adding them (taking care of the fact that the dimensions of the corresponding subspaces add up to N ), that is, just consider (− ∆) s (j1,i1)...(j k ,i k ) (u)(x) = − k l=1 W + j l ,i l (u)(x).
It should be interesting to know if there is a comparison principle for operators like this.
Then, we consider ψ k ± in such a way ψ k ± (x) → ±∞ as k → ∞ for all x ∈ Ω and denoting ū(x) = lim sup k→∞,y→xu k (y); u(x) = lim inf k→∞,y→x u k (y),which are well defined for all x ∈ R N , we clearly have that ū ≥ u in R N .Thus, we have u = ū = g in R N \ Ω and ū and u respectively are a viscosity subsolution and a viscosity supersolution to our problem.Thus, by comparison we get ū ≤ u in R N , and therefore we conclude that ū and u coincide and that u := ū = u is a continuous viscosity solution that satisfies the boundary condition in the classical sense.Uniqueness of solutions follows from the comparison principle.

Theorem 4 . 1 .
Let u s be viscosity subsolutions to our Dirichlet problem for the trace fractional Laplacian (4.3), and define u as the half-relaxed upper limit, u(x) := lim sup s→1 − y→x u s (y).

Theorem 4 . 2 (
Harmonic convergence).Let u s be a viscosity solution to our problem for the trace fractional Laplacian, (4.3).Then u s converge uniformly to the solution u of the Dirichlet problem for the local Laplacian, (4.4).