OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM. 1. Introduction

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM ARSHAK PETROSYAN AND TUNG TO Abstract. We study the regularity of solutions of the obstacle problem when the obstacle is smooth on each half of the unit ball but only Lipschitz across the shared boundary. We prove that the optimal regularity of these solutions is C 1, 1 up to the shared boundary on each half of the unit ball. The proof uses a modification of Almgren s frequency formula. 1. Introduction 1.1. The obstacle problem. Given functions ϕ : B 1 R and g : B 1 R satisfying g ϕ on B 1, consider the problem of minimizing the Dirichlet integral B 1 u on the closed convex set { u W 1, (B 1 ) u = g on B 1, u ϕ in B 1 }. This problem is called the obstacle problem with ϕ being the obstacle. It is well-known that the problem has a unique solution u which enjoys the following properties (see [5]): i) u is superharmonic. ii) u = 0 on the set {u > ϕ}. iii) u is locally as regular as ϕ up to C 1,1. In this work, we study the regularity of u when the obstacle ϕ is assumed to be smooth (C 1,1 ) up to the boundary on each half-ball B 1 ± (see Fig. 1 for illustration). One of the motivations for the study of piecewise-smooth obstacles comes from applications to math finance, where the time-dependent version of this problem appears in the valuation of so-called American options on multiple assets, see e.g. [3, 7]. Since ϕ is Lipschitz in B 1, the result mentioned above says that u is also Lipschitz in B 1 and that regularity is optimal unless extra condition is imposed on ϕ. However, since ϕ is smooth up to boundary on each B 1 ±, it is reasonable to expect that u enjoys better regularity when restricted to these subsets. Indeed, in this work, we will prove that u is C 1, 1 in B ± 1 B 1 (Theorem 6.7). It can be easily seen that C 1, 1 is the optimal regularity, similarly to the thin obstacle problem (see []): the function u(x) = Re(x 1 + i x n ) 3 000 Mathematics Subject Classification. Primary 35R35. Key words and phrases. free boundary, optimal regularity, obstacle problem, frequency formula, monotonicity formula. A. Petrosyan is supported in part by NSF grant DMS-0701015. 1

ARSHAK PETROSYAN AND TUNG TO u ϕ Figure 1. A solution of the obstacle problem with rooftop-like obstacle is a solution of the obstacle problem with the obstacle ϕ(x) = C x n for sufficiently large C. 1.. Notation. To proceed, we fix the notations that we are going to use throughout the paper. For each x = (x 1, x,..., x n ) R n, we denote For any r > 0 and x R n, define x = (x 1, x,..., x n 1 ). B r (x) = {y R n y x < r} B + r (x) = {y R n y x < r, y n > x n } B r (x) = {y R n y x < r, y n < x n } (x) = {y R n y x < r, y n = x n }. When x = 0, we normally omit it and write B r, B ± r and. Partial derivative in the direction τ is denoted by τ. Higher-order derivatives are written as τ1τ,...; ν denotes the outward normal derivative. Since we have to work with functions which are not differentiable in the classical sense on B 1, we introduce following notations, + e n v(x, 0) = e n v(x, 0) = lim x n 0 + e n v(x, x n ) lim e x n 0 n v(x, x n ) + ν v(x, 0) = + e n v(x, 0), ν v(x, 0) = e n v(x, 0). Here + ν and ν should be understood as outward normal derivative with respect to the sets {x n > 0} and {x n < 0} respectively.

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 3 We also use to denote ( e1, e,..., en 1 ). The frequently used set B 1 {u = ϕ} is denoted as Σ. 1.3. Assumptions. We assume that ϕ is C 1,1 up to boundary in B 1 ± and the boundary value g is bounded. This condition implies that u is C 0,1 locally in B 1 with the norm u C 0,1 (B ρ) depending only on n, ϕ C 1 (B 1), g L ( B 1) and ρ for any 0 < ρ < 1 (see [5]). Moreover, the optimal regularity theorem of [5] also implies that u is locally C 1,1 in B 1 ± (but not up to B 1). 1.4. Outline of the proof. - In Section, we show that u is almost convex in every tangential direction τ B 1, in the sense that ττ u is bounded below. The proof is in the spirit of []. - In Section 3, the almost-convexity is used to derive a C 1,α estimate for u. The proof follows ideas of [, 4]. This result is used in two places. First, the extra smoothness gained enables us to prove a monotonicity formula. Secondly, it gives a uniform C 1,α estimate for appropriately defined scalings in our study of blowups of u. For this second use that we have to prove the C 1,α estimate on slightly weaker assumptions about u and ϕ. Precise conditions are stated at the beginning of Section 3. - In Section 4, we obtain a monotonicity formula which, in a sense, is a truncated version of Almgren s frequency formula [1] (see Theorem 4.3). This type of monotonicity formulas has been first used in [6] and subsequently in [10] in the study of the thin obstacle problem. - In Section 5, we derive the existence of a constant β as a consequence of the monotonicity formula such that u grows as a C 1,β functions from the free-boundary Σ. - Finally, in Section 6, we show that the limit of appropriately chosen scalings of u solves a thin obstacle problem. Moreover, it is homogeneous of degree 1 + β. Since the optimal regularity for solutions of thin obstacle problem is known to be C 1, 1 (see []), we conclude that β 1. This in turn leads to the desired C 1, 1 regularity of u up to the boundary on B ± 1.. Almost convexity The first step in our proof is the almost convexity estimate on u in tangential directions τ B 1. We mainly follow the ideas from [, 4], suitably modified for our problem. Lemma.1. There exists a constant C depending on ϕ C 1,1 (B ± 1 B 1 ), g L ( B 1) and dimension n such that inf B 1 ττ u C for any unit vector τ B 1. Remark.. The inequality is understood in the sense of distributions. Namely, if ψ is any nonnegative function in W 1,1 0 (B 1 ), then ( τ u) ( τ ψ) C ψ. B 1 B 1 In fact, since u is locally C 1,1 in B 1 \ B 1, the inequality is also equivalent to saying that ττ u C a.e. in B 1.

4 ARSHAK PETROSYAN AND TUNG TO Proof. Without loss of generality we may assume that ϕ is defined in a slightly larger ball, say B 1+ɛ0. Then for 0 < ɛ < ɛ 0 define the mollifications of ϕ as follows. Fix a smooth radial function ζ on R n with support in B 1 such that ζ = 1. R n and let ϕ ɛ (x) = ζ(y)ϕ(x + ɛy) dy. B 1 Then the following properties are immediate: i) ϕ ɛ is smooth in B 1. ii) ϕ ɛ converges to ϕ uniformly on B 1 as ɛ goes to 0. iii) ϕ ɛ L (B 1) and ττ ϕ ɛ L (B 1) are bounded uniformly by respective norms of ϕ (here τ is any unit vector in B 1). Let u ɛ be the solution for the obstacle problem with the obstacle ϕ ɛ and boundary data g on B 1. Then it is standard that u ɛ u L (B 1) ϕ ɛ ϕ L (B 1) 0. Moreover, u ɛ C 0,1 (B 1 ) are uniformly bounded by ϕ C 0,1 (B 1) and g L ( B 1). Thus, it will be enough to prove the almost convexity estimate for u ɛ with a bound from below independent of ɛ. We denote by A ɛ the set {u ɛ = ϕ ɛ } and by d ɛ (x) the distance from x to A ɛ. consequence of a result in [5] that u ɛ C 1,1 and for almost every x B 1 (.1) ττ u ɛ (x) ττ ϕ ɛ (x) σ(d ɛ (x)) C σ(d ɛ (x)), We have as a where σ is a modulus of continuity depending on ϕ and ɛ. Let ζ be a nonnegative smooth cut-off function with support in B 3 such that ζ = 1 in B 1. Consider the function 4 M(x) = ζ(x) ττ u ɛ (x) λ u ɛ (x) where λ is some constant to be specified later. We will estimate an essential lower bound for M in B 1, independent of ɛ. This will imply our desired estimate, since u ɛ are equibounded on B 1 because of the equiboundedness of ϕ ɛ on B 1. Further, we only need to estimate M(x) in the set {u ɛ > ϕ ɛ }. Note that u ɛ is harmonic, hence smooth there. Since as a consequence of the estimate (.1), lim inf ττ u ɛ (x) C, x A ɛ x {u ɛ>ϕ ɛ} we only need to consider the case M(x) attains its minimum value at some point in the interior of {u ɛ > ϕ ɛ }. Call that minimum point x 0. We can also assume that ζ(x 0 ) 0, otherwise the lower bound for M(x 0 ) is trivial. One then has or We also have 0 = M = ζ( ζ) ττ u ɛ + ζ ττ u ɛ λ xi u ɛ xi u ɛ, ττ u ɛ = ζ ζ ττ u ɛ + λ ζ x i u ɛ xi u ɛ. 0 1 M = (ζ ζ + ζ ) ττ u ɛ + ζ ζ ττ u ɛ λ xix j u ɛ xix j u ɛ = (ζ ζ + ζ ) ττ u ɛ + ζ ζ ττ u ɛ λ D u ɛ.

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 5 Here, we have used that u ɛ = 0 in a neighborhood of x 0. or Combining the computations above, we obtain ( 0 ζ ζ 3 ζ ) ττ u ɛ + 4 λ ζ ( ζ x i u ɛ ) xi u ɛ λ D u ɛ, where C 0 and C 1 depend on ζ only. inequality simplifies to λ D u ɛ C1 λ ζ x i u ɛ xi u ɛ + C 0 ττ u ɛ, Since xi u ɛ D u ɛ and ττ u ɛ D u ɛ, the above λ D u ɛ C1 λ ζ x i u ɛ + C 0. Now, if D u ɛ < 1, then ττ u ɛ < 1 and we have an lower bound for M(x 0 ). Otherwise, choosing λ > C 0, we have ζ D u ɛ C1 u ɛ and so, M C 1 u ɛ λ u ɛ. Remark.3. Since u is superharmonic in B 1, Lemma.1 implies immediately that sup B 1 ene n u C. Again, this should be understood in the sense of distributions. Next, we derive some basic facts about u as consequences of this almost-convexity property. Lemma.4. Let C 0 be a constant such that in B 1, Then we have ττ u > C 0, τ B 1 ene n u < C 0. Proof. i) en u(x, t) en u(x, s) C 0 (t s) for any (x, t), (x, s) B 1 ii) lim ɛ 0 + en u(x, ɛ) and lim ɛ 0 + en u(x, ɛ) exist. iii) For any (x, s) B + 1, en u(x, s) + e n u(x, 0) C 0 s e n u(x, 0) en u(x, s) C 0 s. iv) ν + u(x) + ν u(x) 0 (or equivalently, e n u(x, 0) e + n u(x, 0)) on B 1. v) + ν u(x) + ν u(x) = 0 on B {u > ϕ}. with t > s and t, s 0. vi) τ u(x +sτ, x n ) τ u(x, x n ) > C 0 s for any (x +sτ, x n ), (x, x n ) B 1 with x n 0, s > 0, and τ B 1.

6 ARSHAK PETROSYAN AND TUNG TO i) Recall that the inequality ene n u < C 0 in B 1 one has with support in B 1 If the test function η has the form means that for any nonnegative smooth function η en u en η C 0 η. B 1 B 1 η(x, x n ) = G(x )F (x n ), where G C0 (B ɛ (x )) and F C0 (s ɛ, t + ɛ) for some ɛ small, then we have G(x )F (x n ) en u C 0 G(x )F (x n ). B 1 B 1 Since u C 0,1 (B 1) < and en u(x) is continuous in B 1 \ B 1, we can apply this inequality to a sequence of smooth functions G i (x ) converging to the Dirac s delta function concentrated at x, and then pass to the limit to obtain t s t F (x n ) en u(x, x n )dx n C 0 F (x n )dx n. Applying this inequality to a sequence of smooth functions F i converging to χ (s,t) in W 1,1 0 (s ɛ, t+ɛ) such that F i (0) = 0 (it is possible since s, t 0) and pass to the limit to obtain en u(x, t) en u(x, s) C 0 (t s). ii) From i) we have en u(x, x n ) Cx n is a decreasing function of x n in the set {x n 0}. The existence of those limits follows trivially. iii) Just apply the result of i) for t = x n and s = ɛ (or s = x n and t = ɛ) and let ɛ go to 0. iv) For any ɛ > 0, we have en u(x, ɛ) en u(x, ɛ) C 0 ɛ. s Passing ɛ to 0 we obtain the inequality. v) When u(x) > ϕ(x), u is harmonic in a neighborhood of x and hence smooth at x. vi) The proof is similar to i). 3. C 1,α estimate The next step in the proof is the C 1,α regularity of u up to the boundary in B 1 ±. We will mainly follows the ideas in []. Alternatively, one may use the approach in [1]. Later, in Section 6, we are going to apply the result of this section not just to u but also to some rescalings of u to obtain a uniform C 1,α bound. Since the respective rescaled obstacles are not necessarily uniformly bounded in C 1,1 (B 1 ± B 1), we need to relax our requirement on u and ϕ a little bit. More precisely, we will prove the C 1,α estimate using only the following conditions on u and ϕ that have been established in the previous section: there exists a positive constant C 0 such

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 7 that (3.1) (3.) (3.3) (3.4) (3.5) (3.6) (3.7) (3.8) (3.9) u 0 in B 1 u C 0 in B ± 1 ττ u > C 0 and ene n u < C 0 in B 1 D ϕ L (B ± 1 ) < C 0 u(x) ϕ on B 1 ν + u(x) + ν u(x) 0 on B 1 ν + u(x) + ν u(x) = 0 on B 1 {u > ϕ} u C 0,1 (B 1 ) C 0 ϕ B 1 C 0. It is very important that we do not assume any bound on en ϕ. The regularity exponent α and the norm C 1,α (B ± 1 ) will depend only on C 0 and the dimension n. In the rest of this section, unless stated otherwise, all constants C are understood to depend on C 0 and n only. We observe that if u satisfies these properties, then so does u (x, x n ) = u(x, x n ) + u(x, x n ). On the other hand, if u is C 1,α up to the boundary in B ± 1, then u(., 0) is C 1,α in B 1 agree on B 1. Together with the fact that 0 u C 0 in B ± 1 (see e.g. [11, Theorem 8.34]) implies a C 1,α estimate for u in B ± 1 since u and u, the boundary regularity theorem. Thus, for the purpose of this section, we may assume without loss of generality that u is an even function with respect to x n. A consequence of this assumption is that properties (3.6) and (3.7) above can be simplified to (3.6 ) (3.7 ) ν + u(x) 0 on B 1 ν + u(x) = 0 for all x on B 1 {u > ϕ}. As the first step, we will show that u is C 1,α at any point in the set Σ. To simplify the presentation, we will assume 0 Σ and prove that for some positive constants C and α. Since we can subtract the linear function u(x) u(0) x u(0) C x 1+α L(x, x) = ϕ(0) + x ϕ(0) from both ϕ and u without affecting the validity of (3.1) (3.9) (with possibly C 0 in the place of C 0 ), we can assume without loss of generality that We start with an auxiliary lemma. ϕ(0) = 0 and ϕ(0) = 0.

8 ARSHAK PETROSYAN AND TUNG TO Lemma 3.1. For any (x, x n ) B 1 and r > 0 such that x n + (r + x ) 1/4, there exists a half-ball in R n 1 of radius r with center at (x, x n ), denoted by HB r (x, x n ), such that for all y HB r (x, x n ), u(y, x n ) u(x, x n ) C 0 r Proof. We consider the case x n 0 first. If u(x, x n ) 0, define HB r (x, x n ) = { y R n 1 y x r, (y x ) u(x, x n ) 0 }. On the other hand, if u(x, x n ) = 0, let HB r (x, x n ) be any half-ball of radius r with center at (x, x n ). In either cases, the inequality u(y, x n ) u(x, x n ) C 0 r is an elementary consequence of the fact that τ u(x, x n ) 0 and ττ u C 0 where τ is the unit vector parallel to y x. When x n = 0, we choose a sequence of x i n 0 such that the sequence of symmetry axes of HB r (x, x i n) converges. Then, we can take as HB r (x, 0) the limit of HB r (x, x j n) as j. For each r > 0, define Γ + r = [0, r/n] Γ r = B r [ r/n, 0] Γ r = B r [ r/n, r/n]. Our next result says that a lower bound on en u is enough to imply a bound on u around 0. Lemma 3.. There exists a constant C such that for any r > 0 and δ > C 0 r, the condition inf en u δ Γ + r implies osc u Crδ. Γ + r/ Proof. From + ν u(x, 0) 0 and ene n u C 0, one has for all (x, x n ) Γ + r. Therefore, Consequently, for any fixed x B 1, Thus, it is enough to show that en u(x, x n ) C 0 x n < δ sup en u δ. Γ + r osc 0 x u(x, x n ) 1 n r/n n rδ. Crδ u(x, 0) Crδ

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 9 in / for some C. Due to the normalizing assumption ϕ(0, 0) = 0, ϕ(0, 0) = 0 and D ϕ < C 0, we have on, u(x, 0) ϕ(x, 0) C 0 r 1 rδ. Thus, we only need to prove the right hand side inequality u(x, 0) Crδ. Assume that u(x 0, 0) > Crδ for some x 0 B r/ and C > 0. We want to derive a contradiction when C is large. First, by Lemma 3.1, we have that for all y HB r (x 0, 0), and consequently u(y, 0) Crδ C 0 r > 1 Crδ u(y, x n ) u(y, 0) x n u L 1 Crδ x nδ 1 Crδ 1 n rδ 1 4 Crδ for all 0 x n r/n, as long as C > max {C 0, }. Next, denote by V the solution of the problem { x V (x ) = δ(x ) in B r V = 0 on B r. Applying Green s formula for V and u in B r(0, x n ), one has that for any x n (0, r/n) u(0, x n ) ( x u(x, x n ))V dx = u(x, x n ) ds. Since one has u(0, x n ) u(0) + 1 n rδ = 1 n rδ, u(x, x n ) u(x, 0) 1 n rδ C 0 r 1 n rδ, u(y, x n ) 1 4 Crδ, for all y HB r (x 0, 0), ( x u)v dx ( 1 n + 1 ) a(n)c rδ, where a(n) is a positive dimensional constant. Therefore, ( 1 1 n n + C ) r/n 0 a(n)c r δ ( x u) V dx dx n 0 B r r/n = ( u) V dx dx n 0 r/n 0 = b(n)r δ, r/n 0 ( ene n u) V dx dx n ( u, V 0) V [ en u(x, 0) en u(x, r/n)]dx V ( δ)dx ( ene n u) V dx dx n

10 ARSHAK PETROSYAN AND TUNG TO where b(n) is another positive dimensional constant. We can choose C large enough to obtain a contradiction here. That contradiction implies that for that value of C, we cannot have u(x, 0) > Crδ. The desired conclusion then follows. Next, we show a lower bound on en u. Lemma 3.3. There exist C > 0 and µ (0, 1) such that for any positive integer k, inf en u Cµ k. Γ + k Proof. We prove by induction. Clearly, since u < C 0 in B 1, given any µ (0, 1), we can find C such that the conclusion holds for k = 1. Assuming that it holds for k = k 0, we will prove that it holds for k = k 0 + 1. To simplify the notation, we also use r in the place of k0 in the rest of the proof. Let { x j} be a sequence in B {u > ϕ} such that x j 0. For each j, define the auxiliary function ( ) w j (x, x n ) = u(x, x n ) ϕ(x j, 0) (x x j) ϕ(x j, 0) C 0 x x j nx n. Consider the function w j in Γ + r/. Since w j = u + 4C 0 > 0 when x n > 0, w j must attain its maximum value on the boundary Γ + r/. On the other hand, at the point (x j, 0) we have w(x j, 0) = 0 en w(x j, 0) = 0, and therefore, by the Hopf boundary principle, 0 cannot be the maximum value for w j. Hence the maximum value of w j must be positive. Since ϕ(0) = 0 and D ϕ C0, we have ϕ(x, 0) C 0 x. Thus, on the set B r/ {u = ϕ}, we have w j (x, 0) = (x x j) ϕ(x j, 0) C 0 x x j < 0, while on / {u > ϕ}, + ν w j (x, 0) = + ν u(x, 0) = 0, hence w j cannot attain its maximum value on B r. In other words, w j must attain its positive maximum value on Γ + r/ \ B r/. Passing j to, we derive that the function ) w = u(x, x n ) C 0 ( x nx n must be nonnegative at some point on We consider two cases separately. Γ r/ \ / = B r/ [0, r/4n] / {r/4n}.

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 11 1) w is nonnegative at some point (x 0, r/4n) B r/ {r/4n}. Then, at that point For all y HB r/ (x 0, r/4n) we then have u(x 0, r/4n) C 0 ( x n(r/4n) ) C 0 8n r. (3.10) u(y, r/4n) C 0 8n r C 0 8 r. On the other hand, for any x, if en u(x, 0) = c < 0, then u(x, 0) = ϕ(x, 0) C 0 x C 0 8 r and so, together with ene n u < C 0, one obtains (3.11) u(x, r/n) C 0 8 r + c r n + C 0 (r/n). Combining (3.10) and (3.11), we have for any y HB r/ (x 0, r/4n), en u(y, 0) n ( C 0 r 8n r C 0 8 r C 0 8 r C ) 0 (r/n) = (n + n + )C 0 k0. 4n ) w is nonnegative at some point (x 0, x n ) B r/ [0, r/4n]. It is equivalent to u(x 0, x n ) C 0 ((r/) nx n) C 0 ((r/) n(r/4n) ) = C 0(4n 1) r 8n and consequently, for y HB r/ (x, x n ), As shown in (3.11), if en u(x, 0) < 0, then u(y, x n ) C 0(4n 1) r C 0 8n 8 r = C 0(3n 1) r. 8n u(x, x n ) C 0 8 r + C 0 (r/n) < C 0(3n 1) r. 8n Hence, we must have en u(y, 0) = 0 for all y HB r/ (x 0, x n ). In both cases, we have reached the conclusion that there exists a point (x 0, x n ) Γ + r/ such that for all y HB r/ (x 0, x n ), If C is chosen so that en u(y, 0) (n + n + )C 0 k0. 4n C > (n + n + )C 0, n then as long as µ > 1, we have for all y HB r/ (x 0, x n ), en u(y, 0) (n + n + )C 0 k0 > 1 4n Cµk0. Denote by G 1 (x, y) the Green s function for Γ + 1, then the Green s function for Γ+ r G r (x, y) = 1 ( x r n G 1 r, y ). r is given by

1 ARSHAK PETROSYAN AND TUNG TO Given any fixed x B r/ {r/4n}, we apply the Representation Formula to e n u en u(x) = en u(y) ν G r (x, y) dy G r (x, y) en u dy Γ + r Γ + r = en u(y) ν G r (x, y) dy + en G r (x, y) u dy Γ + r Γ + r θ(n)cµ k0 C 0 en G r (x, y) dy where Γ + r = θ(n)cµ k0 C 0 C (n)r, C (n) = max en G 1 (x, y) dy x B 1 {1/4n} Γ + 1 and θ(n) ( 1, 1) is a dimensional constant. Since en u(x, r/4n) en u(x, x n ) < C 0 (r/4n x n ), we have for all x / and x n (0, r/4n) en u(x, x n ) θ(n)cµ k0 C 0 C (n)r C 0 (r/4n) ( = Cµ k0 θ(n) + C 0C (n) C (µ) k0 + C 0 4nC (µ) k0 Fix µ to be some number between θ(n) and 1, then we could choose C large enough so that θ(n) + C 0C (n) C Since µ > 1, it immediately leads to in Γ + r/. + C 0 4nC < µ. en u(x) > Cµ k0+1 Lemma 3.4. There exist C and α such that for all x B 1, en u(x) C x α u(x) C x 1+α. ). Proof. The proof is elementary, following Lemmas 3. and 3.3. Remark 3.5. A consequence of this lemma is that u is differentiable at 0. Theorem 3.6. Suppose u, ϕ and C 0 satisfy (3.1) (3.9). Then there exist α (0, 1) and C > 0 depending only on C 0 and dimension n such that u C 1,α (B + 1 S B 1 ) < C. Proof. The proof is similar to the one in [], except for a small change. We may apply the result of Lemma 3.4 to any point in the set B 1 Σ by translating, scaling, and subtracting a linear function. Namely, at any z B 1 Σ, u is well-defined in the classical sense and for any x B 1 + with x z < 1 4, we can write that (3.1) u(x) u(z) (x z) u(z) C x z 1+α.

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 13 In order to show that u is C 1,α up to boundary in B + 1, by the boundary regularity theorem (see e.g. [11, Theorem 8.34]), it is enough to show that u(x, 0) is a C α function of x since u is bounded and C 0 u 0 in B + 1. Let y 1 and y be any two points in B 1. For i = 1,, let z i Σ and d i 0 be such that We also denote d = dist(y 1, y ). z i y i = d i = dist(y i, Σ). To show that u(y 1 ) u(y ) Cd α, we consider several possibilities: 1) If both y 1, y {u = ϕ}, then u(y 1 ) u(y ) = ϕ(y 1 ) ϕ(y ) Cd. ) If y 1 {u > ϕ} and y {u = ϕ}, then we argue as follows. Consider the auxiliary function w(x) = u(x) u(z 1 ) (x z 1 ) u(z 1 ). Then by (3.1) w Cd 1+α 1 in B d1 (y 1 ), since B d1 (y 1 ) B d1 (z 1 ). Further, from to the definition of d 1 we have that u > ϕ on the set B d1 (y 1 ) B 1. Consequently, we have C 0 w = u 0 on B d1 (y 1 ). Applying the result of [11, Theorem 3.9] to the function w in B d1 (y 1 ), one obtains u(y 1 ) u(z 1 ) = w(y 1 ) Cd α 1. Now, to finish this case, note that d > d 1 and estimate u(y 1 ) u(y ) u(y 1 ) u(z 1 ) + u(z 1 ) u(y ) u(y 1 ) u(z 1 ) + ϕ(z 1 ) ϕ(y ) Cd α 1 + C(d + d 1 ) Cd α. 3) If y 1, y {u > ϕ}, we consider two subcases. a) If 4d > max {d 1, d }, then similarly to case ) we have the estimates u(y i ) u(z i ) Cd α i and therefore u(y 1 ) u(y ) u(y 1 ) u(z 1 ) + u(y ) u(z ) + u(z 1 ) u(z ) Cd α 1 + Cd α + C(d 1 + d + d) Cd α. b) If 4d d 1 = max {d 1, d }, then the auxiliary function satisfies w(x) = u(x) u(z 1 ) (x z 1 ) u(z 1 ) C 0 w = u 0 in B d1 (y 1 ).

14 ARSHAK PETROSYAN AND TUNG TO Applying [11, Theorem 3.9] to w in B d1 (y 1 ), we have d 1 w(y 1 ) w(y ) d C(n)(sup w + d 1 sup w ) log d 1 d Cd 1+α 1 log d 1 d u(y 1 ) u(y ) Cdd1 α 1 log d 1 d ( ) α 1 Cd α d1 log d 1 d d Cd α. In the last step, we have used the fact that the function t α 1 log t is bounded in [4, ] for all α > 0. Remark 3.7. From now on, α will be reserved for the C 1,α regularity exponent in Theorem 3.6. We also denote by C α a constant such that u C 1,α (B + 1 B 1 ) C α. 4. Monotonicity Formula We are going back to our original problem as stated in Introduction. Our aim in the rest of this work is to show that u is C 1, 1 up to the boundary in B ± 1. We will need the full strength of the definition that u is the solution to an obstacle problem. In particular, we need the fact that u = 0 in {u > ϕ}. Substituting it by the conditions that u is superharmonic and u C in B ± 1, for example, is not enough for our proof to work. Therefore, while conditions (3.1) (3.9) are enough to guarantee a C 1,α estimate, they are not enough for our C 1, 1 result. In this section we prove a monotonicity formula that will be our main tool in establishing the optimal C 1, 1 regularity of u. This type of formulas goes back to Almgren [1] in the study of multivalued harmonic functions and to Garofalo-Lin [8, 9] in the context of unique continuation. Recently, this formula has been adapted to the thin obstacle problem in [6, 10] by the means of the truncation of the growth rate of the function. As demonstrated in the proof of Theorem 3.6, in order to show that u is C 1, 1 in B + 1, it is enough to assume that 0 Σ and to show that u(x) u(0) x u(0) C x 3 x B 1. Furthermore, since subtracting a linear function from both u and ϕ does not effect the hypotheses or conclusions of the problem, we can assume that (4.1) u(0) = 0 and u(0) = 0. Note that this assumption implies that ϕ(0) = 0 and ϕ(0) = 0 (but not en ϕ(0) = 0).

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 15 We start by defining the operators involved in the monotonicity formula. For each function v in B 1 and r (0, 1) let D(r, v) = v B r H(r, v) = v B r G(r, v) = v B r I(r, v) = vv ν = v + v v v ( ν + v ) B r B r B r + Br B r rd(r, v) N(r, v) = H(r, v), whenever those quantities are well defined. Then we have the following differentiation formulas. Lemma 4.1. For any r (0, 1) we have D (r, u) = n D(r, u) + u ν r B r r H (r, u) = n 1 H(r, u) + I(r, u) r G (r, u) = H(r, u) I (r, u) = n I(r, u) + u ν r B r r n (n ) u u r r B + r B r B + r B r B + r B r (x u) u 4 r (x u) + ν u (x u) u 4 (x u) ν + u r B r u ν + u + u u u ν + u B r Proof. All computations are standard (see for example [8, 9]). The only difference here is that we have used + ν u in place of + ν u + ν u on. Recall that u (x, x n ) = u(x, x n ) + u(x, x n ). Using the C 1,α result for u, we obtain some estimates for H(r, u) and G(r, u) in terms of D(r, u). Lemma 4.. There exists a positive constant C depending only on C 0 and n such that for all r (0, 1) H(r, u) C(rD(r, u) + r n+1+α ) G(r, u) C(r D(r, u) + r n++α ). Recall that α denotes the regularity exponent in C 1,α result of Theorem 3.6 and C 0 is the constant defined in (3.1) (3.9). Proof. For each r (0, 1) define ū(r) = B r u.

16 ARSHAK PETROSYAN AND TUNG TO Since u is superharmonic, one has ū(r) u(0) = 0. On the other hand, from the C 1,α estimate we also have that ū(r) C α r 1+α. Applying Poincaré s inequality to u on B r, we obtain (u ū(r)) a(n)r u B r B r u ū(r) u + ū(r) a(n)r u B r B r B r u + (1 b(n)r n 1 )ū(r) a(n)r u B r B r u a(n)r u + b(n)cα r n+1+α, B r B r where a(n) is a dimensional constant. This gives us the first inequality. Integrating it, we obtain the second inequality. Finally, we can prove the monotonicity formula. Lemma 4.3. There exist r 0 > 0 and C m > 0 depending only on C 0 and n such that Φ(r, u) = re Cmr α d dr log max { H(r, u), r n++ } α is increasing for r (0, r 0 ). Proof. Note that Φ is absolutely continuous (or equivalently W 1,1 ) in (0, 1) and therefore it is enough to show that Φ(r, u) = re Cmr α H ( ) (r, u) H(r, u) = α recr n 1 I(r, u) + r H(r, u) is nondecreasing in each component of the set U = { r (0, r 0 ) H(r, u) > r n++ α }. We will prove the monotonicity of re Cmr α I(r, u) H(r, u), since (n 1)e Cmr α r log which is further reduced to ( E(r) = 1 I(r, u) r is clearly increasing. We do it by showing that ( ) re Cmr α I(r, u) H(r, u) B + r B r n r Cr 1+ α = 1 r + α C mr 1+ α + I (r, u) I(r, u) H (r, u) H(r, u) 0 (x u) u (x u) ν + u n r B r r ) u ν + u + u u u ν + u B r B + r B r on the set U for some positive constant C. This reduction is pretty standard and can be seen for example in [6, 8, 10]. u u

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 17 Using Hölder s inequality and the estimates in Lemma 4., we have ( ) 1 ( ) 1 (x u) u x u u B r + Br B r + Br + Br Br CD(r, u) 1 n r +1 ( ) 1 ( ) 1 u u u u B r + Br B r + Br + Br Br CG(r, u) 1 n r CD(r, u) 1 n r +1 + Cr n+1+α ( ) 1 ( ) 1 u u u u B r B r Br CH(r, u) 1 n 1 r CD(r, u) 1 r n + Cr n+α. Next, we estimate terms involving + ν u. First, one has + ν u = 0 on {u > ϕ}. On the other hand, on {u = ϕ} one has that, + ν u Cr α u = ϕ C 0 r u = ϕ C 0 r, as consequences of the C 1,α estimate for u. Combining all these estimates, we obtain D(r, u) 1 r n + r n+α E(r) < C D(r, u) CD(r, u) 1 r n +1 Cr. n+1+α When H(r, u) > r n++ α, the first inequality in Lemma 4. gives Therefore, for some r 0 > 0 and C > 0 one has on the set D(r, u) > Cr n+1+ α Cr n+. E(r) C D(r, u) 1 r n + r n+α Cr 1+ α D(r, u) { r (0, r0 ) H(r, u) > r n++ } α. As noted before, this completes the proof of the lemma. 5. C 1,β Regularity Due to the monotonicity of Φ(r, u) for r (0, r 0 ), under the normalization conditions (4.1), there must exist some β such that lim Φ(r, u) = n + 1 + β. r 0 + Our objective in this section is to show that u grows as x 1+β near the origin. In other words, u(x) C x 1+β.

18 ARSHAK PETROSYAN AND TUNG TO In fact, we are going to see in the next section that β 1, and combined with this growth estimate it will ultimately imply the C 1, 1 regularity. We start with an observation that the truncation of the growth of H(r, u) in the formula for Φ(r, u) provides a natural bound for β from above. Lemma 5.1. We have the inequality Moreover, if β 1 + α 4. β < 1 + α 4 then there exists r u > 0 such that H(r, u) r n++ α for 0 < r < r u. Proof. If there exists a sequence r j 0 such that H(r j, u) < r n++ α j, then α Φ(r j, u) = e Cr ( j n + + α ) and consequently, β = 1 + α 4. Therefore, we only need to consider the possibility that for r close to 0. This assumption implies and thus, H(r, u) r n++ α Φ(r, u) = re Cr α H (r, u) H(r, u), lim r H (r, u) r 0 + H(r, u) = n + 1 + β. Given any ɛ > 0, there must exist some r ɛ > 0 such that r H (r, u) H(r, u) > n + 1 + β ɛ for all r (0, r ɛ ). Following a straightforward computation we can derive that Combining that with the assumption that H(r, u) H(r ɛ) rɛ n+1+β ɛ H(r, u) r n++ α r n+1+β ɛ. when r is close to 0 we have n + + α n + 1 + β ɛ. Since this inequality holds for any ɛ > 0, passing to the limit one obtains β 1 + α 4.

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 19 Lemma 5.. There exists C depending only on C 0 and n such that for all r (0, 1 ), H(r, u) = u Cr n+1+β B r G(r, u) = u Cr n++β B r D(r, u) = u Cr n+β B r Proof. We start with the estimate on H(r, u). First, we can choose C large enough depending on u L (B 1) and r 0 such that H(r, u) Cr n+1+β for all r r 0. Here r 0 is the constant from Lemma 4.3. For all r in the set { r H(r, u) r n+1+ α }, the estimate for H(r, u) is trivially satisfied since n + 1 + β n + + α. Any other value of r must belong to some maximal open interval (r 1, r ) { r (0, r 0 ) H(r, u) > r n++ α }. Note that because either r = r 0 or H(r, u) = r n++ α, one always has We have for any r (r 1, r ), and so where c is a constant such that H(r, u) Cr n+1+β. Φ(r, u) = re Cmr α H (r, u) H(r, u) lim Φ(r, u) = n + 1 + β r 0 + H (r, u) H(r, u) (n + 1 + β) 1 re Cmr α α (n + 1 + β) ecmr c α C mr α (n + 1 + β) e x 1 + c α x re Cmr α ( 1 r cα C mr 1+ α e C mr x (0, C m). Integrating both sides from r to r one obtains log H(r (, u) (n + 1 + β) log r ) α H(r, u) r + ce Cmr ce Cmr α ( (n + 1 + β) log r ) r c or H(r, u) H(r, u) r n+1+β e c(n+1+β) r n+1+β Ce c(n+1+β) r n+1+β. α )

0 ARSHAK PETROSYAN AND TUNG TO This proves the estimate for H(r, u). Integrating, one obtains the estimate for G(r, u). To prove the estimate for D(r, u), fix a smooth radial cut-off function η such that 0 η 1 in R n For each r > 0, denote by η r the scaled function η = 1 in B 1 η = 0 in R n \ B 3 4. η r (x) = η ( x r ). Applying Green s formula to η r and u one has η r u u η r = or equivalently, B + r Doing the same for Br B + r B r B + r η r u u + B r + B + r η r u 1 B + r η r + ν u, u η r = and summing up, one obtains η r u u + η r u 1 u η r = B r B r η r u ( + ν u). η r u ( + ν u ). Rearranging and applying Cauchy-Schwartz inequality to the first term, we have η r u 1 u η r + η r u ( ν + u ) + 1 B r B r r Br ηu + r Using assumptions on η and the fact that B + r B r η( u). η r = 1 r η, we can derive u η + 1 L B r r Br u + r Cr n+β + Cr n+ + Cr n+1+α Cr n+β. Here we have used the estimate proved in Lemma 4.3. B + r B r u( + ν u ) Cr n+1+α ( u) + u ( ν + u ) B r Lemma 5.3. There exists C depending only on C 0 and n such that for any r > 0, in B r. u(x) Cr 1+β

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 1 Proof. First, since u is superharmonic, u is subharmonic. Together with the integral estimate one has In other words, G(r, u) Cr n++β u Cr 1+β. u Cr 1+β. To obtain the estimate from above, we cannot use that u + is subharmonic, since it is not generally true. However, we claim that the function v = max { C 0 r, u } + C 0 x is subharmonic in B r. Away from {B 1 Σ}, u C 0 and so u + C 0 x is subharmonic. Therefore, we only need to study the behavior of v near B r Σ. If x B r Σ, then u(x) = ϕ(x) < C 0 r (ϕ(0) = 0, ϕ(0) = 0, ττ ϕ < C 0 ) and so, v = C 0 r + C 0 x in a neighborhood of x. Thus, v is subharmonic in B r. On the other hand, we easily have from the definition of v that v u + + Cr n+ B r B r Therefore, and so, C(r n G(r, u)) 1 + Cr n+ Cr n+1+β. v Cr 1+β u Cr 1+β. 6. Blowups and C 1, 1 regularity To complete the proof of C 1, 1 regularity, we will show in this section that β 1. We will always assume that since otherwise, there is nothing else to do. β < 1 + α 4 Under the normalizing conditions (4.1), for each r > 0, define the rescaling factor ( ) 1 1 d r = H(r, u) rn 1 and rescaled functions u r (x) = u(rx) d r ϕ r (x) = ϕ(rx) d r. The inequality β 1 will follow from the study of the limits of these rescalings over sequences r = r j 0 +, which are called blowups. In fact, we will show that such blowups solve a thin obstacle

ARSHAK PETROSYAN AND TUNG TO problem and are homogeneous of degree 1+β. From the known C 1, 1 regularity for the thin obstacle problem, we will immediately obtain β 1. First, in order to show that such limits exist, we start with a uniform C 1,λ estimate on u r as an application of Theorem 3.6. To apply that theorem, we need to verify the existence of a constant C, independent of r such that the triple (u r, ϕ r, C) satisfies (3.1) (3.9). The only non-trivial condition to check is (3.8). We start with relatively simple estimates for second derivatives of u r and ϕ r. Lemma 6.1. Define γ = 1 α 4. There exists some positive constant r u (possibly depending on u) such that for 0 < r < r u, we have i) u r is superharmonic in B 1. ii) u r C 0 r γ in B ± 1. iii) ττ u r C 0 r γ and ene n u r C 0 r γ in B 1. iv) D ϕ r C 0 r γ in B ± 1 B 1. Proof. Since we must have when r is small. Thus, β < 1 + α 4 H(r, u) r n++ α d r r 3 + α 4 or r r γ d r for r small. All inequalities are consequences of this fact and the corresponding inequalities for u and ϕ. Lemma 6.. For all 0 < r < 1 i) u r ϕ r on B 1. we have ii) + ν u r (x) + ν u r (x) 0 on B 1. iii) + ν u r (x) + ν u r (x) = 0 on B 1 {u r > ϕ r }. iv) ϕ r C 0 r γ on B 1. Proof. All are trivial consequences of the corresponding properties of u and ϕ. In the last one, we have used ϕ r (0) = 0 and D ϕ r C 0 r γ in B + 1 B 1. In order to estimate u r, we first show a uniform W 1, estimate for u r. Lemma 6.3. There exists some r u > 0 such that for all 0 < r < r u, i) G(r, u) rh(r, u) for r small.

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 3 Proof. i) Since we have ii) lim r 0 + N(r, u) = 1 + β. iii) G(1, u r ) 1 for r small. iv) lim r 0 + D(1, u r ) = 1 + β. β < 1 + α 4 Φ(r, u) = r H (r, u) H(r, u) for r small (say r < r u for some r u > 0). The definition of β therefore becomes lim r H (r, u) r 0 + H(r, u) = n + 1 + β. In particular, H (r, u) > 0 or H(r, u) is an increasing function. Consequently, G(r, u) = r 0 H(s, u) ds rh(r, u). ii) Plugging the formula (Lemma 4.1), H (r, u) = n 1 H(r, u) + I(r, u) r ( = n 1 H(r, u) + u + r B r + into the formula for Φ(r, u) above one has D(r, u) + B r lim r + B u u u ( r B r ν + u ) r 0 + H(r, u) B + r ) u u u ν + u B r = 1 + β. We will show that lim r B r + B u u u ( r B r ν + u ) = 0. r 0 + H(r, u) First, in order to estimate the numerator, we use the inequalities in the proof of Lemma 4.3 and the results of Lemma 5. to obtain u u CG(r, u) 1 n r B + r B r Cr n++β r n = Cr n+1+β and On the other hand, from u ( ν + u ) Crn+1+α. lim r 0 + r H (r, u) H(r, u) = n + 1 + β,

4 ARSHAK PETROSYAN AND TUNG TO we can deduce by using an argument similar to the one in the proof of Lemma 4.3 that for any ɛ > 0, there exist C > 0 and r ɛ > 0 such that Combining all these estimates, one then has Therefore, H(r, u) Cr n+1+β+ɛ, 0 < r < r ɛ. lim r r 0 + lim r 0 B + r B r u u u ( + ν u ) H(r, u) = 0. D(r, u) N(r, u) = lim r + r 0 + H(r, u) = 1 + β. iii) A restatement of i), since G(1, u r ) = G(r, u)/(rh(r, u)). iv) A restatement of ii), since D(1, u r ) = N(r, u). Lemma 6.4. There exists r u such that u r C 0,1 (B 1) are uniformly bounded for 0 < r < r u. Proof. First, by using the same argument as in the proof of Lemma 5.3, the bounds for second derivatives of u r in Lemma 6.1 and W 1, estimates in Lemma 6.3, one obtains in B 3 4 where C can be made independent of r. u r C 1 In the next step, we bound en u r from below in B + 1 Since ene n u r C 0, we have by using the estimate for u r and ene n u r. Therefore, u r (x, x n + 1 4 ) u r(x, x n ) 1 4 ( e n u r (x, x n )) + 1 3 C 0. C 1 1 4 ( e n u r (x, x n )) + 1 3 C 0 which implies en u r C in B + 1. On the other hand, the same argument for the negative side B 1 yields an upper bound en u r (x) < C there. For the other direction of the inequality, recall that + ν u r + ν u r 0 or e n u r + e n u r on B 1. From the upper bound for en u r in B 1, we immediately have Together with ene n u r C 0, this leads to + e n u r (x, 0) C. en u r (x) C + 1 C 0. Combining together the lower and upper bounds, we can write in B 1, where C is independent of r. en u r C As the last step, we estimate u r. Since ϕ r (0) = 0 and D ϕ r C 0, we have ϕ r C 0

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 5 in B 1. Fix any τ B 1. Consider the function v = max {C 0, τ u r }. We want to show that it is subharmonic. If x { τ u r > C 0 }, then It implies that Therefore or consequently, τ u r > τ ϕ r. u r (x) ϕ r (x). u r (x) = 0 ( τ u r ) = 0 in a neighborhood of x. It is enough to say that v is a subharmonic function in B 1. On the other hand, since lim D(1, u r ) = lim u r = 1 + β, r 0 + r 0 B 1 D(1, u r ) is uniformly bounded when r is small. Together, they imply that τ u r C 3 in B 1 for some C 3 independent of r and τ. Since τ is an arbitrary unit vector in B 1, we obtain u r C 3. We have demonstrated a uniform C 1,λ estimate on u r. It guarantees the existence of at least one blowup. In the next step, we study the properties of this blowup. We want to show that it solves a thin obstacle problem and is also a homogeneous function. Lemma 6.5. Assume that β < 1 + α 4. Let {r j} be any sequence converging to 0. If there exists a function u 0 on B 1 such that u rj u 0 in C 1,λ loc (B± 1 B 1), then u 0 possesses the following properties: Proof. i) u 0 is not identically 0 in B 1. ii) u 0 = 0 in B ± 1. iii) u 0 0, + ν u 0 + ν u 0 0, u 0 ( + ν u 0 + ν u 0 ) = 0 on B 1. iv) u 0 is homogeneous of degree 1 + β. i) We can derive directly from the definition of u r that u r j = 1 B 1 for any j. Since we have a uniform W 1, bound on u r in B 1, see Lemma 6.3, we may assume that u rj u 0 strongly in L ( B 1 ) and passing to the limit we obtain u 0 = 1. B 1 Therefore u 0 cannot be identically 0 in B 1.

6 ARSHAK PETROSYAN AND TUNG TO ii) From Lemma 6.5 we have in B 1 ± 0 u r C 0 r γ for some γ > 0. Passing r to 0, we immediately obtain u 0 = 0 in B 1 ±. iii) The facts that u 0 0 and + ν u + ν u 0 on B 1 are elementary consequences of inequalities It is then enough for us to show that u(x, 0) ϕ(x, 0) C 0 x + ν u(x, 0) + ν u(x, 0) 0. + ν u 0 (x) + ν u 0 (x) = 0 when x B 1 {u 0 > 0}. For any such x, there must exist some c > 0 and K such that for all j K, u rj (x) > c which is equivalent to u(r j x) > cr n 1 j H(r j, u) 1 > cr 3 + α 4 j. By choosing a bigger K if necessary, one has for all j K, Therefore, Passing j to, one has u(r j x) > cr 3 + α 4 j > C 0 r j C 0 r j x ϕ(r j x). + ν u(r j x) + ν u(r j x) = 0. + ν u 0 (x) + ν u 0 (x) = 0. iv) From the definition of N(r, v) and u 0 we observe that N(r, u 0 ) = lim rj 0 N(r, u r j ) = lim rj 0 N(rr j, u). Thus, we have N(r, u 0 ) = 1 + β for any r > 0. The conclusion then follows immediately from the lemma below. Lemma 6.6. Assume that v is a function on B 1 with the following properties i) v is not identically 0 in B 1. ii) v = 0 in B + 1 B 1. iii) v 0, + ν v + ν v 0, v( + ν v + ν v) = 0 on B 1. Then N(r, v) is nondecreasing as a function of r. Moreover, if N(r, v) = k for all r (0, 1), then v is homogeneous of degree k. Proof. The proof in [10] works for symmetric function v but can be easily modified for this nonsymmetric case. We just have to substitute ν + v by ν + v + ν v on B 1. Lemma 6.7. We must have β 1.

OPTIMAL REGULARITY IN ROOFTOP-LIKE OBSTACLE PROBLEM 7 Proof. Combining Lemmas 6.1, 6., and 6.4 with Theorem 3.6 we can deduce that u r are uniformly bounded in C 1,λ (B 1 ± B 1) for some λ > 0 when r is small. Because of this, we can always find a sequence {r j } converging to 0 such that u rj converges to some function u 0 in C 1,λ (B 1 ± B 1). Properties of u 0 as stated in Lemma 6.5 means that u 0 is a solution to a thin obstacle problem as studied in []. We know from that work that u 0 is C 1, 1 in B 1. However, our u 0 is also non-zero and homogeneous of degree 1 + β. Therefore, we must have β 1. Theorem 6.8. Let u be a solution of the obstacle problem in B 1 with rooftop-like obstacle ϕ C 1,1 (B ± 1 B 1) and boundary values g L ( B 1 ). Then there exists C > 0 depending only on n, ϕ C 1,1 (B ± 1 B 1 ) and g L ( B 1) such that Proof. From Lemmas 5.3 and 6.7 we have u C 1, 1 (B ± 1 S C. B 1 ) u(x) C x 3 for all x B 1 where C is a constant depending on C 0 and n. Combining this growth estimate with arguments used in Theorem 3.6, we obtain the C 1, 1 estimate for u in B ± 1 B 1. Finally, notice that the constant C 0 depends only on n, ϕ C 1,1 (B ± 1 B 1 ) and g L ( B 1). References [1] Frederick J. Almgren Jr., Almgren s big regularity paper, World Scientific Monograph Series in Mathematics, vol. 1, World Scientific Publishing Co. Inc., River Edge, NJ, 000. Q-valued functions minimizing Dirichlet s integral and the regularity of area-minimizing rectifiable currents up to codimension ; With a preface by Jean E. Taylor and Vladimir Scheffer. [] I. Athanasopoulos and L. A. Caffarelli, Optimal regularity of lower dimensional obstacle problems, Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 310 (004), no. Kraev. Zadachi Mat. Fiz. i Smezh. Vopr. Teor. Funkts. 35 [34], 49 66, 6 (English, with English and Russian summaries); English transl., J. Math. Sci. (N. Y.) 13 (006), no. 3, 74 84. [3] Mark Broadie and Jérôme Detemple, The valuation of American options on multiple assets, Math. Finance 7 (1997), no. 3, 41 86. [4] L. A. Caffarelli, Further regularity for the Signorini problem, Comm. Partial Differential Equations 4 (1979), no. 9, 1067 1075. [5], The obstacle problem revisited, J. Fourier Anal. Appl. 4 (1998), no. 4-5, 383 40. [6] Luis A. Caffarelli, Sandro Salsa, and Luis Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for the fractional Laplacian, Invent. Math. 171 (008), no., 45 461. [7] Jerome Detemple, Shui Feng, and Weidong Tian, The valuation of American call options on the minimum of two dividend-paying assets, Ann. Appl. Probab. 13 (003), no. 3, 953 983. [8] Nicola Garofalo and Fang-Hua Lin, Monotonicity properties of variational integrals, A p weights and unique continuation, Indiana Univ. Math. J. 35 (1986), no., 45 68. [9], Unique continuation for elliptic operators: a geometric-variational approach, Comm. Pure Appl. Math. 40 (1987), no. 3, 347 366. [10] Nicola Garofalo and Arshak Petrosyan, Some new monotonicity formulas and the singular set in the lower dimensional obstacle problem, Invent. Math. 177 (009), no., 415 461. [11] David Gilbarg and Neil S. Trudinger, Elliptic partial differential equations of second order, Classics in Mathematics, Springer-Verlag, Berlin, 001. Reprint of the 1998 edition. [1] N. N. Ural tseva, Hölder continuity of gradients of solutions of parabolic equations with boundary conditions of Signorini type, Dokl. Akad. Nauk SSSR 80 (1985), no. 3, 563 565 (Russian).

8 ARSHAK PETROSYAN AND TUNG TO Department of Mathematics, Purdue University, West Lafayette, IN 47907, USA E-mail address: arshak@math.purdue.edu Department of Mathematics, Purdue University, West Lafayette, IN 47907, USA E-mail address: tto@math.purdue.edu