HIGHER REGULARITY OF THE FREE BOUNDARY IN THE OBSTACLE PROBLEM FOR THE FRACTIONAL LAPLACIAN

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1 HIGHER REGULARITY OF THE FREE BOUNDARY IN THE OBSTACLE PROBLEM FOR THE FRACTIONAL LAPLACIAN YASH JHAVERI AND ROBIN NEUMAYER Abstact. We pove a highe egulaity esult fo the fee bounday in the obstacle poblem fo the factional Laplacian via a highe ode bounday Hanack estimate. 1. Intoduction and Main Results In this pape, we investigate the highe egulaity of the fee bounday in the factional obstacle poblem. We pove a highe ode bounday Hanack estimate, building on ideas developed by De Silva and Savin in [12, 13, 14]. As a consequence, we show that if the obstacle is C m,β, then the fee bounday is C m 1,α nea egula points fo some 0 < α β. In paticula, smooth obstacles yield smooth fee boundaies nea egula points The Factional Obstacle Poblem. Fo a given function (obstacle ϕ C(R n decaying apidly at infinity and s (0, 1, a function v is a solution of the factional obstacle poblem if v(x ϕ(x in R n lim x v(x = 0 on R n ( s v(x 0 in R n (1.1 ( s v(x = 0 in v > ϕ} whee the s-laplacian ( s of a function u is defined by ˆ ( s u(x u(x + z u(x := c n,s PV R z n+2s dz. n The sets P := v = ϕ} and Γ := v = ϕ} ae known as the contact set and the fee bounday espectively. The factional obstacle poblem appeas in many contexts, including the picing of Ameican options with jump pocesses (see [11] and the Appendix of [3] fo an infomal discussion and the study of the egulaity of minimizes of nonlocal inteaction enegies in kinetic equations (see [10]. While the obstacle poblem fo the factional Laplacian is nonlocal, it admits a local fomulation thanks to the extension method (see [9, 24]. Specifically, one consides the a-hamonic 1 extension ṽ of v to the uppe half-space R n+1 + := R n (0, : L a ṽ(x, y = 0 in R n+1 + ṽ(x, 0 = v(x on R n whee L a u(x, y := div( y a u(x, y and a := 1 2s ( 1, 1. The function ṽ is obtained as the minimize of the vaiational poblem ˆ min R n+1 + u 2 y a dx dy : u H 1 (R n+1 +, y a, u(x, 0 = v(x } 1 We say a function u is a-hamonic if Lau = 0. 1

2 2 Y. JHAVERI AND R. NEUMAYER and satisfies lim y a ṽ(x, y = ( s v(x x R n. y 0 Afte an even eflection acoss the hypeplane y = 0}, (1.1 is equivalent to the following local poblem. Fo a given a ( 1, 1 and a function ϕ C(R n decaying apidly at infinity, a function ṽ is a solution of the localized factional obstacle poblem if it is even in y and satisfies ṽ(x, 0 ϕ(x on R n lim (x,y ṽ(x, y = 0 on R n+1 L a ṽ(x, y 0 in R n+1 (1.2 L a ṽ(x, y = 0 in R n+1 \ ṽ(x, 0 = ϕ(x}. When a = 0, i.e., s = 1/2, the opeato L a is the Laplacian, and (1.2 is the well-known Signoini (thin obstacle poblem, which can be stated not only in all of R n+1, but in suitable bounded domains of R n+1. Fo example, a typical fomulation of the Signoini poblem is in B 1 R n+1 : ṽ(x, 0 ϕ(x on B 1 y = 0} ṽ(x, y = g(x, y on B 1 (1.3 ṽ(x, y 0 in B 1 ṽ(x, y = 0 in B 1 \ ṽ(x, 0 = ϕ(x}. Pimay questions in obstacle poblems ae the egulaity of the solution and the stuctue and egulaity of the fee bounday. The local fomulation of the factional obstacle poblem, (1.2, allows the use of local PDE techniques to study the egulaity of the solution and the fee bounday. Unde mild conditions on the obstacle 2, Caffaelli, Salsa, and Silveste show, in [8], that the solution of (1.1 is optimally C 1,s using the almost-optimal egulaity of the solution shown via potential theoetic techniques in [28]. Futhemoe, studying limits of appopiate escalings of the solution (blowups at points on the fee bounday, they show the blowup at x 0 Γ must eithe have (a homogeneity 1 + s o (b homogeneity at least 2. The points at which the blowup is (1 + s-homogeneous ae known as egula points of Γ. In [8], they show that the egula points fom a elatively open subset of Γ and that Γ is C 1,σ nea these egula points fo some 0 < σ < 1. Fo the case s = 1/2, analogous esults wee fist shown in [1, 2]. The stuctue of the fee bounday away fom egula points is investigated, fo example, in [3] and [18] Main Result and Cuent Liteatue. Ou main esult is the following: Theoem 1.1. Let ϕ C m,β (R n with m 4 and β (0, 1 o m = 3 and β = 1. Suppose x 0 is a egula point of the fee bounday Γ = Γ(v of the solution v to (1.1. Then, Γ C m 1,α in a neighbohood of x 0 fo some α (0, 1 depending on s, n, m, and β. In paticula, if ϕ C (R n, then Γ C nea egula points. Stating fom the C 1,σ egulaity obtained in [8], the Hölde exponent α obtained in Theoem 1.1 is the minimum of β and σ. In ode to pove Theoem 1.1, we establish a highe ode bounday Hanack estimate fo the opeato L a. We pove this estimate in slit domains, that is, domains in R n+1 fom which an n-dimensional slit P x n+1 = 0} is emoved. Recall that the classical bounday Hanack pinciple states that the quotient of two positive hamonic functions that vanish continuously on a potion of the bounday of a Lipschitz domain is Hölde continuous up to the bounday (see [7, 17]. In [13], De Silva and Savin emakably extend this idea to the highe ode bounday Hanack pinciple, poving that the quotient of two positive hamonic functions that vanish continuously on a potion of the bounday of a C k,α domain is C k,α. Motivated by applications to the Signoini poblem, in [14], they pove such a highe ode bounday Hanack pinciple on slit domains. To do so, they assume Γ := R np is locally the gaph of a function of the fist n 1 vaiables, and they conside a coodinate system (x, on R n+1 whee x R n and is the distance to Γ. They also define a 2 In [8], the obstacle is assumed to be C 2,1, though in [6], this is elaxed, and ϕ is only assumed to be C 1,s+ε.

3 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 3 coesponding notion of Hölde egulaity Cx, k,α C k,α on Γ when Γ C k,α. (see Section 2 that esticts to the standad notion of Theoem 1.2 (De Silva and Savin, [14]. Let u and U > 0 be hamonic functions in B 1 \ P R n+1 that ae even in x n+1 and vanish continuously on the slit P. Suppose 0 Γ := R np with Γ C k,α fo k 1 and Γ C k,α 1. If u L (B 1 1 and U(ν(0/2 = 1, whee ν is the oute unit nomal to P, then u C U fo some C = C(n, k, α > 0. C k,a x, (Γ B 1/2 Hee, Γ C k,α is defined as in (2.1. Theoem 1.2 is used to pove that the fee bounday is smooth nea egula points fo (1.3 with ϕ 0 in the same way that the bounday Hanack pinciple is used to impove the egulaity of the fee bounday nea egula points fom Lipschitz to C 1,α in, fo example, the classical obstacle poblem. Indeed, if Γ is locally the gaph of a C k,α function γ of the fist n 1 vaiables, then Theoem 1.2 implies that i ṽ/ n ṽ is C k,α on Γ whee ṽ is the solution to the Signoini poblem. On the othe hand, Γ is also the zeo level set of ṽ, and so i γ, fo each i = 1,..., n 1, is given by i ṽ/ n ṽ on Γ. Hence, γ C k+1,α. Stating with k = 1 and poceeding iteatively, one poves that the fee bounday is smooth nea egula points. The poof of the highe ode bounday Hanack estimate in this pape is motivated by the global stategy developed by De Silva and Savin to pove Theoem 1.2. Howeve, some delicate aguments ae needed to adapt these ideas to ou setting, which we biefly descibe hee. The poof involves a petubative agument, the coe egulaity esult being one in which Γ is flat (Poposition 3.3. When a = 0, the flat case follows fom capitalizing on the stuctual symmety of the Laplacian; bounday egulaity is inheited fom inteio egulaity fo a eflection of the solution. Instead, to handle the case a 0, we pove the necessay egulaity of ou solutions by hand. Fist, a eduction agument allows us to focus on the two-dimensional case. Hee, the specific degeneacy of the opeato L a, fo each a 0, foces a specific degeneacy in solutions that vanish on the negative x-axis to the equation L a u = 0. We obseve this fist in global homogeneous solutions. Then, we pove that ou solutions have a well-defined powe seies-like decomposition in tems of these homogeneous solutions, which in tun yields the egulaity esult. (See Section 6. Anothe new difficulty we encounte comes fom consideing nonzeo obstacles in (1.1. As discussed above, Theoem 1.2 implies C egulaity of the fee bounday nea egula points in the Signoini poblem with zeo obstacle, that is, taking ϕ 0 in (1.3. Yet, taking ϕ 0 in the nonlocal poblem (1.1 is athe uninteesting: the solution is identically zeo. While the esults of [14] do extend to (1.1 when s = 1/2 so long as an extension of ϕ can be subtacted off without poducing a ight-hand side 3, such an extension is not geneally possible. The new featue of the highe ode bounday Hanack estimates hee, necessay fo teating geneal obstacles, is that we allow both L a u and L a U to be nonzeo when consideing the quotient u/u. Still, handling geneal obstacles in (1.1 is quite involved. Even afte constucting a suitable extension of the obstacle fom R n to R n+1, one finds a lage gap between having and applying these popositions, anothe consequence of having to wok with a degeneate elliptic opeato. (See Section 7. Poposition 3.2 and Poposition 5.4 ae the highe ode bounday Hanack estimates in the geneality needed to pove Theoem 1.1. The simplest case of these estimates, howeve, takes the following fom. Theoem 1.3 (Highe Ode Bounday Hanack Estimate. Let u and U > 0 be a-hamonic functions in B 1 \ P R n+1 that ae even in x n+1 and vanish continuously on the slit P. Suppose 0 Γ := R np with Γ C k,α fo k 1 and Γ C k,α 1. If u L (B 1 1 and U(ν(0/2 = 1, whee ν is the oute 3 Fo instance, this can be done if m ϕ = 0 fo some m N.

4 4 Y. JHAVERI AND R. NEUMAYER unit nomal to P, then fo some C = C(a, n, k, α > 0. u U C k,a x, (Γ B 1/2 The oiginal appoach to poving highe egulaity in obstacle poblems, pioneeed in [20], was to use the hodogaph-legende tansfom. These techniques have been extended to pove highe egulaity in the Signoini poblem with zeo obstacle in [21] and in thin obstacle poblems with vaiable coefficients and inhomogeneities in [22]. On the othe hand, at the same time as [21], De Silva and Savin used the highe ode bounday Hanack pinciple to show highe fee bounday egulaity in the Signoini poblem with zeo obstacle in [14], as we discussed above. They also used these techniques to give a new poof of highe egulaity of the fee bounday in the classical obstacle poblem (see [13]. The highe ode bounday Hanack appoach has been adapted to the paabolic setting, poving highe egulaity of the fee bounday fo the paabolic obstacle poblem in [4] and fo the paabolic Signoini poblem with zeo obstacle in [5]. We mention that an advantage of the hodogaph-legende tansfom appoach is that it allows one to pove analyticity of the fee bounday nea egula points when the obstacle is analytic. Upon completion of this wok, we leaned that Koch, Rüland, and Shi in [23] independently and at the same time have poven an analogous esult to ou Theoem 1.1. In contast to the methods used heein, they employ the patial hodogaph-legende tansfom techniques mentioned above Oganization. In Section 2, we intoduce some notation and pesent some useful popeties of the opeato L a. In Section 3, we state and pove a pointwise highe ode bounday Hanack estimate. Section 4 is dedicated to poving a pointwise Schaude estimate, while Section 5 extends the esults of the pevious two sections to a lowe egulaity setting. In Section 6, we pove a egulaity esult fo a- hamonic functions vanishing continuously on a hypeplane. Finally, we pove Theoem 1.1 in Section 7. Acknowledgments. We wish to thank Alessio Figalli fo suggesting this poblem. Pat of this wok was done while the authos wee guests of the École Nomale Supéioe de Lyon in the fall of 2015; the hospitality of ENS Lyon is gatefully acknowledged. RN is suppoted by the NSF Gaduate Reseach Fellowship unde Gant DGE C 2. Peliminaies 2.1. Notation and Teminology. Let X R n+1 be given by X = (x, x n, y = (x, z = (x, y whee x R n 1, x n R is the nth component of X, and y R is the (n + 1st component of X. In this way, x = (x, x n R n and z = (x n, y R 2. Futhemoe, define B λ := X < λ}, B λ := x < λ}, and B λ := x < λ}. Let Γ be a codimension two suface in R n+1 of class C k+2,α with k 1. Then, up to tanslation, otation, and dilation, Γ is locally given by whee γ : B 1 R is such that We let Define the n-dimensional slit P by Γ = (x, γ(x, 0 : x B 1} γ(0 = 0, x γ(0 = 0, and γ C k+2,α (B 1 1. Γ C k,α := γ C k,α (B 1. (2.1 P := x n γ(x, y = 0},

5 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 5 and notice that R np = Γ. Let d = d(x denote the signed distance in R n fom x to Γ with d > 0 in the e n -diection, and let = (X be the distance to Γ in R n+1 : Then, := (y 2 + d 2 1/2. x = d ν, xd = ν, and x d = κ whee κ = κ(x denotes the mean cuvatue and ν = ν(x epesents the unit nomal in R n of the paallel suface to Γ passing though x. Moeove, set ( s + d U a :=. (2.2 2 Obseve that one can expess U a as y 2s U a = 2 s ( d s, (2.3 and when Γ is flat, that is, γ 0, U a is equal to ( s z + d Ū a :=. (2.4 2 As shown in [8], Ūa is (up to multiplication by a constant the only positive a-hamonic function vanishing on x n 0, y = 0}. Thus, if ṽ is a global homogeneous solution of (1.2 with ϕ 0, then, up to a otation, Ūa = ν ṽ, whee ν is the unit nomal in R n to the fee bounday Γ. When Γ is not flat, the function U a is not a-hamonic. Howeve, it appoximates a-hamonic functions in the sense of the Schaude estimates of Poposition 4.1. We wok in the coodinate system detemined by x and. Fo a polynomial we let P = P (x, = p µm x µ m, P := max p µm }. Hee, µ is a multi-index, µ = µ µ n, µ i 0, and x µ = x µ1 1 xµn n. It is useful to think that the coefficients p µm ae defined fo all indices (µ, m by extending by zeo. Fequently, we use the convention of summation ove epeated indices. A function f : B 1 R is pointwise Cx, k,α at X 0 Γ if thee exists a (tangent polynomial P (x, of degee k such that f(x = P (x, + O( X X 0 k+α. We will wite f Cx, k,α (X 0, and f C k,α x, (X 0 will denote the smallest constant M > 0 fo which P M and f(x P (x, M X X 0 k+α. We will call objects univesal if they depend only on any o all of a, n, k, o α. Thoughout, unless othewise stated, C and c will denote positive univesal constants that may change fom line to line Basic Regulaity Results fo L a. Let us collect some egulaity esults fo weak solutions u H 1 (B λ, y a of the equation L a u = y a f, beginning with inteio egulaity. Thoughout the section, we assume that 0 < λ 1. As the weight y a is an A 2 -Muckenhoupt weight, we obtain the following local boundedness popety fo subsolutions L a fom [16]. Poposition 2.1 (Local Boundedness. Let u H 1 (B λ, y a be such that L a u y a f in B λ with f L (B λ. Then, ( ˆ 1/2 1 sup u C B λ/2 λ n+1+a u 2 y a dx + Cλ 2 f (Bλ. L (2.5 B λ

6 6 Y. JHAVERI AND R. NEUMAYER Poof. The inequality follows fom [16, Theoem 2.3.1] applied to u = c(λ 2 X 2 and u + c X 2 ; see [3, Lemma 3.4] fo details. 4 The opeato L a also enjoys a Hanack inequality [16, Lemma 2.3.5] and a bounday Hanack inequality [15, p. 585]. By a standad agument (see, fo instance, [19, Theoem 8.22], the Hanack inequality implies that solutions of L a u = y a f fo bounded f ae Hölde continuous. Poposition 2.2 (Hölde Continuity. Let u H 1 (B λ, y a be such that L a u = y a f in B λ with f L (B λ. Then, fo some α (0, 1. u C 0,α (B λ/2 Cλ α u L (B λ + Cλ 2 α f L (B λ If u is such that L a u = 0, then L a ( i u = 0 fo i = 1,..., n. Moeove, as pointed out in [9], the function y a y u satisfies L a ( y a y u = 0. If instead L a u = y a f fo f bounded, then one can show that L a ( i u = y a i f i = 1,..., n and L a ( y a y u = y f. Hee, the patial deivatives of f ae undestood in the distibutional sense. And so, we have the following egulaity esult fo x u and y a y u. Poposition 2.3 (Inteio Regulaity of x u and y a y u. Let u H 1 (B λ, y a be such that L a u = y a f in B λ with f L (B λ. Then, Futhemoe, if g := y a y f L (B λ, then x u L (B λ/4 Cλ 1 u L (B λ + C f L (B λ. y a y u L (B λ/4 Cλ a 1 u L (B λ + Cλ a f L (B λ + Cλ 2 g L (B λ. In fact, i u and y a y u ae Hölde continuous, but boundedness is all we need. Poof. By [16, Theoems and ], x u has the following local boundedness popety: ( ˆ 1/2 1 sup x u C B λ/4 λ n+1+a x u 2 y a dx + Cλ δ f L (Bλ/2 B λ/2 fo some δ > 0. Using the enegy inequality ˆ ˆ u 2 y a dx C f 2 y a dx + C ˆ B λ/2 B λ λ 2 u 2 y a dx (2.6 B λ fo u and ecalling λ 1, the fist estimate follows. Let w := y a y u and note that L a w = y a g. Since g L (B λ, (2.5 implies that ( ˆ 1/2 1 w L (B λ/4 C λ n+1 a w 2 y a dx + Cλ 2 g L (Bλ/2 B λ/2 ( ˆ 1/2 1 C λ n+1 a u 2 y a dx + Cλ 2 g L (Bλ/2. B λ/2 Applying (2.6 once moe concludes the poof. The following bounday egulaity esult is a consequence of Poposition 2.3 applied to odd eflections. Moe specifically, let B + λ := B λ y > 0} and u H 1 (B + λ, ya be such that L a u = y a f in B + λ (2.7 u = 0 on y = 0}, 4 We caution the eade that the authos in [3] define La with the opposite sign convention, that is, they conside L au(x, y := div( y a u(x, y.

7 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 7 and let ū and f be the odd extensions acoss y = 0} of u and f espectively. Then, notice that L a ū = y a f in B λ. Applying Poposition 2.3 to ū yields the following. Coollay 2.4 (Bounday Regulaity fo x u and y a y u. Suppose u H 1 (B + λ, ya satisfies (2.7 whee f C(B + λ. Then, x u L (B + λ/4 Cλ 1 u L (B + λ + C f L (B + λ. Futhemoe, if f vanishes on y = 0} and y a y f L (B + λ, then letting g := ya y f, y a y u L (B + λ/4 Cλa 1 u L (B + λ + Cλa f L (B + λ + Cλ2 g L (B + λ. If f does not vanish on y = 0} yet depends only on x, then we have the following. Coollay 2.5 (Bounday Regulaity of y a y u. Suppose u H 1 (B + λ, ya satisfies (2.7 whee f = f(x and f L (B + λ. Then, y a y u L (B + λ/4 Cλa 1 u L (B + λ + Cλa f L (B + λ. Poof. Letting w := y a y ū, whee again ū is the odd extension of u acoss y = 0}, we have Let M := f L (B + λ and conside the baies which satisfy L a w = 2fH n y = 0}. v ± := w ± M 1 + a y a+1, L a v ± = (2f ± 2MH n y = 0}. Theefoe, L a v + 0 and L a v 0. Applying Poposition 2.1 and aguing as in the poof of Poposition 2.3, we see that sup v + Cλ a 1 u L (B λ + Cλ a f L (B λ. B λ/4 The same bound holds fo sup Bλ/4 v. As v w v +, this concludes the poof. As a consequence of Coollaies 2.4 and 2.5, we have the following gowth estimate on x u when f = f(x is Lipschitz. Coollay 2.6 (Bounday Gowth of i u. Suppose u H 1 (B + λ, ya satisfies (2.7 whee f = f(x and f C 0,1 (B + λ. Then, fo any i 1,..., n, i u(x Cy 2s in B + λ/4. Hee, the constant C > 0 depends on a, n, λ, u L (B + λ, and f C 0,1 (B + λ. Poof. Fo any i = 1,..., n, we have that L a ( i u = y a h whee h := i f. Since f C 0,1 (B + λ, it follows that h L (B + λ. Applying Coollay 2.5 to iu implies that y ( i u(x Cy a, whee, using Poposition 2.3, we see that C depends on a, n, λ, u L (B +, f λ C 0,1 (B +. λ i u(x, 0 = 0, we detemine that ˆ y i u(x C t 2s 1 dt = Cy2s. 0 Since We have the same gowth estimate fo x u when f is less egula and unconstained to depend only on x so long as it vanishes on y = 0}.

8 8 Y. JHAVERI AND R. NEUMAYER Coollay 2.7 (Bounday Gowth of i u. Suppose u H 1 (B + λ, ya satisfies (2.7 whee f C(B + λ and f vanishes on y = 0}. If g := y a y f L (B + λ, then fo any i 1,..., n, i u(x Cy 2s in B + λ/4. Hee, the constant C > 0 depends on a, n, λ, u L (B + λ, f L (B + λ, and g L (B + λ. Poof. Let w := y a y ū and note that L a w = y a ḡ in B λ whee ḡ := y a y f L (B λ. Fom Poposition 2.3, we find that In othe wods, i w L (B + λ/4 Cλ 1 w L (B + λ + C ḡ L (B + λ i 1,..., n. y ( i u(x Cy a, whee we see fom Poposition 2.3 that C depends on a, n, λ, u L (B +, f λ L (B +, and g λ L (B +. λ Aguing as in the poof of Coollay 2.6 completes the poof. 3. A Highe Ode Bounday Hanack Estimate: Γ C k+2,α fo k 0 In this section, we pove a pointwise highe ode bounday Hanack estimate when Γ is at least C 2,α. This estimate, Poposition 3.2, and its C 1,α countepat, Poposition 5.4, will play key oles in poving highe egulaity of the fee bounday in (1.1, as sketched in the intoduction. We efe the eade to Section 7 fo the details of how exactly this is accomplished. Let U C(B 1 be even in y and nomalized so that U(e n /2 = 1. Suppose futhe that U 0 on P and U > 0 on B 1 \ P, and assume U satisfies ( L a U = y a Ua T (x, + G(X whee T (x, is a polynomial of degee k + 1 and T 1 and G(X s 1 X k+1+α. in B 1 \ P (3.1 In Poposition 4.1, we show that if Γ C k+2,α with Γ C k+2,α 1 and U is as above, then U takes the fom U = U a (P 0 + O( X k+1+α (3.2 fo some polynomial P 0 (x, of degee k + 1 with P 0 C and U a as defined in (2.2. Fomally, if we diffeentiate (3.2, we find that x U = U ( a sp 0 ν + x P 0 + ( P 0 dν + O( X k+1+α (3.3 and U = U ( a sp 0 + ( P 0 + x P 0 (dν + O( X k+1+α. (3.4 Rigoously justifying these expansions in ou application to the factional obstacle poblem will equie a delicate agument, which we pesent in Poposition 7.1. That said, in Poposition 3.2, we simply make the assumption that (3.3 and (3.4 hold. Remak 3.1. When T G 0, (3.3 and (3.4 follow by aguing as in Section 5 of [12] and the Appendix of [14], using the egulaity esults in Section 2. Poposition 3.2. Let Γ C k+2,α with Γ C k+2,α 1. Let U, T, G, and P 0 be as in (3.1, (3.3, and (3.4. Suppose that u C(B 1 is even in y with u L (B 1 1, vanishes on P, and satisfies ( L a u = y a Ua R(x, + F (X in B 1 \ P whee R(x, is a polynomial of degee k + 1 with R 1 and F (X s 1 X k+1+α.

9 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 9 Then, thee exists a polynomial P (x, of degee k + 2 with P C such that u U P C X k+2+α fo some constant C = C(a, n, k, α > 0. Poposition 3.2 is poved via a petubation agument that elies on following esult, whee we conside the specific case that T, G, R, F 0 and Γ is flat. Poposition 3.3. Suppose u C(B 1 is even in y with u L (B 1 1 and satisfies L a u = 0 in B 1 \ x n 0, y = 0} u = 0 on x n 0, y = 0}. Then, fo any k 0, thee exists a polynomial P (x, of degee k with P C such that Ū a P is a-hamonic in B 1 \ x n 0, y = 0} and fo some constant C = C(a, n, k > 0. (3.5 u Ūa P CŪa X k+1 (3.6 Recall that Ūa, defined in (2.4, is U a when Γ is flat. We postpone the poof of Poposition 3.3 until Section 6. In ode to poceed with poof of Poposition 3.2, we need to adapt the notion of appoximating polynomials fo u/u, intoduced in [14], to ou setting. Obseve that L a (Ux µ m = x µ m L a U + UL a (x µ m + 2 y a (x µ m U = y a (I + II + III whee, letting ī denote the n-tuple with a one in the ith position and zeos eveywhee else, I = U a xµ m T (x, + x µ m G(X, II = U (m(a + m dκxµ m 1 + µ i (µ i 1 m+1 x µ 2ī + 2dm m 1 (x µ ν, III = 2( m x U (x m + mx µ m 1 U. Up to a dilation, we can assume that (3.7 Γ C k+2,α ε, T, R ε, and G(X, F (X ε s 1 X k+1+α (3.8 fo any ε > 0. Fo ε > 0 sufficiently and univesally small, the constant tem of P 0 in (3.2 is nonzeo (see Remak 3.6 below. So, up to multiplication by a constant, (3.2 takes the fom U = U a (1 + εq 0 + εo( X k+1+α, (3.9 whee Q 0 (x, is a degee k + 1 polynomial with zeo constant tem and Q 0 1. Taylo expansions of ν, κ, and d yield ν i = δ in + + εo( X k+1+α, κ = κ(0 + + εo( X k+α, and d = x n + + εo( X k+2+α. Hence, using (3.3 and (3.4 to expand III, we find that I, II, and III become I = U a sµm σl xσ l + εo( s 1 X k+1+α, II = U ( m(a + m + 2µ n x µ m 1 + µ i (µ i 1 m+1 x µ 2ī + a µm σl xσ l + εo( X k+1+α, (3.10 III = U ( a 2s m µ n x µ n + 2smx µ m 1 + b µm σl xσ l + εo( X k+1+α. Hee, s µm σl, aµm σl, and bµm σl ae coefficients of monomials of degee at least µ + m and at most k + 1; that is, s µm σl, aµm σl, bµm σl 0 only if µ + m σ + l k + 1.

10 10 Y. JHAVERI AND R. NEUMAYER Futhemoe, since the monomials a µm σl xσ l and b µm σl xσ l come fom the Taylo expansions of ν, κ, and d (which vanish when Γ is flat, ecalling (3.8, we have that s µm σl, aµm σl, bµm σl Cε. Theefoe, fom (3.7, (3.9, and (3.10, we detemine that L a (Ux µ m = y a ( Ua ( mx µ m 1 (1 + m + 2µ n + 2sµ n m x µ n + µ i (µ i 1 m+1 x µ 2ī + c µm + εo( s 1 X k+1+α whee c µm σl 0 only if µ + m σ + l k + 1 and c µm σl Cε. Thus, given a polynomial P (x, = p µm x µ m of degee k + 2, whee σ + l k + 1, and L a (UP = y a ( Ua A σlx σ l + h(x σl xσ l h(x Cε P s 1 X k+1+α, (3.11 A σl = (l + 1(l σ n p σ,l+1 + 2s(σ n + 1p σ+ n,l + (σ i + 1(σ i + 2p σ+2ī,l 1 + c µm σl p µm. (3.12 Fom (3.12, we see that p σ,l+1 can be expessed in tems of A σl, a linea combination of p µm fo µ + m σ + l, and a linea combination of p µm fo µ + m σ + l and m l. Consequently, the coefficients p µm ae uniquely detemined by the linea system (3.12 given A σl and p µ0. Definition 3.4. Let u and U be as in Poposition 3.2. A polynomial P (x, of degee k + 2 is appoximating fo u/u if the coefficients A σl coincide with the coefficients of R. Befoe we pove Poposition 3.2, let us make two emaks and state a lemma. Remak 3.5. While U a is not a-hamonic in B 1 \ P, it is compaable in B 1 to a function V a that is a-hamonic in B 1 \ P. Indeed, using the uppe and lowe baies V ± := (1 ± /2U a, one can constuct such a function V a by Peon s method. Remak 3.6. Up to an initial dilation, taking Γ C k+2,α ε fo a univesally small ε > 0, the constant tem of P 0 in (3.2 is nonzeo. If U and U a wee a-hamonic in B 1 \ P, this would follow diectly fom the bounday Hanack estimate applied to U a /U without a dilation. By Remak 3.5, U a is compaable to the a-hamonic function V a. Fo ε > 0 sufficiently small (univesally so, we will find that U is also compaable to an a-hamonic function W ; hence, we can effectively apply the bounday Hanack estimate to U a /U passing though the quotient V a /W to conclude. Moe specifically, afte dilating, let us nomalize so that U(e n /2 = 1. Fist, let W satisfy L a W = 0 in B 1 \ P W = U on B 1 P. The stong maximum pinciple ensues that W is positive in B 1 \ P. Applying the bounday Hanack estimate to W and V a, Remak 3.5 implies that cu a W W (e n /2 CU a. Second, let V := U W, and obseve that L a V Cε y a s 1 in B 1 \ P V = 0 on B 1 P.

11 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 11 Lemma 3.7 then shows that V CεU a. So, 1 Cε W (e n /2 1 + Cε and fo ε > 0 small, cu a U CU a. Now, if the constant tem of P 0 wee zeo, then (3.2 would yield cu a U CU a X, which is impossible. In addition to its use in Remak 3.6 above, the following lemma will be used at seveal othe points. The poof follows by consideing the uppe and lowe baies v ± := ± C(U a U 1/s a. Lemma 3.7. Let v C(B 1 satisfy L a v y a s 1 in B 1 \ P v = 0 on B 1 P. Then, fo some constant C = C(a > 0. v CU a We ae now in a position to pove Poposition 3.2. Poof of Poposition 3.2. Fist, let ε > 0 in (3.8 be sufficiently small so that Remak 3.6 holds. Then, solving a system of linea equations as discussed above, we obtain an initial appoximating polynomial Q 0 (x, of degee k + 2 fo u/u. Up to multiplying u by a small constant and futhe deceasing ε > 0 (ecall that Q 0 Cε, we can assume that Q 0 1, u UQ 0 L (B 1 1, and L a [u UQ 0 ](X y a ε s 1 X k+1+α. Step 1: Thee exists 0 < ρ < 1, depending on a, n, k, and α, such that, up to futhe deceasing ε > 0, the following holds. If thee exists a polynomial Q of degee k + 2 that is appoximating fo u/u with Q 1 and u UQ L (B λ λ k+2+α+s, then thee exists a polynomial Q of degee k + 2 that is appoximating fo u/u with u UQ L (B ρλ (ρλ k+2+α+s and Q Q L (B λ Cλ k+2+α. fo some constant C = C(a, n, k, α > 0. and Fo any λ > 0, define the escalings P λ := 1 λ P, λ(x := (λx, U a,λ (X := U a(λx λ λ s, U λ (X := U(λX λ s, (3.13 [u UQ](λX ũ(x := λ k+2+α+s. Thus, ũ L (B 1 1, and by (3.11, we have that Let w be the unique solution to L a ũ(x Cε y a s 1 λ X k+1+α. (3.14 L a w = 0 in B 1 \ P λ w = 0 on P λ w = ũ on B 1. (3.15 Notice that w is even in y by the symmety of the domain and bounday data and w L (B 1 1 by the maximum pinciple. Since P λ has unifomly positive L a -capacity independently of ε and λ, w

12 12 Y. JHAVERI AND R. NEUMAYER is unifomly Hölde continuous in compact subsets of B 1. So, letting w be the solution of (3.15 in B 1 \ x n 0, y = 0}, by compactness, w is unifomly close to w if ε is sufficiently small (univesally so. Indeed, ecall that Γ x n = 0, y = 0} in C k+2,α as ε 0. Futhemoe, thanks to (3.9, we have that U λ Ūa unifomly as ε 0. Poposition 3.3 ensues that thee exists a polynomial P (X := p µm x µ z m of degee k + 2 such that P C, Ūa P is a-hamonic in the set B 1 \ x n 0, y = 0}, and w Ūa P L (B ρ Cρ k+3+s. Notice that the a-hamonicity of Ūa P implies that (l + 1(l σ n p σ,l+1 + 2s(σ n + 1 p σ+ n,l + (σ i + 1(σ i + 2 p σ+2ī,l 1 = 0 (σ, l. (3.16 Theefoe, choosing ρ and then ε sufficiently small depending on a, n, k, and α, we find that w U λ P L (B ρ w w L (B 1/2 + U λ P Ū a P L (B 1/2 + w Ūa P L (B ρ 1 4 ρk+2+α+s (3.17 whee P (X := p µm x µ λ m has the same coefficients as P. Now, set v := ũ w. Fom (3.14, we find that L a v ε y a s 1 λ in B 1 \ P λ v = 0 on B 1 P λ. Fom Lemma 3.7 and (3.9, we deduce that v CεU a,λ CεU λ. (3.18 Then, combining (3.18 and (3.17 and futhe deceasing ε depending on ρ, a, n, k, and α, we find that Rescaling implies that ũ U λ P L (B ρ v L (B ρ + w U λ P L (B ρ 1 2 ρk+2+α+s. u U Q L (B ρλ 1 2 (ρλk+2+α+s with Q(X := Q(X + λ k+2+α P (X/λ. To conclude, we must alte Q to make it an appoximating polynomial fo u/u by eplacing P (X/λ with anothe polynomial P (X/λ. As Q is aleady an appoximating polynomial fo u/u, we need the coefficients p µm of P to satisfy the system (l +1(l +2+2σ n p σ,l+1 +2s(σ n +1p σ+ n,l +(σ i +1(σ i +2p σ+2ī,l 1 + c µm σl p µm = 0 (σ, l (3.19 whee c µm σl := λ σ +l+1 µ m c µm σl. Notice that c µm σl cµm σl Cε. Futhemoe, subtacting (3.19 fom (3.16, we see that P P solves the system (3.19 with ight-hand side A σl = c µm σl p µm. Hence, A σl Cε, and choosing p µ0 = p µ0, we uniquely detemine P and find that P P Cε. Setting Q (X := Q(X + λ k+2+α P (X/λ concludes Step 1. Step 2: Iteation and Upgade.

13 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 13 Iteating Step 1, letting λ = ρ j fo j = 0, 1, 2,..., we find a limiting appoximating polynomial P such that P C and To upgade this inequality to we ague as in Step 1 in B 1 \ P λ. Setting we have that u UP L (B ρ j Cρ j(k+2+α+s j N. u UP CU X k+2+α, (3.20 ũ(x := [u UP ](λx λ k+2+α+s, ũ w + v CU a,λ CU λ in B 1/2. Indeed, that v and w ae contolled by U a,λ comes fom Lemma 3.7 and an application of the bounday Hanack estimate (cf. Remak 3.5, while the last inequality comes fom (3.9. Thus, afte escaling, we deduce that (3.20 holds since 0 < λ 1 was abitay. Keeping Remak 3.1 in mind, if U is a-hamonic, then (3.3 and (3.4 hold. So, a consequence of Poposition 3.2 and Poposition 5.4, its C 1,α analogue, is the following full genealization of [14, Theoem 2.3]. Theoem 3.8. Suppose 0 Γ := R np with Γ C k,α fo k 1 and Γ C k,α 1. If u and U ae even in x n+1, u L (B 1 1, L a u = y a Ua f in B 1 \ P u = 0 on P fo f Cx, k 1,α (Γ B 1 with f C k 1,α x, (Γ B 1, 1 and U > 0 is a-hamonic in B 1 \ P with U(ν(0/2 = 1, whee ν is the oute unit nomal to P, then fo some C = C(a, n, k, α > 0. u U C k,a x, (Γ B 1/2 C 4. Schaude Estimates: Γ C k+2,α fo k 0 In Poposition 3.2, we wee cucially able to appoximate U in tems of U a P 0, whee P 0 = P 0 (x, is a polynomial of degee k + 1. This appoximation, (3.2, is a consequence of the Schaude estimates of Poposition 4.1 below. These Schaude estimates oughly say if u satisfies L a u = y a U a f in B 1 \ P and u = 0 on P, then u gains egulaity in tems of the egulaity of f and Γ. Moe pecisely, we find that u/u a C k+1,α x, (0 if f Cx, k,α (0 and Γ C k+2,α. Poposition 4.1. Let Γ C k+2,α with Γ C k+2,α 1. Suppose u C(B 1 is even in y with u L (B 1 1, vanishes on P, and satisfies ( L a u(x = y a Ua R(x, + F (X in B 1 \ P whee R(x, is a polynomial of degee k with R 1 and F (X s 1 X k+α. Then, thee exists a polynomial P 0 (x, of degee k + 1 with P 0 C such that u U a P 0 CU a X k+1+α

14 14 Y. JHAVERI AND R. NEUMAYER and fo some constant C = C(a, n, k, α > 0. L a (u U a P 0 C y a s 1 X k+α in B 1 \ P To pove Poposition 4.1, we must fist extend the appopiate notion of appoximating polynomial to this setting. We compute that L a (U a x µ m = y a U ( a (dm + sκx µ m 1 + m(m + 1x µ m m 1 (dm + sν x x µ + µ i (µ i 1x µ 2ī m+1. Each of the functions ν i, κ, and d can be witten as the sum of a degee k polynomial in x and a C k,α function in x whose deivatives vanish up to ode k. The lowest degee tems in the Taylo expansion at zeo of ν i, κ, and d ae δ in, κ(0, and x n espectively. Hence, gouping tems by degee up to ode k and the emainde, we see that Hee, c µm σl L a (U a x µ m = y a U a 0 only if µ + m σ + l k. Also, ( m(m µ n x µ m 1 + 2sµ n x µ n m + µ i (µ i 1x µ 2ī m+1 + c µm σl xσ l + h µm (x, h µm (x, := m h µ m(x + m m 1 h µ m 1 (x, and h µ m, h µ m 1 Ck,α (B1 have vanishing deivatives up to ode k m and k (m 1 at zeo espectively. The coefficients c µm σl ae all linea combinations of the Taylo coefficients at the oigin of κ, dκ, ν i, and dν i, which vanish if Γ is flat. Afte a dilation making Γ C k+2,α ε, we may assume that c µm σl ε, hµ m C k,α (B 1 ε, and h µ m 1 C k,α (B 1 ε. Theefoe, if P (x, = p µm x µ m is a polynomial of degee k + 1, then L a (U a P = y a U ( a A σl x σ l + h(x, whee σ + l k, and A σl = (l + 1(l σ n p σ,l+1 + 2s(σ n + 1p σ+ n,l + (σ i + 1(σ i + 2p σ+2ī,l 1 + c µm σl p µm, (4.1 h(x, := k m h m (x m=0 fo h m C k,α (B 1 with vanishing deivatives up to ode k m at zeo. Assuming that Γ C k+2,α ε, we have h(x ε P X k+α. Consideing (4.1, we see that p σ,l+1 can be expessed in tems of A σl, a linea combination of p µm fo µ + m σ + l, and a linea combination of p µm fo µ + m σ + l and m l. Thus, the coefficients p µm ae uniquely detemined by the linea system (4.1 given A σl and p µ0. Definition 4.2. Let u be as in Poposition 4.1. We call a polynomial P (x, of degee k + 1 appoximating fo u/u a if the coefficients A σl coincide with the coefficients of R. Remak 4.3. Obseve that L a U a = y a U a sκ(x and L au = y a ( Ua T (x, + G(X..

15 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 15 Since Γ C k+2,α, the mean cuvatue κ = κ(x does not possess enough egulaity to yield the same ode eo as G afte being expanded. Indeed, letting g := κ T whee T is the kth ode Taylo polynomial of κ at the oigin, we see that U a g(x s 1 X k+α while G(X s 1 X k+1+α. This discepancy lies at the heat of the diffeence in appoximating u/u a and u/u. With the coect notion of appoximating polynomial in hand, the poof of Poposition 4.1 is now identical to that of Poposition 3.2 upon eplacing U with U a ; it is theefoe omitted. 5. The Low Regulaity Case: Γ C 1,α The goal of this section is to pove Poposition 5.4, which extends the highe ode bounday Hanack estimate of Poposition 3.2 to the case when Γ is only of class C 1,α. In this case, the functions and U a intoduced in Section 2.1 do not possess enough egulaity to diectly extend the poof of Poposition 4.1 o the notion of appoximating polynomial fo u/u in Definition 3.4. Following [14], this is ectified by woking with egulaizations of and U a, denoted by and U a, espectively. The following lemma contains estimates which will allow us to eplace and U a by thei egulaizations when needed. The constuction and the poofs of these estimates can be found in the Appendix. Lemma 5.1. Let Γ C 1,α 1. Thee exist smooth functions and U a, such that 1 C α, U a, 1 U a C α, C α, y y C y a U a s 1+α, U a, 1 U a C α, L 2(1 s y a a C y a α 1, and L a U a, C y a s 2+α whee C = C (a, n, α > 0. If Γ C 1,α ε, then each inequality holds with the ight-hand side multiplied by ε. The following pointwise Schaude estimate plays the ole of Poposition 4.1 in the case when Γ is C 1,α. Poposition 5.2. Let Γ C 1,α with Γ C 1,α 1. Suppose u C(B 1 is even in y with u L (B 1 1, vanishes on P, and satisfies L a u y a s 2+α in B 1 \ P. (5.1 Then, thee exists a constant p with p C such that fo some constant C = C(a, n, α > 0. u p U a CU a X α Note that even though Poposition 5.2 is stated just at the oigin, it holds unifomly at all points Γ B 1/2 since the assumption on the ight-hand side in (5.1 does not distinguish the oigin. The poof of Poposition 5.2 is quite simila to that of Popositions 3.2 and 4.1, but we include it to demonstate how U a, is used. In the poof, we will make use of the following lemma, whose poof via a baie agument is given in the Appendix. Lemma 5.3. Assume Γ C 1,α ε with α (0, 1 s, and suppose u satisfies L a u y a α 2+s in B 1 \ P u = 0 on B 1 P. If ε > 0 is sufficiently small, depending on a, n, and α, then fo some C = C(a, n, α > 0. u CU a

16 16 Y. JHAVERI AND R. NEUMAYER Poof of Poposition 5.2. Up to a dilation, we may assume that Γ C 1,α ε and L a u ε y a s 2+α fo any ε > 0, in paticula, fo ε small enough to apply Lemma 5.3. Step 1: Thee exists 0 < ρ < 1, depending on a, n, and α, such that, up to futhe deceasing ε > 0, the following holds. If thee exists a constant q such that q 1 and u qu a L (B λ λ α+s, then thee exists a constant q with q C such that and fo some constant C = C(a, n, α > 0. u Uq L (B ρλ (ρλ α+s q q Cλ α Define P λ, λ, and U a,λ as in (3.13, and conside the escaling ũ(x := [u qu a, ](λx 2C λ α+s. Note that ũ L (B 1 1 by Lemma 5.1. Let w be the unique solution of L a w = 0 in B 1 \ P w = 0 on P w = ũ on B 1. Obseve that w is even in y and w L (B 1 1. By compactness, w w locally unifomly as ε 0 whee w vanishes on x n 0, y = 0} and is such that L a w = 0 in B 1 \x n 0, y = 0}. Poposition 3.3 ensues the existence of a constant p with p C such that, choosing ρ and then ε sufficiently small, depending on a, n, and α, w pu a,λ L (B ρ w pūa L (B ρ + p(ūa U a,λ L (B ρ 1 8C ρ α+s. Since v := ũ w satisfies L a v ε y a s 2+α λ in B 1 v = 0 on B 1 P λ, Lemma 5.3 shows that v CεU a,λ. Then, up to futhe deceasing ε, we deduce that In tems of u, this implies that ũ pu a,λ L (B ρ 1 4C ρ α+s. u qu a, 2C λ a+s pu a L (B ρλ 1 2 (ρλα+s. Consequently, by Lemma 5.1, futhe deceasing ε if necessay, we find that whee q := q + 2C λ α+s p. Step 2: Iteation and Upgade. u q U a L (B ρλ (ρλ α+s and q q Cλ α+s Iteating Step 1, letting λ = ρ j, we find that thee exists a limiting constant p such that u p U a L (B ρ j Cρ j(α+s j N.

17 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 17 To conclude, we must upgade this inequality to Aguing as in Step 1, in B 1 \ P λ, with we have that u p U a CU a X α. ũ(x := [u p U a, ](λx λ α+s, ũ w + v CU a,λ in B 1/2. Indeed, the bound on v comes fom Lemma 5.3, and the bound on w comes fom an application of the bounday Hanack estimate (see Remak 3.5 and Lemma Thus, afte escaling, since 0 < λ 1 was abitay, we find that as desied. u p U a p U a p U a, + u p U a, CU a X α, We now poceed with the highe ode bounday Hanack estimate. Let U C(B 1 be even in y with U 0 on P and U > 0 on B 1 \ P, nomalized so that U(e n /2 = 1, and satisfy ( L a U = y a t U a + G(X in B 1 \ P (5.2 whee t is a constant with t 1 and G(X s 1 X α. (5.3 If Γ C 1,α with Γ C 1,α 1, then Poposition 5.2 implies that fo a constant p C. As befoe, fomally diffeentiating (5.4 yields and U = U a (p + O( X α (5.4 x U p x U a C U a X α (5.5 y U p y U a C y a s X α ; (5.6 the justification of these deivative estimates fo ou application is somewhat delicate and is given in Poposition 7.3. Again, in the simplest case, taking t G 0, these deivative estimates can be shown by aguing as in Section 5 of [12] and the Appendix of [14], using the egulaity esults in Section 2 (cf. Remak 3.1. Poposition 5.4. Let Γ C 1,α with Γ C 1,α 1. Let U, t, G, and p be as in (5.2, (5.3, (5.5, and (5.6. Suppose that u C(B 1 is even in y with u L (B 1 1, vanishes on P, and satisfies ( L a u = y a b U a + F (X in B 1 \ P whee b is a constant such that b 1 and F (X s 1 X α. Then, thee exists a polynomial P (x, of degee 1 with P C such that u U P C X 1+α fo some constant C = C(a, n, α > 0. 5 Hee, we use the uppe and lowe baies Ua, Ua, 1+α/s vanishes on P and is compaable to U a, and U a. in Peon s method to build an a-hamonic function that

18 18 Y. JHAVERI AND R. NEUMAYER To pove Poposition 5.4, we extend the notion of appoximating polynomial to this low egulaity setting by consideing polynomials in (x, athe than in (x, ; that is, P (x, = p 0 + p i x i + p n+1. Afte pefoming an initial dilation, as befoe, using Lemma 5.1, (5.5, and (5.6, one can show that with L a (UP = y a U a (tp 0 + 2sp n + 2p n+1 + h(x h(x ε P s 1 X α. Definition 5.5. Let u and U be as in Poposition 5.4. A polynomial P (x, of degee 1 is appoximating fo u/u if b = tp 0 + 2sp n + 2p n+1. With this definition of appoximating polynomial, the poof of Poposition 5.4 is identical to that of Poposition 3.2 and is theefoe omitted. 6. Poof of Poposition 3.3 In this section, we pove Poposition 3.3. That is, if u is a-hamonic in B 1 \ x n 0, y = 0} and vanishes continuously on x n 0, y = 0}, then the quotient u/u a is C x,(γ B 1/2. The petubative aguments of Sections 3 though 5 all ely on this coe egulaity esult. The idea of the poof is the following. The domain B 1 \ x n 0, y = 0} and the opeato L a ae tanslation invaiant in the e i diection fo any i = 1,... n 1, so diffeentiating the equation L a u = 0 shows that u is smooth in these diections. We can then educe the poof of Poposition 3.3 to the two-dimensional case, but with a ight-hand side. A final eduction (Lemma 6.7 leaves us with the main task of this section, which is poving Poposition 3.3 in the case n = 2 with zeo ight-hand side. This is Poposition 6.1 below. It will be convenient to fix the following additional notation. Fo x R n 1, we let D λ,x := (x, z R n+1 : z < λ}. We sometimes suppess the dependence on x and view D λ,x = D λ as a subset of R 2. Poposition 6.1. Let u C(B 1 be even in y with u L (D 1 1 and satisfy L a u = 0 in D 1 \ x 0, y = 0} u = 0 on x 0, y = 0}. Then, fo any k 0, thee exists a polynomial P (x, of degee k with P C such that U a P is a-hamonic in D 1 \ x 0, y = 0} and fo some constant C = C(a, k > 0. u U a P CU a z k+1 The geomety of ou domain D 1 \ x 0, y = 0} is simplified though the change of coodinates x(z 1, z 2 = z 2 1 z 2 2 and y(z 1, z 2 = 2z 1 z 2, which identifies the ight-half unit disk D 1 + := z R2 : z < 1, z 1 > 0} and D 1 \ x 0, y = 0}. 6 If we let ū denote u afte this change of coodinates, then ū is even in z 2 and La ū = 0 in D 1 + ū = 0 on z 1 = 0} whee the opeato L a (which is L a in these coodinates is given by L a u := 1 4 z 2 div( 2z 1z 2 a u. 6 This can be seen as the complex change of coodinates z z 2, i.e., ū(z = u(z 2. Abusing notation, we let z denote points in this new coodinate system: z = (z 1, z 2. Similaly, we set D λ := z R 2 : z < λ}.

19 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 19 The odd extension of ū satisfies the same equation in D 1. In this new coodinate system and afte an odd extension in z 1, the function U a becomes z 1 a z 1. Thus, Poposition 6.1 is equivalent to the following poposition. Poposition 6.2. Let u C(D 1 be odd in z 1 and even in z 2 with u L (D 1 1 and satisfy La u = 0 in D 1 u = 0 on z 1 = 0}. (6.1 Fo any k 0, thee exists a polynomial Q of degee 2k with Q C such that L a ( z 1 a z 1 Q = 0 in D 1 and u z 1 a z 1 Q C z 1 2s z 2k+2 fo some constant C = C(a, k > 0. If a = 0, then u is hamonic and Poposition 6.2 follows easily. Instead, when a 0, we pove the esult fom scatch in thee steps. Fist, we constuct a set homogeneous solutions of (6.1. Second, we show that these homogeneous solutions fom an othonomal basis fo L 2 ( D 1 with an appopiate weight. Thid, we expand u D1 in this basis, extend this expansion to the inteio of D 1, and compae u to the extension. Let ω a (z 1, z 2 := 2z 1 z 2 a, and obseve that ω a is an A 2 -Muckenhoupt weight. Remak 6.3 (Homogeneous Solutions. Fo evey j N 0}, define the function Hee, Qj (z 1, z 2 := b i z i 1z j i 2, ū j (z 1, z 2 := z 1 a z 1 Qj (z 2 1, z 2 2. (j i + 1(j i + 1 s b i := b i 1, i(i + s and b 0 = b 0 (j, a is chosen so that ū j L 2 ( D 1, ω a = 1. Each ū j is odd in z 1, even in z 2, vanishes on the z 2 -axis, and satisfies L a ū j = 0 in R 2. The two Geen s identities below will be used in what follows. The fist is applied to pove Lemma 6.5, an impotant estimate fo the poof of Poposition 6.2. The second is utilized in Lemma 6.6 to show that the homogeneous solutions of Remak 6.3 fom an othonomal basis fo L 2 ( D 1, ω a. Remak 6.4 (Geen s Identities. If u, v H 1 (D 1, ω a ae such that L a u = L a v = 0 in D 1, then u and v satisfy the following Geen s identities fo L a : ˆ ˆ v u ω a dz = u ν v ω a dσ λ < 1 (6.2 D λ D λ and ˆ ˆ u ν v ω a dσ D λ v ν u ω a dσ = 0 D λ λ < 1 (6.3 The following lemma shows that the (weighted L 2 -nom of u on the bounday of D 1 contols the (weighted L 2 -nom of u inside D 1. Lemma 6.5. Suppose u H 1 (D 1, ω a satisfies L a u = 0 in D 1. Then, thee exists a positive constant C, depending only on a, such that Poof. Let φ defined by u L 2 (D 1, ω a C u L 2 ( D 1, ω a. φ(λ, u := D λ u 2 ω a dσ,

20 20 Y. JHAVERI AND R. NEUMAYER whee the aveage is taken with espect to ω a. Using (6.2, we compute that φ is inceasing in λ. Hence, ˆ ˆ 1 ˆ ˆ u 2 ω a dz φ(1, u ω a dσ dλ = C u 2 ω a dσ, D 1 0 D λ D 1 as desied. Now, let us demonstate that the homogeneous solutions of Remak 6.3 ae an othonomal basis fo L 2 ( D 1, ω a. Lemma 6.6. Let ι a := ω a 1 L 1 ( D 1 and ū j be as in Remak 6.3. The set ι a, ū j : j = 0, 1,... } is an othonomal basis fo L 2 ( D 1, ω a. Poof. By constuction, ι a L2 ( D 1, ω a = ū j L2 ( D 1, ω a = 1. We show that ι a, ū j : j = 0, 1,... } is an othogonal, dense set in L 2 ( D 1, ω a. We fist teat the question of density. By symmety, letting A := D 1 z 1, z 2 0}, it suffices to show that spanι a, ū j : j = 0, 1,... } is dense in L 2 (A, ω a. Futhemoe, via the locally Lipschitz change of vaiables Φ : [0, 1] A given by Φ(t := (t, 1 t 2 with Jacobian J Φ (t = (1 t 2 1/2, this educes to showing that span1, w j : j = 0, 1,... } is dense in L 2 ([0, 1], µ a whee w j := ū j Φ and µ a := ( ω a ΦJ Φ. To this end, obseve that fo evey l N 0}, thee exist constants c j R fo j = 0, 1,..., l such that l t 2s+2l = c j w j (t. j=0 The Müntz-Szász theoem 7 implies that the family span1, t 2s+2l : l = 0, 1,... } is dense in C([0, 1]. Then, since C([0, 1] is dense in L 2 ([0, 1], µ a, the question of density is settled. We now show the set ι a, ū j : j = 0, 1,... } is othogonal in L 2 ( D 1, ω a. Fist, since the functions ū j ae odd in z 1 and ω a is even, the inne poduct of ū j and ι a in L 2 ( D 1, ω a is zeo fo evey j N 0}. Moeove, since ū j is homogeneous of degee 2s + 2j, one computes fo any z D 1, ν ū j (z = d dλ ū j (λz = (2s + 2jū j (z. λ=1 Theefoe, fo ū k and ū j, (6.3 becomes ˆ 0 = (2k 2j ū j ū k ω a dσ, D 1 as desied. We now pove Poposition 6.2 (and thus, Poposition 6.1 using Lemmas 6.5 and 6.6 and the local boundedness popety fo L a (cf. (2.5. Poof of Poposition 6.2. By Lemma 6.6, u D1 = c a ι a + c j ū j D1 as a function in L 2 ( D 1, ω a, and c a = 0 as u is odd in z 1. Fo evey k N 0}, let v k := k j=0 c jū j and note that L a v k = 0 in D 1. So, applying Lemma 6.5 and [16, Coollay 2.3.4] to w k+1 := u v k, lim u v k L k (D 1/2 C lim u v k L k 2 (D 1, ω a lim C u v k L k 2 ( D 1, ω a = 0. 7 The Müntz-Szász theoem (see [25] and [27] says that span1, t p i : p i > 0} fo i = 0, 1,... is dense in C([0, 1] povided that i=0 j=0 1 p i =.

21 HIGHER FREE BOUNDARY REGULARITY IN THE FRACTIONAL OBSTACLE PROBLEM 21 Theefoe, u = c j ū j in D 1/2. Since u L (D 1 1, it follows that sup j c j 1. Consequently, j=0 w k+1 u + v k C fo a constant C depending only on k and a. Fo any 0 < λ < 1, define the escaling w(z := w k+1(λz λ 2s+2k+2, and obseve that w C in D 1/2 thanks to the homogeneity of each tem in w k+1. Futhemoe, the functions w and z 1 a z 1 vanish on z 1 = 0} and satisfy L a w = L a ( z 1 a z 1 = 0 in D 1/2. Applying the bounday Hanack estimate in D 1/2 z 1 > 0} and ecalling that w and z 1 a z 1 ae odd in z 1, we find that w C z 1 2s in D 1/4. Rescaling and letting Q(z := k j=0 c j Q j (z 2 1, z 2 2, we deduce that concluding the poof. u z 1 a z 1 Q C z 1 2s z 2k+2, With Poposition 6.1 in hand, we use a petubative agument to pove the following, which in tun will allow us to conclude the poof of Poposition 3.3. Lemma 6.7. Suppose u C(B 1 is even in y with u L (D 1 1, vanishes on x 0, y = 0}, and satisfies ( L a u(z = y a Ua R(x, + F (z in D 1 \ x 0, y = 0} whee R(x, is a polynomial of degee k with R 1 and F (z s 1 z k+α. Then, thee exists a polynomial P (x, of degee k + 1 such that P C and fo some constant C = C(a, k, α > 0. u U a P CU a z k+1+α The poof of Lemma 6.7 follows the two step impovement of flatness and iteation pocedue given in the poof of Poposition 3.2, and so we omit it. 8 We now pove Poposition 3.3. Poof of Poposition 3.3. The equation (3.5 is invaiant with espect to x, so any patial deivative of u in an x -diection also satisfies (3.5. Futhemoe, the set x n 0, y = 0}, whee u vanishes, has unifomly positive L a -capacity. It follows that solutions of (3.5 ae unifomly Hölde continuous in compact subsets of B 1. In paticula, as u L (B 1 1, we see that fo any multi-index µ, D µ x u C 0,τ (B 1/2 C (6.4 fo some constant C = C(a, µ > 0; that is, u C x (B 1/2. With the egulaity of u in x undestood, we tun to undestanding the egulaity of u in z. To this end, notice that L a u = y a x u + div z ( y a z u. 8 Since Ua and ae homogeneous when Γ is flat and U a is a-hamonic away fom the set x n 0, y = 0}, the notion of appoximating polynomials and poof ae much simple.

22 22 Y. JHAVERI AND R. NEUMAYER So, fo any fixed x, u satisfies L a u = y a x u in D 1,x \ x n 0, y = 0} u = 0 on x n 0, y = 0} as a function of z. (Hee, L a is seen as a two-dimensional opeato. Let f := x u. By (6.4, f is C 0,τ in compact subsets of B 1. Up to multiplying by a constant, we may assume that f C 0,τ (B 1/2 1. Then, since f(x, 0 = 0 fo evey x, viewed just as a function in z, we see that f(z z τ s 1 z τ. In paticula, u satisfies the hypotheses of Lemma 6.7 with k = 0 taking R(x n, 0 and F (z = f(z. Applying Lemma 6.7, we find a degee 1 polynomial P 0 (x n, such that u U a P 0 CU a z 1+τ. As u and f have the same egulaity in z, it follows that f(z = U a Q 0 (x n, + U a O( z 1+τ fo a polynomial Q 0 (x n, of degee 1. Equivalently, f(z = U a R(x n, + F (z whee R(x n, = Q 0 (x n, is a degee 2 polynomial and, up to multiplication by a constant, R 1 and F (z s 1 z 2+τ. In this way, we can bootstap Lemma 6.7 with k = 2j to find polynomials P j (x n, of degee 2j + 1 such that u U a P j CU a z 2j+1+τ. In othe wods, fo each x, thee exists φ x C (D 1/2,x such that u(x, z = U a (zφ x (z. Since u is smooth in x and U a is independent of x, φ x (z = Φ(x, z is a smooth function of x as well. So, u(x, z = U a (zφ(x, z. This poves (3.6. Finally, that U a P is a-hamonic in B 1 \ P follows by induction: decompose P into its homogeneous pats P (x, = k m=0 P m(x,, whee each P m (x, is a homogeneous polynomial of degee m. If l m = 0, then L a (U a P = 0 in B 1 \ x n 0, y = 0}. Assuming U a m=0 P m(x, is a-hamonic in B 1 \ x n 0, y = 0} fo all l < k, we find that v := u U a l ( P m (x, = U a Pl+1 (x, + o( X l+1 (6.5 m=0 is a-hamonic in B 1 \ x n 0, y = 0}. Now, conside the escalings of v defined by ṽ(x := v(λx λ l+1+s. Obseve that L a ṽ = 0 in B 1 \ x n 0, y = 0} and that ṽ U a P l+1 as λ 0 by (6.5. Theefoe, L a (U a P l+1 = 0 in B 1 \ x n 0, y = 0}, which concludes the poof.