Regularity of the singular set in the fully nonlinear obstacle problem

For the obstacle problem involving a convex fully nonlinear elliptic operator, we show that the singular set in the free boundary stratifies. The top stratum is locally covered by a $C^{1,\alpha}$-manifold, and the lower strata are covered by $C^{1,\log^\varepsilon}$-manifolds. This essentially recovers the regularity result obtained by Figalli-Serra when the operator is the Laplacian.


Introduction
The classical obstacle problem describes the equilibrium shape of an elastic membrane being pushed towards an impenetrable barrier. In its most basic form, the height of the membrane satisfies the following equations ∆u = χ {u>0} and u ≥ 0 in Ω.
Here Ω is a given domain in R d , and χ E denotes the characteristic function of the set E. The right-hand side of the first equation has a jump across the a priori unknown interface ∂{u > 0}, often called the free boundary.
Apart from its various industrial applications, many ideas and techniques developed for the classical obstacle problem have been crucial in the study of other free boundary problems. In this sense, the classical obstacle problem is the prototypical free boundary problem. As a result, it has been studied extensively during the past few decades. For many applications of the classical obstacle problem and some related problems, see Petrosyan-Shahgholian-Uraltseva [PSU] and Ros-Oton [R].
As already observed by Brézis-Kinderlehrer [BK], the solution u enjoys the optimal C 1,1 loc regularity. The interesting problem is to understand the regularity of the free boundary ∂{u > 0}. In that direction, Sakai first gave some results for the planar case in [Sak1,Sak2]. The theory in higher dimensions was developed by Caffarelli in [C1, C2], where he showed that around points on ∂{u > 0} the solution u has two possible types of behaviour. Either it behaves like a half-space solution, in Caffarelli-Serra-Ros-Oton [CSR], and a very general class of unconstrained free boundary problems in Figalli-Shahgholian [FSh] and Indrei-Minne [IM].
The points where the solution behaves like quadratic polynomials are called singular points. As shown by Schaeffer [Sch], the free boundary can form cusps near these points. Nevertheless, certain structural results can be established for singular points.
To be precise, let 1 2 x · A x0 x denote the polynomial modelling the behaviour of u around a singular point x 0 . Depending on the dimension of the kernel of A x0 , the collection of singular points can be further divided into d classes (strata), the kth stratum being Σ k (u) = {x 0 | x 0 is a singular point with dim ker(A x0 ) = k}.
The structural theorem by Caffarelli says that Σ k is locally covered by C 1 -manifolds of dimension k [C2]. His proof was based on the Alt-Caffarelli-Friedman formula in [ACF]. An alternative proof was later found by Monneau [M], using the monotonicity formula bearing his name.
Recently there has been quite some interest in improving this result. In two dimension, Weiss improved the regularity of the manifolds to C 1,α by introducing the Weiss monotonicity formula [W]. Based on the same formula, Colombo-Spolaor-Velichkov [CSV] showed that in higher dimensions the manifolds are C 1,log ε . The best result so far is in Figalli-Serra [FSe]. By applying Almgren's monotonicity formula [Alm], they improved C 1,log ε to C 1,α for the manifolds covering the top stratum Σ d−1 (u). They also showed that each stratum can be further divided into a 'good' part and a 'bad' part, where the former is covered by C 1,1 manifolds, and the latter is of lower dimension.
Despite these exciting new results, almost nothing is known about singular points for obstacle problems involving operators other than the Laplacian. Comparing with the robust argument for regular points, all developments on singular points depend on various monotonicity formulae. These are powerful but restricted, in the sense that they are not expected to hold for nonlinear operators or even for linear operators with coefficients of low regularity. This same obstruction lies behind the lack of understanding of singular points in many other free boundary problems. Consequently, it is important to develop new tools when monotonicity formulae are not available.
In this work, we develop a method for the study of singular points without relying on monotonicity formulae. In particular, this method works for the following obstacle problem involving a convex fully nonlinear elliptic operator F whose derivatives are Hölder continuous: Here Ω is a domain in R d , and the solution u is understood in the viscosity sense, see [CC, L]. For a given boundary data, the solution u is unique and can be obtained either as the least nonnegative supersolution to the the equation F (D 2 u) ≤ 1, or as the largest subsolution to F (D 2 u) ≥ χ {u>0} . Even for the case when F is the Laplacian, our method is interesting as it provides a new approach to the regularity of the singular set. At first reading, it might relieve many technical complications if the reader takes F to be the Laplacian.
Theorem 1.1 states that the singular set of the free boundary in the nonlinear obstacle problem setting enjoys similar regularity properties as in the linear case. The methods developed here rely on linearization techniques, and the hypothesis that F ∈ C 1 is essential in our analysis.
Let us briefly recall the strategy when the operator is the Laplacian. For each point x 0 in the singular set, we study the resclaings u x0,r (·) = u(r·+x0) r 2 as r → 0. Up to a subsequence, they converge to a quadratic polynomial, called the blow-up profile at x 0 . When the operator is the Laplacian, this polynomial is unique in the sense that it is independent of the subsequence r → 0. It models the behaviour of our solution 'at the point x 0 '. A uniform rate of convergence allows the comparison of blow-up profiles at different points. This gives the desired regularity of the covering manifolds.
Up to now, however, even the proof for the uniqueness of the blow-up profile requires monotonicity formulae. Due to the unstable nature of singular points, it is not obvious that the solution cannot behave like completely different polynomials at different scales. This can be ruled out by monotonicity formulae. Once the solution is close to a parabola at a certain scale, a monotone quantity shows that the solution stays close to the same parabola at all smaller scales, leading to uniqueness of the blow-up profile.
Since no monotonicity formula is expected for our problem, we do not have access to the behavior of u at all small scales. Instead, we proceed using an iterative scheme. Suppose the solution is very close to a parabola in B 1 , we need to show that for some ρ < 1, it is even closer to a similar parabola in B ρ . Iterating this argument gives a rate of convergence to the blow-up profile, which in particular gives its uniqueness. Such scheme has been applied to study regularity of solutions of elliptic equations [Sa] as well as regular points along free boundaries [D]. To our knowledge this is the first time it has been applied to singular points along free boundaries.
To be precise, suppose that 0 is a singular point along ∂{u > 0}, and that u is very close to a parabola p in B 1 , in the sense that |u − p| < ε in B 1 for some small ε. Our goal is to show that in B ρ , the solution u can be better approximated. It is natural to look at the normalized solution which solves an obstacle problem withÔ = − 1 ε p as the obstacle. Assume that p takes the form with the coefficients satisfying (1.2) a 1 ≥ a 2 ≥ · · · ≥ a d , and a k ε, then the contact set betweenû and the obstacle concentrates around the subspace From here we need to separate two cases depending on the dimension of this subspace.
When k = 1, this subspace is of codimension 1. In the limit as ε → 0,û effectively solves the thin obstacle problem with 0 as the obstacle along {x 1 = 0}. Letū denote the solution to this problem. After developing new technical tools concerning the directional monotonicity and convexity of solutions, we can show thatū is C 2 at the origin, and the second-order Taylor polynomial ofū gives the approximation of u in B ρ with an error of the order (1 − β)ερ 2 .
When k ≥ 2, in the limit as ε → 0, the effective obstacle lives on a subspace of codimension strictly larger than 1. Here it is more natural to approximate u with the solution to the unconstraint problem F (D 2 h) = 1 in B 1 , and h = u along ∂B 1 .
We show that h 'almost' solves the constrained problem, and its second order Taylor expansion gives the next approximation of u in B ρ with an error of the order (ε − ε µ )ρ 2 for some µ > 1. For ε small, this improvement is much slower than ε → (1 − β)ε. Consequently, we need a much more delicate argument to keep track of the change in the polynomials at each step, essentially saying that if the improvement of error is small, then the change in the polynomials is even smaller.
Combining these two cases together, we get a rate of convergence to the blow-up profile, which allows us to establish the main result Theorem 1.1.
To our knowledge, this is the first structural result for singular points in the obstacle problem with nonlinear operators. We hope that the ideas and techniques developed here can be applied to other types of free boundary problems. This paper is structured as follows. In the next section we provide some preliminary material and introduce some notations. In Section 3 we establish the main new observations of this paper, the improvement of monotonicity and convexity of the solution. With these we prove two lemmata concerning the iterative scheme. In Section 4 we deal with the case when k = 1 as in (1.2). In Section 5 we deal with the case when k ≥ 2. In the last section, we combine these to prove the main result.

Preliminaries and notations
This section is divided into three subsections. In the first subsection we discuss some regularity properties of convex elliptic operators. The main reference for these is Caffarelli-Cabré [CC]. In the next subsection we include some known results on the obstacle problem, mostly from Lee [L]. In the last subsection we recall an expansion of solutions to the thin obstacle problem.
2.1. Fully nonlinear convex elliptic operators. Let S d denote the space of d-by-d symmetric matrices. Our assumptions on the operator F : S d → R are: for all M, P ∈ S d and P ≥ 0.
We call a constant universal if it depends only on the dimension d, the elliptic constant Λ and C F , α F . For a C 2 function ϕ, define the linearized operator L ϕ : where F ij denotes the derivative of F in the (i, j)-entry, and D 2 ϕ is the Hessian of ϕ. One consequence of convexity is As a result, we have the following comparison principle: Proposition 2.1. Let u be a solution to (1.1). Suppose the functions Φ and Ψ satisfy the following equations.
Proof. Define U = Ω ∩ {Φ > 0}. Then (2.4) implies that inside U , we have To see the comparison between u and Ψ, we define V = Ω ∩ {u > 0}. Then (2.4) implies that inside V , we have One corner stone of the regularity theory of fully nonlinear elliptic operators is the Evans-Krylov estimate [CC]: In particular, if u solves (1.1), then in {u > 0} we have enough regularity to differentiate the equation and use convexity of F to get Here e ∈ S d−1 is a unit vector. Here and in later parts of the paper, D e denotes the differentiation in the e-direction. D ee denotes the pure second order derivative in the e-direction. When differentiating along directions of a standard orthonormal basis of R d , we also write D i = D ei and D ij = D ei D ej , where e i is the ith vector in the standard basis. A direct application of the previous theorem gives the following estimate: Proposition 2.2. Let F satisfy the assumptions (2.1), (2.2) and (2.3). If for some universal α ∈ (0, 1) and 0 < C < +∞.
If w also solves F (D 2 w) = 1 in B 1 , then for a universal constant α ∈ (0, 1) we have where C further depends on v L ∞ (B1) and w L ∞ (B1) .
Proof. For the first statement in the proposition, we directly apply the previous theorem to the operator This satisfies all assumptions in Theorem 2.1. The difference v − p solves For the second statement of the proposition, we first apply Theorem 2.1 to v and w, which gives In B 1 the difference v − w solves the linear equation By the previous estimate, this is Hölder continuous. We apply the standard Schauder theory to get the desired estimate.
Next we give an estimate for solutions to linear equations with coefficients which are close to being constant in a large portion of the domain. This is relevant in our analysis since often the linearized operators considered are perturbations of the constant coefficient operator L p . Proposition 2.3. Let Ω be a Lipschitz domain. Suppose v,w are C 2 solutions to the uniformly elliptic linear equations with ϕ Hölder continuous, and with coefficients a ij (x) measurable, b ij constant. If where ω(·) is a modulus of continuity, which depends on the ellipticity constants, the domain Ω and ϕ C α .
Proof. The proposition follows from the perturbative methods developed in [CC].
Here we only sketch a proof by compactness.
The global version of Harnack inequality implies that v, w are uniformly Hölder continuous in Ω. Now we consider a sequence of η k → 0, and the corresponding solutions v k , w k (for equations with coefficients a ij k (x), b ij k ) . Then, up to subsequences, they must converge uniformly to a solution of the same constant coefficient equation. The limiting solutions must coincide since they have the same boundary data, and the conclusion follows.

2.2.
Known results for the obstacle problem. In this subsection we include some classical results concerning the obstacle problem (1.1). Most of the results here can be found in Lee [L].
We begin with the optimal regularity of the solution: Proposition 2.4. Let u be a solution to (1.1). Then for a compact set K ⊂ Ω, for some C depending on universal constants, K, and u L ∞ (Ω) .
A direct consequence is that in the contact set {u = 0}, we have (2.6) ∇u = 0 and D 2 u ≥ 0 in the viscosity sense.
We have the following almost convexity estimate Proposition 2.5. Let u be a solution to (1.1) in Ω = B 1 with u(0) = 0. Then for some universal constants δ 0 > 0 and C, The free boundary decomposes into the regular part and the singular part with Reg(u) given locally by a C 1,α surface which separates the 0 set from the positivity set.
Define the thickness function of a set E, δ E (·), as is the infimum of distances between two pairs of parallel hyperplanes such that E ∩ B r is contained in the strip between them. Geometrically the singular set Σ(u) is characterized by the vanishing thickness of the zero set: Proposition 2.6. Let u be a solution to (1.1) in B 1 with 0 ∈ Σ(u). There is a universal modulus of continuity σ 1 such that In particular, if 0 ∈ Σ(u), the zero set {u = 0} cannot contain a nontrivial cone with vertex at 0.
Another characterization of the singular set is that at points in Σ(u) the solution is approximated by quadratic polynomials.
For this, we define the following class of polynomial solutions to the obstacle problem. We also define the class of convex polynomials that do not necessarily satisfy the non-negative constraint.
Definition 2.1. The class of quadratic solutions is defined as The class of unconstraint convex quadratic solutions is defined as Here and in later parts of the paper, x · y denotes the standard inner product between two vectors x and y.
Note that for a polynomial p ∈ UQ, D 2 p ≥ 0. Ellipticity (2.3) then gives for some universal C. For points in Σ(u), we have the following uniform approximation by quadratic solutions: Proposition 2.7. Let u be a solution to (1.1) in B 1 with 0 ∈ Σ(u). There is a universal modulus of continuity σ 2 such that for each r ∈ (0, 1/2), there is p r ∈ Q satisfying u − p r L ∞ (Br) ≤ σ 2 (r)r 2 . Combining Proposition 2.5 and Proposition 2.7, we know that after some rescaling, our solution is in the following class: Definition 2.2. Given ε, r ∈ (0, 1) and p ∈ UQ, we say that u is ε-approximated by the polynomial p in B r , and use the notation where c 0 = 1 16Λ 2 . The universal bound 0 ≤ D 2 p ≤ C I for p ∈ U Q immediately gives a universal bound on the size of u whenever u ∈ S(p, ε, r) : where C is universal.
2.3. The thin obstacle problem. In this subsection we discuss solutions to the thin obstacle problem. In certain cases, our solution converges to them after normalization. Readers interested in the thin obstacle problem may consult Athanasopoulos-Caffarelli-Salsa [ACS] or Petrosyan-Shahgholian-Uraltseva [PSU]. In its most basic form, the thin obstacle problem is the following system: Here x 1 denotes the first coordinate function of R d . For solutions to this problem, we have the following effective expansion according to frequencies at 0: Proposition 2.8. Let v be a non-trivial solution to (2.9) with v(0) = 0.
Then one of the following three possibilities happens for v: (2) For some r > 0 and e ∈ S d−1 ∩ {x 1 = 0}, For a real number x, x + and x − denote the positive and negative parts of x respectively. Recall that D e denotes the differentiation in the e-direction.
If ϕ = 1, then v blows up to a 1-homogeneous solution to (2.9). In this case, possibility (1) as in the statement of the lemma holds.
For details, the reader may consult [ACS] or [PSU].

Improvement of monotonicity and convexity
In this section are some new observations concerning the directional monotonicity and convexity of solutions to the obstacle problem. They are at the heart of the further development of the theory.
Roughly speaking, if the solution is 'almost' monotone/ convex in B 1 and strictly monotone/ convex away from the free boundary, then the results here imply that the solution is indeed monotone/ convex in B 1/2 . As already evident in the classical work of Caffarelli [C1], it is of fundamental importance to develop such tools to transfer information away from the free boundary to the full domain.
Before we state the main results of this section, we begin with the construction of a barrier function. In the following lemma, γ is the constant such that Here I is the identity matrix. By (2.3), 1 Λ ≤ γ ≤ Λ. Lemma 3.1. For 0 < η < r < 1 and N > 8γ r 2 , let w be the solution to the following system There isη depending on r, N and universal constants, such that if η <η, then for all x 0 ∈ B r/2 , w x0 satisfies , for some C r depending on universal constants and r.
Consequently, there is a modulus of continuity ω, depending on universal constants, N and r, such that C r ϕ L ∞ (B 7 8 r ) ≤ ω(η) whenever η <η. Thus the previous estimate gives In order to prove the claim (3.2) we notice that ϕ satisfies a linear elliptic equation with ellipticity constant Λ. Also, ϕ vanishes on ∂B r except on ∂B r ∩ {|x 1 | ≤ η} where N ≥ ϕ ≥ 3 4 N . We extend ϕ = 0 outside B r , and by the weak Harnack inequality it follows that max ϕ decreases geometrically on the outward dyadic regions centered around a point y ∈ ∂B r ∩ {x 1 = 0}, We easily obtain the claim (3.2) as we let η → 0.
. The conclusion follows from (3.3) by using With this we prove the following improvement of monotonicity lemma. Recall the class of solutions S(p, ε, r) is defined in Definition 2.2, and that D e is the differentiation along direction e.
Lemma 3.2. Suppose u ∈ S(p, ε, r) satisfies the following for some constants K, σ, and 0 < η < r, and a direction e ∈ S d−1 : Proof. Choose c > 0 small, depending on universal constants and σ, such that Then define N = max{4K/c, 10 u L ∞ (Br) }, depending only on K, σ and universal constants since we have the universal bound (2.8).
Letη be the constant given in Lemma 3.1, depending on N and r. Let w x0 be the barrier as in that lemma. Assume η <η.
If we define U = B r ∩ {u > 0} and pick and cε(u − w x0 )(x) ≤ 0 along ∂{u > 0}. Our assumptions on D e u and (2.6) imply Since this is true for all A slightly different version is also useful: Lemma 3.3. Suppose u ∈ S(p, ε, r) satisfies the following for some constants K, σ, and 0 < η < r, and a direction e ∈ S d−1 : There isη, depending on universal constants, r, σ and K, such that if η ≤η, Proof. The proof is almost the same as the previous proof. The only difference happens for the comparison along the boundary Thus along this piece of the boundary we still have Finally we have the following improvement of convexity estimate: Lemma 3.4. Suppose u ∈ S(p, ε, 1). There is a universal constant C such that if D ee p ≥ Cε along some direction e ∈ S d−1 , then Proof. Let γ be the constant as in (3.1), and c 0 be the constant as in Definition 2.2. For Note that by (2.4) and (2.5), Thus once the claim is proved, h ≥ 0 in U by maximum principle. In particular, we have D ee u(x 0 ) ≥ 0. Together with (2.6), we have D ee u ≥ 0 in the entire B 1/2 . Therefore, it suffices to prove the claim. First we note that along ∂{u > 0}, D ee u ≥ 0 and u = 0, thus h ≥ 0 along this part of ∂U .
We divide the other part ∂B 3/4 ∩ {u > 0} into two pieces It remains to deal with y 0 ∈ ∂B 3/4 ∩ {u > 1 64 γ}. Firstly the universal bound (2.8) and Proposition 2.4 give a universal r 0 > 0 such that dist(y 0 , {u = 0}) ≥ r 0 . In particular F (D 2 u) = 1 in B r0 , and we can apply Proposition 2.2 to get Again note the universal bound on max u as in (2.8), if we choose C universally large, then h ≥ 0 on this last piece of ∂U . This completes the proof for the claim.

Quadratic approximation of solution: Case 1
In this section and the next, we use the technical tools developed in the previous sections to study the behaviour of our solution near a singular point, say, 0 ∈ Σ(u).
The classical approach is to study the rescales of u, u r (x) = 1 r 2 u(rx) as r → 0. Proposition 2.4 gives enough compactness to get convergence of u rj to some quadratic polynomial, say p, along a subsequence r j → 0. If the limit does not depend on the particular subsequence, then there is a well-defined stratification of Σ(u) depending on the dimension of ker(D 2 p). If there is a rate of convergence of u r → p, then we get regularity of the singular set near 0 ∈ Σ(u).
With the help of monotonicity formulae, this program has been executed with various degrees of success in [C2], [CSV], [FSe], [M] and [W]. One idea behind these works is that once u r0 is close p for a particular r 0 , then monotonicity formulae imply u r remains close to p for all r < r 0 .
Since no monotonicity formula is available in our problem, we do not have access to all small scales. Instead, we proceed by performing an iterative scheme. Let ρ ∈ (0, 1), the building block of this scheme is to study the following question: If u is close to p in B 1 , can we approximate u better in B ρ ?
Quantitatively, we seek to prove the following: If |u − p| < ε in B 1 for some small ε, then we can find a quadratic polynomial q such that |u − q| < ε ρ 2 in B ρ , where ε < ε.
The rate of decay ε → ε is linked to the rate of convergence in the blow-up procedure.
To this end, we need to separate two different cases. Let λ 1 ≥ λ 2 ≥ · · · ≥ λ d ≥ 0 denote the eigenvalues of D 2 p. Depending on their sizes, the contact set {u = 0} concentrates along subspaces of various dimensions. When λ 2 ≤ Cε, p ∼ 1 2 (x · e) 2 and the contact set concentrates along a (d − 1)dimensional subspace {x · e = 0}. When λ 2 ε, the contact set concentrates along a subspace with higher co-dimension.
In this section, we deal with the first case when λ 2 ≤ Cε.
In this caseû ε converges to the solution of the thin obstacle problem (2.9), and in particular Proposition 2.8 applies to the limitû 0 . To showû 0 is C 2 near 0, we need to rule out possibilities (1) and (2) as in the statement of Proposition 2.8. This can be achieved using explicit barriers and the lemmata in the previous section.
The class of quadratic solutions Q and the class of well-approximated solution S(p, ε, 1) are defined in Definition 2.1 and Definition 2.2.
Here and in later parts of the paper, λ j (M ) denotes the jth largest eigenvalue of the matrix M .
Remark 4.1. The parameter κ will be chosen in the final section, depending only on universal constants. After that, all constants in this lemma become universal.
We begin with some preparatory lemmata. Then where L depends only on universal constants and κ.
Proof. Define the normalizationû Since F (D 2 u) ≤ 1 and F (D 2 p) = 1 in B 1 , by (2.4) we have Up to a rotation, the polynomial p is of the form with a 1 ≥ a 2 ≥ · · · ≥ a d ≥ 0 and a 2 ≤ Cκε for some universal constant C. Then for some C depending only on universal constants and κ. Now the result easily follows from this and the fact thatû ∈ C 1,1 satisfies (4.1). Indeed, after a linear deformation, we can assume that L p = and the inequality on D eeû is still satisfied after relabeling the constant C. Then ∆û ≤ 0 implies that we also have D 11û ≤ C in B 1 .
This lemma provides us with enough compactness for a family of normalized solutions. Actually it even allows us to consider a family of nomalized solutions to the obstacle problem involving a family of different operators. This is necessary to get uniform estimates.
To fix ideas, let {F j } be a sequence of operators satisfying the same assumptions that we have on our operator F , namely, (2.1), (2.2) and (2.3).
For each F j , there is a unique γ j such that F j (γ j e 1 ⊗ e 1 ) = 1.
Ellipticity implies γ j ∈ [1/Λ, Λ]. Define the associated polynomial Then we have the following lemma, that identifies the problem solved by the limit of nomalized solutions: Lemma 4.3. Let F j be a sequence of operators satisfying the same assumptions as in (2.1), (2.2) and (2.3). Let u j solve the obstacle problem (1.1) with operator F j in B 1 .
Suppose for some constant κ > 0 and a sequence ε j → 0, there are polynomials Then up to a subsequence, the normalized solution converges locally uniformly in B 1 to someû ∞ , where q j is the polynomial as in (4.2). Moreover, up to scaling,û ∞ solves the thin obstacle problem as in (2.9).
Proof. Lemma 4.2 gives locally uniform C 0,1 bound on the family {û j }. Consequently, up to a subsequence they converge to someû ∞ locally uniformly in B 1 . Define the operator G j by By uniform C 1,α F estimate on the family {F j }, up to a subsequence G j locally uniformly converges to some linear elliptic operator. Up to a scaling, we assume this limiting operator is the Laplacian.
If x 0 ∈ {û ∞ > 0}, thenû j > 0 in a neighborhood of x 0 for large j. Note that q j ≥ 0, thus u j > 0 in a neighborhood of x 0 for large j. Thus G j (D 2û j ) = 0 in a neighborhood of x 0 for all large j. Consequently, ∆û ∞ (x 0 ) = 0. That is, It remains to show thatû ∞ ≥ 0 along {x 1 = 0}. For this, simply note that u j ≥ 0 and q j = 0 for all j along {x 1 = 0}. Now we start the proof of Lemma 4.1. As explained at the beginning of this section, the normalized solutions converge to a solution to the thin obstacle problem.
The key to the improvement in approximation is the show this limit is C 2 at 0, that is, possibilities (1) and (2) as in Proposition 2.8 cannot happen.
Suppose there is noε > 0 satisfying the statement of the lemma. For a sequence of ε j → 0, a sequence of operators F j satisfying the assumptions (2.1), (2.2) and (2.3), we have a sequence of solutions to (1.1) with these operators such that u j ∈ S(p j , ε j , 1) for some p j ∈ Q with λ 2 (D 2 p j ) ≤ κε j , and 0 ∈ Σ(u j ), but u j ∈ S(q, (1 − β)ε j , r) for any q ∈ Q and r ∈ (r, 1/2).
Up to a rotation, we assume where q j (x) = 1 2 γ j x 2 1 with F j (γ j e 1 ⊗ e 1 ) = 1. Then up to a scaling, Lemma 4.3 shows that up to a subsequence, u j →û locally uniformly in B 1 , whereû solves the thin obstacle problem (2.9).
We show that possibilities (1) and (2) of Proposition 2.8 cannot happen forû.
Suppose it happens, then we havê Assume that a + > 0, and then we use a barrier to show that u(0) > 0, contradicting 0 ∈ Σ(u).
For this we choose r small such that This means thatû j satisfies the same inequality (4.3) above for all j large enough. For notational simplicity, we omit the subscript j in the computations below. Define the barrier function Φ(x 1 , x ) = 1 2 (γ + Λ 2 dε)(x 1 + ε 2 ) 2 − 1 2 ε|x | 2 . and notice thatΦ We compare u and Φ on the boundary of the set In conclusion, Φ ≤ u along the boundary, and Proposition 2.1 gives u ≥ Φ in the interior of the domain.
In particular u(0) ≥ Φ(0) > 0, contradicting 0 ∈ Σ(u). Therefore we have a ± ≤ 0. Next we show that a ± cannot be negative. Suppose that a + < 0, and in this case, we use a barrier to prove that {u = 0} contains a cone with positive opening and with vertex at 0. With Proposition 2.6, this contradicts 0 ∈ Σ(u). Since a − ≤ 0, we can choose r small such that near ∂U r . We compare u and Φ on the boundary of the set U r where with A := a + /(2γ) and |ξ | ≤ r/2. Sincê we find thatΨ >û, hence Ψ > u on ∂U r for all ε small. From here Proposition 2.1 becomes applicable and gives u ≤ Ψ in U .
In particular this gives u(−Aε, ξ ) = 0 for all |ξ | ≤ r/2. Now note that with u ∈ S(p, ε, 1), we have D 2 u ≥ −c 0 ε in B 1 . Also since e 1 is the direction corresponding to the largest eigenvalue of D 2 p, there is a cone of directions around e 1 , say, K ⊂ S d−1 with a universal positive opening such that D ee p > c > 0 for all e ∈ K. For small ε we can then apply Lemma 3.4 to get Together with u(0) = 0 and u(−Aε, ξ ) = 0 for all |ξ | ≤ r/2, this implies that the coincidence set {u = 0} contains a cone of positive opening with vertex at 0, contradicting Proposition 2.6.
This finishes the proof of Step 1.
Step 2: Possibility (2) in Proposition 2.8 does not happen forû. Suppose it happens, then for some r > 0 and ν ∈ S d−1 ∩ {x 1 = 0}, we have Therefore, there is some σ > 0 such that Consequentlyû j →û in C 1,α loc (B 1 \{x 1 = 0}). Therefore, for large j, That is, Thus, we have established Also, by Lemma 4.2, we have D ν u j ≥ −Lε j in B r . Now take η depending on K = L, r and σ as in Lemma 3.3. Note the convergence ofû j →û in C 1,α loc (B 1 \{x 1 = 0}) implies By the C 1,α -regularity of u j , there is a cone of directionsK ⊂ S d−1 around ν with positive opening such that for all e ∈K, we have

Thus Lemma 3.3 applies and gives
D e u j ≥ 0 in B r/2 for all e ∈K.
After the previous two steps, we have that the limiting profileû falls into possibility (3) as in Proposition 2.8. Consequently, for some δ > 0 to be chosen later, there is r > 0 such that where trace(A) = 0, and e · Ae ≥ 0 for all e ∈ S d−1 ∩ {x 1 = 0}. Locally uniform convergence ofû j →û gives for large j Here we omit the index j for the sake of simplicity, then we have With Cauchy-Schwarz inequality and e · Ae ≥ 0 for all e ∈ S d−1 ∩ {x 1 = 0}, we see that there is a constant C, depending on |A|, such that (4.5) D 2 q + Aε + Cε 2 ≥ 1 4 γe 1 ⊗ e 1 for ε small. Now note that we are assuming, after necessary scaling, that F ij (D 2 Consequently, there is t ∈ [−C, C] such that the polynomial Meanwhile, by (4.5) we have for ε small. Thus D 2 p ≥ 0 and p ∈ Q. Finally (4.4) implies that in B r , for all ε small.
It remains to show that (see (2.7)) where c 0 is the constant as in Definition 2.2.
For fixed e ∈ S d−1 , define w = D ee u + c 0 ε. Then u ∈ S(p, ε, 1) implies w ≥ 0 in B 1 . Now for small ε, u > 0 in B 1 4 r ( 1 2 re 1 ). Thus Proposition 2.2 implies for a universal C. Now fix δ, depending on universal constants and κ, such that the right-hand side is less than 1 2 c 0 ε. Then In particular Λ is the maximal Pucci operator [CC], then Meanwhile, along ∂{u > 0}, (4.7) w ≥ c 0 ε by (2.6). Thus w ≥ εΦ along ∂{u > 0}. In conclusion, Together with (4.6) and (4.7), this implies Meanwhile, there is a constant β ∈ (0, 1), depending on universal constants and κ, such that This contradicts our construction of u at the beginning of this proof.

Quadratic approximation of solution: Case 2
In this section, we prove a version of Lemma 4.1 but for u ∈ S(p, ε, 1) where λ 2 (D 2 p) ε. Here the situation is different since the zero set {u = 0} concentrates around subspaces of codimension at least 2, say This brings technical challenges as the normalized solutionû = 1 ε (u − p) now solves an obstacle problem with an obstacleÔ = − 1 ε p whose capacity converges to 0 as ε → 0.
We define h to be the solution to the unconstrained problem We will show thatû is well approximated in L ∞ by the corresponding functionĥ, but only away from a tubular neighborhood around the (d−k)-dimensional subspace above (see Lemma 5.2). Inside this neighborhood, the difference betweenĥ andû could be of order 1, andû has no longer a uniform modulus of continuity (as ε → 0) in B 1/2 as in the codimension 1 case. Heuristically, as ε → 0, we end up with limiting functionsū,Ō andh, that satisfy that |h|, |ū| and maxŌ are all bounded by 1 in B 1 , and 1)h is a solution to a constant coefficient elliptic equation, 2) the obstacleŌ is a concave quadratic polynomial supported on the x -subspace, extended to −∞ outside its support, 3)ū = max{h,Ō}, which can be discontinuous.
The improved quadratic error forû cannot be deduced right away from the C 2,α estimate ofh at the origin. This will follow after we show that 0 ∈ Σ(u) essentially implies thath andŌ are tangent of order 1 at the origin in the x direction andŌ can only separate on top ofh in this direction by a small quadratic amount.
It turns out that the improvement in convexity and approximation is much slower. Instead of ε → (1 − β)ε as in Lemma 4.1, we only have an improvement of the form ε → (ε − ε µ ), where µ > 1, universal. This is consistent with C 1,log εregularity of covering for lower strata in the classical obstacle problem [CSV]. This slow rate of improvement could a priori break the convergence of the polynomials p k and the uniqueness of the blow-up profile, as well as the iteration scheme. Suppose p k is the approximating quadratic polynomial in the kth iteration. Then a rate of ε → (1 − β)ε implies The summability of this sequence implies the convergence of D 2 p k . When the rate is ε → (ε − ε µ ), this is not true anymore.
In the next section, we establish the convergence of D 2 p k by working instead with the corresponding approximations D 2 h k (0). These are not necessarily positive definite, but still approximate u quadratically with error proportional to ε k . The main point is that the series is convergent, which is a consequence of the main result of this section, Lemma 5.1 below. This lemma provides a dichotomy concerning the rate of the quadratic improvement between two consecutive balls. Essentially it says that either we have a fast improvement as in Lemma 4.1, or the difference between consecutive errors ε k is bounded below by the difference between u and h at some point away from the x subspace, which could be as small as ε µ . We recall that by Definition 2.2, u ∈ S(p, ε, 1) means that u solves (1.1), and |u − p| ≤ εr 2 and D 2 u ≥ −c 0 εI in B r .
The dichotomy is dictated by the behavior of the matrix D 2 h(0) along the x subspace. If D 2 x h(0) ≥ − c0 8 ε I, then we end up in alternative (1), otherwise we end up in (2).
Suppose that we are in the slow improvement situation (2). Let h denote the solution to (5.2) in the ball B ρ . By maximum principle, we have u ≥ h ≥ h in the common domain. The Harnack inequality for the difference h − h and Proposition 2.2 imply for some universal C 1 . Iteratively, the series in (5.3) is bounded from above by a telescoping sum. Thus its convergence is justified.
Recall that λ 2 (M ) denotes the second largest eigenvalue of a matrix M . A technical point is that we are working in the class of quadratic polynomials p ∈ UQ defined in Definition 2.1 which have a linear part as well.
Up to a rotation, p takes the form with a 1 ≥ a 2 ≥ · · · ≥ a d ≥ 0, a 2 ≥ κ 0 ε and F ( a j e j ⊗ e j ) = 1. Throughout this section we assume that p is of this form.
For a positive constant η, we define the following cylinder We first show that u is well approximated by h outside this cylinder.
From here, we first choose η such that C(η)ω(η ) < 1 2 η, and then choose ε such that C(η)ω(ε) < 1 2 η. This gives the desired estimate. We now give the proof of the main result in this section: Proof of Lemma 5.1. As discussed above, we define the normalizationŝ Then in B 1 , we have −1 ≤ĥ ≤û ≤ 1,û(0) =Ô(0) = 0, and Proposition 2.2 implies (5.6) ĥ C 2,α (B 3/4 ) ≤ C for some universal constant C. We divide the technical proof into 6 Steps. Here we give an outline first. We decompose the space x = (x , x ) according to the curvatures of the obstaclê O. The curvatures are very negative along the directions in the x -subspace, and are uniformly bounded in the x -subspace. In Steps 1-2 we show thatĥ andÔ are "essentially tangent" in the x direction at the origin, and deduce thatÔ can only slightly separate quadratically on top ofĥ near the origin. In Step 3, we show that the same is true forû. In Step 4, we use the C 2,α estimate forĥ to approximate u quadratically in B ρ by a polynomial p ∈ UQ with an improved error ε 2 . The convexity estimate for D 2 u in B ρ (see (2.7) in Definition 2.2) is given in Steps 5 and 6, according to whether or not the obstacleÔ separates quadratically on top ofĥ along some direction in the x subspace. This leads to our dichotomy.
Throughout this proof, there are several parameters to be fixed in the end. The radius ρ ∈ (0, 1/2) depends only on universal constants. The parameter δ > 0 can be made arbitrarily small, and will be chosen to be universal. The parameter η from Lemma 5.2, which depends on δ, allows us to makeû andĥ very close to each other. This η imposes the choice of κ 0 = κ η as in Lemma 5.2. The parameterε is chosen after all these.
The obstacleÔ is changing rapidly in the x direction, and we denote by x the point in this direction where its maximum is achieved, which is the same as the minimum point for p in the x direction.
We write p as the sum of two quadratic polynomials in the x and x variables where p 1 ≥ 0 is the homogenous of degree 2 polynomial The obstacleÔ satisfies Step 1: If η and ε are small depending on δ, then The idea is to show that otherwise u is monotone in a cone of directions near the x subspace, and we contradict 0 ∈ Σ(u).
Suppose there is i > k such that D i (ĥ −Ô)(0) > δ. By |D iiÔ | ≤ C δ and the universal estimate (5.6), we have for some r > 0 depending only on δ.
Together with |û| ≤ 1, this implies By continuity, there exists of a cone of directions,K ⊂ S d−1 , with positive opening around e i , such that for all e ∈K, Define the constantη as in Lemma 3.2 depending on r, K = −2C δ , and σ = 1 8 δ. If we choose η < min{η, 1 8 δ}, then Lemma 5.2 gives for all e ∈K. This implies that {u = 0} contains a cone of positive opening with vertex at 0, contradicting Proposition 2.6.
Step 2: If η and ε are small depending on δ, then IfÔ is a bit larger thanĥ at (x , 0) we show thatû coincides with the obstaclê O in a small neighborhood of (x , 0) hence u = 0 in this neighborhood. On the other hand, by Lemma 3.4, u is convex in the directions close to the x subspace, and we obtain that {u = 0} contains a cone with vertex at the origin and reach a contradiction. Next we provide the details.
We show that u = 0 in a neighborhood of (x , 0) by using barriers. Sinceĥ has universal Lipschitz norm, and |D x Ô 2 | ≤ C δ , there is r > 0 depending on δ, such thatĥ Consequently for η small, Lemma 5.2 implies Let Ω := {|x 1 | < η} ∩ B r (x , 0). We define the barrier for some B depending on δ and r. Note that Ψ(x , 0) = 0, and thus Ψ ≥ 0 if ε is small. We choose B large such that by ellipticity (2.3) and on ∂Ω ∩ ∂B r (x , 0), for η sufficiently small, Thus if η is small, then Consequently, we can apply Proposition 2.1 to Ψ in Ω to get u ≤ Ψ in Ω.
In particular, we have u(x , 0) = 0. Now note that along ∂Ω, we have u < Ψ − 1 8 δε. Thus we can translate Ψ a small amount and still preserve the comparison u ≤ Ψ along ∂Ω. This gives for a small r > 0.
The inequality holds outside C η by Lemma 5.2. It remains to establish it in C η . First we use Steps 1 and 2 to show that a similar inequality holds forÔ: (5.11)Ô ≤ĥ + c 0 |x | 2 + 3δ =: g in B 1/2 .
Then we use barriers to extended the inequality fromÔ toû. Note that in B 1/2 \C η , Lemma 5.2 gives By choosing η small, and using that D 2Ô is constant together with the Hölder continuity of D 2ĥ (see (5.6)), we extend the estimate to the full ball Also, the conclusion (5.8) of Step 1 together with the second estimate in (5.13) and |x | ≤ δ 2 give Now it is easy to check that the estimates (5.12)-(5.15) together with the conclusion (5.9) of Step 2 imply the claim (5.11).
To this end, pick a point x * ∈ B 1/4 ∩ C η and r > 0 depending on δ such that We compare u and the barrier function in the set Ω. We have v ≥ g + 1 2 δ − 2Λ 2 Bη 2 ≥ g if η is small, thus Ψ ≥ p + εg ≥ p + εÔ = 0.
Next we improve the convexity of u. There are two cases to consider, corresponding to the two alternatives as in Lemma 5.1.
The inequality at the origin can be extended to a fixed neighborhood by continuity and then, by Lemma 5.2 transferred to D 2 u away from the cylinder C η . This can be further extended to the whole domain by using that pure second derivatives of u are global supersolutions.
More precisely, our hypothesis together with (5.6) and the fact that D 2Ô is constant imply that the inequality holds in a small ball B c , c universal, with − 1 4 c 0 I as the right hand side. Using (5.13), we can extend the inequality to the full Hessian, By choosing η small, Lemma 5.2 gives From here we can apply the same argument as in Step 4 of the proof for Lemma 4.1 to get D 2 u ≥ −(1 − β)c 0 ε I in B ρ . This corresponds to the first alternative as in Lemma 5.1.
The key observations are that u − h is a subsolution and D ee u + c 0 ε is a supersolution for the same linearized operator L u , and that the two functions can be compared in the domain B 1 ∩ {u > 0}. On the other hand u − h is a global supersolution for L h , and then its minimum in B 1/2 is controlled below by its value at any given point that is not too close to C η . The hypothesis at the origin is used to guarantee that this minimum value for u − h in B 1/2 is at least ε µ . Now we provide the details.
By (5.6) and the fact that D ξξÔ is constant, we conclude D ξξ (ĥ −Ô) < − 1 16 c 0 in B c for a universal c > 0. Together with Step 2, this implies the existence of some With the universal Lipschitz regularity of h, we get for some small universal c, c > 0. Note that L h (u − h) ≤ 0 in B 1 as in (2.4), u = h on ∂B 1 . We compare u − h to the corresponding solution of the maximal Pucci operator in B 1 \ B c ε (x * ) and obtain as a consequence of Harnack inequality u − h ≥ ε µ in B 1/2 , for some universal µ > 1. Moreover, since u − h solves a linear equation away from C η , the same argument combined with Harnack inequality imply that

As in
Step 4 of the proof for Lemma 4.1, for e ∈ S d−1 , we define This is a nonnegative function satisfying L u (w) ≤ 0 in B 1 ∩ {u > 0}. Note that w ≥ c 0 ε along ∂{u > 0}, and 2ε ≥ u − h in B 1 , hence Combining this with (5.17) we find Define the right-hand side to be −c 0 ε , then Also, (u − h)( 1 2 ρe 0 ) = C(ε − ε ) as in the second alternative in Lemma 5.1.

Iteration scheme and proof of main result
Lemma 4.1 and Lemma 5.1 form the basic building blocks of the iteration scheme that we perform to prove the main result. As mentioned in Introduction and at the beginning of Section 4, such iteration scheme compensates the absence of monotonicity formulae.
In the following proposition, we give the details of this iteration when the approximating polynomial p satisfies λ 2 (D 2 p) ε. Again λ 2 (M ) denotes the second largest eigenvalue of the matrix M . The proposition implies that once this condition is satisfied, it holds true for all approximating polynomials in the iteration.
Proposition 6.1. Suppose u ∈ S(p, ε, 1) for some p ∈ U Q. There are universal constantsε, c > 0 small and κ, C large, such that if ε <ε and λ 2 (D 2 p) ≥ κε, then there is q ∈ Q with |D 2 q − D 2 p| < Cε such that Proof. We will takeε as in Lemma 5.1, and take κ to be much larger than κ 0 as in that lemma.
Define u 0 = u, p 0 = p, and ε 0 = ε. Let h 0 be the solution to We apply Lemma 5.1 to get a p ∈ UQ and ε such that u 0 ∈ S(p , ε , ρ).
In general, once u k , p k , ε k are found satisfying u k ∈ S(p k , ε k , 1) with ε k <ε and λ 2 (D 2 p k ) ≥ κ 0 ε as in Lemma 5.1, we apply that lemma to get p ∈ UQ such that u k ∈ S(p , ε , ρ). Then we update and define ε k+1 = ε , This gives u k+1 ∈ S(p k+1 , ε k+1 , 1). And we solve In particular, Proposition 2.2 gives (6.1) for universal C. This is called a step of the iteration. Suppose the assumptions in Lemma 5.1 are always satisfied at each step. Then we get a sequence of {ε k }. We then divide all steps into different stages depending on ε k .
Also, suppose k and k + 1 are two steps within the same stage. Definẽ h k (x) = 1 ρ 2 h k (ρx).
Therefore, |D 2 p n − D 2 p 0 | ≤ Cε 0 for some universal C for all n. Consequently, if we choose κ universally large, then λ 2 (D 2 p n ) ≥ λ 2 (D 2 p 0 ) − Cε 0 ≥ (κ − C)ε 0 ≥ κ 0 ε n for all n, and the assumptions in Lemma 5.1 are always satisfied. Similar estimate gives |D 2 p n − D 2 p m | ≤ Cε m whenever n ≥ m. As a result, there will be a quadratic polynomial q defined as where D 2 p n → M and |M − D 2 p n | ≤ Cε n . Note in particular that q ∈ Q and |D 2 q − D 2 p| ≤ Cε. Inside B ρ m , |u − q − ∇p m (0) · x| ≤ Cε m ρ 2m .
Now the parameter κ is fixed depending on universal constants, constants in Lemma 4.1 become universal. See Remark 4.1.
We can now give a full description of the iteration scheme: For u ∈ S(p, ε, 1) with ε <ε, whereε is the smaller constants between the ones in Lemma 4.1 and Proposition 6.1.
Define u 0 = u, p 0 = p and ε 0 = ε. Once we have u k ∈ S(p k , ε k , 1) we apply Lemma 4.1 or Lemma 5.1, depending on the comparison between λ 2 (D 2 p k ) and κε k , to get p k+1 such that u k ∈ S(p k+1 , ε , r k ).
If for all k, λ 2 (D 2 p k ) ≤ κε k , then we are always in the case described by Lemma 4.1, where each time the improvement is ε → (1 − β)ε. Here standard argument gives a polynomial q ∈ Q such that (6.5) |u − q| ≤ C|x| 2+α in B 1/2 for a universal α ∈ (0, 1). Moreover, in this case, we have λ 2 (D 2 q) = 0. Once we have the explicit rates of approximation as in (6.4) and (6.5), it is standard that we have the uniqueness of blow-up: Theorem 6.1. Suppose u solves (1.1) and x 0 ∈ Σ(u). Then there is a unique quadratic solution, denoted by p x0 ∈ Q, such that 1 r 2 u(x 0 + r·) → p x0 locally uniformly in R d as r → 0.
Here p x is the blow-up profile at point x as in Theorem 6.1. Now we can give the proof of the main result: Proof of Theorem 1.1. Let K ⊂ Ω be a compact set. With Proposition 2.7 and Proposition 2.5 we know that there is r K > 0, such that for any x 0 ∈ Σ(u) ∩ K |u(x 0 + ·) − p| ≤εr 2 K in B r K for some p ∈ Q, and D 2 u(x 0 + ·) ≥ −c 0ε I in B r K .
Defineũ(x) = 1 r 2 K u(x 0 + r K x), then we start the iteration as described before Theorem 6.1.
We have that x 0 ∈ Σ d−1 (u) if and only if λ 2 (D 2 p k ) ≤ κε k for all k in the iteration. In this case we have |ũ − p x0 | ≤ C|x| 2+α in B 1/2 . Scaling back, we have |u(x 0 + ·) − p x0 | ≤ C|x| 2+α in B 1 2 r K for some C depending on r K but nevertheless uniform on the set K.
After this, it is standard to apply Whitney's extension theorem and get the C 1,α -covering of Σ d−1 (u) ∩ K. For details of this argument, see Theorem 7.9 in Petrosyan-Shahgholian-Uraltseva [PSU].