Free Boundary Regularity for Almost Every Solution to the Signorini Problem

We investigate the regularity of the free boundary for the Signorini problem in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathbb {R}}^{n+1}$$\end{document}Rn+1. It is known that regular points are \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(n-1)$$\end{document}(n-1)-dimensional and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^\infty $$\end{document}C∞. However, even for \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^\infty $$\end{document}C∞ obstacles \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi $$\end{document}φ, the set of non-regular (or degenerate) points could be very large—e.g. with infinite \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {H}}^{n-1}$$\end{document}Hn-1 measure. The only two assumptions under which a nice structure result for degenerate points has been established are when \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi $$\end{document}φ is analytic, and when \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Delta \varphi < 0$$\end{document}Δφ<0. However, even in these cases, the set of degenerate points is in general \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(n-1)$$\end{document}(n-1)-dimensional—as large as the set of regular points. In this work, we show for the first time that, “usually”, the set of degenerate points is small. Namely, we prove that, given any \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^\infty $$\end{document}C∞ obstacle, for almost every solution the non-regular part of the free boundary is at most \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(n-2)$$\end{document}(n-2)-dimensional. This is the first result in this direction for the Signorini problem. Furthermore, we prove analogous results for the obstacle problem for the fractional Laplacian \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(-\Delta )^s$$\end{document}(-Δ)s, and for the parabolic Signorini problem. In the parabolic Signorini problem, our main result establishes that the non-regular part of the free boundary is \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(n-1-\alpha _\circ )$$\end{document}(n-1-α∘)-dimensional for almost all times t, for some \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha _\circ > 0$$\end{document}α∘>0. Finally, we construct some new examples of free boundaries with degenerate points.


Introduction
The Signorini problem (also known as the thin or boundary obstacle problem) is a classical free boundary problem that was originally studied by Antonio Signorini in connection with linear elasticity [27,39,40]. The problem gained further This work has received funding from the European Research Council (ERC) under the Grant Agreements No 721675 and No 801867. In addition, X. F. was supported by the SNF Grant 200021_182565 and X.R. was supported by the SNF Grant 200021_178795 and by the MINECO grant MTM2017-84214-C2-1-P. attention in the seventies due to its connection to mechanics, biology, and even finance-see [11,14,34], and [17,37]-, and since then it has been widely studied in the mathematical community; see [2,3,7,9,10,12,18,20,22,26,29,30,36,38] and references therein.
The main goal of this work is to better understand the size and structure of the non-regular part of the free boundary for such problem.
In particular, our goal is to prove for the first time that, for almost every solution (see Remark 1.2), the set of non-regular points is small. As explained in detail below, this is completely new even when the obstacle ϕ is analytic or when it satisfies ϕ < 0.

The Signorini Problem
Let us denote x = (x , x n+1 ) ∈ R n × R and B + 1 = B 1 ∩ {x n+1 > 0}. We say that u ∈ H 1 (B + 1 ) is a solution to the Signorini problem with a smooth obstacle ϕ defined on B 1 (1.1) in the weak sense, for some boundary data g ∈ C 0 (∂ B 1  (u − ϕ) Cr 2 , ∀r ∈ (0, r • ) , (1.2) (see [3]). Alternatively, each of the subsets can be defined according to the order of the blow-up at that point. Namely, the set of regular points are those whose blow-up is of order 3 2 , and the set of degenerate points are those whose blow-up is of order κ for some κ ∈ [2, ∞].
Let us denote κ the set of free boundary points of order κ. That is, those points whose blow-up is homogeneous of order κ (we will be more precise about it later on, in Section 2; the definition of ∞ is slightly different). Then, it is well known that the free boundary can be divided as where • 3/2 = Reg(u) is the set of regular points. They are an open (n−1)-dimensional subset of (u), and it is C ∞ (see [3,13,29]). • even = m≥1 2m (u) denotes the set of points whose blow-ups have even homogeneity. Equivalently, they can also be characterised as those points of the free boundary where the contact set has zero density, and they are often called singular points. They are contained in the countable union of C 1 (n − 1)dimensional manifolds; see [18,22]. • odd = m≥1 2m+1 (u) is, a priori, also an at most (n − 1)-dimensional subset of the free boundary and it is (n − 1)-rectifiable (see [19][20][21]31]), although it is not actually known whether it exists. • half = m≥1 2m+3/2 (u) corresponds to those points with blow-up of order 7 2 , 11 2 , etc. They are much less understood than regular points. The set half is an (n − 1)-dimensional subset of the free boundary and it is (n − 1)-rectifiable (see [20,21,31]). • * is the set of all points with homogeneities κ ∈ (2, ∞), with κ / ∈ N and κ / ∈ 2N − 1 2 . This set has Hausdorff dimension at most n − 2, so it is always small, see [20,21,31]. • ∞ is the set of points with infinite order (namely, those points at which u − ϕ vanishes at infinite order, see (2.11)). For general C ∞ obstacles it could be a huge set, even a fractal set of infinite perimeter with dimension exceeding n −1.
When ϕ is analytic, instead, ∞ is empty.
Overall, we see that, for general C ∞ obstacles, the free boundary could be really irregular.
The only two assumptions under which a better regularity is known are • ϕ < 0 on B 1 and u = 0 on ∂ B 1 ∩ {x n+1 > 0}. In this case, (u) = 3/2 ∪ 2 and the set of degenerate points is locally contained in a C 1 manifold; see [5]. • ϕ is analytic. In this case, ∞ = ∅ and is (n − 1)-rectifiable, in the sense that it is contained in a countable union of C 1 manifolds, up to a set of zero H n−1 -measure, see [20,31].
The goal of this paper is to show that, actually, for most solutions, all the sets even , odd , half , and ∞ are small, namely, of dimension at most n − 2. This is new even in case that ϕ is analytic and ϕ < 0.

Our Results
We will prove here that, even if degenerate points could potentially constitute a large part of the free boundary (of the same dimension as the regular part, or even higher), they are not common. More precisely, for almost every obstacle (or for almost every boundary datum), the set of degenerate points is small. This is the first result in this direction for the Signorini problem, even for zero obstacle.
Our main result reads as follows: where Deg(u λ ) is defined by (1.2).
This result is completely new even for analytic obstacles, or for ϕ = 0. No result of this type was known for the Signorini problem.
The results we prove (see Theorem 4.4 and Proposition 4.8) are actually more precise and concern the Hausdorff dimension of ≥κ (u λ ), the set of points of order greater or equal than κ. We will show that, if 3 κ n + 1, then ≥κ (u λ ) has dimension n − κ + 1, while for κ > n + 1, then ≥κ (u λ ) is empty for almost every λ ∈ [0, 1]. We refer to [32,Chapter 4] for the definition of Hausdorff dimension.

Remark 1.2.
In the context of the theory of prevalence, [25] (see also [35]), Theorem 1.1 says that the set of solutions satisfying that the free boundary has a small degenerate set is prevalent within the set of solutions (say, given by C 0 or L ∞ boundary data). Alternatively, the set of solutions whose degenerate set is not lower dimensional is shy.
In particular, we can say that for almost every boundary data (see [35,Definition 3.1]) the corresponding solution has a lower dimensional degenerate set. This is because adding a constant as in (1.5) is a 1-probe (see [35,Definition 3.5]) for the set of boundary data, thanks to Theorem 1.1.
We will establish a finer result regarding the set ∞ (u λ ). While it is known that it can certainly exist for some solutions u λ (see Proposition 1.9), we show that it will be empty for almost every λ ∈ [0, 1].
We remark that in the previous result, dim H denotes the Hausdorff dimension, whereas dim M denotes the Minkowski dimension (we refer to [32,Chapters 4 and 5]). As such, the second part of the result is much stronger than the first one (e.g., Let us briefly comment on the condition (1.5). Notice that such condition can be reformulated in many ways. In the simplest case, one could simply take g λ = g 0 ±λ. Alternatively, one could take a family of obstacles ϕ λ = ϕ 0 ±λ (with fixed boundary conditions); this is equivalent to fixing the obstacle ϕ 0 and moving the boundary data g λ = g ∓ λ. Furthermore, one could also consider g λ = g 0 + λ for any ≥ 0, ≡ 0. Then, even if the second condition in (1.5) is not directly fulfilled, a simple use of strong maximum principle makes it true in some smaller ball B 1−ρ , so that g λ+ε ≥ g λ +c(ρ)ε on ∂ B 1−ρ ∩{x n+1 ≥ 1 2 −ρ/2}. By rescaling the function and the domain, we can rewrite it as (1.5).
Regularity results for almost every solution have been established before in the context of the classical obstacle problem by Monneau in [33]. In such problem, however, all free boundary points have homogeneity 2, and non-regular points are characterised by the density of the contact set around them: non-regular points are those at which the contact set has density zero. In the Signorini problem, instead, the structure of non-regular points is quite different, and they are characterised by the growth of u around them (recall (1.2) and the definition of even , odd , half , and ∞ ). This is why the approach of [33] cannot work in the present context. More recently, the results of Monneau for the classical obstacle problem have been widely improved by Figalli, the second author, and Serra in [19]. The results in [19] are based on very fine higher order expansions at singular points, which then lead to a better understanding of solutions around them, combined with new dimension reduction arguments and a cleaning lemma to get improved bounds on higher order expansions.
Here, due to the different nature of the problem, we do not need any fine expansion at non-regular points nor any dimension reduction. Most of our arguments require only the growth of solutions at different types of degenerate points, combined with appropriate barriers, and Harnack-type inequalities. The starting point of our results is to use a simple (but key) GMT lemma from [19] (see Lemma 4.1 below).

Parabolic Signorini Problem
The previous results use rather general techniques that suitably modified can be applied to other situations. We show here that using a similar approach as in the elliptic case, one can deduce results regarding the size of the non-regular part of the free boundary for the parabolic version of the Signorini problem, for almost every time t.
We say that a function in the weak sense (cf. (1.1)). A thorough study of the parabolic Signorini problem was made by Danielli, Garofalo, Petrosyan, and To, in [12]. The parabolic Signorini problem is a free boundary problem, where the free boundary belongs to B 1 × (−1, 0] and is defined by where ∂ B 1 ×(−1,0] denotes the boundary in the relative topology of B 1 × (−1, 0]. Analogously to the elliptic Signorini problem, the free boundary can be divided into regular points and degenerate (or non-regular) points: The set of regular points are those where parabolic blow-ups are parabolically 3 2homogeneous. On the other hand, degenerate points are those where parabolic blowups of the solution are parabolically κ-homogeneous, with κ ≥ 2 (alternatively, the solution detaches at most quadratically from the obstacle in parabolic cylinders, B r ×(−r 2 , 0]). Further stratifications according to the homogeneity of the parabolic blow-ups can be done in an analogous way to the elliptic problem, see [12].
The set of regular points Reg(u) is a relatively open subset of (u) and the free boundary is smooth (C 1,α ) around them (see [12,Chapter 11]). The set of degenerate points, however, could be even larger than the set of regular points.
In this manuscript we show that, under the appropriate conditions, for a.e. time t ∈ (−1, 0] the set of degenerate points has dimension (n −1−α • ) for some α • > 0 depending only on n. That is, for a.e. time, the free boundary is mostly comprised of regular points, and therefore, it is smooth almost everywhere.
In order to be able to get results of this type we must impose some conditions on the solution. We will assume that that is, wherever the solution u is not in contact with the obstacle ϕ, it is strictly monotone. Alternatively, by the strong maximum principle, the condition can be rewritten as up to a constant multiplicative factor. Condition (1.7) is somewhat necessary. If the strict monotonicity was not required, we could be dealing with a bad solution (with large non-regular set) of the elliptic problem for a set of times of positive measure, and therefore, we could not expect a result like the one we prove. On the other hand, if one allowed changes in the sign of u t (alternatively, one allowed non-stationary obstacles), then the result is also not true (see, for instance, the example discussed in [12, Figure  12.1]).
Condition (1.7) is actually quite natural. One of the main applications of the parabolic Signorini problem is the study of semi-permeable membranes (see [14,Section 2.2]): We consider a domain (B + 1 ) and a thin membrane (B 1 ), which is semipermeable: that is, a fluid can pass through B 1 into B + 1 freely, but outflow of the fluid is prevented by the membrane. If we suppose that there is a given liquid pressure applied to the membrane B 1 given by ϕ, and we denote u(x, t) the inside pressure of the liquid in B + 1 , then the parabolic Signorini problem (1.6) describes the evolution of the inside pressure with time. In particular, since liquid can only enter B + 1 (and we assume no liquid can leave from the other parts of the boundary), pressure inside the domain can only become higher, and the solution will be such that u t > 0. The same condition also appears in volume injection through a semi-permeable wall ( for some α • > 0 depending only on n. In particular, for a.e. t • ∈ (−1, 0] the free boundary (u) ∩ {t = t • } is a C 1,α (n − 1)-dimensional manifold, up to a closed subset of Hausdorff dimension n − 1 − α 0 .
When ϕ is analytic, then the free boundary is actually C ∞ around regular points. Higher regularity of the free boundary is also expected for smooth obstacles, but so far it is only known when ϕ is analytic; see [4].
It is important to remark that the parabolic case presents some extra difficulties with respect to the elliptic one, and in fact we do not know if a result analogous to Theorem 1.3 holds in this context. This means that points of order ∞ could a priori still appear for all times (even though by Theorem 1.4 they are lower-dimensional for almost every time).

The Fractional Obstacle Problem
The Signorini problem in R n+1 can be reformulated in terms of a fractional obstacle problem with operator (− ) 1 2 in R n . Conversely, fractional obstacle problems (with the operator (− ) s , s ∈ (0, 1)) can also be reformulated in terms of thin obstacle problems with weights. In this work we will generally deal with the thin obstacle problem with a weight, so that the results from Section 1.3 can also be formulated for the fractional obstacle problem.
Given an obstacle ϕ ∈ C ∞ (R n ) such that (1.9) Solutions to the fractional obstacle problem are C 1,s (see [8]). We denote (v) = {v = ϕ} the contact set, and (v) = ∂ (v) the free boundary. As in the Signorini problem (which corresponds to s = 1 2 ) the free boundary can be partitioned into regular points and non-regular (or degenerate) points, More precisely, if we denote by κ (v) the free boundary points of order κ, then the free boundary (v) can be further stratified analogously to (1.3) as (1.11) Here, 1+s = Reg(v) is the set of regular points ( [8,41]). Again, it is an open subset of the free boundary, which is smooth. Similarly, 2m for m ≥ 1 are often called singular points, and are those where the contact set has zero measure (see [23]). Together with the sets 2m+2s and 2m+1+s for m ≥ 1, they are an (n − 1)dimensional rectifiable subset of the free boundary, [21,23]. Finally, * denotes the set containing the remaining homogeneities (except infinite), and has dimension n − 2; and ∞ denotes those boundary points where the solution is approaching the obstacle faster than any power (i.e., at infinite order). As before, the set ∞ could have dimension even higher than n − 1.
The type of result we want to prove in this setting regarding regularity for most solutions is concerned with global perturbations of the obstacle (rather than boundary perturbations, as before). That is, we will consider obstacles fulfilling (1.8).
We define the set of solutions indexed by λ ∈ [0, 1] to the fractional obstacle problem as (1.12) Then, our main result reads as follows: As before, we actually prove more precise results (see Theorem 4.4 and Proposition 4.8). We establish an estimate for the Hausdorff dimension of ≥κ (v λ ). We show that, for 2 κ − 2s n, then dim H ≥κ (v λ ) n − κ + 2s, and if κ > n + 2s, then ≥κ (v λ ) is empty for almost every λ ∈ [0, 1]. Similarly, we can also reduce the regularity of the obstacle to ϕ ∈ C 4,α so that, for a.e. λ ∈ [0, 1], dim H (Deg(v λ )) n − 2 (in particular, the free boundary (v λ ) is C 3,α up to a subset of dimension n − 2 for a.e. λ ∈ [0, 1]; see [1,26] That is, analogously to Theorem 1.3, we can also control the size of λ for which the free boundary points of infinite order exist.

Examples of Degenerate Free Boundary Points
Let us finally comment on the non-regular part of the free boundary, that is, The main open questions regarding each of the subsets of the degenerate part of the free boundary are Q1: Are there non-trivial examples (e.g., the limit of regular points) of singular points in even ? Q2: Do points in odd exist? Q3: Can one construct arbitrary contact sets with free boundary formed entirely of half (alternatively, do they exist apart from the homogeneous solutions)? Q4: Do points in * exist? Q5: How big can the set ∞ be? In this paper, we answer questions Q1, Q3, and Q5. (Questions Q2 and Q4 remain open.) Let us start with Q1. The set even = m≥1 2m , often called the set of singular points, is an (n − 1)-dimensional subset of the free boundary. Examples of free boundary points belonging to even are easy to construct as level sets of homogeneous harmonic polynomials, such as x 2 1 − x 2 n+1 , in which case we have = even = {x 1 = 0}. They are also expected to appear in less trivial situations but, as far as we know, none has been constructed so far that appears as limit of regular points (i.e., on the boundary of the interior of the contact set). Here, we show Proposition 1.7. There exists a boundary data g such that the free boundary of the solution to the Signorini problem (1.1) with ϕ = 0 has a sequence of regular points (of order 3/2) converging to a singular point (of order 2).
The proof of the previous result is given in Section 5. In contrast to what occurs with the classical obstacle problem, the construction of singular points does not seem to immediately arise from continuous perturbations of the boundary value under symmetry assumptions. Instead, one has to be aware that there could appear other points (different from regular, but not in even ). Thus, our strategy is based on being in a special setting that avoids the appearance of higher order free boundary points.
On the other hand, regarding question Q3, it is known that examples of such points can be constructed through homogeneous solutions, in which case they can even appear as limit of regular (or lower frequency) points (see [10,Example 1]). Until now, however, it was not clear whether such points could appear in non-trivial (say, non-homogeneous) situations.
We show that, given any smooth domain ⊂ R n , one can find a solution to the Signorini problem whose contact set is exactly given by , and whose free boundary is entirely made of points of order 7 2 (or 11 2 , etc.). More generally, we show that given , the contact set for the fractional obstacle problem can be made up entirely of points belonging to m≥1 2m+1+s (the case s = 1 2 corresponding to the Signorini problem). Proposition 1.8. Let ⊂ R n be any given C ∞ bounded domain, and let m ∈ N. Then, there exists an obstacle ϕ ∈ C ∞ (R n ) with ϕ → 0 at ∞, and a global solution to the obstacle problem such that the contact set is (u) = {u = ϕ} = , and all the points on the free boundary ∂ (u) have frequency 2m + 1 + s.
The proof of the previous proposition is constructive: we show a way in which such solutions can be constructed, using some results from [1,24].
Finally, we also answer question Q5, that deals with the set ∞ . Not much has been discussed about it in the literature, though its lack of structure was somewhat known by the community. For instance, the following result is not difficult to prove: Proposition 1.9. For any ε > 0 there exists a non-trivial solution u and an obstacle and the boundary of the contact set, (u) = {u = ϕ}, fulfils This shows that, in general, there is no hope to get nice structure results for the full free boundary for C ∞ obstacles. However, thanks to Theorem 1.6 above we know that such behaviour is extremely rare. As before, we are answering question Q5 in the generality of the fractional obstacle problem; the Signorini problem corresponds to the case s = 1 2 .

Organization of the Paper
The paper is organised as follows: In Section 2 we study the behaviour of degenerate points under perturbation. In particular, we show how the free boundary moves around them when perturbing monotonically the solution to the obstacle problem. We treat separately general degenerate points, and those of order 2. In Section 3 we study the dimension of the set 2 by means of an appropriate application of Whitney's extension theorem. In Section 4 we prove the main results of this work, Theorems 1.1, 1.3, 1.5, and 1.6. In Section 5 we construct the examples of degenerate points introduced in Section 1.6, proving Propositions 1.7, 1.8, and 1.9 . Finally, in Section 6 we deal with the parabolic Signorini problem and prove Theorem 1.4.

Behaviour of Non-regular Points Under Perturbations
be our obstacle on the thin space. Let us consider the fractional operator with a ∈ (−1, 1), and (0, 1) s = 1−a 2 . We will interchangeably use both a and s depending on the situation. (In general, we will use a for the weight exponent, and s for all the other situations.) Let us suppose that we have a family of increasing even solutions u λ for 0 λ 1 to the fractional obstacle problem for a given obstacle ϕ satisfying (2.1). In particular, {u λ } 0 λ 1 satisfy for some constant M independent of λ, that will depend on the obstacle (see (2.6)-(2.7) below). Notice that solutions are C 1,s in B 1/2 (or in B + 1/2 ), but only C 2s in B 1 (C 0,1 when s = 1 2 ). We denote (u λ ) := {x : u λ (x , 0) = ϕ(x )}×{0} ⊂ R n the contact set, and its boundary in the relative topology of R n , ∂ (u λ ) = ∂{x : is the free boundary. Note that, from the monotonicity assumption, (2.4)

Lemma 2.1. Let u λ denote the family of solutions to (2.2)-(2.3). Then, for any
for some constant c > 0 depending only on n, s, and h. In particular, for some constant c > 0 depending only on n, s, and h.
Proof. Fix some λ > 0 and ε > 0, and define We will show that the result holds for δ λ,ε u λ for some constant c independent of ε > 0, and in particular, it also holds after taking the lim inf.
and we define the barrier function ψ : Notice that, by the boundary Harnack inequality for Muckenhoupt weights A 2 (see [15]), ψ is comparable to |x n+1 | 2s (since both vanish continuously at x n+1 = 0, and both are a-harmonic), and in particular, there exists some c > 0 small depending only on n, s, and h, such that ψ ≥ c |x n+1 | 2s in B r (x • ). We have that for some c depending only on n, s, and h. Thus, as we wanted to see.
Let 0 ∈ ∂ (u λ ) be a free boundary point for u λ . Let us denote Q τ (x ) the Taylor expansion of ϕ(x ) around 0 up to order τ , and we denote Q a τ (x) its unique even a-harmonic extension (see [23,Lemma 5 In particular, notice that since Q τ (x ) is the Taylor approximation of ϕ up to order τ , we have that for some M > 0 depending only on ϕ. We take M larger if necessary, so that it coincides with the one of (2.3). We consider the generalized frequency formula, for θ ∈ (0, α), and for some C θ (that is independent of the point around which is taken) Then, by [23, Proposition 6.1] (see also [8,22]) we know that τ,α,θ (r,ū λ ) is nondecreasing for 0 < r < r • for some r • . In particular, τ,α,θ (0 + ,ū λ ) is well defined, and by [ We say that 0 ∈ ∂ (u λ ) is a point of order κ if τ,α,θ (0 + ,ū λ ) = n+1−2s+2κ. In particular, by the previous inequalities Thanks to [23,Lemma 6.4] (see, also, [5, Lemma 7.1]) we know that for a point of order greater or equal than κ, for κ < τ + α − θ , then we have for some constant C M depending only on M, τ , α, θ . In general, for any point x • ∈ ∂ (u λ ), we can defineū x • λ analogously to before, as follows: . With this, we can define the free boundary points of u λ of order κ, with 1 + s κ < τ + α − θ , as Equivalently, one can define λ ≥κ as those points where (2.9) occurs. Notice that the previous sets are consistently defined, in the sense that if x • is a free boundary point for u λ , and τ ∈ N, α ∈ (0, 1) are such that τ + α τ + α, then Proof. Suppose that it is not true. In particular, suppose that there exists some z = (z , z n+1 ) ∈ B 1/2 \ {x n+1 = 0} such that w(z) = 0. Let us define the cylinder and let That is, v is supera-harmonic and is negative at z ∈ Q, then it must be negative somewhere on ∂ Q. Let us check that this is not the case, to reach a contradiction. First, notice that, assuming On the other hand, on if ε is small enough depending only on n and a. Thus, v ≥ 0 on ∂ Q and on , and is supera-harmonic in Q \ , so we must have v ≥ 0 in Q, contradicting v(z) < 0.
Let us now show the following proposition: for some C * depending only on n, s, M, κ, τ , α, and h.
From the previous inequality applied at x ∈ B r (0)∩{|x n+1 | ≥ r σ }, for some σ > 0 to be chosen, for r < h 4 , and with ε = C * r κ−2s for some C * to be chosen, On the other hand, notice that 0 is a free boundary point ofū x • λ of order greater or equal than κ. In particular, from the growth estimate (2.9), we know that for some C depending only on n, M, s, τ , α, θ , and h. By choosing, for example, } in the definition of the generalized frequency function, (2.8), we can get rid of the dependence on θ . That is, Notice, also, that Let us rescale in domain. We denote Then w is a solution to a thin obstacle problem with right-hand side and with zero obstacle in the ball (Notice that τ + α − a − κ ≥ 0 by assumption.) We now want to apply Lemma 2.3. We need to choose σ < ε • (n, a), and C * such that we get that such C * exists independently of r , depending only on n, M, s, κ, τ , α, and h.
Finally, notice that thanks to the optimal regularity of solutions, if x • ∈ λ , then x • ∈ λ ≥1+s , so that applying the previous result we are done.
In particular, we have We also define which is uniquely defined on .
A direct consequence of Proposition 2.4 is that is continuous: 3), and let ϕ satisfy (2.1). The function is continuous in the C 0 -norm.
Proof. Let us start with the first statement. If x 1 , x 2 ∈ are such that |x 1 − x 2 | δ 2 for δ > 0 small enough, and λ(x 1 ) ≥ λ(x 2 ), then by Proposition 2.4. In particular, λ(y) < λ(x 2 ) + C * δ 1−s for any y ∈ B δ (x 2 ), so that λ(x 1 ) < λ(x 2 ) + C * δ 1−s . That is, Let us now show that is also continuous (in the C 0 -norm). From the definition ofū x • λ(x • ) , Definition 2.2, and since ϕ is continuous, it is enough to show that is continuous. Moreover, since each u λ is continuous (and in fact, they are uniformly C 2s ), we will show that x • → u λ(x • ) is continuous, in the sense that, for every ε > 0, there exists some δ > 0 such that if x, z ∈ ∩ B 1−h (for some h > 0), |x − z| δ, then Let us argue by contradiction. Suppose that it is not true, and that there exist for some ε • > 0. In particular, let us assume that λ( After taking a subsequence (by compactness, using also that u λ C 2s (B 1 ) M), we can assume that there exists some ball B ρ (y) ⊂ B 1 such that (The radius ρ depends only on n, ε • , and M.) By interior Harnack's inequality, we have that for some constant c depending on ρ and h. After translating and scaling, we are in a situation to apply Corollary 2.5. In particular, for some δ > 0 (depending on ε • and h), The following lemma improves Lemma 2.1 in case x • ∈ 2 (we denote here that a − := max{0, −a}): for some constant c > 0 independent of λ and μ (but possibly depending on everything else).
Proof. Fix some μ > 0 and ε > 0 small, and define As in the proof of Lemma 2.1, we know that δ λ,εū Let us start by showing that, for every A > 0, there exists some ρ A > 0 (independent of μ) such that, after a rotation, In particular, we will show that, for every A > 0, there exists some ρ A > 0 such that, after a rotation, for some 2-homogeneous, a-harmonic polynomial, such that p 2 ≥ 0 on {x n+1 = 0} (recall that we are assuming that x • ∈ λ 2 ) and p 2 ≡ 0. After a rotation, thus, we may assume that p 2 (x , 0) ≥ cx 2 1 . That is, if ρ A is small enough (depending on A, but also on the point x • , and the function u x • λ ). That is, (2.17), and in particular, (2.16), holds. Considering again the x n+1 direction, we know that for every A > 0 there exists some ρ A such that, after a rotation, Notice that ρ A ↓ 0 as A → ∞. Let us suppose that we are always in the rotated setting so that the previous inclusion holds. Let us denote ψ A the unique homogeneous solution to . It corresponds to the first eigenvalue on the sphere S n of L a with zero boundary condition on C A/2 . Alternatively, it corresponds to the infimum of the corresponding Rayleigh quotient among functions with the same boundary values. Notice that, as A → ∞, C A/2 → {x 1 = x n+1 = 0} locally uniformly in the Hausdorff distance, and {x 1 = x n+1 = 0} has zero a-harmonic capacity when s 1 2 (see [28,Corollary 2.12]). Thus, when s 1 2 the infimum of the Rayleigh quotient converges to the first eigenvalue of L a on the sphere without boundary conditions (namely, 0), and thus, η • ↓ 0 as A → ∞ if a ≥ 0. Alternatively, if s > 1 2 the first eigenvalue corresponds to the homogeneity −a (attained by the function (x 2 1 + x 2 n+1 ) −a/2 ), so that η • ↓ −a as A → ∞ if a < 0. In all, η • ↓ a − , with a − = max{0, −a}.
Let us choose some A large enough such that η • < η + a − . Now, let and let ψ A,r for r < ρ A /2 denote the solution to Letc small enough (depending on ρ A , A, h, n, s, M) such thatcψ A δ λ,εū By Harnack's inequality, there exists a constant C depending only on n and s such that where in the last inequality we are using the η • -homogeneity of ψ A , and c depends only on n and a. Thus, for some c > 0 that might depends on everything, but it is independent of μ and λ, where we assumed r < ρ A /2. We can reach all r > 0 by taking a smaller c > 0 (independent of λ and μ), thanks to Lemma 2.1. Recalling η • < η + a − , and letting ε ↓ 0, this gives the desired result.
Proof. We use Lemma 2.8. We know that, for each η > 0 small, On the other hand, from the optimal regularity for the thin obstacle problem, we know thatū Solving the ODE between λ and μ, this yields Let us now suppose that there exists some z has quadratic growth around zero (since z • is a singular point of order 2), that is . In particular, whenever μ − λ > Cδ Taking δ and η small enough we get the desired result.
for any x • ∈ K .
Proof. This follows exactly as the proof of [ Singular points (that is, points of order 2m < τ + α) have a non-degeneracy property. Namely, as proved in [23,Lemma 8.1], if x • ∈ λ 2m , then there exists some constant C > 0 (depending on the point x • ) such that In particular, we can further divide the set 2m according to the degree of degeneracy of the singular point. That is, let us define and each 2m, j ⊂ 2m is compact (see [22,Lemma 2.8.2], which only uses the upper semi-continuity of the frequency formula with respect to the point).
In the next proposition we are going to use a Monneau-type monotonicity formula. In particular, we will use that, if we define for m ∈ N, for any 2m-homogeneous, a-harmonic, even polynomial p 2m with p 2m (x , 0) ≥ 0, such that p 2m C for some universal bound C, then for some constant C M independent of λ.
Proof. Suppose it is not true. That is, suppose that there exist sequences x k , z k ∈ 2m, j with k ∈ N, such that |x k − z k | → 0 and for some δ > 0. Suppose also that λ(x k ) λ(z k ). We have that On the other hand, let us study the convergence of the degree τ polynomials P k since Q x k τ and Q z k τ are the Taylor expansions of ϕ of order τ at x k and z k respectively, and |x k − z k | = ρ k . Similarly, for any multi-index β = (β 1 , . . . , β n−1 ) with |β| τ , Thus, the P k τ = o(ρ τ k ) (say, in any norm in B 1 ), and so the same occurs with the a-harmonic extension. Notice, also, that by assumption, 2m τ . In all, we have that On the other hand, we have thanks to Proposition 3.1 with K = 2m, j , and for some modulus of continuity σ K , j depending on j. Similarly, if we denote From the definition of 2m, j we know that In particular, up to subsequences, p x k 2m → p x uniformly for some 2m-homogeneous polynomial p x , a-harmonic, such that p x (x , 0) ≥ 0, and (3.8) Notice that both bounds (3.7) are crucial: the bound from above allows a convergence, and the bound from below avoid getting as a limit the zero polynomial. We similarly have that p z k 2m → p z for some p z 2m-homogeneous polynomial, a-harmonic, with p z (x , 0) ≥ 0 and such that (3.8) holds for p z .
Combining the convergences of p x k 2m and p z k 2m to p x and p z with (3.5)-(3.6) we obtain that On the other hand, from (3.4), we know that p x ≥ p z (· − ξ • ). Thus, p x − p z (·−ξ • ) ≥ 0, and is a-harmonic, therefore by Lioville's theorem is constant. Moreover, both terms are non-negative on the thin space, and both attain the value 0 (since they are homogeneous), therefore, p x = p z (· − ξ • ). Since both p x and p z are homogeneous of the same degree, this implies that p x = p z .
Let us now use the Monneau-type monotonicity formula, (3.1)-(3.2), with polynomials p x and p z : where we are using that On the other hand, proceeding analogously, Thus, since p x = p z , we obtain that which is a contradiction with (3.3).
Proof. The proof is now standard, and it follows applying the Whitney extension theorem, which can be applied thanks to Proposition 3.2. We refer the reader to the proof of [22,Theorem 1.3.8], which we summarise here for completeness. Indeed, if x • ∈ 2m , and β = (β 1 , . . . , β n+1 ) is a multi-index, we denote for x ∈ 2m , fulfils the compatibility conditions to apply Whitney's extension theorem on 2m, j . That is, there exists some for any |β| 2m. Now, for any x • ∈ 2m, j , since p x • 2m = 0, there exists some ν ∈ R n such that In particular, for some multi-index β • with |β • | = 2m − 1, so that, thanks to (3.9), by the implicit function theorem 2m, j is locally contained in a (n − 1)-dimensional C 1 manifold. Thus, 2m is contained in a countable union of (n − 1)-dimensional C 1 manifolds.

Proof of Main Results
Finally, in this section we prove the main results. To do this, the starting point is the following GMT lemma from [19]: 1] with E λ ⊂ R n . and let us denote Suppose that for some β ∈ (0, n] and γ ≥ 1, we have • for any ε > 0, and for any x Then, We will also use the following lemma, analogous to the first part of Lemma 4.1 but dealing with the upper Minkowski dimension instead (which we denote dim M ). We refer to [32,Chapter 5] Then, dim M ({λ : E λ = ∅}) β/γ < 1. (see [32]). Notice that the definition of upper Minkowski dimension does not change if we assume that the balls B r (x i ) from (4.1) are centered at points in A (by taking, for instance, balls with twice the radius).

Proof. Given
Since dim M E β, we have that for any δ > 0, N (E, r ) = o(r β+δ ). Let us consider N (E, r ) balls of radius r centered at E, B r (x i ), with x i ∈ E. Thanks to our second hypothesis we have that where the intervals are balls of radius r γ −ε . In particular, using that N (E, r ) = o(r β+δ ), we deduce that Since this works for any δ, ε > 0, we deduce the desired result.
If 2 + 2s κ n + 2s, then, On the other hand, if κ > n + 2s, then Proof. The proof of this result follows applying Lemmas 4.1 and 4.2 to the right sets. Indeed, we consider the sets Notice that E = ≥κ , and we can take β = n in Lemma 4.1. On the other hand, we know that for any thanks to Proposition 2.4. That is, for any ε > 0 there exists some ρ = ρ(ε, x • , λ • ) > 0 such that In particular, we can also deal with the set of free boundary points of infinite order.
And we get that the free boundary points of order greater or equal than 2 + 2s are at most (n − 2)-dimensional, for almost every λ ∈ [0, 1]. Proof. This is simply Theorem 4.4 with κ = 2 + 2s.
On the other hand, combining the results from Sections 2 and 3 with Lemma 4.1, we get the following regarding the free boundary points of order 2: Notice that E has dimension H(E) = n − 1 by Proposition 3.3, so that we can take β = n − 1 in Lemma 4.1. On the other hand, we know that for any λ • ∈ [0, 1], thanks to Proposition 2.9 (notice that 2 2−s 1+s > 1 for all s ∈ (1/2, 1)). That is, the hypotheses of Lemma 4.1 are fulfilled, with β = n − 1 and γ = 1. The result now follows by Lemma 4.1.
In fact, the previous theorem is a particular case of the more general statement involving singular points given by the following proposition (we give it for completeness, although we do not need it in our analysis): Finally, if m ∈ N is such that 2m τ , Proof. This proof simply follows by analysing the previous results more carefully. The first part follows exactly as Theorem 4.7, using Proposition 2.9 and looking at each case separately.
Finally, regarding general singular points of order 2m, the proof follows exactly as Theorem 4.4 using that 2m has dimension n − 1 instead of n thanks to Proposition 3.3.
Finally, in order to control the size of points of homogeneity in the interval (2, 2 + 2s), we refer to the following result by Focardi-Spadaro, that establishes that points in * are lower dimensional with respect to the free boundary. The result in [21] involves higher order points as well, but we state it in the explicit form in which it will be used below. Proposition 4.9. [21] Let u be a solution to the fractional obstacle problem with obstacle ϕ ∈ C 4,α for some α ∈ (0, 1), Let θ ∈ (0, α) and let us denotẽ * := κ∈(2,2+2s) Then dim H˜ * n − 2.
Combining the previous results, we obtain the following: for almost every λ ∈ [0, 1].
Proof. This follows by combining the previous results. Notice that The result now follows thanks to Proposition 4.9, Corollary 4.6, and Theorem 4.7.
Remark 4.11. Following the proofs carefully, one can see that the previous result holds true for obstacles ϕ ∈ C 3,1 if s 1 2 . The condition ϕ ∈ C 4,α is only used whenever s > 1 2 , since otherwise, in this case the previous methods do not imply the smallness of˜ * .
We can now prove the main results.
Proof of Theorem 1.3. With the same transformation as in the previous proof, the result now follows from Corollary 4.5.
Proof of Theorem 1.6. With the same transformation as in the previous proof, the result follows from Corollary 4.5.

Examples of Degenerate Free Boundary Points
Let us consider the thin obstacle problem in a domain ⊂ R n+1 , with zero obstacle defined on x n+1 = 0; that is, for some continuous boundary values g ∈ C 0 (∂ ) such that g > 0 on ∂ ∩{x n+1 = 0}.
Proof of Proposition 1.7. We will show that there exists some domain and some boundary data g such that the solution to (5.1) has a sequence of regular points (of order 3/2) converging to a non-regular (singular) point (of order 2). Then, the solution from Proposition 1.7 will be the solution here constructed restricted to any ball inside containing such singular point, with its own boundary data (and appropriately rescaled, if necessary). In order to build such a solution we will use [5,Lemma 3.2], which says that solutions to with x ϕ −c 0 < 0 and convex and even in x n+1 have a free boundary containing only regular points (frequency 3/2) and singular points of frequency 2. In particular, they establish a non-degeneracy result stating that, for any for some r 1 , c 1 that do not depend on the point x • . More precisely, they show it around points x ∈ {u > ϕ} and then take the limit x → x • ∈ (u).
On the other hand, from their proof one can also show that in fact, the convexity on can be weakened to convexity in in the e n+1 direction.
Let us fix n = 2. Up to subtracting the right obstacle, we consider the problem for some analytic obstacle ϕ t , and some domain smooth, convex and even in x 3 , to be chosen.
Notice that, in the thin space, x ϕ t = −12x 2 1 − 4 −4, so that, by the result in [5], under the appropriate domain , the points on the free boundary (u t ) are either regular (with frequency 3/2) or singular (with frequency 2), and we have non-degeneracy (5.3). Let 2}, and take any bounded, convex in x 3 , and even in x 3 extension of , . Then, if t = 2 and ⊂ {|x 3 | 1}, the solution u 2 to (5.4) is exactly equal to the solution to so that, in particular, the contact set is full. 1 Notice that, when t < 0, the contact set is empty, (u t ) = ∅, and when t = 0 the contact set is two points, p ± = (±1, 0, 0) (which, in particular, are singular points). Notice, also, that the contact set is always closed and is monotone in t, in the sense that Let us say that a set is p ± -connected if the points p + and p − belong to the same connected component. Then, there exists some t * ∈ (0, 2] such that (u t ) is not p ± -connected for t < t * , and is p ± -connected for t > t * . Notice, also, that since (u t ) ⊂ {x : ϕ t ≥ 0} then t * > 1.
We claim that (u t * ) is p ± -connected and has a set of regular points converging to a singular point.
Let us first show that (u t * ) is p ± -connected. Suppose it is not. That is, (u t * ) is a closed set with p ± on different connected components. On the other hand, 1 To see this, we compare u 2 with the harmonic extension of ϕ 2 ,φ 2 (x 1 , (u t ) is compact and p ± -connected for t > t * , and nested ( (u t ) ⊂ (u t ) for t < t ). Take˜ then˜ t * is p ± -connected (being the intersection of compact p ± -connected nested sets), and (u t * ) ˜ t * , since (u t * ) is not p ± -connected. In particular, there exists some x • ∈ (u t ) for all t > t * such that x • ∈ (u t * ). But, by continuity, this is not possible: Take p (u t * ) to be the connected component containing both p + and p − . Then, ∂ p (u t * ) must contain at least one singular point. Indeed, suppose it is not true. In this case, all points in ∂ p (u t * ) are regular, and in particular, p (u t * ) is a compact connected set with smooth boundary, with all points of the boundary having positive density (in {x 3 = 0}), and therefore ( p (u t * )) • is also connected. Let us denote p ± (u t ) the corresponding connected components of (u t ) containing p ± for t < t * (notice that, by definition of t * , given that the left-hand side is not connected, and the right-hand side is. Take t<t * , so that around y • the non-degeneracy (5.3) holds for any t < t * . Then, there exists some r • > 0, which is a contradiction. That is, not all points on ∂ p (u t * ) are regular. By [5], then there exist some degenerate (singular) point of frequency 2, x D ∈ ∂ p (u t * ). Now consider D , the connected component in ∂ p (u t * ) containing x D . Since the density of the contact set around singular points is zero, if D consist exclusively of singular points, then D itself is the whole connected component p (u t ), and p ± ∈ D are singular points. Nonetheless, for small t > 0, (u t ) contains a neighbourhood of p ± , which contradicts the singularity of p ± . Therefore, D is not formed exclusively of singular points, and then there exists a sequence of regular points converging to a singular point. Now, before proving Proposition 1.8, let us show the following lemma: Proof. We consider the extension problem from R n to R n+1 . Namely, let us denote u 1 the extension of (x 1 ) 2m+1+s where a = 1 − 2s. Then, we know that for x ∈ R n . On the other hand, let u 2 be the unique a-harmonic extension of (x 1 ) 2m+1+s + from R n to R n+1 . That is, u 2 is homogeneous (of degree 2m + 1 + s), and fulfils The fact that such solution exists, and that lim y↓0 y a ∂ x n+1 u 2 (x , y) = 0 if x 1 > 0, follows, for example, from [20,Proposition A.1]. On the other hand, notice that, since u 2 is (2m + 1 + s)-homogeneous, we have that, lim y↓0 y a ∂ x n+1 u 2 (x , y) = C m,s |x 1 | 2m+1−s for x 1 < 0, so that, in all, Again, by [20, Proposition A.1] u 2 is a solution to the thin obstacle problem with operator L a , so C m,s > 0 (otherwise, it would not be a supersolution for L a ).
We are now in disposition to give the proof of Proposition 1.8.
Proof of Proposition 1.8. We divide the proof into two steps. In the first step, we show the results holds up to an intermediate claim, that will be proved in the second step.
Step 1. Thanks to [24,Theorem 4] or [1, Section 2], we have that (− ) s (d s η) ∈ C ∞ ( c ) for any η ∈ C ∞ with sufficient decay at infinity. Here, d denotes any C ∞ function (with at most polynomial growth at infinity) such that in a neighbourhood of coincides with the distance to , and d| ≡ 0.
In particular, once d is fixed, we know that for any k ∈ N, and, if we make sure that d > 0 in c , with exponential decay at infinity, we get Define, for some g with the previous decay, |g(x)| C(1 + |x| n+2s ) −1 , ϕ g such that that is, one can take ϕ g (x) = I 2s g(x) := c R n g(y) |x − y| n−2s dy. In all, also using that ϕ(x) is bounded around the origin, we obtain that

Notice that
wheref is some appropriate C ∞ extension of f inside . Then, if we define and notice that, since v > 0 in c and v = 0 in , by definition, we have that the contact set is exactly equal to . Moreover, by the growth of v at the boundary, the free boundary points are of frequency k + s. Also, by the decay at infinity of v and ϕ −f , u → 0 at infinity.
Step 2. We still have to show that, for an appropriate choice off , (5.5) holds for k = 2m + 1. Notice that, in fact, in c we know that f is C ∞ . Moreover, we only have to show the claim for a neighbourhood of ∂ inside , given that exactly at the boundary we expect a unique extension of f (that is, all derivatives are prescribed at the boundary).
That is, if we let δ := {x ∈ : dist(x, ∂ ) < δ}, we have to show that there exists some δ > 0 small enough such that (− ) s v ≥f in δ , where we recall that f is a C ∞ extension of f ∈ C ∞ ( c ) inside .
Let z • ∈ ∂ . After a translation and a rotation, we assume that z • = 0 and ν(0, ∂ ) = e 1 , where ν(0, ∂ ) denotes the outward normal to ∂ at 0. After rescaling if necessary, let us assume that we are working in B 1 , that each point in B 1 has a unique projection onto ∂ , and that d| B 1 ∩ c = dist(·, ). Moreover, again after a rescaling if necessary (since is a C ∞ domain), let us assume that so that, in particular, {−t e 1 : t ∈ (0, 1)} ⊂ .
Let us start by noticing that, from (5.7), together with the fact that (− ) s v is smooth in c , we already know that w 1 ∈ C ∞ ([0, 1/2)), so that we only care about the case z 1 < 0.
To finish, we study the points of order infinity. To do this, we start with the following proposition: Proof. Take any obstacle ψ ∈ C ∞ (R n ) such that supp ψ ⊂⊂ B 1 (2e 1 ), with ψ > 0 somewhere, and take the non-trivial solution to Notice that u > ψ in B 1 (in particular, u ∈ C ∞ (B 1 )). Let f C be any C ∞ function such that 0 f C 1 and C = { f C = 0}. Now let η ∈ C ∞ c (B 3/2 ) such that η ≥ 0 and η ≡ 1 in B 1 . Consider, as new obstacle, ϕ = ψ + η(u − ψ)(1 − f C ) ∈ C ∞ (B 1 ). Notice that u − ϕ ≥ 0. Notice, also, that for x ∈ B 1 , (u − ϕ)(x) = 0 if and only if x ∈ C. Thus, u with obstacle ϕ gives the desired result.
And now we can provide the proof of Proposition 1.9: Proof of Proposition 1.9. The proof is now immediate thanks to Proposition 5.2, since we can choose as contact set any closed set with boundary of dimension greater or equal than n − ε for any ε > 0, and points of finite order are at most (n − 1)-dimensional.
In order to study (6.1), one also needs to add some boundary condition on (∂ B 1 × (−1, 0]) ∩ {x n+1 > 0}. Instead of doing that, we will assume the additional hypothesis u t > 0 on (∂ B 1 × (−1, 0]) ∩ {x n+1 > 0}. That is, there is actually some time evolution, and it makes the solution grow. Recall that such hypothesis is (somewhat) necessary, and natural in some applications (see Section 1.4).
Notice, also, that if u t > 0 on the spatial boundary, by strong maximum principle applied to the caloric function u t in Q 1 ∩ {x n+1 > 1 2 }, we know that u t > c > 0 for x n+1 > 1 2 . Thus, after dividing u by a constant, we may assume c = 1, and thus, our problem reads as In order to deal with the order of free boundary points, one requires the introduction of heavy notation, analogous to what has been presented in the elliptic case, but for the parabolic version. We will avoid this boundary by focusing on the main property we require about the order of the extended free boundary points. π x 2 is (n − 1)-dimensional.
Step 2. Thanks to Step 1, and by Proposition 6.2 with κ = 2, proceeding analogously to Theorem 4.4 by means of Lemma 4.1, we reach the desired result. We can now give the proof of the main result regarding the parabolic Signorini problem.
Proof of Theorem 1.4. Is a direct consequence of (6.4), Proposition 6.3, and Proposition 6.4 with a = α • depending only on n, given by [38,Proposition 4.5]. The regularity of the free boundary follows from [12,Theorem 11.6].
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