Obstacle Problems Involving The Fractional

Laplacian

Donatella Danielli∗ and Sandro Salsa

January 27, 2017

1 Introduction

Obstacle problems involving a fractional power of the Laplace operator appearin many contexts, such as in pricing of American options governed by assetsevolving according to jump processes [25], or in the study of local minimizers ofsome nonlocal energies [24].

In the first part of this expository paper we are concerned with the stationarycase, that can be stated in several ways. Given a smooth function ϕ : Rn → R,n > 1, with bounded support (or at least rapidly decaying at infinity), we lookfor a continuous function u satisfying the following system:

u ≥ ϕ in Rn(−∆)

su ≥ 0 in Rn

(−∆)su = 0 when u > ϕ

u (x)→ 0 as |x| → +∞.

(1)

Here we consider only the case s ∈ (0, 1). The set Λ (u) = u = ϕ is called thecontact or coincidence set. The free boundary is the set

F (u) = ∂Λ (u) .

The main theoretical issues in a constrained minimization problem are opti-mal regularity of the solution and the analysis of the free boundary.

If s = 1 and Rn is replaced by a bounded domain Ω our problem correspondsto the usual obstacle problem for the Laplace operator. The existence of a uniquesolution satisfying some given boundary condition u = g can be obtained byminimizing the Dirichlet integral in H1 (Ω) under the constraint u ≥ ϕ. Thesolution is the least superharmonic function greater or equal to ϕ in Ω, withu ≥ g on ∂Ω, and inherits up to a certain level the regularity of ϕ ([32]). In fact,even if ϕ is smooth, u is only C1,1

loc , which is the optimal regularity. A classical

∗D.D. was supported in part by NSF grant DMS-1101246

1

reference for the obstacle problem, including the regularity and the completeanalysis of the free boundary is [18]. See also the recent book [55].

Analogously, the existence of a solution u for problem (1) can be obtainedby variational methods as the unique minimizer of the functional

J (v) =

∫Rn

∫Rn

|v (x)− v (y)|2

|x− y|n+2s dxdy

over a suitable set of functions v ≥ ϕ. We can also obtain u via a Perron typemethod, as the least supersolution of (−∆)

ssuch that u ≥ ϕ. By analogy (see

also later the Signorini problem), when ϕ is smooth, we expect the optimalregularity for u to be C1,s. This is indeed true, as it is shown by Silvestre in[62] when the contact set u = ϕ is convex and by Caffarelli, Salsa, Silvestrein [22] in the general case.

The case s = 1/2 is strongly related to the so called thin (or lower di-mensional) obstacle problem for the Laplace operator. To keep a connectionwith the obstacle problem for (−∆)

s, it is better to work in Rn+1, writing

X = (x, y) ∈ Rn × R. The thin obstacle problem concerns the case in whichthe obstacle is not anymore n + 1 dimensional, but supported instead on asmooth n−dimensional manifold M in Rn+1. This problem and variations ofit also arise in many applied contexts, such as flow through semi-permeablemembranes, elasticity (known as the Signorini problem, see below), boundarycontrol temperature or heat problems (see [28]).

More precisely, let Ω be a domain in Rn+1 divided into two parts Ω+ andΩ− by M. Let ϕ :M→R be the (thin) obstacle and g be a given function on∂Ω satisfying g > ϕ on M∩∂Ω.

The problem consists in the minimization of the Dirichlet integral

J (v) =

∫Ω

|∇v|2

over the closed convex set

K =v ∈ H1 (Ω) : v = g on ∂Ω and v ≥ ϕ on M∩ ∂Ω

.

Since we can perturb the solution u upwards and freely away from M, it isapparent that u is superharmonic in Ω and harmonic in Ω\M. One expects thecontinuity of the first derivatives along the directions tangential to M, and theone sided continuity of normal derivatives ([32]). In fact (see [16]), on M, usatisfies the following complementary conditions

u ≥ ϕ, uν+ + uν− ≤ 0, (u− ϕ) (uν+ + uν−) = 0

where ν± are the interior unit normals toM from the Ω± side. The free bound-ary here is given by the boundary of the set Ω\Λ (u) in the relative topology ofM, and in general, we expect it is a (n− 1)−dimensional manifold.

2

As mentioned above, a related problem is the Signorini problem1 (or bound-ary thin obstacle problem), in which the manifoldM is part of ∂Ω and one hasto minimize the Dirichlet integral over the closed convex set

K =v ∈ H1 (Ω) : v = g on ∂Ω\M and v ≥ ϕ on M

.

In this case, u is harmonic in Ω and on M it satisfies the complementary con-ditions

u ≥ ϕ, uν+ ≤ 0, (u− ϕ)uν+ = 0.

IfM is a hyperplane (say y = 0) and Ω is symmetric with respect toM, thenthe thin obstacle in Ω and the boundary obstacle problems in Ω+ or Ω− areequivalent.

Let us see how these problems are related to the obstacle problem for the

(−∆)1/2

. This is explained through the following remarks.

(a) Reduction to a global problem. Let Ω = B1 be the unit ball in Rn+1

and B′1 = B1 ∩ y = 0. Let ϕ : Rn → R be a smooth obstacle, ϕ < 0 on ∂B′1and positive somewhere inside B′1. Consider the following Signorini problem inB+

1 = B1 ∩ y > 0:−∆u = 0 in B+

1

u = 0 on ∂B1 ∩ y > 0u (x, 0) ≥ ϕ (x) in B′1uy (x, 0) ≤ 0 in B′1uy (x, 0) = 0 when u (x, 0) > ϕ (x) .

(2)

We want to convert the above problem in B1 into a global one, that is inRn × (0,+∞). To do this, let η be a radially symmetric cut-off function in B′1such that

ϕ > 0 b η = 1 and supp (η) ⊂ B′1.

Extending ηu by zero outsideB1, we have η (x)u (x, 0) ≥ ϕ (x) and also (ηu)y (x, 0) ≤0 for every x ∈ Rn. Moreover, (ηu)y (x, 0) = 0 if η (x)u (x, 0) > ϕ (x).

Let now v be the unique solution of the following Neumann problem in theupper half space, vanishing at infinity:

∆v = ∆ (ηu) in Rn × y > 0vy (x, 0) = 0 in Rn.

Then w = ηu − v is a solution of a global Signorini problem with ϕ − v as theobstacle. Thus, the regularity of u in the local setting can be inferred from theregularity for the global problem.

The opposite statement is obvious.

(b) Realization of (−∆)1/2

as a Dirichlet-Neumann map. Consider a smoothfunction u0 : Rn −→ R with rapid decay at infinity. Let u : Rn× (0,+∞) −→ Rbe the unique solution of the Dirichlet problem

∆u = 0 in Rn × (0,+∞)u (x, 0) = u0 (x) in Rn

1After Fichera, see ([30]), 1963.

3

vanishing at infinity. We call u the harmonic extension of u0 to the upper halfspace.

Consider the Dirichlet-Neumann map T : u0 (x) 7−→ −uy (x, 0). We have:

(Tu0, u0) =

∫Rn−uy (x, 0)u (x, 0) dx

=

∫Rn×(0,+∞)

∆u (X)u (X) + |∇u (X)|2

dX

=

∫Rn×(0,+∞)

|∇u (X)|2 dX ≥ 0

so that T is a positive operator. Moreover, since u0 is smooth and uy is har-monic, we can write:

T Tu0 = −∂y(−∂y)u (x, 0) = uyy (x, 0) = −∆u0.

We conclude thatT = (−∆)

1/2.

As a consequence:

1. If u = u (X) is a solution of the Signorini problem in Rn × (0,+∞), thatis ∆u = 0 in Rn+1, u (x, 0) ≥ ϕ, uy (x, 0) ≤ 0, and (u− ϕ)uy (x, 0) = 0 in Rn,

then u0 = u (·, 0) solves the obstacle problem for (−∆)1/2

.

2. If u0 is a solution of the obstacle problem for (−∆)1/2

, then its harmonicextension to Rn × (0,+∞) solves the corresponding Signorini problem.

Therefore, the two problems are equivalent and any regularity result for oneof them can be carried to the other one. More precisely, consider the optimal

regularity for the solution u0 of the obstacle problem for (−∆)1/2

, which isC1,1/2. If we can prove a C1,1/2 regularity of the solution u of the Signoriniproblem up to y = 0, then the same is true for u0.

On the other hand, the C1,α regularity of u0 extends to u, via boundaryestimates for the Neumann problem. Similarly, the analysis of the free boundary

in the Signorini problem carries to the obstacle problem for (−∆)1/2

as well andvice versa.

Although the two problem are equivalent, there is a clear advantage in favorof the Signorini type formulation. This is due to the possibility of avoiding the

direct use of the non local pseudodifferential operator (−∆)1/2

, by localizingthe problem and using local PDE methods, such as monotonicity formulas andclassification of blow-up profiles.

At this point it is a natural question to ask whether there exists a PDErealization of (−∆)

sfor every s ∈ (0, 1), s 6= 1

2 .The answer is positive as it is shown by Caffarelli and Silvestre in [23]. Indeed

in a weak sense we have that

(−∆)su0 (x) = −κa lim

y→0+yauy (x, y)

4

for a suitable constant κa, where u is the solution of the problemLau = div (ya∇u) = 0 in Rn × (0,+∞)u (x, 0) = u0 (x) in Rn

vanishing at infinity, where a = 2s− 1. Coherently, we call u the La−harmonicextension of u0.

Thus problem (1) is equivalent to the following Signorini problem for theoperator La = div(ya∇),

u (x, 0) ≥ ϕ in Rn (3)

Lau = div (ya∇u) = 0 in Rn × (0,+∞) (4)

limy→0+ yauy (x, y) = 0 when u (x, 0) > ϕ (x) (5)

limy→0+ yauy (x, y) ≤ 0 in Rn. (6)

For y > 0, u is smooth so that (4) is understood in the classical sense. Theequations at the boundary (5) and (6) should be understood in a weak sense.Since in [62] it is shown that u (x, 0) ∈ C1,α for every α < s, for the range ofvalues 2s − 1 < α < s, limy→0+ y

auy (x, y) can be understood in the classicalsense too.

The solution u of the above Signorini problem can be extended to the wholespace by symmetrization, setting u (x,−y) = u (x, y). Then, by the results in[23], condition (5) holds if and only if the extended u is a solution of Lau = 0across y = 0, where u (x, 0) > ϕ (x). On the other hand, condition (6) isequivalent to Lau ≤ 0 in the sense of distributions. Thus, for the extended u,the Signorini problem translates into the following system:

u (x, 0) ≥ ϕ (x) in Rnu (x,−y) = u (x, y) in Rn+1

Lau = 0 in Rn+1\ (x, 0) : u (x, 0) = ϕ (x)Lau ≤ 0 in Rn+1, in the sense of distributions.

with u vanishing at infinity. Again, we can exploit the advantages to analyzethe obstacle problem for a nonlocal operator in PDE form by considering a localversion of it. Indeed, to study the optimal regularity properties of the solutionwe will focus on the following local version, where ϕ : B′ −→ R:

u (x, 0) ≥ ϕ (x) in B′1u (x,−y) = u (x, y) in B1

Lau = 0 in B1\ (x, 0) : u (x, 0) = ϕ (x)Lau ≤ 0 in B1, in the sense of distributions.

The above problem can be thought of as the minimization of the weightedDirichlet integral

Ja (v) =

∫B1

|y|a |∇u (X)|2 dX

5

over the setKa =

v ∈W 1,2 (B1, |y|a) : u (x, 0) ≥ ϕ (x)

.

In a certain sense, this corresponds to an obstacle problem, where the obstacleis defined in a set of codimension 1 + a, where a is not necessarily an integer.

The operator La is degenerate elliptic, with a degeneracy given by the weight|y|a. This weight belongs to the Muckenhoupt class A2

(Rn+1

). We recall that

a positive weight function w = w (X) belongs to A2

(RN)

if(1

|B|

∫B

w

)(1

|B|

∫B

w−1

)≤ C

for every ball B ⊂ RN . For the class of degenerate elliptic operators of the formLu =div(A (X)∇u), where

λw (X) |ξ|2 ≤ A (X) ξ · ξ ≤ Λw (X) |ξ|2 ,

there is a well established potential theory for solutions in the weighted Sobolevspace W 1,2 (Ω, w) (Ω bounded domain in RN ), defined as the closure of C∞

(Ω)

in the norm [∫Ω

v2w +

∫Ω

|∇v|2 w]1/2

(see [29]). Since w ∈ A2, the gradient of a function in W 1,2 (Ω, w) is well definedin the sense of distributions and belongs to the weighted space L2 (Ω, w).

The outline of the first part. We intend here to give a brief review ofthe results concerning the analysis of the solution and the free boundary of theobstacle for the fractional Laplacian, mainly based on the extension method.

For the thin obstacle problem, Caffarelli in [16] proves that u is C1,α up toy = 0, for some α ≤ 1/2. Subsequently Athanasopoulos and Caffarelli achievethe optimal regularity of the solution in [7]. In the case of zero obstacle, theregularity of the free boundary around a so called nondegenerate (or stableor regular) point is analyzed by Athanasopoulos, Caffarelli and Salsa in [10].Indeed these last two papers opened the door to all subsequent developments.

In [36], Garofalo and Petrosyan start the analysis of F (u) around non regularpoints (also for non zero obstacles). They obtain a stratification result forsingular points, i.e. points of F (u) of vanishing density for Λ (u), in terms ofhomogeneity of suitable blow-ups of the solution.

The analysis of the obstacle problem for the operator (−∆)s, 0 < s < 1,

starts with Silvestre in ([62]), which shows C1,α estimates for the solution forany α < s and also α = s if the interior of the coincidence set is convex. Notably,Silvestre does not use any extension properties; his methods are purely nonlocal.A few years later, Caffarelli, Salsa and Silvestre ([22]) extend to the fractionalLaplacian case the results in [10] on the optimal regularity and the analysis ofthe regular part of the free boundary.

Recently, in [12], Barrios, Figalli and Ros-Oton continued the work of [36],giving a complete picture of the free boundary under two basic assumptions.

6

The first one is a strict concavity of the obstacle, the same assumption neededin the case of the classical obstacle problem. The second one prescribes zeroboundary values of the solution and it turns out to be crucial.

Here, we shall focus mainly on the optimal regularity of the solution and onthe analysis and structure of the free boundary, only mentioning, for brevityreasons, recent important results on higher regularity and extension to moregeneral operators. In particular, we will give here an outline of the strategyused in the papers ([22]) and [12].

A few comments on the key concepts and tools that will repeatedly appearare in order.

Semi-convexity: it is a peculiarity of solutions of the obstacle problem. Moreprecisely, semi-convexity along tangential directions τ (i.e. parallel to the planey = 0) and semiconcavity along the y−direction consistently play a key role.It is noteworthy that, for global solution of the zero thin obstacle case, thetangential semi-convexity of u comes for free, since u (X + hτ) and u (X − hτ)are admissible nonnegative superharmonic functions, and therefore

1

2(u (X + hτ) + u (X − hτ)) ≥ u (X) .

Asymptotic profiles. From semi-convexity, one deduces that suitable globalasymptotic profiles coming from blow-ups of u around a free boundary point(say, the origin) are tangentially convex and can be classified according to theirhomogeneity degree. From this it is an easy matter to deduce optimal regularity.

Frequency and monotonicity formulas. Frequency formulas of Almgren type,first introduced in the case s = 1/2 in ([10]) are key tools in carrying optimalregularity from global to local solutions. Other types of monotonicity formulas,such as Weiss or Monneau-types, first introduced in ([36]) for s = 1/2, play acrucial role in the analysis of non-regular points of the free boundary.

Carleson estimates and boundary Harnack principles are by now standardtools in the study of the optimal regularity of the free boundary, in our casearound the so called regular points. Due to the non homogeneous right handside in the equation, the Carleson estimate and boundary Harnack principleproved here are somewhat weaker than the usual ones. More recently, De Silvaand Savin ([27]) have applied these principles to prove higher regularity of thefree boundary.

We will always assume that the origin belongs to the free boundary.

The outline of the second part. In Section 3 we consider two time-dependent models, which can be thought of as parabolic counterparts of thesystems (1) and (2). In the first part, Section 3.1, we discuss the regularity ofsolutions to the parabolic fractional obstacle problem

min−vt + (−∆)sv, v − ψ = 0 on [0, T ]× Rn

v(T ) = ψ on Rn,

following [19]. In particular, under some assumptions on the obstacle ψ, thesolution v is shown to be globally Lipschitz continuous in space-time. Moreover,

7

vt and (−∆)sv belong to suitable Holder and logLipschitz spaces. The regularityin space is optimal, whereas the regularity in time is almost optimal in the casess = 1/2 and s→ 1−.

In Section 3.2 we give an overview of the parabolic Signorini problem∆v − ∂tv = 0 in ΩT := Ω× (0, T ],

v ≥ φ, ∂νv ≥ 0, (v − φ)∂νv = 0 on MT :=M× (0, T ],

v = g on ST := S × (0, T ],

v(·, 0) = φ0 on Ω0 := Ω× 0.

Here S = ∂Ω \ M. Similarly to the elliptic case, we are interested in theregularity properties of v, and the structure and regularity of the free boundaryΓ(v) = ∂MT

(x, t) ∈ MT | v(x, t) > φ(x, t), where ∂MTindicates the bound-

ary in the relative topology of MT . Following [26], the analysis comprises themonotonicity of a generalized frequency function, the study of blow-ups and theensuing regularity of solutions, the classification of free boundary points, andthe regularity of the free boundary at so-called regular points.

2 The obstacle problem for the fractional Lapla-cian

This section is devoted to the study of the fractional Laplacian obstacle problemthat we recall below.

Given a smooth function ϕ : Rn → R, with bounded support (or rapidlyvanishing at infinity), we look for a continuous function u satisfying the followingconditions:

• u ≥ ϕ in Rn

• (−∆)su ≥ 0 in Rn

• (−∆)su = 0 when u > ϕ

• u (x)→ 0 as |x| → +∞.

We list below the main steps in the analysis of the problem that we are goingto describe.

1. Construction of the solution and basic properties.

2. Lipschitz continuity, semiconvexity and C1,α estimates.

3. Reduction to the thin obstacle for the operator La.

4. Optimal regularity for tangentially convex global solution.

5. Classification of asymptotic blow-up profiles around a free boundary point.

8

6. Optimal regularity of the solution.

7. Analysis of the free boundary at stable points: Lipschitz continuity.

8. Boundary Harnack Principles and C1,α regularity of the free boundary atstable points.

9. Structure of the free boundary.

Steps 1, 2 and 3 are covered in [62]. In particular from [62] we borrow theC1,α estimates without proof, for brevity. The steps 3-8 follow basically thepaper [22], except for step 2 and part of 5. In particular, the optimal regularityof global solutions follows a different approach, similar to the correspondingproof for the zero obstacle problem in [10]. Step 8 is taken from [12]. Finally,step 9 comes from [10] (Carleson estimate) and [22] (Boundary Harnack).

2.1 Construction of the solution and basic properties

We start by proving the existence of a solution. Observe that the proof fails forn = 1 and s > 1/2, because in this case it is impossible to have (−∆)

su ≥ 0 in

R with u vanishing at infinity.Let S be the set of rapidly decreasing C∞ functions in Rn. We denote by

Hs the completion of S in the norm

‖f‖2Hs =

∫Rn

∫Rn

|f (x)− f (y)|2

|x− y|n+2s dxdy ∼∫Rn|ξ|2s

∣∣∣f (ξ)∣∣∣2 dξ.

Equipped with the inner product

(f, g)Hs =

∫Rn

∫Rn

(f (x)− f (y)) (g (x)− g (y))

|x− y|n+2s dxdy

= 2

∫Rnf (x) (−∆)

sg (x) dx ∼

∫Rn|ξ|2s f (ξ) g (ξ)dξ,

Hs is a Hilbert space. Since we are considering n ≥ 2 and s < 1 ≤ n/2, itturns out that Hs coincides with the set of functions in L2n/(n−2s), for whichthe Hs−norm is finite.

The solution u0 of the obstacle problem is constructed as the unique mini-mizer of the strictly convex functional

J (v) = ‖v‖2Hs

over the closed, convex set Ks =v ∈ Hs : v ≥ ϕ

.

In the following proposition we gather some basic properties of u.

Proposition 1 Let u0 be the minimizer of the functional J over Ks. Then:

9

(a) The function u0 is a supersolution, that is (−∆)su0 ≥ 0 in Rn in the

sense of measures. Thus, it is lower semicontinuous and, in particular, the setu0 > ϕ is open.

(b) u0 is actually continuous in Rn.(c) If u0 (x) > ϕ (x) in some ball B then (−∆)

su0 = 0 in B.

Proof. (a) Let h ≥ 0 be any smooth function with compact support. Ift > 0, u0 + th ≥ ϕ so that

(u0, u0)Hs ≤ (u0 + th, u0 + th)Hs

or0 ≤ 2t (u0, h)

Hs+ t2 (h, h)Hs = ((−∆)

sh)L2 + t2 (h, h)Hs

from which(u0, (−∆)

sh)L2 = ((−∆)

su0, h)L2 ≥ 0.

Therefore (−∆)su0 is a nonnegative measure and therefore is lower semicontin-

uous by Proposition A1.

(b) The continuity follows from Proposition A2.

(c) For any test function h ≥ 0, supported in B the proof in (a) holds alsofor t < 0. Therefore (−∆)

su0 = 0 in B.

Corollary 2 The minimizer u0 of the functional J over Ks is a solution of theobstacle problem.

2.2 Lipschitz continuity and semiconvexity and C1,α esti-mates

Following our strategy, we first show that, if ϕ is smooth enough, then thesolution of our obstacle problem is Lipschitz continuous and semiconvex. Weare mostly interested in the case ϕ ∈ C1,1. When ϕ has weaker regularity,u0 inherits corresponding weaker regularity (see [62]). We emphasize that theproof in this subsection depends only on the maximum principle and translationinvariance.

Lemma 3 The function u0 is the least supersolution of (−∆)s

such that u0 > ϕand lim inf |x|→∞ u0 (x) ≥ 0.

Proof. Let v such that (−∆)sv ≥ 0, v > ϕ and lim inf |x|→∞ v (x) ≥ 0.

Let w = min u0, v. Then w is lower-semicontinuous in Rn and is anothersupersolution above ϕ (by Propositions A1 and A4). We show that w ≥ u0.

Since ϕ ≤ w ≤ u0, we have w (x) = u0 (x) for every x in the contact setΛ (u0) = u0 = ϕ . In Ω = u0 > ϕ, u0 solves (−∆)

su0 = 0 and w is a

supersolution. By Proposition 1 (b) , u0 is continuous. Then w − u0 is lower-semicontinuous and w ≥ u0 from comparison.

10

Corollary 4 The function u0 is bounded and supu0 ≤ supϕ. If the obstacle ϕhas a modulus of continuity ω, then u0 has the same modulus of continuity. Inparticular, if ϕ is Lipschitz, then u0 is Lipschitz and Lip(u0) ≤Lip(ϕ).

Proof. By hypothesis u0 ≥ ϕ. The constant function v (x) = supϕ is asupersolution that is above ϕ. By Lemma 3, u0 ≤ v in Rn.

Moreover, since ω is a modulus of continuity for ϕ, for any h ∈ Rn,

ϕ (x+ h) + ω (|h|) ≥ ϕ (x)

for all x ∈ Rn. Then, the function u0 (x+ h) + ω (|h|) is a supersolution aboveϕ (x). Thus u0 (x+ h) + ω (|h|) ≥ u0 (x) for all x, h ∈ Rn. Therefore u0 has amodulus of continuity not larger than ω.

Lemma 5 Let ϕ ∈ C1,1 and assume that inf ∂ττϕ ≥ −C, for any unit vector τ .Then ∂ττu0 ≥ −C too. In particular, u0 is semiconvex.

Proof. Since ∂ττϕ ≥ −C, we have

ϕ (x+ hτ) + ϕ (x− hτ)

2+ Ch2 ≥ ϕ (x)

for every x ∈ Rn and h > 0. Therefore:

V (x) ≡ u0 (x+ hτ) + u0 (x− hτ)

2+ Ch2 ≥ ϕ (x)

and V is also a supersolution: (−∆)sV ≥ 0. Thus, by Lemma 3, V ≥ u so that:

u0 (x+ hτ) + u0 (x− hτ)

2+ Ch2 ≥ u0 (x)

for every x ∈ Rn and h > 0. This implies ∂ττu0 ≥ −C. From the results in [62] we can prove a partial regularity result, under the

hypothesis that ϕ is smooth.

Theorem 6 Let ϕ ∈ C2. Then u0 ∈ C1,α for every α < s and (−∆)su0 ∈ Cβ

for every β < 1− s.

The proof is long and very technical, so we refer to the original paper [62].However, an idea of the proof in the case of the zero thin obstacle problem canbe given without much effort. Indeed, from tangential semi-convexity we deduce(here u is harmonic outside Λ (u))

uyy ≤ C in B1\Λ (u) .

In particular, the function uy − Cy is monotone and bounded. Thus we areallowed to define in B′1 : σ (x) = limy→0+ uy (x, y) . Since, in this case

0 ≥ ∆u = 2uyHn|Λ(u)

11

in the sense of measures, we have σ (x) ≤ 0 in Λ (u) and by symmetry, σ (x) = 0in B′1\Λ (u). We may summarize the properties of the solution of a zero thinobstacle problem in complementary form as follows:

∆u ≤ 0, u∆u = 0 in B1

∆u = 0 in B1\Λ (u)u (x, 0) ≥ 0, σ (x) ≤ 0, u (x, 0)σ (x) = 0 in B′1σ (x) = 0 in B′1\Λ (u)

Now, let u be a solution of the zero thin obstacle problem in B1, normalizedby ‖u‖L2(B1) = 1. To prove local C1,α estimates, it is enough to show that

σ ∈ C0,α near the free boundary F (u).In fact, in the interior of Λ (u), u (x, 0) is smooth and so is σ. On the

other hand, on Ω′ = B′\Λ′ (u) , σ = 0. Thus, if we show that σ is C0,α in aneighborhood of F (u), then u ∈ C1,α from both sides of the free boundary bystandard estimates for the Neumann problem.

In particular it is enough to show uniform estimates around a free boundarypoint, say the origin. This can be achieved by a typical iteration procedure, inthe De Giorgi style. We distinguish two steps:

Step 1: To show that near the free boundary we can locate large regionswhere −σ grows at most linearly (estimates in measure of the oscillation of −σ).

Step 2: Using Poisson representation formula and semi-concavity, we con-vert the estimate in average of the oscillation of −σ, done in step 1, into point-wise estimates, suitable for a dyadic iteration of the type

uy (X) ≥ −βk in B′γk (x0)×[0, γk

](x0 ∈ F (u) )

for some 0 < γ < 1, 0 < β < 1, and any k ≥ 0.The details can be found in the paper of Caffarelli [16] (see also the review

paper [60]).

2.3 Thin obstacle for the operator La. Local C1,α estimates

To achieve optimal regularity we now switch to the equivalent thin obstacleproblem for the operator La as mentioned in the introduction and that werestate here:

u (x, 0) ≥ ϕ (x) in B′1u (x,−y) = u (x, y) in B1

Lau = 0 in B1\Λ (u)Lau ≤ 0 in B1, in the sense of distributions.

(7)

In the global setting (i.e. with B1 replaced by Rn), u0 (x) = u (x, 0) is thesolution of the global obstacle problem for (−∆)

sand

(−∆)su0 = κa lim

y→0+yauy (x, y) .

12

The estimates in Corollary 4 and Lemma 5 translate, after an appropriatelocalization argument and the use of boundary estimates for the operator La,into corresponding estimates for the solution of u. Namely:

Lemma 7 Let ϕ ∈ C2,1 (B′1) and u be the solution of (7). Then

1. ∇xu (X) ∈ Cα(B1/2

)for every α < s;

2. |y|a uy (X) ∈ Cα(B1/2

)for every α < 1− s;

3. uττ (X) ≥ −C in B1/2.

Proof. From Corollary 4 and Lemma 5 we have that the above estimatesholds on y = 0. Since ∂xju and uττ also solve the equation Law = 0 in B1\Λ (u),the estimates 1 and 3 extend to the interior. On the other hand w (x, y) =|y|a uy (X) solves the conjugate equation div(|y|−a∇w (X)) and we obtain 2.

Remark 1 Observe that u can only be C1,α in both variables up to y = 0 onlyif a ≤ 0. If a > 0, since yauy (X) has a non-zero limit for some x in the contactset, it follows that uy cannot be bounded.

We close this subsection with a compactness result, useful in dealing withblow-up sequences.

Lemma 8 Let vj be a bounded sequence of functions in W 1,2 (B1, |y|a). As-sume that there exists a constant C such that, in B1:

|∇xvj (X)| ≤ C and |∂yvj (X)| ≤ C |y|−a (8)

and that, for each small δ > 0, vj is uniformly C1,α in B1−δ ∩ |y| > δ.

Then, there exists a subsequence vjk strongly convergent in W 1,2(B1/2, |y|

a).

Proof. From the results in [42], there is a subsequence, that we still call vj,that converges strongly in L2

(B1/2, |y|

a). Since for each δ > 0, vj is uniformly

bounded in C1,α in the set B1−δ∩|y| > δ, we can extract a subsequence so that∇vj converges uniformly in B1−δ ∩|y| > δ . Thus, ∇vj converges pointwise inB1\ y = 0.

Now, from (8) and the fact that C and |y|−a both belong to L2(B1/2, |y|

a),

the convergence of each partial derivative of vj in L2(B1/2, |y|

a)follows from

the dominated convergence theorem.

2.4 Minimizers of the weighted Rayleigh quotient and amonotonicity formula

The next step towards optimal regularity is to consider tangentially convexglobal solutions.

13

Lemma 9 Let ∇θ denote the surface gradient on the unit sphere ∂B1. Set, for−1 < a < 1,

λ0,a = inf

∫∂B+

1|∇θw|2 yadS∫

∂B+1w2yadS

: w ∈W 1,2(∂B+

1 , yadS

): w = 0 on (∂B′1)

+

where (∂B′1)

+= (x′, xn) ∈ ∂B′1, xn > 0 . Then the first eigenfunction, up to

a multiplicative factor, is given by

w (x, y) =(√

x2n + y2 − xn

)ss = (1− a) /2

and 2

λ0,a =1− a

4(2n+ a− 1) .

The following lemma gives a first monotonicity result.

Lemma 10 Let w be continuous in B1, w (0) = 0, w (x, 0) ≤ 0, w (x, 0) = 0 onΛ ⊂ y = 0, Law = 0 in B1\Λ. Assume that the set

x ∈ B′r : w (x, 0) < 0

is non empty and convex. Set

β (r) = β (r;w) =1

r1−a

∫B+r

ya |∇w (X)|2

|X|n+a−1 dX.

Then, β (r) is bounded and increasing for r ∈ (0, 1/2].

Proof. We have Law2 = 2wLaw + 2ya |∇w|2 = 2ya |∇w|2 , so that

β (r) =1

r1−a

∫B+r

ya |∇w (X)|2

|X|n+a−1 dX =1

2r1−a

∫B+r

La(w2)

|X|n+a−1 dX.

Now:

β′ (r) =a− 1

2r2−a

∫B+r

La(w2)

|X|n+a−1 dX +1

rn

∫∂B+

r

ya |∇w|2 dS.

Since w (0, 0) = 0 and yawy (x, y)w (x, y) → 0 as y → 0+, after simple compu-tations, we obtain:

β′ (r) ≥ − (1− a)(2n+ a− 1)

4rn+2

∫∂B+

r

yaw2dS +1

rn

∫∂B+

r

ya |∇θw|2 dS.

The convexity of x ∈ B′r : w (x, 0) < 0 implies that the Rayleigh quotient mustbe greater than λ0,a and therefore we conclude β′ (r) ≥ 0 and, in particular,β (r) ≤ β (1/2).

2Formally, the first eigenvalue can be obtained plugging α = s = (1− a) /2 and n + ainstead of n into the formula α (α− 1) + nα.

14

2.5 Optimal regularity for tangentially convex global so-lutions

In this section we consider global solutions that represent possible asymptoticprofiles, obtained by a suitable blow-up of the solution at a free boundary point.

First of all we consider functions u : Rn ×R→ R, homogeneous of degree k,solutions of the following problem:

u (x, 0) ≥ 0 in Rnu (x,−y) = u (x, y) in Rn × RLau = 0 in (Rn × R) \ΛLau ≤ 0 in the sense of distributions in Rn × Ruττ ≥ 0 in Rn × R, for every tangential unit vector τ

(9)

where Λ = Λ (u) = (x, 0) : u (x, 0) = 0. The following proposition gives alower bound for the degree k, which implies the optimal regularity of the solu-tion.

Lemma 11 If there exists a solution u of problem (9), then k ≥ 1 + s =(3− a) /2.

Proof. Apply the monotonicity formula in Lemma 10 to w = uτ . Then,Law = 0 in (Rn × R) \Λ and, by symmetry, w (x, 0)wy (x, 0) = 0. Moreover,the contact set where w = 0 is convex, since uττ ≥ 0.Therefore w satisfies allthe hypotheses of that lemma. Recall that we always assume that (0, 0) ∈ F (u)so that w (0, 0) = 0. Thus

β (r;w) =1

r1−a

∫B+r

ya |∇w (X)|2

|X|n+a−1 dX ≤ β (1, w) .

On the other hand, since ∇w is homogeneous of degree k − 2, we have

β (r;w) =r2k−2

r1−a

∫B+

1

ya |∇w (X)|2

|X|n+a−1 dX =r2k−2

r1−a β (1, w) .

This implies r2k−2 ≤ r1−a = r2s or k ≥ 1 + s.

From Lemma 11 it would be possible to deduce the optimal regularity of thesolution u to (7). However, to study the free boundary regularity we need toclassify precisely the solutions to problem (9). For the operator La, we need tointroduce the following subset of the coincidence set. Let

Λ∗ =

(x, 0) ∈ Rn : lim

y→0+yauy (x, y) < 0

.

Notice that Λ∗ is the support of Lau and since Lau = 0 in (Rn × R) \Λ, we haveΛ∗ ⊂ Λ (Λ is closed).

The analysis depend on whether Λ∗ has positive Hn measure or not. Firstexamine the case Hn(Λ∗) = 0.

15

Lemma 12 Let u be a solution of problem (9). If Hn(Λ∗) = 0, then u is apolynomial of degree k.

Proof. We know from Lemma 7 that |y|a uy (x, y) is locally bounded. IfHn(Λ∗) = 0, then

limy→0|y|a uy (x, y) = 0 a.e. x ∈ Rn.

Thus limy→0 |y|a uy (x, y) = 0 weak∗ in L∞ and from [23] we infer that u is aglobal solution of Lau = 0 in Rn ×R. Using Lemma A3 below we conclude theproof.

Lemma 13 Let u be a solution of problem (9). If Hn(Λ) 6= 0 then, either u ≡ 0or k = 1 + s and Λ is a half n−dimensional space.

Proof (sketch). First observe that if Hn(Λ∗) = 0, then u ≡ 0, otherwise,from Lemma 12, u (x, 0) would be a polynomial vanishing on a set of positivemeasure and therefore identically zero. Thus, the polynomial u must have theform

u (x, y) = p1 (x) y2 + ...+ pj (x) y2j

and iterating the computation of Lau one deduce p1 = p2 = ... = pj = 0.Consider now the case Hn(Λ∗) 6= 0. Then Λ∗ is a thick convex cone. Assume

that en is a direction inside Λ∗ such that a neighborhood of en is contained inΛ∗. Using the convexity in the en direction, we infer that w = uxn cannot bepositive at any point X. Moreover, w = 0 on Λ and

Law (X) = 0 in (Rn × R) \Λ∗ ⊇ (Rn × R) \Λ.

Thus w must coincide with the first eigenfunction of the weighted sphericalLaplacian, minimizer of

∫S1|∇θv|2 |y|a dS over all v vanishing on Λ and such

that∫S1v2 |y|a dS = 1.

Since Λ is convex, Λ∩B1 is contained in half of the sphere B1∩y = 0. If itwere exactly half of the sphere then it would be given by the first eigenfunctiondefined in Lemma 9, up to a multiplicative constant, by the explicit expression

w (x, y) =(√

x2n + y2 − xn

)ss = (1− a) /2.

On the other hand, the above function is not a solution across y = 0, xn ≥ 0.Therefore, if Λ ∩ B1 is strictly contained in half of the sphere B1 ∩ y = 0 ,there must be another eigenfunction corresponding to a smaller eigenvalue andconsequently to a degree of homogeneity smaller than s. This would implyk < 1 + s, contradicting Lemma 11. The only possibility is therefore k = 1 + s,with Λ = y = 0, xn ≥ 0.

The next theorem gives the classification of asymptotic profiles.

Theorem 14 Let u be a non trivial solution of problem (9). There are onlytwo possibilities:

16

(1) k = 1 + s, Λ is a half n−dimensional space and u depends only on twovariables. Up to rotations and multiplicative constants u is unique and there isa unit vector τ such that Λ = (x, 0) : x · τ ≥ 0 and

uτ (x, y) =

(√(x · τ)

2+ y2 − (x · τ)

)s(2) k is an integer greater than equal to 2, u is a polynomial and Hn(Λ) = 0.

Proof. If Hn(Λ) 6= 0, from Lemma 13 we deduce that, up to rotations andmultiplicative constants, there is a unique solution of problem (9), homogeneousof degree k = 1 + s. Moreover, for this solution the free boundary F (u) is flat,that is there is a unit vector (say) en such that

Λ = (x, 0) : xn ≥ 0

and

uxn (x, y) =(√

x2n + y2 − xn

)s.

Integrating uxn from F (u) along segments parallel to en we uniquely determineu (x, 0) = u (xn, 0). If we had another solution v, homogeneous of degree 1 + s,with v (x, 0) = u (x, 0), then necessarily (see the proof of the Liouville-typeLemma A3), for some constant c and y 6= 0, we have

v (x, y)− u (x, y) = c |y|s y.

But the constant c must be zero, otherwise v − u cannot be solution acrossy = 0 \Λ.

As a consequence, if u is a solution homogeneous of degree 1 + s, with ennormal to F (u), then u = u (xn, y). Indeed, translating in any direction orthog-onal to xn and y we get another global solution with the same free boundary.By uniqueness, u must be invariant in those directions.

If Hn(Λ) = 0, then Hn(Λ∗) = 0 and from Lemma 12 we conclude that u isa polynomial and k ≥ 2.

2.6 Frequency formula

As we have already stated, a crucial tool in order to achieve optimal regularityis given by a frequency formula of Almgren type. If the obstacle were zero, thenthe frequency formula states that the quantity

Da (r;u) =r∫Br|y|a |∇u|2 dX∫

∂Br|y|a u2dS

is bounded and monotonically increasing. The conclusion is the following.

Theorem 15 Let u be a solution of the zero thin obstacle for the operator Lain B1. Then Da (r;u) is monotone nondecreasing in r for r < 1. Moreover,Da (r;u) is constant if and only if u is homogeneous.

17

When the obstacle ϕ is non zero we cannot reduce to that case. Instead,assuming that ϕ ∈ C2,1, we let

u (x, y) = u (x, y)− ϕ (x) +∆ϕ (0)

2 (1 + a)y2

so that Lau (0) = 0. Moreover Λ = Λ (u) = u = 0. The function u is asolution of the following system:

u (x, 0) ≥ 0 in B′1

u (x,−y) = u (x, y) in B1

Lau (x, y) = |y|a g (x) in B1\ΛLau (x, y) ≤ |y|a g (x) in B1, in the sense of distributions

(10)

whereg (x) = ∆ϕ (x)−∆ϕ (0)

is Lipschitz. Notice that |y|a g (x)→ 0 as x→ 0 and in B1\Λ

|Lau (x, y)| ≤ C |y|a |x| .

What we expect is a small variation of Almgren’s formula. Since u− u is a C2,1

function, it is enough to prove any regularity result for u instead of u. In orderto simplify the notation we will still write u for u. Define

F (r) = F (r;u) =

∫∂Br

u2 |y|a dσ = rn+a

∫∂B1

(u (rX))2 |y|a dS.

We have:F ′ (r) =

= (n+ a) rn+a−1

∫∂B1

(u (rX))2 |y|a dS + rn+a

∫∂B1

2u (rX)∇u (rX) ·X |y|a dS

= (n+ a) r−1

∫∂Br

(u (X))2 |y|a dS +

∫∂Br

2u (X)uν (X) |y|a dS

Thus logF (r) is differentiable for r > 0 and:

d

drlogF (r) =

F ′ (r)

F (r)=n+ a

r+

∫∂Br

2uuν |y|a dS∫∂Br

u2 |y|a dS.

Note that the monotonicity of Da (r;u) when ϕ = 0 amounts to say that thefunction

r 7−→ rd

drlogF (r) = 2D (r;u) + n+ a

is increasing, since in this case∫∂Br

2uuν |y|a dS =

∫Br

La(u2) =

∫Br

(|y|a |∇u|2 + 2uLau)dX

=

∫Br

2 |y|a |∇u|2 dX.

18

Due to presence of a nonzero right hand side, we need to prevent the possibilitythat F (r) become too small under rescaling when compared to the terms in-volving Lau. It turns out that this can be realized by introducing the followingmodified formula:

Φ (r) = Φ (r;u) =(r + c0r

2) ddr

log max[F (r) , rn+a+4

]. (11)

Then:

Theorem 16 (Monotonicity formula). Let u be a solution of problem (10).Then, there exists a small r0 and a large c0, both depending only on a, n, ‖ϕ‖C2,1, suchthat Φ (r;u) is monotone nondecreasing for r < r0.

For the proof, we need a Poincare type estimate and a Rellich type identity.Recall that u (0, 0) = 0 since the origin belongs to the free boundary.

Lemma 17 Let u be a solution of problem (10), u (0, 0) = 0. Then∫∂Br

(u (X))2 |y|a dS ≤ Cr

∫Br

|∇u (X)|2 |y|a dX + c (a, n) r6+a+n

and by integrating in r,∫Br

(u (X))2 |y|a dX ≤ Cr2

∫Br

|∇u (X)|2 |y|a dX + c (a, n) r7+a+n

where c, C depend only on a, n and ‖ϕ‖C2,1 .

Proof. See [22].

Lemma 18 The following identity holds for any r ≤ 1.

r

∫∂Br

(|∇θu|2 − u2

ν

)|y|a dS =

∫Br

[(n+a−1) |∇u (X)|2−2 〈X,∇u〉 g (X)] |y|a dX,

(12)where ∇θu denotes the tangential gradient.

Proof. Consider the vector field

F =1

2ya |∇u|2X − ya 〈X,∇u〉∇u (y > 0) .

We have:

divF =1

2(n+ a− 1)ya |∇u|2 − 〈X,∇u〉Lau.

Since 〈X,∇u〉 is a continuous function on B′r that vanishes on Λ = u = 0, wehave that 〈X,∇u〉Lau has no singular part and coincides with 〈X,∇u〉 |y|a g (x).An application of the divergence theorem gives (12).

19

Proof of Theorem 16. First we observe that by taking the maximum in (11) itmay happen that we get a non differentiable functions. However, max

[F (r) , rn+a+4

]is absolutely continuous (it belongs to W 1,1

loc (0, 1)) and in any case, the jump inthe derivative will be in the positive direction.

When F (r) ≤ rn+a+4 we have

Φ (r) =(r + c0r

2) ddr

log rn+a+4

and Φ′ (r) = (n+ a+ 4)c0 > 0.Thus we can concentrate on the case F (r) > rn+a+4 where

Φ (r) =(r + c0r

2) ddr

logF (r) .

We have:

Φ (r) =(r + c0r

2) ∫∂Br 2uuν |y|a dS∫

∂Bru2 |y|a dS

+ (1 + c0r) (n+ a)

≡ 2Ψ (r) + (1 + c0r) (n+ a).

We show that the first term is increasing, by computing its logarithmic deriva-tive. We find:

d

drlog Ψ (r) =

1

r+

c01 + c0r

+ddr

∫∂Br

uuν |y|a dS∫∂Br

uuν |y|a dS−∫∂Br

2uuν |y|a dS∫∂Br

u2 |y|a dS− n+ a

r.

We estimate ddr

∫∂Br

uuν |y|a dS from below. Since∫∂Br

uuν |y|a dS =

∫Br

(|y|a |∇u|2 + uLau)dX

we can write, recalling that |Lau| ≤ c |y|a |x| ,

d

dr

∫∂Br

uuν |y|a dS ≥∫∂Br

|y|a |∇u|2 dS − cr(n+a+2)/2 [F (r)]1/2

.

We now use Lemma 18 to estimate∫∂Br|y|a |∇u|2 dS from below. We find∫

∂Br

|y|a |∇u|2 dS =

∫∂Br

(|uθ|2 + u2

ν

)|y|a dS =

= 2

∫∂Br

u2ν |y|

adS +

1

r

∫Br

[(n+ a− 1) |∇u (X)|2 − 2 〈X,∇u〉 g (X)] |y|a dX =

= 2

∫∂Br

u2ν |y|

adS +

n+ a− 1

r

∫∂Br

uuν |y|a dS

−1

r

∫Br

[(n+ a− 1)u− 2 〈X,∇u〉]g (X) |y|a dX.

20

Therefore

d

dr

∫∂Br

uuν |y|a dS ≥ 2

∫∂Br

u2ν |y|

adS +

n+ a− 1

r

∫∂Br

uuν |y|a dS

−crn+a+1[√

G (r) + r√H (r) +

√rF (r)

]where

G (r) =

∫Br

u2 |y|a dX and H (r) =

∫Br

|∇u|2 |y|a dX.

Collecting all the above estimates, we can write:

d

drlog Ψ (r) = P (r) +Q (r)

with

P (r) =2∫∂Br

u2ν |y|

adS∫

∂Bruuν |y|a dS

−∫∂Br

2uuν |y|a dS∫∂Br

u2 |y|a dS≥ 0

and

Q (r) =c0

1 + c0r− cr(n+a+1)/2

√G (r) +

√rF (r) + r

√H (r)∫

∂Bruuν |y|a dσ

≥ c01 + c0r

− cr(n+a+1)/2

√G (r) +

√rF (r) + r

√H (r)

H (r)− r(n+a+2)/2√G (r)

.

First we estimate F,G,H. Since F (r) > rn+4+a, from the Poincare Lemma 17we have:

rn+4+a < F (r) ≤ CrH (r) + c (a, n) r6+a+n.

Integrating the above inequalities in r, we get:

G (r) =

∫ r

0

F (s) ds ≤ Cr2H (r) + c (a, n) r7+a+n.

This means that, for small enough r0 and r < r0 :

F (r) ≤ crH (r) and G (r) ≤ Cr2H (r)

so that:

Q (r) ≥ c01 + cr

− cr(n+a+1)/2 r√H (r)

H (r)− r(n+a+4)/2√Hr

.

Since rn+4+a < F (r) ≤ CrH (r), we also have H (r) ≥ crn+a+3 and for r0

small:

Q (r) ≥ c01 + c0r

− cr(n+a+1)/2 r√H (r)− r(n+a+4)/2

≥ c01 + c0r

− crn+a+1

2 +1−n+a+32 =

c01 + cr

− c

which is positive if c0 is large and r0 small.

21

2.7 Blow-up sequences and optimal regularity

The optimal regularity of the solution can be obtained by a careful analysis ofthe possible values of Φ (0+). When Φ is constant and the obstacle is zero,[Φ (0+)− n− a] /2 represents the degree of homogeneity at the origin. Thus,by a suitable blow-up of the solution, we will be able to classify the possibleasymptotic behaviors at the origin, using the results of Section 2.5. The firstresult is the following.

Theorem 19 Let u be a solution of problem (10), u (0, 0) = 0. Then eitherΦ (0+;u) = n+ a+ 2 (1 + s) or Φ (0+;u) ≥ n+ a+ 4.

To prove the theorem, guided by the zero obstacle case in [10], a key pointis to introduce the following rescaling:

ur (X) =u (rX)

dr(13)

where

dr =

(r−(n+a)

∫∂Br

u2 |y|a dσ)1/2

=(r−(n+a)F (r)

)1/2

.

Notice that the ”natural” rescaling u (rX) /rµ, where µ = Φ (0+;u) − n − a,would not be appropriate, because on this kind of rescaling we have precisecontrol of its behavior as r → 0 merely from one-side. Rescaling by an averageover smaller and smaller balls provides the necessary adjustments for controllingthe oscillations of u around the origin.Two things can occur:

lim infr→0

drr2

= +∞ first case< +∞ second case.

(14)

The next lemma takes care of the first case.

Lemma 20 Let u be as in Theorem 19. Assume that

lim infr→0

drr2

= +∞.

Then, there is a sequence rk → 0 and a function U0 : Rn+1 → R, non identicallyzero, such that:

1. urk → U0 in W 1,2(B1/2, |ya|

)2. urk → U0 uniformly in B1/2.

3. ∇xurk → ∇xU0 uniformly in B1/2.

4. |y|a ∂yurk → |y|a∂yU0 uniformly in B1/2.

22

Moreover, U0 is a solution of system (9) and its degree of homogeneity is[Φ (0+;u)− n− a]/2.

Proof. First of all, observe that ‖ur‖L2(∂B1,|y|a) = 1 for every r. Using thefrequency formula, the Poincare type Lemma A5 below and the semiconvexityof ur, one can show that ur is bounded in W 1,2

(B1/2, |ya|

).

Now, u+r and u−r are subsolutions of the equation

Law ≥ −cr |y|a |x|

and therefore (see [29]) ur is bounded in L∞(B3/4

).

The semiconvexity of u in the x variable gives, for every tangential directionτ ,

∂ττur =r2

druττ (rX) ≥ −C r

2

dr. (15)

From Lemma 8 we obtain a subsequence urk strongly convergent inW 1,2(B1/2, |ya|

)to some function U0 as rk 0. On the other hand (Theorem 16) we know thatΦ (r;u) is monotone and converges to Φ (0+;u) as r 0. We have:

Φ (rs;u) ∼ rs

∫Brs|∇u (X)|2 |y|a dX∫∂Brs

u2 |y|a dS+ (n+ a) = r

∫Br|∇us (X)|2 |y|a dX∫∂Br

u2s |y|

adS

+ (n+ a) .

We want to set s = rk 0 and pass to the limit in the above expression toobtain:

Φ (0+;u)− (n+ a) = r

∫Br|∇U0 (X)|2 |y|a dX∫∂Br

U20 |y|

adS

. (16)

This is possible since one can show that∫∂Br

u2s |y|

adS ≥ c > 0. From (15), we

have

∂ττurk ≥ −Cr2k

drk→ 0

and therefore U0 is tangentially convex. On the other hand, each ur satisfiesthe following conditions:

(a)ur (x, 0) ≥ 0 in B′1.

(b)

Laur =r2−a

drLau (rX) =

r2

dr|y|a g (rx) in B1\Λ (ur)

(c)

|Laur| ≤r2

dr|y|a |g (rx)| in B1.

Sincer2

dr|y|a |g (rx)| ≤ c r

2

dr|y|a r |x| → 0 as r → 0

23

it follows that U0 is a solution of the homogeneous problemU0 (x, 0) ≥ 0 in B′1

U0 (x,−y) = U0 (x, y) in B1

LaU0 = 0 in B1\ΛLaU0 ≤ 0 in B1, in the sense of distributions

For this problem, the frequency formula holds without any error correction.Thus we conclude that U0 is homogeneous in B1/2 and its degree of homogeneityis exactly (Φ (0+;u)− (n+ a)) /2. Since it is homogeneous, then it can beextended to Rn+1 as a global solution of the homogeneous problem.

Finally, from the a priori estimates in Lemma 7, it follows that we can chooserk so that the sequences urk , ∇urk and |y|a ∂urk converge uniformly in B1/2.

Proof of Theorem 19. In the first case of (14), we use Lemma 20 andTheorem 14 to find the blow-up profile U0 and to obtain that the degree ofhomogeneity of U0 is 1 + s or at least 2. Therefore

Φ (0+;u) = Φ (0+;U0) = n+a+2 (1 + s) or Φ (0+;u) = Φ (0+;U0) ≥ n+a+4.

Consider now the second case of (14).

If F (rk;u) < rn+a+4k for some sequence rk → 0, then, for these values of rk,

Φ (rk;u) = (n+ a+ 4) (1 + c0rk)

so that Φ (0+;U0) = n+ a+ 4.On the other hand, assume that F (r;u) ≥ rn+a+4 for r small. Since we are

in the second case, for some sequence rj 0 we have drj/r2j ≤ C so that

rn+a+4j ≤ F (rj ;u) ≤ Crn+a+4

j .

Taking logs in the last inequality, we get

(n+ a+ 4) log rj ≤ logF (rj ;u) ≤ C + (n+ a+ 4) log rj .

We want to show that Φ (0+;u) ≥ n+ a+ 4. By contradiction, assume that forsmall rj we have Φ (rj ;u) ≤ n+ a+ 4− ε0. Take rm < rj 1 and write:

(n+ a+ 4) (log rj − log rm)− C ≤ logF (rj ;u)− logF (rm;u)

=

∫ rj

rm

d

drlogF (s;u) ds ≤

∫ rj

rm

(r + c0r

2)−1

Φ (s;u) ds ≤∫ rj

rm

r−1Φ (rj ;u) ds

≤ (n+ a+ 4− ε0) (log rj − log rm)

which gives a contradiction if we make (log rj − log rm)→ +∞.

From the classification of the homogeneity of a global profile, we may proceedto identify the unique limiting profile U0, along the whole sequence ur, modulusrotations. precisely, we have:

24

Proposition 21 Let u be as in Theorem 19. Assume that Φ (0+;u) = n+ a+2 (1 + s) . There is a family of rotations Ar, with respect to x, such that ur Arconverges to the unique profile U0 of homogeneity degree 1 + s. More precisely:

1. ur Ar → U0 in W 1,2(B1/2, |ya|

).

2. ur Ar → U0 uniformly in B1/2.

3. ∇x(ur Ar)→ ∇xU0 uniformly in B1/2.

4. |y|a ∂y(ur Ar)→ |y|a ∂yU0 uniformly in B1/2.

We are now ready to prove the optimal 1 + s−decay of u at (0, 0). This isdone in two steps. First, we control the decay of u in terms of the decay ofF (r;u). In turn, the frequency formula provides the precise control of F (r)from above. This is the content of the next two lemmas.

Lemma 22 IfF (r;u) ≤ crn+a+2(1+α) (17)

for every r < 1, then u (0, 0) = 0, |∇u (0, 0)| = 0 and u is C1,α at the origin inthe sense that

|u (X)| ≤ C1 |X|1+α

for |X| ≤ 1/2 and C1 = C1 (C, n, a).

Proof. Consider u+ and u−, the positive and negative parts of u. We havealready noticed that

Lau+, Lau

− ≥ −C |y|a |x| .

For some r > 0, let h be the La−harmonic replacement of u+ in Br. Note that

0 = Lah ≤ La

(u+ + C

|X|2 − r2

2 (n+ a+ 1)

).

Hence, by comparison principle

h ≥ u+ − C ′r2.

From (17) we have:∫∂Br

h2 |y|a dσ =

∫∂Br

(u+)2 |y|a dσ ≤ crn+a+2(1+α).

Since w (X) = |y|a is a A2 weight, from the local L∞ estimates (see [29]) weconclude

supBr/2

|h (X)| ≤ C1 |r|1+α.

Then u+ ≤ h + r2 ≤ Cr1+α in Br/2. A similar estimate holds for u− and theproof is complete.

25

Lemma 23 If Φ (0+;u) = µ then

F (r;u) ≤ crµ

for any r < 1 and c = c (F (1;u) , c0) .

Proof. Let f (r) = maxF (r) , rn+a+4

≥ F (r). Since Φ is nondecreasing:

µ = Φ (0+) ≤ Φ (r) =(r + c0r

2) ddr

log f (r)

and thend

drlog f (r) ≥ µ

(r + c0r2).

An integration gives:

log f (1)− log f (r) ≥∫ 1

r

µ

(s+ c0s2)ds

≥ −µ log r − µ∫ 1

r

[1

s− µ

(s+ c0s2)

]ds

≥ −µ log r − C1µ

so thatlog f (r) ≤ log f (1) + µ log r + C1µ.

Taking the exponential of the two sides we infer

F (r) ≤ f (r) ≤ Crµ

with C = f (1) eC1µ.

The optimal decay of a solution u of (10) follows easily. Precisely:

Theorem 24 Let u be a solution of (10), with u (0, 0) = 0. Then

|u (X)| ≤ C |X|1+ssupB1

|u| ,

where C = C(n, a, ‖g‖Lip

).

Proof. From Theorem 19, µ = Φ (0+) ≥ n+a+ 2 (1 + s) and the conclusionfollows from Lemmas 22 and 23.

The optimal regularity of the solution of the obstacle problem for the frac-tional laplacian is now a simple corollary.

Corollary 25 (Optimal regularity of solutions) Let ϕ ∈ C2,1. Then the solutionu of the obstacle problem for the operator (−∆)

sbelongs to C1,s (Rn).

26

Proof. Using the equivalence between the obstacle problem for (−∆)s

andthe thin obstacle for La, Theorem 24 shows that u − ϕ has the right decay atfree boundary points. This is enough to prove that u ∈ C1,s (Rn).

Remark 2 Observe that it is not true that the solution of the thin obstacleproblem for La is C1,s in both variables x and y. It is quite interesting however,that the optimal decay takes place in both variables at a free boundary point. Inany case, we have that a solution u of one of the systems (7) or (10) belongs toC1,s(B′1/2), for every y0 ∈ (0, 1/2).

2.8 Nondegenerate case. Lipschitz continuity of the freeboundary

In analogy with what happens in the zero obstacle problem, the regularity ofthe free boundary can be inferred for points around which u has an asymptoticprofile corresponding to the optimal homogeneity degree Φ (0+;u) = n + a +2 (1 + s). Accordingly, we say that X0 ∈ F (u) is regular or stable if

µ (X0) = Φ (0+;u) = n+ a+ 2 (1 + s) .

As always, we refer to the origin (X0 = (0, 0)). The strategy to prove regular-ity of F (u) follows the well known pattern first introduced by Athanasopou-los and Caffarelli in [6] and further developed by Caffarelli in [18]. The firststep is to prove that in a neighborhood of (0, 0) there is a cone of tangentialdirections (cone of monotonicity) along which the derivatives of u are nonneg-ative and have nontrivial growth. In particular, the free boundary is the graphxn = f (x1, ..., xn−1) of a Lipschitz function f.The second step is to prove aboundary Harnack principle assuring the Holder continuity of the quotient oftwo nonnegative tangential derivatives. This implies that F (u) is locally a(n− 2)−dimensional manifold of class C1,α.

The following theorem establishes the existence of a cone of monotonicity.

Theorem 26 Assume µ < n + a + 4. Then, there exists a neighborhood Bρof the origin and a tangential cone Γ′ (θ, en) ⊂ Rn × 0 such that, for everyτ ∈ Γ′ (θ, en), we have ∂τu ≥ 0 in Bρ. In particular, the free boundary is thegraph xn = f (x1, ..., xn−1) of a Lipschitz function f.

The theorem follows by applying the following lemma to a tangential deriva-tive h = ∂τur, where ur is the blow-up family (13) that defines the limitingprofile U0, for r small.

Lemma 27 Let Λ be a subset of Rn ×0 . Assume h is a continuous functionwith the following properties:

1. Lah ≤ γ |y|a in B1\Λ.

2. h ≥ 0 for |y| ≥ σ > 0, h = 0 on Λ.

27

3. h ≥ c0 for |y| ≥√

1 + a/8n.

4. h ≥ −ω (σ) for |y| < σ, where ω is the modulus of continuity of h.

There exist σ0 = σ0 (n, a, c0, ω) and γ0 = γ0 (n, a, c0, ω) such that, if σ < σ0

and γ < γ0, then h ≥ 0 in B1/2.

Proof. Suppose by contradiction X0 = (x0, y0) ∈ B1/2 and h (X0) < 0. Let

Q =

(x, y) : |x− x0| <

1

3, |y| ≥

√1 + a

4n

and P (x, y) = |x− x0|2 − n

a+1y2. Observe that LaP = 0. Define

v (X) = h (X) + δP (X)− γ

2(a+ 1)y2.

Then:

• v (X0) = h (X0) + δP (X0)− γ2(a+1)y

20 < 0

• v (X) ≥ 0 on Λ

• Lav = Lah+ δLaP − γ |y|a ≤ 0 outside Λ.

Thus, v must have a negative minimum on ∂Q. On the other hand, if δ, γare small enough, then v ≥ 0 on ∂Q and we have a contradiction. Thereforeh ≥ 0 in B1/2.

Proof of Theorem 26. Since µ = Φ (0; u) < n + a + 4, Theorem 19 givesµ = n + a + 2 (1 + s). Moreover, the blow-up sequence ur converges (modulussubsequences) to the global profile U0, whose homogeneity degree is 1 + s andwhose free boundary is flat.

Let us assume that en is the normal to the free boundary of U0. Then

∂nU0 (x, y) = c(√

x2n + y2 − xn

)s.

For some θ0 > 0, let σ any vector orthogonal to y and xn such that |σ| < θ0.From Theorem 15 we know that U0 is constant in the direction of σ and therefore,if τ = en + σ, ∂τU0 = ∂nU0.

On the other hand, ∇xur → ∇xU0 uniformly in every compact subset ofRn+1. Thus, for every δ0, there is an r for which

|∂τU0 − ∂τur| ≤ δ0

where τ = en + σ. If we differentiate the equation

Laur (X) =r2

dr|y|a g (rX)

28

we get

La [∂τur (X)] =r2

dr|y|a r∂τg (rX) ≤ Cr |y|a in B1\Λ (ur) (18)

and the right hand side tends to zero as r → 0.Thus, for r small enough, ∂τur satisfies all the hypotheses of Lemma 27 and

therefore is nonnegative in B1/2. This implies that near the origin, the freeboundary is a Lipschitz graph.

2.9 Boundary Harnack principles and C1,α regularity ofthe free boundary

2.9.1 Growth control for tangential derivatives

We continue to examine the regularity of F (u) at stable points. As we haveseen, at these point we have an exact asymptotic picture and this fact allowsus to get a minimal growth for any tangential derivative when ur is close to theblow-up limit u0. This is needed in extending the Carleson estimate and theBoundary Harnack principle in our non-homogeneous setting.

First, we need to refine Lemma 27.

Lemma 28 Let δ0 = (12n)−1/2s

. There exists ε0 = ε0 (n, a) > 0 such that if vis a function satisfying the following properties:

1. Lav (X) ≤ ε0 for X ∈ B′1 × (0, δ0) ,

2. v (X) ≥ 0 for X ∈ B′1 × (0, δ0) ,

3. v (x, δ0) ≥ 14n for x ∈ B′1,

thenv (x, y) ≥ C |y|2s in B′1/2 × [0, δ0] .

Proof. Compare v with

w (x, y) =(

1 +ε0

2

)y2 − |x− x0|2

n+ y2s.

Inside B′1 × (0, δ0), to show that w ≤ v in B′1 × (0, δ0) .

Corollary 29 Let u be a solution of (10) with |g| , |∇g| ≤ ε0. Let u0 the usualasymptotic nondegenerate profile and assume that |∇xu−∇u0| ≤ ε0.

Then, if ε0 is small enough, there exists c = c (n, a) such that

uτ (X) ≥ c dist (X,Λ)2s

for every X ∈ B1/2 and every tangential direction τ such that |τ − en| < 1/2.

29

Proof. From (18) and Theorem 26, we know that uτ is positive in B1/2.Applying Lemma 28 we get

uτ ≥ c |y|2s in B1/4.

Let now X = (x, y) ∈ B1/8 and d =dist(X,Λ) . Consider the ball Bd/2 (X). Atthe top point of this ball, say (xT , yT ) we have yT ≥ d/2. Therefore

uτ (xT , yT ) ≥ cd2s.

By Harnack’s inequality,

uτ (x, y) ≥ cuτ (xT , yT ) ≥ cd2s.

2.9.2 Boundary Harnack

Using the growth control from below provided by Lemma 28 it is easy to extendthe Carleson estimate to our nonhomogeneous setting.

Lemma 30 (Carleson estimate). Let D = B1\Λ where Λ ⊂ y = 0. Assumethat ∂Λ ∩ B′1 is given by a Lipschitz (n − 2)−manifold with Lipschitz constantL. Let w ≥ 0 in D, vanishing on Λ. Assume in addition that:

1. |Law|≤ c |y|a in D;

2. nondegeneracy:w (X) ≥ CdβX

for some β ∈ (0, 2), where dX =dist(X,Λ).

Then, for every Q ∈ Λ ∩B1/2 and r small:

supBr(Q)∩D

w ≤ C (n, a, L)w (Ar (Q)) ,

where Ar (Q) is a point such that Bηr (Ar (Q)) ⊂ Br (Q) ∩ D for some η de-pending only on n and L.

Proof. Let w∗ be the harmonic replacement of w in B2r (Q) ∩ D, r small.Standard arguments (see e.g. [21]) give

w∗ (X) ≤ Cw∗ (Ar (Q)) in Br (Q) ∩D.

On the other hand, comparing w with the function w∗ (X)+C(|X −Q|2 − r2

)we get

|w∗ − w| ≤ cr2 in B2r (Q) ∩D.

Thusw (X) ≤ C

[w (Ar (Q)) + cr2

]in Br (Q) ∩D.

30

From the nondegeneracy condition we infer w (Ar (Q)) ≥ crβ and since β < 2,the theorem follows.

The following theorem expresses a boundary Harnack principle valid in ournonhomogeneous setting.

Theorem 31 (Boundary Harnack principle). Let D = B1\Λ where Λ ⊂ y = 0.Let v, w positive functions in D satisfying the hypotheses 1 and 2 of Lemma 30and symmetric in y. Then there is a constant c = c (n, a, L) such that

v (X)

w (X)≤ c

v(0, 1

2

)w(0, 1

2

) in B1/2.

Moreover, the ratio v/w is Holder continuous in B1/2, uniformly up to Λ.

Proof. Let us normalize v, w setting v(0, 1

2

)= w

(0, 1

2

)= 1. From the

Carleson estimate and Harnack inequality, for any δ > 0 we get:

v (X) ≤ C in B3/4

andw (X) ≥ c in B3/4 ∩ |y| > δ .

This implies that, for a constant s small enough, v − sw fulfills the conditionsof Lemma 26. Therefore v − sw ≥ 0 in B1/2 or, in other words:

v (X)

w (X)≤ s in B1/2.

At this point, the rest of proof follows by standard iteration.

2.9.3 C1,α regularity of the free boundary

As in the case of the thin-obstacle for the Laplace operator, the C1,α regularityof the free boundary follows by applying Theorem 31 to the quotient of twopositive tangential derivatives. Precisely we have:

Theorem 32 Let u be a solution of (10). Assume ϕ ∈ C2,1 and Φ (0) <n+a+4. Then the free boundary is a C1,α (n− 1)−dimensional surface aroundthe origin.

Remark 3 As we have already noted, the boundary Harnack principle in The-orem 31 is somewhat weaker than the usual one. Notice the less than quadraticdecay to zero of the solution at the boundary, necessary to control the effect ofthe right hand side.

31

2.10 Non regular points on the free boundary

As we have seen, the regularity of the free boundary can be achieved around regu-lar or stable points, corresponding to the optimal homogeneity [Φ (0+;u)− n− (a+ 4)] /2 =1 + s.

On the other hand, there are solutions of the zero-thin obstacle problem like

ρk+1/2 cos2k + 1

2θ or ρ2k cos 2kθ, k ≥ 2,

vanishing of higher order at the origin. In these cases we cannot expect anyregularity of the free boundary.

The non regular points of the free boundary can be divided in two classes:the set Σ (u) at which Λ (u) has a vanishing density (singular points), that is

Σ (u) =

(x0, 0) ∈ F (u) : lim

r→0+

Hn (Λ (u) ∩B′r (x0))

rn= 0

,

and the set of non regular, non singular points. The following example (see[36]), given by the harmonic polynomial

p (x1, x2, y) = x21x

22 − (x2

1 + x22)y2 +

1

3y4

shows that the entire free boundary of the zero-thin obstacle problem (2) couldbe composed by singular points. In fact F (p) = Λ (p) is given by the union ofthe lines x1 = y = 0 and x2 = y = 0.

We shall see that, as in this example, the singular set is contained in theunion of C1−manifold of suitable dimension ([36]).

2.10.1 Structure of the singular set (zero thin obstacle)

In this section we describe the main results and ideas from ([36]). Their tech-niques works also for the fractional Laplacian but, for the sake of simplicity, wepresent them in the (important case) of the thin obstacle problem. Considerthe following problem:

u− ϕ ≥ 0 in B′1∆u = 0 in B1\ (x, 0) : u (x, 0) = ϕ (x)(u− ϕ)uxn = 0, uxn ≤ 0 in B′1u (x,−y) = u (x, y) in B1.

where ϕ : B′ −→ R. For better clarity we outline the proofs in the special caseϕ = 0.

As we shall see, around the singular points a precise analysis of the behaviorof u and the structure of the free boundary can be carried out. The analysis ofthe free boundary around the other kind of points is still an open question ingeneral. However, in some important special cases, complete information canbe given, as we shall see in the sequel.

32

It is convenient to classify a point on F (u) according to the degree of ho-mogeneity of u, given by the frequency formula centered at that point. In otherwords, set

ΦX0 (r;u) = r

∫Br(X0)

|∇u|2∫∂Br(X0)

u2dS

and define

Fκ (u) =X0 ∈ F (u) : ΦX0 (0+;u) = κ

,

Σκ (u) = Σ (u) ∩ Fκ (u) .

According to these notations, X0 is a regular point if it belongs to F3/2 (u) .Since r 7−→ ΦX0 (r;u) is nondecreasing, it follows that the mapping

X0 7−→ ΦX0 (0+;u)

is upper-semicontinuous. Moreover, since ΦX0 (0+;u) misses all the values inthe interval (3/2, 2), it follows that F3/2 (u) is a relatively open subset of F (u).

Before stating the structure theorems of Σ (u), it is necessary to exam-ine the asymptotic profiles obtained at a singular point from the rescalingsvr (X) = v (rX) /(r−n

∫∂Br(X0)

u2)1/2; indeed, saying that X0 = (x0, y) ∈ Σ (u)

is equivalent to state that

limr→0+

Hn (Λ (vr) ∩B′1 (x0)) = 0. (19)

As we see immediately, this implies that any blow-up v∗ at a singular point isharmonic in B1 (X0). Moreover, it is possible to give a complete characterizationof these blow-ups in terms of the value κ = ΦX0 (r;u) . In particular

Σκ (u) = Fκ (u) for κ = 2m, m ∈ N.

Theorem 33 (Blow-ups at singular points). Let (0, 0) ∈ Fκ (u). The followingstatements are equivalent:

(i) (0, 0) ∈ Σκ (u).

(ii) Any blow-up of u at the origin is a non zero homogeneous polynomialpκ of degree κ satisfying

∆pκ = 0, pκ (x, 0) ≥ 0, pκ (x,−y) = pκ (x, y) .

(iii) κ = 2m for some m ∈ N.

Proof. (i) =⇒ (ii). Since u is harmonic in B±1 , we have:

∆vr = 2(∂yvr)Hn|Λ(vr) in D′ (B1) . (20)

Then vr is equibounded in H1loc (B1), and (19) says that

Hn (Λ (vr) ∩B′1)→ 0

33

as r → 0. Thus (20) implies that ∆vr → 0 in D′ (B1), and therefore any blow-upv∗ must be harmonic in B1.

From Section 2.7 we know that v∗ is homogeneous and non trivial, and thusit can be extended to a harmonic function in all of Rn+1. Being homogeneous,v∗ has at most a polynomial growth at infinity, hence Liouville Theorem impliesthat v∗ is a non trivial homogeneous harmonic polynomial pκ of integer degreeκ. The properties of u imply that pκ (x, 0) ≥ 0, and pκ (x,−y) = pκ (x, y) inRn+1.

(ii) =⇒ (iii). We must show that κ is an even integer. If κ is odd, thenonnegativity of pκ on y = 0 implies that pκ vanishes on the hyperplane y = 0.On the other hand, from the even symmetry in y we infer that ∂ypκ (x, 0) ≡ 0in Rn. Since pκ is harmonic, the Cauchy-Kowaleskaya Theorem implies thatpκ ≡ 0 in Rn+1. Thus κ = 2m, for some m ∈ N.

(ii) =⇒ (i) . Suppose (0, 0) is not a singular point. Then, there exists asequence rj → 0 such that

Hn (Λ (vr) ∩B′1) ≥ δ > 0.

We may assume that vrj converges to a blow-up p∗. We claim that

Hn (Λ (p∗) ∩B′1) ≥ δ > 0.

Indeed, otherwise, there exists an open set U ⊂ Rn with Hn (U) < δ such that

Λ (p∗) ∩ B′1 ⊂ U. Then, for j large, we must have Λ (vr) ∩ B′1 ⊂ U which is a

contradiction, since Hn(Λ(vrj)∩B′1) ≥ δ > Hn (U).

This implies that p∗ (x, 0) ≡ 0 in Rn and consequently in Rn+1, by theCauchy-Kowaleskaya theorem. Contradiction to (ii).

(iii) =⇒ (ii) . From Almgren’s formula, any blow-up is a κ−homogeneoussolution of the zero thin obstacle problem in Rn+1. Then ∆v = 2vyHn|Λ(v) in

Rn+1, with vy ≤ 0 on y = 0. Since κ = 2m, the following auxiliary lemmaimplies that ∆v = 0 in Rn+1 and therefore v is a polynomial.

Lemma 34 Let v ∈ H1loc

(Rn+1

)satisfy ∆v ≤ 0 in Rn+1 and ∆v = 0 in

Rn+1\ y = 0. If v is homogeneous of degree κ = 2m then ∆v = 0 in Rn+1.

Proof. By assumption, µ = ∆v is a nonpositive measure, supported ony = 0 . We have to show that µ = 0.

Let q be a 2m−homogeneous harmonic polynomial, which is positive ony = 0 \ (0, 0). For instance:

q (X) =

n∑j=1

Re(xj + iy)2m.

Take ψ ∈ C∞0 (0,+∞) such that ψ ≥ 0 and let Ψ (X) = ψ (|X|). Then, we have:

34

−〈µ,Ψq〉 = −〈∆v,Ψq〉 =

∫Rn+1

(Ψ∇v · ∇q + q∇v · ∇Ψ) dX

=

∫Rn+1

(−Ψv∆q − v∇q · ∇Ψ + q∇v · ∇Ψ) dX

=

∫Rn+1

[−Ψv∆q − vψ′ (|X|)|X|

(X · ∇q) + qψ′ (|X|)|X|

(X · ∇v)]dX

= 0

since ∆q = 0, X ·∇q = 2mq, X ·∇v = 2mv. This implies that µ is supported atX = 0, that is µ = cδ(0,0). On the other hand, δ(0,0) is homogeneous of degree− (n+ 1) while µ is homogeneous of degree 2m− 2 and therefore µ = 0.

Definition 1 We denote by Pκ the class of homogeneous harmonic polynomialsof degree κ = 2m defined in Theorem 33, that is:

Pκ = pκ : ∆pκ = 0,∇pκ ·X = κpκ, pκ (x, 0) ≥ 0, pκ (x,−y) = pκ (x, y) . (21)

Via the Cauchy-Kovaleskaya Theorem, it is easily shown that the polynomi-als in Pκ can be uniquely determined from their restriction to the hyperplaney = 0. Thus, if pκ ∈ Pκ is not trivial, then also its restriction to y = 0 must benon trivial.

The next theorem gives an exact asymptotic behavior of u near a pointX0 ∈ Σκ (u) .

Theorem 35 (κ− differentiability at singular points) Let X0 ∈ Σκ (u), withκ = 2m, m ∈ N. Then there exists a non trivial pX0

κ ∈ Pκ such that

u (X) = pX0κ (X −X0) + o (|X −X0|κ) . (22)

Moreover, the mapping X0 7−→ pX0κ is continuous on Σκ (u) .

The proof is given in Subsection B... Note that, since Pκ is a convex subsetof the finite dimensional space of the homogeneous polynomial of degree κ, allthe norms on Pκ are equivalent. Thus, the continuity in Theorem 35 can beunderstood, for instance, in the L2 (∂B1) norm.

The structure of F (u) around a singular point X0 depends on the dimensionof the singular set at that point, as defined below in terms of the polynomialpX0κ :

Definition 2 (Dimension at a singular point) Let X0 ∈ Σκ (u) . The dimensionof Σκ (u) at X0 is defined as the integer

dX0κ = dim

ξ ∈ Rn : ξ · ∇xpX0

κ (x, 0) = 0 for all x ∈ Rn.

Since pX0κ (x, 0) is not identically zero on Rn, we have

0 ≤ dX0κ ≤ n− 1.

For d = 0, 1, ..., n− 1 we define

Σdκ (u) =X0 ∈ Σκ (u) : dX0

κ = d.

35

(0, 0) x1

x2

Σ04

Σ12

Σ12

Σ12

Σ12

6

-t

Figure 1: Free boundary for u(x1, x2, y) = x21x

22−(x2

1 + x22

)y2 + 1

3y4 in R3 with

zero thin obstacle on R2 × 0. (Credit: Garofalo-Petrosyan, 2009)

Here is the structure theorem:

Theorem 36 (Structure of the singular set) Every set Σdκ (u), κ = 2m, m ∈ N ,d = 0, 1, ..., n − 1, is contained in a countable union of d− dimensional C1

manifolds.

For the harmonic polynomial

p (x1, x2, y) = x21x

22 − (x2

1 + x22)y2 +

1

3y4

considered above, it is easy to check that (0, 0) ∈ Σ04 (u) and the rest of the

points on F (u) belongs to Σ12 (u) (see Figure 1).

As in the classical obstacle problem, the main difficulty in the analysis con-sists in establishing the uniqueness of the Taylor expansion (22), which in turnis equivalent to establish the uniqueness of the limiting profile obtained by thesequence of rescalings ur.

A couple of monotonicity formulas, strictly related to Almgren’s formula andto formulas in [65] and [53], play a crucial role in circumventing these difficulties.

2.10.2 Monotonicity formulas

We introduce here two main tools. We start with a one-parameter family ofmonotonicity formulas (see also citeW) based on the functional:

WX0κ (r;u) =

1

rn−1+2κ

∫Br(X0)

|∇u|2 dX − κ

rn+2κ

∫∂Br(X0)

u2dS

≡ 1

rn−1+2κH (r)− κ

rn+2κK (r) .

36

where κ ≥ 0. If X0 = (0, 0) we simply write Wκ (r;u).The functionals WX0

κ (r;u) and ΦX0 (r;u) are strictly related. Indeed, takingfor brevity X0 = (0, 0), we have:

Wκ (r;u) =K (r)

rn+2κ[Φ (r;u)− κ] . (23)

This formula shows that WX0κ (r;u) is particularly suited for the analysis of

asymptotic profiles at points X0 ∈ Fκ (u). Moreover, for these points, sincefrom Almgren’s frequency formula we have ΦX0 (r;u) ≥ ΦX0 (0+;u) = κ, wededuce that

WX0κ (r;u) ≥ 0. (24)

The next theorem shows the main properties of Wκ (r;u) .

Theorem 37 (W−type monotonicity formula) Let u be a solution of our zeroobstacle problem in B1. Then, for 0 < r < 1,

d

drWκ (r;u) =

1

rn+2κ

∫∂Br

(X · ∇u− κu)2dS.

As a consequence, the function r 7−→Wκ (r;u) is nondecreasing in (0, 1) . More-over, Wκ (·;u) is constant if and only if u is homogeneous of degree κ.

Proof. Using the identities

H ′ (r) =

∫∂Br

|∇u|2 dS and K ′ (r) =n

rK (r) + 2

∫∂Br

uuνdS (25)

and ∫∂Br

|∇u|2 dS =(n− 1)

r

∫Br

|∇u|2 + 2

∫∂Br

(uν)2dS. (26)

we get:

d

drWκ (r;u)

=1

rn−1+2κ

H ′ (r)− n− 1 + 2κ

rV (r)− κ

rK ′ (r) +

κ (n+ 2κ)

r2K (r)

=

1

rn−1+2κ

2

∫∂Br

u2νdS −

4κ

r

∫∂Br

uuνdS +2κ2

r2

∫∂Br

u2dS

=

2

rn+1+2κ

∫∂Br

(X · ∇u− κu)2dS.

The next one is a generalization of a formula in [53], based on the functional

MX0κ (r;u, pκ) =

1

rn+2κ

∫∂Br(X0)

(u− pκ)2dS.

37

We set Mκ (r;u, pκ) = M(0,0)κ (r;u, pκ). Here κ = 2m and pκ is a polynomial

in the class Pκ defined in (21). Since it measures the distance of u from anhomogeneous polynomial of even degree, it is apparent that MX0

κ (r;u, pκ) isparticularly suited for the analysis of blow-up profiles at points X0 ∈ Σκ (u).We have:

Theorem 38 (M−type monotonicity formula) Let u be a solution of our zeroobstacle problem in B1. Assume (0, 0) ∈ Σκ (u), κ = 2m, m ∈ N. Then, for0 < r < 1, the function r 7−→Mκ (r;u, pκ) is nondecreasing in (0, 1) .

Proof. We show that

d

drMκ (r;u, pκ) ≥ 2

rWκ (r;u) ≥ 0.

Let w = u− pκ. We have:

d

dr

1

rn+2κ

∫∂Br

w2dS =d

dr

1

r2κ

∫∂B1

w (rY )2dS

=1

r2κ+1

∫∂B1

[w (rY ) [∇w (rY ) · rY − κw(rY )]dS

=2

rn+1+2κ

∫∂Br

w(X · ∇w − κw)dS

On the other hand, since Φ (r; pκ) = κ, it follows that

Wκ (r; pκ) = 0

and we can write:Wκ (r;u) = Wκ (r;u)−Wκ (r; pκ)

=2

rn−1+2κ

∫Br

(|∇w|2 + 2∇w · ∇pκ)dX − κ

rn+2κ

∫∂Br

(w2 + 2wpκ)dS

=2

rn−1+2κ

∫Br

|∇w|2 dX − κ

rn+2κ

∫∂Br

w2dS +

∫∂Br

w(X · ∇pκ − κpκ)dS

=2

rn−1+2κ

∫Br

|∇w|2 dX − κ

rn+2κ

∫∂Br

w2dS

= − 2

rn−1+2κ

∫Br

w∆wdX +1

rn+2κ

∫∂Br

w(X · ∇w − κw)dS

≤ 1

rn+2κ

∫∂Br

w(X · ∇w − κw)dS =r

2

d

drMκ (r;u, pκ)

since w∆w = (u− pκ) (∆u−∆pκ) = −pκ∆u ≥ 0.

38

2.10.3 Proofs of Theorems 35 and 36

With the above monotonicity formulas at hands we are ready to prove Theorems35 and 36. Recall that, from the frequency formula, we have the estimate

|u (X)| ≤ c |X|κ in B2/3 (27)

for any solution of our free boundary problem. At a singular point, we have alsoa control from below.

Lemma 39 Let u be a solution of our zero obstacle problem in B1. Assume(0, 0) ∈ Σκ (u). Then,

sup∂Br

|u| ≥ crκ (0 < r < 2/3). (28)

Proof. Suppose that (28) does not hold. Then, for a sequence rj → 0 wehave

hj =

(∫∂Brj

u2dS

)1/2

= o(rκj)

.

We may also assume that (see Lemma 33)

vj (X) =u (rjX)

hj→ qκ (X)

uniformly on ∂B1, for some qκ ∈ Pκ. Since∫∂B1

q2κdS = 1, it follows that qκ is

non trivial.Under our hypotheses, we have

Mκ (0+;u, qκ) = limj→∞

1

rn+2κj

∫∂Brj

(u−qκ)2dS =

∫∂B1

q2κdS =

1

rn+2κj

∫∂Brj

q2κdS.

Hence, ∫∂Brj

(u− qκ)2dS ≥∫∂Brj

q2κdS

or ∫∂Brj

(u2 − 2uqκ)dS ≥ 0.

Rescaling, we obtain

hjrκj

∫∂Brj

(hjrκjv2j − 2vjqκ)dS ≥ 0.

Dividing by hjrκj and letting j →∞ we get

−∫∂Brj

q2κdS ≥ 0

39

which gives a contradiction, since qκ is non trivial.

Given the estimates (27) and (28) around a point X0 ∈ Σκ (u) , it is naturalto introduce the family of homogeneous rescalings given by

u(κ)r (X) =

u (rX +X0)

rκ.

From the estimate (27) we have that, along a sequence r = rj , uκr → u0 in

C1,αloc (Rn). We call u0 homogeneous blow-up. Lemma 39 assures that u0 is non

trivial. The next results establishes the uniqueness of these asymptotic profilesand proves the first part of Theorem 35.

Theorem 40 (Uniqueness of homogeneous blow-up at singular points) Assume(0, 0) ∈ Σκ (u) . Then there exists a unique non trivial pκ ∈ Pκ such that

u(κ)r (X) =

u (rX)

rκ→ pκ (X) .

As a consequence, (22) holds.

Proof. Consider a homogeneous blow-up u0. For any r > 0 we have:

Wκ (r;u0) = limrj→0

Wκ(r;u(κ)j ) = lim

rj→0Wκ(rrj ;u) = lim

rj→0Wκ(0+;u).

From Theorem 35 we infer that u0 is homogeneous of degree κ. The samearguments in the proof of Lemma 34 give that u0 must be a polynomial pκ ∈ Pκ.

To prove the uniqueness of u0, apply the M−monotonicity formula to u andu0. We have:

Mκ (0+;u, u0) = cn limj→∞

∫∂B1

(u(κ)j − u0)2dS = 0.

In particular, by monotonicity, we obtain also that

cn

∫∂B1

(u(κ)r − u0)2dS = Mκ (r;u, u0)→ 0

as r → 0, and not just over a subsequence rj . Thus, if u′0 is a homogeneousblow-up, obtained over another sequence r′j → 0, we deduce that∫

∂B1

(u′0 − u0)2dS = 0.

Since u0 and u′0 are both homogeneous of degree κ, they must coincide in Rn+1.

The next lemma gives the second part of Theorem 35.

40

Lemma 41 (Continuous dependence of the blow-ups) For X0 ∈ Σκ (u) denoteby pX0

κ the blow-up of u obtained in Theorem 40 so that:

u (X) = pX0κ (X −X0) + o (|X −X0|κ) .

Then, the mapping X0 7−→ pX0κ from Σκ (u) to Pκ is continuous. Moreover,

for any compact K ⊂ Σκ (u) ∩ B1, there exists a modulus of continuity σK ,σK (0+) = 0, such that∣∣u (X)− pX0

κ (X −X0)∣∣ ≤ σK (|X −X0|) |X −X0|κ

for every X0 ∈ K.

Proof. As we have already observed, we endow Pκ with the L2 (∂B1) norm.The first part of the lemma follows as in Theorem 40. Indeed, fix ε > 0 and rεsuch that

MX0κ (rε;u, p

X0κ ) =

1

rn+2κε

∫∂Brε

(u (X +X0)− pX0κ )2dS < ε.

There exists δε such that if X ′0 ∈ Σκ (u) and |X0 −X ′0| < δε, then

MX′0κ (rε;u, p

X0κ ) =

1

rn+2κε

∫∂Brε

(u (X +X ′0)− pX0κ )2dS < 2ε.

By monotonicity, we deduce that, for 0 < r < rε,

MX′0κ (r;u, pX0

κ ) =1

rn+2κ

∫∂Br

(u (X +X ′0)− pX0κ )2dS < 2ε.

Letting r → 0 we obtain

MX′0κ (0+;u, pX0

κ ) = cn

∫∂B1

(pX′0κ − pX0

κ )2dS < 2ε

and therefore the first part of the lemma is proved.To show the second part, note that if |X0 −X ′0| < δε and 0 < r < rε, we

have:∥∥∥u (·+X ′0)− pX′0

κ

∥∥∥L2(∂Br)

≤∥∥u (·+X ′0)− pX0

κ

∥∥L2(∂Br)

+∥∥∥pX0

κ − pX′0κ

∥∥∥L2(∂Br)

≤ 2 (2ε)1/2

rκ+(n−1)/2.

This is equivalent to ∥∥∥wX′0r − pX′0κ ∥∥∥L2(∂B1)

≤ 2 (2ε)1/2

(29)

where

wX′0r (X) =

u (rX +X ′0)

rκ.

41

Covering K with a finite number of balls Bδε(X′0)(X ′0) for some X ′0 ∈ K, we

obtain that (29) holds for all X ′0 ∈ K with r ≤ rKε .We claim that, if X ′0 ∈ K and 0 < r < rKε then∥∥∥wX′0r − pX′0κ ∥∥∥

L∞(B1/2)≤ Cε with Cε → 0 as ε→ 0. (30)

To prove the claim, observe that the two functions wX′0r and p

X′0κ are both solution

of our zero thin obstacle problem, uniformly bounded in C1,α(B±1 ). If (30) were

not true, by compactness, we can construct a sequence of solutions convergingto a non trivial zero trace solutions (from (29)). The uniqueness of the solutionof the thin obstacle problem with Dirichlet data implies a contradiction.

It is easy to check that the claim implies the second part of the lemma.

We are now in position to prove Theorem 36. The proof uses the Whitney’sextension theorem (see [66] or [67]) and the implicit function theorem. We recallthat the extension theorem prescribes the compatibility conditions under whichthere exists a Ck function f in RN having prescribed derivatives up to the orderk on a given closed set.

Since our reference set is Σκ (u), we first need to show that Σκ (u) is acountable union of closed sets (an Fσ set). This is done in the next Lemma.

Lemma 42 (Topological structure of Σκ (u)) Σκ (u) is a Fσ set.

Proof. Let Ej be the set of points X0 ∈ Σκ (u) ∩B1−1/j such that

1

jρκ ≤ sup

|X−X0|=ρ|u (X)| < jρκ (31)

for 0 < ρ < 1− |X0|. By non degeneracy and (27) we know that

Σκ (u) ⊂ ∪j≥1Ej .

We want to show that Ej is a closed set. Indeed, if X0 ∈ Ej then X0 satisfies(31), and we only need to show that X0 ∈ Σκ (u), i.e., from Theorem 2.9.1, thatΦX0 (0+;u) = κ.

Since the function X 7→ ΦX (0+;u) is upper-semicontinuous we deduce thatΦX0 (0+;u) = κ′ ≥ κ. If we had κ′ > κ, we would have

|u (X)| ≤ |X −X0|κ′

in B1−|X0| (X0) ,

which contradicts the estimate from below in (31). Thus κ′ = κ and X0 ∈Σκ (u).

We are now in position for the proof of Theorem 36.

Proof of Theorem 36. We divide the proof into two steps. Recall thatΣκ (u) = Fκ (u) if κ = 2m.

42

Step 1. Whitney’s extension. For simplicity it is better to make a slightchange of notations, letting y = xn+1 and X = (x1, ..., xn, xn+1) . Let K = Ejbe one of the compact subsets of Σκ (u) constructed in Lemma 42. We can write

pX0κ (X) =

∑|α|=κ

aα (X0)

α!Xα.

The coefficients aα (X) are continuous on Σκ (u) by Theorem 35. Since u = 0on Σκ (u) we have∣∣pX0

κ (X −X0)∣∣ ≤ σ (|X −X0|) |X −X0|κ X ∈ K.

For every multi-index α, 0 ≤ |α| ≤ κ, define:

fα (X) =

0 if 0 < |α| < κaα (X) if |α| = κ

X ∈ Σκ (u) .

We want to construct a function f ∈ Cκ(Rn+1

), whose derivatives ∂αf up

to the order κ are prescribed and equal to fα on K. The Whitney extensiontheorem states that this is possible if, for all X,X0 ∈ K, the following coherenceconditions hold for every multi-index α, 0 ≤ |α| ≤ κ :

fα (X) =∑

|β|≤κ−|α|

fα+β (X0)

β!(X −X0)β +Rα (X,X0) (32)

with|Rα (X,X0)| ≤ σKα (|X −X0|) |X −X0|κ−|α| , (33)

where σKα is a modulus of continuity.

Claim: (32) and (33) hold in our case.

Proof. Case |α| = κ. Then we have

Rα (X,X0) = aα (X)− aα (X0)

and therefore |Rα (X,X0)| ≤ σα (|X −X0|) by the continuity on K of the mapX 7→ pXκ .

Case 0 ≤ |α| < κ. We have

R0 (X,X0) = −∑

γ>α,|γ|=κ

aγ (X0)

(γ − α)!(X −X0)γ−α = −∂αpX0

κ (X −X0) .

Now, suppose that there exists no modulus of continuity σα such that (33)holds for all X,X0 ∈ K. Then, there is δ > 0 and sequences Xi, Xi

0 ∈ K with∣∣Xi −Xi0

∣∣ = ρi → 0

and such that ∣∣∂αpX0κ (X −X0)

∣∣ ≥ δ ∣∣Xi −Xi0

∣∣κ−|α| . (34)

43

Consider the rescalings

wi (X) =u(Xi

0 + ρiX)

ρκi, ξi =

Xi −Xi0

ρi.

We may assume that Xi0 → X0 ∈ K and ξi → ξ0 ∈ ∂B1. From Lemma 41 we

have that ∣∣∣wi (X)− pXi0

κ (X)∣∣∣ ≤ σ (ρi |X|) |X|

κ

and therefore wi (X) converges to pXi0κ (X) , uniformly in every compact subset

of Rn+1.Note that, since Xi, Xi

0 ∈ K, the inequalities (31) are satisfied there. More-over, we also have that similar inequalities are satisfied for the rescaled functionwi at 0 and ξi.

Thus, passing to the limit, we deduce that

1

jρκ ≤ sup

|X−X0|=ρ

∣∣pX0κ (X)

∣∣ < jρκ, 0 < ρ < +∞.

This implies that ξ0 is a point of frequency κ = 2m for the polynomial pX0κ so

that, from Theorem 35, we infer that ξ0 ∈ Σκ(pξ0κ ). In particular,

∂αpξ0κ = 0 for |α| < κ.

However, dividing both sides of (34) by ρκ−|α|i and passing to the limit, we

obtain ∣∣∂αpξ0κ ∣∣ ≥ δ,a contradiction.

This ends the proof of the claim.

Step 2. Implicit function theorem. Applying Whitney’s Theorem we deducethe existence of a function f ∈ Cκ

(Rn+1

)such that

∂αf = fα on Ej

for every |α| ≤ κ. Suppose now X0 ∈ Σdκ (u). This means that

d = dimξ ∈ Rn : ξ · ∇xpX0

κ (x, 0) ≡ 0.

Then there are n− d linearly independent unit vectors νi ∈ Rn, such that

νi · ∇xpX0κ (x, 0) is not identically zero.

This implies that there exist multi-indices βi of order∣∣βi∣∣ = κ− 1 such that

∂νi(∂βipX0

κ (X0)) 6= 0.

This can be written as

∂νi(∂βif (X0)) 6= 0, i = 1, ..., n− d. (35)

44

On the other hand, we have

Σdκ (u) ∩ Ej ⊂ ∪i=1,...,n−d

∂β

i

f = 0.

From (35) and the implicit function theorem, we deduce that Σdκ (u) ∩ Ej iscontained in a d−dimensional C1 manifold in a neighborhood of X0. SinceΣκ (u) = ∪Ej we conclude the proof.

2.11 Non zero obstacle

The above differentiability and the structure theorems can be extended to thenon zero obstacle case, if ϕ ∈ Ck,1 (B′1) for some integer k ≥ 2 ([36]). A crucialtool is once more a frequency formula which generalizes the one in Theorem 16.First, we introduce a useful change of variable to reduce to the case in whichthe Laplacian is very small outside the coincidence set. Let u be a solution ofthe obstacle problem and set:

v (x, y) = u (x, y)− Qk (x, y)− (ϕ (x)−Qk (x)),

where Qk is the k-th Taylor polynomial of ϕ at the origin and Qk is its evenharmonic extension to all of Rn+1. Now v can be evenly extended to y < 0 andthen

|∆v| ≤M |x|k−1+ 2 |uy|Hn|B′1 in D′(B1).

The generalized frequency formula takes the following form: If ‖v‖C1

(B

+1

) ≤M/2, then there exists rM and CM such that the function

r 7−→ Φk (r;u) =(r + CMr

2) ddr

log maxK (r) , rn+2k

is nondecreasing for 0 < r < rM .

Also the Weiss and Monneau monotonicity formulas have to be modifiedaccordingly to take into account the perturbation introduced by the non-zeroobstacle. Indeed we have: For κ ≤ k, 0 < r < r′M , C

′M > 0,

d

drWκ (r;u) ≥ −C ′M

and for κ = 2m < k, 0 < r < r′′m, C′′M > 0,

d

drMκ (r;u, pκ) ≥ −C

′′

M

(1 + ‖pκ‖L2(B1)

).

Coherently, the Differentiability Theorem 35 and the Structure Theorem 36 holdfor κ = 2m < k. This limitation is due to the fact that for points in Fκ (u) eventhe blow-ups are not properly defined.

45

Thus, the analysis of the free boundary around a singular point is rathersatisfactory. It remains open the study of the nonregular, nonsingular points,i.e. the set

Ξ (u) = ∪Fκ (u) : κ >

3

2, κ 6= 2m,m ≥ 1, integer

.

It should be noted that Ξ (u) could in principle be a large part of F (u) . Anotherimportant issue that remains open is whether F (u) has Hausdorff dimensionn− 1.

Further analysis of F (u) clearly depends on the possible values that the fre-quency κ may attain. A partial classification of convex global solutions excludesthe interval (3/2, 2) from the range of possible values of κ.

As observed in ([36]) it is plausible that the only possible values are

κ = 2m− 1

2or κ = 2m, m ≥ 1, integer.

This is indeed true in dimension 2 (n = 1), see remark 1.2.8. in [36].

2.12 A global regularity result (fractional Laplacian)

In ([12]), Barrios, Figalli and Ros-Oton consider both the global and the localversion of the Signorini problem for the operator La, thus covering also thecase of the obstacle problem for the fractional Laplacian. We limit ourselves tobriefly describe their main results and ideas in the case of the local Signoriniproblem.

Under basically two main assumptions the authors can give a complete pic-ture of the free boundary, recovering a result completely analogous to the clas-sical case (s = 1). The first assumption is a strict concavity of the obstacle,the same assumption needed in the case of the classical obstacle problem. Thesecond one prescribes zero boundary values of the solution and it turns out tobe a crucial assumption. Precisely their main result in the case of the Signoriniproblem is the following. As in Section 2.7 we define the blow-ups of v at X0

by

vX0r (X) =

v (r(X −X0))

drdr =

(r−n−a

∫∂Br(X0)

u2 |y|a dσ

)1/2

. (36)

Theorem 43 Let ϕ : B′1 → R with ϕ|∂B′1 < 0, and u : B1 ⊂ Rn+1 → R be a

solution of the following problem:u (x, 0) ≥ ϕ (x) on B′1Lau = 0 in B1\ (y = 0 ∩ u = ϕ)Lau ≥ 0 in B1

u = 0 on ∂B

46

with u (x,−y) = u (x, y). Assume that

ϕ ∈ C3,γ (B′1) , ∆ϕ ≤ −c0 < 0 in ϕ > 0 , ∅ 6= ϕ > 0 ⊂⊂ B′1 (37)

for some c0 > 0 and γ > 0. Then, at every singular point the blow-up of u is ahomogeneous polynomial of degree 2, and the free boundary can be decomposedas

F (u) = F1+s (u) ∪ F2 (u)

where F1+s (u) (resp. F2 (u)) is an open (resp. closed) subset of F (u) . More-over, F1+s (u) is a (n− 1)-dimensional manifold of class C1,α, while F2 (u) canbe stratified as the union of sets

F k2 (u)

k=0,1,...,n−1

, where F k2 (u) is contained

in a k-dimensional manifold of class C1.

The regularity of F1+s (u) comes from [22]. Here the main points are thatthe nonregular points are only singular points, that the blow-up at these pointshas homogeneity 2, that F2 (u) can be stratified into C1 manifolds and that eachF k2 (u) is completely contained in a k-dimensional C1−manifold.

The key lemma is the following nondegeneracy result. It says that u detachesquadratically from the obstacle putting out of game all possible frequenciesgreater than 2.

Lemma 44 Let u be as in Theorem 43. Then there exists constants c1, r1 > 0such that the following holds: for any X0 ∈ F (u) we have

supx∈B′r(x0)

(u (x, 0)− ϕ (x)) ≥ c1r2, 0 < r < r1.

Proof. Since u > 0 on the contact set, compactly contained in B1, we deducethat

ϕ ≥ h0 > 0 on u = ϕ .

For r1 > 0 small, from the properties of ϕ, in the set

U1 = x : u (x, 0) > ϕ (x) ,dist (x, F (u)) ≤ 2r1 ⊂⊂ B′1,

we have ϕ > 0. Take x1 ∈ U1, with dist(x, F (u)) ≤ r1 and consider the barrierfunction given by

w (x, y) = u (x, y)− ϕ (x)− c02n+ 2 (1 + a)

(|x− x1|+ y2

)where c0 is as in (37). Note that w (x1, 0) > 0 and w < 0 on u = ϕ∩y = 0 .We want to apply the maximum principle in the set

U = Br (x1, 0) \ (u = ϕ ∩ y = 0) ⊂⊂ B1.

Since Lau = 0 and ∆ϕ (x) ≤ −c0, we have

Law (x, y) = Lau (x, y)− |y|a (∆ϕ (x) + c0) ≥ 0.

47

Noticing that ∂U = ∂Br (x1, 0)∪u = ϕ∩y = 0, by the maximum principlewe infer that

0 < w (x1, 0) ≤ supUw = sup

∂Uw = sup

∂Br(x1,0)

w

= sup∂Br(x1,0)

(u− ϕ)− c02n+ 2 (1 + a)

r2.

Letting x1 → x0 we get

supBr(x0,0)

(u− ϕ) ≥ sup∂Br(x0,0)

(u− ϕ) ≥ c02n+ 2 (1 + a)

r2. (38)

To conclude the proof we must show that the above supremum is attained ony = 0. To prove it we show that uy ≤ 0 in B+

1 . Here comes into play the zeroboundary condition.

Since Lau ≥ 0 and Lau = 0 outside the contact set u = ϕ , by the max-imum principle it follows that u ≥ 0 in B1 and u attains its maximum onu = ϕ . Using again that u = 0 on ∂B1 and that u (x,−y) = u (x, y) , wededuce that

|y|a uy ≤ 0 and limy→0+

|y|a uy (x, y) ≤ 0 on u = ϕ .

Moreoverlimy→0+

|y|a uy (x, y) = 0 on u > ϕ .

Now, by direct computation, one can check that L−a(yauy) = 0 in B+1 . By the

maximum principle we infer that yauy ≤ 0 in B+1 . Thus, u (x, y) is decreasing

with respect to y in B+1 . Since u is even in y, this yields

u (x, y) ≤ u (x, 0) in B1

and (38) finally gives

supx∈B′r(x0)

(u (x, 0)− ϕ (x)) = supBr(x0,0)

(u− ϕ) ≥ c02n+ 2 (1 + a)

r2.

Having the above nondegeneracy at hands, to proceed further once again

the main tools are frequency and monotonicity formulas, adapted to take intoaccount the homogeneity of the solution. For a free boundary point x0 ∈ F (u),the change of variable

vx0 (x, y) = u (x, y)− ϕ (x) +1

2 (1 + a)

∆ϕ (x0) y2 +∇∆ϕ (x0) · (x− x0) y2

reduces to the case

|Lavx0 | ≤ C |y|a |x− x0|1+γ(39)

48

outside the coincidence set and takes care of the errors in the frequency/monotonicityformulas, due to the presence of a non-zero obstacle. Note that

vx0 (x0, 0) = u (x, 0)− ϕ (x)

and that vx0 depends continuously on x0 since ϕ ∈ C3,γ . The non-degeneracyof u translates into

supB′r(x0)

vx0 (x, 0) ≥ cr2.

Moreover, exploiting (39) and a weak Harnack inequality (see [29]) one also gets,for small r, ∫

Br(x0,0)

|ya| |vx0 (x, y)|2 dxdy ≥ c1rn+a+5. (40)

Setting x0 = 0, v = v0 and

K (r) =

∫∂Br(0,0)

∣∣y2∣∣ v2,

the function

r 7−→ Φ (r; v) =(r + C0r

2) ddr

log maxK (r) , rn+a+4+2γ

is monotone non-decreasing for small r, and a suitable constant C0 > 0. Let

Φ (0+; v) = n+ a+ 2m.

Now, the above non-degeneracy estimates implies that either m = 1 + s orm = 2. It is clearly enough to show that m ≤ 2. Indeed, the monotonicity of Φgives n+ a+ 2m ≤ Φ (r; v) and integrating we get

K (r) ≤ Crn+a+2m.

A further integration gives∫Br(0,0)

|ya| |vx0 (x, y)|2 dxdy ≤ c1rn+a+2m+1

that together with (40) implies m ≤ 2. At this point, following the strategyin [22], it is possible to show that, up to a subsequence, the blow-ups vr in(36) converge as r → 0+ to a function, which is nonnegative on y = 0 andhomogeneous of degree m. Moreover:

a) either there exist c > 0 and a unit vector ν such that

m = 1 + s and v0 (x, 0) = c (x · ν)1+s+

b) orm = 2 and v0 (x, 0) is a polynomial of degree 2

49

and the origin is a singular point.Uniqueness of the blow-ups can be proved by a suitable modification of

the Weiss and Monneau monotonicity formulas. As consequence, there existsa modulus of continuity ω : R+ → R+ such that, for all x0 belonging to thesingular set F2 (u) of the free boundary, we have

u (x)− ϕ (x) = px02 (x− x0) + ω (|x− x0|)

∣∣x− x20

∣∣ (41)

for some polynomial px02 (x) = (Ax0x, x), A ∈ Rn×n, symmetric and nonnega-

tive, A 6= 0. In addition, the mapping F2 (u) 3 x0 7→ px02 is continuous with∫

∂B1

|y|a (px12 − p

x02 ) ≤ ω (|x1 − x0|) ∀x1, x0 ∈ F2 (u) .

Concerning the regularity of F2 (u), given any point x0 in this set, from (41),the blow-up of u − ϕ coincides with the polynomial px0

2 (x) = (Ax0x, x) . Westratify F2 (u) according to the dimension of kerA :

F k2 (u) = x0 ∈ F2 (u) : dim kerAx0 = k k = 0, 1, ..., n− 1.

Then, reasoning as in the classical obstacle case (see [18]), one can show thatfor any x0 ∈ F k2 (u) there exists r = rx0

> 0 such that F k2 (u) ∩ Br (x0) iscontained in a connected k−dimensional C1 manifold. This concludes the proofof Theorem 43.

3 Comments and further reading

In recent time and while we were writing this survey several important resultshave been established. We briefly mention some of the more strictly related toour presentation.• Different approaches to the regularity of the free boundary. When the thin

manifold is non-flat, by a standard procedure one is lead to a Signorini problem,but for a divergence form operator with variable coefficient matrix A(x) =[aij(x)]. For instance, when M satisfies the minimal smoothness assumptionC1,1 = W 2,∞, then A(x) is C0,1 = W 1,∞, i.e., Lipschitz continuous.

The approach described above in the case A(x) ≡ In to establish the C1,α

regularity of the regular free boundary, based on differentiating the equationfor the solution v in tangential directions e ∈ Rn−1 and establishing directionalmonotonicity, does not work well for variable coefficients, particularly when theobstacle ϕ 6= 0. For coefficients A(x) ∈ W 1,∞, and the obstacle ϕ ∈ W 2,∞, the

optimal C1,1/2loc (Ω± ∪M) regularity of the solutions was established by Garofalo

and Smit Vega Garcia in [39] by means of monotonicity formulas of Almgren’stype. The C1,α smoothness of the regular free boundary has been obtainedby Garofalo, Petrosyan and Smit Vega Garcia in [37]. The two central toolsare a Weiss type monotonicity formula and an epiperimetric inequality, whichallow to control the homogeneous blow-ups. The latter results, in the special

50

case A(x) ≡ In, have also been established by Focardi and Spadaro ([31]). Adifferent approach, based on Carleman estimates, to the optimal regularity ofthe solutions and C1,α regularity of the free boundary for A(x) ∈W 1,p, p > 2n,and vanishing obstacle is used by Koch, Ruland and Shi in [44] and [45] (morerecently the authors were able to extend these results to non-zero obstacles inW 2,p, p > 2n).• Higher regularity. Real analyticity of the regular part of the free boundary

for the thin obstacle (Signorini) problem has been proved by Koch, Petrosyanand Shi in [43] via a partial hodograph-Legendre transformation, subsequentlyextended by Koch, Ruland and Shi to the fractional Laplacian operator in ([46]).The lack of regularity of this map can be overcome by providing a precise asymp-totic behavior at a regular free boundary point. The Legendre transforms (onwhich one reads the regularity of the free boundary) satisfies a subelliptic equa-tion of Baouendi-Grushin type and its analyticity is achieved by using the Lp

theory available for this kind of operator. A different approach to higher regu-larity for the thin obstacle and also to one-phase free boundary problems is dueto De Silva and Savin in [27] and is based on a higher order Boundary Harnackprinciple. Jhaveri and Neumayer in [41] extend this approach to the fractionalLaplacian obstacle problem. The higher regularity of the free boundary in thevariable coefficient case has been established in [47].• More general problems and operators. Allen, Lindgren and Petrosyan in [1]

consider the two phase problem for the fractional Laplacian and prove optimalregularity of the solution and separation of the positive and negative phase.

Operators with drift in the subcritical regime s ∈ (0, 1/2) are considered byPetrosyan and Pop in [54] where optimal regularity of the solution is proved.The regularity of the free boundary is addressed by Garofalo et al. in [38]. Weemphasize that the presence of the drift the extension operator exhibits somesingularities and makes the problem quite delicate.

The results in [22] have been extended to obstacle problem for integro-differential operators by Caffarelli, Ros-Oton and Serra in [20]. Here the authorsdevelop an entirely non local powerful approach, independent of any monotonic-ity formulas.

Korvenpaa, Kuusi and Palatucci in [48] consider the obstacle problem for aclass of nonlinear integro-differential operators including the fractional p−Laplacian.The solution exists, it is unique and inherits regularity properties (such as Holdercontinuity) from the obstacle.

4 Parabolic obstacle problems

In this section we will focus on time-dependent models, which can be thoughtof as parabolic counterparts of the systems (1) and (2). We emphasize that,although their time-independent versions are locally equivalent (for s = 1/2),the problems we are about to describe are very different from each other.

51

4.1 The parabolic fractional obstacle problem

As mentioned above, one of the motivations behind the recent increased interestin studying constrained variational problems with a fractional diffusion comesfrom mathematical finance. Jump-diffusion processes allow, in fact, to take intoaccount large price changes, and appear to be better suited to model marketfluctuations. An American option gives its holder the right to buy a stock orbasket of stocks at a given price prior to, but not later than, a given time T > 0from the time of inception of the contract. If v(x, t) denotes the price of anAmerican option with a payoff ψ at time T , then v will be a viscosity solutionto the obstacle problem

minLv, v − ψ = 0

v(T ) = ψ.(42)

Here L is a backward parabolic integro-differential operator of the form

Lv = −vt − rv − b · ∇v + (−∆)sv +Hv, s ∈ (0, 1),

where r > 0, b ∈ Rn, and H is a non-local operator of lower order with respectto (−∆)s.

This problem was studied by Caffarelli and Figalli in the paper [19], whereregularity properties are established for the model equation

min−vt + (−∆)sv, v − ψ = 0 on [0, T ]× Rn

v(T ) = ψ on Rn.(43)

The advantage of considering (43) is twofold. On the one hand, the absenceof the transport term allows to prove that solutions have the same regularityas in the stationary case for all values of s ∈ (0, 1), even if, when s ∈ (0, 1/2),the time derivative is of higher order with respect to the elliptic term (−∆)sv.In addition, it is feasible that the regularity theory established for the modelequation (43) could be adapted to the solutions to (42) when s > 1/2. It shouldbe noted that when s < 1/2, the leading term becomes b · ∇v, and there is noexpectation of a regularity theory.

In order to proceed, we need to introduce the relevant function spaces.

Definition 3 Given α, β ∈ (0, 1) and [a, b] ⊂ R, we say that

• w ∈ Cα,βt,x ([a, b]× Rn) if

‖w‖Cα,βt,x ([a,b]×Rn) := ‖w‖L∞([a,b]×Rn) + [w]Cα,βt,x ([a,b]×Rn)

= ‖w‖L∞([a,b]×Rn) + sup[a,b]×Rn

|w(t, x)− w(t′, x′)||t− t′|α + |x− x′|β

<∞.

• w ∈ logLiptCβx ([a, b]× Rn) if

‖w‖logLiptCβx ([a,b]×Rn) := ‖w‖L∞([a,b]×Rn)+ sup

[a,b]×Rn

|w(t, x)− w(t′, x′)||t− t′|

(1 +

∣∣ log |t− t′|∣∣)+ |x− x′|β

<∞.

52

We will also say that

• w ∈ Cα−0+,βt,x ([a, b]× Rn) if w ∈ Cα−ε,βt,x ([a, b]× Rn) for all ε > 0;

• w ∈ Cα,βt,x ((a, b]× Rn) if w ∈ Cα,βt,x ([a+ ε, b]× Rn) for all ε > 0.

In what follows, ψ : Rn → R+ will be a globally Lipschitz function of classC2 satisfying

∫Rn

ψ(1+|x|)n+2s <∞ and (−∆)sψ ∈ L∞(Rn). For s ∈ (0, 1) we let

u : [0, T ]× Rn → R be a continuous viscosity solution to the obstacle problemminut + (−∆)su, u− ψ = 0 on [0, T ]× Rn

u(0) = ψ on Rn.(44)

Existence and uniqueness of solutions can be shown either by probabilistictechniques, or by approximating the equation via a penalization method.

We are now ready to state the main result from [19].

Theorem 45 Assume that ψ ∈ C2(Rn) satisfies

‖∇ψ‖L∞(Rn) + ‖D2ψ‖L∞(Rn) + ‖(−∆)sψ‖C1−sx (Rn) <∞,

and that u solves (44). Then u is globally Lipschitz in space-time on [0, T ]×Rn,and satisfies

ut ∈ logLiptC1−sx ((0, T ]× Rn),

(−∆)su ∈ logLiptC1−sx ((0, T ]× Rn) if s ≤ 1/3,

ut ∈ C1−s2s −0+

t,x ((0, T ]× Rn),

(−∆)su ∈ C1−s2st,x ((0, T ]× Rn) if s > 1/3.

Some remarks are in order. First of all, comparison with Corollary 25 showsthat, at least in terms of spacial regularity, this result is optimal. Secondly,the criticality of s = 1/3 is a consequence of the invariance of the operator∂t + (−∆)s under the scaling (t, x) → (λ2st, λx). Hence, the spacial regularity

C1−sx naturally corresponds to time regularity C

1−s2st when 1−s

2s < 1, or s > 1/3.In addition, the time regularity is almost optimal in the case s = 1/2 (as it ispossible to construct traveling waves which are C1+1/2 both in space and time),as well as in the limit s→ 1 (since, when s = 1, it is well known that solutionsare C1,1 in space and C1 in time).

Several basic properties of the solution u, such as global regularity in space-time, semi-convexity in space, and the boundedness of (−∆)su, follow from acomparison principle, which in turn is established using a penalization method.Combining the semi-convexity of u(t) with the L∞ bound on (−∆)su(t), it ispossible then to deduce the C1 regularity in time of solutions when s ≥ 1/2.The next step toward the proof of Theorem 45 is to prove a Cα+2s-regularityresult in space, which, roughly speaking, says the following. Let v : Rn → R bea semiconvex function touching from above an obstacle ψ : Rn → R of class C2.

53

If (−∆)sv is non-positive outside the contact set and non-negative on it, thenv detaches from ψ in a Cα+2s fashion, with α > 0 depending exclusively on s.More precisely, assume that v, ψ : Rn → R are two globally Lipschitz functionswith v ≥ ψ satisfying the following conditions:

A1. ‖D2ψ‖L∞(Rn) := C0 <∞;

A2. ‖(−∆)sψ‖C1−sx (Rn) <∞;

A3. v − ψ, (−∆)sv ∈ L∞(Rn);

A4. The function v + C0|x2|/2 is convex (we say that v is C0-semiconvex );

A5. v is smooth and (−∆)sv ≤ 0 inside the open set v > ψ;

A6. ‖(−∆)sψ‖L∞(Rn) ≥ (−∆)sv ≥ 0 on v = ψ.

One has then the following:

Theorem 46 There exist C > 0 and α ∈ (0, 1), depending only on C0, ‖(−∆)sψ‖C1−sx (Rn),

‖v − ψ‖L∞(Rn), and ‖(−∆)sv‖L∞(Rn), such that

supBr(x)

|v − ψ| ≤ Crα+2s, supBr(x)

|(−∆)svχv=ψ| ≤ Crα (45)

for all r ≤ 1 and for every x ∈ ∂u = ψ.

The strategy of the proof is analogous to the one used in [7]. It is based ongrowth estimates for the La-harmonic extension of v, i.e. the solution to (4),satisfying (with slight abuse of notation) v(x, 0) = v(x), with v(x) as above.

Having shown that (−∆)svχv=ψ grows at most as rα near any free bound-ary point, it is not difficult to show that (−∆)svχv=ψ ∈ Cαx (Rn):

Corollary 47 There exist C ′ > 0 and α ∈ (0, 1 − s], depending only on C0,‖(−∆)sψ‖C1−s

x (Rn), ‖v − ψ‖L∞(Rn), and ‖(−∆)sv‖L∞(Rn), such that

‖(−∆)svχv=ψ‖Cαx (Rn) ≤ C ′.

At this point we consider the function w : Rn × R+ → R, which solves theDirichlet problem

L−aw = 0 on Rn × R+,

w(x, 0) = (−∆)sv(x)χv=ψ(x) on Rn.

Since w(x, 0) ≥ 0, the maximum principle implies w ≥ 0 everywhere. As-sume now that 0 is a free boundary point. Given that (−∆)sv(x) is globallybounded by (A3) above, it follows from the Poisson representation formula forw (see [23]) and Corollary 47 that

sup|x|2+y2≤r2

w(x, y) ≤ Crα,

for some uniform constant C and some α ∈ (0, 1 − s]. Our next goal is toestablish the following sharp growth estimate:

54

Proposition 48 There exists C > 0 depending only on C0, ‖(−∆)sψ‖C1−sx (Rn),

‖v − ψ‖L∞(Rn), and ‖(−∆)sv‖L∞(Rn), such that

sup|x|2+y2≤r2

w(x, y) ≤ Cr1−s.

A crucial ingredient in the proof of Proposition 48 is the following mono-tonicity formula (compare with Lemma 10).

Lemma 49 Let w be as above and define

φ (r) =1

r2(1−s)

∫B+r

y−a |∇w (X)|2

|X|n−a−1 dX.

Then there exists a constant C ′′, depending only on C0, ‖(−∆)sψ‖C1−sx (Rn),

‖v − ψ‖L∞(Rn), and ‖(−∆)sv‖L∞(Rn), such that

φ(r) ≤ C ′′[1 + r2α+δα−a−1]

for all r ≤ 1. Here, δα = 14

(α

α+2s −α2

).

Proof of Proposition 48. Using an approximation argument, it is possible toshow that the function

w(x, y) =(w(x, |y|)− rα+δα

)++

(1 +

nC0

1 + α

)|y|1+a

is globally L−a-subharmonic. Moreover, because of Lemma 49, it vanishes onmore than half of the n-dimensional disc Br × 0. One can then apply theweighted Poincare inequality established in [29] to show∫

B+r

(w)2y−a dX ≤ Crn+2[φ(r) + 1]

for all r ≤ 1. Combining this estimate with the L−a-subharmonicity of w, andLemma 49 again, we infer

supB+r/2

(w)2 ≤ C

rn+1−a

∫B+r

(w)2y−a dX ≤ C[r1+α + r2α+δα ].

But 1 + α = 2(1− s), and therefore

supB+r/2

w ≤ C

supB+r/2

w + rα+δα + r1+a

≤ C[r1−s + rα+δα/2].

Arguments similar to the ones in the proof of Corollary 47 yield

‖w‖Cβαx (Rn)

≤ C,

55

with βα = minα+ δα/2, 1− s. An iteration gives the desired conclusion.

Arguing as in the proof of Corollary 47, we obtain

‖(−∆)svχv=ψ‖C1−sx (Rn) ≤ C

′′. (46)

Next, we apply (46) to v = u(t), obtaining the following result.

Proposition 50 Let u be a solution to (44), with ψ ∈ C2(Rn) satisfying (A1)and (A2) above. Then there exists a constant CT > 0, depending on T , ‖D2ψ‖L∞(Rn),and ‖(−∆)sψ‖C1−s

x (Rn), such that

supt∈[0,T ]

‖(−∆)su(t)χu(t)=ψ‖C1−sx (Rn) ≤ CT .

Finally, to conclude the proof of Theorem 45 one exploits the fact that u isa solution of the parabolic equation

ut + (−∆)su =((−∆)su

)χu=ψ on (0, T ]× Rn. (47)

Proposition 50 ensures that the right-hand side of (47) is in L∞((0, T ];C1−s

x (Rn)).

By parabolic regularity theory we infer

ut, (−∆)su ∈ L∞((0, T ];C1−s−0+

x (Rn)).

The desired Holder regularity in time follows from a bootstrap argument whichuses equation (47) again.

4.2 The parabolic Signorini problem

In this section we will give an overview of the time-dependent analogue of (2),i.e., the parabolic Signorini problem, following the ideas in [26].

4.2.1 Statement of the problem

Given a domain Ω in Rn, n ≥ 2, with a sufficiently regular boundary ∂Ω, letMbe a relatively open subset of ∂Ω (in its relative topology), and set S = ∂Ω\M.We consider the solution of the problem

∆v − ∂tv = 0 in ΩT := Ω× (0, T ], (48)

v ≥ φ, ∂νv ≥ 0, (v − φ)∂νv = 0 on MT :=M× (0, T ], (49)

v = g on ST := S × (0, T ], (50)

v(·, 0) = φ0 on Ω0 := Ω× 0, (51)

where ∂ν is the outer normal derivative on ∂Ω, and φ :MT → R, φ0 : Ω0 → R,and g : ST → R are prescribed functions satisfying the compatibility condi-tions: φ0 ≥ φ on M× 0, g ≥ φ on ∂S × (0, T ], and g = φ on S × 0, seeFig. 2. Classical examples where Signorini-type boundary conditions appear are

56

ΩT

MT

v > φ

∂νv = 0v = φ

∂νv ≥ 0

v = φ0

v = g

Γ(v)HY

Figure 2: The parabolic Signorini problem

the problems with unilateral constraints in elastostatics (including the originalSignorini problem [61, 30]), problems with semipermeable membranes in fluidmechanics (including the phenomenon of osmosis and osmotic pressure in bio-chemistry), and the problems on the temperature control on the boundary inthermics. We refer to the book of Duvaut and Lions [28] for further details.

Another historical importance of the parabolic Signorini problem is that itserves as one of the prototypical examples of evolutionary variational inequali-ties. We thus say that a function v ∈W 1,0

2 (ΩT ) solves (48)–(51) if∫ΩT

∇v∇(w − v) + ∂tv(w − v) ≥ 0 for every w ∈ K,

v ∈ K, ∂tv ∈ L2(ΩT ), v(·, 0) = φ0,

where K = w ∈ W 1,02 (ΩT ) | w ≥ φ on MT , w = g on ST (please see below

for the definitions of the relevant parabolic functional classes). The existenceand uniqueness of such v, under some natural assumptions on φ, φ0, and g canbe found in [14,28,3, 4].

Similarly to the elliptic case, we are interested in:

• the regularity properties of v;

• the structure and regularity of the free boundary

Γ(v) = ∂MT(x, t) ∈MT | v(x, t) > φ(x, t),

where ∂MTindicates the boundary in the relative topology of MT .

Concerning the regularity of v, it has long been known that the spatialderivatives ∂xiv, i = 1, . . . , n, are α-Holder continuous on compact subsets ofΩT ∪MT , for some unspecified α ∈ (0, 1). In the parabolic case, this was firstproved by Athanasopoulos [5], and subsequently by Uraltseva [64] (see also [3]),

57

under certain regularity assumptions on the boundary data, which were furtherrelaxed by Arkhipova and Uraltseva [4].

One of the main objectives of this section is to establish, in the parabolic

Signorini problem, and for a flat thin manifoldM, that v ∈ H3/2,3/4loc (ΩT ∪MT ),

see Theorem 56 below. The proof, which we only sketch here, is inspired by theworks [10] and [22] described in Section 2. For further details, we refer to [26].

Before proceeding, we introduce the relevant parabolic function spaces andnotations. We use notations similar to those in the classical book of Ladyzhen-skaya, Solonnikov, and Uraltseva [49]. The class C(ΩT ) = C0,0(ΩT ) is theclass of functions continuous in ΩT with respect to parabolic (or Euclidean)distance. Further, given for m ∈ Z+ we say u ∈ C2m,m(ΩT ) if for |α|+ 2j ≤ 2m∂αx ∂

jt u ∈ C0,0(ΩT ), and define the norm

‖u‖C2m,m(ΩT ) =∑

|α|+2j≤2m

sup(x,t)∈ΩT

|∂αx ∂jt u(x, y)|.

The parabolic Holder classes H`,`/2(ΩT ), for ` = m+ γ, m ∈ Z+, 0 < γ ≤ 1 aredefined as follows. First, we let

〈u〉(0)ΩT

= |u|(0)ΩT

= sup(x,t)∈ΩT

|u(x, t)|,

〈u〉(m)ΩT

=∑

|α|+2j=m

|∂αx ∂jt u|

(0)ΩT,

〈u〉(β)x,ΩT

= sup(x,t),(y,t)∈ΩT0<|x−y|≤δ0

|u(x, t)− u(y, t)||x− y|β

, 0 < β ≤ 1,

〈u〉(β)t,ΩT

= sup(x,t),(x,s)∈ΩT

0<|t−s|<δ20

|u(x, t)− u(x, s)||t− s|β

, 0 < β ≤ 1,

〈u〉(`)x,ΩT =∑

|α|+2j=m

〈∂αx ∂jt u〉

(γ)x,ΩT

,

〈u〉(`/2)t,ΩT

=∑

m−1≤|α|+2j≤m

〈∂αx ∂jt u〉

((`−|α|−2j)/2)t,ΩT

,

〈u〉(`)ΩT= 〈u〉(`)x,ΩT + 〈u〉(`/2)

t,ΩT.

Then, we defineH`,`/2(ΩT ) as the space of functions u for which the followingnorm is finite:

‖u‖H`,`/2(ΩT ) =

m∑k=0

〈u〉(k)ΩT

+ 〈u〉(`)ΩT.

The parabolic Lebesgue space Lq(ΩT ) indicates the Banach space of thosemeasurable functions on ΩT for which the norm

‖u‖Lq(ΩT ) =(∫

ΩT

|u(x, t)|qdxdt)1/q

58

is finite. The parabolic Sobolev spaces W 2m,mq (ΩT ), m ∈ Z+, denote the spaces

of those functions in Lq(ΩT ), whose distributional derivative ∂αx ∂jt u belongs to

Lq(ΩT ), for |α|+ 2j ≤ 2m.For x, x0 ∈ Rn, t0 ∈ R we let

x′ = (x1, x2, . . . , xn−1), x′′ = (x1, x2, . . . , xn−2), x = (x′, xn), x′ = (x′′, xn−1),

and

Br(x0) = x ∈ Rn | |x| < r (Euclidean ball)

B±r (x0) = Br(x0) ∩ Rn± (Euclidean halfball)

B′r(x0) = Br(x) ∩ Rn−1 (“thin” ball)

Qr(x0, t0) = Br(x0)× (t0 − r2, t0] (parabolic cylinder)

Q′r(x0, t0) = B′r(x0)× (t0 − r2, t0] (“thin” parabolic cylinder)

Q±r (x0, t0) = B±r (x0)× (t0 − r2, t0] (parabolic halfcylinders)

Q′′r (x0, t0) = B′′r (x0)× (t0 − r2, t0]

Sr = Rn × (−r2, 0] (parabolic strip)

S±r = Rn± × (−r2, 0] (parabolic halfstrip)

S′r = Rn−1 × (−r2, 0] (“thin” parabolic strip)

When x0 = 0 and t0 = 0, we omit the centers x0 and (x0, t0) in the abovenotations.

Since we are mostly interested in local properties of the solution v of theparabolic Signorini problem and of its free boundary, we focus our attention onsolutions in parabolic (half-)cylinders.

Definition 4 Given φ ∈ H2,1(Q′1), we say that v ∈ Sφ(Q+1 ) if v ∈W 2,1

2 (Q+1 )∩

L∞(Q+1 ), ∇v ∈ Hα,α/2(Q+

1 ∪Q′1) for some 0 < α < 1, and v satisfies

∆v − ∂tv = 0 in Q+1 , (52)

v − φ ≥ 0, −∂xnv ≥ 0, (v − φ)∂xnv = 0 on Q′1, (53)

and

(0, 0) ∈ Γ∗(v) := ∂Q′1(x′, t) ∈ Q′1 | v(x′, 0, t) = φ(x′, t), ∂xnv(x′, 0, t) = 0,

(54)where ∂Q′1 is the boundary in the relative topology of Q′1.

Our very first step is the reduction to vanishing obstacle. The differencev(x, t)− φ(x′, t) satisfies the Signorini conditions on Q′1 with zero obstacle, butat an expense of solving a nonhomogeneous heat equation instead of the homoge-neous one. This difference may then be extended to the strip S+

1 by multiplyingit by a suitable cutoff function ψ. The resulting function will satisfy

∆u− ∂tu = f(x, t) in S+1 ,

59

with

f(x, t) = −ψ(x)[∆′φ− ∂tφ] + [v(x, t)− φ(x′, t)]∆ψ + 2∇v∇ψ. (55)

Remark 4 It is important to observe that, for smooth enough φ, the functionf is bounded in S+

1 .

With this process, we arrive at the following notion of solution.

Definition 5 A function u is in the class Sf (S+1 ), for f ∈ L∞(S+

1 ), if u ∈W 2,1

2 (S+1 ), ∇u ∈ Hα,α/2(S+

1 ∪ S′1), u has compact support and solves

∆u− ∂tu = f in S+1 , (56)

u ≥ 0, −∂xnu ≥ 0, u∂xnu = 0 on S′1, (57)

(0, 0) ∈ Γ(u) = ∂(x′, t) ∈ S′1 : u(x′, 0, t) > 0. (58)

4.2.2 Monotonicity of the generalized frequency function

Similarly to the elliptic case, one of the central results towards the study ofregularity properties (both of the solution and of the free boundary) is a gen-eralization of Almgren’s frequency formula, see Theorem 51 below. As it iswell known, the parabolic counterpart of Almgren’s formula was established byPoon [58], for functions which are caloric in an infinite strip Sρ = Rn× (−ρ2, 0].Poon’s parabolic frequency function is given by

Nu(r) =iu(−r2)

hu(−r2), (59)

where

hu(t)=

∫Rn+u(x, t)2G(x, t)dx,

iu(t)= −t∫Rn+|∇u(x, t)|2G(x, t)dx,

for any function u in the parabolic half-strip S+1 for which the integrals involved

are finite. Here G denotes the backward heat kernel on Rn × R

G(x, t) =

(−4πt)−

n2 e

x2

4t , t < 0,

0, t ≥ 0.

We explicitly remark that Poon’s monotonicity formula cannot be directlyapplied in the present context, since functions in the class Sf (S+

1 ) (see Defini-tion 5) are not caloric functions. It should also be noted that the time dependentcase of the Signorini problem presents substantial novel challenges with respectto the stationary setting. These are mainly due to the lack of regularity ofthe solution in the t-variable, a fact which makes the justification of differen-tiation formulas and the control of error terms quite difficult. To overcome

60

these obstructions, (Steklov-type) averaged versions of the quantities involvedare introduced in the main monotonicity formulas. This basic idea allows tosuccessfully control the error terms. More precisely, we introduce the quantities

Hu(r)=1

r2

∫ 0

−r2hu(t)dt =

1

r2

∫S+r

u(x, t)2G(x, t)dxdt,

Iu(r)=1

r2

∫ 0

−r2iu(t)dt =

1

r2

∫S+r

|t||∇u(x, t)|2G(x, t)dxdt,

One further obstruction is represented by the fact that the above integralsmay become unbounded near the endpoint t = 0, where G becomes singular.To remedy this problem we introduce truncated versions of Hu and Iu:

Hδu(r) =

1

r2

∫ −δ2r2−r2

hu(t)dt =1

r2

∫S+r \S+

δr

u(x, t)2G(x, t)dxdt,

Iδu(r) =1

r2

∫ −δ2r2−r2

iu(t)dt =1

r2

∫S+r \S+

δr

|t||∇u(x, t)|2G(x, t)dxdt

for 0 < δ < 1.The idea at this point is to obtain differentiation formulas for Hδ

u(r) andIδu(r), which - by means of a delicate limiting process - will yield correspondingones for Hu(r) and Iu(r). In turn, such formulas will allow to establish almost-monotonicity of a suitably defined frequency function. To state this result weneed the following notion.

Definition 6 We say that a positive function µ(r) is a log-convex function oflog r on R+ if logµ(et) is a convex function of t. In other words

µ(e(1−λ)s+λt) ≤ µ(es)1−λµ(et)λ, 0 ≤ λ ≤ 1.

This is equivalent to saying that µ is locally absolutely continuous on R+ andrµ′(r)/µ(r) is nondecreasing. For instance, µ(r) = rκ is a log-convex function oflog r for any κ. The importance of this notion in our context is that Almgren’sand Poon’s frequency formulas can be regarded as log-convexity statements inlog r for the appropriately defined quantities Hu(r).

Theorem 51 (Monotonicity of the truncated frequency) Let u ∈ Sf (S+1 ) with

f satisfying the following condition: there is a positive monotone nondecreasinglog-convex function µ(r) of log r, and constants σ > 0 and Cµ > 0, such that

µ(r) ≥ Cµr4−2σ

∫Rnf2(·,−r2)G(·,−r2) dx.

Then, there exists C > 0, depending only on σ, Cµ and n, such that the function

Φu(r) =1

2reCr

σ d

drlog maxHu(r), µ(r)+ 2(eCr

σ

− 1)

is nondecreasing for r ∈ (0, 1).

61

Remark 5 On the open set where Hu(r) > µ(r) we have Φu(r) ∼ 12rH

′u(r)/Hu(r),

which coincides, when f = 0, with 2Nu (Nu as in (59)). The purpose of the“truncation” of Hu(r) with µ(r) is to control the error terms in computationsthat appear from the right-hand-side f .

Proof. [Proof of Theorem 51] First, we observe that the functions Hu(r) andµ(r) are absolutely continuous and therefore so is maxHu(r), µ(r). It followsthat Φu is uniquely identified only up to a set of measure zero. The mono-tonicity of Φu should be understood in the sense that there exists a monotoneincreasing function which equals Φu almost everywhere. Therefore, without lossof generality we may assume that

Φu(r) =1

2reCr

σ µ′(r)

µ(r)+ 2(eCr

σ

− 1)

on F = Hu(r) ≤ µ(r) and

Φu(r) =1

2reCr

σH ′u(r)

Hu(r)+ 2(eCr

σ

− 1)

in O = Hu(r) > µ(r). Following an idea introduced in [33, 34] we nownote that it will be enough to check that Φ′u(r) > 0 in O. Indeed, from theassumption on µ, it is clear that Φu is monotone on F . Next, if (r0, r1) is amaximal open interval in O, then Hu(r0) = µ(r0) and Hu(r1) = µ(r1) unlessr1 = 1. Besides, if Φu is monotone in (r0, r1), it is easy to see that the limitsH ′u(r0+) and H ′u(r1−) will exist and satisfy

µ′(r0+) ≤ H ′u(r0+), H ′u(r1−) ≤ µ′(r1−) (unless r1 = 1)

and therefore we will have

Φu(r0) ≤ Φu(r0+) ≤ Φu(r1−) ≤ Φu(r1),

with the latter inequality holding when r1 < 1. This will imply the monotonicityof Φu in (0, 1).

Therefore, we will concentrate only on the set O = Hu(r) > µ(r), wherethe monotonicity of Φu(r) is equivalent to that of(

rH ′u(r)

Hu(r)+ 4)eCr

σ

= 2Φu(r) + 4.

The latter will follow, once we show that

d

dr

(rH ′u(r)

Hu(r)

)≥ −C

(rH ′u(r)

Hu(r)+ 4)r−1+σ

in O. Now, one can show that

rH ′u(r)

Hu(r)= 4

Iu(r)

Hu(r)− 4

r2

∫S+rtufG

Hu(r).

62

The desired result will be obtained by direct differentiation of this expression,and using the aforementioned formulas for the derivatives of Hu(r) and Iu(r).We omit the details.

4.2.3 Blow-ups and regularity of solutions

Similarly to the elliptic case, the generalized frequency formula in Theorem 51can be used to study the behavior of the solution u near the origin. The centralidea is again to consider some appropriately normalized rescalings of u, indi-cated with ur (see Definition 7), and then pass to the limit as r → 0+ (seeTheorem 53).

Henceforth, we assume that u ∈ Sf (S+1 ), and that µ(r) be such that the

conditions of Theorem 51 are satisfied. In particular, we assume that

r4

∫Rnf2(·,−r2)G(·,−r2) dx ≤ r2σµ(r)

Cµ.

Consequently, Theorem 51 implies that the function

Φu(r) =1

2reCr

σ d

drlog maxHu(r), µ(r)+ 2(eCr

σ

− 1)

is nondecreasing for r ∈ (0, 1). Hence, there exists the limit

κ := Φu(0+) = limr→0+

Φu(r). (60)

It is possible to show that κ is independent of the choice of the cut-off ψ intro-duced in the extension procedure. Since we assume that rµ′(r)/µ(r) is nonde-creasing, the limit

κµ :=1

2limr→0+

rµ′(r)

µ(r)(61)

also exists. We then have the following basic proposition concerning the valuesof κ and κµ.

Lemma 52 Let u ∈ Sf (S+1 ) and µ satisfy the conditions of Theorem 51. With

κ, κµ as above, we haveκ ≤ κµ.

Moreover, if κ < κµ, then there exists ru > 0 such that Hu(r) ≥ µ(r) for0 < r ≤ ru. In particular,

κ =1

2limr→0+

rH ′u(r)

Hu(r)= 2 lim

r→0+

Iu(r)

Hu(r).

We now define the appropriate notion of rescalings that works well with thegeneralized frequency formula.

63

Definition 7 For u ∈ Sf (S+1 ) and r > 0 define the rescalings

ur(x, t) :=u(rx, r2t)

Hu(r)1/2, (x, t) ∈ S+

1/r = Rn+ × (−1/r2, 0].

It is easy to see that the function ur solves the nonhomogeneous Signorini prob-lem

∆ur − ∂tur = fr(x, t) in S+1/r,

ur ≥ 0, −∂xnur ≥ 0, ur∂xnur = 0 on S′1/r,

with

fr(x, t) =r2f(rx, r2t)

Hu(r)1/2.

In other words, ur ∈ Sfr (S+1/r). We next show that, unless we are in the

borderline case κ = κµ, we will be able to study the blowups of u at the origin.The condition κ < κµ below can be understood, in a sense, that we can “detect”the growth of u near the origin.

Theorem 53 (Existence and homogeneity of blowups) Let u ∈ Sf (S+1 ), µ sat-

isfy the conditions of Theorem 51, and

κ := Φu(0+) < κµ =1

2limr→0+

rµ′(r)

µ(r).

Then, we have:

i) For any R > 0, there is rR,u > 0 such that∫S+R

(u2r + |t||∇ur|2 + |t|2|D2ur|2 + |t|2(∂tur)

2)G ≤ C(R), 0 < r < rR,u.

ii) There is a sequence rj → 0+, and a function u0 in S+∞ = Rn+ × (−∞, 0],

such that ∫S+R

(|urj − u0|2 + |t||∇(urj − u0)|2)G→ 0.

We call any such u0 a blowup of u at the origin.

iii) u0 is a nonzero global solution of Signorini problem:

∆u0 − ∂tu0 = 0 in S+∞

u0 ≥ 0, −∂xnu0 ≥ 0, u0∂xnu0 = 0 on S′∞,

in the sense that it solves the Signorini problem in every Q+R.

iv) u0 is parabolically homogeneous of degree κ:

u0(λx, λ2t) = λκu0(x, t), (x, t) ∈ S+∞, λ > 0

64

In addition to Theorem 51 and Lemma 52, the main ingredients in the proofof Theorem 53 are growth estimates for Hur (ρ) and log-Sobolev inequalities.

Remark 6 Using growth estimates for Hu(r) (where u ∈ Sf (S+1 )), it is possible

to show that necessarily κ > 1.

In addition, we have the following:

Proposition 54 Let u0 be a nonzero κ-parabolically homogeneous solution ofthe Signorini problem in S+

∞ = Rn+ × (−∞, 0] with 1 < κ < 2. Then, κ = 3/2and

u0(x, t) = C Re(x′ · e+ ixn)3/2+ in S+

∞

for some tangential direction e ∈ ∂B′1.

Proof. Extend u0 by even symmetry in xn to the strip S∞, i.e., by putting

u0(x′, xn, t) = u0(x′,−xn, t).

Take any e ∈ ∂B′1, and consider the positive and negative parts of the directionalderivative ∂eu0

v±e = max±∂eu0, 0.

It can be shown that they satisfy the following conditions

(∆− ∂t)v±e ≥ 0, v±e ≥ 0, v+e · v−e = 0 in S∞.

Hence, we can apply Caffarelli’s monotonicity formula [17] to the pair v±e , ob-taining that the functional

φ(r) =1

r4

∫Sr

|∇v+e |2G

∫Sr

|∇v−e |2G,

is monotone nondecreasing in r. On the other hand, from the homogeneity ofu, it is easy to see that

φ(r) = r4(κ−2)φ(1), r > 0.

Since κ < 2, φ(r) can be monotone increasing if and only if φ(1) = 0 andconsequently φ(r) = 0 for all r > 0. It follows that one of the functions v±e isidentically zero, which is equivalent to ∂eu0 being either nonnegative or nonpos-itive on the entire Rn × (−∞, 0]. Since this is true for any tangential directione ∈ ∂B1, it thus follows that u0 depends only on one tangential direction, andis monotone in that direction. Without loss of generality, we may thus as-sume that n = 2 and that the coincidence set at t = −1 is an infinite intervalΛ−1 = (x′, 0) ∈ R2 | u0(x′, 0,−1) = 0 = (−∞, a]× 0 =: Σ−a . Repeating themonotonicity formula argument above for the pair of functions max±w, 0,where

w(x, t) =

−∂x2u0(x1, x2, t), x2 ≥ 0

∂x2u0(x1, x2, t), x2 < 0

65

we obtain that also w does not change sign. Hence, we get

∂x1u0 ≥ 0, −∂x2

u0(x1, x2, t) ≥ 0 in R2+ × (−∞, 0].

Let g1(x) = ∂x1u0(x,−1), and g2(x) = −∂x2

u0(x1, x2,−1) in x2 ≥ 0 andg2(x) = ∂x2

u0(x1, x2,−1) for x2 < 0. Exploiting the fact that g1 and g2 arethe ground states for the Ornstein-Uhlenbeck operator in R2 \Σ−a and R2 \Σ+

a ,(with Σ+

a := [a,∞)×0) respectively, one reaches the conclusion that κ = 3/2and that g1(x) must be a multiple of Re(x1 + i|x2|)1/2. From this, the desiredconclusion easily follows.

From Remark 6 and Proposition 54, we immediately obtain the following

Theorem 55 Let u ∈ Sf (S+1 ) and µ satisfies the conditions of Theorem 51.

Assume also κµ = 12 limr→0+ rµ

′(r)/µ(r) ≥ 3/2. Then

κ := Φu(0+) ≥ 3/2.

More precisely, we must have

either κ = 3/2 or κ ≥ 2.

We are now ready to state the optimal regularity of solutions of the parabolicSignorini problem with sufficiently smooth obstacles.

Theorem 56 Let φ ∈ H2,1(Q+1 ), f ∈ L∞(Q+

1 ). Assume that v ∈W 2,12 (Q+

1 ) besuch that ∇v ∈ Hα,α/2(Q+

1 ∪Q′1) for some 0 < α < 1, and satisfy

∆v − ∂tv = f in Q+1 , (62)

v − φ ≥ 0, −∂xnv ≥ 0, (v − φ)∂xnv = 0 on Q′1. (63)

Then, v ∈ H3/2,3/4(Q+1/2 ∪Q

′1/2) with

‖v‖H3/2,3/4(Q+1/2∪Q′

1/2) ≤ Cn

(‖v‖W 1,0

∞ (Q+1 ) + ‖f‖L∞(Q+

1 ) + ‖φ‖H2,1(Q′1)

).

The proof of Theorem 56 will follow from the interior parabolic estimatesand the following growth bound of u away from the free boundary Γ(v) (which,in turn, can be deduced from Theorem 51 by fixing σ = 1/4 and choosingµ(r) = M2r4−2σ) .

Lemma 57 Let u ∈ Sf (S+1 ) with ‖u‖L∞(S+

1 ), ‖f‖L∞(S+1 ) ≤M . Then,

Hr(u) ≤ CnM2r3.

66

4.2.4 Regular free boundary points

At this point we turn our attention to the study of points on the free boundaryhaving minimal frequency κ = 3/2.

Definition 8 Let v ∈ Sφ(Q+1 ) with φ ∈ H`,`/2(Q′1), ` ≥ 2. We say that

(x0, t0) ∈ Γ∗(v) is a regular free boundary point if it has a minimal homogeneityκ = 3/2. The collection R(v) of regular free boundary points will be called theregular set of v.

The following basic property about R(v) is a consequence of the fact that κdoes not take any values between 3/2 and 2 .

Proposition 58 The regular set R(v) is a relatively open subset of Γ(v). Inparticular, for any (x0, t0) ∈ R(v) there exists δ0 > 0 such that

Γ(v) ∩Q′δ0(x0, t0) = R(v) ∩Q′δ0(x0, t0).

Our goal is to show that, if the thin obstacle φ is sufficiently smooth, thenthe regular set can be represented locally as a (n − 2)-dimensional graph of aparabolically Lipschitz function. Further, such function can be shown to haveHolder continuous spatial derivatives. We begin with the following basic result.

Theorem 59 (Lipschitz regularity of R(v)) Let v ∈ Sφ(Q+1 ) with φ ∈ H`,`/2(Q′1),

` ≥ 3 and that (0, 0) ∈ R(v). Then, there exist δ = δv > 0, and g ∈ H1,1/2(Q′′δ )(i.e., g is a parabolically Lipschitz function), such that possibly after a rotationin Rn−1, one has

Γ(v) ∩Q′δ = R(v) ∩Q′δ = (x′, t) ∈ Q′δ | xn−1 = g(x′′, t),Λ(v) ∩Q′δ = (x′, t) ∈ Q′δ | xn−1 ≤ g(x′′, t),

The proof of the space regularity follows the same circle of ideas illustrated,for the elliptic case, in Section 2.8. To show the 1/2-Holder regularity in t (actu-ally better than that), we will use the fact that the 3/2-homogeneous solutionsof the parabolic Signorini problem are t-independent (see Proposition 54).

However, in order to carry out the program outlined above, in additionto (i) and (ii) in Theorem 53 above, we will need a stronger convergence ofthe rescalings ur to the blowups u0. This will be achieved by assuming a slightincrease in the regularity assumptions on the thin obstacle φ, and, consequently,on the regularity of the right-hand side f in (55).

Lemma 60 Let u ∈ Sf (S+1 ), and suppose that for some `0 ≥ 2

|f(x, t)| ≤M‖(x, t)‖`0−2 in S+1 ,

|∇f(x, t)| ≤ L‖(x, t)‖(`0−3)+ in Q+1/2,

andHu(r) ≥ r2`0 , for 0 < r < r0.

67

Λ(v)

v = φ

xn−1 = g(x′′, t)

R(v)

@I

Figure 3: The regular set R(v) in Q′δ, given by the graph xn−1 = g(x′′) with

g ∈ H1,1/2(Q′′δ ) and ∇′′α,α/2(Q′′δ ) by Theorems 59 and 61

Then, for the family of rescalings ur0<r<r0 we have the uniform bounds

‖ur‖H3/2,3/4(Q+R∪Q′R) ≤ Cu, 0 < r < rR,u.

In particular, if the sequence of rescalings urj converges to u0 as in Theorem 53,then over a subsequence

urj → u0, ∇urj → ∇u0 in Hα,α/2(Q+R ∪Q

′R),

for any 0 < α < 1/2 and R > 0.

Proof. [Proof of Theorem 59] We only sketch the main ideas of the proof,beginning with the space regularity. For u ∈ Sf (S+

1 ) and r > 0 define therescalings

ur(x, t) :=u(rx, r2t)

Hu(r)1/2, (x, t) ∈ S+

1/r = Rn+ × (−1/r2, 0].

Theorem 53 guarantees the existence of u0 such that∫S+R

(|urj − u0|2 + |t||∇(urj − u0)|2)ρ→ 0.

Since κ = 3/2, Proposition 54 ensures that u0(x, t) = C Re(x′ ·e+ixn)3/2+ in S+

∞.For given η > 0, define now the thin cone

C ′(η) = x′ = (x′′, xn−1) ∈ Rn−1| xn−1 ≥ η|x′′|.

A direct computation shows that for any unit vector e ∈ C ′(η)

∂eu0 ≥ 0 in Q+1 , ∂eu0 ≥ δn,η > 0 in Q+

1 ∩ xn ≥ cn.

68

Thanks to Lemma 60, we know that ∂eurj ≥ 0 in Q+1/2, and thus, undoing the

scaling, ∂eu ≥ 0 in Q+rη for any unit vector e ∈ C ′(η). The Lipschitz continuity

in space follows in a standard fashion.As for the regularity in time, our goal is to show

|g(x, t)− g(x, s)| = o(|t− s|1/2).

uniformly in Q′′rη/2. Arguing by contradiction, we assume

|g(x′′j , tj)− g(x′′j , sj)| ≥ C|tj − sj |1/2.

Let x′j = (x′′j , g(x′′j , tj)), y′j = (x′′j , g(x′′j , sj)) and

δj = max|g(x′′j , tj)− g(x′′j , sj)|, |tj − sj |1/2.

We consider the rescalings of u at (xj , tj) by the factor of δj :

wj(x, t) =u(xj + δjx, tj + δ2

j t)

Hu(xj,tj)(δj)1/2

. (64)

The sequence wj converges to a homogeneous global solution in S∞, time-independent, of homogeneity 3/2. Combining this information with the factthat ∂eu ≥ 0 in a thin cone, we obtain a contradiction.

We next observe that the regularity of the function g can be improved withan application of a boundary Harnack principle.

Theorem 61 (Holder regularity of ∇′′g)In the conclusion of Theorem 59, one can take δ > 0 so that ∇′′g ∈ Hα,α2 (Q′′δ )

for some α > 0.

The proof of this result relies on two crucial ingredients. The first one is thefollowing non-degeneracy of ∂eu:

∂eu ≥ cd(x, t) in Q+δ ,

where d(x, t) denotes the parabolic distance from coincidence set Λ(v) ∩ Q′δ.This property, in turn, relies on a suitable parabolic version of Lemma 43. Thesecond ingredient in the proof of Theorem 61 is the following version of theparabolic boundary Harnack principle for domains with thin Lipschitz com-plements established in [56, Section 7]. To state the result, we will need thefollowing notations. For a given L ≥ 1 and r > 0 denote

Θ′′r = (x′′, t) ∈ Rn−2 × R | |xi| < r, i = 1, . . . , n− 2,−r2 < t ≤ 0,Θ′r = (x′, t) ∈ Rn−1 × R | (x′′, t) ∈ Θ′′r , |xn−1| < 4nLr,Θr = (x, t) ∈ Rn × R | (x′′, t) ∈ Θ′r, |xn| < r.

69

Lemma 62 (Boundary Harnack principle) Let

Λ = (x′, t) ∈ Θ′1 | xn−1 ≤ g(x′′, t)

for a parabolically Lipschitz function g in Θ′′1 with Lipschitz constant L ≥ 1 suchthat g(0, 0) = 0. Let u1, u2 be two continuous nonnegative functions in Θ1 suchthat for some positive constants c0, C0, M , and i = 1, 2,

i) 0 ≤ ui ≤M in Θ1 and ui = 0 on Λ,

ii) |(∆− ∂t)ui| ≤ C0 in Θ1 \ Λ,

iii) ui(x, t) ≥ c0 d(x, t) in Θ1 \ Λ, where d(x, t) = supr | Θr(x, t) ∩ Λ = ∅.

Assume additionally that u1 and u2 are symmetric in xn. Then, there existsα ∈ (0, 1) such that

u1

u2∈ Hα,α/2(Θ1/2).

Furthermore, α and the bound on the corresponding norm ‖u1/u2‖Hα,α/2(Θ1/2)

depend only on n, L, c0, C0, and M .

Remark 7 Lemma 62 is the parabolic version of Theorem 31. Unlike the ellip-tic case, it cannot be reduced to the other known results in the parabolic setting.We also note that this version of the boundary Harnack is for functions withnonzero right-hand side and therefore the nondegeneracy condition as in iii) isnecessary.

4.2.5 Singular free boundary points

The main goal of this section is to establish a structural theorem for the set ofthe so-called singular points, i.e. the points where the coincidence set v = φhas zero Hn-density in the thin manifold with respect to the thin paraboliccylinders. This corresponds to free boundary points with frequency κ = 2m,m ∈ N. We will show that the blowups at those points are parabolically κ-homogeneous polynomials.

As in the approach in [36], described in Sections 2.10.1-2.10.3, the main toolsare parabolic versions of monotonicity formulas of Weiss and Monneau type.These are instrumental in proving the uniqueness of the blowups at singularfree boundary points (x0, t0), and consequently obtain a Taylor expansion ofthe type

v(x, t)− φ(x′, t) = qκ(x− x0, t− t0) + o(‖(x− x0, t− t0)‖κ), t ≤ t0,

where qκ is a polynomial of parabolic degree κ that depends continuously onthe singular point (x0, t0) with frequency κ. We note explicitely that suchexpansion holds only for t ≤ t0 and may fail for t > t0 (see Remark 8 below).Nevertheless, this expansion essentially holds when restricted to singular points(x, t), even for t ≥ t0. This is necessary in order to verify the compatibility

70

condition in a parabolic version of the Whitney’s extension theorem. Using thelatter we are then ready to prove a structural theorem for the singular set. Itshould be mentioned at this moment that one difference between the paraboliccase treated in this section and its elliptic counterpart is the presence of newtypes of singular points, which we call time-like. At such points the blowupmay become independent of the space variables x′. We show that such singularpoints are contained in a countable union of graphs of the type

t = g(x1, . . . , xn−1),

where g is a C1 function. The other singular points, which we call space-like,are contained in countable union of d-dimensional C1,0 manifolds (d < n − 1).After a possible rotation of coordinates in Rn−1, such manifolds are locallyrepresentable as graphs of the type

(xd+1, . . . , xn−1) = g(x1, . . . , xd, t),

with g and ∂xig, i = 1, . . . , d, continuous.We now proceed to make these statements more precise. Since the overall

strategy is similar to the one described in Sections 2.10.1-2.10.3, we will omit theproofs and focus only on the main differences between the elliptic and paraboliccases. For futher details, we refer the interested reader to [26].

Definition 9 Let v ∈ Sφ(Q+1 ) with φ ∈ H`,`/2(Q′1), ` ≥ 2. We say that

(x0, t0) ∈ Γ∗(v) is singular if

limr→0+

Hn(Λ(v) ∩Q′r(x0, t0))

Hn(Q′r)= 0.

We will denote the set of singular points by Σ(u) and call it the singular set.We can further classify singular points according to the homogeneity of theirblowup, by defining

Σκ(v) := Σ(v) ∩ Γ(`)κ (v), κ ≤ `.

The following proposition gives a complete characterization of the singularpoints in terms of the blowups and the generalized frequency. In particular, itestablishes that

Σκ(u) = Γκ(u) for κ = 2m < `, m ∈ N.

Proposition 63 Let u ∈ Sf (S+1 ) with |f(x, t)| ≤M‖(x, t)‖`−2 in S+

1 , |∇f(x, t)| ≤L‖(x, t)‖`−3 in Q+

1/2, ` ≥ 3 and 0 ∈ Γ(`)κ (u) with κ < `. Then, the following

statements are equivalent:

(i) 0 ∈ Σκ(u).

(ii) any blowup of u at the origin is a nonzero parabolically κ-homogeneouspolynomial pκ in S∞ satisfying

∆pκ − ∂tpκ = 0, pκ(x′, 0, t) ≥ 0, pκ(x′,−xn, t) = pκ(x′, xn, t).

We denote this class by Pκ.

71

(iii) κ = 2m, m ∈ N.

We now state the two main results of this section.

Theorem 64 Let u ∈ Sf (S+1 ) with |f(x, t)| ≤M‖(x, t)‖`−2 in S+

1 , |∇f(x, t)| ≤L‖(x, t)‖`−3 in Q+

1/2, ` ≥ 3, and 0 ∈ Σκ(u) for κ = 2m < `, m ∈ N. Then,

there exists a nonzero pκ ∈ Pκ such that

u(x, t) = pκ(x, t) + o(‖(x, t)‖κ), t ≤ 0.

Moreover, if v ∈ Sφ(Q+1 ) with φ ∈ H`,`/2(Q′1), (x0, t0) ∈ Σκ(v) and we let

v(x0,t0) = v(x0 + ·, t0 + ·), then the mapping (x0, t0) 7→ p(x0,t0)κ from Σκ(v) to

Pκ is continuous.

Remark 8 We want to emphasize here that the asymptotic development, asstated in Theorem 64, does not generally hold for t > 0. Indeed, consider thefollowing example. Let u : Rn × R→ R be a continuous function such that

• u(x, t) = −t− x2n/2 for x ∈ Rn and t ≤ 0.

• In xn ≥ 0, t ≥ 0, u solves the Dirichlet problem

∆u− ∂tu = 0, xn > 0, t > 0,

u(x, 0) = −x2n, xn ≥ 0,

u(x′, 0, t) = 0 t ≥ 0.

• In xn ≤ 0, t ≥ 0, we extend the function by even symmetry in xn:

u(x′, xn, t) = u(x′,−xn, t).

It is easy to see that u solves the parabolic Signorini problem with zero obstacleand zero right-hand side in all of Rn×R. Moreover, u is homogeneous of degreetwo and clearly 0 ∈ Σ2(u). Now, if p(x, t) = −t − x2

n/2, then p ∈ P2 and wehave the following equalities:

u(x, t) = p(x, t), for t ≤ 0,

u(x′, 0, t) = 0, p(x′, 0, t) = −t for t ≥ 0.

So for t ≥ 0 the difference u(x, t)− p(x, t) is not o(‖(x, t)‖2), despite being zerofor t ≤ 0.

In order to state the aforementioned structural theorem, we need the follow-ing definitions.

Definition 10 For a singular point (x0, t0) ∈ Σκ(v) we define

d(x0,t0)κ = dimξ ∈ Rn−1 | ξ · ∇x′∂α

′

x′ ∂jt p

(x0,t0)κ = 0

for any α′ = (α1, . . . , αn−1) and j ≥ 0 such that |α′|+ 2j = κ− 1,

72

Σ22

Σ22

Σ110

6

Σ14

Σ14

t = −x21

t = 0

Figure 4: Structure of the singular set Σ(v) ⊂ R2 × (−∞, 0] for the solution vwith v(x1, x2, 0, t) = −t(t + x2

1)4, t ≤ 0 with zero thin obstacle. Note that thepoints on Σ1

4 and Σ110 are space-like, and the points on Σ2

2 are time-like.

which we call the spatial dimension of Σκ(v) at (x0, t0). Clearly, d(x0,t0)κ is an

integer between 0 and n− 1. Then, for any d = 0, 1, . . . , n− 1 define

Σdκ(v) := (x0, t0) ∈ Σκ(v) | d(x0,t0)κ = d.

Definition 11 We say that a (d+ 1)-dimensional manifold S ⊂ Rn−1×R, d =0, . . . , n− 2, is space-like of class C1,0, if locally, after a rotation of coordinateaxes in Rn−1 one can represent it as a graph

(xd+1, . . . , xn−1) = g(x1, . . . , xd, t),

where g is of class C1,0, i.e., g and ∂xig, i = 1, . . . , d are continuous.We say that (n−1)-dimensional manifold S ⊂ Rn−1×R is time-like of class

C1 if it can be represented locally as

t = g(x1, . . . , xn−1),

where g is of class C1.

Theorem 65 (Structure of the singular set) Let v ∈ Sφ(Q+1 ) with φ ∈ H`,`/2(Q′1),

` ≥ 3. Then, for any κ = 2m < `, m ∈ N, we have Γκ(v) = Σκ(v). Moreover,for every d = 0, 1, . . . , n − 2, the set Σdκ(v) is contained in a countable unionof (d+ 1)-dimensional space-like C1,0 manifolds and Σn−1

κ (v) is contained in acountable union of (n− 1)-dimensional time-like C1 manifolds.

For a small illustration, see Fig. 4.

The following two monotonicity formulas of Weiss and Monneau type playa crucial role in the proofs of Theorems 64 and 65.

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Theorem 66 (Weiss-type monotonicity formula) Let u ∈ Sf (S+1 ) with |f(x, t)| ≤

M‖(x, t)‖`−2 in S+1 , ` ≥ 2. For any κ ∈ (0, `), define the Weiss energy func-

tional

Wκu (r) :=

1

r2κ+2

∫S+r

(|t||∇u|2 − κ

2u2)G

=1

r2κ

(Iu(r)− κ

2Hu(r)

), 0 < r < 1.

Then, for any σ < `− κ there exists C > 0 depending only on σ, `, M , and n,such that

d

drWκu ≥ −Cr2σ−1, for a.e. r ∈ (0, 1).

In particular, the function

r 7→Wκu (r) + Cr2σ

is monotonically nondecreasing for r ∈ (0, 1).

The proof is by direct computation. Note that in Theorem 66 we do not

require 0 ∈ Γ(`)κ (u). However, if we do so, then we will have the following fact.

Lemma 67 Let u be as in Theorem 66, and assume additionally that 0 ∈Γ

(`)κ (u), κ < `. Then,

Wκu (0+) = 0.

The proof of this result uses the following growth estimate.

Lemma 68 Let u ∈ Sf (S+1 ) with |f(x, t)| ≤ M‖(x, t)‖`−2, ` ≥ 2, and 0 ∈

Γ(`)κ (u) with κ < ` and let σ < `− κ. Then

Hr(u) ≤ C(‖u‖2

L2(S+1 ,G)

+M2)r2κ, 0 < r < 1,

with C depending only on σ, `, n.

Theorem 69 (Monneau-type monotonicity formula) Let u ∈ Sf (S+1 ) with |f(x, t)| ≤

M‖(x, t)‖`−2 in S+1 , |∇f(x, t)| ≤ L‖(x, t)‖`−3 in Q+

1/2, ` ≥ 3. Suppose that

0 ∈ Σκ(u) with κ = 2m < `, m ∈ N. Further, let pκ be any parabolically κ-homogeneous caloric polynomial from class Pκ as in Proposition 63. For anysuch pκ, define Monneau’s functional as

Mκu,pκ(r) : =

1

r2κ+2

∫S+r

(u− pκ)2G, 0 < r < 1,

=Hw(r)

r2κ, where w = u− pκ.

Then, for any σ < `− κ there exists a constant C, depending only on σ, `, M ,and n, such that

d

drMκu,pκ(r) ≥ −C

(1 + ‖u‖L2(S+

1 ,G) + ‖pκ‖L2(S+1 ,G)

)rσ−1.

74

In particular, the function

r 7→Mκu,pκ(r) + Crσ

is monotonically nondecreasing for r ∈ (0, 1) for a constant C depending σ, `,M , n, ‖u‖L2(S+

1 ,G), and ‖pκ‖L2(S+1 ,G).

The proof of Theorem 69 relies on Theorem 66, and Lemmas 67 and 68. Toproceed with the proofs of Theorems 64 and 65, we observe that, similarly tothe elliptic case, it will be more convenient to work with a slightly different typeof rescalings and blowups than the ones used up to now. Namely, we will workwith the following κ-homogeneous rescalings

u(κ)r (x, t) :=

u(rx, r2t)

rκ

and their limits as r → 0+. The next lemma, which is the parabolic analogueof Lemma 39, shows the viability of this approach.

Lemma 70 Let u ∈ Sf (S+1 ) with |f(x, t)| ≤ M‖(x, t)‖`−2 in S+

1 , |∇f(x, t)| ≤L‖(x, t)‖`−3 in Q+

1/2, ` ≥ 3, and 0 ∈ Σκ(u) for κ < `. Then, there exists

c = cu > 0 such that

Hu(r) ≥ c r2κ, for any 0 < r < 1.

We explicitely observe that, by combining Lemmas 68 and 70, we obtain thatthe κ-homogeoneous rescalings are essentially equivalent to the ones introducedin Definition 7. Using this fact and Theorem 69, it is possible to show theuniqueness of the homogeneous blow-ups. The proofs of Theorems 64 and 65,at this point, proceed along the lines of their elliptic counterparts (see Section2.10.3), and therefore we omit them. We only point out that, in the first step ofthe proof of Theorem 65, one needs the following parabolic version of Whitney’sextension theorem.

Theorem 71 Let fα,j|α|+2j≤m be a family of functions on E, with f0,0 ≡ f ,satisfying the following compatibility conditions: there exists a family of moduliof continuity ωα,j|α|+2j≤2m, such that

fα,j(x, t) =∑

|β|+2k≤2m−|α|−2j

fα+β,j+k(x0, t0)

β!k!(x−x0)β(t−t0)k+Rα,j(x, t;x0, t0)

and

|Rα,j(x, t;x0, t0)| ≤ ωα,j(‖(x− x0, t− t0)‖)‖(x− x0, t− t0)‖2m−|α|−2j .

Then, there exists a function F ∈ C2m,m(Rn × R) such that F = f on E andmoreover ∂αx ∂

jtF = fα,j on E, for |α|+ 2j ≤ 2m.

75

Finally, in the second step of the proof of Theorem 65, some care needs tobe applied in the use of the implicit function theorem, as two cases need to beconsidered. When d ∈ 0, 1, . . . , n − 2, Σdκ(v) is contained in the countableunion of d-dimensional space-like C1,0 manifolds. When instead d = n − 1,Σdκ(v) is contained in a time-like (n− 1)-dimensional C1 manifold, as required.

Comments and further readings

• The continuity of the temperature, in space and time variables, in bound-ary heat control problems (both of Stefan and porous media type) hasbeen established by Athanasopoulos and Caffarelli in [8]. In addition, theauthors extend this result to the fractional order case. The proof is basedon a combination of penalization and De Giorgi’s method.

• Higher regularity of the time derivative in the parabolic thin obstacleproblem has recently been established by Petrosyan and Zeller in [57] andAthanasopoulos, Caffarelli and Milakis in [9]. It was already known fromArkhipova A, Uraltseva ([4]) that if the initial data φ0 ∈ W 2

∞(B+1 ), then

the time derivative ∂tv of the solution to (48)-(51) is locally bounded inQ+

1 ∩Q′1. This assumption on the initial data, however, is rather restric-tive and excludes time-independent solutions as in Proposition 54. Thefirst main result in [57] shows that ∂tv is in fact bounded, without anyextra assumptions on the initial data, even though some more regularityon the thin obstacle φ is required. The key observation in the proof isthat incremental quotients of the solution in the time variable satisfy adifferential inequality. In addition, the authors prove the Holder continu-ity of ∂tv at regular free boundary points, which in turn allows them toshow that the free boundary, at regular points, is a C1,α surface both inspace and time. The Holder continuity in time for solutions (and the ensu-ing C1,α regularity of the free boundary) has also been established in [9],with a different approach based on quasi-convexity properties of the solu-tion. The regularity of the time derivative plays also a crucial role in [11],where Banerjee, Smit Vega Garcia and Zeller show that the free bound-ary near a regular point is C∞ in space and time when the obstacle is zero.

A Some auxiliary tools

In this section we collect some results on the fractional Laplacian and the ex-tension operator La from [23] and [22]. Standard reference for the fractionalLaplacian is Landkof’s book [50]. See also the recent book by Molica Bisci,Radulescu and Servadei [52].

76

Supersolutions and comparison for the fractional Laplacian The def-inition of supersolution is given for u ∈ P ′s the dual of

Ps =f ∈ C∞ (Rn) :

(1 + |x|n+2s

)Dkf is bounded, ∀k ≥ 0

,

endowed with the topology induced by the seminorms

|f |k,s = sup∣∣∣(1 + |x|n+2s

)Dkf

∣∣∣ .and the meaning is that it is a nonnegative measure:

Let u ∈ P ′s. We say that (−∆)su ≥ 0 in an open set Ω if 〈(−∆)

su, ϕ〉 ≥ 0,

for every nonnegative test function ϕ ∈ C∞ (Rn) , rapidly decreasing at infinity.

Every supersolution shares some properties of superharmonic functions. Forinstance, u is lower-semicontinuous.

Proposition A1. Let (−∆)su ≥ 0 in an open set Ω, then u is lower-

semicontinuous in Ω.

Moreover, if the restriction of u on supp((−∆)su) is continuous then, u is

continuous everywhere. Precisely, we have:

Proposition A2. Let v be a bounded function in Rn such that (−∆)sv ≥ 0.

If E =supp((−∆)sv) and v|E is continuous, then v is continuous in Rn.

Due to the nonlocal nature of (−∆)s, a comparison theorem in a domain Ω

must take into account what happens outside Ω. Indeed we have:

Proposition A3. (Comparison) Let Ω b Rn be an open set. Let (−∆)su ≥

0 and (−∆)sv ≤ 0 in Ω, such that u ≥ v in Rn\Ω and u − v is lower-

semicontinuous in Ω. Then u ≥ v in Rn. Moreover, if x ∈ Ω and u (x) = v (x)then u = v in Rn.

Also, the set of supersolutions is a directed set, as indicated by the followingproposition.

Proposition A4. Let Ω b Rn be an open set. Let (−∆)su1 ≥ 0 and

(−∆)su2 ≥ 0 in Ω, such that u ≥ v in Rn\Ω. Then u = min u1, u2 is a

supersolution in Ω.

Estimates for the operator La As we have already noted, the operatorLa is a particular case of the class of degenerate elliptic operators considered in[29]. For the following result see Theorems 2.3.8 and 2.3.12 in that paper.

(2) v ∈ C0,α(Br/2

)for some α ≤ 1 and if f = 0

‖v‖C0,α(Br/2) ≤C

rαoscBrv.

Using the translational invariance of the equation in the x variable, we obtainthe following result.

77

Lemma A1. (Schauder type estimates) Assume Lav = 0 in Br.Then, forevery k ≥ 1, integer: ∥∥Dk

xv∥∥L∞(Br/2)

≤ C

rkoscBrv

and3 ∣∣Dkxv∣∣C0,α(Br/2)

≤ C

rk+αoscBrv.

Using the above Theorem and the equation in the form

∆xv = −vyy −a

yvy

we get:

Lemma A2. Assume Lav = 0 in Br.Then, for every r ≤ 1:∥∥∥∥vyy +a

yvy

∥∥∥∥L∞(Br/2)

≤ C

r2oscBrv.

The next is a Liouville type result.

Lemma A3. Let v a solution of Lav (X) = 0 in Rn+1. Assume that

v (x, y) = v (x,−y) and |v (X)| ≤ C (1 + |X|γ) , γ ≥ 0.

Then v is a polynomial.

We now state a mean value property for supersolution of a nonhomogeneoussolution:

Lemma A4. (Mean value property) Let v be a solution of

Lav (X) ≤ C |ya| |X|k in B1.

Then, for every r ≤ 1,

v (0) ≥ 1

ωn+arn+a

∫∂Br

v (X) |y|a dS − Crk+2

where

ωn+a =

∫∂B1

|y|a dS.

Poincare type inequalities in the context of weighted Sobolev spaces can befound in [29]. In our case, letting v = 1

ωn+arn+a

∫∂Br

v (X) |y|a dσ,we have4:

3|w|C0,α(D) denotes the seminorm

supx,y∈D

|w (x)− w (y)| .|x− y|α

4A function v ∈ W 1,2 (B1, |y|a) has a trace in L2 (∂B1, |y|a) and the trace operator iscompact.

78

Lemma A5. (Poincare inequalities) Let v ∈ W 1,2 (B1, |y|a) . Then, forr ≤ 1 : ∫

∂Br

(v (X)− v)2 |y|a dσ ≤ C (a, n) r

∫Br

|∇v (X)| |y|a dX.

and ∫∂B1

(v (X)− v (rX))2 |y|a dσ ≤ C (a, n, r)

∫B1

|∇v (X)| |y|a dX. (65)

The first inequality is standard, The second one can be proved by integrating∇v along the lines sX with s ∈ (r, 1) .

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Donatella DanielliDepartment of MathematicsPurdue UniversityE-mail: danielli@math.purdue.edu

Sandro SalsaDepartment of MathematicsPolitecnico di MilanoE-mail: sandro.salsa@polimi.it

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