On the fine structure of the free boundary for the classical obstacle problem

Abstract

In the classical obstacle problem, the free boundary can be decomposed into “regular” and “singular” points. As shown by Caffarelli in his seminal papers (Caffarelli in Acta Math 139:155–184, 1977; J Fourier Anal Appl 4:383–402, 1998), regular points consist of smooth hypersurfaces, while singular points are contained in a stratified union of $C^1$ manifolds of varying dimension. In two dimensions, this $C^1$ result has been improved to $C^{1,\alpha }$ by Weiss (Invent Math 138:23–50, 1999). In this paper we prove that, for $n=2$ singular points are locally contained in a $C^2$ curve. In higher dimension $n\ge 3$, we show that the same result holds with $C^{1,1}$ manifolds (or with countably many $C^2$ manifolds), up to the presence of some “anomalous” points of higher codimension. In addition, we prove that the higher dimensional stratum is always contained in a $C^{1,\alpha }$ manifold, thus extending to every dimension the result in Weiss (1999). We note that, in terms of density decay estimates for the contact set, our result is optimal. In addition, for $n\ge 3$ we construct examples of very symmetric solutions exhibiting linear spaces of anomalous points, proving that our bound on their Hausdorff dimension is sharp.

Appendix A: Examples of \((m-1)\)-dimensional anomalous sets \(\Sigma ^a_m\)

In this “Appendix”, for any $1\le m \le n-2$ we construct an example of solution to the obstacle problem in $\mathbb {R}^n$ for which the anomalous set $\Sigma _m^a$ is $(m-1)$-dimensional. The existence of such examples shows that the assertion $\mathrm{dim}_\mathcal {H}(\Sigma ^a_m) \le m-1$ in Theorem 1.1(b) is optimal.

These solutions are constructed as follows: given $1\le m \le n-2$ we consider functions

$$\begin{aligned} u(x_1,x_2, \dots , x_n) = u^\star (z,r), \qquad z:= x_{m}, \quad r:= \sqrt{x_{m+1}^2 +x_{m+2}^2+\cdots + x_{n}^2},\nonumber \\ \end{aligned}$$

(A.1)

which are independent of the first $(m-1)$ variables and are axially symmetric with respect to the last $m-n$ variables. Then our goal is to find solutions to the obstacle problem which are of the form (A.1) and for which all points in the $(m-1)$ dimensional affine space

$$\begin{aligned} {\mathcal {Z}} := \{x_m= x_{m+1} =\cdots =x_n=0 \} \end{aligned}$$

are anomalous points in $\Sigma _m^a$. To build these examples, we rely on some ideas introduced in [23].

Fix $\phi :[-1,1] \rightarrow \mathbb {R}$ a nonnegative $C^2$ function satisfying

$$\begin{aligned} \phi (0)>0,\quad \phi (1)=0, \quad \phi '(z)\le 0 \quad \forall \,z\in (0,1),\quad \text {and} \quad \phi (-z)=\phi (z). \end{aligned}$$

Then, for any real number $k>0$ we consider the solution $u^k\ge 0$ to the obstacle problem in $(-1,1)\times (0,1)\subset \mathbb {R}^2$

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathrm{div}( r^{n-m-1}\nabla u^k) = k\,r^{n-m-1}\,\chi _{\{u^k>0\}} \quad &{}\text{ in } (-1,1)\times (0,1),\\ u^k(\pm 1,r) =0&{} r\in (0,1),\\ u^k(z,1) = \phi (z) &{} z\in (-1,1). \end{array}\right. } \end{aligned}$$

In other words, $\frac{1}{k} u^k(x_{m}, \sqrt{x_{m+1}^2+\cdots + x_{n}^2})$ is a solution of the classical obstacle problem in the cylinder $\mathbb {R}^{m-1} \times (-1,1)\times B_1^{(n-m)}$ satisfying the symmetries in (A.1).

Note that, when we think of this obstacle problem as a two dimensional problem (in the variables z, r), we do not prescribe “boundary” values at $r=0$ since this line has zero capacity for the operator $\mathrm{div}( r^{n-m-1}\nabla \,\cdot \,)$ when $n-m-1\ge 1$ (this is equivalent to saying that the set $\{x_{m+1} =\cdots =x_n=0 \}$ has zero harmonic capacity in $\mathbb {R}^n$).

We now claim that:

• (i) $u^k$ is even z, namely $u^k(z,r)=u^k(-z,r)$;

• (ii) $\partial _z u^k \le 0$ in $(0,1)\times (0,1)$;

• (iii) the set $\{u^k>0\}$ is convex in the direction z, and is symmetric with respect to $z=0$.

Indeed, (i) follows from the symmetry of the boundary data (and the uniqueness of solution to the obstacle problem).

To show (ii) we consider the open set $U := (0,1)\times (0,1) {\setminus } \{u^k=0\}$ and observe that $\partial _z u^k \le 0$ on $\partial U$. Indeed:

• $\partial _z u^k =0$ on $\{z=0\}$, by symmetry;

• $\partial _z u^k =0$ on $\partial \{u^k=0\}$, since $u^k$ is nonnegative and of class $C^{1,1}$;

• $\partial _z u^k \le 0$ on $\{z=1\}$, since $u^k(r,1) =0$ while $u^k(r,z) \ge 0$ for $z<1$;

• $\partial _z u^k =\phi ' \le 0$ on $\{r=1\}$.

As a consequence, since $\mathrm{div}( r^{n-m-1}\nabla (\partial _z u^k))=0$ inside U (this follows by differentiating the equation for u with respect to z), we deduce by the maximum principle that $\partial _z u^k \le 0$ in U. Also, we note that $\partial _z u^k = 0$ in $(0,1)\times (0,1) {\setminus } U$ since $u^k\equiv 0$ there (recall that $u^k$ is nonnegative and of class $C^{1,1}$), proving (ii).

Finally, (iii) is an immediate consequence of (i) and (ii).

We now observe that, for k sufficiently large, the contact set contains a neighborhood of the origin, and hence it must contain a cylindrical neighborhood of $\{r=0\}$ [thanks to (iii)]. On the other hand, for $k\ll 1$ we have $u^k(0)>0$. Therefore, by continuity, there exists (a unique) $k_{\star }>0$ such that $\partial \{u^{k_\star }>0\}$ touches tangentially the line $\{r=0\}$.

Set $u^\star := u^{k_\star }$. Observe that that, with this definition, the function

$$\begin{aligned} u(x):=\frac{1}{k_\star }u^{\star }(x_{m}, \sqrt{x_{m+1}^2 +\cdots + x_{n}^2}) \end{aligned}$$

(A.2)

has a full $(m-1)$-dimenional space of singular points on ${\mathcal {Z}}=\{x_m=x_{m+1} = \cdots =x_n=0\}$. Also, by the given symmetry, these singular points belong to the stratum $\Sigma _{m}$.

We now prove the following:

Proposition A.1

Let u be the symmetric solution defined in (A.2). Then the set ${\mathcal {Z}}\subset \Sigma _m$ consists of anomalous points, that is ${\mathcal {Z}}\subset \Sigma _m^a$.

Proof

The proof consists of three steps.

- Step 1. We show that u has no other singular points in a neighborhood of ${\mathcal {Z}}$ (except of course the points in ${\mathcal {Z}}$).

To prove this, assume by contradiction that there exists a sequence of singular points $x_\ell \rightarrow 0$. Note that, since u is invariant in the first $m-1$ variable, we can assume that $x_\ell \in \{x_1=\cdots =x_{m-1}=0\}$. Then, by the symmetries of u and Lemma 3.2, a blow-up q of $u-p_*$ at 0 is a homogeneous harmonic polynomial that vanishes on the $x_m$-axis and that enjoys the same symmetries as u (note that $p_*$ has the same symmetries as u and vanishes on $r=0$). Thus, $q=q^\star (z,r)$, where $q^\star$ solves

$$\begin{aligned} \mathrm{div}( r^{n-m-1} \nabla q^\star )=0, \qquad q^\star (z,0) = 0 \quad \forall \, z, \qquad \text{ and }\quad q^\star \not \equiv 0. \end{aligned}$$

Let $\ell$ be the degree of q. Since q is smooth and $q^\star (z,0) = 0$, $q^\star$ must be a polynomial of the form $\sum _{1\le k\le \ell /2}a_kz^{\ell -2k}r^{2k}$. Let $k_0\ge 1$ be the first index such that $a_{k_0}\ne 0$. Then

$$\begin{aligned} 0=\mathrm{div}( r^{n-m-1} \nabla q^\star ) =\Bigl (2k_0[n-m+2(k_0-1)]a_{k_0}z^{\ell -2k_0}+r^{2}Q(z,r)\Bigr )r^{n-m-1+2(k_0-1)}, \end{aligned}$$

where Q is a polynomial of degree $\ell -2k_0-2$. In particular, the terms inside the parenthesis cannot be identically zero, giving the desired contradiction.

As a consequence, there is a neighborhood of ${\mathcal {Z}}$ which is free of singular points. (except the points in ${\mathcal {Z}}$). In particular, as a consequence of [4, 5], there exists $\varepsilon >0$ such that $(\partial \{u^{\star }>0\} {\setminus } \{0\})\cap B_\varepsilon$ is a smooth curve contained inside $(-\varepsilon ,\varepsilon )\times (0,\varepsilon ){\setminus }\{0\}$.

- Step 2. We show that, for any $\alpha >0$ small, there exists $R =R(\alpha )>0$ small such that the following holds:

$$\begin{aligned} \forall \,\varrho _\circ \in (0,R), \ \exists \, Y_{\varrho _\circ } \in \partial \{u^\star >0\}\cap B_{2\varrho _\circ } \quad \text{ s.t. } \quad Y_{\varrho _\circ }\cdot {\varvec{e}}_r \ge \varrho _\circ ^{1+\alpha }, \end{aligned}$$

(A.3)

where $Y_{\varrho _\circ }\cdot {\varvec{e}}_r$ denotes the r-component of the point $Y_{\varrho _\circ }$.

Let $\rho := \sqrt{r^2+z^2}$ and $\theta := \arctan (z/r) \in (-\frac{\pi }{2},\frac{\pi }{2})$ be polar coordinates in ${(0,1)\times (-1,1)}$. We now state the following fact, whose proof is postponed to the end of the Step 2.

Claim For any $\alpha >0$ small, there exists $\delta = \delta (\alpha )>0$, and a function $\Theta _\alpha : [\delta ,\frac{\pi }{2} -\delta ] \rightarrow \mathbb {R}$ smooth, such that $S_\alpha (r,z) := \rho ^{1+\alpha /4} \Theta _\alpha (\theta )$ satisfies

$$\begin{aligned} \mathrm{div}( r^{n-m-1} \nabla S_\alpha )=0 \quad \text{ in } \text{ the } \text{ cone } {\mathcal {C}}_\delta := \left\{ (\rho ,\theta )\in (0,1)\times (\delta ,\frac{\pi }{2} -\delta )\right\} , \end{aligned}$$

and

$$\begin{aligned} \Theta _\alpha (\delta )= \Theta _\alpha \Big (\frac{\pi }{2}-\delta \Big ) =0,\qquad \Theta _\alpha >0\quad \text {for }\theta \in \Big (\delta ,\frac{\pi }{2}-\delta \Big ). \end{aligned}$$

(A.4)

Note that, since 0 is a singular point and recalling the symmetries of u, $u(x)=p_*(x)+o(|x|^2)$ where $p_*(x)=\frac{1}{2(n-m)}\sum _{i=m+1}^nx_i^2$. This implies that $u^\star (z,r) \ge \frac{k_\star }{4(n-m)}r^2+o(z^2)$ near the origin. Thus, given $\alpha >0$, there exists $\eta =\eta (\alpha )>0$ small enough such that the inclusion

$$\begin{aligned} {\mathcal {C}}_\delta \cap B_\eta \subset {\mathcal {C}}_{\delta /2} \cap B_\eta \subset \{u^\star >0\} \end{aligned}$$

(A.5)

holds. Note that, up to reducing $\eta$, we can assume that $\eta <\varepsilon ,$ so that (by Step 1) $\partial \{u^\star >0\}{\setminus } \{0\}$ is a smooth curve inside $B_\eta$.

Let us consider the function $h :=-\partial _z u^{\star }$, and recall that $h\ge 0$ (by property (iii) in the construction of $u^\star$). Also, differentiating the equation

$$\begin{aligned} \mathrm{div}( r^{n-m-1}\nabla u^\star ) = k_\star \,r^{n-m-1}\qquad \text {in }\{u^\star >0\} \end{aligned}$$

with respect to z, we obtain

$$\begin{aligned} \mathrm{div}( r^{n-m-1}\nabla h) = 0\qquad \text {in } \{u^\star >0\}. \end{aligned}$$

(A.6)

Since $h>0$ inside $\{u^\star >0\}$ (by the strong maximum principle), it follows that

$$\begin{aligned} \partial \{h>0\}\cap B_\eta \cap \{z>0\}=\partial \{u^\star>0\}\cap B_\eta \cap \{z>0\} \end{aligned}$$

(A.7)

and ${\mathcal {C}}_\delta \cap \partial B_\eta \subset \subset \{u^\star >0\}$. Thus there exists a constant $c_0=c_0(\alpha )>0$ such that $h\ge c_1S_\alpha$ on ${\mathcal {C}}_\delta \cap \partial B_\eta$. In addition $h\ge 0=S_\alpha$ on $\partial {\mathcal {C}}_\delta \cap B_\eta$. Hence, since h and $S_\alpha$ solve the same equation, it follows by the maximum principle that $h\ge S_\alpha$ inside ${\mathcal {C}}_\delta \cap B_\eta$. Recalling (A.4), this implies that

$$\begin{aligned} h \ge c_1\rho ^{1+\alpha /4} \qquad \text{ in } \left\{ (\rho ,\theta )\in (0,\eta )\times [2\delta ,\textstyle \frac{\pi }{2} -2\delta ]\right\} \end{aligned}$$

(A.8)

for some $c_1= c_1(\alpha )>0$.

We now want to find a lower bound on the normal derivative of h at points on $\partial \{u^\star >0\}$. For this we use a Hopf-type argument, constructing suitable barriers for our operator $\mathrm{div}(r^{n-m-1}\nabla \,\cdot \,)$. These are given by the family of functions

$$\begin{aligned} S^{H}_{b}(r,z) := 2(r-b) - (n-m-1)(z-1)^2,\quad \text{ where } b\in \mathbb {R}. \end{aligned}$$

Note that

$$\begin{aligned} \mathrm{div}( r^{n-m-1} \nabla S^{H}_b)=0\quad \text{ in } \{r>0\}, \end{aligned}$$

and $S^{H}_b<0$ outside of the parabolic region ${\mathcal {P}}_b:=\left\{ b +\frac{n-m-1}{2} (z-1)^2 \le r\right\}$.

Now, given $0<\varrho _\circ \ll \eta$ we consider the rescaled function

$$\begin{aligned} h_{\varrho _\circ } := \varrho _\circ ^{-1-\alpha /2} h(\varrho _\circ \,\cdot \,) \end{aligned}$$

and we note that [thanks to (A.8)]

$$\begin{aligned} h_{\varrho _\circ }\bigr (\rho ,\textstyle \frac{\pi }{2} -2\delta \bigl )\ge c_1\varrho _\circ ^{-\alpha /4}\rho ^{1+\alpha /4}\qquad \forall \,\rho \in (0,\textstyle \frac{\eta }{\varrho _\circ }). \end{aligned}$$

(A.9)

Consider now our barriers $S_b^H$. For $b>1/2$ the parabolic region ${\mathcal {P}}_b$ is contained inside the set $\left\{ \theta <\textstyle \frac{\pi }{2} -2\delta \right\}$. In particular $S_b^H<0$ inside the cone $\hat{{\mathcal {C}}}_{2\delta } := \left\{ \theta \in (\frac{\pi }{2}-2\delta ,\frac{\pi }{2}) \right\}$. Then, we start decreasing b until the first value $b_\circ$ such that $\partial {\mathcal {P}}_{b_\circ }$ touches $\partial \{h_{\varrho _\circ } >0\}$. Note that, thanks to (A.7) and Step 1, $b_\circ >0$ and the contact point will happen for some ${\widetilde{Y}}_{\varrho _\circ }\in \{r>0\}$.

Since $h_{\varrho _\circ }\ge 0=S_{b_\circ }^H$ on $\partial {\mathcal {P}}_{b_\circ }\cap \hat{{\mathcal {C}}}_{2\delta }$ and $h_{\varrho _\circ }\ge c_2(\alpha )\varrho _\circ ^{-\alpha /4}\ge S_{b_\circ }^H$ on ${\mathcal {P}}_{b_\circ }\cap \partial \hat{{\mathcal {C}}}_{2\delta }$ for $\varrho _\circ$ sufficiently small [see (A.9)], it follows by the maximum principle that $h_{\varrho _\circ }\ge S_{b_\circ }^H$ inside ${\mathcal {P}}_{b_\circ }\cap \hat{{\mathcal {C}}}_{2\delta }$. Hence, since both $h_{\varrho _\circ }$ and $S_{b_\circ }^H$ vanish at ${\widetilde{Y}}_{\varrho _\circ }$, we deduce that

$$\begin{aligned} \partial _\nu h_{\varrho _\circ } ({\widetilde{Y}}_{\varrho _\circ }) \ge \partial _\nu S_{b_\circ }^H({\widetilde{Y}}_{\varrho _\circ }) \ge c_2>0, \end{aligned}$$

(A.10)

where $\nu$ is the unit inwards normal to $\partial \{h_{\varrho _\circ } >0\}$, and $c_2=c_2(\alpha )$ is independent of $\varrho _\circ$ (here we use $b_\circ \le 1/2$). Observe also that, provided $\delta$ is small enough, ${\mathcal {P}}_{b_\circ }\cap \hat{{\mathcal {C}}}_{2\delta }\subset B_2$.

Rescaling (A.10), we obtain that for all $\varrho _\circ \in (0,\eta )$ small enough there is a point $Y_{\varrho _\circ } \in \partial \{u^\star >0\} \cap (0,1)\times (0,1)$ such that,

$$\begin{aligned} |Y_{\varrho _\circ }|\le 2\varrho _\circ \qquad \text{ and } \qquad \partial _\nu h(Y_{\varrho _\circ }) \ge c_2 \varrho _\circ ^{\alpha /2}, \end{aligned}$$

(A.11)

where $\nu$ denotes the unit inwards normal to $\{u^\star >0\}$ [see (A.7)].

To conclude the proof we observe that differentiating the equation

$$\begin{aligned} \mathrm{div}(r^{n-m-1} \nabla u^\star ) = k^\star \, r\, \chi _{\{u^\star >0\}} \end{aligned}$$

with respect to z we obtain $-\mathrm{div}(r^{n-m-1}\nabla h) = k^\star \, r^{n-m-1} \nu _z\, {\mathcal {H}}^1|_{\partial \{u^\star >0\}}$, thus

$$\begin{aligned} -r^{n-m-1} \partial _\nu h = k^\star \, r^{n-m-1} \nu _z \quad \text{ on } {\partial \{u^\star >0\}} \end{aligned}$$

where $\nu _z=\nu \cdot {\varvec{e}}_z$ denotes the z-component of $\nu$. Recalling (A.11), this proves that

$$\begin{aligned} \nu _z( Y_{\varrho _\circ }) \le -\frac{c_2}{k^\star } \varrho _\circ ^{\alpha /4}\qquad \text{ and } \qquad |Y_{\varrho _\circ }| \le 2| \varrho _\circ |. \end{aligned}$$

(A.12)

Note that (A.12) already implies that $\partial \{u^\star >0\}$ cannot be a $C^{1,\alpha }$ curve at 0. We now prove the more precise estimate (A.3).

Since $b_\circ >0$ and the rescaled parabolic cap

$$\begin{aligned} \left\{ (\varrho _\circ z,\varrho _\circ r)\,:\,{b_\circ } +\frac{n-m-1}{2} (z-1)^2 \le r \le 1 \right\} \end{aligned}$$

touches $\partial \{u^\star =0\}$ at $Y_{\varrho _\circ }=(r_{\varrho _\circ },z_{\varrho _\circ })$, it follows that

$$\begin{aligned} \frac{n-m-1}{2\varrho _\circ } (z_{\varrho _\circ }-\varrho _\circ )^2<\varrho _\circ {b_\circ } +\frac{n-m-1}{2\varrho _\circ } (z_{\varrho _\circ }-\varrho _\circ )^2= r_\circ \end{aligned}$$

(A.13)

and that

$$\begin{aligned} \nu ( Y_{\varrho _\circ })=\frac{\left( 1,\frac{n-m-1}{\varrho _\circ }(z_{\varrho _\circ }-\varrho _\circ )\right) }{\sqrt{1+\left( \frac{n-m-1}{\varrho _\circ }(z_{\varrho _\circ }-\varrho _\circ )\right) ^2}}. \end{aligned}$$

(A.14)

Combining (A.13) with (A.12), we get

$$\begin{aligned} \frac{\frac{n-m-1}{\varrho _\circ }(z_{\varrho _\circ }-\varrho _\circ )}{\sqrt{1+\left( \frac{n-m-1}{\varrho _\circ }(z_{\varrho _\circ }-\varrho _\circ )\right) ^2}}\le & {} -\frac{c_2}{k^\star } \varrho _\circ ^{\alpha /4} \\\Rightarrow & {} \qquad \frac{n-m-1}{\varrho _\circ }(z_{\varrho _\circ }-\varrho _\circ )\le -\frac{c_2}{k^\star }\varrho _\circ ^{\alpha /4}. \end{aligned}$$

Recalling (A.13), this yields

$$\begin{aligned} \frac{1}{2(n-m-1)}\Bigl (\frac{c_2}{k^\star }\Bigr )^2 \varrho _\circ ^{1+\alpha /2}\le r_\circ =Y_{\varrho _\circ }\cdot {\varvec{e}}_r. \end{aligned}$$

Since $\varrho _\circ ^{1+\alpha }\ll \varrho _\circ ^{1+\alpha /2}$, this proves (A.3).

We conclude Step 2 proving the claim.

Proof of Claim. The linear function $z = \rho \sin \theta$ satisfies $\mathrm{div}( r^{n-m-1} \nabla z)=0$. Also $\Theta _0(\theta ) = \sin \theta$ is positive on $(0,\pi /2)$ and satisfies $\Theta _0(0)=0$, hence $\Theta _0$ must be the minimizer of the Rayleigh quotient

$$\begin{aligned} \min \bigg \{ \frac{ \int _0^{\pi /2} (\cos \theta )^{n-m-1} (\Theta ')^2}{\int _0^{\pi /2} (\cos \theta )^{n-m-1} (\Theta )^2} \ :\ \Theta (0)=0 \bigg \} ={n-m} \end{aligned}$$

[equivalently, recalling (A.1), the restriction to the harmonic function $x_m$ to $\mathbb {R}^{m-1}\times {\mathbb {S}}^{n-m}\subset \mathbb {R}^{m-1}\times \mathbb {R}^{n-m+1}$ is a minimizer of the Rayleigh quotient for the classical Dirichlet integral on ${\mathbb {S}}^{n-m}\cap \{x_m\ge 0\}$].

Since $n-m-1\ge 1$ (this is where we crucially use this assumption), we have

$$\begin{aligned} \lim _{\delta \downarrow 0} \min \bigg \{ \frac{ \int _0^{\pi /2-\delta } (\cos \theta )^{n-m-1} (\Theta ')^2}{\int _0^{\pi /2-\delta } (\cos \theta )^{n-m-1} (\Theta )^2} \ :\ \Theta (\delta )=\Theta (\pi /2-\delta )=0 \bigg \}\downarrow n-m \end{aligned}$$

(this is equivalent to saying that the north pole in ${\mathbb {S}}^{n-m}\cap \{x_m\ge 0\}$ has harmonic capacity 0, so the boundary condition $\Theta (\pi /2-\delta )=0$ disappears in the limit $\delta \rightarrow 0$). Thus, by continuity, for any $\alpha >0$ small there exists $\delta =\delta (\alpha )>0$ small enough such that the previous Rayleigh quotient in $(\delta ,\pi /2-\delta )$ will give the value

$$\begin{aligned} \mu _\alpha :=(n-m+\alpha /4)(1+\alpha /4). \end{aligned}$$

This implies that if we denote by $\Theta _\alpha$ the first eigenfunction (i.e., the function attaining the minimal quotient value $\mu _\alpha$), then $-( (\cos \theta )^{n-m-1}\Theta _\alpha ')'=\mu _\alpha (\cos \theta )^{n-m-1}\Theta _\alpha$, and it follows by a direct computation (or by classical spectral theory) that $S_\alpha (r,z) := \rho ^{1+\alpha /4} \Theta _\alpha (\theta )$ satisfies $\mathrm{div}( r^{n-m-1} \nabla S_\alpha )=0$, as desired. Finally, the strict positivity of $\Theta _\alpha$ inside $\left( \delta ,\frac{\pi }{2}-\delta \right)$ is a classical property of the first eigenfunction.

- Step 3. We conclude the proof of the proposition by showing that (A.3) is incompatible with $0\in \Sigma ^g_m$. Recall that, by definition, 0 belongs to $\Sigma ^g_m$ (resp. $\Sigma ^a_m$) if $\phi (0^+,u-p_*) \ge 3$ (resp. $\phi (0^+,u-p_*)<3$), see (3.13).

Assume by contradiction that $\phi (0^+,u-p_*) \ge 3$. Then by Lemma 2.6 we have

We now note that

$$\begin{aligned} \Delta (u-p_*) = -\chi _{\{u=0\}} \le 0, \end{aligned}$$

so it follows by the mean value formula for superharmonic functions that

$$\begin{aligned} u-p_* \ge -CR^3 \quad \text{ in } B_R. \end{aligned}$$

(A.15)

Recalling that $p_* = p_*(z,r) = \frac{1}{2(n-m)} r^2$ in the variables (z, r), it follows by (A.15) that

$$\begin{aligned} -\frac{1}{n-m} r^2 \ge -C(r^2+ z^2)^{3/2}\quad \text{ on } \partial \{u=0\}, \end{aligned}$$

therefore

$$\begin{aligned} r \le C |(z,r)|^{3/2}\quad \text{ on } \partial \{u =0\}, \end{aligned}$$

which clearly contradicts (A.3) if we choose $\alpha <1/2$. As a consequence 0 (and by symmetry all points on ${\mathcal {Z}}$) must belong to $\Sigma ^a_m$. $\square$