1. Introduction Boundary estimates for the second derivatives of the solution to the Dirichlet problem for the Monge-Ampere equation

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1 POINTWISE C 2,α ESTIMATES AT THE BOUNDARY FOR THE MONGE-AMPERE EQUATION O. SAVIN Abstract. We prove a localization property of boundary sections for solutions to the Monge-Ampere equation. As a consequence we obtain pointwise C 2,α estimates at boundary points under appropriate local conditions on the right h side boundary data. 1. Introduction Boundary estimates for the second derivatives of the solution to the Dirichlet problem for the Monge-Ampere equation { det D 2 u = f in Ω, u = ϕ on Ω, were first obtained by Ivockina [I] in A few years later independently Krylov [K] Caffarelli-Nirenberg-Spruck [CNS] obtained the global C 2,α estimates in the case when Ω, ϕ f are sufficiently smooth this led to the solvability of the classical Dirichlet problem for the Monge-Ampere equation. When the right h side f is less regular, i.e f C α, the global C 2,α estimates were obtained recently by Trudinger Wang in [TW] for ϕ, Ω C 3. In this paper we discuss pointwise C 2,α estimates at boundary points under appropriate local conditions on the right h side boundary data. Our main result can be viewed as a boundary Schauder estimate for the Monge-Ampere equation which extends up to the boundary the pointwise interior C 2,α estimate of Caffarelli [C2] (see also [JW]). These sharp estimates play an important role for example when dealing with fourth order Monge-Ampere type equations arising in geometry, (see [TW], [LS]) or when the right h side f depends also on the second derivatives. We start with the following definition (see [CC]). Definition: Let 0 < α 1. We say that a function u is pointwise C 2,α at x 0 write u C 2,α (x 0 ) if there exists a quadratic polynomial P x0 such that We say that u C 2 (x 0 ) if u(x) = P x0 (x) + O( x x 0 2+α ). u(x) = P x0 (x) + o( x x 0 2 ). Similarly one can define the notion for a function to be C k C k,α at a point for any integer k 0. The author was partially supported by NSF grant

2 2 O. SAVIN It is easy to check that if u is pointwise C 2,α at all points of a Lipschitz domain Ω the equality in the definition above is uniform in x 0 then u C 2,α ( Ω) in the classical sense. Precisely, if there exist M δ such that for all points x 0 Ω then u(x) P x0 (x) M x x 0 2+α [D 2 u] C α ( Ω) C(δ, Ω)M. if x x 0 δ, x Ω Caffarelli showed in [C2] that if u is a strictly convex solution of det D 2 u = f f C α (x 0 ), f(x 0 ) > 0 at some interior point x 0 Ω, then u C 2,α (x 0 ). Our main theorem deals with the case when x 0 Ω. Theorem 1.1. Let Ω be a convex domain let u : Ω R convex, continuous, solve the Dirichlet problem for the Monge-Ampere equation { det D (1.1) 2 u = f in Ω, u = ϕ on Ω, with positive, bounded right h side i.e 0 < λ f Λ, for some constants λ, Λ. Assume that for some point x 0 Ω we have f C α (x 0 ), ϕ, Ω C 2,α (x 0 ), for some α (0, 1). If ϕ separates quadratically on Ω from the tangent plane of u at x 0, then u C 2,α (x 0 ). The way ϕ separates locally from the tangent plane at x 0 is given by the tangential second derivatives of u at x 0. Thus the assumption that this separation is quadratic is in fact necessary for the C 2,α estimate to hold. Heuristically, Theorem 1.1 states that if the tangential pure second derivatives of u are bounded below then the boundary Schauder estimates hold for the Monge-Ampere equation. A more precise, quantitative version of Theorem 1.1 is given in section 7 (see Theorem 7.1). Given the boundary data, it is not always easy to check the quadratic separation since it involves some information about the slope of the tangent plane at x 0. However, this can be done in several cases (see Proposition 3.2). One example is when Ω is uniformly convex ϕ, Ω C 3 (x 0 ). The C 3 condition of the data is optimal as it was shown by Wang in [W]. Other examples are when Ω is uniformly convex ϕ is linear, or when Ω is tangent of second order to a plane at x 0 ϕ has quadratic growth near x 0. As a consequence of Theorem 1.1 we obtain a pointwise C 2,α estimate in the case when the boundary data the domain are pointwise C 3. As mentioned above, the global version was obtained by Trudinger Wang in [TW]. Theorem 1.2. Let Ω be uniformly convex let u solve (1.1). Assume that f C α (x 0 ), ϕ, Ω C 3 (x 0 ), for some point x 0 Ω, some α (0, 1). Then u C 2,α (x 0 ).

3 POINTWISE C 2,α ESTIMATES 3 We also obtain the C 2,α estimate in the simple situation when Ω C 2,α ϕ is constant. Theorem 1.3. Let Ω be a uniformly convex domain assume u solves (1.1) with ϕ 0. If f C α ( Ω), Ω C 2,α, for some α (0, 1) then u C 2,α ( Ω). The key step in the proof of Theorem 1.1 is a localization theorem for boundary points (see also [S]). It states that under natural local assumptions on the domain boundary data, the sections S h (x 0 ) = {x Ω u(x) < u(x 0 ) + u(x 0 ) (x x 0 ) + h}, with x 0 Ω are equivalent to ellipsoids centered at x 0. Theorem 1.4. Let Ω be convex u satisfy (1.1), assume Ω, ϕ C 1,1 (x 0 ). If ϕ separates quadratically from the tangent plane of u at x 0, then for each small h > 0 there exists an ellipsoid E h of volume h n/2 such that with c, C constants independent of h. ce h Ω S h (x 0 ) x 0 CE h Ω, Theorem 1.4 provides useful information about the geometry of the level sets under rather mild assumptions it extends up to the boundary the localization theorem at interior points due to Caffarelli in [C1]. The paper is organized as follows. In section 2 we discuss briefly the compactness of solutions to the Monge-Ampere equation which we use later in the paper (see Theorem 2.7). For this we need to consider also solutions with possible discontinuities at the boundary. In section 3 we give a quantitative version of the Localization Theorem (see Theorem 3.1). In sections 4 5 we provide the proof of Theorem 3.1. In section 6 we obtain a version of the classical Pogorelov estimate in half-domain (Theorem 6.4). Finally, in section 7 we use the previous results together with a stard approximation method prove our main theorem. 2. Solutions with discontinuities on the boundary Let u : Ω R be a convex function with Ω R n bounded convex. Denote by U := {(x, x n+1 ) Ω R x n+1 u(x)} the upper graph of u. Definition 2.1. We define the values of u on Ω to be equal to ϕ i.e if the upper graph of ϕ : Ω R { } u Ω = ϕ, Φ := {(x, x n+1 ) Ω R is given by the closure of U restricted to Ω R, Φ := Ū ( Ω R). x n+1 ϕ(x)}

4 4 O. SAVIN From the definition we see that ϕ is lower semicontinuous. If u : Ω R is a viscosity solution to det D 2 u = f(x), with f 0 continuous bounded on Ω, then there exists an increasing sequence of subsolutions, continuous up to the boundary, with u n : Ω R, lim u n = u det D 2 u n f(x) in Ω, where the values of u on Ω are defined as above. Indeed, let us assume for simplicity that 0 Ω, u(0) = 0, u 0. Then, on each ray from the origin u is increasing, hence v ε : Ω R, v ε (x) = u((1 ε)x) is an increasing family of continuous functions as ε 0, with lim v ε = u in Ω. In order to obtain a sequence of subsolutions we modify v ε as with w ε ε, convex, so that thus u ε (x) := v ε (x) + w ε (x), det D 2 w ε f(x) (1 ε) 2n f((1 ε)x), det D 2 u ε (x) = det(d 2 v ε + D 2 w ε ) det D 2 v ε + det D 2 w ε f(x). The claim is proved since as ε 0 we can choose w ε to converge uniformly to 0. Proposition 2.2 (Comparison principle). Let u, v be defined on Ω with in the viscosity sense Then det D 2 u f(x) det D 2 v u Ω v Ω. u v in Ω. Proof. Since u can be approximated by a sequence of continuous functions on Ω it suffices to prove the result in the case when u is continuous on Ω u < v on Ω. Then, u < v in a small neighborhood of Ω the inequality follows from the stard comparison principle. A consequence of the comparison principle is that a solution det D 2 u = f is determined uniquely by its boundary values u Ω. Next we define the notion of convergence for functions which are defined on different domains.

5 POINTWISE C 2,α ESTIMATES 5 Definition 2.3. a) Let u k : Ω k R be a sequence of convex functions with Ω k convex. We say that u k converges to u : Ω R i.e if the upper graphs converge Ū k Ū u k u in the Haudorff distance. In particular it follows that Ω k Ω in the Hausdorff distance. b) Let ϕ k : Ω k R { } be a sequence of lower semicontinuous functions. We say that ϕ k converges to ϕ : Ω R { } i.e if the upper graphs converge Φ k Φ ϕ k ϕ in the Haudorff distance. c) We say that f k : Ω k R converge to f : Ω R if f k are uniformly bounded f k f uniformly on compact sets of Ω. Remark: When we restrict the Hausdorff distance to the nonempty closed sets of a compact set we obtain a compact metric space. Thus, if Ω k, u k are uniformly bounded then we can always extract a convergent subsequence u km u. Similarly, if Ω k, ϕ k are uniformly bounded we can extract a convergent subsequence ϕ km ϕ. Proposition 2.4. Let u k : Ω k R be convex If then det D 2 u k = f k, u k Ωk = ϕ k. u k u, ϕ k ϕ, f k f, (2.1) det D 2 u = f, u = ϕ on Ω, where ϕ is the convex envelope of ϕ on Ω i.e Φ is the restriction to Ω R of the convex hull generated by Φ. Remark: If Ω is strictly convex then ϕ = ϕ. Proof. Since Ū k Ū, Φ k Φ, Φ k Ūk, we see that Φ Ū. Thus, if K denotes the convex hull generated by Φ, then Φ K Ū. It remains to show that Ū ( Ω R) K. Indeed consider a hyperplane x n+1 = l(x) which lies strictly below K. Then for all large k {u k l 0} Ω k by Alexrov estimate we have that u k l Cd 1/n k

6 6 O. SAVIN where d k represents the distance to Ω k. By taking k we see that u l Cd 1/n thus no point on Ω R below the hyperplane belongs to Ū. Proposition 2.4 says that given any ϕ bounded lower semicontinuous, f 0 bounded continuous we can always solve uniquely the Dirichlet problem { det D 2 u = f in Ω, u = ϕ on Ω by approximation. Indeed, we can find sequences ϕ k, f k of continuous, uniformly bounded functions defined on strictly convex domains Ω k such that ϕ k ϕ f k f. Then the corresponding solutions u k are uniformly bounded continuous up to the boundary. Using compactness the proposition above we see that u k must converge to the unique solution u in (2.1). In view of Proposition 2.4 we extend the Definition 2.1 in order to allow boundary data that is not necessarily convex. Definition 2.5. Let ϕ : Ω R be a lower semicontinuous function. When we write that a convex function u satisfies we underst u = ϕ on Ω u Ω = ϕ where ϕ is the convex envelope of ϕ on Ω. Whenever ϕ ϕ do not coincide we can think of the graph of u as having a vertical part on Ω between ϕ ϕ. It follows easily from the definition above that the boundary values of u when we restrict to the domain Ω h := {u < h} are given by ϕ h = ϕ on Ω {ϕ h} Ω h ϕ h = h on the remaining part of Ω h. then By Proposition 2.2, the comparison principle still holds. Precisely, if u = ϕ, v = ψ, ϕ ψ on Ω, det D 2 u f det D 2 v in Ω, u v in Ω. The advantage of introducing the notation of Definition 2.5 is that the boundary data is preserved under limits. Proposition 2.6. Assume with Ω k, ϕ k uniformly bounded det D 2 u k = f k, u k = ϕ k on Ω k, ϕ k ϕ, f k f.

7 POINTWISE C 2,α ESTIMATES 7 Then u satisfies u k u det D 2 u = f, u = ϕ on Ω. Proof. Using compactness we may assume also that ϕ k ψ. Since ϕ k ϕ we find ϕ ψ ϕ. the conclusion follows from Proposition 2.4. Finally, we state a version of the last proposition for solutions with bounded right-h side i.e λ det D 2 u Λ, where the two inequalities are understood in the viscosity sense. Theorem 2.7. Assume λ det D 2 u k Λ, u k = ϕ k on Ω k, Ω k, ϕ k uniformly bounded. Then there exists a subsequence k m such that with u km u, ϕ km ϕ λ det D 2 u Λ, u = ϕ on Ω. 3. The Localization Theorem In this section we state the quantitative version of the localization theorem at boundary points (Theorem 3.1). Let Ω be a bounded convex set in R n. We assume that (3.1) B ρ (ρe n ) Ω {x n 0} B 1 ρ, for some small ρ > 0, that is Ω (R n ) + Ω contains an interior ball tangent to Ω at 0. Let u : Ω R be continuous, convex, satisfying (3.2) det D 2 u = f, 0 < λ f Λ in Ω. We extend u to be outside Ω. After subtracting a linear function we assume that (3.3) x n+1 = 0 is the tangent plane to u at 0, in the sense that u 0, u(0) = 0, any hyperplane x n+1 = εx n, ε > 0, is not a supporting plane for u. We investigate the geometry of the sections of u at 0 that we denote for simplicity of notation S h := {x Ω : u(x) < h}. We show that if the boundary data has quadratic growth near {x n = 0} then, as h 0, S h is equivalent to a half-ellipsoid centered at 0. Precisely, our theorem reads as follows.

8 8 O. SAVIN Theorem 3.1 (Localization Theorem). Assume that Ω, u satisfy (3.1)-(3.3) above for some µ > 0, (3.4) µ x 2 u(x) µ 1 x 2 on Ω {x n ρ}. Then, for each h < c(ρ) there exists an ellipsoid E h of volume h n/2 such that ke h Ω S h k 1 E h Ω. Moreover, the ellipsoid E h is obtained from the ball of radius h 1/2 by a linear transformation A 1 h (sliding along the x n = 0 plane) with A h E h = h 1/2 B 1 A h (x) = x νx n, ν = (ν 1, ν 2,..., ν n 1, 0), ν k 1 log h. The constant k above depends on µ, λ, Λ, n c(ρ) depends also on ρ. The ellipsoid E h, or equivalently the linear map A h, provides information about the behavior of the second derivatives near the origin. Heuristically, the theorem states that in S h the tangential second derivatives are bounded from above below the mixed second derivatives are bounded by log h. The hypothesis that u is continuous up to the boundary is not necessary, we just need to require that (3.4) holds in the sense of Definition 2.5. Given only the boundary data ϕ of u on Ω, it is not always easy to check the main assumption (3.4) i.e that ϕ separates quadratically on Ω (in a neighborhood of {x n = 0}) from the tangent plane at 0. Proposition 3.2 provides some examples when this is satisfied depending on the local behavior of Ω ϕ (see also the remarks below). Proposition 3.2. Assume (3.1),(3.2) hold. Then (3.4) is satisfied if any of the following holds: 1) ϕ is linear in a neighborhood of 0 Ω is uniformly convex at the origin. 2) Ω is tangent of order 2 to {x n = 0} ϕ has quadratic growth in a neighborhood of {x n = 0}. 3) ϕ, Ω C 3 (0), Ω is uniformly convex at the origin. Proposition 3.2 is stard (see [CNS], [W]). We sketch its proof below. Proof. 1) Assume ϕ = 0 in a neighborhood of 0. By the use of stard barriers, the assumptions on Ω imply that the tangent plane at the origin is given by x n+1 = µx n for some bounded µ > 0. Then (3.4) clearly holds. 2) After subtracting a linear function we may assume that µ x 2 ϕ µ 1 x 2 on Ω in a neighborhood of {x n = 0}. Using a barrier we obtain that l 0, the tangent plane at the origin, has bounded slope. But Ω is tangent of order 2 to {x n = 0},

9 POINTWISE C 2,α ESTIMATES 9 thus l 0 grows less than quadratic on Ω in a neighborhood of {x n = 0} (3.4) is again satisfied. 3) Since Ω is uniformly convex at the origin, we can use barriers obtain that l 0 has bounded slope. After subtracting this linear function we may assume l 0 = 0. Since ϕ, Ω C 3 (0) we find that ϕ = Q 0 (x ) + o( x 3 ) with Q 0 a cubic polynomial. Now ϕ 0, hence Q 0 has no linear part its quadratic part is given by, say µ i 2 x2 i, with µ i 0. i<n We need to show that µ i > 0. If µ 1 = 0, then the coefficient of x 3 1 is 0 in Q 0. Thus, if we restrict to Ω in a small neighborhood near the origin, then for all small h the set {ϕ < h} contains for some c > 0 with { x 1 r(h)h 1/3 } { x ch 1/2 } r(h) as h 0. Now S h contains the convex set generated by {ϕ < h} thus, since Ω is uniformly convex, S h c (r(h)h 1/3 ) 3 h (n 2)/2 c r(h) 3 h n/2. On the other h, since u satisfies (3.2) we obtain (see (4.4)) 0 u h in S h S h Ch n/2, for some C depending on λ n, we contradict the inequality above as h 0. Remark 3.3. The proof easily implies that if Ω, ϕ C 3 (Ω) Ω is uniformly convex, then we can find a constant µ which satisfies (3.4) for all x Ω. Remark 3.4. From above we see that we can often verify (3.4) in the case when ϕ, Ω C 1,1 (0) Ω is uniformly convex at 0. Indeed, if l ϕ represents the tangent plane at 0 to ϕ : Ω R (in the sense of (3.3)), then (3.4) holds if either ϕ separates from l ϕ quadratically near 0, or if ϕ is tangent to l ϕ of order 3 in some tangential direction. Remark 3.5. Given ϕ, Ω C 1,1 (0) Ω uniformly convex at 0, then (3.4) holds if λ is sufficiently large. 4. Proof of Theorem 3.1 (I) We prove Theorem 3.1 in the next two sections. In this section we obtain some preliminary estimates reduce the theorem to a statement about the rescalings of u. This statement is proved in section 5 using compactness. Next proposition was proved by Trudinger Wang in [TW]. It states that the volume of S h is proportional to h n/2 after an affine transformation (of controlled norm) we may assume that the center of mass of S h lies on the x n axis. Since our setting is slightly different we provide its proof.

10 10 O. SAVIN Proposition 4.1. Under the assumptions of Theorem 3.1, for all h c(ρ), there exists a linear transformation (sliding along x n = 0) with such that the rescaled function satisfies in the following: A h (x) = x νx n, ν n = 0, ν C(ρ)h n 2(n+1) ũ(a h x) = u(x), S h := A h S h = {ũ < h} (i) the center of mass of S h lies on the x n -axis; (ii) k 0 h n/2 S h = S h k 1 0 hn/2 ; (iii) the part of S h where {ũ < h} is a graph, denoted by that satisfies G h = S h {ũ < h} = {(x, g h (x ))} g h C(ρ) x 2 µ 2 x 2 ũ 2µ 1 x 2 on G h. The constant k 0 above depends on µ, λ, Λ, n the constants C(ρ), c(ρ) depend also on ρ. In this section we denote by c, C positive constants that depend on n, µ, λ, Λ. For simplicity of notation, their values may change from line to line whenever there is no possibility of confusion. Constants that depend also on ρ are denote by c(ρ), C(ρ). Proof. The function v := µ x 2 + Λ µ n 1 x2 n C(ρ)x n is a lower barrier for u in Ω {x n ρ} if C(ρ) is chosen large. Indeed, then v u on Ω {x n ρ}, In conclusion, hence v 0 u v u on Ω {x n = ρ}, det D 2 v > Λ. in Ω {x n ρ}, (4.1) S h {x n ρ} {v < h} {x n > c(ρ)(µ x 2 h)}.

11 POINTWISE C 2,α ESTIMATES 11 Let x h be the center of mass of S h. We claim that (4.2) x h e n c 0 (ρ)h α, α = n n + 1, for some small c 0 (ρ) > 0. Otherwise, from (4.1) John s lemma we obtain S h {x n C(n)c 0 h α h α } { x C 1 h α/2 }, for some large C 1 = C 1 (ρ). Then the function w = εx n + h ( x 2 C 1 h α/2 ) 2 + ΛC 2(n 1) 1 h is a lower barrier for u in S h if c 0 is sufficiently small. Indeed, for all small h, Hence ( xn h α ) 2 w h 4 + h 2 + ΛC2(n 1) 1 (C(n)c 0 ) 2 h < h in S h, w εx n + h1 α C1 2 x 2 x n + C(ρ)hc 0 h α µ x 2 u det D 2 w = 2Λ. w u in S h, on Ω, we contradict that 0 is the tangent plane at 0. Thus claim (4.2) is proved. Now, define The center of mass of S h = A h S h is A h x = x νx n, ν = x h x h e, n ũ(a h x) = u(x). x h = A h x h lies on the x n -axis from the definition of A h. Moreover, since x h S h, we see from (4.1)-(4.2) that ν C(ρ) (x h e n) 1/2 (x h e n) C(ρ)h α/2, this proves (i). If we restrict the map A h on the set on Ω where {u < h}, i.e. on we have S h Ω {x n x 2 ρ } { x < Ch 1/2 } A h x x = ν x n C(ρ)h α/2 x 2 C(ρ)h 1 α 2 x, part (iii) easily follows. Next we prove (ii). From John s lemma, we know that after relabeling the x coordinates if necessary, (4.3) D h B 1 S h x h C(n)D h B 1

12 12 O. SAVIN where Since d d 2 0 D h = d n ũ 2µ 1 x 2 on G h = {(x, g h (x ))}, we see that the domain of definition of g h contains a ball of radius (µh/2) 1/2. This implies that d i c 1 h 1/2, i = 1,, n 1, for some c 1 depending only on n µ. Also from (4.2) we see that x h e n = x h e n c 0 (ρ)h α which gives d n c(n) x h e n c(ρ)h α. We claim that for all small h, n d i k 0 h n/2, i=1 with k 0 small depending only on µ, n, Λ, which gives the left inequality in (ii). To this aim we consider the barrier, n ( ) 2 xi w = εx n + ch. We choose c sufficiently small depending on µ, n, Λ so that for all h < c(ρ), i=1 d i w h on S h, on the part of the boundary G h, we have w ũ since w εx n + c c 2 x 2 + ch 1 µ 4 x 2 + chc(n) x n d n ( ) 2 xn d n µ 4 x 2 + ch 1 α C(ρ) x 2 µ 2 x 2. Moreover, if our claim does not hold, then det D 2 w = (2ch) n ( d i ) 2n > Λ, thus w ũ in S h. By definition, ũ is obtained from u by a sliding along x n = 0, hence 0 is still the tangent plane of ũ at 0. We reach again a contradiction since ũ w εx n the claim is proved. Finally we show that (4.4) S h Ch n/2 for some C depending only on λ, n. Indeed, if v = h on S h,

13 POINTWISE C 2,α ESTIMATES 13 then Since we obtain the desired conclusion. det D 2 v = λ v u 0 in S h. h h min S h v c(n, λ) S h 2/n In the proof above we showed that for all h c(ρ), the entries of the diagonal matrix D h from (4.3) satisfy d i ch 1/2, i = 1,... n 1 d n c(ρ)h α, α = n n + 1 ch n/2 d i Ch n/2. The main step in the proof of Theorem 3.1 is the following lemma that will be completed in Section 5. Lemma 4.2. There exist constants c, c(ρ) such that (4.5) d n ch 1/2, for all h c(ρ). Using Lemma 4.2 we can easily finish the proof of our theorem. Proof of Theorem 3.1. Since all d i are bounded below by ch 1/2 their product is bounded above by Ch n/2 we see that Ch 1/2 d i ch 1/2 for all h c(ρ). Using (4.3) we obtain Moreover, since S h Ch 1/2 B 1. i = 1,, n x h e n d n ch 1/2, ( x h) = 0, the part G h of the boundary S h contains the graph of g h above x ch 1/2, we find that ch 1/2 B 1 Ω S h, with Ω = A h Ω, S h = A h S h. In conclusion We define the ellipsoid E h as hence ch 1/2 B 1 Ω A h S h Ch 1/2 B 1. E h := A 1 h (h1/2 B 1 ), ce h Ω S h CE h.

14 14 O. SAVIN Comparing the sections at levels h h/2 we find we easily obtain the inclusion If we denote then the inclusion above implies which gives the desired bound for all small h. ce h/2 Ω CE h A h A 1 h/2 B 1 CB 1. A h x = x ν h x n ν h ν h/2 C, ν h C log h In order to prove Lemma 4.2 we introduce a new quantity b(h) which is proportional to d n h 1/2 is appropriate when dealing with affine transformations. Notation. Given a convex function u we define b u (h) = h 1/2 sup S h x n. Whenever there is no possibility of confusion we drop the subindex u use the notation b(h). Below we list some basic properties of b(h). 1) If h 1 h 2 then 2) A rescaling ( h1 h 2 ) 1 2 b(h 1 ) b(h 2 ) ũ(ax) = u(x) ( h2 h 1 ) 1 2. given by a linear transformation A which leaves the x n coordinate invariant does not change the value of b, i.e bũ(h) = b u (h). 3) If A is a linear transformation which leaves the plane {x n = 0} invariant the values of b get multiplied by a constant. However the quotients b(h 1 )/b(h 2 ) do not change values i.e bũ(h 1 ) bũ(h 2 ) = b u(h 1 ) b u (h 2 ). then 4) If we multiply u by a constant, i.e. ũ(x) = βu(x) bũ(βh) = β 1/2 b u (h),

15 POINTWISE C 2,α ESTIMATES 15 bũ(βh 1 ) bũ(βh 2 ) = b u(h 1 ) b u (h 2 ). From (4.3) property 2 above, c(n)d n b(h)h 1/2 C(n)d n, hence Lemma 4.2 will follow if we show that b(h) is bounded below. We achieve this by proving the following lemma. Lemma 4.3. There exist c 0, c(ρ) such that if h c(ρ) b(h) c 0 then (4.6) for some t [c 0, 1]. b(th) b(h) > 2, This lemma states that if the value of b(h) on a certain section is less than a critical value c 0, then we can find a lower section at height still comparable to h where the value of b doubled. Clearly Lemma 4.3 property 1 above imply that b(h) remains bounded for all h small enough. The quotient in (4.6) is the same for ũ which is defined in Proposition 4.1. We normalize the domain S h ũ by considering the rescaling v(x) = 1 hũ(h1/2 Ax) where A is a multiple of D h (see (4.3)), A = γd h such that Then the diagonal entries of A satisfy The function v satisfies det A = 1. ch 1/2 γ Ch 1/2, a i c, i = 1, 2,, n 1, cb u (h) a n Cb u (h). λ det D 2 v Λ, v 0, v(0) = 0, is continuous it is defined in Ω v with Then for some x, Ω v := {v < 1} = h 1/2 A 1 Sh. x + cb 1 Ω v CB + 1, ct n/2 S t (v) Ct n/2, t 1, where S t (v) denotes the section of v. Since ũ = h in S h {x n C(ρ)h},

16 16 O. SAVIN then v = 1 on Ω v {x n σ}, σ := C(ρ)h 1 α. Also, from Proposition 4.1 on the part G of the boundary of Ω v where {v < 1} we have (4.7) n µ n 1 a 2 i x 2 i v 2µ 1 a 2 i x 2 i. i=1 In order to prove Lemma 4.3 we need to show that if σ, a n are sufficiently small depending on n, µ, λ, Λ then the function v above satisfies (4.8) b v (t) 2b v (1) for some 1 > t c 0. Since α < 1, the smallness condition on σ is satisfied by taking h < c(ρ) sufficiently small. Also a n being small is equivalent to one of the a i, 1 i n 1 being large since their product is 1 a i are bounded below. In the next section we prove property (4.8) above by compactness, by letting σ 0, a i for some i (see Proposition 5.1). i=1 5. Proof of Theorem 3.1 (II) In this section we consider the class of solutions v that satisfy the properties above. After relabeling the constants µ a i, by abuse of notation writing u instead of v, we may assume we are in the following situation. Fix µ small λ, Λ. For an increasing sequence with we consider the family of solutions a 1 a 2... a n 1 a 1 µ, u D µ σ(a 1, a 2,..., a n 1 ) of convex functions u : Ω R that satisfy (5.1) λ det D 2 u Λ in Ω, 0 u 1 in Ω; (5.2) 0 Ω, B µ (x 0 ) Ω B + 1/µ for some x 0 ; (5.3) µh n/2 S h µ 1 h n/2. Moreover we assume that the boundary Ω has a closed subset G (5.4) G {x n σ} Ω which is a graph in the e n direction with projection π n (G) R n 1 along e n n 1 n 1 (5.5) { µ 1 a 2 i x 2 i 1 } π n (G) { µ a 2 i x 2 i 1 }, 1 (see Definition 2.5), the boundary values of u = ϕ on Ω satisfy (5.6) ϕ = 1 on Ω \ G; 1

17 POINTWISE C 2,α ESTIMATES 17 n 1 (5.7) µ a 2 i x 2 i ϕ min{ 1, In this section we prove 1 a 2 i x 2 i } on G. µ 1 n 1 Proposition 5.1. For any M > 0 there exists C depending on M, µ, λ, Λ, n such that if u D µ σ(a 1, a 2,..., a n 1 ) with a n 1 C, σ C 1 then b(h) = (sup x n )h 1/2 M S h for some h with C 1 h 1. Property (4.8) (hence Theorem 3.1), easily follows from this proposition. Indeed, by choosing M = 2µ 1 2b(1) in Proposition 5.1 we prove the existence of a section S h with h c 0 such that b(h) 2b(1). Clearly the function v of the previous section satisfies the hypotheses above (after renaming the constant µ) provided that σ, a n are sufficiently small. We prove Proposition 5.1 by compactness. We introduce the limiting solutions from the class D µ σ(a 1,..., a n 1 ) when a k+1 σ 0. If µ a 1... a k, we denote by D µ 0 (a 1,..., a k,,,..., ), 0 k n 2 the class of functions u that satisfy properties (5.1)-(5.2)-(5.3) with, (5.8) G {x i = 0, i > k} Ω if we restrict to the space generated by the first k coordinates then k k (5.9) { µ 1 a 2 i x 2 i 1 } G { µ a 2 i x 2 i 1 }. Also, u = ϕ on Ω with (5.10) ϕ = 1 on Ω \ G; (5.11) µ 1 k k a 2 i x 2 i ϕ min{ 1, µ 1 a 2 i x 2 i } on G. 1 The compactness theorem (Theorem 2.7) implies that if u m D µ σ m (a m 1,..., a m n 1) is a sequence with σ m 0 a m k+1 for some fixed 0 k n 2, then we can extract a convergent subsequence to a function u (see Definition 2.3) with for some l k a 1... a l. u D µ 0 (a 1,.., a l,,.., ), 1 1 1

18 18 O. SAVIN Proposition 5.1 follows easily from the next proposition. Proposition 5.2. For any M > 0 0 k n 2 there exists c k depending on M, µ, λ, Λ, n, k such that if (5.12) u D µ 0 (a 1,..., a k,,..., ) then b(h) = (sup x n )h 1/2 M S h for some h with c k h 1. Indeed, if Proposition 5.1 fails for a sequence of constants C then we obtain a limiting solution u as in (5.12) for which b(h) M for all h > 0. This contradicts Proposition 5.2 (with M replaced by 2M). We prove Proposition 5.2 by induction on k. notation. Denote We start by introducing some x = (y, z, x n ), y = (x 1,..., x k ) R k, z = (x k+1,..., x n 1 ) R n 1 k. Definition 5.3. We say that a linear transformation T : R n R n is a sliding along the y direction if with T x := x + ν 1 z 1 + ν 2 z ν n k 1 z n k 1 + ν n k x n ν 1, ν 2,..., ν n k span{e 1,..., e k } We see that T leaves the (z, x n ) components invariant together with the subspace (y, 0, 0). Clearly, if T is a sliding along the y direction then so is T 1 det T = 1. The key step in the proof of Proposition 5.2 is the following lemma. Lemma 5.4. Assume that u p( z qx n ), for some p, q > 0 assume that for each section S h of u, h (0, 1), there exists T h a sliding along the y direction such that for some constant C 0. Then T h S h C 0 h 1/2 B + 1, u D µ 0 (1,..., 1,,..., ). Proof. Assume by contradiction that u D µ 0 q q 0 for some q 0. We show that (5.13) u p ( z q x n ), q = q η, it satisfies the hypotheses with for some 0 < p p, where the constant η > 0 depends only on q 0 µ, C 0, Λ, n. Then, since q q 0, we can apply this result a finite number of times obtain u ε( z + x n ), for some small ε > 0. This gives S h {x n ε 1 h} hence T h S h {x n ε 1 h}

19 POINTWISE C 2,α ESTIMATES 19 by the hypothesis above we contradict (5.3). Now we prove (5.13). Since u D µ 0 such that on the subspace (y, 0, 0) S h = T h S h = O(h (n+1)/2 ) as h 0, as above, there exists a closed set G h S h {z = 0, x n = 0} {µ 1 y 2 h} G h {µ y 2 h}, the boundary values ϕ h of u on S h satisfy (see Section 2) Let w be a rescaling of u, ϕ h = h on S h \ G h ; µ y 2 ϕ h min { h, µ 1 y 2} on G h. for some small h p. Then w(x) := 1 h u(h1/2 T 1 h x) S 1 (w) := Ω w = h 1/2 T h S h B + C 0 our hypothesis becomes (5.14) w p h ( z qx n). 1/2 Moreover the boundary values ϕ w of w on Ω w satisfy ϕ w = 1 on Ω w \ G w µ y 2 ϕ w min{1, µ 1 y 2 } on G w := h 1/2 G h. Next we show that ϕ w v on Ω w where v is defined as v := δ x 2 + Λ δ n 1 (z 1 qx n ) 2 + N(z 1 qx n ) + δx n, δ is small depending on µ C 0, N is chosen large such that Λ δ n 1 t2 + Nt is increasing in the interval t (1 + q 0 )C 0. From the definition of v we see that det D 2 v > Λ. On the part of the boundary Ω w where z 1 qx n we use that Ω w B C0 obtain v δ( x 2 + x n ) ϕ w. On the part of the boundary Ω w where z 1 > qx n we use (5.14) obtain 1 = ϕ w C( z qx n ) C(z 1 qx n ) with C arbitrarily large provided that h is small enough. We choose C such that the inequality above implies Λ δ n 1 (z 1 qx n ) 2 + N(z 1 qx n ) < 1 2.

20 in B + 1/µ. Then v ϕ on Ω, det D 2 v > Λ, 20 O. SAVIN Then ϕ w = 1 > δ( x 2 + x n ) v. In conclusion ϕ w v on Ω w hence the function v is a lower barrier for w in Ω w. Then w N(z 1 qx n ) + δx n, since this inequality holds for all directions in the z-plane, we obtain Scaling back we get w N( z (q η)x n ), η := δ N. u p ( z (q η)x n ) in S h. Since u is convex u(0) = 0, this inequality holds globally, (5.13) is proved. Lemma 5.5. Proposition 5.2 holds for k = 0. Proof. By compactness we need to show that there does not exist u D µ 0 (,..., ) with b(h) M for all h. If such u exists then G = {0}. Let v := δ( x x 2 ) + Λ δ n 1 x2 n Nx n with δ small depending on µ, N large so that Λ δ n 1 x2 n Nx n 0 hence v u in Ω. This gives u δ x Nx n, we obtain S h { x C(x n + h)}. Since b(h) M we conclude S h Ch 1/2 B 1 +, we contradict Lemma 5.4 for k = 0. Now we prove Proposition 5.2 by induction on k. Proof of Proposition 5.2. In this proof we denote by c, C positive constants that depend on M, µ, λ, Λ, n k. We assume that the proposition holds for all nonnegative integers up to k 1, 1 k < n 2, we prove it for k. Let u D µ 0 (a 1,..., a k,,..., ). By the induction hypotheses compactness we see that there exists a constant C k (µ, M, λ, Λ, n)

21 POINTWISE C 2,α ESTIMATES 21 such that if a k C k then b(h) M for some h C 1 k. Thus, it suffices to consider only the case when a k < C k. If no c k+1 exists then we can find a limiting solution that, by abuse of notation, we still denote by u such that (5.15) u D µ 0 (1, 1,..., 1,,..., ) with (5.16) b(h) Mh 1/2, h > 0 where µ depends on µ C k. We show that such a function u does not exist. Denote as before x = (y, z, x n ), y = (x 1,..., x k ) R k, z = (x k+1,..., x n 1 ) R n 1 k. On Ω we have ϕ(x) δ x 2 + δ z + Λ δ n 1 x2 n Nx n where δ is small depending on µ, N is large so that Λ δ n 1 x2 n Nx n 0 in B + 1/ µ. As before we obtain that the inequality above holds in Ω, hence (5.17) u(x) δ z Nx n. From (5.16)-(5.17) we see that the section S h of u satisfies (5.18) S h { z < δ 1 (Nx n + h)} {x n Mh 1/2 }. From John s lemma we know that S h is equivalent to an ellipsoid E h of the same volume i.e (5.19) c(n)e h S h x h C(n)E h, E h = S h, with x h the center of mass of S h. For any ellipsoid E h in R n of positive volume we can find T h, a sliding along the y direction (see Definition 5.3), such that (5.20) T h E h = E h 1/n AB 1, with a matrix A that leaves the (y, 0, 0) (0, z, x n ) subspaces invariant, det A = 1. By choosing an appropriate system of coordinates in the y z variables we may assume in fact that with with 0 < β 1 β k, A(y, z, x n ) = (A 1 y, A 2 (z, x n )) β β 2 0 A 1 = β k

22 22 O. SAVIN γ k θ k+1 0 γ k+2 0 θ k+2 A 2 = γ n 1 θ n θ n with γ j, θ n > 0. The h section S h = T h S h of the rescaling ũ(x) = u(t 1 h x) satisfies (5.18) since u D µ 0, there exists G h = G h, such that on the subspace (y, 0, 0) G h {z = 0, x n = 0} S h {µ 1 y 2 h} G h {µ y 2 h}, the boundary values ϕ h of ũ on S h satisfy Moreover, using that ϕ h = h on S h \ G h ; µ y 2 ϕ h min { h, µ 1 y 2} on G h. S h h n/2 in (5.19), (5.20) that 0 S h, we obtain (5.21) x h + ch 1/2 AB 1 S h Ch 1/2 AB 1, det A = 1, for the matrix A as above with x h the center of mass of S h. Next we use the induction hypothesis show that S h is equivalent to a ball. Lemma 5.6. There exists C 0 such that Proof. We need to show that Since S h satisfies (5.18) we see that T h S h = S h C 0 h n/2 B + 1. A C. S h { (z, x n ) Ch 1/2 }, which together with the inclusion (5.21) gives A 2 C hence Also, since we find from (5.21) We define the rescaling γ j, θ n C, θ j C. G h S h, β i c > 0, i = 1,, k. w(x) = 1 hũ(h1/2 Ax) defined in a domain Ω w = S 1 (w). Then (5.21) gives B c (x 0 ) Ω w B + C,

23 POINTWISE C 2,α ESTIMATES 23 w = ϕ w on Ω w with ϕ w = 1 on Ω w \ G w, µ k βi 2 x 2 i ϕ w min{1, µ 1 1 This implies that k βi 2 x 2 i } on G w := h 1/2 A 1 Gh. 1 w D µ 0 (β 1, β 2,..., β k,,..., ) for some small µ depending on µ, M, λ, Λ, n, k. We claim that b u (h) c. First we notice that Since b u (h) = bũ(h) θ n. θ n βi γj = det A = 1 γ j C, we see that if b u (h) ( therefore θ n ) becomes smaller than a critical value c then β k C k ( µ, M, λ, Λ, n), with M := 2 µ 1, by the induction hypothesis b w ( h) M 2b w (1) for some h > C 1 k. This gives b u (h h) b u (h) = b w( h) b w (1) 2, which implies b u (h h) 2b u (h) our claim follows. Next we claim that γ j are bounded below by the same argument. Indeed, from the claim above θ n is bounded below if some γ j is smaller than a small value c then β k C k ( µ, M 1, λ, Λ, n) with By the induction hypothesis hence M 1 := 2M µc. b w ( h) M 1 2M c b w (1), b u (h h) b u (h) 2M c which gives b u (h h) 2M, contradiction. In conclusion θ n, γ j are bounded below which implies that β i are bounded above. This shows that A is bounded the lemma is proved.

24 24 O. SAVIN End of the proof of Proposition 5.2. The proof is finished since Lemma 5.6, (5.15), (5.17) contradict Lemma Pogorelov estimate in half-domain In this section we obtain a version of Pogorelov estimate at the boundary (Theorem 6.4 below). A similar estimate was proved also in [TW]. We start with the following a priori estimate. Proposition 6.1. Let u : Ω R, u C 4 ( Ω) satisfy the Monge-Ampere equation Assume that for some constant k > 0, det D 2 u = 1 in Ω. B + k Ω B+ k 1, { u = 1 2 x 2 on Ω {x n = 0} u = 1 on Ω {x n > 0}. Then u C 3,1 ({u< 1 16 k2 }) C(k, n). Proof. We divide the proof into four steps. Step 1: We show that u C(k, n) in the set D := {u < k 2 /2}. For each we consider the barrier x 0 { x k, x n = 0}, w x0 (x) := 1 2 x x 0 (x x 0 ) + δ x x δ 1 n (x 2 n k 1 x n ), where δ is small so that Then thus in Ω w x0 1 in B + k 1. w x0 (x 0 ) = u(x 0 ), w x0 u on Ω {x n = 0}, w x0 1 = u on Ω {x n > 0}, det D 2 w x0 > 1, u w x0 u(x 0 ) + x 0 (x x 0 ) δ 1 n k 1 x n. This gives a lower bound for u n (x 0 ). Moreover, writing the inequality for all x 0 with x 0 = k we obtain D {x n c( x k)}. From the values of u on {x n = 0} the inclusion above we obtain a lower bound on u n on D in a neighborhood of {x n = 0}. Since Ω contains the cone generated by ke n { x 1, x n = 0} u 1 in Ω, we can use the convexity of u obtain also an upper bound for u n all u i, 1 i n 1, on D in a neighborhood of {x n = 0}. We find u C on D {x n c 0 },

25 POINTWISE C 2,α ESTIMATES 25 where c 0 > 0 is a small constant depending on k n. We obtain a similar bound on D {x n c 0 } by bounding below by a small positive constant. Indeed, if dist( D {x n c 0 }, Ω) y Ω {x n c 0 /2}, then there exists a linear function l y with bounded gradient so that u(y) = l y (y), u l y on Ω. Then, using Alexrov estimate for (u l y ) we obtain u(x) l y (x) Cd(x) 1/n, hence D stays outside a fixed neighborhood of y. Step 2: We show that d(x) := dist(x, Ω) D 2 u C(k, n) on E := {x n = 0} { x k/2}. It suffices to prove that u in are bounded in E with i = 1,.., n 1. Let L ϕ := u ij ϕ ij denote the linearized Monge-Ampere operator for u. Then L u i = 0, u i = x i on {x n = 0}, L u = n, if we define P (x) = δ x 2 + δ 1 n x 2 n then L P = T r ( (D 2 u) 1 D 2 P ) n ( det(d 2 u) 1 det D 2 P ) 1 n n. Fix x 0 E. We compare u i v x0 (x) := x i + γ 1 [ δ x x δ 1 n (x 2 n γ 2 x n ) (u l x0 ) ], where l x0 denotes the supporting linear function for u at x 0, δ = 1/4, γ 1, γ 2 0. Clearly, L v x0 0,, since u is Lipschitz in D we can choose γ 1, γ 2 large, depending only on k n such that v x0 u i on D. This shows that the inequality above holds also in D we obtain a lower bound on u in (x 0 ). Similarly we obtain an upper bound. Step 3: We show that D 2 u C on {u < k 2 /8}. We apply the classical Pogorelov estimate in the set F := {u < k 2 /4}.

26 26 O. SAVIN Precisely if the maximal value of ( ) 1 log 4 k2 u + log u ii u2 i occurs in the interior of F then this value is bounded by a constant depending only on n max F u (see [C2]). From step 2, the expression is bounded above on F the estimate follows. Step 4: The Monge-Ampere equation is uniformly elliptic in {u < k 2 /8} by Evans-Krylov theorem Schauder estimates we obtain the desired C 3,1 bound. Remark 6.2. Assume the boundary values of u are given by { u = p(x ) on Ω {x n = 0} u = 1 on Ω {x n > 0}, with p(x ) a quadratic polynomial that satisfies for some ρ > 0. Then ρ x 2 p(x ) ρ 1 x 2, u C 3,1 ({u< 1 16 k2 }) C(ρ, k, n). Indeed, after an affine transformation we can reduce the problem to the case p(x ) = x 2 /2. Remark 6.3. Proposition 6.1 holds as well if we replace the half-space {x n 0} with a large ball of radius ε 1 B ε := { x ε 1 e n ε 1 }. Precisely, if B k B ε Ω B k 1 B ε, the boundary values of u satisfy { u = 1 2 x 2 on B 1 B ε Ω u [1, 2] on Ω \ (B 1 B ε ), then for all small ε, u C 3,1 ({u<k 2 /16}) C, with C depending only on k n. The proof is essentially the same except that in the barrier functions w x0, v x0 we need to replace x n by (x x 0 ) ν x0 where ν x0 denotes the inner normal to Ω at x 0, in step 2 we work (as in [CNS]) with the tangential derivative instead of xi. T i := (1 εx n ) xi + εx i xn, As a consequence of the Proposition 6.1 the remarks above we obtain Theorem 6.4. Let u : Ω R satisfy the Monge-Ampere equation Assume that for some constants ρ, k > 0, det D 2 u = 1 in Ω. B + k Ω B+ k 1,

27 POINTWISE C 2,α ESTIMATES 27 (see Definition 2.5) the boundary values of u are given by { u = p(x ) on {p(x ) 1} {x n = 0} Ω u = 1 on the rest of Ω, where p is a quadratic polynomial that satisfies Then ρ x 2 p(x ) ρ 1 x 2. (6.1) u C 3,1 (B + c 0 ) c 1 0, with c 0 > 0 small, depending only on k, ρ n. Proof. We approximate u on Ω by a sequence of smooth functions u m on Ω m, with Ω m smooth, uniformly convex, so that u m, Ω m satisfy the conditions of Remark 6.3 above. Notice that u m is smooth up to the boundary by the results in [CNS], thus we can use Proposition 6.1 for u m. We let m obtain (6.1) since by convexity. B + c 0 {u < k 2 /16}, 7. Pointwise C 2,α estimates at the boundary Let Ω be a bounded convex set with (7.1) B ρ (ρe n ) Ω {x n 0} B 1 ρ, for some small ρ > 0, that is Ω (R n ) + Ω contains an interior ball tangent to Ω at 0. Let u : Ω R be convex, continuous, satisfying (7.2) det D 2 u = f, 0 < λ f Λ in Ω, (7.3) x n+1 = 0 is a tangent plane to u at 0, in the following sense: u 0, u(0) = 0, any hyperplane x n+1 = εx n, ε > 0 is not a supporting plane for u. We also assume that on Ω, in a neighborhood of {x n = 0}, u separates quadratically from the tangent plane {x n+1 = 0}, (7.4) ρ x 2 u(x) ρ 1 x 2 on Ω {x n ρ}. Our main theorem is the following. Theorem 7.1. Let Ω, u satisfy (7.1)-(7.4) above with f C α at the origin, i.e f(x) f(0) M x α in Ω B ρ, for some M > 0, α (0, 1). Suppose that Ω u Ω are C 2,α at the origin, i.e we assume that on Ω B ρ we satisfy x n q(x ) M x 2+α, u p(x ) M x 2+α, where p(x ), q(x ) are quadratic polynomials.

28 28 O. SAVIN Then u C 2,α at the origin, that is there exists a quadratic polynomial P 0 with det D 2 P 0 = f(0), D 2 P 0 C(M), such that u P 0 C(M) x 2+α in Ω B ρ, where C(M) depends on M, ρ, λ, Λ, n, α. From (7.1) (7.4) we see that p, q are homogenous of degree 2 D 2 p, D 2 q ρ 1. A consequence of the proof of Theorem 7.1 is that if f C α near the origin, then u C 2,α in any cone C θ of opening θ < π/2 around the x n -axis i.e C θ := {x (R n ) + x x n tan θ}. Corollary 7.2. Assume u satisfies the hypotheses of Theorem 7.1 f C α ( Ω) M. Given any θ < π/2 there exists δ(m, θ) small, such that u C 2,α (C θ B δ ) C(M, θ). We also mention the global version of Theorem 7.1. Theorem 7.3. Let Ω be a bounded, convex domain let u : Ω R be convex, Lipschitz continuous, satisfying det D 2 u = f, 0 < λ f Λ in Ω. Assume that Ω, u Ω C 2,α, f C α ( Ω), for some α (0, 1) there exists a constant ρ > 0 such that u(y) u(x) u(x) (y x) ρ y x 2 x, y Ω, where u(x) is understood in the sense of (7.3). Then u C 2,α ( Ω) u C 2,α ( Ω) C, with C depending on Ω C 2,α, u Ω C 2,α, u C 0,1 ( Ω), f Cα ( Ω), ρ, λ, Λ, n, α. In general, the Lipschitz bound is easily obtained from the boundary data u Ω. We can always do this if for example Ω is uniformly convex. The proof of Theorem 7.1 is similar to the proof of the interior C 2,α estimate from [C2], it has three steps. First we use the localization theorem to show that after a rescaling it suffices to prove the theorem only in the case when M is arbitrarily small (see Lemma 7.4). Then we use Pogorelov estimate in half-domain (Theorem 6.4) reduce further the problem to the case when u is arbitrarily close to a quadratic polynomial (see Lemma 7.5). In the last step we use a stard iteration argument to show that u is well-approximated by quadratic polynomials at all scales. We assume for simplicity that f(0) = 1, otherwise we divide u by f(0). Constants depending on ρ, λ, Λ, n α are called universal. We denote them by C, c they may change from line to line whenever there is no possibility of

29 POINTWISE C 2,α ESTIMATES 29 confusion. Constants depending on universal constants other parameters i.e M, σ, δ, etc. are denoted as C(M, σ, δ). We denote linear functions by l(x) quadratic polynomials which are homogenous convex we denote by p(x ), q(x ), P (x). The localization theorem says that the section S h is comparable to an ellipsoid E h which is obtained from B h 1/2 by a sliding along {x n = 0}. Using an affine transformation we can normalize S h so that it is comparable to B 1. In the next lemma we show that, if h is sufficiently small, the corresponding rescaling u h satisfies the hypotheses of u in which the constant M is replaced by an arbitrary small constant σ. Lemma 7.4. Given any σ > 0, there exist small constants h = h 0 (M, σ), k > 0 depending only on ρ, λ, Λ, n, a rescaling of u u h (x) := u(h1/2 A 1 h x) h where A h is a linear transformation with so that a) b) det A h = 1, A 1 h, A h k 1 log h, B k Ω h S 1 (u h ) B + k 1, S 1 (u h ) := {u h < 1}, det D 2 u h = f h, f h (x) 1 σ x α in Ω h B k 1, c) On Ω h B k 1 we have where q h is a quadratic polynomial. x n q h (x ) σ x 2+α, q h (x ) σ, u h p(x ) σ x 2+α, Proof. By the localization theorem Theorem 3.1, for all h c, satisfies with S h := {u < h} Ω, ke h Ω S h k 1 E h, E h = A 1 h B h 1/2, A hx = x ν h x n ν h e n = 0, A 1 h, A h k 1 log h. Then we define u h as above obtain hence where Then S 1 (u h ) = h 1/2 A h S h, B k Ω h S 1 (u h ) B + k 1, Ω h := h 1/2 A h Ω. det D 2 u h = f h (x) = f(h 1/2 A 1 h x),

30 30 O. SAVIN f h (x) 1 M h 1/2 A 1 h x α M(h 1/2 k 1 log h ) α x α σ x α if h 0 (M, σ) is sufficiently small. Next we estimate x n h 1/2 q(x ) u h p(x ) on Ω h B k 1. We have or If x k 1 then x Ω h y := h 1/2 A 1 h x Ω, h 1/2 x n = y n, h 1/2 x = y ν h y n. y k 1 h 1/2 log h x h 1/4, if h 0 is small hence, since Ω has an interior tangent ball of radius ρ, we have Then thus We obtain y n ρ 1 y 2. ν h y n k 1 log h y 2 y /2, 1 2 y h 1/2 x 3 2 y. x n h 1/2 q(x ) h 1/2 y n q(y ) + h 1/2 q(h 1/2 y ) q(x ) Mh 1/2 y 2+α + Ch 1/2 ( x ν h x n + ν h x n 2) 2Mh (α+1)/2 x 2+α + Ch 1/2 ( h 1/2 log h x 3 + h log h 2 x 4) σ x 2+α, if h 0 is chosen small. Hence on Ω h B k 1, x n q h (x ) σ x 2+α, q h := h 1/2 q(x ), q h σ, also u h p(x ) h 1 u(y) p(y ) + p(h 1/2 y ) p(x ) Mh 1 y 2+α + C ( x ν h x n + ν h x n 2) 2Mh α/2 x 2+α + C (h 1/2 log h x 3 + h log h 2 x 4) σ x 2+α. In the next lemma we show that if σ is sufficiently small, then u h can be wellapproximated by a quadratic polynomial near the origin.

31 POINTWISE C 2,α ESTIMATES 31 Lemma 7.5. For any δ 0, ε 0 there exist σ 0 (δ 0, ε 0 ), µ 0 (ε 0 ) such that for any function u h satisfying properties a), b), c) of Lemma 7.4 with σ σ 0 we can find a rescaling with that satisfies a) in Ω B 1, ũ(x) := (u h l h )(µ 0 x) µ 2, 0 l h (x) = γ h x n, γ h C 0, C 0 universal, det D 2 ũ = f, f(x) 1 δ 0 ε 0 x α in Ω B1, for some P 0, quadratic polynomial, ũ P 0 ε 0 in Ω B1, det D 2 P 0 = 1, D 2 P 0 C 0 ; b) On Ω B 1 there exist p 0, q 0 such that x n q 0 (x ) δ 0 ε 0 x 2+α, q 0 (x ) δ 0 ε 0, ũ p 0 (x ) δ 0 ε 0 x 2+α, p 0 (x ) = P 0 (x ρ ), 2 x 2 p 0 (x ) 2ρ x 2. Proof. We prove the lemma by compactness. Assume by contradiction that the statement is false for a sequence u m satisfying a), b), c) of Lemma 7.4 with σ m 0. Then, we may assume after passing to a subsequence if necessary that converges to (see Definition 2.3) Then, by Theorem 2.6, u satisfies p m p, q m 0 uniformly on B k 1, u m : S 1 (u m ) R u : Ω R. B + k Ω B + k 1, det D 2 u = 1, { u = p (x ) on {p (x ) < 1} {x n = 0} Ω u = 1 on the rest of Ω. From Pogorelov estimate in half-domain (Theorem 6.4) there exists c 0 universal such that u l P c 1 0 x 3 in B + c 0, where l := γ x n, γ c 1 0, P is a quadratic polynomial such that P (x ) = p (x ), det D 2 P = 1, D 2 P c 1 0.

32 32 O. SAVIN Choose µ 0 small such that hence c 1 0 µ 0 = ε 0 /32, u l P 1 4 ε 0µ 2 0 in B + 2µ 0, which together with p m p implies that for all large m u m l P 1 2 ε 0µ 2 0 in S 1 (u m ) B + µ 0. Then, for all large m, ũ m := (u m l )(µ 0 x) µ 2 0 satisfies in Ω m B 1 ũ m P ε 0 /2, We define clearly On Ω m B 1 we have det D 2 ũ m = f m (x) = f m (µ 0 x), f m (x) 1 σ m (µ 0 x ) α δ 0 ε 0 x α. q m := µ 0 q m, p m := p m γ q m, p m p, q m 0 uniformly in B 1. x n q m (x ) = µ 1 0 µ 0x n q m (µ 0 x ) µ 1 0 σ m µ 0 x 2+α δ 0 ε 0 x 2+α, ũ m p m (x ) = µ 2 0 (u m l )(µ 0 x) p m (µ 0 x ) + γ q m (µ 0 x ) µ 2 0 ( (u m p m )(µ 0 x ) + γ µ 0 x n q m (µ 0 x ) ) σ m µ α 0 (1 + γ ) x 2+α δ 0 ε 0 x 2+α. Finally, we let P m be a perturbation of P such that P m (x ) = p m (x ), det D 2 P m = 1, P m P uniformly in B 1. Then ũ m, P m, p m, q m satisfy the conclusion of the lemma for all large m, we reached a contradiction. From Lemma 7.4 Lemma 7.5 we see that given any δ 0, ε 0 there exist a linear transformation T := µ 0 h 1/2 0 A 1 h 0 a linear function l(x) := γx n with γ, T 1, T C(M, δ 0, ε 0 ),

33 POINTWISE C 2,α ESTIMATES 33 such that the rescaling ũ(x) := (u l)(t x) (det T ) 2/n, defined in Ω R n satisfies 1) in Ω B 1 det D 2 ũ = f, f 1 δ 0 ε 0 x α, for some P 0 with 2) on Ω B 1 we have p, q so that ũ P 0 ε 0, det D 2 P 0 = 1, D 2 P 0 C 0 ; x n q(x ) δ 0 ε 0 x 2+α, q(x ) δ 0 ε 0, ũ p(x ) δ 0 ε 0 x 2+α ρ, 2 x 2 p(x ) = P 0 (x ) 2ρ x 2. By choosing δ 0, ε 0 appropriately small, universal, we show in Lemma 7.6 that there exist l, P such that ũ l P C x 2+α in Ω B 1, l, D 2 P C, with C a universal constant. Rescaling back, we obtain that u is well approximated by a quadratic polynomial at the origin i.e u l P C(M) x 2+α in Ω B ρ, l, D 2 P C(M) which, by (7.3), proves Theorem 7.1. Since α (0, 1), in order to prove that ũ C 2,α (0) it suffices to show that ũ is approximated of order 2 + α by quadratic polynomials l m + P m in each ball of radius r0 m for some small r 0 > 0, then l + P is obtained in the limit as m (see [C2], [CC]). Thus Theorem 7.1 follows from the next lemma. Lemma 7.6. Assume ũ satisfies the properties 1), 2) above. There exist ε 0, δ 0, r 0 small, universal, such that for all m 0 we can find l m, P m so that ũ l m P m ε 0 r 2+α in Ω Br, with r = r m 0. Proof. We prove by induction on m that the inequality above is satisfied with l m = γ m x n, γ m 1, P m (x ) = p(x ) γ m q(x ), det D 2 P m = 1, D 2 P m 2C 0. From properties 1),2) above we see that this holds for m = 0 with γ 0 = 0. Assume the conclusion holds for m we prove it for m + 1. Let define Then v(x) := (ũ l m)(rx) r 2, with r := r m 0, ε := ε 0 r α. (7.5) v P m ε in Ω v B 1, Ω v := r 1 Ω, det D 2 v 1 = f(rx) 1 δ 0 ε.

34 34 O. SAVIN On Ω v B 1 we have which also gives x n r q(x ) = r 2 rx n q(rx ) δ 0 ε x 2+α, δ 0 ε, (7.6) x n 2δ 0 ε on Ω v B 1. From the definition of v the properties of P m we see that in B 1 v P m r 2 (ũ p)(rx) + γ m x n /r q + 2nC 0 x n, the inequalities above property 2) imply (7.7) v P m C 1 δ 0 ε in Ω v B 1, with C 1 universal constant (depending only on n C 0 ). We want to compare v with the solution w : B + 1/8 R, det D2 w = 1, which has the boundary conditions { w = v on B + 1/8 Ω v w = P m on B + 1/8 \ Ω v. In order to estimate u w we introduce a barrier φ defined as φ : B 1/2 \ B 1/4 R, φ(x) := c(β) ( 4 β x β), where c(β) is chosen such that φ = 1 on B 1/2 φ = 0 on B 1/4. We choose the exponent β > 0 depending only on C 0 n such that for any symmetric matrix A with we have (2C 0 ) 1 n I A (2C 0 ) n 1 I, T r A(D 2 φ) η 0 < 0, for some η 0 small, depending only on C 0 n. For each y with y n = 1/4, y 1/8 the function φ y (x) := P m + ε(c 1 δ + φ(x y)) satisfies det D 2 φ y 1 η 0 2 ε in B 1/2(y) \ B 1/4 (y), if ε ε 0 is sufficiently small. From (7.5), (7.7) we see that if δ 0 η 0 /2, This gives using the definition of w we obtain v φ y on (Ω v B 1/2 (y)), det D 2 v det D 2 φ y. v φ y in Ω v B 1/2 (y), w φ y on B + 1/8.