Fractional elliptic equations in nondivergence form: definition, applications and Harnack inequality
FFRACTIONAL ELLIPTIC EQUATIONS IN NONDIVERGENCE FORM:DEFINITION, APPLICATIONS AND HARNACK INEQUALITY
PABLO RA ´UL STINGA AND MARY VAUGHAN
Abstract.
We define the fractional powers L s = ( − a ij ( x ) ∂ ij ) s , 0 < s <
1, of nondiver-gence form elliptic operators L = − a ij ( x ) ∂ ij in bounded domains Ω ⊂ R n , under minimalregularity assumptions on the coefficients a ij ( x ) and on the boundary ∂ Ω. We show thatthese fractional operators appear in several applications such as fractional Monge–Amp`ereequations, elasticity, and finance. The solution u to the nonlocal Poisson problem (cid:40) ( − a ij ( x ) ∂ ij ) s u = f in Ω u = 0 on ∂ Ωis characterized with a local degenerate/singular extension problem. We develop the methodof sliding paraboloids in the Monge–Amp`ere geometry and prove the interior Harnack in-equality and H¨older estimates for solutions to the extension problem when the coefficients a ij ( x ) are bounded, measurable functions. This in turn implies the interior Harnack in-equality and H¨older estimates for solutions u to the fractional problem. Introduction
In this paper, we inaugurate the analysis of equations driven by fractional powers ofnondivergence form uniformly elliptic operators(1.1) L s = ( − a ij ( x ) ∂ ij ) s in Ω for 0 < s < , under minimal regularity assumptions on the coefficients a ij ( x ) and the boundary of thedomain Ω ⊂ R n , n ≥
1. We show in Section 2 that fractional power operators as in (1.1) insuch minimal regularity regime arise in applications to fractional Monge–Amp`ere equations,elasticity, and finance, among others.The very first difficulty we need to overcome when considering (1.1) is that of giving ameaningful definition of the fractional power operator L s when(1.2) L = − a ij ( x ) ∂ ij ≡ − n (cid:88) i,j =1 a ij ( x ) ∂ ij x ∈ Ωis an elliptic operator in nondivergence form with nonsmooth coefficients. As in other well-known cases, this is not immediately obvious. For example, the fractional Laplacian ( − ∆) s in R n can be defined using the Fourier transform as (cid:92) ( − ∆) s u = | ξ | s (cid:98) u . However, the nondiver-gence form operator (1.2) has nonsmooth coefficients in a bounded domain Ω, so the Fouriertransform does not seem to be the most convenient tool. If − ∆ D denotes the Laplacian ina bounded domain Ω subject to homogeneous Dirichlet boundary conditions on ∂ Ω, then
Mathematics Subject Classification.
Primary: 35R11, 35B65, 35J96. Secondary: 35B45, 35J70,47D06.
Key words and phrases.
Fractional nondivergence form elliptic equations, Monge–Amp`ere equations, reg-ularity estimates.Research supported by Simons Foundation grant 580911. a r X i v : . [ m a t h . A P ] D ec P. R. STINGA AND M. VAUGHAN ( − ∆ D ) s is naturally defined in a spectral way using the basis of eigenfunctions and the cor-responding eigenvalues of − ∆ D in the Sobolev space H (Ω). In contrast, there is no naturalHilbert space structure for nondivergence form operators as in (1.2). The spectral method isalso used to define fractional powers of divergence form elliptic operators ( − ∂ i ( a ij ( x ) ∂ j )) s ,see [4]. Nevertheless, our operator (1.2) has nonsmooth coefficients so it cannot be writtenin divergence form. We further remark that these definitions, though adequate from theoperator theory point of view, do not immediately give explicit pointwise, nonlocal formulas.Our idea to define (1.1) is to apply the method of semigroups . The main ingredient in thisapproach is the semigroup { e − tL } t ≥ generated by L . With this, we define(1.3) L s u ( x ) = 1Γ( − s ) ˆ ∞ (cid:0) e − tL u ( x ) − u ( x ) (cid:1) dtt s for 0 < s < x ∈ Ω, where Γ is the Gamma function. Using the semigroup, we can alsogive formulas for the solution u to the Poisson problem L s u = f as u = L − s f and for thesolution U to local extension problems. Moreover, if L has a heat kernel, then one can deriveexplicit pointwise expressions for L s u ( x ), L − s f ( x ) and U ( x, z ). These results are presentedin Section 3. For details about the semigroup method applied to the fractional Laplacian inthe whole space and to other different contexts, see [28] and the references therein.We then consider the following fractional elliptic equation in nondivergence form:(1.4) (cid:40) ( − a ij ( x ) ∂ ij ) s u = f in Ω u = 0 on ∂ Ω . Here Ω ⊂ R n , n ≥
1, is a bounded domain satisfying the uniform exterior cone condition.The coefficients a ij ( x ) : Ω → R are symmetric a ij ( x ) = a ji ( x ), i, j = 1 , . . . , n , a ij ( x ) ∈ C (Ω) ∩ L ∞ (Ω), and uniformly elliptic, meaning that there exist constants 0 < λ ≤ Λ suchthat(1.5) λ | ξ | ≤ a ij ( x ) ξ i ξ j ≤ Λ | ξ | for all ξ ∈ R n and a.e. x ∈ Ω . Under these conditions, the operator L = − a ij ( x ) ∂ ij generates a uniformly bounded C -semigroup with exponential decay. Therefore, L s u is well-defined by means of (1.3). SeeSection 3 for these notions and the necessary notation.The main regularity result for (1.4) in this paper is the following interior Harnack inequalityand H¨older regularity estimate. Theorem 1.1.
Assume that Ω ⊂ R n is a bounded domain that satisfies the uniform exteriorcone condition, a ij ( x ) ∈ C α (Ω) ∩ L ∞ (Ω) , for some < α < , are symmetric, satisfy (1.5) ,and f ∈ C (Ω) . There exist positive constants C H = C H ( n, λ, Λ , s ) > , κ = κ ( n, s ) < , and ˆ K = ˆ K ( n, s ) > such that for every ball B ˆ KR = B ˆ KR ( x ) satisfying B ˆ KR ⊂⊂ Ω and everynonnegative u ∈ Dom( L ) satisfying (1.6) ( − a ij ( x ) ∂ ij ) s u = f in B ˆ KR , we have that (1.7) sup B κR u ≤ C H (cid:18) inf B κR u + (cid:107) f (cid:107) L ∞ ( B ˆ KR ) R s (cid:19) . Furthermore, there exist positive constants α = α ( n, λ, Λ , s ) < and ˆ C = ˆ C ( n, λ, Λ , s ) such that for any u ∈ Dom( L ) satisfying (1.6) , we have that, for every x ∈ B R ( x ) , (1.8) | u ( x ) − u ( x ) | ≤ ˆ CR α | x − x | α (cid:18) sup Ω | u | + (cid:107) f (cid:107) L ∞ ( B ˆ KR ) R s (cid:19) . RACTIONAL ELLIPTIC EQUATIONS 3
We mention that Grubb [12, 13] and Seeley [26] studied properties of fractional powers ofnondivergence form elliptic operators with smooth coefficients and smooth domains from theoperator theory and pseudo-differential operators points of view. In particular, their resultsdo not include the Harnack inequality and H¨older estimate in Theorem 1.1.Our proof of Theorem 1.1 is based on the extension problem characterization of fractionalpower operators in Banach spaces given by the method of semigroups in [10] (see [29] forthe case of Hilbert spaces). In our particular case, the extension result of [10] allows us torewrite the nonlocal equation (1.4) in an equivalent way as a local PDE problem.
Theorem 1.2 (Particular case of [10]) . Assume that the bounded domain Ω ⊂ R n satisfiesthe uniform exterior cone condition and that a ij ( x ) ∈ C (Ω) ∩ L ∞ (Ω) are symmetric andsatisfy (1.5) . If u ∈ Dom( L ) , then a solution U ∈ C ∞ ((0 , ∞ ); Dom( L )) ∩ C ([0 , ∞ ); C (Ω)) to the extension problem (1.9) a ij ( x ) ∂ ij U + z − s ∂ zz U = 0 in Ω × { z > } U ( x,
0) = u ( x ) on Ω × { z = 0 } U = 0 on ∂ Ω × { z ≥ } is given by (1.10) U ( x, z ) = s s z Γ( s ) ˆ ∞ e − s z /s /t e − tL u ( x ) dtt s and satisfies (cid:107) U ( · , z ) (cid:107) L ∞ (Ω) ≤ M (cid:107) u (cid:107) L ∞ (Ω) for some M > . Furthermore, U z + ∈ C ([0 , ∞ ); C (Ω)) and − ∂ z + U ( x,
0) = d s L s u ( x ) ∈ C (Ω) where d s = s s Γ(1 − s )Γ(1+ s ) > and ∂ z + U ( x,
0) = lim z → + U ( x, z ) − U ( x, z for all x ∈ Ω . If, in addition, a ij ( x ) ∈ C α (Ω) for some < α < , then the solution U in (1.10) is the uniqueclassical solution U ∈ C (Ω × (0 , ∞ )) ∩ C (Ω × [0 , ∞ )) such that lim z →∞ (cid:107) U ( · , z ) (cid:107) L ∞ (Ω) = 0 . Hence, to prove Theorem 1.1, we will show interior Harnack inequalities and H¨older esti-mates for solutions U to the local, degenerate/singular elliptic equation in (1.9) subject tothe Neumann boundary condition − ∂ z + U ( x,
0) = f ( x ) up to { z = 0 } , and then take thetrace at { z = 0 } . Towards this end, we define the even reflection of U in the variable z by˜ U ( x, z ) = U ( x, | z | ), for x ∈ Ω, z ∈ R . For convenience, we continue to use the notation U instead of ˜ U and notice that U , being symmetric across { z = 0 } , satisfies the equation(1.11) a ij ( x ) ∂ ij U + | z | − s ∂ zz U = 0 in Ω × { z (cid:54) = 0 } . Furthermore, if f ( x ) (cid:54) = 0 then ∂ z U is discontinuous across ( x, < s <
1, theequation (1.11) either degenerates or blows up at z = 0, unless s = 1 / ψ is givenby det D ψ = G . By taking the directional derivative ∂ e in a unit direction e to the equationand defining v = ∂ e ψ and g = ∂ e G , we find that v satisfies the linearized Monge–Amp`ereequation(1.12) tr( A ψ ( x ) D v ) = g. P. R. STINGA AND M. VAUGHAN
Here, A ψ ( x ) = det( D ψ ( x ))( D ψ ( x )) − is the matrix of cofactors of D ψ ( x ). Notice that(1.12) is a linear equation in nondivergence form that is elliptic as soon as D ψ > G >
0. However, it is not uniformly elliptic in general since the eigenvalues of A ψ ( x ) are not a priori controlled.For our degenerate equation (1.11), we consider the strictly convex function Φ = Φ( x, z ) : R n +1 → R given by Φ( x, z ) = 12 | x | + s − s | z | s , < s < . Then Φ is in C ( R n +1 ) but, when s > /
2, is not in C ( R n +1 ). Since the Hessian of Φ is D Φ( x, z ) = (cid:18) I | z | s − (cid:19) , where I denotes the identity matrix of size n × n , the linearized Monge–Amp`ere equationassociated with Φ is(1.13) tr(( D Φ) − D U ) = ∆ x U + | z | − s ∂ zz U = 0 for z (cid:54) = 0 . As the coefficients a ij ( x ) are uniformly elliptic, see (1.5), we see that the coefficients in (1.11)are comparable to the coefficients in (1.13).An important feature of the linearized Monge–Amp`ere equation is its intrinsic geometrythat was first discovered by Caffarelli–Guit´errez [3]. They proved Harnack inequality forclassical nonnegative solutions v to (1.12) when ψ ∈ C and g ≡
0, where the Euclideanballs and distance are replaced by Monge–Amp`ere sections and the Monge–Amp`ere quasi-distance, respectively. The Monge–Amp`ere sections associated to a convex, C function ψ arethe sublevel sets of ψ − (cid:96) where (cid:96) is any linear function, while the corresponding quasi-distanceis given by δ ψ ( x , x ) = ψ ( x ) − ψ ( x ) − (cid:104)∇ ψ ( x ) , x − x (cid:105) .We show that the geometry for our degenerate/singular equation (1.11) with Neumannboundary condition at { z = 0 } is given by the Monge–Amp`ere sections S R associated tothe strictly convex function Φ, that is, the sublevel sets of Φ − (cid:96) , and the Monge–Amp`erequasi-distance δ Φ in R n +1 . See Section 4 for more details. We prove the following Harnackinequality and H¨older regularity estimate for the extension equation in such sections. Theorem 1.3.
Let Ω be a bounded domain, a ij ( x ) : Ω → R be bounded, measurable functionsthat satisfy (1.5) and let f ∈ L ∞ (Ω) . There exist positive constants C H = C H ( n, λ, Λ , s ) > , κ = κ ( n, s ) < , and ˆ K = ˆ K ( n, s ) > such that for every section S ˆ K R = S ˆ K R ( x , z ) ⊂⊂ Ω × R and every nonnegative solution U ∈ C ( S ˆ K R \ { z = 0 } ) ∩ C ( S ˆ K R ) such that U issymmetric across { z = 0 } and U z + ∈ C ( S ˆ K R ∩ { z ≥ } ) to (1.14) (cid:40) a ij ( x ) ∂ ij U + | z | − s ∂ zz U = 0 in S ˆ K R ∩ { z (cid:54) = 0 }− ∂ z + U ( x,
0) = f on S ˆ K R ∩ { z = 0 } , we have that (1.15) sup S κ R U ≤ C H (cid:18) inf S κ R U + (cid:107) f (cid:107) L ∞ ( S ˆ K R ∩{ z =0 } ) R s (cid:19) . Consequently, there exist constants < α = α ( n, λ, Λ , s ) < / and ˆ C = ˆ C ( n, λ, Λ , s ) > such that, for every solution U ∈ C ( S ˆ K R \ { z = 0 } ) ∩ C ( S ˆ K R ) such that U is symmetric RACTIONAL ELLIPTIC EQUATIONS 5 across { z = 0 } and U z + ∈ C ( S ˆ K R ∩ { z ≥ } ) to (1.14) , (1.16) | U ( x , z ) − U ( x, z ) |≤ ˆ C R α [ δ Φ (( x , z ) , ( x, z ))] α (cid:32) sup S R ( x ,z ) | U | + (cid:107) f (cid:107) L ∞ ( S ˆ K R ( x ,z ) ∩{ z =0 } ) R s (cid:33) for every ( x, z ) ∈ S R ( x , z ) . Regularity estimates, such as Harnack inequalities, for the linearized Monge–Amp`ere equa-tion (1.12) have been studied by Caffarelli–Guti´errez [3], Forzani–Maldonado [8], Le [18], Mal-donado [19, 21], Savin [24], among others. In each case, they either assume that det( D ψ ) isbounded away from zero and infinity (that is, the Monge–Amp`ere measure µ ψ ( E ) = |∇ ψ ( E ) | , E ⊂ R n , is comparable to the Lebesgue measure), or that ψ is sufficiently regular, e.g. ψ ∈ C .For our function Φ, we have that D Φ either degenerates or blows up near { z = 0 } when s (cid:54) = 1 /
2, and, moreover, Φ / ∈ C when s > /
2. Therefore, (1.11) is not covered by suchprevious results. On the other hand, Maldonado proved Harnack inequality for degenerateelliptic equations associated with convex functions of the form ψ ( x ) = | x | p , p ≥
2, see [20].However, not only are his techniques different than the ones presented here but also his workdoes not include the singular case in which s > /
2. Moreover, when we write (1.14) in Ω × R as a single equation, we see that U satisfies (1.11) in Ω × R with a right hand side that is asingular measure with density f ( x ) supported on { z = 0 } .We develop a method of sliding paraboloids inspired by the work of Savin for fully nonlinearuniformly elliptic equations [25]. For our setting, we work with a Neumann problem in aMonge–Amp`ere geometry that brings additional challenges because Φ is only C and D Φis degenerate/singular. For this, we define paraboloids P : R n +1 → R of opening a > x v , z v ) by P ( x, z ) = − aδ Φ (( x v , z v ) , ( x, z )) + c where c is a constant. We lift these paraboloids from below until they touch the graphof U in a section S R for the first time. We estimate the Monge–Amp`ere measure of theresulting set of contact points by the Monge–Amp`ere measure of the set of vertices. Observethat, since our equation (1.11) is degenerate/singular and − ∂ z + U ( x,
0) = f ( x ), we needto be able to control the contact points ( x, z ) for which z = 0 in terms of the size of f .Next, we show that, by increasing the opening of these paraboloids, they almost cover thesection S R in measure. This relies on explicit barriers whose construction is very delicatebecause of Neumann boundary condition and the degeneracy/singularity of (1.11). Then, webuild a refined geometric argument to obtain a localization estimate. Thus, using a coveringargument, we can conclude the proof of Theorem 1.3 and deduce Theorem 1.1.Our function Φ was also considered in [22] to study the fractional nonlocal linearizedMonge–Amp`ere equation. They established Harnack inequality and H¨older estimates forsolutions to (1.4) when the coefficients a ij ( x ) are given by the matrix of cofactors of D ψ ,where ψ is a C strictly convex function and Ω is a section of ψ . Observe that in [22] the weakHarnack inequality is proved using the divergence form structure of the equation. Whereas,in (1.4), we not only consider general elliptic coefficients a ij ( x ), but also the equation cannotbe written in divergence form. Nevertheless, since the proof of the local boundedness of thesolution to the extension problem in [22] uses purely nondivergence form techniques, one caneasily check that solutions to our extension problem (1.9) satisfy the same local boundednessestimate as that of [22, Theorem 11.3].We additionally mention that Le in [18] proved Harnack inequality for the linearizedMonge–Amp`ere equation (1.12) when ψ ∈ C and 0 < λ ≤ det( D ψ ( x )) ≤ Λ, by using
P. R. STINGA AND M. VAUGHAN sliding paraboloids within the roadmap of the proof of Caffarelli–Guti´errez [3]. Again, ourmethods (inspired by Savin [25]) and results are different and independent of [18] (in partic-ular, Φ is not smooth when s > / D Φ is degenerate/singular, and we have the Neumannboundary condition − ∂ z + U ( x,
0) = f ( x )).Theorem 1.3 holds for bounded domains Ω and bounded, measurable coefficients a ij ( x ). InTheorem 1.1 we additionally require that Ω satisfies the uniform exterior cone condition andthat a ij ( x ) are H¨older continuous. There are several reasons for these technical assumptions.First, the uniform exterior cone condition and the hypothesis a ij ( x ) ∈ C (Ω) ∩ L ∞ (Ω) give usthe existence of an appropriate C -semigroup generated by L , so the fractional power operator L s can be defined using (1.3). Furthermore, under these conditions, the extension problemcharacterization in Theorem 1.2 holds. Second, our proof of Theorem 1.3 is for classical solutions U to the extension problem and does not require any continuity assumptions on a ij ( x ) nor geometric conditions on Ω. Third, to apply Theorem 1.3, we need to ensure thatthe solution U given in Theorem 1.2 is classical, and for this we must require a ij ( x ) ∈ C α (Ω).It is an open problem and will be the object of future work to define ( − a ij ( x ) ∂ ij ) s in boundeddomains when the coefficients are only bounded, measurable and to establish a correspondingextension equation and Harnack inequality for viscosity solutions to (1.4).The paper is organized as follows. First, in Section 2, we show several applications offractional powers of nondivergence form operators (1.1). Then, in Section 3, we preciselydefine the fractional operator ( − a ij ( x ) ∂ ij ) s and prove the extension characterization. InSection 4, we provide the necessary Monge–Amp`ere background associated to our functionΦ. We prove a sequence of reductions of Theorem 1.3 in Section 5. Section 6 containspreliminary results on the Monge–Amp`ere paraboloids P associated to Φ. Next, we establishseveral key results that will be used to prove the final reduction of Theorem 1.3. In Section 7,we estimate the Monge–Amp`ere measure of the set of contact points for sliding paraboloids offixed opening by the measure of the set of vertices. The delicate construction of the barriersis done in Section 8. These are used in Section 9 to prove a localization estimate by meansof a refined geometric argument. A Calder´on–Zygmund-type covering lemma is proved inSection 10. Finally, in Section 11, we present the proof of the final reduction of Theorem 1.3and the proof of Theorem 1.1. 2. Applications
In this section we present some applications where fractional powers of nondivergence formelliptic operators naturally arise.2.1.
Fractional Monge–Amp`ere equations. If u = u ( x ) is a convex, C function, thenone can check that the Monge–Amp`ere operator acting on u at a point x can be written as n det( D u ( x )) /n = inf (cid:8) ∆( u ◦ B )( B − x ) : B ∈ M (cid:9) = inf (cid:8) a ij ∂ ij u ( x ) : ( a ij ) = B , B ∈ M (cid:9) , where the infimum is taken over the class M of all positive definite, symmetric matrices B of size n × n such that det( B ) = 1. Motivated by these identities, Caffarelli–Charro definedin [2] the fractional Monge–Amp`ere operator by(2.1) D s u ( x ) = inf (cid:8) − ( − ∆) s ( u ◦ B )( B − x ) : B ∈ M (cid:9) , / < s < . On the other hand, it was shown in [16] that the operator in (2.1) can also be written as(2.2) D s u ( x ) = inf (cid:8) − ( − a ij ∂ ij ) s u ( x ) : ( a ij ) = B , B ∈ M (cid:9) , where ( − a ij ∂ ij ) s is the fractional power of the constant coefficients operator − a ij ∂ ij . RACTIONAL ELLIPTIC EQUATIONS 7
The fractional Monge–Amp`ere operator (2.1) is degenerate elliptic because the eigenvaluesof the matrices B ∈ M are not a priori controlled from below or above. Nevertheless, it isproved in [2] that if u is Lipschitz, semiconcave, and D s u ≥ η > D s becomes uniformly elliptic in u , that is, there is a constant λ > D s u ( x ) = D λs u ( x ) := inf (cid:8) − ( − ∆) s ( u ◦ B )( B − x ) : B ∈ M , B ≥ λI (cid:9) for all x ∈ Ω. Equivalently, in the description of (2.2), D λs u ( x ) = inf (cid:8) − ( − a ij ∂ ij ) s u ( x ) : ( a ij ) = B , B ∈ M , B ≥ λI (cid:9) . It was observed in [30] that, for each x ∈ Ω, the infimum above is attained at some matrix a ij = a ij ( x ). Therefore, the fractional Monge–Amp`ere operator in the uniformly ellipticregime is in fact given by D λs u ( x ) = − ( − a ij ( x ) ∂ ij ) s u ( x ) for every x ∈ Ω . In other words, D λs u ( x ) is the fractional power of the nondivergence form uniformly ellipticoperator L = − a ij ( x ) ∂ ij , where a ij ( x ) are bounded, measurable coefficients.2.2. Elasticity.
Consider an anisotropic elastic membrane represented by the graph of afunction U ( x, z ), for ( x, z ) ∈ Ω × [0 , ∞ ). Suppose that we place a thin obstacle φ : Ω → R onthe hyperplane { z = 0 } , such that φ ≤ ∂ Ω, which pushes U from below at { z = 0 } . Byfixing U = 0 on ∂ Ω × [0 , ∞ ), this problem is modeled by the following thin obstacle problem:(2.3) a ij ( x ) ∂ ij U + ∂ zz U = 0 in Ω × { z > } U ( x, z ) = 0 on ∂ Ω × { z ≥ } U ( x, ≥ φ ( x ) on Ω − ∂ z + U ( x, ≥ − ∂ z + U ( x,
0) = 0 on { U ( x, > φ ( x ) } . The last two conditions are called the
Signorini complementary conditions . They follow fromthe fact that φ is pushing U upwards, while U is actually free in the noncoincidence set { U ( x, > φ ( x ) } . The coefficients a ij ( x ) encode the heterogeneity of the membrane. Thethin obstacle problem (2.3) is equivalent to the problem of semipermeable cell membranes inbiology (see [7]), where a ij ( x ) are a model for the cytoplasm inside the cell.It follows from the extension problem characterization (see Theorem 1.2) with s = 1 / u ( x ) := U ( x,
0) satisfies − ∂ z + U ( x,
0) = ( − a ij ( x ) ∂ ij ) / u ( x ) in Ω . Therefore, U solves the thin obstacle problem (2.3) if and only if its trace u is the solutionto the following fractional obstacle problem ( − a ij ( x ) ∂ ij ) / u ≥ − a ij ( x ) ∂ ij ) / u = 0 in Ω ∩ { u > φ } u ≥ φ in Ω u = 0 on ∂ Ω . P. R. STINGA AND M. VAUGHAN
Finance.
Consider a particle moving randomly in a heterogeneous domain Ω that iskilled when it reaches the boundary ∂ Ω. This random behavior can be modeled by a dif-fusion process X t whose infinitesimal generator is a nondivergence form elliptic operator L = − a ij ( x ) ∂ ij in Ω, subject to homogeneous Dirichlet boundary conditions on ∂ Ω. In thissituation, the coefficients a ij ( x ) serve as a measure of the anisotropy of the medium, or thepreferred directions the particle chooses at every point x . A model for particles randomlyjumping inside a heterogeneous medium that are killed as soon as they reach or try to crossthe boundary can be given by subordinating the process X t with a 2 s -stable L`evy subordi-nator T t , for 0 < s <
1. The resulting subordinated process Y t = X T t is then generated bythe fractional power operator L s = ( − a ij ( x ) ∂ ij ) s , 0 < s <
1. See [15] for the case of smoothcoefficients and domains, and [27] for the case when X t is a Wiener process.Next, let τ be the optimal stopping time that maximizes the function u ( x ) = sup τ E [ φ ( Y τ ); τ < + ∞ ] , where φ ∈ C (Ω) (see (3.3)), E denotes the expected value, and the process Y is set to startat x ∈ Ω. It turns out that u is the solution to the following obstacle problem:(2.4) ( − a ij ( x ) ∂ ij ) s u ≥ − a ij ( x ) ∂ ij ) s u = 0 in Ω ∩ { u > φ } u ≥ φ in Ω u = 0 on ∂ Ω . These free boundary problems appear in financial models (see [6]) where u is the value of aperpetual American option in which the asset prices are modeled by Y t and φ is the payofffunction.3. Fractional powers of elliptic operators and extension problem
Here, we give the precise definition of the fractional power operator L s = ( − a ij ( x ) ∂ ij ) s in (1.4) and present the extension problem characterization, i.e. Theorem 1.2. For this, weapply the method of semigroups of [10, 29] (see also [28]) which we describe next.3.1. Method of semigroups for fractional power operators.
A family { T t } t ≥ of bounded,linear operators on a Banach space X is a semigroup on X if T = Id and T t ◦ T t = T t + t for every t , t ≥ , where Id denotes the identity operator. We say that a semigroup { T t } t ≥ is a C -semigroup if T t u → u as t → + for all u ∈ X . A semigroup { T t } t ≥ is uniformly bounded if its operatornorm is uniformly bounded in t , that is, there is a constant M ≥ (cid:107) T t (cid:107) ≤ M forall t ≥
0. The infinitesimal generator A of a semigroup { T t } t ≥ is the closed linear operatordefined as(3.1) − Au = lim t → + T t u − ut in the domain Dom( A ) = { u ∈ X : the limit in (3.1) exists } ⊂ X . In this case, we write T t = e − tA . Hence, if A is the infinitesimal generator of a C -semigroup { e − tA } t ≥ on X , thenthe function v = e − tA u , for u ∈ X , satisfies the heat equation for A : (cid:40) ∂ t v = − Av for t > v = u for t = 0 . RACTIONAL ELLIPTIC EQUATIONS 9
Conversely, a linear operator ( A, Dom( A )) on X is said to generate a semigroup if there isa semigroup { T t } t ≥ for which A is its infinitesimal generator, that is, T t = e − tA . Givena uniformly bounded C -semigroup { T t = e − tA } t ≥ on X , the fractional power A s of itsinfinitesimal generator is defined as A s u = 1Γ( − s ) ˆ ∞ (cid:0) e − tA u − u (cid:1) dtt s , for all u ∈ Dom( A ) ⊂ X, where 0 < s <
1. If the semigroup { e − tA } t ≥ has exponential decay , that is, (cid:107) e − tA (cid:107) ≤ M e − εt ,for some ε >
0, for all t ≥
0, then the negative power A − s , s >
0, is given by A − s f = 1Γ( s ) ˆ ∞ e − tA f dtt − s , for all f ∈ X. Thus, under the exponential decay assumption on { e − tA } t ≥ , given f ∈ X , the solution u ∈ Dom( A s ) to the fractional problem A s u = f is u = A − s f . Here Dom( A s ) is defined asthe range of A − s . For all these details, see [23, 31].Fractional powers A ± s of infinitesimal generators A of uniformly bounded C -semigroupscan be characterized by extension problems. For the case when X is a Hilbert space see [29],while for the case when X is a general Banach space see [10]. Theorem 3.1 (See [10, Theorems 1.1 and 2.1, Remark 2.2]) . Let ( A, Dom( A )) be the in-finitesimal generator of a uniformly bounded C -semigroup { e − tA } t ≥ on a Banach space X .Let < s < . Define, for y > and any u ∈ X , (3.2) U ( y ) = y s s Γ( s ) ˆ ∞ e − y / (4 t ) e − tA u dtt s . Then U ∈ C ∞ ((0 , ∞ ) , Dom( A )) ∩ C ([0 , ∞ ) , X ) is a solution to the extension problem (cid:40) − AU + − sy ∂ y U + ∂ yy U = 0 for y > y → + U ( y ) = u in X. Moreover, (cid:107) U ( y ) (cid:107) X ≤ M (cid:107) u (cid:107) X , for all y ≥ . Furthermore, if u ∈ Dom( A ) then − lim y → + y − s ∂ y U ( y ) = c s A s u = − s lim y → + U ( y ) − uy s in X where c s = Γ(1 − s )4 s − / Γ( s ) > . If, in addition, { e − tA } t ≥ has exponential decay and u ∈ Dom( A ) satisfies A s u = f , for some f ∈ X , then the solution U in (3.2) can also be written as U ( y ) = 1Γ( s ) ˆ ∞ e − y / (4 t ) e − tA f dtt − s and, in particular, − lim y → + y − s ∂ y U ( y ) = c s f and U (0) = u . Fractional powers of nondivergence form elliptic operators.
To give the defini-tion of (1.1), we need conditions on a ij ( x ) and Ω so that L as in (1.2) generates a uniformlybounded C -semigroup with exponential decay in an appropriate Banach space X . For this,we assume that the bounded domain Ω ⊂ R n satisfies the uniform exterior cone condition ,namely, that there is a right circular cone C such that for all x ∈ ∂ Ω there is a cone C x with vertex x that is congruent to C such that Ω ∩ C x = { x } . We define the Banach space X = C (Ω) by(3.3) C (Ω) = { u ∈ C (Ω) : u ≡ ∂ Ω } , endowed with the L ∞ -norm. Let L be the linear operator on C (Ω) given by(3.4) L = − a ij ( x ) ∂ ij , Dom( L ) = { u ∈ C (Ω) ∩ W ,n loc (Ω) : Lu ∈ C (Ω) } , where the coefficients a ij ( x ) ∈ C (Ω) ∩ L ∞ (Ω) are symmetric and satisfy (1.5). Under thesehypotheses, L generates a uniformly bounded C -semigroup on C (Ω) with exponential decay. Proposition 3.2 (See [1, Proposition 4.7]) . Assume that Ω ⊂ R n is a bounded domainthat satisfies the uniform exterior cone condition and that a ij ( x ) ∈ C (Ω) ∩ L ∞ (Ω) are sym-metric and satisfy (1.5) . The operator L defined by (3.4) generates a uniformly bounded C -semigroup, denoted by { e − tL } t ≥ , on C (Ω) , such that if u ∈ C (Ω) satisfies u ≥ , then e − tL u ≥ , for all t ≥ . Moreover, there are constants M ≥ and ε > such that (3.5) (cid:107) e − tL u (cid:107) C (Ω) ≤ M e − εt (cid:107) u (cid:107) C (Ω) , for all t ≥ . In other words, by Proposition 3.2 and the maximum principle for parabolic equations(see [9]), for any u ∈ C (Ω), the function v ( x, t ) = e − tL u ( x ) ∈ C ((0 , ∞ ) , Dom( L )) ∩ C ([0 , ∞ ) , C (Ω)) is the unique solution to the heat equation driven by L with initial data u :(3.6) ∂ t v ( x, t ) = a ij ( x ) ∂ ij v ( x, t ) in Ω × { t > } v ( x, t ) = 0 on ∂ Ω × { t ≥ } v ( x,
0) = u ( x ) on Ω × { t = 0 } . Now we can formalize the definition of the fractional power operator (1.1).
Definition 3.3.
Assume that the bounded domain Ω ⊂ R n satisfies the uniform exteriorcone condition and that a ij ( x ) ∈ C (Ω) ∩ L ∞ (Ω) are symmetric and satisfy (1.5). Considerthe Banach space C (Ω) and the operator L = − a ij ( x ) ∂ ij given by (3.4). We define thefractional power operator L s = ( − a ij ( x ) ∂ ij ) s : Dom( L ) → C (Ω) by(3.7) ( − a ij ( x ) ∂ ij ) s u ( x ) = 1Γ( − s ) ˆ ∞ (cid:0) e − tL u ( x ) − u ( x ) (cid:1) dtt s , < s < , x ∈ Ω . Remark 3.4 (Pointwise formula) . It is known that, under certain conditions on L , thesemigroup { e − tL } t ≥ has a heat kernel, namely, there is a function H t ( x, y ) such that e − tL u ( x ) = ˆ Ω H t ( x, y ) u ( y ) dy for all t > , x ∈ Ω . For example, the heat kernel exists and satisfies the Gaussian estimate(3.8) 0 ≤ H t ( x, y ) ≤ C e − c | x − y | /t t n/ for all t > , x, y ∈ Ω , for some constants C, c >
0, whenever the coefficients a ij ( x ) are H¨older continuous, see [9].In this situation, it follows from (3.7) that for any smooth function u ∈ Dom( L ),( − a ij ( x ) ∂ ij ) s u ( x ) = ˆ Ω (cid:0) u ( x ) − u ( y ) (cid:1) K s ( x, y ) dy + B s ( x ) u ( x ) for all x ∈ Ωwhere 0 ≤ K s ( x, y ) ≤ C n,s | x − y | − ( n +2 s ) , for x, y ∈ Ω, x (cid:54) = y , and B s ( x ) ∈ L ∞ (Ω). Therefore,the fractional operator L s is a nonlocal, integro-differential operator in Ω. Remark 3.5 (Negative fractional powers) . Let f ∈ C (Ω) and assume that u ∈ Dom( L s )is a solution to (1.4), that is, ( − a ij ( x ) ∂ ij ) s u = f in Ω. By Proposition 3.2, the semigroup { e − tL } t ≥ has exponential decay. Then u can be written as(3.9) u ( x ) = ( − a ij ( x ) ∂ ij ) − s f ( x ) = 1Γ( s ) ˆ ∞ e − tL f ( x ) dtt − s for all x ∈ Ω . RACTIONAL ELLIPTIC EQUATIONS 11
If the coefficients a ij ( x ) are H¨older continuous then we can use the heat kernel from Remark3.4 into (3.9) to write u ( x ) = ( − a ij ( x ) ∂ ij ) − s f ( x ) = ˆ Ω K − s ( x, y ) f ( y ) dy for all x ∈ Ωwhere, by the estimate (3.8), 0 ≤ K − s ( x, y ) ≤ C n, − s | x − y | − ( n − s ) , for all x, y ∈ Ω, x (cid:54) = y . Proof of Theorem 1.2.
We choose X = C (Ω) and A = L as in (3.4) so that, by Proposition3.2, L generates a uniformly bounded C -semigroup on C (Ω) with exponential decay. Then,the solution U ( y ) in (3.2) satisfies the properties stated in Theorem 3.1. With the changeof variables z = ( y/ (2 s )) s , we obtain that U ( z ) ≡ U ( x, z ) verifies the formulas and prop-erties of Theorem 1.2. If the coefficients a ij ( x ) are also H¨older continuous then, by interiorSchauder estimates (see [11, Theorem 9.19]), the solution U is classical. Moreover, by theweak maximum principle (see [11, Theorem 3.1]), there is at most one classical solution to(1.9) such that lim z →∞ (cid:107) U ( · , z ) (cid:107) L ∞ (Ω) = 0 . Using (3.5) it is easy to check that the solution U given by (1.10) has such decay at infinity and hence is the unique solution. (cid:3) Monge–Amp`ere setting
We present the necessary background for the Monge–Amp`ere geometry associated to equa-tion (1.11) as well as the notation that will be used throughout the rest of this work. Wereference the reader to [8, 14] for details about the Monge–Amp`ere geometry associated togeneral convex functions.Given 0 < s <
1, we define the functions ϕ : R n → R and h : R → R by(4.1) ϕ ( x ) = 12 | x | and h ( z ) = s − s | z | s . Notice that ϕ ∈ C ∞ ( R ) and h ∈ C ( R ) ∩ C ( R \ { } ) are strictly convex. Set(4.2) Φ( x, z ) = ϕ ( x ) + h ( z ) for all ( x, z ) ∈ R n +1 . We note that h (cid:48) ( z ) = s − s | z | s − z, h (cid:48)(cid:48) ( z ) = | z | s − , D Φ( x, z ) = (cid:18) I | z | s − (cid:19) . It is clear that h (cid:48) ( − z ) = − h (cid:48) ( z ) and h (cid:48) (0) = 0.The Monge-Amp`ere measure associated to a strictly convex function ψ ∈ C ( R n ) is theBorel measure given by µ ψ ( E ) = |∇ ψ ( E ) | for every Borel set E ⊂ R n , where | A | denotes the Lebesgue measure of a measurable set A ⊂ R n . Since ∇ ϕ ( x ) = x , itclear that µ ϕ ( E ) = | E | . Lemma 4.1.
For a Borel set I ⊂ R , µ h ( I ) = ˆ I h (cid:48)(cid:48) ( z ) dz. Consequently, for a Borel set E ⊂ R n +1 , µ Φ ( E ) = ˆ E h (cid:48)(cid:48) ( z ) dz dx. Proof.
Consider an interval ( a, b ) ⊂ R such that 0 ∈ ( a, b ). Note that h (cid:48) is monotoneincreasing, injective, and h (cid:48) ( z ) = 0 if and only if z = 0. Since h is C and strictly convex in R \ { z = 0 } , we have that µ h (( a, b )) = (cid:12)(cid:12) h (cid:48) (( a, b )) (cid:12)(cid:12) = (cid:12)(cid:12) h (cid:48) (( a, (cid:12)(cid:12) + (cid:12)(cid:12) h (cid:48) (0) (cid:12)(cid:12) + (cid:12)(cid:12) h (cid:48) ((0 , b )) (cid:12)(cid:12) = ˆ a h (cid:48)(cid:48) ( z ) dz + 0 + ˆ b h (cid:48)(cid:48) ( z ) dz = ˆ ba h (cid:48)(cid:48) ( z ) dz. The result follows for any interval and hence for any Borel set I ⊂ R . (cid:3) The
Monge-Amp`ere (quasi)-distance associated to a strictly convex function ψ ∈ C ( R n )is given by δ ψ ( x , x ) = ψ ( x ) − ψ ( x ) − (cid:104)∇ ψ ( x ) , x − x (cid:105) . By convexity, δ ψ ≥
0, and δ ψ ( x , x ) = 0 if and only if x = x . We use the terminologyquasi-distance when there exists a constant K ≥ δ ψ ( x , x ) ≤ K (min { δ ψ ( x , x ) , δ ψ ( x , x ) } + min { δ ψ ( x , x ) , δ ψ ( x , x ) } )for all x , x , x ∈ R n . For our functions ϕ , h , and Φ in (4.1) and (4.2), we have δ ϕ ( x , x ) = 12 | x | − | x | − (cid:104) x , x − x (cid:105) = 12 | x − x | δ h ( z , z ) = h ( z ) − h ( z ) − h (cid:48) ( z )( z − z ) δ Φ (( x , z ) , ( x, z )) = δ ϕ ( x , x ) + δ h ( z , z ) . We will later show that δ h , δ ϕ , and δ Φ are indeed quasi-distances (see Lemma 4.6).A Monge–Amp`ere section of radius
R >
0, centered at x ∈ R n associated to a strictlyconvex function ψ ∈ C ( R n ) is defined as S ψ ( x , R ) = { x ∈ R n : δ ψ ( x , x ) < R } . The supporting hyperplane to ψ at x is given by (cid:96) ( x ) = ψ ( x ) + (cid:104)∇ ψ ( x ) , x − x (cid:105) . Then, S ψ ( x , R ) = { x : ψ ( x ) − (cid:96) ( x ) < R } , and we see that the Monge–Amp`ere sections for ψ arethe sublevel sets of ψ − (cid:96) . In the case of ϕ in (4.1), the sections correspond to Euclidean ballswith the same center(4.3) S ϕ ( x , R ) = (cid:110) x : 12 | x − x | < R (cid:111) = B √ R ( x ) . The sections for h in (4.1) with radius R > R . Moreover, we willsee that they are comparable to intervals of radius R s (see Lemma 4.8).We say that the Monge–Amp`ere measure µ ψ is doubling with respect to the center of mass on the sections of ψ , written µ ψ ∈ ( DC ) ψ , if there is a constant C d > µ ψ ( S ψ ( x, R )) ≤ C d µ ψ (cid:18) S ψ ( x, R ) (cid:19) for all sections S ψ ( x, R ) . Here, we use the notation αS ψ ( x, R ) = { α ( y − x ∗ ) + x ∗ : y ∈ S ψ ( x, R ) } , for α >
0, where x ∗ is the center of mass of S ψ ( x, R ). On the other hand, we say that µ ψ is doubling with respectto the parameter on the sections of ψ if there is a constant C (cid:48) d > µ ψ ( S ψ ( x, R )) ≤ C (cid:48) d µ ψ (cid:18) S ψ (cid:18) x, R (cid:19)(cid:19) for all sections S ψ ( x, R ) . It can be seen that (4.4) implies (4.5), but the converse is not true in general, see [14].Finally, we say that ψ satisfies the engulfing property if there is a constant θ ≥ S ψ ( x, R ), if x ∈ S ψ ( x, R ), then S ψ ( x, R ) ⊂ S ψ ( x , θR ). RACTIONAL ELLIPTIC EQUATIONS 13
For the next result, see Theorem 5 in [8] and the comments following it.
Theorem 4.2.
Let ψ ∈ C ( R n ) be a strictly convex function. The following are equivalent.(1) µ ψ ∈ ( DC ) ψ ;(2) ψ satisfies the engulfing property;(3) there are constants c, C > such that cR n ≤ | S ψ ( x, R ) | µ ψ ( S ψ ( x, R )) ≤ CR n for all sections S ψ ( x, R ) ;(4) δ ψ is a quasi-distance.All the statements are equivalent in the sense that the constants in each property only dependon each other. In order to show that our convex function Φ in (4.2) satisfies Theorem 4.2, we need tointroduce the notion of Monge–Amp`ere cubes associated with Φ. Many of our proofs willrely on the fact that Φ( x, z ) = ϕ ( x ) + h ( z ) has separated variables. Definition 4.3. A Monge–Amp`ere cube of radius
R >
0, centered at ( x, z ) ∈ R n +1 , associ-ated to Φ is given by Q R ( x, z ) = S ϕ ( x , R ) × · · · × S ϕ n ( x n , R ) × S h ( z, R )where x = ( x , . . . , x n ) and ϕ i : R → R is defined by ϕ i ( x i ) = | x i | for i = 1 , . . . , n . Notation 4.4.
We will always use the following notation. • x = ( x , . . . , x n ) ∈ R n , z ∈ R . • S R ( x ) ⊂ R n is a section of radius R > ϕ centered at x . • S R ( z ) ⊂ R is a section of radius R > h , centered at z . • S R ( x, z ) ⊂ R n +1 is a section of radius R > x, z ). • Q R ( x ) ⊂ R n is a cube of radius R > ϕ centered at x . • Q R ( z ) ⊂ R is a cube of radius R > h centered at z . • Q R ( x, z ) ⊂ R n +1 is a cube of radius R > x, z ).The relation between Monge–Amp`ere cubes and sections is given by the following result.
Lemma 4.5 (Lemma 6 in [8]) . Fix m ∈ N . For each j = 1 , . . . , m , let ψ j : R n j → R bestrictly convex, differentiable functions. Set n = (cid:80) mj =1 n j and define ψ ( x ) = m (cid:88) j =1 ψ j ( x j ) , x = ( x , . . . , x m ) ∈ R n , x j ∈ R n j . Then S ψ ( x, R ) ⊂ m (cid:89) j =1 S ψ j ( x j , R ) ⊂ S ψ ( x, mR ) for all x = ( x , . . . , x m ) ∈ R n and R > .In particular, if ψ j satisfy the engulfing property with corresponding constants θ j for all j = 1 , . . . m , then φ satisfies the engulfing property with θ = m max j { θ j } . Conversely, if ψ satisfies the engulfing property for some θ > , then ψ j satisfies the engulfing property withconstant θ for all j = 1 , . . . , m . As a consequence of Lemma 4.5, S R ( x, z ) ⊂ S R ( x ) × S R ( z ) ⊂ S R ( x, z ) and S R ( x, z ) ⊂ Q R ( x, z ) ⊂ S ( n +1) R ( x, z )for all ( x, z ) ∈ R n +1 and R > ψ = Φ. Corollary 4.6.
We have µ ϕ ∈ ( DC ) ϕ and µ h ∈ ( DC ) h . Hence, the following statementshold and are equivalent.(1) µ Φ ∈ ( DC ) Φ with corresponding doubling constant C d = C d ( n, s ) > ;(2) Φ satisfies the engulfing property with corresponding constant θ = θ ( n, s ) > ;(3) there are positive constants C = C ( n, s ) , c = c ( n, s ) such that cR n +1 ≤ | S R ( x, z ) | µ Φ ( S R ( x, z )) ≤ CR n +1 for all sections S R ( x, z ) ;(4) δ Φ is a quasi-distance with constant K = K ( n, s ) ≥ .Proof. By (4.3), we can write µ ϕ ( S R ( x )) = | B √ R ( x ) | ≤ n (cid:12)(cid:12) B √ R ( x ) (cid:12)(cid:12) = 2 n µ ϕ (cid:18) S R ( x ) (cid:19) Hence ϕ ∈ ( DC ) ϕ with doubling constant C ϕd = C ϕd ( n ). As discussed in [22, Section 7.1], h (cid:48)(cid:48) ( z ) is a Muckenhoupt A ∞ ( R ) weight for all 0 < s <
1. Consequently, h (cid:48)(cid:48) is doubling on thereal line, which is equivalent to µ h ∈ ( DC ) h with doubling constant C hd = C hd ( s ). It followsfrom Theorem 4.2 that µ ϕ and µ h satisfy the engulfing property and, by Lemma 4.5, so does µ Φ . Hence, the conclusion follows from Theorem 4.2. (cid:3) As a consequence of the doubling property (see [22, Equation 7.8]), there are positiveconstants K d = K d ( n, s ) and ν = ν ( n, s ) such that(4.6) µ Φ ( S r ( x, z )) ≤ K d (cid:18) r r (cid:19) ν µ Φ ( S r ( x, z )) for all 0 < r < r , Notation 4.7.
We will always use the following notation. • θ is the engulfing constant associated with Φ. • K is the quasi-triangle constant associated with Φ. • K d and ν are the constants in (4.6).Next, we make precise the size and measure of the Monge–Amp`ere sections associated to h .The following follows immediately from [20, Section 11] (see also [17]) and, since µ h ∈ ( DC ) h ,from Theorem 4.2(3). Lemma 4.8.
There exist constants c s , C s > such that, for all R > and all z ∈ R , B c s R s ( z ) ⊂ S R ( z ) ⊂ B C s R s ( z ) . Consequently, there exist constants c (cid:48) s , C (cid:48) s > such that, for all sections S R ( z ) , c s R s ≤ | S R ( z ) | ≤ C s R s and c (cid:48) s R − s ≤ µ h ( S R ( z )) ≤ C (cid:48) s R − s . We end this section by presenting two lemmas that will be used later in the proofs.
Lemma 4.9.
Let ψ ∈ C ( R ) be a strictly convex function. If x < x < x , then δ ψ ( x , x ) < δ ψ ( x , x ) and δ ψ ( x , x ) < δ ψ ( x , x ) . Consequently, for any x , x ∈ R and R > , if r > is such that S ψ ( x , r ) ⊂ S ψ ( x , R ) ,then r ≤ R . In particular, for any ( x , z ) , ( x , z ) ∈ R n +1 and R > , if r > is such that Q r ( x , z ) ⊂ Q R ( x , z ) , then r ≤ R . RACTIONAL ELLIPTIC EQUATIONS 15
Proof.
By the convexity of ψ , ψ (cid:48) ( x ) < ψ ( x ) − ψ ( x ) x − x , so that δ ψ ( x , x ) = ψ ( x ) − ψ ( x ) − ψ (cid:48) ( x )( x − x ) < ψ ( x ) − ψ ( x ) − ψ (cid:48) ( x )( x − x ) = δ ψ ( x , x ) . Next, define a function Ψ : R → R byΨ( x ) := [ ψ ( x ) + ψ (cid:48) ( x )( x − x )] − [ ψ ( x ) + ψ (cid:48) ( x )( x − x )] . By the convexity of ψ , Ψ (cid:48) ( x ) = ψ (cid:48) ( x ) − ψ (cid:48) ( x ) >
0, so Ψ is increasing. SinceΨ( x ) = ψ ( x ) − ψ ( x ) + ψ (cid:48) ( x )( x − x ) = δ ψ ( x , x ) > , we know that0 < Ψ( x ) = [ ψ ( x ) + ψ (cid:48) ( x )( x − x )] − [ ψ ( x ) + ψ (cid:48) ( x )( x − x )]= [ − ψ ( x ) + ψ ( x ) + ψ (cid:48) ( x )( x − x )] − [ − ψ ( x ) + ψ ( x ) + ψ (cid:48) ( x )( x − x )] . Hence, δ ψ ( x , x ) = ψ ( x ) − ψ ( x ) − ψ (cid:48) ( x )( x − x ) < ψ ( x ) − ψ ( x ) − ψ (cid:48) ( x )( x − x ) = δ ψ ( x , x ) . Lastly, fix x , x ∈ R and R >
0. Suppose that r > S ψ ( x , r ) ⊂ S ψ ( x , R ).Write S ψ ( x , R ) = ( x L , x R ) , x L < x < x R S ψ ( x , r ) = ( x L , x R ) , x L < x < x R . Without loss of generality, assume that x ≤ x so that x ≤ x < x R ≤ x R . Then r = δ ψ ( x , x R ) ≤ δ ψ ( x , x R ) ≤ δ ψ ( x , x R ) = R. (cid:3) Lemma 4.10 (Theorem 3.3.10 in [14]) . There exist constants C > , p ≥ , depending onlyon n and s , such that for < r < r ≤ , t > and ( x , z ) ∈ S r t ( x , z ) , we have that S C ( r − r ) p t ( x , z ) ⊂ S r t ( x , z ) . Reductions of Theorem 1.3
In this section we show that, after a series of reductions, Theorem 1.3 will follow fromTheorem 5.3. Hence, the rest of the paper will be devoted to proving Theorem 5.3 andTheorem 1.1.
First reduction.
We first show that it is enough to prove Theorem 1.3 in Monge–Amp`ere cubes and with f ≥ Theorem 5.1.
Let Ω be a bounded domain, a ij ( x ) : Ω → R be bounded, measurable functionsthat satisfy (1.5) and let f ∈ L ∞ (Ω) be nonnegative. There exist positive constants C H = C H ( n, λ, Λ , s ) > , κ = κ ( n, s ) < , and ˆ K = ˆ K ( n, s ) > such that for every cube Q ˆ K R = Q ˆ K R ( x , z ) ⊂ Ω × R and every nonnegative solution U ∈ C ( Q ˆ K R \ { z = 0 } ) ∩ C ( Q ˆ K R ) such that U is symmetric across { z = 0 } and U z + ∈ C ( Q ˆ K R ∩ { z ≥ } ) to (cid:40) a ij ( x ) ∂ ij U + | z | − s ∂ zz U = 0 in Q ˆ K R ∩ { z (cid:54) = 0 }− ∂ z + U ( x,
0) = f on Q ˆ K R ∩ { z = 0 } , we have that (5.1) sup Q κ R U ≤ C H (cid:18) inf Q κ R U + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) R s (cid:19) . Proof of Theorem 1.3 from Theorem 5.1.
Observe that Q ˆ K R ( x , z ) ⊂ Q θ ˆ K R ( x , z ) ⊂ S ( n +1) θ ˆ K R ( x , z )and S κ R ( x , z ) ⊂ Q κ R ( x , z )Let ˆ K = ( n + 1) θ ˆ K and κ = κ . Case 1 : Q ˆ K R ( x , z ) ∩ { z = 0 } = ∅ . By Theorem 5.1 and the inclusion above, we getsup S κ R U ≤ sup Q κ R U ≤ C H inf Q κ R U ≤ C H inf S κ R U. Case 2 : Q ˆ K R ( x , z ) ∩ { z = 0 } (cid:54) = ∅ and z = 0. Define V = U − (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) | z | + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) q R s , q = (cid:18) − ss ˆ K (cid:19) s . We claim that V is nonnegative in Q ˆ K R . Indeed, let ( x, z ) ∈ Q ˆ K R , so that z ∈ S ˆ K R (0) ⊂ R . In particular, s − s | z | s = h ( z ) = δ h (0 , z ) < ˆ K R which implies | z | < (cid:18) − ss ˆ K R (cid:19) s = q R s . Consequently, for any ( x, z ) ∈ Q ˆ K R , we have that − (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) | z | + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) q R s ≥ , so that V ≥ U ≥ Q ˆ K R .Next, notice that V is symmetric across { z = 0 } and that V ∈ C ( Q ˆ K R \ { z = 0 } ) ∩ C ( Q ˆ K R ), V z + ∈ C ( Q ˆ K R ∩ { z ≥ } ). Moreover, for ( x, z ) ∈ Q ˆ K R ∩ { z (cid:54) = 0 } , it is clear that a ij ( x ) ∂ ij V + | z | − s ∂ zz V = a ij ( x ) ∂ ij U + | z | − s ∂ zz U = 0and for ( x, ∈ Q ˆ K R ∩ { z = 0 } , − ∂ z + V ( x,
0) = f ( x ) + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) := g ( x ) ≥ . RACTIONAL ELLIPTIC EQUATIONS 17
Therefore, V is a nonnegative solution to (cid:40) a ij ( x ) ∂ ij V + | z | − s ∂ zz V = 0 in Q ˆ K R ∩ { z (cid:54) = 0 }− ∂ z + V = g on Q ˆ K R ∩ { z = 0 } . Therefore, by Theorem 5.1 applied to V ,sup S κ R U ≤ sup Q κ R U ≤ sup Q κ R V ≤ C H (cid:18) inf Q κ R V + (cid:107) g (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) R s (cid:19) = C H (cid:18) inf Q κ R (cid:16) U − (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) | z | + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) q R s (cid:17) + (cid:13)(cid:13)(cid:13) f + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) (cid:13)(cid:13)(cid:13) L ∞ ( Q ˆ K R ∩{ z =0 } ) R s (cid:19) ≤ C H (cid:18) inf Q κ R U + ( q + 2) (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) R s (cid:19) ≤ C (cid:48) H (cid:18) inf S κ R U + (cid:107) f (cid:107) L ∞ ( S ˆ K R ∩{ z =0 } ) R s (cid:19) . Case 3 : Q ˆ K R ( x , z ) ∩ { z = 0 } (cid:54) = ∅ and z (cid:54) = 0. In this case, 0 ∈ S ˆ K R ( z ). Then, by theengulfing property, Q ˆ K R ( x , z ) = Q ˆ K R ( x ) × S ˆ K R ( z ) ⊂ Q θ ˆ K R ( x ) × S θ ˆ K R (0) = Q θ ˆ K R ( x , . Again, applying the engulfing property, Q θ ˆ K R ( x ,
0) = Q θ ˆ K R ( x ) × S θ ˆ K R (0) ⊂ Q θ ˆ K R ( x ) × S θ ˆ K R ( z ) = Q θ ˆ K R ( x , z ) . Define V = U − (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , ∩{ z =0 } ) | z | + (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , ∩{ z =0 } ) q R s where q = (cid:18) − ss θ ˆ K (cid:19) s = θ s q . We claim that V is nonnegative in Q θ ˆ K R ( x , x, z ) ∈ Q θ ˆ K R ( x , z ∈ S θ ˆ K R (0) ⊂ R . In particular, s − s | z | s = h ( z ) = δ h (0 , z ) < θ ˆ K R which implies | z | < (cid:18) − ss θ ˆ K R (cid:19) s = q R s . Consequently, for any ( x, z ) ∈ Q ˆ K R ( x , z ), we have that − (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , ∩{ z =0 } ) | z | + (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , ∩{ z =0 } ) q R s ≥ , so that V ≥ U ≥ Q θ ˆ K R ( x , V is symmetric across { z = 0 } and that V ∈ C ( Q θ ˆ K R ( x , \ { z =0 } ) ∩ C ( Q θ ˆ K R ( x , V z + ∈ C ( Q θ ˆ K R ( x , ∩{ z ≥ } ). Moreover, for ( x, z ) ∈ Q θ ˆ K R ( x , ∩{ z (cid:54) = 0 } , it is clear that a ij ( x ) ∂ ij V + | z | − s ∂ zz V = a ij ( x ) ∂ ij U + | z | − s ∂ zz U = 0and for ( x, ∈ Q θ ˆ K R ( x , ∩ { z = 0 } , − ∂ z + V ( x,
0) = f ( x ) + (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , ∩{ z =0 } ) := g ( x ) ≥ . Therefore, V is a nonnegative solution to (cid:40) a ij ( x ) ∂ ij V + | z | − s ∂ zz V = 0 in Q ˆ K R ( x , z ) ∩ { z (cid:54) = 0 }− ∂ z + V = g on Q ˆ K R ( x , z ) ∩ { z = 0 } . By applying Theorem 5.1 to V , we getsup S κ R ( x ,z ) U ≤ sup Q κ R ( x ,z ) U ≤ sup Q κ R ( x ,z ) V ≤ C H (cid:32) inf Q κ R ( x ,z ) V + (cid:107) g (cid:107) L ∞ ( Q ˆ K R ( x ,z ) ∩{ z =0 } ) R s (cid:33) = C H (cid:18) inf Q κ R ( x ,z ) (cid:16) U − (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , | z | + (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , q R s (cid:17) + (cid:13)(cid:13)(cid:13) f + (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , ∩{ z =0 } ) (cid:13)(cid:13)(cid:13) L ∞ ( Q ˆ K R ( x ,z ) ∩{ z =0 } ) R s (cid:19) ≤ C H (cid:18) inf Q κ R ( x ,z ) U + ( q + 2) (cid:107) f (cid:107) L ∞ ( Q θ ˆ K R ( x , ∩{ z =0 } ) R s (cid:19) ≤ C (cid:48) H (cid:18) inf S κ R ( x ,z ) U + (cid:107) f (cid:107) L ∞ ( S ˆ K R ( x ,z ) ∩{ z =0 } ) R s (cid:19) . Therefore, (1.15) holds in all cases.It remains to prove the H¨older estimate (1.16). The proof follows by a standard argument(see, for example, [11, Sections 8.9 and 9.8]). We provide the details for completeness. Let0 < r ≤ R and define M ( r ) = sup S r ( x ,z ) U and m ( r ) = inf S r ( x ,z ) U. Apply (1.15) to M ( r ) − U ≥ S r ( x , z ) to obtainsup S κ r ( x ,z ) ( M ( r ) − U ) ≤ C H (cid:18) inf S κ r ( x ,z ) ( M ( r ) − U ) + (cid:107) f (cid:107) L ∞ ( S ˆ K r ( x ,z ) ∩{ z =0 } ) r s (cid:19) . Therefore,(5.2) M ( r ) − m ( κ r ) ≤ C H (cid:16) M ( r ) − M ( κ r ) + (cid:107) f (cid:107) L ∞ ( S ˆ K r ( x ,z ) ∩{ z =0 } ) r s (cid:17) . RACTIONAL ELLIPTIC EQUATIONS 19
Similarly, apply (1.15) to U − m ( r ) ≥ S r ( x , z ) to obtainsup S κ r ( x ,z ) ( U − m ( r )) ≤ C H (cid:18) inf S κ r ( x ,z ) ( U − m ( r ) + (cid:107) f (cid:107) L ∞ ( S ˆ K r ( x ,z ) ∩{ z =0 } ) r s (cid:19) , so that(5.3) M ( κ r ) − m ( r ) ≤ C H (cid:16) m ( κ r ) − m ( r ) + (cid:107) f (cid:107) L ∞ ( S ˆ K r ( x ,z ) ∩{ z =0 } ) r s (cid:17) . Let ω ( r ) = M ( r ) − m ( r ). Adding (5.2) and (5.3) together, we get ω ( r ) + ω ( κ r ) ≤ C H (cid:16) ω ( r ) − ω ( κ r ) + 2 (cid:107) f (cid:107) L ∞ ( S ˆ K R ( x ,z ) ∩{ z =0 } ) r s (cid:17) . After rearranging, ω ( κ r ) ≤ γω ( r ) + σ ( r ) , γ = C H − C H + 1 < σ ( r ) = 2 C H C H + 1 (cid:107) f (cid:107) L ∞ ( S ˆ K R ( x ,z ) ∩{ z =0 } ) r s is a non-decreasing function of r . Note that γ = γ ( n, λ, Λ , s ). By [11, Lemma 8.23], for any µ ∈ (0 , C = C ( n, λ, Λ , s ) > α = (1 − µ ) log γ/ log κ such that ω ( r ) ≤ C (cid:16)(cid:16) rR (cid:17) α ω ( R ) + σ ( r µ R − µ ) (cid:17) . Choose µ = µ ( n, λ, Λ , s ) so that 2 α < µs . Then, ω ( r ) ≤ C (cid:16)(cid:16) rR (cid:17) α ω ( R ) + σ ( r µ R − µ ) (cid:17) = C (cid:16) rR (cid:17) α (cid:18) ω ( R ) + σ ( r µ R − µ ) (cid:18) Rr (cid:19) α (cid:19) ≤ C (cid:16) rR (cid:17) α (cid:32) S R ( x ,z ) | U | + 2 C H C H + 1 (cid:107) f (cid:107) L ∞ ( S ˆ K R ( x ,z ) ∩{ z =0 } ) r µs R (1 − µ ) s R α r − α (cid:33) ≤ ˆ C (cid:16) rR (cid:17) α (cid:32) sup S R ( x ,z ) | U | + (cid:107) f (cid:107) L ∞ ( S ˆ K R ( x ,z ) ∩{ z =0 } ) R s (cid:33) . By taking r = δ Φ (( x , z ) , ( x, z )), the estimate in (1.16) follows. (cid:3) Second reduction.
Next, we show that Theorem 5.1 follows from the following result.
Theorem 5.2.
Let Ω be a bounded domain, a ij ( x ) : Ω → R be bounded, measurable functionsthat satisfy (1.5) and let f ∈ L ∞ (Ω) be nonnegative. There exist positive constants C H = C H ( n, λ, Λ , s ) > , κ = κ ( n, s ) < , and ˆ K = ˆ K ( n, s ) > such that for every cube Q ˆ K R = Q ˆ K R (˜ x, ˜ z ) ⊂⊂ Ω × R and every nonnegative solution U ∈ C ( Q ˆ K R \ { z = 0 } ) ∩ C ( Q ˆ K R ) such that U is symmetric across { z = 0 } and U z + ∈ C ( Q ˆ K R ∩ { z ≥ } ) to (cid:40) a ij ( x ) ∂ ij U + | z | − s ∂ zz U = 0 in Q ˆ K R ∩ { z (cid:54) = 0 }− ∂ z + U ( x,
0) = f on Q ˆ K R ∩ { z = 0 } , we have that sup Q κ R U ≤ C H (cid:16) U (˜ x, ˜ z ) + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) R s (cid:17) . Proof of Theorem 5.1 from Theorem 5.2.
Let ˆ K = ˆ K ( n, s ) and κ = κ ( n, s ) be such that1 < θ ˆ K ≤ ˆ K and θκ ≤ κ < . Let (˜ x, ˜ z ) ∈ Q κ R ( x , z ). By the engulfing property, Q κ R ( x , z ) ⊂ Q θκ R (˜ x, ˜ z ) ⊂ Q κ R (˜ x, ˜ z ) . Again applying the engulfing property, we have Q ˆ K R (˜ x, ˜ z ) ⊂ Q θ ˆ K R ( x , z ) ⊂ Q ˆ K R ( x , z ) ⊂⊂ Ω × R . By Theorem 5.2, we getsup Q κ R ( x ,z ) U ≤ sup Q κ R (˜ x, ˜ z ) U ≤ C H (cid:16) U (˜ x, ˜ z ) + (cid:107) f (cid:107) L ∞ ( Q ˆ K R (˜ x, ˜ z ) ∩{ z =0 } ) R s (cid:17) ≤ C H (cid:16) U (˜ x, ˜ z ) + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ( x ,z ) ∩{ z =0 } ) R s (cid:17) . Taking the infimum over all (˜ x, ˜ z ) ∈ Q κ R ( x , z ), the Harnack inequality (5.1) holds. (cid:3) Third reduction.
Here we will see that Theorem 5.2 follows from the next, and final,reduction.
Theorem 5.3.
Fix a > . Let Ω be a bounded domain, a ij ( x ) : Ω → R be bounded,measurable functions that satisfy (1.5) and let f ∈ L ∞ (Ω) be nonnegative. Let ˆ K be as inTheorem 5.2. There exist positive constants C H = C H ( n, λ, Λ , s ) > , κ = κ ( n, s ) < , and K = K ( n, s ) such that for any cube Q ˆ K R = Q ˆ K R (˜ x, ˜ z ) ⊂⊂ Ω × R and every nonnegativesolution U ∈ C ( Q ˆ K R \ { z = 0 } ) ∩ C ( Q ˆ K R ) such that U is symmetric across { z = 0 } and U z + ∈ C ( Q ˆ K R ∩ { z ≥ } ) to (cid:40) a ij ( x ) ∂ ij U + | z | − s ∂ zz U = 0 in Q ˆ K R ∩ { z (cid:54) = 0 }− ∂ z + U ( x,
0) = f on Q ˆ K R ∩ { z = 0 } , if U (˜ x, ˜ z ) ≤ aR K and (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) ≤ aR − s , then (5.4) U ≤ C H aR in Q κ R . Proof of Theorem 5.2 from Theorem 5.3 .
Let ε >
0. Define the nonnegative function W ε in Q ˆ K R by W ε ( x, z ) = aR K U (˜ x, ˜ z ) + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) R s + ε U ( x, z ) . Then W ε is symmetric across { z = 0 } and W ε ∈ C ( Q ˆ K R \ { z = 0 } ) ∩ C ( Q ˆ K R ), ( W ε ) z + ∈ C ( Q ˆ K R ∩ { z ≥ } ). Moreover, in Q ˆ K R ∩ { z (cid:54) = 0 } , we have a ij ( x ) ∂ ij W ε + | z | − s ∂ zz W ε = 0 RACTIONAL ELLIPTIC EQUATIONS 21 and, in Q ˆ K R ∩ { z = 0 } , − ∂ z + W ε ( x,
0) = aR K U (˜ x, ˜ z ) + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) R s + ε f ( x ) =: g ( x ) ≥ . Therefore, W ε is a nonnegative solution to (cid:40) a ij ( x ) ∂ ij W ε + | z | − s ∂ zz W ε = 0 in Q ˆ K R ∩ { z (cid:54) = 0 }− ∂ z + W ε ( x,
0) = g on Q ˆ K R ∩ { z = 0 } . Clearly, (cid:107) g (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) ≤ aR − s and W ε (˜ x, ˜ z ) ≤ aR K . Whence, by Theorem 5.3 applied to W ε , W ε ≤ C H aR in Q κ R , which implies sup Q κ R U ≤ C (cid:48) H (cid:16) U (˜ x, ˜ z ) + (cid:107) f (cid:107) L ∞ ( Q ˆ K R ∩{ z =0 } ) R s + ε (cid:17) . As ε > (cid:3) Paraboloids associated to
ΦIn this section, we define the Monge–Amp`ere paraboloids associated with Φ in (4.2) andstudy their basic properties and relations with respect to solutions to the extension problem.
Definition 6.1.
Let a >
0. A paraboloid P of opening a in R n +1 is defined as(6.1) P ( x, z ) = − a Φ( x, z ) + (cid:104) ( y, w ) , ( x, z ) (cid:105) + b ( x, z ) ∈ R n +1 for some ( y, w ) ∈ R n +1 and b ∈ R .Since Φ ∈ C ( R n +1 ) is strictly convex, the point where the maximum of P occurs, whichwe call the vertex ( x v , z v ) of P , is the unique solution to ∇ P ( x v , z v ) = 0.We say that P touches a continuous function U : R n +1 → R from below at ( x , z ) in aconvex set S ⊂ R n +1 if P ( x , z ) = V ( x , z ) and P ( x, z ) ≤ V ( x, z ) for all ( x, z ) ∈ S. Lemma 6.2.
A paraboloid P of opening a > with vertex ( x v , z v ) given by (6.1) can bewritten as (6.2) P ( x, z ) = − aδ Φ (( x v , z v ) , ( x, z )) + c for some constant c ∈ R . Moreover, ∇ P ( x, z ) = − a ( ∇ Φ( x, z ) − ∇ Φ( x v , z v )) = − a ( x − x v , h (cid:48) ( z ) − h (cid:48) ( z v )) and ∂ z P ( x,
0) = ah (cid:48) ( z v ) for all ( x, z ) ∈ R n +1 . If P coincides with a continuous function U : R n +1 → R at a point ( x , z ) , i.e. P ( x , z ) = U ( x , z ) , then P ( x, z ) = − aδ Φ (( x v , z v ) , ( x, z )) + aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z ) . Proof.
Since 0 = ∇ P ( x v , z v ) = − a ∇ Φ( x v , z v ) + ( y, w ), we can write P ( x, z ) = − a Φ( x, z ) + a (cid:104)∇ Φ( x v , z v ) , ( x, z ) (cid:105) + b. Consequently, ∇ P ( x, z ) = − a ( x − x v , h (cid:48) ( z ) − h (cid:48) ( z v ))and ∂ z P ( x,
0) = − a (cid:0) h (cid:48) ( z ) − h (cid:48) ( z v ) (cid:1) (cid:12)(cid:12) ( x, = ah (cid:48) ( z v ) . Moreover, we have P ( x, z ) = − a Φ( x, z ) + a (cid:104)∇ Φ( x v , z v ) , ( x, z ) (cid:105) + b + a Φ( x v , z v ) − a Φ( x v , z v ) − a (cid:104)∇ Φ( x v , z v ) , ( x v , z v ) (cid:105) + a (cid:104)∇ Φ( x v , z v ) , ( x v , z v ) (cid:105) = − a (Φ( x, z ) − Φ( x v , z v ) − (cid:104)∇ Φ( x v , z v ) , ( x, z ) − ( x v , z v ) (cid:105) ) − a Φ( x v , z v ) + a (cid:104)∇ Φ( x v , z v ) , ( x v , z v ) (cid:105) + b = − aδ Φ (( x v , z v ) , ( x, z )) + c. If P ( x , z ) = U ( x , z ), then U ( x , z ) = − aδ Φ (( x v , z v ) , ( x , z )) + c and, after solving for c , we conclude that P ( x, z ) = − aδ Φ (( x v , z v ) , ( x, z )) + aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z ). (cid:3) For the remainder of the paper, we use the terminology paraboloids to reference those givenby (6.1), or equivalently, (6.2).
Lemma 6.3.
Suppose that P is a paraboloid of opening a > that touches a continuousfunction U : R n +1 → R from below at ( x , z ) in a convex set S ⊂ R n +1 . For any ˜ a ≥ a ,there exists a paraboloid ˜ P of opening ˜ a > that touches U from below at ( x , z ) in S .Proof. Begin by writing P ( x, z ) = − aδ Φ (( x v , z v ) , ( x, z )) + aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z )= − a (Φ( x, z ) − Φ( x v , z v ) − (cid:104)∇ Φ( x v , z v ) , ( x, z ) − ( x v , z v ) (cid:105) )+ a (Φ( x , z ) − Φ( x v , z v ) − (cid:104)∇ Φ( x v , z v ) , ( x , z ) − ( x v , z v ) (cid:105) ) + U ( x , z )= − a Φ( x, z ) + a (cid:104)∇ Φ( x v , z v ) , ( x, z ) (cid:105) + a Φ( x , z ) − a (cid:104)∇ Φ( x v , z v ) , ( x , z ) (cid:105) + U ( x , z )= − a (Φ( x, z ) − Φ( x , z ) − (cid:104)∇ Φ( x , z ) , ( x, z ) − ( x , z ) (cid:105) )+ a (cid:104)∇ Φ( x v , z v ) , ( x, z ) − ( x , z ) (cid:105) − a (cid:104)∇ Φ( x , z ) , ( x, z ) − ( x , z ) (cid:105) + U ( x , z )= − aδ Φ (( x , z ) , ( x, z )) + a (cid:104)∇ Φ( x v , z v ) − ∇ Φ( x , z ) , ( x, z ) − ( x , z ) (cid:105) + U ( x , z ) . Define ˜ P by˜ P ( x, z ) = − ˜ aδ Φ (( x , z ) , ( x, z )) + a (cid:104)∇ Φ( x v , z v ) − ∇ Φ( x , z ) , ( x, z ) − ( x , z ) (cid:105) + U ( x , z ) . Note that ˜ P is a paraboloid of opening ˜ a > P ( x, z ) = − ˜ a Φ( x, z ) + ˜ a Φ( x , z ) + ˜ a (cid:104)∇ φ ( x , z ) , ( x, z ) − ( x , z ) (cid:105) + a (cid:104)∇ Φ( x v , z v ) − ∇ Φ( x , z ) , ( x, z ) − ( x , z ) (cid:105) + U ( x , z )= − ˜ a Φ( x, z ) + (cid:104) ˜ a ∇ Φ( x , z ) + a ∇ Φ( x v , z v ) − a ∇ Φ( x , z ) , ( x, z ) (cid:105) + ˜ a Φ( x , z ) − (cid:104) ˜ a ∇ φ ( x , z ) + a ∇ Φ( x v , z v ) − a ∇ Φ( x , z ) , ( x , z ) (cid:105) + U ( x , z ) . Since ˜ P ( x , z ) = U ( x , z ) and˜ P ( x, z ) ≤ − aδ Φ (( x , z ) , ( x, z )) + a (cid:104)∇ Φ( x v , z v ) − ∇ Φ( x , z ) , ( x, z ) − ( x , z ) (cid:105) + U ( x , z ) RACTIONAL ELLIPTIC EQUATIONS 23 = P ( x, z ) ≤ U ( x, z ) , for every ( x, z ) ∈ S , we conclude that ˜ P touches U from below at ( x , z ) in S . (cid:3) The next two lemmas provide some observations regarding how the symmetry of U across { z = 0 } effects the geometry of the paraboloids that touch U from below. Lemma 6.4.
Let S ⊂ R n +1 be an open, convex set that is symmetric across { z = 0 } .Consider a continuous function U : S → R which is symmetric across { z = 0 } . Let P be aparaboloid of opening a > with vertex ( x v , z v ) that touches U from below at ( x , z ) in S . If z > , then z v ≥ , and if z < , then z v ≤ . Moreover, the paraboloid ˜ P ( x, z ) = P ( x, − z ) of opening a > with vertex ( x v , − z v ) that touches U from below at ( x , − z ) in S .Proof. Assume that z >
0. Write P ( x, z ) = − aδ Φ (( x v , z v ) , ( x, z )) + aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z )and note that P ( x , − z ) = − aδ Φ (( x v , z v ) , ( x , − z )) + aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z )= − aδ h ( z v , − z ) + aδ h ( z v , z ) + U ( x , − z ) . Then 0 ≤ U ( x , − z ) − P ( x , − z )= aδ h ( z v , − z ) − aδ h ( z v , z )= a ( h ( − z ) − h ( z )) + 2 ah (cid:48) ( z v ) z = 2 ah (cid:48) ( z v ) z . Since z >
0, it follows that h (cid:48) ( z v ) ≥
0. Hence, z v ≥
0, as desired. The case for z < P by ˜ P ( x, z ) = P ( x, − z ). Since h (cid:48) ( − z ) = − h (cid:48) ( z ) and(6.3) δ h ( z , − z ) = h ( − z ) − h ( z ) − h (cid:48) ( z )( − z − z )= h ( z ) − h ( − z ) − h (cid:48) ( − z )( z − ( − z ))= δ h ( − z , z )for all z , z ∈ R , we may write˜ P ( x, z ) = P ( x, − z )= − aδ Φ (( x v , z v ) , ( x, − z )) + aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z )= − aδ ϕ ( x v , x ) + aδ ϕ ( x v , x ) − aδ h ( z v , − z ) + aδ h ( z v , z ) + U ( x , − z )= − aδ ϕ ( x v , x ) + aδ ϕ ( x v , x ) − aδ h ( − z v , z ) + aδ h ( − z v , − z ) + U ( x , − z )= − aδ Φ (( x v , − z v ) , ( x, z )) + aδ Φ (( x v , − z v ) , ( x , − z )) + U ( x , − z ) . Hence, ˜ P is a paraboloid of opening a > x v , − z v ). Since˜ P ( x, z ) = P ( x, − z ) ≤ U ( x, − z ) = U ( x, z ) for all ( x, z ) ∈ S and ˜ P ( x , − z ) = P ( x , z ) = U ( x , z ) = U ( x , − z ) , we have that ˜ P touches U from below at ( x , − z ) in S . (cid:3) Notation 6.5.
Given f : Ω → R , we define the functions f ± by f − ( x ) = min { , f ( x ) } ≤ f + ( x ) = max { f ( x ) , } ≥ . Lemma 6.6.
Let f ∈ L ∞ (Ω) and let S ⊂⊂ Ω × R ⊂ R n +1 be an open, convex set suchthat S ∩ { z = 0 } (cid:54) = ∅ . Suppose that a continuous function U : Ω × R → R such that U z + ∈ C ([0 , ∞ ); C (Ω)) is symmetric across { z = 0 } and satisfies − ∂ z + U ( x, ≥ f ( x ) on S ∩ { z = 0 } . If f ( x ) > , then U cannot be touched from below at ( x , in S by any paraboloid. If f ( x ) ≤ and P is a paraboloid of opening a > with vertex ( x v , z v ) that touches U frombelow in S at ( x , , then | h (cid:48) ( z v ) | ≤ | f − ( x ) | /a . Consequently, if f ( x ) = 0 , then z v = 0 .Proof. Suppose that P is a paraboloid of opening a > U from below at ( x , S . Write P ( x, z ) = − aδ Φ (( x v , z v ) , ( x, z )) + aδ Φ (( x v , z v ) , ( x , U ( x , . Let ε > U − P attains a local minimum of 0 at ( x , U ( x , ε ) − P ( x , ε )) − ( U ( x , − P ( x , ε = U ( x , ε ) − P ( x , ε ) ε ≥ . Therefore, taking the limit as ε → + , we obtain(6.4) 0 ≤ ∂ z + U ( x , − ∂ z P ( x , ≤ − f ( x ) − ah (cid:48) ( z v ) . We note that, by the symmetry of U across { z = 0 } , we have that ∂ z − U ( x ,
0) = lim h → − U ( x , h ) − U ( x , h = − lim h → + U ( x , − h ) − U ( x , − h = − lim h → + U ( x , h ) − U ( x , h = − ∂ z + U ( x , ≥ f ( x ) . For ε > U ( x , − ε ) − P ( x , − ε )) − ( U ( x , − P ( x , − ε = − U ( x , − ε ) + P ( x , − ε ) ε ≤ . Taking the limit as ε → + , we obtain(6.5) 0 ≥ ∂ z − U ( x , − ∂ z P ( x , ≥ f ( x ) − ah (cid:48) ( z v ) . By combining (6.4) and (6.5), f ( x ) ≤ ah (cid:48) ( z v ) ≤ − f ( x ) . If f ( x ) >
0, then this is a contradiction, so P cannot touch U from below in S at ( x , f ( x ) ≤
0, then − (cid:12)(cid:12) f − ( x ) (cid:12)(cid:12) ≤ ah (cid:48) ( z v ) ≤ (cid:12)(cid:12) f − ( x ) (cid:12)(cid:12) as desired. If f ( x ) = 0, then h (cid:48) ( z v ) = 0 which implies that z v = 0. (cid:3) RACTIONAL ELLIPTIC EQUATIONS 25 Estimate on the Monge–Amp`ere measure of the set of contact points
Our first key result is a measure estimate similar to the Alexandroff–Bakelman–Pucciestimate for fully nonlinear equations. We prove that if we lift paraboloids of fixed opening a > U for the first time, then, by using the equation and the Neumann boundary condition, theMonge–Amp`ere measure of the contact points is a universal proportion of the Monge–Amp`eremeasure of the set of vertices. Theorem 7.1.
Assume that Ω is a bounded domain and that a ij ( x ) : Ω → R are bounded,measurable functions that satisfy (1.5) . Let Q R = Q R (˜ x, ˜ z ) ⊂⊂ Ω × R and f ∈ L ∞ ( Q R ∩ { z =0 } ) . Suppose U ∈ C ( Q R \ { z = 0 } ) ∩ C ( Q R ) such that U is symmetric across { z = 0 } and U z + ∈ C ( Q R ∩ { z ≥ } ) is a supersolution to (cid:40) a ij ( x ) ∂ ij U + | z | − s ∂ zz U ≤ in Q R ∩ { z (cid:54) = 0 }− ∂ z + U ≥ f on Q R ∩ { z = 0 } . Let B ⊂ Q R be a closed set and fix a > . For each ( x v , z v ) ∈ B , we slide paraboloids ofopening a > and vertex ( x v , z v ) from below until they touch the graph of U for the firsttime. Let A denote the set of contact points and assume that A ⊂ Q R . Then A is compactand if µ Φ (cid:32) B ∩ (cid:40) ( x, z ) : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f − (cid:107) L ∞ ( Q R ∩{ z =0 } ) a (cid:41)(cid:33) ≤ (1 − ε ) µ Φ ( B ) , for some ε > , then there is a positive constant c = c ( n, λ, Λ) < such that µ Φ ( A ) ≥ ε cµ Φ ( B ) . Proof.
We first show that A is closed. Let ( x k , z k ) ∈ A be such that ( x k , z k ) → ( x , z ).There exist corresponding polynomials P k with vertices ( x kv , z kv ) ∈ B such that P k touches U from below at ( x k , z k ) in Q R . Since B ⊂ Q R is closed, B is compact. Thus, up to asubsequence, ( x kv , z kv ) → ( x v , z v ) ∈ B . By the continuity of δ Φ and U , as k → ∞ , P k ( x, z ) = − aδ Φ (( x kv , z kv ) , ( x, z )) + aδ Φ (( x kv , z kv ) , ( x k , z k )) + U ( x k , z k ) → − aδ Φ (( x v , z v ) , ( x, z )) + aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z ) =: P ( x, z ) . Since P k ( x, z ) ≤ U ( x, z ), it must be that P ≤ U in Q R . Moreover, P ( x , z ) = U ( x , z ).Therefore, P is a paraboloid of opening a > x v , z v ) ∈ B that touches U frombelow at ( x , z ). This shows that ( x , z ) ∈ A , so that A is closed and, moreover, compact.Define the sets B = B ∩ (cid:40) ( x, z ) : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f − (cid:107) L ∞ ( Q R ∩{ z =0 } ) a (cid:41) B = B \ (cid:40) ( x, z ) : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f − (cid:107) L ∞ ( Q R ∩{ z =0 } ) a (cid:41) , so that B = B ∪ B and B ∩ B = ∅ . We lift paraboloids of opening a > B and B to form the contact sets A and A , respectively. Note that A = A ∪ A .We will first show that µ Φ ( B ) ≤ Cµ Φ ( A ) for some positive constant C = C ( n, λ, Λ).Let ( x , z ) ∈ A . There exists a paraboloid P of opening a > x v , z v ) ∈ B that touches U from below at ( x , z ). If z = 0, then, by Lemma 6.6, it must be that f ( x ) ≤ (cid:12)(cid:12) h (cid:48) ( z v ) (cid:12)(cid:12) ≤ | f − ( x ) | a ≤ (cid:107) f − (cid:107) L ∞ ( Q R ∩{ z =0 } ) a which contradicts that ( x v , z v ) ∈ B . Hence, z (cid:54) = 0.Since U − P attains a local minimum at ( x , z ), ∇ U ( x , z ) = ∇ P ( x , z ) = − a (cid:0) x − x v , h (cid:48) ( z ) − h (cid:48) ( z v ) (cid:1) which implies ( x v , h (cid:48) ( z v )) = ( x , h (cid:48) ( z )) + 1 a ∇ U ( x , z ) . This is how the vertices ( x v , z v ) ∈ B are uniquely determined by ( x , z ) ∈ A . Notice thatthis identity is equivalent to ∇ Φ( x v , z v ) = ∇ (cid:18) Φ + 1 a U (cid:19) ( x , z ) for all ( x v , z v ) ∈ B . Consider the map T : A → T ( A ) = ∇ Φ( B ) given by T ( x , z ) = ∇ (cid:18) Φ + 1 a U (cid:19) ( x , z ) . For ε >
0, let A ε ⊂ A be given by A ε = A \ { ( x, z ) : | z | < ε } . Then, T is Lipschitz and injective on A ε , so that, by the area formula for Lipschitz maps, | T ( A ε ) | = ˆ T ( A ε ) dy dw = ˆ A ε | det ( ∇ T ( x, z )) | dz dx = ˆ A ε (cid:12)(cid:12)(cid:12)(cid:12) det (cid:18) D (cid:18) Φ + 1 a U (cid:19) ( x, z ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dz dx. We claim that there is a constant C = C ( n, λ, Λ) > x , z ) ∈ A (7.1) − aD Φ( x , z ) ≤ D U ( x , z ) ≤ CaD Φ( x , z ) . The first inequality is straightforward because P touches U from below at ( x , z ). To provethe second inequality in (7.1), suppose, by way of contradiction, that(7.2) D U ( x , z ) > CaD Φ( x , z ) for all C > . Then D U ( x , z ) > Ca (cid:18) e k ⊗ e k
00 0 (cid:19) > Ca (cid:18) e k ⊗ e k
00 0 (cid:19) − a (cid:18) I | z | s − (cid:19) where e k , k = 1 , . . . , n are the standard basis vectors in R n . Since ˜ A = (cid:18) A ( x ) 00 0 (cid:19) ≥ D U ( x , z ) − Ca (cid:18) e k ⊗ e k
00 0 (cid:19) + a (cid:18) I | z | s − (cid:19) ≥ , we have that tr (cid:18) ˜ AD U ( x , z ) − Ca ˜ A (cid:18) e k ⊗ e k
00 0 (cid:19) + a ˜ A (cid:18) I | z | s − (cid:19)(cid:19) ≥ . By ellipticity (see (1.5)),(7.3) a ij ( x ) ∂ ij U ( x , z ) ≥ ( Ca ) a kk ( x ) − a tr( A ( x )) ≥ Caλ − an Λ . RACTIONAL ELLIPTIC EQUATIONS 27
Similarly, from (7.2), D U ( x , z ) > Ca (cid:18) | z | s − (cid:19) > Ca (cid:18) | z | s − (cid:19) − a (cid:18) I | z | s − (cid:19) . From the definition of positive definite matrices, ∂ zz U ( x , z ) − Ca | z | s − + a | z | s − > . Therefore,(7.4) | z | − s ∂ zz U ( x , z ) > Ca − a. By (7.3) and (7.4), it follows that0 ≥ a ij ( x ) ∂ ij U ( x , z ) + | z | − s ∂ zz U ( x , z ) > Caλ − an Λ + Ca − a = [ C ( λ + 1) − ( n Λ + 1)] a, which is a contradiction when C = C ( n, λ, Λ) > ≤ D (cid:18) Φ + 1 a U (cid:19) ( x , z ) ≤ D Φ( x , z ) + CD Φ( x , z ) = ( C + 1) D Φ( x , z )for all ( x , z ) ∈ A . Hence, | T ( A ε ) | = ˆ A ε det (cid:18) D (cid:18) Φ + 1 a U (cid:19) ( x, z ) (cid:19) dx dz ≤ ˆ A ε det (cid:0) ( C + 1) D Φ( x, z ) (cid:1) dx dz = ( C + 1) n +1 µ Φ ( A ε ) ≤ ( C + 1) n +1 µ Φ ( A ) . As this holds for all ε > µ Φ ( B ) = |∇ Φ( B ) | = | T ( A ) | ≤ ( C + 1) n +1 µ Φ ( A ) . Thus, µ Φ ( B ) = µ Φ ( B ) + µ Φ ( B ) ≤ (1 − ε ) µ Φ ( B ) + ( C + 1) n +1 µ Φ ( A )from which it follows that µ Φ ( A ) ≥ µ Φ ( A ) ≥ ε C + 1) n +1 µ Φ ( B ) = cε µ Φ ( B ) . (cid:3) The following is a parallel result to that of Theorem 7.1 for subsolutions when paraboloidsof opening a < U for the first time.The proof is straightforward. We will apply this lemma in the proof of Theorem 5.3. Theorem 7.2.
Assume that Ω is a bounded domain and that a ij ( x ) : Ω → R are bounded,measurable functions that satisfy (1.5) . Let Q R = Q R (˜ x, ˜ z ) ⊂⊂ Ω × R and f ∈ L ∞ ( Q R ∩ { z = } ) . Suppose U ∈ C ( Q R \ { z = 0 } ) ∩ C ( Q R ) such that U is symmetric across { z = 0 } and U z + ∈ C ( Q R ∩ { z ≥ } ) is a subsolution to (cid:40) a ij ( x ) ∂ ij U + | z | − s ∂ zz U ≥ in Q R ∩ { z (cid:54) = 0 }− ∂ z + U ≤ f on Q R ∩ { z = 0 } . Let B ⊂ Q R be a closed set and fix a < . For each ( x v , z v ) ∈ B , we slide paraboloids ofopening a and vertex ( x v , z v ) from above until they touch the graph of U for the first time.Let A denote the set of contact points and assume that A ⊂ Q R . Then A is compact and if µ Φ (cid:32) B ∩ (cid:40) ( x, z ) : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f + (cid:107) L ∞ ( Q R ∩{ z =0 } ) | a | (cid:41)(cid:33) ≤ (1 − ε ) µ Φ ( B ) , for some ε > , then there is a positive constant c = c ( n, λ, Λ) < such that µ Φ ( A ) ≥ ε cµ Φ ( B ) . Remark 7.3.
By checking the proofs, it is easy to see that Theorems 7.1 and 7.2 are stillvalid when the cube Q R is replaced by a section S R .8. Explicit barriers
This section contains the construction of the barriers that will be used in Section 9 to provea localization estimate. This is a quite delicate task due to the degeneracy/singularity of theextension equation and the presence of the Neumann boundary condition. The following isa preliminary result that will be used in the case when 0 < s ≤ / Lemma 8.1.
Let < s ≤ / and z > be fixed. Define the function Q : R → R by Q ( z ) = ( h (cid:48) ( z ) − h (cid:48) ( z )) δ h ( z , z ) h (cid:48)(cid:48) ( z ) . Then Q is a continuous function of z > , and Q ( z ) ≥ for all z > .Proof. By L’H¨opital’s rule, lim z → z Q ( z ) = 2, so that Q ( z ) is continuous for z >
0. Also, for s = 1 / z (cid:54) = z , we have Q ( z ) = 2. Hence, let us assume for the remainder of theproof that 0 < s < / z → + Q ( z ) = ∞ and that lim z →∞ Q ( z ) = − s ≥
1. Therefore, itis enough to prove that Q ( z ) is decreasing for z > z (cid:54) = z . To this end, we will show that Q (cid:48) ( z ) < z (cid:54) = z . First, observe that(8.1) Q (cid:48) ( z ) = ( h (cid:48) ( z ) − h (cid:48) ( z )) h (cid:48)(cid:48) ( z )( δ h ( z , z ) h (cid:48)(cid:48) ( z )) · I ( z )where I ( z ) = 2 δ h ( z , z ) h (cid:48)(cid:48) ( z ) − ( h (cid:48) ( z ) − h (cid:48) ( z )) − δ h ( z , z )( h (cid:48) ( z ) − h (cid:48) ( z )) h (cid:48)(cid:48)(cid:48) ( z ) h (cid:48)(cid:48) ( z ) . We can write I ( z ) = − s − s z s − + s − s z s z s − − s (1 − s )1 − s z s − z s − + s (1 − s )1 − s z s − z − . It follows that I ( z ) > z > ψ ( z ) := − z s − z + z s +10 z s − − (1 − s ) z s z s + (1 − s ) z s > , RACTIONAL ELLIPTIC EQUATIONS 29 for all z >
0. Note that ψ ( z ) = 0 and ψ (0) = (1 − s ) z s >
0. We claim that ψ is decreasingas function of z >
0. Indeed, ψ (cid:48) ( z ) < − z s − + (cid:18) s − (cid:19) z s +10 z s − − (1 − s ) s z s z s − < . Multiplying both sides by z − /s s/ (1 − s ) > z z s − < (cid:18) s − s (cid:19) z s − + (cid:18) − s − s (cid:19) z s − , which is true by Young’s inequality, and the claim follows. Thus, we conclude that ψ ( z ) > < z < z and ψ ( z ) < z > z . This gives that I ( z ) > < z < z and I ( z ) < z > z . Since, in addition, h (cid:48) ( z ) − h (cid:48) ( z ) < < z < z and h (cid:48) ( z ) − h (cid:48) ( z ) > z > z , we deduce from (8.1) that Q (cid:48) ( z ) < z (cid:54) = z . This completes the proof. (cid:3) We now construct the barriers φ . For a set S ⊂ R n +1 , we introduce the notation S + = S ∩ { z ≥ } and S − = S ∩ { z ≤ } . To deal with the singularity at z = 0, we define φ in either the positive or negative halfspaces. In particular, if z ≥
0, then we consider the partial ring [ S r ( x , z ) \ S γr ( x , z )] + . If z <
0, then we consider the partial ring [ S r ( x , z ) \ S γr ( x , z )] − . We will use the condensednotation [ S r ( x , z ) \ S γr ( x , z )] ± = (cid:40) [ S r ( x , z ) \ S γr ( x , z )] + if z ≥ S r ( x , z ) \ S γr ( x , z )] − if z < . Lemma 8.2.
Fix < γ < and consider a section S r ( x , z ) ⊂ R n +1 .If z ≥ , then there exists a classical subsolution φ = φ ( x, z ) to (8.2) (cid:40) a ij ( x ) ∂ ij φ + | z | − s ∂ zz φ > a ( n Λ + 1) in [ S r ( x , z ) \ S γr ( x , z )] + ∩ { z (cid:54) = 0 }− ∂ z + φ ( x, < on [ S r ( x , z ) \ S γr ( x , z )] + ∩ { z = 0 } . If z ≤ , then there exist a classical subsolution φ = φ ( x, z ) to (8.3) (cid:40) a ij ( x ) ∂ ij φ + | z | − s ∂ zz φ > a ( n Λ + 1) in [ S r ( x , z ) \ S γr ( x , z )] − ∩ { z (cid:54) = 0 }− ∂ z − φ ( x, > on [ S r ( x , z ) \ S γr ( x , z )] − ∩ { z = 0 } . In each case, φ > in [ S r ( x , z ) \ S γr ( x , z )] ± , φ ≤ on [ ∂S r ( x , z )] ± , and there is aconstant C = C ( n, λ, Λ , γ ) > such that φ ≤ Car on [ ∂S γr ( x , z )] ± .Proof. The proof of (8.3) will follow from (8.2) at the end by symmetry. The construction ofthe subsolution in (8.2) will depend on whether z > z = 0 and on whether 0 < s ≤ / / < s < Case 1 : z > < s ≤ / δ Φ (( x , z ) , ( x, z ))) − α for a large constant α = α ( γ, n, λ, Λ , s ) > Q ( z ) be the function defined in Lemma x, z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] + \ { z = 0 } , we use ellipticity and Lemma8.1 to estimate a ij ( x ) ∂ ij ( δ Φ (( x , z ) , ( x, z ))) − α + | z | − s ∂ zz ( δ Φ (( x , z ) , ( x, z ))) − α = α ( δ Φ (( x , z ) , ( x, z ))) − α − (cid:20) ( α + 1) (cid:0) a ij ( x )( x − x ) i ( x − x ) j + | z | − s ( h (cid:48) ( z ) − h (cid:48) ( z )) (cid:1) − (tr( A ( x )) + 1) δ Φ (( x , z ) , ( x, z )) (cid:21) ≥ α ( δ Φ (( x , z ) , ( x, z ))) − α − (cid:20) ( α + 1) (cid:0) λ | x − x | + | z | − s ( h (cid:48) ( z ) − h (cid:48) ( z )) (cid:1) − ( n Λ + 1) δ Φ (( x , z ) , ( x, z )) (cid:21) = α ( δ Φ (( x , z ) , ( x, z ))) − α − (cid:20) ( α + 1) (cid:18) λδ ϕ ( x , x ) + ( h (cid:48) ( z ) − h (cid:48) ( z )) h (cid:48)(cid:48) ( z ) δ h ( z , z ) δ h ( z , z ) (cid:19) − ( n Λ + 1) ( δ ϕ ( x , x ) + δ h ( z , z )) (cid:21) = α ( δ Φ (( x , z ) , ( x, z ))) − α − (cid:20) (2 λ ( α + 1) − ( n Λ + 1)) δ ϕ ( x , x ) + ( Q ( z )( α + 1) − ( n Λ + 1)) δ h ( z , z ) (cid:21) ≥ α ( δ Φ (( x , z ) , ( x, z ))) − α − (cid:20) (2 λ ( α + 1) − ( n Λ + 1)) δ ϕ ( x , x ) + (( α + 1) − ( n Λ + 1)) δ h ( z , z ) (cid:21) . Choose α = α ( γ, n, λ, Λ) large so that2 λ ( α + 1) − ( n Λ + 1) > γ − ( n Λ + 1) and ( α + 1) − ( n Λ + 1) > γ − ( n Λ + 1) . Since γr ≤ δ Φ (( x , z ) , ( x, z )) = δ ϕ ( x , x ) + δ h ( z , z ), it must be that δ ϕ ( x , x ) ≥ γr/ δ h ( z , z ) ≥ γr/
2. If δ ϕ ( x , x ) ≥ γr/
2, then a ij ( x ) ∂ ij ( δ Φ (( x , z ) , ( x, z ))) − α + | z | − s ∂ zz ( δ Φ (( x , z ) , ( x, z ))) − α > α ( δ Φ (( x , z ) , ( x, z ))) − α − (cid:20) γ − ( n Λ + 1) δ ϕ ( x , x ) + 0 (cid:21) ≥ α ( n Λ + 1)(2 r ) − α − . If δ h ( z , z ) ≥ γr/
2, then a ij ( x ) ∂ ij ( δ Φ (( x , z ) , ( x, z ))) − α + | z | − s ∂ zz ( δ Φ (( x , z ) , ( x, z ))) − α > α ( δ Φ (( x , z ) , ( x, z ))) − α − (cid:20) γ − ( n Λ + 1) δ h ( z , z ) (cid:21) ≥ α ( n Λ + 1)(2 r ) − α − . Combing the previous two estimates, we have that, for all ( x, z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] + \{ z = 0 } , a ij ( x ) ∂ ij ( δ Φ (( x , z ) , ( x, z ))) − α + | z | − s ∂ zz ( δ Φ (( x , z ) , ( x, z ))) − α > α ( n Λ + 1)(2 r ) − α − . Define φ in [ S r ( x , z ) \ S γr ( x , z )] + by φ ( x, z ) = α − a (2 r ) α +1 [( δ Φ (( x , z ) , ( x, z ))) − α − r − α ] . RACTIONAL ELLIPTIC EQUATIONS 31
Then a ij ( x ) ∂ ij φ ( x, z ) + | z | − s φ ( x, z ) > a ( n Λ + 1). We next check the Neumann condition.Let ( x, ∈ [ S r ( x , z ) \ S γr ( x , z )] + ∩ { z = 0 } and observe that ∂ z + φ ( x,
0) = − a (2 r ) α +1 ( δ Φ (( x , z ) , ( x, z )) − α − ( h (cid:48) ( z ) − h (cid:48) ( z )) (cid:12)(cid:12) z =0 = a (2 r ) α +1 ( δ Φ (( x , z ) , ( x, − α − h (cid:48) ( z ) ≥ ah (cid:48) ( z ) > z >
0. Therefore, φ defined in [ S r ( x , z ) \ S γr ( x , z )] + is a subsolution to (8.2). Itis easy to check that φ ≤ S r ( x , z ) \ S r ( x , z )] + and φ > S r ( x , z )] + . Lastly,for ( x, z ) ∈ [ ∂S γr ( x , z )] ± , we have that φ ( x, z ) = α − a α +1 ( γ − − r = Car , where C = C ( γ, n, λ, Λ) > Case 2 : z ≥ / < s < | z | − s is small with respect to the size of thesection S r ( z ) ⊂ R . Let ε > H ε by H ε = (cid:110) z ∈ S r ( z ) : | z | − s ≤ ε s − − s r s − (cid:111) = (cid:110) z ∈ S r ( z ) : | z | ≤ ε s − s r s (cid:111) . We first show that the measure of H ε is small with respect to the measure of the section S r ( z ). Indeed, by Lemmas 4.1 and 4.8, µ h ( H ε ) = ˆ H ε h (cid:48)(cid:48) ( z ) dz ≤ ˆ ε s − s r s − ε s − s r s h (cid:48)(cid:48) ( z ) dz = 2 h (cid:48) (cid:16) ε s − s r s (cid:17) = Cεr − s ≤ Cεµ h ( S r ( z ))for a constant C = C ( s ).We will construct a function h ε in [ S r ( z )] + that bypasses the points in H ε . Let ˜ H ε bean open interval such that H ε ⊂ ˜ H ε ⊂ S r ( z ) , µ h ( ˜ H ε \ H ε ) ≤ ε µ h ( S r ( z )) , and let ψ ε = ψ ε ( z ) be a smooth function satisfying ψ ε = 1 in H ε , ψ ε = ε in S r ( z ) \ ˜ H ε , ε ≤ ψ ε ≤ S r ( z ) . We use the notation[ S r ( z )] + = ( z L , z R ) , where 0 ≤ z L ≤ z < z R . Note that z L = 0 if 0 ∈ S r ( z ).In [ S r ( z )] + , let h ε = h ε ( z ) be the strictly convex solution to h (cid:48)(cid:48) ε = 2( n Λ + 1) ψ ε h (cid:48)(cid:48) in [ S r ( z )] + h ε ( z R ) = 0 h (cid:48) ε ( z L ) = εr − s . We remark that h ε ∈ C ∞ (( z L , z R )) and, since h ∈ C ( R ), we have h ε ∈ C ([ S r ( z )] + ).Since h ε is strictly convex in [ S r ( z )] + and h ε ∈ C ([ S r ( z )] + ), it follows that h (cid:48) ε > S r ( z )] + . Moreover, since h ε is strictly increasing, h ε achieves its maximum at z = z R , sothat h ε ≤ S r ( z )] + . To bound h ε and h (cid:48) ε , we first estimate ˆ S r ( z ) ψ ε dµ h = ˆ H ε ψ ε dµ h + ˆ ˜ H ε \ H ε ψ ε dµ h + ˆ S r ( z ) \ ˜ H ε ψ ε dµ h ≤ ˆ H ε dµ h + ˆ ˜ H ε \ H ε dµ h + ˆ S r ( z ) \ ˜ H ε ε dµ h = µ h ( H ε ) + µ h ( ˜ H ε \ H ε ) + εµ h ( S r ( z ) \ ˜ H ε ) ≤ Cεµ h ( S r ( z )) + εµ h ( S r ( z )) + εµ h ( S r ( z )) = Cεµ h ( S r ( z )) . for a constant C = C ( s ) >
0. Let δ >
0. For z ∈ [ S r ( z )] + , by the previous estimate andLemma 4.8, (cid:12)(cid:12) h (cid:48) ε ( z ) (cid:12)(cid:12) = h (cid:48) ε ( z ) = ˆ zz L + δ h (cid:48)(cid:48) ε ( w ) dw + h (cid:48) ε ( z L + δ )= ˆ zz L + δ n Λ + 1) ψ ε h (cid:48)(cid:48) ( w ) dw + h (cid:48) ε ( z L + δ ) ≤ n Λ + 1) ˆ S r ( z ) ψ ε dµ h + h (cid:48) ε ( z L + δ ) ≤ n Λ + 1)
Cεµ h ( S r ( z )) + h (cid:48) ε ( z L + δ ) ≤ Cεr − s + h (cid:48) ε ( z L + δ )for a constant C = C ( n, Λ , s ) >
0. Taking the limit as δ →
0, we have h (cid:48) ε ( z ) ≤ Cεr − s + h (cid:48) ε ( z L ) = Cεr − s + εr − s = C εr − s for a constant C = C ( n, Λ , s ).Again, let δ >
0. For z ∈ [ S r ( z )] + , by Lemma 4.8, | h ε ( z ) | = − h ε ( z ) = ˆ z R − δz h (cid:48) ε ( w ) dw − h ε ( z R − δ ) ≤ C εr − s ˆ z R − δz dw − h ε ( z R − δ ) ≤ C εr − s | S r ( z ) | − h ε ( z R − δ ) ≤ C εr − s r s − h ε ( z R − δ )for a constant C = C ( n, Λ , s ) >
0. Taking the limit as δ →
0, we have | h ε ( z ) | ≤ C εr − h ε ( z R ) = C εr. Suppose that γr/ ≤ δ h ( z , z ) < r . By the convexity of δ h ( z , z ) in the variable z , weobtain 0 = δ h ( z , z ) ≥ δ h ( z , z ) + ∂ z δ h ( z , z ) · ( z − z ) . Since S r ( z ) ⊂ B C s (2 r ) s ( z ) by Lemma 4.8, this implies | ∂ z δ h ( z , z ) | ≥ δ h ( z , z ) | z − z | ≥ γr/ C s (2 r ) s = C r − s for a constant C = C ( γ, s ). Choose ε = ε ( γ, n, Λ , s ) > C ε < C . Then, (cid:12)(cid:12) ∂ z δ h ( z , z ) − h (cid:48) ε ( z ) (cid:12)(cid:12) ≥ | ∂ z δ h ( z , z ) | − (cid:12)(cid:12) h (cid:48) ε ( z ) (cid:12)(cid:12) ≥ ( C − C ε ) r − s > ∂ z δ h ( z , z ) − h (cid:48) ε ( z )) ≥ ( C − C ε ) r − s = C r − s RACTIONAL ELLIPTIC EQUATIONS 33 for a constant C = C ( γ, n, Λ , s ) > α = α ( γ, n, λ, Λ , s ) >
0, we define the function ˜ φ on [ S r ( x , z ) \ S γr ( x , z )] + by ˜ φ ( x, z ) = ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α . Let ( x, z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] + \ { z = 0 } . Since h ε ≤
0, we first note that(8.5) γr ≤ δ Φ (( x , z ) , ( x, z )) ≤ δ Φ (( x , z ) , ( x, z )) − h ε ( z ) < r + C εr = (2 + C ε ) r. The equation for ˜ φ in [ S r ( x , z ) \ S γr ( x , z )] + \ { z = 0 } is a ij ( x ) ∂ ij ˜ φ + | z | − s ∂ zz ˜ φ = α ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α − (cid:18) ( α + 1) (cid:20) a ij ( x )( x − x ) i ( x − x ) j + | z | − s ( ∂ z ( δ Φ (( x , z ) , ( x, z ))) − h (cid:48) ε ( z )) (cid:21) − ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) (cid:20) tr( A ( x )) + 1 − n Λ + 1) ψ ε (cid:21)(cid:19) . Using ellipticity, we estimate a ij ( x ) ∂ ij ˜ φ + | z | − s ∂ zz ˜ φ ≥ α ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α − (cid:18) ( α + 1) (cid:20) λδ ϕ ( x , x ) + | z | − s ( ∂ z ( δ Φ (( x , z ) , ( x, z ))) − h (cid:48) ε ( z )) (cid:21) − ( δ Φ (( x , z ) , ( x, z )) − h ε ( z ))(1 − ψ ε )( n Λ + 1) (cid:19) . Suppose that z ∈ H ε . Since ψ ε ( z ) = 1, we can use (8.5) to estimate(8.6) a ij ( x ) ∂ ij ˜ φ + | z | − s ∂ zz ˜ φ ≥ α ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α − (cid:18) δ Φ (( x , z ) , ( x, z )) − h ε ( z ))( n Λ + 1) (cid:19) ≥ α ( n Λ + 1)(2 + C ε ) − α − r − α − . Next, suppose that z / ∈ H ε . Since ψ ε ( z ) > | z | − s > ε s − − s r s − , we estimate(8.7) a ij ( x ) ∂ ij ˜ φ + | z | − s ∂ zz ˜ φ ≥ α ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α − (cid:18) ( α + 1) (cid:20) λδ ϕ ( x , x ) + ε s − − s r s − ( ∂ z ( δ Φ (( x , z ) , ( x, z )) − h ε ( z ))) (cid:21) − ( δ Φ (( x , z ) , ( x, z )) − h ε ( z ))( n Λ + 1) (cid:19) . Since δ Φ (( x , z ) , ( x, z )) ≥ γr , we have that δ ϕ ( x , x ) ≥ γr/ δ h ( z , z ) ≥ γr/
2. Supposefirst that δ ϕ ( x , x ) ≥ γr/
2. Then2 λδ ϕ ( x , x ) + ε s − − s r s − ( ∂ z ( δ Φ (( x , z ) , ( x, z ))) − h ε ( z ))) ≥ λδ ϕ ( x , x ) ≥ λγr. Choose α = α ( γ, n, λ, Λ , s ) large enough to guarantee that( α + 1) λγ − ( n Λ + 1)(2 + C ε ) > ( n Λ + 1)(2 + C ε ) . Then, from (8.7) and (8.5),(8.8) a ij ( x ) ∂ ij ˜ φ + | z | − s ∂ zz ˜ φ ≥ α ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α − (cid:18) ( α + 1) λγr − ( n Λ + 1)(2 + C ε ) r (cid:19) > α ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α − ( n Λ + 1)(2 + C ε ) r ≥ α ( n Λ + 1)(2 + C ε ) − α − r − α − . Next, suppose that δ h ( z , z ) ≥ γr/
2. Since S r ( x , z ) ⊂ S r ( x ) × S r ( z ), we know that γr/ ≤ δ h ( z , z ) < r . By (8.4), we obtain2 λδ ϕ ( x , x ) + ε s − − s r s − ( ∂ z ( δ Φ (( x , z ) , ( x, z ))) − h ε ( z ))) ≥ ε s − − s r s − ( ∂ z ( δ Φ (( x , z ) , ( x, z ))) − h (cid:48) ε ( z )) ≥ ε s − − s r s − C r − s = C ε s − − s r. Let α = α ( γ, n, λ, Λ , s ) > α + 1) C ε s − − s − ( n Λ + 1)(2 + C ε ) > ( n Λ + 1)(2 + C ε ) . Then, from (8.7), we use (8.5) to obtain(8.9) a ij ( x ) ∂ ij ˜ φ + | z | − s ∂ zz ˜ φ ≥ α ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α − (cid:18) ( α + 1) C ε s − − s r − ( n Λ + 1)(2 + C ε ) r (cid:19) > α ( δ Φ (( x , z ) , ( x, z )) − h ε ( z )) − α − ( n Λ + 1)(2 + C ε ) r ≥ α ( n Λ + 1)(2 + C ε ) − α − r − α − . From (8.6), (8.8), and (8.9), there is an α = α ( γ, n, λ, Λ , s ) > x, z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] + \ { z = 0 } , we have a ij ( x ) ∂ ij ˜ φ + | z | − s ∂ zz ˜ φ > α ( n Λ + 1)(2 + C ε ) − α − r − α − . We define the barrier φ on [ S r ( x , z ) \ S γr ( x , z )] + by φ ( x, z ) = aα − (2 + C ε ) α +1 r α +1 (cid:16) ˜ φ ( x, z ) − (1 + C ε ) − α r − α (cid:17) . For ( x, z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] + \{ z = 0 } , it then follows that a ij ( x ) ∂ ij φ + | z | − s ∂ zz φ >a ( n Λ + 1). For the Neumann condition, let ( x, ∈ [ S r ( x , z ) \ S γr ( x , z )] + ∩ { z = 0 } .Using (8.5), we see that ∂ z + φ ( x,
0) = a (2 + C ε ) α +1 r α +1 ( δ Φ (( x , z ) , ( x, − h ε (0)) − α − ( h (cid:48) ( z ) + εr − s ) > a (2 + C ε ) α +1 r α +1 (2 + C ε ) − α − r − α − ( h (cid:48) ( z ) + εr − s )= a ( h (cid:48) ( z ) + εr − s ) > . since z ≥
0. Therefore, φ is a subsolution to (8.2). In [ S r ( x , z ) \ S γr ( x , z )] + , we have γr ≤ δ Φ (( x , z ) , ( x, z )) − h ε ( z ) < (1 + C ε ) r, so that φ > S r ( x , z ) \ S γr ( x , z )] + . Choose ε > > C ε . Then, φ ≤ ∂S r ( x , z )] + . Indeed, for ( x, z ) ∈ [ ∂S r ( x , z )] + , we have that − h ε ( z ) ≥ > (1 + C ε − r RACTIONAL ELLIPTIC EQUATIONS 35 which implies δ Φ (( x , z ) , ( x, z )) − h ε ( z ) = 2 r − h ε ( z ) > (1 + C ε ) r. Thus, φ ( x, z ) ≤
0. Lastly, let ( x, z ) ∈ [ ∂S γr ( x , z )] + and observe that φ ( x, z ) = aα − (2 + C ε ) α +1 r α +1 (cid:0) ( γr − h ε ( z )) − α − (1 + C ε ) − α r − α (cid:1) ≤ aα − (2 + C ε ) α +1 r α +1 (cid:0) ( γr + 0) − α − (cid:1) = Car for C = C ( γ, n, λ, Λ , s ) > Case 3 : z = 0 and 0 < s ≤ / z = 0. We will add a function g ε to the quasi-distance function δ Φ to adjust thebarrier as we did in Case 2.Let ( x, z ) ∈ [ S r ( x , \ S γr ( x , + . Since ( x, z ) ∈ S r ( x , ⊂ S r ( x ) × S r (0), we knowthat s − s | z | s = δ h (0 , z ) < r. Therefore,(8.10) | z | < ¯ C r s , ¯ C := (cid:18) − s ) s (cid:19) s . which means that S r (0) = B ¯ C r s (0). Also, since 2 − s ≤ | z | − s ≥ ¯ C s − s r s − . Given ε >
0, define g ε in [ S r (0)] + by g ε ( z ) = εr − s z − ε ¯ C r. For all z ∈ [ S r (0)] + , we have that g ε ≤ | g ε ( z ) | = ε ¯ C r − εr − s z ≤ ¯ C εr, and g (cid:48) ε ( z ) = εr − s > . Let z be such that γr/ ≤ δ h (0 , z ) < r . As in Case 2 above, since z ∈ S r (0) = B ¯ C r s (0),we can use the convexity of δ h (0 , z ) in the variable z to obtain | ∂ z δ h (0 , z ) | ≥ δ h (0 , z ) | z | ≥ γr/ C r s = ¯ C r − s for a constant ¯ C = ¯ C ( γ, s ). Choose ε = ε ( γ, n, Λ , s ) > ε < ¯ C . Then, (cid:12)(cid:12) ∂ z δ h (0 , z ) − g (cid:48) ε ( z ) (cid:12)(cid:12) ≥ | ∂ z δ h (0 , z ) | − (cid:12)(cid:12) g (cid:48) ε ( z ) (cid:12)(cid:12) ≥ ( ¯ C − ε ) r − s > ∂ z δ h (0 , z ) − g (cid:48) ε ( z )) ≥ ( ¯ C − ε ) r − s = ¯ C r − s for a constant ¯ C = ¯ C ( γ, n, Λ , s ) > x, z ) ∈ [ S r ( x , \ S γr ( x , + . Since − g ε ≥
0, we have that(8.13) γr ≤ δ Φ (( x , , ( x, z )) ≤ δ Φ (( x , , ( x, z )) − g ε ( z ) < r + ε ¯ C r = (2 + ε ¯ C ) r. We define a function ¯ φ on [ S r ( x , \ S γr ( x , + by¯ φ ( x, z ) = ( δ Φ (( x , , ( x, z )) − g ε ( z )) − α . Let ( x, z ) ∈ [ S r ( x , \ S γr ( x , + \ { z = 0 } . Using ellipticity, (8.11), and (8.13), weestimate the equation for ¯ φ as follows:(8.14) a ij ( x ) ∂ ij ¯ φ + | z | − s ∂ zz ¯ φ ≥ α ( δ Φ (( x , , ( x, z )) − g ε ( z )) − α − (cid:18) ( α + 1) (cid:20) λδ ϕ ( x , x ) + ¯ C s − s r s − ( ∂ z ( δ Φ (( x , , ( x, z ))) − g (cid:48) ε ( z )) (cid:21) − ( δ Φ (( x , , ( x, z )) − g ε ( z ))( n Λ + 1) (cid:19) ≥ α ( δ Φ (( x , , ( x, z )) − g ε ( z )) − α − (cid:18) ( α + 1) (cid:20) λδ ϕ ( x , x ) + ¯ C s − s r s − ( ∂ z ( δ Φ (( x , , ( x, z ))) − g (cid:48) ε ( z )) (cid:21) − ( n Λ + 1)(2 + ε ¯ C ) r (cid:19) . Since δ Φ (( x , , ( x, z )) ≥ γr , we know that δ ϕ ( x , x ) ≥ γr/ δ h (0 , z ) ≥ γr/
2. Supposefirst that δ ϕ ( x , x ) ≥ γr/
2. Then2 λδ ϕ ( x , x ) + ¯ C s − s r s − ( ∂ z ( δ Φ (( x , , ( x, z ))) − g (cid:48) ε ( z )) ≥ λδ ϕ ( x , x ) ≥ λγr. Choose α = α ( γ, n, λ, Λ , s ) large enough to guarantee that( α + 1) λγ − ( n Λ + 1)(2 + ¯ C ε ) > ( n Λ + 1)(2 + ¯ C ε ) . Then, from (8.14) and using (8.13), we have that(8.15) a ij ( x ) ∂ ij ¯ φ + | z | − s ∂ zz ¯ φ ≥ α ( δ Φ (( x , , ( x, z )) − g ε ( z )) − α − (cid:0) ( α + 1) λγr − ( n Λ + 1)(2 + ¯ C ε ) r (cid:1) > α ( δ Φ (( x , , ( x, z )) − g ε ( z )) − α − ( n Λ + 1)(2 + ¯ C ε ) r ≥ α ( n Λ + 1)(2 + ¯ C ε ) − α − r − α − . Next, suppose that δ h (0 , z ) ≥ γr/
2. We further know that γr/ ≤ δ h (0 , z ) < r , so, by(8.12), 2 λδ ϕ ( x , x ) + ¯ C s − s r s − ( ∂ z ( δ Φ (( x , , ( x, z ))) − g ε ( z ))) ≥ C s − s r s − ¯ C r − s = ¯ C ¯ C s − s r. Let α = α ( γ, n, λ, Λ , s ) > α + 1) ¯ C ¯ C s − s − ( n Λ + 1)(2 + ¯ C ε ) > ( n Λ + 1)(2 + ¯ C ε ) . Then, from (8.14) and using (8.13),(8.16) a ij ( x ) ∂ ij ¯ φ + | z | − s ∂ zz ¯ φ ≥ α ( δ Φ (( x , , ( x, z )) − g ε ( z )) − α − (cid:18) ( α + 1) ¯ C ¯ C s − s r − ( n Λ + 1)(2 r + ¯ C ε ) r (cid:19) > α ( δ Φ (( x , , ( x, z )) − g ε ( z )) − α − ( n Λ + 1)(2 + ¯ C ε ) r ≥ α ( n Λ + 1)(2 + ¯ C ε ) − α − r − α − . RACTIONAL ELLIPTIC EQUATIONS 37
From (8.15) and (8.16), there is an α = α ( γ, n, λ, Λ , s ) > x, z ) ∈ [ S r ( x , \ S γr ( x , + \ { z = 0 } , we have a ij ( x ) ∂ ij ¯ φ + | z | − s ∂ zz ¯ φ > α ( n Λ + 1)(2 + ¯ C ε ) − α − r − α − . We define the barrier φ on [ S r ( x , \ S γr ( x , + by φ ( x, z ) = aα − (2 + ¯ C ε ) α +1 r α +1 (cid:0) ¯ φ ( x, z ) − (1 + ¯ C ε ) − α r − α (cid:1) . For ( x, z ) ∈ [ S r ( x , \ S γr ( x , + \ { z = 0 } , it follows that a ij ( x ) ∂ ij φ + | z | − s ∂ zz φ >a ( n Λ + 1). For ( x, ∈ [ S r ( x , \ S γr ( x , + ∩ { z = 0 } , by (8.13), ∂ z + φ ( x,
0) = a (2 + ¯ C ε ) α +1 r α +1 ( δ Φ (( x , , ( x, − g ε (0)) − α − εr − s ≥ a (2 + ¯ C ε ) α +1 r α +1 (2 + ¯ C ε ) − α − r − α − εr − s = aεr − s > . Therefore, φ defined in [ S r ( x , \ S γr ( x , + is a subsolution to (8.2). One can also checkthat φ > S r ( x , \ S γr ( x , + and that φ ≤ ∂S r ( x , + when ε = ε ( γ, s ) is smallenough to guarantee that 2 > C ε . Moreover, there is a constant C = C ( γ, n, λ, Λ , s ) > φ ( x, z ) ≤ Car on [ ∂S γr ( x , + . Case 4 : z ≤ < s < x, z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] − , then ( x, − z ) ∈ [ S r ( x , − z ) \ S γr ( x , − z )] + .Define ψ in [ S r ( x , z ) \ S γr ( x , z )] − to be the even reflection across { z = 0 } of the solution φ to (8.2) in [ S r ( x , − z ) \ S γr ( x , − z )] + : ψ ( x, z ) = φ ( x, − z ) , for ( x, z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] − . Since D ψ ( x, z ) = D φ ( x, − z ), we know, for ( x, z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] − \ { z = 0 } ,that a ij ( x ) ∂ ij ψ ( x, z ) + | z | − s ∂ zz ψ ( x, z ) = a ij ( x ) ∂ ij φ ( x, − z ) + | z | − s ∂ zz φ ( x, − z ) > a ( n Λ + 1) . For ( x, ∈ [ S r ( x , z ) \ S γr ( x , z )] − ∩ { z = 0 } , we have − ∂ z − ψ ( x,
0) = ∂ z + φ ( x, >
0. Therefore, ψ is a subsolution to (8.3). It is straightforward to check that ψ > S r ( x , z ) \ S γr ( x , z )] − and that ψ ≤ ∂S r ( x , z )] − . Lastly, if ( x, z ) ∈ [ ∂S γr ( x , z )] − ,then ( x, − z ) ∈ [ ∂S γr ( x , − z )] + . This gives the desired estimate ψ ( x, z ) = φ ( x, − z ) ≤ Car for ( x, z ) ∈ [ ∂S γr ( x , z )] − . (cid:3) Localization lemma
Before stating the main lemma of this section, we need to introduce some notation. First,we define a constant ˆ K to be large enough so that for any ( x , z ) , (˜ x, ˜ z ) ∈ R n +1 and R > Q r ( x , z ) ⊂ Q R (˜ x, ˜ z ), then Q n +1) r ( x , z ) ⊂ Q ˆ K R (˜ x, ˜ z ). By Lemma 4.9, we know thatif Q r ( x , z ) ⊂ Q R (˜ x, ˜ z ) then r ≤ R . If ( x, z ) ∈ Q n +1) r ( x , z ) then, by the quasi-triangleinequality (see Notation 4.7), δ ϕ (˜ x, x ) ≤ K ( δ ϕ (˜ x, x ) + δ ϕ ( x , x )) < K ( R + 2( n + 1) r ) < K (1 + 2( n + 1)) Rδ h (˜ z, z ) ≤ K ( δ h (˜ z, z ) + δ h ( z , z )) < K (1 + 2( n + 1)) R. We then take ˆ K = ˆ K ( n, s ) as(9.1) ˆ K = (2 n + 3) K. Let ˆ K = ˆ K ( n, s ) be given by(9.2) ˆ K = θ ˆ K . If Q ˆ K R (˜ x, ˜ z ) ∩ { z = 0 } (cid:54) = ∅ , then 0 ∈ S ˆ K R (˜ z ) and, by the engulfing property, Q ˆ K R (˜ x, ˜ z ) = Q ˆ K R (˜ x ) × S ˆ K R (˜ z ) ⊂ Q θ ˆ K R (˜ x ) × S θ ˆ K R (0) = Q θ ˆ K R (˜ x, Q θ ˆ K R (˜ x,
0) = Q θ ˆ K R (˜ x ) × S θ ˆ K R (0) ⊂ Q θ ˆ K R (˜ x ) × S θ ˆ K R (˜ z ) = Q ˆ K R (˜ x, ˜ z ) . We define a vertex set B v ⊂ Q ˆ K R (˜ x, ˜ z ) by B v = (cid:40) Q ˆ K R (˜ x, ˜ z ) if ˜ z = 0 or if Q ˆ K R (˜ x, ˜ z ) ∩ { z = 0 } = ∅ Q θ ˆ K R (˜ x,
0) if ˜ z (cid:54) = 0 and Q ˆ K R (˜ x, ˜ z ) ∩ { z = 0 } (cid:54) = ∅ , so that B v is symmetric with respect to { z = 0 } if Q ˆ K R (˜ x, ˜ z ) ∩ { z = 0 } (cid:54) = ∅ .Define the contact set A a,R for a continuous function U on Q ˆ K R (˜ x, ˜ z ) by(9.3) A a,R := (cid:26) ( x, z ) ∈ Q ˆ K R (˜ x, ˜ z ) : U ( x, z ) ≤ aR and there is ( x v , z v ) ∈ B v such that U can be touched from below at ( x, z ) in Q ˆ K R (˜ x, ˜ z )by a paraboloid of opening a > x v , z v ) (cid:27) . Lemma 9.1.
The contact set A a,R is closed in Q ˆ K R (˜ x, ˜ z ) .Proof. Let ( x k , z k ) ∈ A a,R and ( x , z ) ∈ Q ˆ K R (˜ x, ˜ z ) be such that ( x k , z k ) → ( x , z ). Since U ( x k , z k ) ≤ aR and U is continuous, U ( x , z ) ≤ aR . By the same argument as in the proofof Theorem 7.1 with B = B v , we can touch U from below in Q ˆ K R (˜ x, ˜ z ) at ( x , z ) by aparaboloid P of opening a > x v , z v ) ∈ B v . Therefore, ( x , z ) ∈ A a,R whichshows that A a,R is closed in Q ˆ K R (˜ x, ˜ z ). (cid:3) The following is the main localization lemma of this section. We show that if a supersolu-tion U can be touched from below with a paraboloid P of opening a > Q r , thenthe set in which U can be touched from below by paraboloids of increased opening Ca > C = C ( n, λ, Λ , s ) >
0, in a smaller cube Q ηr makes up a universal proportion of Q r .To state the result, we define positive constants K > η < K = 2 K + 2 K and η = 1 K (2 KK + 1) . Lemma 9.2.
Fix a > . Assume that Ω is a bounded domain and that a ij ( x ) : Ω → R are bounded, measurable functions that satisfy (1.5) . For a cube Q R = Q R (˜ x, ˜ z ) ⊂ R n +1 ,consider Q ˆ K R = Q ˆ K R (˜ x, ˜ z ) where ˆ K is as in (9.2) . Let f ∈ L ∞ ( Q ˆ K R ∩ { z = 0 } ) benonnegative. Suppose U ∈ C ( Q ˆ K R \ { z = 0 } ) ∩ C ( Q ˆ K R ) such that U is symmetric across { z = 0 } and U z + ∈ C ( Q ˆ K R ∩ { z ≥ } ) is a supersolution to (cid:40) a ij ( x ) ∂ ij U + | z | − s ∂ zz U ≤ in Q ˆ K R ∩ { z (cid:54) = 0 }− ∂ z + U ≥ f on Q ˆ K R ∩ { z = 0 } . Let Q r ( x , z ) be such that Q r ( x , z ) ⊂ Q R and Q r ( x , z ) ∩ A a,R (cid:54) = ∅ . RACTIONAL ELLIPTIC EQUATIONS 39
There exists positive constants C = C ( n, λ, Λ , s ) > and c = c ( n, λ, Λ , s ) < such that µ Φ ( A Ca,R ∩ Q ηr ( x , z )) ≥ cµ Φ ( Q r ( x , z )) . where η = η ( n, s ) < is as in (9.4) . Remark 9.3.
Once the existence of C = C ( n, λ, Λ , s ) > C larger. Indeed, if C (cid:48) > C then, by Lemma 6.3, we have that A Ca,R ⊂ A C (cid:48) a,R .To prove Lemma 9.2, we first use the barrier φ constructed in Lemma 8.2 to control how U detaches from a touching paraboloid P . Lemma 9.4.
Fix < γ < . Assume that Ω is a bounded domain and that a ij ( x ) : Ω → R are bounded, measurable functions that satisfy (1.5) . For a cube Q R = Q R (˜ x, ˜ z ) ⊂ R n +1 ,consider a cube Q ˆ K R = Q ˆ K R (˜ x, ˜ z ) where ˆ K is as in (9.2) . Let f ∈ L ∞ ( Q ˆ K R ∩ { z = 0 } ) benonnegative. Suppose U ∈ C ( Q ˆ K R \ { z = 0 } ) ∩ C ( Q ˆ K R ) such that U is symmetric across { z = 0 } and U z + ∈ C ( Q ˆ K R ∩ { z ≥ } ) is a supsersolution to (cid:40) a ij ( x ) ∂ ij U + | z | − s ∂ zz U ≤ in Q ˆ K R ∩ { z (cid:54) = 0 }− ∂ z + U ≥ f on Q ˆ K R ∩ { z = 0 } . Assume that Q r ( x , z ) ⊂ Q R for some point ( x , z ) such that z ≥ . Suppose that U istouched from below at ( x , z ) ∈ [ S r ( x , z )] + ∩ A a,R in Q ˆ K R by a paraboloid P of opening a > with vertex ( x v , z v ) such that z v ≥ . Then, there exists a constant C = C ( γ, n, λ, Λ) > and a point ( x , z ) ∈ [ S γr ( x , z )] + such that U ( x , z ) − P ( x , z ) ≤ Car.
Proof.
If ( x , z ) ∈ [ S γr ( x , z )] + , U ( x , z ) − P ( x , z ) = 0 ≤ Car for all
C >
0, so we can take ( x , z ) = ( x , z ). Therefore, we assume for the remainder ofthe proof that ( x , z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] + .Let W = U − P . For ( x, z ) ∈ Q ˆ K R \ { z = 0 } , we have that a ij ( x ) ∂ ij P ( x, z ) + | z | − s ∂ zz P ( x, z ) = − a (tr( A ( x )) + 1) ≥ − a ( n Λ + 1)which implies a ij ( x ) ∂ ij W ( x, z ) + | z | − s ∂ zz W ( x, z ) ≤ a ( n Λ + 1) . Since z v ≥
0, we also have that − ∂ z + W ( x, ≥ f ( x ) + ah (cid:48) ( z v ) ≥ . Let φ be the subsolution to (8.2) in [ S r ( x , z ) \ S γr ( x , z )] + . By the choice of ˆ K in(9.1), we have that Q r ( x , z ) ⊂ Q R implies S r ( x , z ) ⊂ Q r ( x , z ) ⊂ Q ˆ K R ⊂ Q ˆ K R Therefore, W − φ satisfies(9.5) (cid:40) a ij ( x ) ∂ ij ( W − φ ) + | z | − s ∂ zz ( W − φ ) < S r ( x , z ) \ S γr ( x , z )] + ∩ { z (cid:54) = 0 }− ∂ z + ( W − φ )( x, > S r ( x , z ) \ S γr ( x , z )] + ∩ { z = 0 } . Let ( x , z ) ∈ [ S r ( x , z ) \ S γr ( x , z )] + be such that W ( x , z ) − φ ( x , z ) = min [ S r ( x ,z ) \ S γr ( x ,z )] + ( W − φ ) . By the maximum principle (see [11, Theorem 3.1]), the minimum of W − φ occurs on theboundary ∂ [ S r ( x , z ) \ S γr ( x , z )] + . That is,( x , z ) ∈ [ ∂S r ( x , z )] + ∪ [ ∂S γr ( x , z )] + ∪ [( S r ( x , z ) \ S γr ( x , z )) ∩ { z = 0 } ] . We claim that ( x , z ) ∈ [ ∂S γr ( x , z )] + .First, we will show that ( x , z ) / ∈ [ ∂S r ( x , z )] + . Since ( x , z ) ∈ [ S r ( x , z )] + , we knowthat φ ( x , z ) > W ( x , z ) − φ ( x , z ) = 0 − φ ( x , z ) <
0. Moreover, since φ ≤ ∂S r ( x , z )] + , we have that W ( x, z ) − φ ( x, z ) ≥ ∂S r ( x , z )] + . Therefore,the minimum is strictly negative and cannot occur on [ ∂S r ( x , z )] + .If [ S r ( x , z )] + ∩ { z = 0 } = ∅ , then our claim holds. Suppose that [ S r ( x , z )] + ∩{ z = 0 } (cid:54) = ∅ . Assume, by way of contradiction, that the minimum occurs on [ S r ( x , z ) \ S γr ( x , z )] + ∩ { z = 0 } , i.e. z = 0. Then − ∂ z + ( W − φ )( x , ≤
0, which contradicts (9.5).Therefore, it must be that the minimum occurs at ( x , z ) ∈ [ ∂S γr ( x , z )] + ⊂ [ S γr ( x , z )] + .It follows from Lemma 8.2 that φ ( x , z ) ≤ Car for C = C ( n, λ, Λ , γ ) >
0. Since W ( x , z ) − φ ( x , z ) <
0, this implies that U ( x , z ) − P ( x , z ) = W ( x , z ) < φ ( x , z ) ≤ Car. (cid:3)
Remark 9.5.
An analogue of Lemma 9.4 with z , z , z v ≤ φ to (8.3) in [ S r ( x , z ) \ S γr ( x , z )] − . Proof of Lemma 9.2.
Without loss of generality, we can assume that Q r ( x , z ) ∩ A a,R (cid:54) = ∅ .Otherwise, we replace r by r + ε and then take the limit as ε → + at the end. Let( x , z ) ∈ Q r ( x , z ) ∩ A a,R .Since ( x , z ) ∈ A a,R , there is a paraboloid P of opening a > x v , z v ) ∈ B v that touches U from below in Q ˆ K R at ( x , z ). We write P as P ( x, z ) = − aδ Φ (( x v , z v ) , ( x, z )) + aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z ) . As for z , it must be that either z ≥ z <
0. We may assume that z has the samesign as z , meaning that z , z ≥ z , z ≤
0. Indeed, suppose that z ≥ z < Q ˆ K R ∩ { z = 0 } = ∅ , this is a contradiction. If Q ˆ K R ∩ { z = 0 } (cid:54) = ∅ then, by Lemma 6.4,˜ P ( x, z ) = P ( x, − z ) touches U from below in Q ˆ K R at ( x , − z ) with vertex ( x v , − z v ) ∈ B v .Since δ h ( z , − z ) = h ( z ) − h ( z ) + h (cid:48) ( z ) z + h (cid:48) ( z ) z < h ( z ) − h ( z ) − h (cid:48) ( z ) z + h (cid:48) ( z ) z since z ≥ z < < − z = δ h ( z , z ) < r, it follows that ( x , − z ) ∈ Q r ( x , z ) ∩ A a,R . We proceed with the proof of the lemma using˜ P and − z > P and z <
0. The argument for z ≤ z > z , z ≥
0. Then, z v ≥
0. Indeed, if z >
0, then by Lemma 6.4, we know that z v ≥
0. If z = 0, then, since f ≥
0, by Lemma6.6, it must be that f ( x ) = 0 and, consequently, z v = 0. RACTIONAL ELLIPTIC EQUATIONS 41
Let γ = η/ (2 θ ). Note that ( x , z ) ∈ Q r ( x , z ) ⊂ S ( n +1) r ( x , z ). We apply Lemma 9.4with r = ( n + 1) r and γ = γ/ ( n + 1) to find a point( x , z ) ∈ [ S γ r ( x , z )] + = [ S γr ( x , z )] + ⊂ S γr ( x , z )and a constant C = C ( n, λ, Λ , s ) > U ( x , z ) − P ( x , z ) ≤ Car.
Let α = η/ (2 θ ) < C (cid:48) = C (cid:48) ( n, λ, Λ , s ) > P ( x, z ) = P ( x, z ) − C (cid:48) aδ Φ ((¯ x v , ¯ z v ) , ( x, z )) + d, for (¯ x v , ¯ z v ) ⊂ S αr ( x , z )until they touch the graph of U in Q ˆ K R for the first time. It is clear that¯ P ( x, z ) = − aδ ϕ ( x v , x ) − C (cid:48) aδ ϕ (¯ x v , x ) − aδ h ( z v , z ) − C (cid:48) aδ h (¯ z v , z ) + d (cid:48) for some constant d (cid:48) . Let ξ ∈ R be such that h (cid:48) ( ξ ) = h (cid:48) ( z v ) + C (cid:48) h (cid:48) (¯ z v ) C (cid:48) + 1 . It follows that, for some constant b (cid:48) , − aδ h ( z v , z ) − C (cid:48) aδ h (¯ z v , z ) = − ( C (cid:48) + 1) aδ h ( ξ, z ) + b (cid:48) . Since ∇ ϕ ( x v ) + C (cid:48) ∇ ϕ (¯ x v ) C (cid:48) + 1 = x v + C (cid:48) ¯ x v C (cid:48) + 1 = ∇ ϕ (cid:18) x v + C (cid:48) ¯ x v C (cid:48) + 1 (cid:19) , we similarly write, for some constant b (cid:48)(cid:48) , − aδ ϕ ( x v , x ) − C (cid:48) aδ ϕ (¯ x v , x ) = − ( C (cid:48) + 1) aδ ϕ (cid:18) x v + C (cid:48) ¯ x v C (cid:48) + 1 , x (cid:19) + b (cid:48)(cid:48) . Therefore ¯ P ( x, z ) = − ( C (cid:48) + 1) aδ Φ (cid:18)(cid:18) x v + C (cid:48) ¯ x v C (cid:48) + 1 , ξ (cid:19) , ( x, z ) (cid:19) + d (cid:48)(cid:48) , for some constant d (cid:48)(cid:48) . Hence, the opening of ¯ P is ( C (cid:48) + 1) a > (cid:18) x v + C (cid:48) ¯ x v C (cid:48) + 1 , ξ (cid:19) where h (cid:48) ( ξ ) = h (cid:48) ( z v ) + C (cid:48) h (cid:48) (¯ z v ) C (cid:48) + 1 . Let B be the set of these vertices and let A denote the set of corresponding touching points.Since ¯ P ( x , z ) ≤ U ( x , z ), we have that P ( x , z ) − C (cid:48) aδ Φ ((¯ x v , ¯ z v ) , ( x , z )) + d ≤ U ( x , z ). By the engulfing property, S αr ( x , z ) ⊂ S αθr (¯ x v , ¯ z v ), so that δ Φ ((¯ x v , ¯ z v ) , ( x , z )) <αθr . Therefore, d ≤ U ( x , z ) − P ( x , z ) + C (cid:48) aδ ((¯ x v , ¯ z v ) , ( x , z )) ≤ Car + C (cid:48) αθar. Since ( x , z ) ∈ S αθr (¯ x v , ¯ z v ) ⊂ S αθr (¯ x v , ¯ z v ), we again use the engulfing property to see that S αθr (¯ x v , ¯ z v ) ⊂ S αθ r ( x , z ). Suppose that ( x, z ) ∈ Q ˆ K R is such that δ Φ (( x , z ) , ( x, z )) ≥ αθ r . Then δ Φ ((¯ x v , ¯ z v ) , ( x, z )) ≥ αθr and¯ P ( x, z ) ≤ P ( x, z ) − C (cid:48) a (2 αθr ) + (cid:0) Car + C (cid:48) αθar (cid:1) = P ( x, z ) + (cid:0) C − C (cid:48) θα (cid:1) ar < P ( x, z ) ≤ U ( x, z )when C (cid:48) = C (cid:48) ( n, λ, Λ , s ) > C (cid:48) > C/ ( θα ). Hence, the contact points for ¯ P areinside S αθ r ( x , z ). That is, A ⊂ S αθ r ( x , z ). Recall that ( x , z ) ∈ S γr ( x , z ). Since γ = αθ , we use the engulfing property to obtain S γr ( x , z ) = S αθr ( x , z ) ⊂ S αθ r ( x , z ) ⊂ S αθ r ( x , z ) ⊂ S αθ r ( x , z ) = S ηr ( x , z ) ⊂ Q ηr ( x , z ) . Consequently, A ⊂ S αθ r ( x , z ) ⊂ Q ηr ( x , z ).We now estimate¯ P ( x, z ) ≤ P ( x, z ) + d ≤ aδ Φ (( x v , z v ) , ( x , z )) + U ( x , z ) + d ≤ aδ Φ (( x v , z v ) , ( x , z )) + aR + ( Car + C (cid:48) αθar ) ≤ aK ( δ Φ ((˜ x, ˜ z ) , ( x v , z v )) + δ Φ ((˜ x, ˜ z ) , ( x , z ))) + aR + ( CaR + C (cid:48) αθaR ) ≤ aK ( ˆ K R + R ) + aR + ( CaR + C (cid:48) αθaR )= (cid:16) ( ˆ K + 1) K + 1 + C + C (cid:48) αθ (cid:17) aR. If C (cid:48) = C (cid:48) ( n, λ, Λ , s ) > P ( x, z ) ≤ ( C (cid:48) + 1) aR which shows that A ⊂ A ( C (cid:48) +1) a,R .Since f ≥
0, we trivially have that µ Φ (cid:18) B ∩ (cid:26) ( x, z ) : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f − (cid:107) L ∞ ( Q ˆ K R ) ( C (cid:48) + 1) a (cid:27)(cid:19) = µ Φ ( B ∩ { ( x, z ) : z = 0 } ) = 0 . Therefore, by Theorem 7.1,(9.7) µ Φ ( A ( C (cid:48) +1) a,R ∩ Q ηr ( x , z )) ≥ µ Φ ( A ∩ Q ηr ( x , z )) = µ Φ ( A ) ≥ cµ Φ ( B ) . We claim that(9.8) cµ Φ ( B ) ≥ c (cid:48) µ Φ ( Q r ( x , z ))for a positive constant c (cid:48) = c (cid:48) ( n, λ, Λ , s ) < µ Φ ( B ) ≥ (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n +1 µ Φ ( S αr ( x , z )) . Observe that the B can be expressed as B = (cid:26) ( x, z ) : x = x v + C (cid:48) ¯ x v C (cid:48) + 1 , h (cid:48) ( z ) = h (cid:48) ( z v ) + C (cid:48) h (cid:48) (¯ z v ) C (cid:48) + 1 , (¯ x v , ¯ z v ) ∈ S αr ( x , z ) (cid:27) . Define the sets B and B by B = (cid:26) x = x v + C (cid:48) ¯ x v C (cid:48) + 1 : ¯ x v ∈ S αr/ ( x ) (cid:27) B = (cid:26) z = ( h (cid:48) ) − (cid:18) h (cid:48) ( z v ) + C (cid:48) h (cid:48) (¯ z v ) C (cid:48) + 1 (cid:19) : ¯ z v ∈ S αr/ ( z ) (cid:27) . Since S αr/ ( x , z ) ⊂ S αr/ ( x ) × S αr/ ( z ) ⊂ S αr ( x , z ), we know that B × B ⊂ B and(9.10) µ Φ ( B ) ≥ µ Φ ( B × B ) = µ ϕ ( B ) µ h ( B ) . RACTIONAL ELLIPTIC EQUATIONS 43
By a change of variables, µ ϕ ( B ) = ˆ B dx = (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n ˆ S αr/ ( x ) d ¯ x v = (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n µ ϕ (cid:0) S αr/ ( x ) (cid:1) . Notice that the set Z given by Z = (cid:26) ¯ z v ∈ R : h (cid:48) (¯ z v ) = − C (cid:48) h (cid:48) ( z v ) (cid:27) is a singleton. Then, by using a change of variables, µ h ( B ) = ˆ B \{ z =0 } h (cid:48)(cid:48) ( z ) dz = ˆ S αr/ ( z ) \ Z h (cid:48)(cid:48) (cid:18) ( h (cid:48) ) − (cid:18) h (cid:48) ( z v ) + C (cid:48) h (cid:48) (¯ z v ) C (cid:48) + 1 (cid:19)(cid:19) ∂ z ( h (cid:48) ) − (cid:12)(cid:12)(cid:12)(cid:12) h (cid:48) ( zv )+ C (cid:48) h (cid:48) (¯ zv ) C (cid:48) +1 (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) h (cid:48)(cid:48) (¯ z v ) d ¯ z v = C (cid:48) C (cid:48) + 1 ˆ S αr/ ( z ) \ Z h (cid:48)(cid:48) (¯ z v ) d ¯ z v = C (cid:48) C (cid:48) + 1 µ h (cid:0) S αr/ ( z ) (cid:1) . Combining these estimates into (9.10), we obtain µ Φ ( B ) ≥ (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n +1 µ ϕ (cid:0) S αr/ ( x ) (cid:1) µ h (cid:0) S αr/ ( z ) (cid:1) ≥ (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n +1 µ Φ (cid:0) S αr/ ( x , z ) (cid:1) and (9.9) holds.For (9.8), observe that, by the doubling estimate (4.6), µ Φ ( S γθr ( x , z )) ≤ K d (cid:18) θγα (cid:19) ν µ Φ ( S αr ( x , z ))and µ Φ ( S ( n +1) r ( x , z )) ≤ K d (cid:18) n + 1 γ (cid:19) ν µ Φ ( S γr ( x , z )) . Since ( x , z ) ∈ S γr ( x , z ), the engulfing property gives S γr ( x , z ) ⊂ S γθr ( x , z ). Hence,by using (9.9) and the previous two estimates, cµ Φ ( B ) ≥ c (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n +1 µ Φ ( S αr ( x , z )) ≥ c (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n +1 K d (cid:18) α θγ (cid:19) ν µ Φ ( S γθr ( x , z )) ≥ c (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n +1 K d (cid:18) α θγ (cid:19) ν µ Φ ( S γr ( x , z )) ≥ c (cid:18) C (cid:48) C (cid:48) + 1 (cid:19) n +1 K d (cid:18) α θ ( n + 1) (cid:19) ν µ Φ ( S ( n +1) r ( x , z )) ≥ c (cid:48) µ Φ ( Q r ( x , z )) . This completes the proof of (9.8).From (9.7) and (9.8), the lemma follows. (cid:3)
Covering lemma
Here, we establish the following covering lemma that is similar to the Calder´on–Zygmunddecomposition:
Lemma 10.1.
Let K = K ( n, s ) > , η = η ( n, s ) < be as in (9.4) , and fix < c < .Consider a cube Q R/K = Q R/K (˜ x, ˜ z ) . Suppose there is a countable family of closed sets D k ⊂ R n +1 that satisfy the following properties:1) D ⊂ D ⊂ · · · ⊂ D k ⊂ · · · ⊂ Q R/K , D (cid:54) = ∅ ;2) for any ( x, z ) ∈ R n +1 , ρ > such that Q ρ ( x, z ) ⊂ Q R (˜ x, ˜ z ) , Q ηρ ( x, z ) ⊂ Q R/K (˜ x, ˜ z ) , Q ρ ( x, z ) ∩ D k (cid:54) = ∅ , we have µ Φ ( Q ηρ ( x, z ) ∩ D k +1 ) ≥ cµ Φ ( Q ρ ( x, z )) . Then µ Φ ( Q R/K \ D k ) ≤ (1 − c ) k µ Φ ( Q R/K ) . To prove Lemma 10.1, we need the following simple consequence of [5, Theorem 1.2] forMonge–Amp`ere cubes.
Lemma 10.2.
Let E ⊂ R n +1 be a bounded subset. For each ( x, z ) ∈ E , consider a cube Q r ( x,z ) ( x, z ) with radius r ( x,z ) > . Then there is a countable subfamily of such cubes { Q r i ( x i , z i ) } ∞ i =1 such that E ⊂ ∞ (cid:91) i =1 Q r i ( x i , z i ) , with Q r i /K ( x i , z i ) pairwise disjoint . Proof of Lemma 10.1.
For any ( x , z ) ∈ E := Q R/K (˜ x, ˜ z ) \ D k and let r be given by(10.1) r = r ( x ,z ) = inf { r : Q r ( x , z ) ∩ D k (cid:54) = ∅ } . The family { Q r ( x , z ) } covers E . By Lemma 10.2, there is a countable collection of cubes { Q r i ( x i , z i ) } ∞ i =1 such that E = Q R/K \ D k ⊂ (cid:83) i Q r i ( x i , z i ), with Q r i /K ( x i , z i ) pairwisedisjoint. Then, µ Φ ( Q R/K \ D k ) ≤ µ Φ (cid:32)(cid:91) i Q r i ( x i , z i ) ∩ Q R/K (cid:33) ≤ (cid:88) i µ Φ ( Q r i ( x i , z i ) ∩ Q R/K ) . We claim that, for any ( x , z ) ∈ E and r given by (10.1),(10.2) µ Φ ( Q r ( x , z ) ∩ Q R/K ) ≤ c µ Φ ( Q r/K ( x , z ) ∩ D k +1 ) . Suppose for now that (10.2) holds. Then µ Φ ( Q R/K \ D k ) ≤ (cid:88) i µ Φ ( Q r i ( x i , z i ) ∩ Q R/K ) ≤ (cid:88) i c µ Φ ( Q r i /K ( x i , z i ) ∩ D k +1 )= 1 c µ Φ (cid:32)(cid:91) i Q r i /K ( x i , z i ) ∩ ( D k +1 \ D k ) (cid:33) ≤ c µ Φ ( D k +1 \ D k ) . RACTIONAL ELLIPTIC EQUATIONS 45
In the second to last estimate, we used our choice of r in (10.1). Since µ Φ ( Q R/K \ D k +1 ) = µ Φ ( Q R/K \ D k ) − µ Φ ( D k +1 \ D k ) ≤ µ Φ ( Q R/K \ D k ) − cµ Φ ( Q R/K \ D k )= (1 − c ) µ Φ ( Q R/K \ D k ) , by iteration, we finally obtain µ Φ ( Q R/K \ D k ) ≤ (1 − c ) k µ Φ ( Q R/K ), and the lemma is proved.It is left to prove (10.2). We will present the proof for n = 1 for which Q R/K (˜ x, ˜ z ) = S R/K (˜ x ) × S R/K (˜ z ) ⊂ R . The more general case follows similarly and is left to the reader.First, we estimate r . Given any point ( x, z ) ∈ Q R/K and ( x , z ) ∈ Q R/K \ D k , we have δ ϕ ( x , x ) ≤ K ( δ ϕ (˜ x, x ) + δ ϕ (˜ x, x )) < KRK . and, similarly, δ h ( z , z ) ≤ KR/K . Therefore, r < KR/K whenever r is given by (10.1).Let ( x , z ) ∈ Q R/K \ D k and r as in (10.1) be fixed.Next, let ( x, z ) ∈ Q r ( x , z ). By the quasi-triangle inequality, the choice of K in (9.4),and the estimate on r , δ ϕ (˜ x, x ) ≤ K ( δ ϕ (˜ x, x ) + δ ϕ ( x , x )) < K (cid:18) RK + r (cid:19) ≤ R. Similarly, one can show that δ h (˜ z, z ) < R . Therefore, we have that(10.3) S r ( x ) × S r ( z ) = Q r ( x , z ) ⊂ Q R (˜ x, ˜ z ) = S R (˜ x ) × S R (˜ z ) . We will break into cases based on how far (˜ x, ˜ z ) is from ( x , z ). Case 1 . Suppose that ˜ x ∈ S r/K ( x ), ˜ z ∈ S r/K ( z ).We will show that Q r ( x , z ) satisfies the hypothesis 2) in the statement with ρ = r : Q r ( x , z ) ⊂ Q R (˜ x, ˜ z ) , Q ηr ( x , z ) ⊂ Q R/K (˜ x, ˜ z ) , Q r ( x , z ) ∩ D k (cid:54) = ∅ . We have already established (10.3). By the definition of r , we know that Q r ( x , z ) ∩ D k (cid:54) = ∅ .Thus, it is left to show that Q ηr ( x , z ) ⊂ Q R/K (˜ x, ˜ z ). Let ( x, z ) ∈ Q ηr ( x , z ). By the quasi-triangle inequality and by choice of K and η in (9.4), since x ∈ S ηr ( x ), δ ϕ (˜ x, x ) ≤ K ( δ ϕ ( x , ˜ x ) + δ ϕ ( x , x )) < K (cid:18) rK + ηr (cid:19) ≤ RK . We can similarly show that(10.4) δ h (˜ z, z ) < RK since z ∈ S ηr ( z ) . Hence, Q ηr ( x , z ) ⊂ Q R/K (˜ x, ˜ z ).Therefore, since η ≤ /K , by property 2), we obtain the desired estimate: µ Φ ( Q r/K ( x , z ) ∩ D k +1 ) ≥ µ Φ ( Q ηr ( x , z ) ∩ D k +1 ) ≥ cµ Φ ( Q r ( x , z )) ≥ cµ Φ ( Q r ( x , z ) ∩ Q R/K (˜ x, ˜ z )) . Case 2 . Suppose that ˜ x / ∈ S r/K ( x ), ˜ z ∈ S r/K ( z ).It must be that x < ˜ x or ˜ x < x . Without loss of generality, we assume that x < ˜ x . From (10.3) and (10.4), we deduce that S r ( z ) ⊂ S R (˜ z ) , S ηr ( z ) ⊂ S R/K (˜ z ) . We will find x between x and ˜ x such that(10.5) S r/ (2 K K ) ( x ) ⊂ S r/K ( x ) ∩ S R/K (˜ x ) . Let x > x be such that δ ϕ ( x , x ) = r/ (2 KK ). We first show that S r/ (2 KK ) ( x ) ⊂ S r/K ( x ). Indeed, for x ∈ S r/ (2 KK ) ( x ), we have that δ ϕ ( x , x ) ≤ K ( δ ϕ ( x , x ) + δ ϕ ( x , x )) < rK . Since r KK = δ ϕ ( x , x ) ≤ Kδ ϕ ( x , x ) ≤ K δ ϕ ( x , x ) = K r KK , we know that r K K ≤ δ ϕ ( x , x ) ≤ r K . Thus, x / ∈ S r/ (2 K K ) ( x ). Since the sections S r/ (2 K K ) ( x ) and S r/K ( x ) are one-dimensionalintervals, we can write them as S r/ (2 K K ) ( x ) = ( x L , x R ) where x L < x < x R S r/K ( x ) = ( x L , x R ) where x L < x < x R . Since ˜ x / ∈ S r/K ( x ) and x < ˜ x , we know that x L < x < x R < ˜ x. Since x < x and S r/ (2 K K ) ( x ) ⊂ S r/K ( x ), we have that x < x L < x < x R < x R < ˜ x. Thus, for any x ∈ S r/ (2 K K ) ( x ), we know that x < x < ˜ x . By Lemma 4.9, δ ϕ (˜ x, x ) < δ ϕ (˜ x, x ) < RK Hence, S r/ (2 K K ) ( x ) ⊂ S R/K (˜ x ) and we proved (10.5).Define ρ = (cid:18) K + 12 K (cid:19) r. Clearly S r ( z ) ⊂ S ρ ( z ). Let x ∈ S r ( x ). Then, δ ϕ ( x , x ) ≤ K ( δ ϕ ( x , x ) + δ ϕ ( x , x )) ≤ K (cid:18) r KK + r (cid:19) = ρ. Hence, S r ( x ) ⊂ S ρ ( x ). Therefore,(10.6) Q r ( x , z ) = S r ( x ) × S r ( z ) ⊂ S ρ ( x ) × S ρ ( z ) = Q ρ ( x , z ) . Since Q r ( x , z ) ∩ D k (cid:54) = ∅ , we know by (10.6) that Q ρ ( x , z ) ∩ D k (cid:54) = ∅ . Next, in orderto apply property 2) in the statement, we will show that Q ρ ( x , z ) satisfies the following:(10.7) Q ρ ( x , z ) ⊂ Q R (˜ x, ˜ z ) , Q ηρ ( x , z ) ⊂ Q R/K (˜ x, ˜ z ) , Q ηρ ( x , z ) ⊂ Q r/K ( x , z ) . First, let us check that Q ρ ( x , z ) ⊂ Q R (˜ x, ˜ z ). Take ( x, z ) ∈ Q ρ ( x , z ) and observe that δ ϕ (˜ x, x ) ≤ K ( δ ϕ (˜ x, x ) + δ ϕ ( x , x )) < K (cid:18) RK + ρ (cid:19) ≤ R. RACTIONAL ELLIPTIC EQUATIONS 47
We can similarly show that δ h (˜ z, z ) < R . Hence, Q ρ ( x , z ) ⊂ Q R (˜ x, ˜ z ). Next, by the choiceof η in (9.4), we know that(10.8) ηρ = r K K ≤ rK . Then, by (10.5), Q ηρ ( x , z ) = S r/ (2 K K ) ( x ) × S r/ (2 K K ) ( z ) ⊂ S r/K ( x ) × S r/K ( z ) = Q r/K ( x , z ) . Lastly, since ˜ z ∈ S r/K ( z ), for z ∈ S ηρ ( z ), by (9.4), δ h (˜ z, z ) ≤ K ( δ h ( z , ˜ z ) + δ h ( z , z )) < K (cid:18) rK + ηρ (cid:19) ≤ RK . Therefore, S ηρ ( z ) ⊂ S R/K (˜ z ). With this, (10.8), and (10.5), we obtain Q ηρ ( x , z ) = S r/ (2 K K ) ( x ) × S ηρ ( z ) ⊂ S R/K (˜ x ) × S R/K (˜ z ) = Q R/K (˜ x, ˜ z ) . We have shown that Q ρ ( x , z ) satisfies the hypotheses of property 2). Therefore, by using(10.7), the conclusion of 2), and (10.6), we obtain the desired estimate: µ Φ ( Q r/K ( x , z ) ∩ D k +1 ) ≥ µ Φ ( Q ηρ ( x , z ) ∩ D k +1 ) ≥ cµ Φ ( Q ρ ( x , z )) ≥ cµ Φ ( Q r ( x , z )) . Case 3 . Suppose that ˜ x ∈ S r/K ( x ), ˜ z (cid:54)∈ S r/K ( z ).This follows exactly as in Case 2 by switching the roles of ˜ x and ˜ z and using δ h in placeof δ ϕ . Case 4 . Suppose that ˜ x / ∈ S r/K ( x ), ˜ z / ∈ S r/K ( z ).This follows by combining the arguments in Case 2 and Case 3. (cid:3) Proof of Theorem 5.3 and Theorem 1.1
Proof of Theorem 5.3.
We begin by sliding a paraboloid P of opening a > x, ˜ z ) from below until it touches the graph of U for the first time in Q ˆ K R , say at( x , z ) ∈ Q ˆ K R . Then P ( x, z ) = − aδ Φ ((˜ x, ˜ z ) , ( x, z )) + aδ Φ ((˜ x, ˜ z ) , ( x , z )) + U ( x , z ) . If δ Φ ((˜ x, ˜ z ) , ( x , z )) > R/K , then aR K ≥ U (˜ x, ˜ z ) ≥ P (˜ x, ˜ z ) = aδ Φ ((˜ x, ˜ z ) , ( x , z )) + U ( x , z ) > aRK . Hence, ( x , z ) ∈ S R/K = S R/K (˜ x, ˜ z ) ⊂ Q R/K and U ( x , z ) = P ( x , z ) ≤ P (˜ x, ˜ z ) < aR. Thus, if A a,R is defined as in (9.3), A a,R ∩ Q R/K (cid:54) = ∅ . In order to apply Lemma 10.1, we define the closed sets D k ⊂ Q R/K by D k := A aC k ,R ∩ Q R/K , k ≥ where C = C ( n, λ, Λ , s ) > C to guarantee that(11.1) C − K ≥ C − K − Kθ > , see Remark 9.3. As a consequence of Lemma 6.3, we have ∅ (cid:54) = D ⊂ D ⊂ D ⊂ · · · ⊂ D k ⊂ · · · ⊂ Q R/K . Thus, hypothesis 1) of Lemma 10.1 is satisfied. To check that property 2) in Lemma 10.1holds, let ( x, z ) ∈ R n +1 , ρ > Q ρ ( x, z ) ⊂ Q R (˜ x, ˜ z ) , Q ηρ ( x, z ) ⊂ Q R/K (˜ x, ˜ z ) , Q ρ ( x, z ) ∩ D k (cid:54) = ∅ . By Lemma 9.2, there is a positive constant c = c ( n, λ, Λ , s ) < µ Φ ( D k +1 ∩ Q ηρ ( x, z )) = µ Φ ( A aC k +1 ,R ∩ Q ηρ ( x, z )) ≥ cµ Φ ( Q ηρ ( x, z )) . Hence, property 2) is satisfied. It follows from Lemma 10.1 that(11.2) µ Φ ( Q R/K \ D k ) ≤ (1 − c ) k µ Φ ( Q R/K ) . Also, from the definition of A aC k ,R ,(11.3) U ( x, z ) ≤ aRC k for ( x, z ) ∈ D k . For k ≥
0, let ρ k = ρ k ( n, λ, Λ , s ) < ρ k (cid:38) k → ∞ . For convenience in the notation, let β = 13 K . Let k = k ( n, λ, Λ , s ) > Claim . Suppose that, for some k ≥ k , there exists a point ( x k , z k ) ∈ Q βR/ ( n +1) ⊂ S βR = S βR (˜ x, ˜ z ) such that U ( x k , z k ) ≥ aRC k +1 . Then there is a point ( x k +1 , z k +1 ) ∈ ∂S ρ k R ( x k , z k ) such that U ( x k +1 , z k +1 ) ≥ aRC k +2 . Proof of claim.
Suppose, by way of contradiction, that
U < aRC k +2 on ∂S ρ k R ( x k , z k ). Inthe section S k = S ρ k R ( x k , z k )we lower paraboloids of the form(11.4) P ( x, z ) = 2 aKC k +2 ρ k δ Φ (( x v , z v ) , ( x, z )) + c v , ( x v , z v ) ∈ S ρkRθC ( x k , z k )from above until they touch the graph of U for the first time in S k . Let A denote the set ofcontact points. Fix a point ( x , z ) ∈ A and a corresponding paraboloid P as in (11.4) thattouches U from above in S k at ( x , z ).If necessary, slide P further until it intersects U at ( x k , z k ) and let us denote this paraboloidby ˜ P . By Lemma 6.2, we can write˜ P ( x, z ) = 2 aKC k +2 ρ k δ Φ (( x v , z v ) , ( x, z )) − aKC k +2 ρ k δ Φ (( x v , z v ) , ( x k , z k )) + U ( x k , z k ) . RACTIONAL ELLIPTIC EQUATIONS 49
Since ( x v , z v ) ∈ S ρkRθC ( x k , z k ), by the engulfing property, S ρkRθC ( x k , z k ) ⊂ S ρkRC ( x v , z v ). Inparticular, δ Φ (( x v , z v ) , ( x k , z k )) ≤ ρ k RC . Therefore, for ( x, z ) ∈ S k ,(11.5) ˜ P ( x, z ) ≥ aKC k +2 ρ k δ Φ (( x v , z v ) , ( x, z )) − aKC k +2 ρ k ρ k RC + aRC k +1 ≥ aKC k +2 ρ k δ Φ (( x v , z v ) , ( x, z )) + 2 aRC k , where we used (11.1). Therefore, U ( x , z ) = P ( x , z ) ≥ ˜ P ( x , z ) ≥ aRC k which shows that A ⊂ { ( x , z ) ∈ S ρ k R ( x k , z k ) : U ( x , z ) ≥ aRC k } . We will next prove that ( x , z ) ∈ S ρ k R ( x k , z k ); that is, the contact points in A are interiorpoints of the section S ρ k R ( x k , z k ). Assume, by way of contradiction, that δ Φ (( x k , z k ) , ( x , z )) = ρ k R . By the quasi-triangle inequality, ρ k R ≤ K ( δ Φ (( x k , z k ) , ( x v , z v )) + δ Φ (( x v , z v ) , ( x , z ))) < K (cid:18) ρ k RθC + δ Φ (( x v , z v ) , ( x , z )) (cid:19) , so that δ Φ (( x v , z v ) , ( x , z )) > ρ k R (cid:18) K − θC (cid:19) . Since ( x , z ) ∈ S k , from (11.5) and (11.1), we get U ( x , z ) = P ( x , z ) ≥ ˜ P ( x , z ) ≥ aKC k +2 ρ k δ Φ (( x v , z v ) , ( x , z )) + aRC k ( C − K ) > aKC k +2 ρ k ρ k R (cid:18) K − θC (cid:19) + aRC k ( C − K ) > aRC k +2 , which contradicts our assumption that U < aRC k +2 on ∂S ρ k R ( x k , z k ). Therefore, it must bethat ( x , z ) ∈ S ρ k R ( x k , z k ). Consequently,(11.6) A ⊂ { ( x , z ) ∈ S ρ k R ( x k , z k ) : U ( x , z ) ≥ aRC k } . Next, we want to apply Theorem 7.2 with Remark 7.3 in S k with ε = 1 /
2. For this, weneed to choose k = k ( n, λ, Λ , s ) sufficiently large to guarantee that(11.7) µ Φ (cid:32) S ρkRθC ( x k , z k ) ∩ (cid:40) ( x, z ) : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f + (cid:107) L ∞ ( S k ∩{ z =0 } ) (2 aKC k +2 /ρ k ) (cid:41)(cid:33) ≤ µ Φ ( S ρkRθC ( x k , z k ))for all k ≥ k . Indeed, observe that µ Φ (cid:32) S ρkRθC ( x k , z k ) ∩ (cid:40) ( x, z ) : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f + (cid:107) L ∞ ( S k ∩{ z =0 } ) (2 aKC k +2 /ρ k ) (cid:41)(cid:33) ≤ µ ϕ (cid:18) S ρkRθC ( x k ) (cid:19) µ h (cid:32)(cid:40) z ∈ R : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f (cid:107) L ∞ ( S k ∩{ z =0 } ) (2 aKC k +2 /ρ k ) (cid:41)(cid:33) . Notice that, by Lemma 4.8, µ h (cid:32)(cid:40) z ∈ R : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f (cid:107) L ∞ ( S k ∩{ z =0 } ) (2 aKC k +2 /ρ k ) (cid:41)(cid:33) = 2 (cid:107) f (cid:107) L ∞ ( S k ∩{ z =0 } ) (2 aKC k +2 /ρ k ) ≤ ρ k KC k +2 c (cid:48) s (cid:16) ρ k θC (cid:17) s − c (cid:48) s (cid:18) ρ k RθC (cid:19) − s ≤ ρ sk θ − s KC k + s c (cid:48) s µ h (cid:18) S ρkRθC ( z k ) (cid:19) . This and the doubling property (4.6) give µ Φ (cid:32) S ρkRθC ( x k , z k ) ∩ (cid:40) ( x, z ) : (cid:12)(cid:12) h (cid:48) ( z ) (cid:12)(cid:12) ≤ (cid:107) f + (cid:107) L ∞ ( S k ∩{ z =0 } ) (2 aKC k +2 /ρ k ) (cid:41)(cid:33) ≤ ρ sk θ − s KC k + s c (cid:48) s µ ϕ (cid:18) S ρkRθC ( x k ) (cid:19) µ h (cid:18) S ρkRθC ( z k ) (cid:19) ≤ ρ sk θ − s KC k + s c (cid:48) s µ Φ (cid:18) S ρkRθC ( x k , z k ) (cid:19) ≤ ρ sk θ − s KC k + s c (cid:48) s K d ν µ Φ (cid:18) S ρkRθC ( x k , z k ) (cid:19) . Therefore, (11.7) holds if we choose k = k ( n, λ, Λ , s ) large enough so that ρ sk θ − s KC k + s c (cid:48) s K d ν ≤
12 for all k ≥ k . Whence, by Theorem 7.2 with Remark 7.3 for ε = 1 /
2, it follows that(11.8) µ Φ ( A ) ≥ c µ Φ ( S ρkRC θ ( x k , z k )) . Next, we will choose ρ k in order to estimate µ Φ ( S ρkRC θ ( x k , z k )) in (11.8) from below by µ Φ ( Q R/K (˜ x, ˜ z )) and get(11.9) µ Φ ( A ) ≥ − c ) k µ Φ ( Q R/K (˜ x, ˜ z )) . In fact, since β < /K , we have that ( x k , z k ) ∈ Q βR/ ( n +1) (˜ x, ˜ z ) ⊂ S βR (˜ x, ˜ z ) ⊂ S R/K (˜ x, ˜ z ),so that, by the engulfing property, S R/K (˜ x, ˜ z ) ⊂ S θR/K ( x k , z k ) . As a consequence of the doubling property (4.6), µ Φ ( S θR/K ( x k , z k )) ≤ K d (cid:18) C θ ρ k K (cid:19) ν µ Φ ( S ρkRC θ ( x k , z k ))and µ Φ ( S R ( n +1) /K (˜ x, ˜ z )) ≤ K d ( n + 1) ν µ Φ ( S R/K (˜ x, ˜ z )) . Combining these estimates, we obtain µ Φ ( S ρkRC θ ( x k , z k )) ≥ K − d (cid:18) ρ k K C θ (cid:19) ν µ Φ ( S θR/K ( x k , z k )) ≥ K − d (cid:18) ρ k K C θ (cid:19) ν µ Φ ( S R/K (˜ x, ˜ z )) RACTIONAL ELLIPTIC EQUATIONS 51 ≥ K − d (cid:18) ρ k K C θ (cid:19) ν K − d ( n + 1) − ν µ Φ ( S R ( n +1) /K (˜ x, ˜ z )) ≥ K − d (cid:18) ρ k K C θ (cid:19) ν K − d ( n + 1) − ν µ Φ ( Q R/K (˜ x, ˜ z )) . If we take ρ k = c (1 − c ) k/ν , c = C θ ( n + 1) K (cid:18) K d c (cid:19) /ν we arrive at (11.9).We next show, by enlarging k if necessary, that S (( x k , z k ) , ρ k R ) = S ρ k R ( x k , z k ) ⊂⊂ Q R/K (˜ x, ˜ z )so that(11.10) A = A ∩ S ρ k R ( x k , z k ) = A ∩ Q R/K (˜ x, ˜ z ) . Let C > p > x k , z k ) ∈ S βR (˜ x, ˜ z ), we knowby Lemma 4.10 with r = β , r = β + ( ρ k /C ) /p , and t = R , that S (( x k , z k ) , ρ k R ) ⊂ S (cid:32) (˜ x, ˜ z ) , (cid:32) β + (cid:18) ρ k C (cid:19) /p (cid:33) R (cid:33) . If necessary, make k = k ( n, λ, Λ , s ) larger to guarantee that(11.11) ∞ (cid:88) j = k (cid:18) ρ j C (cid:19) /p < K − β. In particular, β + (cid:18) ρ k C (cid:19) /p ≤ K for all k ≥ k . Therefore, S ρ k R ( x k , z k ) ⊂ S R/ (2 K ) (˜ x, ˜ z ) ⊂⊂ S R/K (˜ x, ˜ z ) ⊂ Q R/K (˜ x, ˜ z ), which shows (11.10).By the definition of D k , { ( x, z ) : U ( x, z ) > aRC k } ∩ Q R/K ⊂ Q R/K \ D k . With this, (11.2), (11.9), (11.10), and (11.6), we estimate µ Φ ( { U > aRC k } ∩ Q R/K ) ≤ µ Φ ( Q R/K \ D k ) ≤ (1 − c ) k µ Φ ( Q R/K ) ≤ µ Φ ( A )= 12 µ Φ ( A ∩ Q R/K ) ≤ µ Φ ( { U ≥ aRC k } ∩ Q R/K ) ≤ µ Φ ( { U > aRC k } ∩ Q R/K ) , which is a contradiction. This completes the proof of the claim. (cid:3) We now use the claim to prove (5.4) with κ = β/ ( n + 1) and C H = C k +1 . Suppose, byway of contradiction, that there is a point ( x k , z k ) ∈ Q βR/ ( n +1) such thatsup Q βR/ ( n +1) U ≥ U ( x k , z k ) > aRC k +1 . By the claim, there is a point ( x k +1 , z k +1 ) ∈ ∂S ρ k R ( x k , z k ) such that U ( x k +1 , z k +1 ) > aRC k +2 . Repeating this process, we can find a sequence ( x k +1 , z k +1 ) ∈ ∂S ρ k R ( x k , z k ) such that U ( x k +1 , z k +1 ) > aRC k +2 for k > k . For all k ≥ k , by Lemma 4.10 with r = β + k (cid:88) j = k (cid:18) ρ j C (cid:19) /p , r = β + k +1 (cid:88) j = k (cid:18) ρ j C (cid:19) /p , t = R, and by (11.11), we obtain S (( x k +1 , z k +1 ) , ρ k +1 R ) ⊂ S (˜ x, ˜ z ) , β + k +1 (cid:88) j = k (cid:18) ρ j C (cid:19) /p R ⊂ S (cid:18) (˜ x, ˜ z ) , R K (cid:19) ⊂ Q (cid:18) (˜ x, ˜ z ) , R K (cid:19) . Therefore, ( x k +1 , z k +1 ) ∈ Q R/ (2 K ) for all k ≥ k . In particular, U is unbounded in Q R/ (2 K ) .This is a contradiction and completes the proof. (cid:3) Proof of Theorem 1.1.
Let κ = κ ( n, s ) < K = ˆ K ( n, s ) > κ ≤ √ κ and (cid:113) K ≤ ˆ K where κ and ˆ K are the constants from Theorem 1.3. We recall from (4.3) that B r ( x ) = S r / ( x ) for any r > . By taking r = √ κ R , B κR ( x ) × { z = 0 } ⊂ S κ R / ( x ) × S κ R / (0) ⊂ S κ R ( x , . (11.12)By taking r = (cid:112) K R , S ˆ K R ( x , ⊂ B √ K R ( x ) × S ˆ K R (0) ⊂ B ˆ KR ( x ) × S ˆ K R (0) ⊂⊂ Ω × R . We also note that(11.13) S ˆ K R ( x , ∩ { z = 0 } ⊂ B ˆ KR ( x ) × { z = 0 } . Let U be as in Theorem 1.2. By Proposition 3.2 and (1.10), it follows that U ≥ U be the even reflection of U so that ˜ U is symmetric across { z = 0 } . Notice that˜ U ∈ C ( S ˆ K R ( x , \ { z = 0 } ) ∩ C ( S ˆ K R ( x , U z + ∈ C ( S ˆ K R ( x , ∩ { z ≥ } ) and that˜ U is a nonnegative solution to (cid:40) a ij ( x ) ∂ ij ˜ U + z − s ∂ zz ˜ U = 0 in S ˆ K R ( x , ∩ { z (cid:54) = 0 }− ∂ z + ˜ U ( x,
0) = f ( x ) on S ˆ K R ( x , ∩ { z = 0 } . RACTIONAL ELLIPTIC EQUATIONS 53
Since U ( x,
0) = u ( x ), by (11.12), Theorem 1.3 and (11.13), we have thatsup B κR ( x ) u ≤ sup S κ R ( x , ˜ U ≤ C H (cid:32) inf S κ R ( x , ˜ U + (cid:107) f (cid:107) L ∞ ( S ˆ K R ( x , ∩{ z =0 } ) R s (cid:33) ≤ C H (cid:18) inf B κR ( x ) u + (cid:107) f (cid:107) L ∞ ( B ˆ KR ( x )) R s (cid:19) , which proves (1.7). To prove the H¨older estimate (1.8), note that B R ( x ) × { z = 0 } ⊂ S R ( x , x ∈ B R ( x ), that | u ( x ) − u ( x ) | = | ˜ U ( x , − ˜ U ( x, |≤ ˆ C R α δ Φ (( x , , ( x, α (cid:32) sup S R ( x , | ˜ U | + (cid:107) f (cid:107) L ∞ ( S ˆ K R ( x , ∩{ z =0 } ) R s (cid:33) ≤ ˆ C R α | x − x | α (cid:32) sup B R ( x ) × S R (0) | ˜ U | + (cid:107) f (cid:107) L ∞ ( B ˆ KR ( x )) R s (cid:33) . For each fixed z ≥
0, by (3.5), (cid:107) U ( · , z ) (cid:107) L ∞ ( B R ( x )) ≤ (2 s ) z s Γ( s ) ˆ ∞ e − s t z s (cid:13)(cid:13) e − tL u (cid:13)(cid:13) L ∞ ( B R ( x )) dtt s ≤ M (cid:107) u (cid:107) L ∞ (Ω) . Letting ˆ C = M ˆ C /
2, and α = 2 α <
1, we conclude (1.8). (cid:3)
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Department of Mathematics, Iowa State University, 396 Carver Hall, Ames, IA 50011, UnitedStates of America
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