An Epiperimetric Inequality Approach to the Regularity of the One-Phase Free Boundaries

Abstract

Throughout this section, we will use the notation

$$\displaystyle W_0(u)=\int _{B_1}|\nabla u|{ }^2\,dx-\int _{\partial B_1}u^2\,d\mathcal {H}^{d-1}\qquad \text{and}\qquad W(u)=W_0(u)+|\{u>0\}\cap B_1|, $$

where B ₁ is the unit ball in $\mathbb {R}^d$, d ≥ 2 and u ∈ H ¹(B ₁).

Throughout this chapter, we will use the notation

$$\displaystyle \begin{aligned} W_0(u)=\int_{B_1}|\nabla u|{}^2\,dx-\int_{\partial B_1}u^2\,d\mathcal{H}^{d-1}\qquad \text{and}\qquad W(u)=W_0(u)+|\{u>0\}\cap B_1|, \end{aligned}$$

where B ₁ is the unit ball in $\mathbb {R}^d$, d ≥ 2 and u ∈ H ¹(B ₁).

The aim of this chapter is to prove an epiperimetric inequality for the energy W in dimension two. As a consequence, we will obtain the C ^{1, α} regularity of the one-phase free boundaries in dimension two (see Proposition 12.13). Our main result is the following.

FormalPara Theorem 12.1 (Epiperimetric Inequality for the Flat Free Boundaries)

There are constants δ ₀ > 0 and ε > 0 such that: if c ∈ H ¹(∂B ₁) is a non-negative function on the boundary of the disk $B_1\subset \mathbb {R}^2$ and

$$\displaystyle \begin{aligned} \pi-\delta_0\le\mathcal{H}^1\big(\{c>0\}\cap \partial B_1\big)\le \pi+\delta_0, \end{aligned}$$

then, there exists a (non-negative) function h ∈ H ¹(B ₁) such that h = c on ∂B ₁ and

$$\displaystyle \begin{aligned} W(h)-\frac\pi2\le (1-{\varepsilon})\Big(W(z)-\frac\pi2\Big),\end{aligned} $$

(12.1)

z ∈ H ¹(B ₁) being the one-homogeneous extension of c in B ₁ , that is,

FormalPara Remark 12.2

On the figures in this section, we will use the following convention:

• is the support Ω_h = {h > 0} of the competitor h;

• is the support Ω_z = {z > 0} of the one-homogeneous function z;

• is the boundary ∂ Ω_h;

• is the boundary ∂ Ω_z;

• is the common boundary ∂ Ω_h ∩ ∂ Ω_z.

In Theorem 12.1 the main assumption on the trace c is that the set Ω_c ⊂ ∂B ₁ is close to the half-sphere. In [49, Theorem 1] the epiperimetric inequality was proved under the different assumption that the trace is non-degenerate. In fact, the epiperimetric inequality (12.1) holds without any assumption on the trace $c:\partial B_1\to \mathbb {R}$ or its free boundary ∂ Ω_c ⊂ ∂B ₁. Indeed, in the Appendix, we will prove the following result, which covers both Theorem 12.1 and [49, Theorem 1].

FormalPara Theorem 12.3 (Epiperimetric Inequality)

There is a constant ε > 0 such that: If c ∈ H ¹(∂B ₁) is a non-negative function on the boundary of the disk $B_1\subset \mathbb {R}^2$ then, there exists a (non-negative) function h ∈ H ¹(B ₁) such that (12.1) holds and h = c on ∂B ₁.

FormalPara Remark 12.4 (The Epiperimetric Inequality in Dimension d ≥ 2)

In higher dimension, the epiperimetric inequality for the one-phase energy is still an open problem. We expect that it will still be true under the assumption that the spherical set Ω_c is close to the half-sphere with respect to the Hausdorff distance. Indeed, it is an immediate consequence from the results in [29] that the epiperimetric inequality holds when the free boundary Ω_c is a C ^{2, α} regular graph (in the sphere) over the equator.

We stress that in higher dimension the epiperimetric inequality can hold only under some additional assumption on the distance from the trace to the half-plane solution. Indeed, if this was not the case (and so, the epiperimetric inequality was true in dimension d without any assumption on the trace), then the singular set would be empty in any dimension. This is due to the following remark.

FormalPara Remark 12.5 (The Epiperimetric Inequality Implies Regularity in Any Dimension)

We claim that if u is a local minimizer of $\mathcal F_\Lambda$ in a neighborhood of x ₀ and

$$\displaystyle \begin{aligned} W(u_{r,x_0})-\frac{\omega_d}{2}\le (1-{\varepsilon})\Big(W(z_{r,x_0})-\frac{\omega_d}{2}\Big), \end{aligned} $$

(12.2)

holds, for every r > 0, then x ₀ is a regular point. This is due to the following facts:

• A point x ₀ ∈ ∂ Ω_u is regular, of and only if, the Lebesgue density of Ω_u at x ₀ is precisely equal to (see Lemma 9.22).

• There are no points of Lebesgue density smaller than (Lemma 9.22).

• The function $r\mapsto W(u_{r,x_0})$ is non-decreasing and the limit

  $$\displaystyle \begin{aligned} \displaystyle\lim_{r\to0}W(u_{r,x_0}) \end{aligned}$$

  is precisely the Lebesgue density of Ω_u at x ₀ (see Lemma 9.20); in particular

  

• Suppose that the epiperimetric inequality (12.2) holds for every r > 0. Then, by the Weiss formula (Lemma 9.2) we obtain the following bound on the energy

  

  for some α > 0 depending on ε (this was proved in (12.28), which is the first step of the proof of Lemma 12.14). Since $W(u_{r,x_0})-\frac {\omega _d}{2}$ is non-negative, we get that

  $$\displaystyle \begin{aligned} \displaystyle\lim_{r\to0}W(u_{r,x_0})=\frac{\omega_d}2. \end{aligned}$$

  In particular, x ₀ is a point of Lebesgue density and so, it should be a regular point, as mentioned in the first bullet above.

As a consequence of Remark 12.5 at the singular points of the free boundary (12.2) cannot hold, which means that in higher dimension the epiperimetric inequality can only be true under the additional assumption that the trace on ∂B ₁ is close (in some sense) to a half-plane solution.

In this chapter, we will prove Theorem 12.1 and we will show that it implies the regularity of the free boundary (Proposition 12.13). The proof of Theorem 12.1 will be a consequence of the following two lemmas. The first one (Lemma 12.6) is based on a PDE argument which does not depend on the geometry of the free boundary; this lemma is proved in Sect. 12.5 and holds in any dimension d ≥ 2. The second lemma (Lemma 12.7) reflects the interaction of the free boundary with the Dirichlet energy; we prove it in Sect. 12.3.3 and the proof strongly uses the fact that we work in dimension two, even if the main idea can be used also in dimension d ≥ 2. Precisely, we use the Slicing Lemma (Lemma 12.10) to write the total energy as an integral of an energy defined on the spheres ∂B _r. Then, we prove the epiperimetric inequality by writing the second order expansion of the spherical energy for sets which are graphs over the equator (that is, arcs of length close to π).

FormalPara Lemma 12.6

Let ∂B ₁ be the unit sphere in dimension d ≥ 2. For every κ > 0, there are constants ρ ∈ (0, 1), ε ∈ (0, 1) and α > 1, depending only on κ and d such that:

If ψ ∈ H ¹(∂B ₁) satisfies the inequality

$$\displaystyle \begin{aligned} \int_{\partial B_1}|\nabla_\theta \psi|{}^2\,d\mathcal{H}^{d-1}\ge (d-1+\kappa) \int_{\partial B_1} \psi^2\,d\mathcal{H}^{d-1}, \end{aligned}$$

then, we have

$$\displaystyle \begin{aligned} W_0(h_\rho)\le (1-{\varepsilon})W_0(z)\qquad \mathit{\text{and}}\qquad W(h_\rho)\le (1-{\varepsilon})W(z), \end{aligned} $$

(12.3)

where in polar coordinates the functions $z,\,h_\rho \,:\,B_1\to \mathbb {R}$ are given by

$$\displaystyle \begin{aligned} z(r,\theta)=r\psi(\theta)\qquad \mathit{\text{and}}\qquad h_\rho(r,\theta)=\big(\max\{r-\rho,0\}\big)^\alpha\frac{\psi(\theta)}{(1-\rho)^\alpha}. \end{aligned}$$

Precisely, we can take

$$\displaystyle \begin{aligned} {\varepsilon}=\rho=\left(\frac{\kappa}{32d^2(2\kappa+1)}\right)^{3}. \end{aligned}$$

FormalPara Lemma 12.7 (Epiperimetric Inequality for Principal Modes: The Flat Free Boundary Case)

Let B ₁ be the unit ball in $\mathbb {R}^2$ . There are constants δ ₀ > 0 and ε > 0 such that the following holds.

If the continuous non-negative function $c:\partial B_1\to \mathbb {R}$ , c ∈ H ¹(∂B ₁), is a multiple of the first eigenfunction on {c > 0}⊂ ∂B ₁ and

$$\displaystyle \begin{aligned} \pi-\delta_0\le \mathcal{H}^1(\{c>0\}\cap \partial B_1)\le \pi+\delta_0, \end{aligned}$$

then, there exists a (non-negative) function h ∈ H ¹(B ₁) such that h = c on ∂B ₁ and

$$\displaystyle \begin{aligned} {} W(h)-\frac\pi2\le (1-{\varepsilon})\Big(W(z)-\frac\pi2\Big), \end{aligned}$$

z ∈ H ¹(B ₁) being the one-homogeneous extension of c in B ₁ . Moreover, if we assume that the function c is of the form

then the one-homogeneous extension is given by z(r, θ) = r c(θ) and the competitor h can be chosen as (the support of h is illustrated on Fig. 12.1)

Fig. 12.1

The positivity sets Ω_h and Ω_z. Here, the trace c is a multiple of the first eigenfunction on the arc (0, π + δ), |δ| < δ ₀ (δ < 0 on the left and δ > 0 on the right); the competitor is obtained by moving the free boundary ∂ Ω_z towards the line {x ₂ = 0}

12.1 Preliminary Results

In this section we prove several preliminary results that we will use in the proof of Theorem 12.1 (and also in the proof of Theorem 12.3).

This section is organized as follows:

• In Lemmas 12.8 and 12.9 we discuss the scale-invariance and the decomposition of the energy in orthogonal directions; both these results are implicitly contained in [49].

• The Slicing Lemma (Lemma 12.10) shows how to disintegrate the energy along the different spheres ∂B _r, 0 < r < 1. This result appeared for the first time in [29] and was crucial for the analysis of the free boundary around isolated singularities. We will use it in the proof of Lemma 12.7 (Sect. 12.3) and also in Sect. 12.2.

We start with the following result, which states that once we have a competitor for z in B ₁, then we can rescale it and use it in any ball B _ρ (ρ ≤ 1) by attaching it to z at ∂B _ρ.

Lemma 12.8 (Scaling)

Suppose that $z:B_1\to \mathbb {R}$ , z(r, θ) = rc(θ) is a one-homogeneous function and that h ∈ H ¹(B ₁) is such that h = c = z on ∂B ₁ . For every ρ ∈ (0, 1), we set

then, we have

$$\displaystyle \begin{aligned} {} W(h_\rho)-W(z)=\rho^d\big(W(h)-W(z)\big). \end{aligned}$$

Proof

We first compute

$$\displaystyle \begin{aligned} W_0(h_\rho)-W_0(z) & =\int_{B_1}|\nabla h_\rho|{}^2\,dx-\int_{B_1}|\nabla z|{}^2\,dx =\int_{B_\rho}|\nabla h_\rho|{}^2\,dx-\int_{B_\rho}|\nabla z|{}^2\,dx\\ & =\rho^d\left(\int_{B_1}|\nabla h|{}^2\,dx-\int_{B_1}|\nabla z|{}^2\,dx\right)=\rho^d\big(W_0(h)-W_0(z)\big). \end{aligned} $$

On the other hand, for the measure term, we have

$$\displaystyle \begin{aligned} |\{h_\rho>0\}\cap B_1|-|\{z>0\}\cap B_1| & =|\{h_\rho>0\}\cap B_\rho|-|\{z>0\}\cap B_\rho|\\ & =\rho^d\big(|\{h>0\}\cap B_1|-|\{z>0\}\cap B_1|\big), \end{aligned} $$

which concludes the proof. □

Lemma 12.9 (Decomposition of the Energy)

Suppose that the functions h ₁, h ₂ ∈ H ¹(B ₁) are such that, for every r ∈ (0, 1], we have

$$\displaystyle \begin{aligned} \int_{\mathbb S^{d-1}}\nabla_\theta h_1(r,\theta)\cdot\nabla_\theta h_2(r,\theta)\,d\theta=\int_{\mathbb S^{d-1}} h_1(r,\theta) h_2(r,\theta)\,d\theta=0. \end{aligned}$$

Then

$$\displaystyle \begin{aligned} W_0(h_1+h_2)=W_0(h_1)+W_0(h_2). \end{aligned}$$

Proof

The claim follows directly from the definition of W ₀ and the formula

$$\displaystyle \begin{aligned} \int_{B_1}|\nabla h|{}^2\,dx=\int_0^1r^{d-1}dr\int_{\partial B_1}\big(|\partial_r h|{}^2+r^{-2}|\nabla_\theta h|{}^2\big)\,d\theta, \end{aligned}$$

which holds for any h ∈ H ¹(B ₁). □

Lemma 12.10 (Slicing Lemma)

Let B ₁ be the unit ball in $\mathbb {R}^2$ . Let $\phi :(0,1]\times \mathbb S^1\to \mathbb {R}$ be a function such that $\phi \in H^1((0,1]\times \mathbb S^1)$ . Then, setting ϕ(r, θ) = ϕ _r(θ), we have

$$\displaystyle \begin{aligned} {} W_0(r\phi_r(\theta))=\int_0^1\mathcal F_0(\phi_r)\,rdr+\int_0^1\int_{\mathbb S^1}\big(\partial_r\phi_r(\theta)\big)^2\,r^3dr, \end{aligned}$$

and

$$\displaystyle \begin{aligned} W(r\phi_r(\theta))=\int_0^1\mathcal F(\phi_r)\,rdr+\int_0^1\int_{\mathbb S^1}\big(\partial_r\phi_r(\theta)\big)^2\,r^3dr, \end{aligned} $$

(12.4)

where, for any $\phi \in H^1(\mathbb S^1)$ , we set

$$\displaystyle \begin{aligned} \mathcal F_0(\phi)=\int_{\mathbb S^1}\big(|\partial_\theta\phi|{}^2-\phi^2\big)\,d\mathcal{H}^1\qquad \mathit{\text{and}}\qquad \mathcal F(\phi)=\mathcal F_0(\phi)+\mathcal{H}^1\left(\{\phi>0\}\cap\mathbb S^1\right). \end{aligned}$$

Proof

Let $\phi :]0,1]\times \partial B_1\to \mathbb {R}$. Then,

$$\displaystyle \begin{aligned} W_0(r\phi_r(\theta)) & =\int_0^1\int_{\mathbb S^1}\left(\big(\phi_r+r\partial_r\phi_r\big)^2+\big(\partial_\theta\phi_r\big)^2\right)\,d\theta\,rdr-\int_{\mathbb S^1}\phi_1^2(\theta)\,d\theta\\ & =\int_0^1\int_{\mathbb S^1}\left(\phi_r^2+r\partial_r(\phi_r^2)+r^2(\partial_r\phi_r)^2+\big(\partial_\theta\phi_r\big)^2\right)\,d\theta\,rdr-\int_{\mathbb S^1}\phi_1^2(\theta)\,d\theta \end{aligned} $$

Integrating by parts, we get that

$$\displaystyle \begin{aligned} \int_0^1r^2\partial_r(\phi_r^2)\,dr=\phi_1^2-2\int_0^1\phi_r^2\,rdr, \end{aligned}$$

which implies that

$$\displaystyle \begin{aligned} W_0(r\phi_r(\theta)) & =\int_0^1\mathcal F_0(\phi_r)\,rdr+\int_0^1\int_{\mathbb S^1}\big(\partial_r\phi_r(\theta)\big)^2\,r^3dr. \end{aligned} $$

In order to prove (12.4), it is sufficient to notice that

$$\displaystyle \begin{aligned} |\{h>0\}\cap B_1|=\int_0^1\mathcal{H}^1\big(\{\phi_r>0\}\cap\mathbb S^1\big)\,rdr, \end{aligned}$$

where h(r, θ) = rϕ _r(θ). □

Remark 12.11 (The Energy of a One-Homogeneous Function)

As an immediate consequence of Lemma 12.10, we get that if c ∈ H ¹(∂B ₁) and $z:B_1\to \mathbb {R}$ is the one homogeneous extension of c in B ₁, that is, z(r, θ) = rc(θ), then

$$\displaystyle \begin{aligned} W_0(z)=\frac 12\mathcal F_0(c)\qquad \text{and}\qquad W(z)=\frac 12\mathcal F(c). \end{aligned}$$

12.2 Homogeneity Improvement of the Higher Modes: Proof of Lemma 12.6

Let ρ ∈ (0, 1) be fixed. We will first compute the energy of h _ρ. For this purpose, we will use the Slicing Lemma; for every r ∈ [ρ, 1], we set

$$\displaystyle \begin{aligned} \phi_r(\theta)=\frac{\big(\max\{r-\rho,0\}\big)^\alpha}r\frac{\psi(\theta)}{(1-\rho)^\alpha} \end{aligned}$$

and we compute

$$\displaystyle \begin{aligned} \mathcal F_0(\phi_r)=\frac{(r-\rho)^{2\alpha}}{r^2(1-\rho)^{2\alpha}}\mathcal F_0(\psi)\qquad \text{and}\qquad \end{aligned}$$

$$\displaystyle \begin{aligned} \int_{\mathbb S^1}|\partial_r\phi_r|{}^2\,d\theta= \left(\alpha-1+\frac{\rho}r\right)^2\frac{(r-\rho)^{2\alpha-2}}{r^2(1-\rho)^{2\alpha}}\int_{\mathbb S^1}\psi^2\,d\theta. \end{aligned}$$

Integrating in r, we obtain

$$\displaystyle \begin{aligned} \int_\rho^1\mathcal F_0(\phi_r)\,r^{d-1}dr & =\frac{\mathcal F_0(\psi)}{(1-\rho)^{2\alpha}}\int_\rho^1(r-\rho)^{2\alpha}r^{d-3}dr\notag\\ & \le \frac{\mathcal F_0(\psi)}{(1-\rho)^{2\alpha}}\int_\rho^1r^{2\alpha+d-3}dr\le \frac{1}{2\alpha+d-2}\frac{\mathcal F_0(\psi)}{(1-\rho)^{2\alpha}}.{} \end{aligned} $$

(12.5)

We now compute

$$\displaystyle \begin{aligned} \int_\rho^1\int_{\mathbb S^{d-1}}|\partial_r\phi_r|{}^2\,d\theta\,r^{d+1}dr & =\int_\rho^1 \left(\alpha-1+\frac{\rho}r\right)^2\frac{(r-\rho)^{2\alpha-2}}{r^2(1-\rho)^{2\alpha}}\,r^{d+1}dr\int_{\mathbb S^{d-1}}\psi^2\,d\theta\\ & \le\frac{2}{(1-\rho)^{2\alpha}}\int_\rho^1 \left((\alpha-1)^2+\frac{\rho^2}{r^2}\right)r^{2\alpha+d-3}dr\int_{\mathbb S^{d-1}}\psi^2\,d\theta. \end{aligned} $$

Integrating in r ∈ [ρ, 1] and using that α ≥ 1 and d ≥ 2, we get

$$\displaystyle \begin{aligned} \int_\rho^1 \left((\alpha-1)^2+\frac{\rho^2}{r^2}\right)r^{2\alpha+d-3}dr& \le\frac{(\alpha-1)^2}{2\alpha+d-2}+\frac{\rho^2}{2\alpha+d-4}\\ & \le\frac 12\left((\alpha-1)^2+\frac{\rho^2}{\alpha-1}\right). \end{aligned} $$

Together with the inequality

$$\displaystyle \begin{aligned} \int_{\mathbb S^{d-1}}\psi^2\,d\theta\le \frac 1{\kappa}\mathcal F_0(\psi), \end{aligned}$$

which we have by hypothesis, this implies

$$\displaystyle \begin{aligned} \int_\rho^1\int_{\mathbb S^{d-1}}|\partial_r\phi_r|{}^2\,d\theta\,r^{d+1}dr\le \frac{1}{(1-\rho)^{2\alpha}}\left((\alpha-1)^2+\frac{\rho^2}{\alpha-1}\right)\frac 1{\kappa}\mathcal F_0(\psi).{} \end{aligned} $$

(12.6)

Furthermore, it is immediate to check that for every α ≤ 2 and $\rho \le \frac 12$ we have

$$\displaystyle \begin{aligned} \frac 1{(1-\rho)^{2\alpha}}\le \frac 1{(1-\rho)^{4}}\le 1+128\rho\qquad \text{and}\qquad \frac 1{(1-\rho)^{2\alpha}}\le 16. \end{aligned}$$

In particular,

$$\displaystyle \begin{aligned} \frac{(1-\rho)^{-2\alpha}}{2\alpha+d-2}\le \frac{1+128\rho}{2\alpha+d-2}\le \frac{1}{2\alpha+d-2}+64\rho, \end{aligned}$$

which, together with (12.5) implies:

$$\displaystyle \begin{aligned} \int_\rho^1\mathcal F_0(\phi_r)\,r^{d-1}dr\le \left(\frac{1}{2(\alpha-1)+d}+64\rho\right){\mathcal F_0(\psi)}. \end{aligned} $$

(12.7)

Analogously, from (12.6), we deduce

$$\displaystyle \begin{aligned} \int_\rho^1\int_{\mathbb S^{d-1}}|\partial_r\phi_r|{}^2\,d\theta\,r^{d+1}dr\le \frac{16}{\kappa}\left((\alpha-1)^2+\frac{\rho^2}{\alpha-1}\right)\mathcal F_0(\psi). \end{aligned} $$

(12.8)

We are now in position to estimate the difference W ₀(h _ρ) − W ₀(z). First of all, we set

$$\displaystyle \begin{aligned} \delta:=\alpha-1. \end{aligned}$$

Using the identity (see Remark 12.11)

$$\displaystyle \begin{aligned} W_0(z)=\frac 1d\mathcal F_0(\psi), \end{aligned}$$

and the inequalities (12.7) and (12.8), we estimate

$$\displaystyle \begin{aligned} W_0(h_\rho)-W_0(z)& \le \left(\frac{d}{2\delta+d}+ 64 d\, \rho+\frac{16d}{\kappa}\left(\delta^2+\frac{\rho^2}{\delta}\right)-1\right)W_0(z)\notag\\ & \le \left(-\frac{2\delta}{d}+ 64 d\, \rho+\frac{16d}{\kappa}\left(\delta^2+\frac{\rho^2}{\delta}\right)\right)W_0(z).{} \end{aligned} $$

(12.9)

We now choose

Substituting in (12.9), we obtain

In particular, the first inequality in (12.3) holds for any . In order to prove the second inequality in (12.3), we notice that, by the definition of h _ρ, we have

$$\displaystyle \begin{aligned} \big|\{h_\rho>0\}\cap B_1\big|=(1-\rho)\big|\{z>0\}\cap B_1\big|. \end{aligned}$$

Thus,

$$\displaystyle \begin{aligned} W(h_\rho)-W(z) & =W_0(h_\rho)-W_0(z)+\big|\{h_\rho>0\}\cap B_1\big|-\big|\{z>0\}\cap B_1\big|\\ & \le -\frac{\delta}{d}W_0(z)-\rho \big|\{z>0\}\cap B_1\big|. \end{aligned} $$

Choosing

we have that and so, we obtain

$$\displaystyle \begin{aligned} W(h_\rho)-W(z)& \le -{\varepsilon} W(z), \end{aligned} $$

which concludes the proof of (12.3). □

12.3 Epiperimetric Inequality for the Principal Modes: Proof of Lemma 12.7

We suppose that the spherical set {c > 0} is the arc (0, π + δ), where $\delta \in \mathbb {R}$ (and it might change sign). We recall that in Lemma 12.7 we assume that |δ| < δ ₀. Then, we can write the trace c in the following form

$$\displaystyle \begin{aligned} c(\theta)=c_1\phi_\delta(\theta)\quad \text{where}\quad c_1>0\quad \text{and}\quad \phi_\delta(\theta)=\sin\left(\frac{\theta\pi}{\pi+\delta}\right)\quad \text{for}\quad \theta\in[0,\pi+\delta]. \end{aligned}$$

Next, for every $t\in \mathbb {R}$, we define the function $\phi _t:\mathbb S^1\to \mathbb {R}$ as

$$\displaystyle \begin{aligned} \phi_t(\theta)=\sin\left(\frac{\theta\pi}{\pi+t}\right)\quad \text{for}\quad \theta\in[0,\pi+t]\ ,\qquad \phi_t(\theta)=0\quad \text{for}\quad \theta\notin[0,\pi+t]. \end{aligned}$$

Then set

$$\displaystyle \begin{aligned} f(t):=\int_{\partial B_1} \Big(|\partial_\theta\phi_t(\theta)|{}^2-\phi_{t}^2(\theta)\Big)\,d\theta+\mathcal{H}^{1}(\{\phi_{t}>0\})-\pi \end{aligned}$$

and

$$\displaystyle \begin{aligned} g(t):=\int_{\partial B_1} |\partial_t\phi_t(\theta)|{}^2\,d\theta. \end{aligned}$$

We consider the function

$$\displaystyle \begin{aligned} t(r):=\big(1-3(1-r){\varepsilon}\big)\delta, \end{aligned}$$

and define the competitor h _δ as

$$\displaystyle \begin{aligned} h_\delta(r,\theta)=r\phi_{t(r)}(\theta), \end{aligned} $$

(12.10)

which we will use in both Lemmas 12.7 and A.2.

We will show that for ε > 0 and δ > 0 small enough, we have

$$\displaystyle \begin{aligned} W(c_1h_\delta)-\frac\pi2\le (1-{\varepsilon})\Big(W(c_1z_\delta)-\frac\pi2\Big), \end{aligned} $$

(12.11)

where z _δ is the one-homogeneous extension of ϕ _δ in B ₁

$$\displaystyle \begin{aligned} z_\delta(r,\theta)=r\phi_\delta(\theta). \end{aligned} $$

(12.12)

12.3.1 Reduction to the Case c ₁ = 1

Let h _δ and z _δ be defined by (12.10) and (12.12). We claim that if, for some δ > 0 and ε > 0, we have

$$\displaystyle \begin{aligned} W(h_\delta)-\frac\pi2\le (1-{\varepsilon})\Big(W(z_\delta)-\frac\pi2\Big),\end{aligned} $$

(12.13)

then (12.11) does hold for every c ₁ > 0.

Indeed, using the homogeneity of W ₀, we get that

$$\displaystyle \begin{aligned} W_0(c_1 z_\delta)=c_1^2W_0(z_\delta)\qquad \text{and}\qquad W_0(c_1 h_\delta)=c_1^2W_0(h_\delta). \end{aligned}$$

On the other hand, we have that

$$\displaystyle \begin{aligned} \big|\{u>0\}\cap B_1\big|=\int_0^1 \mathcal{H}^1\big(\{u>0\}\cap\partial B_r\big)\,dr, \end{aligned}$$

for every (continuous) function $u:B_1\to \mathbb {R}$. Thus,

$$\displaystyle \begin{aligned} \big|\{z_\delta>0\}\cap B_1\big| & =\int_0^1 \mathcal{H}^1\big(\{\phi_\delta>0\}\cap\partial B_r\big)\,dr=\int_0^1 \mathcal{H}^1\big(\{\phi_\delta>0\}\cap\partial B_1\big)\,rdr\\ & =\frac 12(\pi+\delta). \end{aligned} $$

The analogous computation for the competitor h _δ gives

$$\displaystyle \begin{aligned} \big|\{h_\delta>0\}\cap B_1\big|=\int_0^1 \mathcal{H}^1\big(\{\phi_{t(r)}>0\}\cap\partial B_1\big)\,rdr=\int_0^1 (\pi+t(r))\,rdr. \end{aligned}$$

Putting together these computations, we obtain

$$\displaystyle \begin{aligned} \left(W(c_1h_\delta)-\frac\pi2\right)-(1-{\varepsilon})\left(W(c_1z_\delta)-\frac\pi2\right) & =c_1^2\left[\left(W(h_\delta)-\frac\pi2\right)-(1-{\varepsilon})\left(W(z_\delta)-\frac\pi2\right)\right]\\ & \qquad +(1-c_1^2)\left(\int_0^1t(r)r\,dr-(1-{\varepsilon})\int_0^1\delta r\,dr\right)\\ & =c_1^2\left[\left(W(h_\delta)-\frac\pi2\right)-(1-{\varepsilon})\left(W(z_\delta)-\frac\pi2\right)\right]\le 0, \end{aligned} $$

where we used that the function t(r) is chosen in such a way that, for any δ and ε, we have

$$\displaystyle \begin{aligned} \int_0^1\big(t(r)-(1-{\varepsilon})\delta \big)r\,dr & =\delta\int_0^1\big((1-3(1-r){\varepsilon})-(1-{\varepsilon})\big) r\,dr\\ & =\delta{\varepsilon}\int_0^1\big(3r^2-2r\big)\,dr=0. \end{aligned} $$

The rest of Sect. 12.3 is dedicated to the proof of (12.13).

12.3.2 An Estimate on the Energy Gain

The Slicing Lemma (Lemma 12.10) implies that

$$\displaystyle \begin{aligned} W(h_\delta)=\int_0^1f(t(r))\,r\,dr+\int_0^1|t^{\prime}(r)|{}^2 g(t(r))\,r^3\,dr\qquad \text{and}\qquad W(z_\delta)=\frac 12f(\delta). \end{aligned}$$

We first notice that the error term $\displaystyle \int _0^1|t^{\prime }(r)|{ }^2 g(t(r))\,r^3dr$ is lower order. Precisely, we have

$$\displaystyle \begin{aligned} \int_0^1|t^{\prime}(r)|{}^2 g(t(r))\,r^3\,dr=9{\varepsilon}^2\delta^2\int_0^1(1-r)^2 g(t(r))\,r^3\,dr\le C{\varepsilon}^2\delta^2, \end{aligned}$$

where C is a universal numerical constant. Thus, we get

$$\displaystyle \begin{aligned} \left(W(h_\delta)-\frac\pi2\right)-(1-{\varepsilon})\left(W(z_\delta)-\frac\pi2\right)\le F(\delta)+C{\varepsilon}^2\delta^2, \end{aligned} $$

(12.14)

where we have set

$$\displaystyle \begin{aligned} F(t):=\int_0^1\Big(f\big((1-3(1-r){\varepsilon})t\big)-(1-{\varepsilon})f(t)\Big)r\,dr. \end{aligned} $$

(12.15)

We will show that F is always negative in a neighborhoods of t = 0. First of all, we notice that the function f can be explicitly computed.

12.3.3 Computation of f

We now compute

where we set for simplicity . In particular, this implies that

$$\displaystyle \begin{aligned} f(0)=f^{\prime}(0)=0\qquad \text{and}\qquad f^{\prime\prime}(0)=\frac 1{\pi}\ . \end{aligned} $$

(12.16)

Moreover, we have that

(12.17)

12.3.4 Conclusion of the Proof of Lemma 12.7

Notice that, by using (12.16) and taking the derivative under the sign of the integral, we get that

$$\displaystyle \begin{aligned} F(0)=F^{\prime}(0)=0. \end{aligned}$$

Moreover, for the second derivative, we obtain

$$\displaystyle \begin{aligned} F^{\prime\prime}(0) & =f^{\prime\prime}(0)\int_0^1\Big(\big(1-3(1-r){\varepsilon} \big)^2-(1-{\varepsilon})\Big)r\,dr\\ & =f^{\prime\prime}(0)\int_0^1\Big(-6(1-r){\varepsilon} +9(1-r)^2{\varepsilon}^2+{\varepsilon}\Big)r\,dr\\ & =f^{\prime\prime}(0)\int_0^1\Big(-5r{\varepsilon} +6r^2{\varepsilon}+9(1-r)^2{\varepsilon}^2 r\Big)\,dr=-C_{\varepsilon} f^{\prime\prime}(0), \end{aligned} $$

where we have set

$$\displaystyle \begin{aligned} C_{\varepsilon}=\frac{{\varepsilon}}{2} \Big(1-\frac{3{\varepsilon}}{2}\Big). \end{aligned}$$

Thus, the second order Taylor expansion of F in zero is given by

$$\displaystyle \begin{aligned} F(0)+F^{\prime}(0)t+\frac 12 F^{\prime\prime}(0)t^2=-\frac{C_{\varepsilon}}{2\pi}\,t^2. \end{aligned}$$

We will next show that

(12.18)

Indeed, using (12.17) we can compute

$$\displaystyle \begin{aligned} \left|F(t)+\frac{C_{\varepsilon}}{2\pi}\,t^2\right|& \le\left|\int_0^1\Big(f\big((1-3(1-r){\varepsilon})t\big)-\frac{t^2}2 f^{\prime\prime}(0)\big(1-3(1-r){\varepsilon} \big)^2\Big)r\,dr\right|\\ & \qquad +(1-{\varepsilon})\left|\int_0^1\Big(f(t)-\frac{t^2}2 f^{\prime\prime}(0)\Big)r\,dr\right|\\ & \le\frac{|t|{}^3}{\pi^2}\left(\int_0^1\big(1-3(1-r){\varepsilon} \big)^3r\,dr +(1-{\varepsilon})\int_0^1r\,dr\right)\le\frac{|t|{}^3}{\pi^2}, \end{aligned} $$

which gives (12.18).

Now, using (12.14), we estimate

$$\displaystyle \begin{aligned} \left(W(h_\delta)-\frac\pi2\right)-(1-{\varepsilon})\left(W(z_\delta)-\frac\pi2\right)& \le F(\delta)+C{\varepsilon}^2\delta^2\\ & \le -\frac{1}{2\pi}\Big(1-\frac{3{\varepsilon}}{2}\Big)\,\frac{\varepsilon}2\delta^2+|\delta|{}^3+C{\varepsilon}^2\delta^2\\ & \le \left(-\frac{1}{2\pi}\Big(1-\frac{3{\varepsilon}}{2}\Big)\frac{\varepsilon}2+\delta_0+C{\varepsilon}^2\right)\delta^2, \end{aligned} $$

where C is the numerical constant from (12.14) and we recall that, by hypothesis, |δ|≤ δ ₀.

We now choose ε and δ ₀.

We set ε = 16πδ ₀. In particular, if $\displaystyle 0<\delta _0\le \frac {1}{48\pi }$, then $\displaystyle 1-\frac {3{\varepsilon }}{2}\ge \frac 12,$ and so

$$\displaystyle \begin{aligned} -\frac{1}{2\pi}\Big(1-\frac{3{\varepsilon}}{2}\Big)\frac{\varepsilon}2+\delta_0+C{\varepsilon}^2& \le -2\delta_0+\delta_0+C{\varepsilon}^2\le -\delta_0+256\pi^2C\delta_0^2, \end{aligned} $$

which is negative, whenever $\displaystyle \delta _0\le \frac {1}{256\pi ^2C}$. This means that in the end, choosing

$$\displaystyle \begin{aligned} \delta_0=\min\left\{\frac{1}{48\pi},\frac{1}{256\pi^2C}\right\}\qquad \text{and}\qquad {\varepsilon}=16\pi\delta_0, \end{aligned}$$

(12.13) holds for every δ such that |δ|≤ δ ₀. This concludes the proof of Lemma 12.7. □

12.4 Proof of Theorem 12.1

Since c ∈ H ¹(∂B ₁), we have that c is continuous and so, the set {c > 0} is a countable union of disjoint intervals (arcs), that is,

$$\displaystyle \begin{aligned} \{c>0\}=\bigcup_{j\ge 1}\mathcal I_j \end{aligned}$$

where, by hypothesis, we have

$$\displaystyle \begin{aligned} \pi-\delta_0\le\sum_{j\ge 1}|\mathcal I_j|\le \pi+\delta_0, \end{aligned}$$

where $|\mathcal I_j|=\mathcal {H}^1(\mathcal I_j)$ denotes the length of the interval $\mathcal I_j$. Now, we consider two cases:

• Case 1. There is one interval, say $\mathcal I_1$, of length $|\mathcal I_1|\ge \pi -\delta _0$. See Fig. 12.2.

  Fig. 12.2

  

  The supports of the one homogeneous extension z (in red) and the competitor h (in blue); the trace c falls in Case 1; the length of $\mathcal I_1$ is smaller than π

• Case 2. All the intervals are shorter than π − δ ₀, that is, $|\mathcal I_j|\le \pi -\delta _0$, for every j ≥ 1. See Fig. 12.3.

  Fig. 12.3

  

  The supports of the one homogeneous extension z (in red) and the competitor h (in blue) in Case 2

We first notice that if $\phi \in H^1_0(\mathcal I_j)$, then

$$\displaystyle \begin{aligned} \int_{\mathcal I_j}|\nabla_\theta\phi|{}^2\,d\theta\ge \frac{\pi^2}{|\mathcal I_j|{}^2}\int_{\mathcal I_j}\phi^2\,d\theta. \end{aligned}$$

In particular, if $|\mathcal I_j|\le \pi -\delta _0$, then

$$\displaystyle \begin{aligned} \int_{\mathcal I_j}|\nabla_\theta\phi|{}^2\,d\theta\ge \left(1+\frac{\delta_0}{\pi}\right)\int_{\mathcal I_j}\phi^2\,d\theta. \end{aligned}$$

Thus, if we are in Case 2, then the epiperimetric inequality is an immediate consequence of Lemma 12.6 with .

Suppose that we are in Case 1. Let {ϕ _j}_j≥1 be a complete orthonormal system of eigenfunctions on the interval $\mathcal I_1$. For every j ≥ 1, we set c _j to be the Fourier coefficient

$$\displaystyle \begin{aligned} c_j:=\int_{\partial B_1}c(\theta)\phi_j(\theta)\,d\theta. \end{aligned}$$

Then, we can decompose the trace c as

$$\displaystyle \begin{aligned} c(\theta)=c_1\phi_1(\theta)+\psi_1(\theta)+\psi_2(\theta), \end{aligned}$$

where

$$\displaystyle \begin{aligned} \displaystyle\psi_1(\theta)=\sum_{j=2}^\infty c_j\phi_j(\theta), \end{aligned}$$

and ψ ₂ is the restriction of c on the set $\displaystyle \bigcup _{j\ge 2}\mathcal I_j$. We first claim that, for i = 1, 2, we have

$$\displaystyle \begin{aligned} \int_{\mathbb S^1}|\nabla_\theta \psi_i|{}^2\,d\theta\ge (1+\kappa)\int_{\mathbb S^1}\psi_i^2\,d\theta, \end{aligned} $$

(12.19)

where κ > 0 is a constant depending only on δ ₀. Indeed, since ψ ₂ is supported on $\displaystyle \bigcup _{j\ge 2}\mathcal I_j$ and since $|\mathcal I_j|\le 2\delta _0$, for j ≥ 2, we have that

$$\displaystyle \begin{aligned} \int_{\mathbb S^1}|\nabla_\theta\psi_2|{}^2\,d\theta\ge \frac{\pi^2}{4\delta_0^2}\int_{\mathbb S^1}\psi_2^2\,d\theta. \end{aligned} $$

(12.20)

On the other hand, ψ ₁ contains only higher modes on the interval $\mathcal I_1$. Thus,

$$\displaystyle \begin{aligned} \int_{\mathcal I_1}|\nabla_\theta\psi_1|{}^2\,d\theta\ge \frac{4\pi^2}{(\pi+\delta_0)^2}\int_{\mathcal I_1}\psi_1^2\,d\theta. \end{aligned} $$

(12.21)

Now, choosing δ ₀ small enough (for instance, ), (12.20) and (12.21) imply (12.19). Let now ρ > 0 and ε _ψ > 0 be the constants from Lemma 12.6 corresponding to the constant κ from (12.19). Let $h_{\psi _1}$ and $h_{\psi _2}$ be the competitors from Lemma 12.6 associated to the traces ψ ₁ and ψ ₂, respectively. Thus, we have

$$\displaystyle \begin{aligned} W_0(h_{\psi_1})\le (1-{\varepsilon}_\psi)W_0(z_{\psi_1})\qquad \text{and}\qquad W(h_{\psi_2})\le (1-{\varepsilon}_\psi)W(z_{\psi_2}), \end{aligned} $$

(12.22)

where $z_{\psi _i}(r,\theta ):=z\psi _i(\theta )$.

Let $\tilde h$ be the competitor from Lemma 12.7, associated to the trace c ₁ ϕ ₁, and let

$$\displaystyle \begin{aligned} \tilde z(r,\theta):=rc_1\phi_1(\theta). \end{aligned}$$

We set

Thus, Lemmas 12.7 and 2.3 imply that

$$\displaystyle \begin{aligned} W(\tilde h_\rho)-\frac\pi2\le (1-\rho^d\tilde {\varepsilon})\Big(W(\tilde z)-\frac\pi2\Big), \end{aligned} $$

(12.23)

$\tilde {\varepsilon }$ being the constant from Lemma 12.7. We now define the competitor $h:B_1\to \mathbb {R}$ as:

• h = z if , where $z=\tilde z+z_{\psi _1}+z_{\psi _2}$ is the 1-homogeneous extension of c;

• $h=\tilde z+h_{\psi _1}+h_{\psi _2}$ if , but ;

• $h=\tilde h+h_{\psi _1}+h_{\psi _2}$ if and .

The first case is trivial and the second one follows directly by (12.22). We will prove the epiperimetric inequality in the most interesting third case. We first notice that the decomposition lemma (Lemma 12.9) implies that

$$\displaystyle \begin{aligned} W(z)=W(\tilde z)+W_0(z_{\psi_1})+W(z_{\psi_1}), \end{aligned}$$

and

$$\displaystyle \begin{aligned} W(h)=W(\tilde h_\rho)+W_0(h_{\psi_1})+W(h_{\psi_1}), \end{aligned}$$

where in the second decomposition, we use the fact that $h_{\psi _1}=h_{\psi _2}=0$ in B _ρ and that $\tilde h=\tilde z$ outside B _ρ. Now, setting

$$\displaystyle \begin{aligned} {\varepsilon}=\min\{\rho^d\tilde {\varepsilon},{\varepsilon}_\psi\}, \end{aligned}$$

the epiperimetric inequality (12.1) follows by (12.22) and (12.23). □

12.5 Epiperimetric Inequality and Regularity of the Free Boundary

In this section we will show how the epiperimetric inequality (12.1) implies the C ^{1, α} regularity of the free boundary. The main result of this section is Proposition 12.13, which we prove under the following assumption.

Condition 12.12 (Epiperimetric Inequality in Dimension d ≥ 2)

We say that the epiperimetric inequality holds in dimension d if there are constants δ _d > 0 and ε _d > 0 such that, for every non-negative one-homogeneous function z ∈ H ¹(B ₁), which is δ _d -flat in the ball B ₁ in some direction ν ∈ ∂B ₁ , that is

$$\displaystyle \begin{aligned} (x\cdot\nu-\delta_d)_+\le z(x)\le (x\cdot\nu+\delta_d)_+\quad \mathit{\text{for every}}\quad x\in B_1, \end{aligned}$$

there exists a non-negative function $h:B_1\to \mathbb {R}$ such that z = h on ∂B ₁ and

$$\displaystyle \begin{aligned} W(h)-\frac{\omega_d}2\le (1-{\varepsilon}_d)\Big(W(z)-\frac{\omega_d}2\Big). \end{aligned} $$

(12.24)

Proposition 12.13 (ε-Regularity via Epiperimetric Inequality)

Suppose that the epiperimetric inequality holds in dimension d (that is, Condition 12.12 holds). Then, there is a constant ε > 0 such that if $u:B_1\to \mathbb {R}$ is a non-negative minimizer of $\mathcal F_1$ in B ₁ and is ε-flat in B ₁ in some direction ν ∈ ∂B ₁

$$\displaystyle \begin{aligned} (x\cdot\nu-{\varepsilon})_+\le u(x)\le (x\cdot\nu+{\varepsilon})_+\quad \mathit{\text{for every}}\quad x\in B_1, \end{aligned}$$

then the free boundary ∂ Ω _u is C ^{1, α} regular in .

Proof

The claim is a consequence of Lemma 12.18, Lemma 12.14 and the results of the previous sections. By Condition 12.12 and Lemma 12.18, we have that the epiperimetric inequality (12.24) holds whenever

is small enough for some half-plane solution h _ν.

Using this, together with the Weiss’ monotonicity formula (Lemma 9.2), we get that the energy

$$\displaystyle \begin{aligned} \mathcal E(u):=W(u)-\frac{\omega_d}{2} \end{aligned}$$

satisfies the hypotheses of Lemma 12.14. Thus, we obtain the uniqueness of the blow-up limit and the decay of the blow-up sequences at every point of the free boundary in , that is, for every , there is a function $u_{x_0}:\mathbb {R}^d\to \mathbb {R}$ such that

$$\displaystyle \begin{aligned} u_{x_0}=\lim_{r\to0}u_{r,x_0}\qquad \text{and}\qquad \|u_{r,x_0}-u_{x_0}\|{}_{L^2(\partial B_1)}=0. \end{aligned}$$

Moreover, $u_{x_0}$ is a global minimizer of $\mathcal F_1$ in $\mathbb {R}^d$ (Proposition 6.2) and is one-homogeneous (Proposition 9.12). Using again Lemma 12.14 (see the energy-decay estimate (12.28) in the first step of the proof), we get that

$$\displaystyle \begin{aligned} \lim_{r\to0}\Big(W(u_{r,x_0})-\frac{\omega_d}{2}\Big)=0. \end{aligned}$$

Thus, the strong convergence of the blow-up sequence $u_{r,x_0}$ (Proposition 6.2) implies that

$$\displaystyle \begin{aligned} \frac{\omega_d}{2}=\lim_{r\to0}W(u_{r,x_0})=W(u_{x_0}). \end{aligned}$$

By Lemma 9.22, we get that $u_{x_0}$ is a half-plane solution. Thus, by Proposition 8.6, we get that the free boundary is a C ^{1, α} regular in . □

The idea that a purely variational inequality as (12.1) encodes the local behavior of the free boundary goes back to Reifenberg [45] who proved the regularity of the area-minimizing surfaces via an epiperimetric inequality for the area functional. Weiss was the first to prove an epiperimetric inequality in the context of a free boundary problem; in [53] he proved such an inequality for the obstacle problem and recovered the C ^{1, α} regularity of the (regular part of the) free boundary in any dimension, which was first proved by Caffarelli [11]. In [49], together with Luca Spolaor, we proved for the first time an epiperimetric inequality for the one-phase problem; in this case the interaction between the geometry of the free boundary and the Dirichlet energy functional is very strong and induced us to introduce the different constructive approach, which was the core of the previous section. In all these different contexts, once we have the epiperimetric inequality, we can obtain the regularity of the free boundary essentially by the same argument that we will describe in this section. The key result of this subsection is Lemma 12.14, which we attribute to Reifenberg, who was also the first to relate the variational epiperimetric inequality to the regularity of the local behavior of the free boundary (or area-minimizing surface).

Vocabulary and Notations

We recall that, for any $u:B_1\to \mathbb {R}$ and r ≤ 1, we use the notation u _r to indicate the one-homogeneous rescaling of u

$$\displaystyle \begin{aligned} u_r(x):=\frac 1r u(rx). \end{aligned}$$

Then, if $\mathcal E:H^1(B_1)\to \mathbb {R}$ is a given energy (for instance, $\displaystyle \mathcal E(u)=W_1(u)-\frac {\omega _d}2$), we will use the following terminology:

• By variation of the energy we mean the variation, with respect to r, of the energy $\mathcal E$ of the rescaling u _r. In other words, the variation of the energy is simply

  $$\displaystyle \begin{aligned} \frac{\partial}{\partial r}\mathcal E(u_r). \end{aligned}$$

• The energy deficit of a function $v:B_1\to \mathbb {R}$ is the difference

  $$\displaystyle \begin{aligned} \mathcal E(v)-\mathcal E(u), \end{aligned}$$

  where $u:B_1\to \mathbb {R}$ is a minimizer of $\mathcal E$ among all functions such that u = v on ∂B ₁.

• The deviation of a function $u:B_1\to \mathbb {R}$ (from being one-homogeneous) is

  $$\displaystyle \begin{aligned} \mathcal D(u):=\int_{\partial B_1}|x\cdot\nabla u(x)-u(x)|{}^2\,d\mathcal{H}^{d-1}(x). \end{aligned}$$

  We notice that a function

  $$\displaystyle \begin{aligned} \begin{array}{rcl} & u\in H^1(B_1)\quad \text{is one-homogeneous}\quad \\ & \Leftrightarrow\quad \mathcal D(u_r)=0\quad \text{for}\ \text{almost-every}\quad r\in(0,1). \end{array} \end{aligned} $$

Lemma 12.14 (Reifenberg [45])

Suppose that the function u ∈ H ¹(B ₁) and the energy functional $\mathcal E:H^1(B_1)\to \mathbb {R}$ are such that:

• (i) Minimality. u _r ∈ H ¹(B ₁) minimizes $\mathcal E$ in B ₁ , for every 0 < r ≤ 1, that is,

  $$\displaystyle \begin{aligned} \mathcal E(u_r)\le \mathcal E(v)\quad \mathit{\text{for every}}\quad v\in H^1(B_1),\quad v=u_r\quad \mathit{\text{on}}\quad \partial B_1. \end{aligned}$$

• (ii) The variation of the energy controls the deviation. The function $r\mapsto \mathcal E(u_r)$ is non-negative, differentiable and there is a constant C ₂ > 0 such that

  $$\displaystyle \begin{aligned} \frac{\partial }{\partial r}\mathcal E(u_r)\ge \frac{C_2}{r} \mathcal D(u_r)\qquad \mathit{\text{for every}}\qquad 0<r<1, \end{aligned}$$

  where $\mathcal D$ is given by

  $$\displaystyle \begin{aligned} \mathcal D(u):=\int_{\partial B_1}|x\cdot\nabla u(x)-u(x)|{}^2\,d\mathcal{H}^{d-1}(x). \end{aligned}$$

• (iii) The variation of the energy controls the energy deficit of the homogeneous extension. There is a constant C ₃ > 0 such that

  $$\displaystyle \begin{aligned} \frac{\partial }{\partial r}\mathcal E(u_r)\ge \frac{C_3}{r}\big(\mathcal E(z_r)-\mathcal E(u_r)\big)\qquad \mathit{\text{for every}}\qquad 0<r<1, \end{aligned}$$

  where $z_r:B_1\to \mathbb {R}$ is the one-homogeneous extension of the trace $u_r|{ }_{\partial B_1}$ , that is,

  

• (iv) Epiperimetric inequality. There is a one-homogeneous function $b:\mathbb {R}^d\to \mathbb {R}$ such that, if u _r is close to b in $\text{IMAGE}$ , then an epiperimetric inequality holds in B ₁ . Precisely, there are constants ε > 0 and δ ₄ > 0 such that:

  For every satisfying

  

  (12.25)

  there is a function h _r ∈ H ¹(B ₁) such that h _r = u _r = z _r on ∂B ₁ and

  $$\displaystyle \begin{aligned} \mathcal E(h_r)\le \big(1-{\varepsilon}\big)\mathcal E(z_r).\end{aligned} $$

  (12.26)

Under the hypotheses (i) , (ii) , (iii) and (iv) , there is δ > 0 such that, if u satisfies

then there is a unique u ₀ ∈ H ¹(B ₁) such that

(12.27)

where the constants γ and C can be chosen as

$$\displaystyle \begin{aligned} \gamma=\frac 12{\varepsilon} C_3\qquad \mathit{\text{and}}\qquad C=\delta_4. \end{aligned}$$

Remark 12.15

If the epiperimetric inequality (12.26) holds without the closeness assumption (12.25), then the Step 4 of the proof of Lemma 12.14 can be omitted.

Remark 12.16

The energy to which we will apply Lemma 12.14 is the Weiss’ boundary adjusted energy

$$\displaystyle \begin{aligned} \mathcal E(u)=W_1(u)=\int_{B_1}|\nabla u|{}^2\,dx-\int_{\partial B_1}u^2\,d\mathcal{H}^{d-1}+|\{u>0\}\cap B_1|. \end{aligned}$$

In this case, both (ii) and (iii) are implied by the Weiss’ formula (Lemma 9.2).

Remark 12.17

In our case, the function b from assumption (iv) is the half-plane solution b(x) = (x ⋅ ν)₊ for some ν ∈ ∂B ₁. Notice, that this does not mean that the blow-up limit u ₀ of u _r is equal to b. In fact, it may happen that the blow-up limit is another half-plane solution $b(x)=(x\cdot \tilde \nu )_+$, with $\tilde \nu$, which is close to ν. More generally, this lemma can be applied to situations in which u ₀ is not just a rotation of b, but is a completely different function. This happens for instance at isolated singularities in higher dimension (see [29]).

Proof of Lemma 12.14

Let now be the smallest non-negative number such that

and so, we can apply the epiperimetric inequality (12.26) for every u _r with . Notice that, since b is 1-homogeneous, a simple change of variables gives that

Thus, by choosing δ < 4^d+2 δ ₄, we get that

for every . Thus, .

We divide the proof in several steps.

Step 1: The Epiperimetric Inequality Implies the Decay of the Energy

Let . By (iii), (iv) and the minimality of u _r (assumption (i)), we have

$$\displaystyle \begin{aligned} \frac{\partial}{\partial r}\mathcal E(u_r)& \ge\frac{C_3}{r}\big(\mathcal E(z_r)-\mathcal E(u_r)\big)\\ & \ge\frac{C_3}{r}\big(\mathcal E(h_r)+{\varepsilon}\,\mathcal E(z_r)-\mathcal E(u_r)\big)\ge \frac{{\varepsilon} C_3}{r}\mathcal E(u_r). \end{aligned} $$

Setting $\displaystyle \gamma =\frac 12{\varepsilon } C_3$, we get that

$$\displaystyle \begin{aligned} \frac{\partial}{\partial r}\left(\frac{\mathcal E(u_r)}{r^{2\gamma}}\right)\ge 0, \end{aligned}$$

and so,

(12.28)

Step 2: The Energy Controls the Deviation

We set

$$\displaystyle \begin{aligned} e(r)=\mathcal E(u_{r})\qquad \text{and}\qquad f(r)=\mathcal D(u_{r}). \end{aligned}$$

By (ii), we get that

which implies that for every , we have the estimate

Step 3: The Deviation Controls the Oscillation of the Blow-up Sequence ur

Let x ∈ ∂B ₁ be fixed. Then, we have

$$\displaystyle \begin{aligned} \frac{\partial}{\partial r} \big(u_r(x)\big)=\frac{\partial}{\partial r} \left(\frac{u(rx)}r\right)=\frac{x\cdot\nabla u(rx)}{r}-\frac{u(rx)}{r^2}=\frac 1r\big(x\cdot\nabla u_r(x)-u_r(x)\big). \end{aligned} $$

Integrating in r, we get that, for every 0 < r ₁ < r ₂ ≤ 1,

$$\displaystyle \begin{aligned} |u_{r_2}(x)-u_{r_1}(x)|\le \int_{r_1}^{r_2}\frac 1r\big|x\cdot\nabla u_r(x)-u_r(x)\big|\,dr. \end{aligned}$$

Integrating in x ∈ ∂B ₁, and taking , we obtain

(12.29)

Step 4: The Blow-up Sequence Remains Close to b

Taking and r ₁ = r ∈ (ρ, r ₂) in (12.29), we get

Now, since

we can choose such that

On the other hand, taking , we obtain

This implies that if u = u ₁ is such that

then ρ = 0, that is, the epiperimetric inequality (12.26) can be applied to every r ∈ (0, 1].

Step 5: Conclusion

As a consequence of the previous step, the decay estimate (12.29) holds on the whole interval (0, 1]:

(12.30)

Thus, there is u ₀ ∈ L ²(∂B ₁), which is the strong L ²(∂B ₁)-limit of the blow-up sequence u _r

$$\displaystyle \begin{aligned} \lim_{r\to0}u_r=u_0. \end{aligned}$$

Finally, taking r ₂ = r ∈ (0, 1) and passing to the limit as r ₁ → 0 in (12.30), we obtain (12.27). This concludes the proof.

□

In order to prove Proposition 12.13 under the Condition 12.12 we will need the following lemma.

Lemma 12.18

For every ε > 0 there is δ > 0 such that the following holds.

If $u:B_2\to \mathbb {R}$ is a (non-negative) minimizer of $\mathcal F_1$ in B ₂ satisfying

where h _ν is the half-plane solution h _ν(x) = (x ⋅ ν)₊,

then u is ε-flat in the direction ν in the ball B ₁ , that is,

$$\displaystyle \begin{aligned} (x\cdot\nu-{\varepsilon})_+\le u(x)\le (x\cdot\nu+{\varepsilon})_+\qquad \mathit{\text{for every}}\qquad x\in B_1. \end{aligned} $$

(12.31)

Proof

We will first prove that there is ε > 0 such that u is ε-flat on , that is,

(12.32)

From this, we will deduce that u is ε-flat in B ₁.

In order to prove (12.32), we start by noticing that that, since u minimizes $\mathcal F_1$ in B ₂, it is L-Lipschitz continuous in , for some L ≥ 1 depending only on the dimension (see Theorem 3.1). Then, also the function

is (2L)-Lipschitz continuous. Thus, there is a dimensional constant C _d such that

We now choose δ > 0 such that

(12.33)

Thus,

(12.34)

Now, using (12.34), we obtain the estimate from below

while from above we only have

Indeed, if , then

Thus, in order to prove that (12.32) it is sufficient to show that

(12.35)

On the other hand, u is also non-degenerate in the annulus $\text{IMAGE}$, that is, there is a dimensional constant 0 < κ < 1 such that (see Proposition 4.1)

$$\text{IMAGE}$$

Suppose by absurd that there is a point

$$\text{IMAGE}$$

Then, taking we get that there is

such that

$$\displaystyle \begin{aligned} \big|u(y_0)-h_\nu(x_0)\big|=u(y_0)\ge \frac 12 \kappa{\varepsilon}. \end{aligned}$$

If we choose δ such that

$$\displaystyle \begin{aligned} C_d\,\delta^{\frac{2}{d+2}}\le \frac 12\kappa{\varepsilon}, \end{aligned} $$

(12.36)

then we reach a contradiction. Notice that, since κ < 1, (12.36) implies (12.33).

This concludes the proof of (12.32). The conclusion now follows by Proposition 12.19. □

12.6 Comparison with Half-Plane Solutions

In this subsection, we prove the following result, which we use in the proof of Lemma 12.18; but is also of general interest.

Proposition 12.19

Let $D\subset \mathbb {R}^d$ be a bounded open set and let $u:\overline D\to \mathbb {R}$ be a non-negative continuous function and a minimizer of the functional $\mathcal F_\Lambda$ in D. Let $c\in \mathbb {R}$ be a constant, $\nu \in \mathbb {R}^d$ be a unit vector and

$$\displaystyle \begin{aligned} h(x)=\sqrt\Lambda\,\big(x\cdot\nu+c\big)_+ \end{aligned}$$

be a half-plane solution. Then, the following claims do hold.

• (i) If u ≤ h on ∂D, then u ≤ h in D.

• (ii) If u ≥ h on ∂D, then u ≥ h in D.

Remark 12.20

Up to replacing u and h by and (which are minimizers of $\mathcal F_1$ in D), we may assume that Λ = 1.

We will give two different proofs to Proposition 12.19. The first one is more natural, but is based on the notion of viscosity solution and so it requires the results from Sect. 7.1. The second proof is direct and is based on a purely variational argument in the spirit of Lemma 2.13.

Proof I of Proposition 12.19

By Proposition 7.1, u is a viscosity solution (see Definition 7.6) of

$$\displaystyle \begin{aligned} \Delta u=0\quad \text{in}\quad \Omega_u\cap D,\qquad |\nabla u|=1\quad \text{on}\quad \partial\Omega_u\cap D. \end{aligned}$$

The conclusion now follows by Lemma 12.21 bellow. □

Proof II of Proposition 12.19

We only prove the first claim, the proof of the second one being analogous. For every t > 0, consider the half-plane solution

$$\displaystyle \begin{aligned} h_t(x)=(x\cdot\nu+c+t)_+. \end{aligned}$$

Then, for every x ∈ ∂D ∩ Ω_u, we have that h(x) ≥ u(x) > 0 and so,

$$\displaystyle \begin{aligned} u(x)\le h(x)\le h_t(x)-t. \end{aligned}$$

Thus, we can apply Lemma 12.22 to u and h _t, obtaining that u ≤ h _t in D. Since t is arbitrary, we obtain claim (i). □

Lemma 12.21 (Comparison of a Viscosity and a Half-Plane Solution)

Let D be a bounded open set in $\mathbb {R}^d$ and let $u:D\to \mathbb {R}$ be a non-negative continuous function and a viscosity solution (see Definition 7.6 ) to

$$\displaystyle \begin{aligned} \Delta u=0\quad \mathit{\text{in}}\quad \Omega_u\cap D,\qquad |\nabla u|=1\quad \mathit{\text{on}}\quad \partial\Omega_u\cap D. \end{aligned}$$

Let $c\in \mathbb {R}$ be a constant, $\nu \in \mathbb {R}^d$ be a unit vector and h(x) = (x ⋅ ν + c) ₊ be a half-plane solution. Then, the following claims do hold.

• (i) If u ≥ h on ∂D, then u ≥ h in D.

• (ii) If u ≤ h on ∂D, then u ≤ h in D.

Proof

We first prove (i). Let $M=\|h\|{ }_{L^\infty (D)}$.

For any t > 0, we define the real function $f_t:\mathbb {R}\to \mathbb {R}$ as

$$\displaystyle \begin{aligned} f_t(s)=(1+t)\max\{s,0\}+t\,\big(\max\{s,0\}\big)^{2}, \end{aligned}$$

for every $s\in \mathbb {R}$. Then, it is immediate to check that the function

$$\displaystyle \begin{aligned} v_{t}(x)=f_t\big(x\cdot\nu+c-M(M+1)t\big) \end{aligned}$$

satisfies the following conditions:

• (1) Δv _t > 0 in the set {v _t > 0};

• (2) |∇v _t| > 1 on $\overline {\{v_{t}>0\}}$;

• (3) v _t(x) ≤ h(x) ≤ u(x) for every x ∈ ∂D.

Indeed, the first two conditions are immediate, since h is the positive part of an affine function. In order to prove (3), we notice that the inequality is trivial whenever x ⋅ ν + c − M(M + 1)t ≤ 0. The case x ⋅ ν + c − M(M + 1)t > 0 is a consequence of the following estimate, which holds for any S := x ⋅ ν + c > M(M + 1)t.

$$\displaystyle \begin{aligned} f_t(S-M(M+1)t) & =(1+t)\big(S-M(M+1)t\big)+t\big(S-M(M+1)t\big)^2\\ & \le (1+t)\big(S-M(M+1)t\big)+Mt\big(S-M(M+1)t\big)\\ & =S+t\Big(-M(M+1)+S-M(M+1)t+MS-M^2(M+1)t\Big)\\ & \le S+t\big(-M(M+1)+S(M+1)\big)\le S. \end{aligned} $$

We next claim that v _t ≤ u on D. Indeed, suppose that this is not the case and let T > 0 be the smallest real number such that (v _t − T)₊ ≤ u on $\overline D$. Then, there is $x_0\in \overline \Omega _u$ such that v _t(x ₀) − T = u(x ₀) and (v _t(x) − T)₊ ≤ u(x), for every other $x\in \overline D$, that is, the test function (v _t − T)₊ touches from below u at x ₀. Since u is a viscosity solution (see Definition 7.6 and Proposition 7.1) of

$$\displaystyle \begin{aligned} \Delta u=0\quad \text{in}\quad \Omega_u\cap D,\qquad |\nabla u|=1\quad \text{on}\quad \partial\Omega_u\cap D, \end{aligned}$$

we have that and x ₀∉ Ω_u. Then, the only possibility is that x ₀ ∈ ∂D, but this is also impossible since (v _t − T)₊ < v _t ≤ u on ∂D. This proves that v _t ≤ u on D. Now, letting t → 0, we get that

$$\displaystyle \begin{aligned} u(x)\ge h(x)\quad \text{in}\quad D, \end{aligned}$$

which concludes the proof of (i).

The proof of claim (ii) is analogous. We give the proof for the sake of completeness. For any t > 0, we define the real function

$$\displaystyle \begin{aligned} g_t(s)=(1-{\varepsilon} t)\max\{s,0\}-{\varepsilon} t\,\big(\max\{s,0\}\big)^{2}\qquad \text{for every}\qquad s\in\mathbb{R}, \end{aligned}$$

where ε > 0 will be chosen below. We set

$$\displaystyle \begin{aligned} M_u=\text{diam}\,(D)+|c|+\|u\|{}_{L^\infty(D)}\qquad \text{and}\qquad M_h=\|h\|{}_{L^\infty(D)}. \end{aligned}$$

The test function

$$\displaystyle \begin{aligned} w_{t}(x)=g_t\big(x\cdot\nu+c+t\big) \end{aligned}$$

satisfies the following conditions:

• 1. w _t ≥ 0 for every 0 < t ≤ M _u and every s ≤ M _h;

• 2. Δw _t < 0 in the open set {w _t > 0};

• 3. |∇w _t| < 1 on the closed set $\overline {\{w_{t}>0\}}$;

• 4. w _t(x) ≥ h(x) ≥ u(x) for every x ∈ ∂D and every t ≤ M _u.

We start with (1_w). We notice that

$$\displaystyle \begin{aligned} g_t(s)=(1-{\varepsilon} t)\max\{s,0\}-{\varepsilon} t\,\big(\max\{s,0\}\big)^{2}\ge 1-{\varepsilon} \big(M_uM_h+M_uM_h^2\big). \end{aligned}$$

Thus, in order to have (1_w), we choose

$$\displaystyle \begin{aligned} {\varepsilon}\le \big(M_uM_h+M_uM_h^2\big)^{-1}. \end{aligned}$$

Again (2_w) and (3_w) are trivial, while for (4_w) we will need the following estimate, which holds for every S > 0 and t > 0.

$$\displaystyle \begin{aligned} g_t(S+t) & =(1-{\varepsilon} t)\big(S+ t\big)-{\varepsilon} t\big(S+ t\big)^2\notag\\ & \ge (1-{\varepsilon} t)\big(S+ t\big)-{\varepsilon} tS\big(S+ t\big)\notag\\ & =S+t\Big(1-{\varepsilon} S-{\varepsilon} t-{\varepsilon} S^2-{\varepsilon} S t\Big)\notag\\ & \ge S+t\Big(1-{\varepsilon}\big(S+t+S^2+ST\big)\Big).{} \end{aligned} $$

(12.37)

In order to have (4_w), we choose

$$\displaystyle \begin{aligned} {\varepsilon}\le\big(M_h+M_u+M_h^2+M_hM_u\big)^{-1}. \end{aligned} $$

(12.38)

We next complete the proof of (4_w). First, notice that the second inequality is always true by hypothesis. Since w _t ≥ 0, the first inequality is trivial whenever x ⋅ ν + c ≤ 0. Thus, we only need to prove that w _t(x) ≥ h(x), whenever x ⋅ ν + c > 0. This follows by (12.37) and the second bound on ε (12.38). This concludes the proof of (1_w) − (4_w).

We now consider the set

$$\displaystyle \begin{aligned} I:=\Big\{t\in[0,M_u]\ :\ w_t\ge u\quad \text{on}\quad D\Big\}. \end{aligned}$$

We notice that I _t is non-empty since M _u ∈ I _t. Let

$$\displaystyle \begin{aligned} T=\inf I. \end{aligned}$$

If T > 0, then there is a point $x_0\in \overline \Omega _u$ such that w _T touches u from above in x ₀. But this contradicts (2_w) − (4_w). Indeed, (2_w) implies that x ₀∉ Ω_u ∩ D, (3_w) gives that x ₀∉∂ Ω_u ∩ D and (4_w) gives that x ₀∉∂D. Thus, T = 0, which concludes the proof. □

Lemma 12.22 (Comparison of Minimizers)

Let D be a bounded open set in $\mathbb {R}^d$ and $u,v:\overline D\to \mathbb {R}$ be continuous non-negative functions and minimizers of $\mathcal F_\Lambda$ in D. Suppose that:

• (a) u ≤ v on ∂D;

• (b) the above inequality is strict on the set $\overline \Omega _u\cap \partial D$ , that is, $\displaystyle \min _{\overline \Omega _u\cap \partial D}(v-u)=m>0.$

Then, u ≤ v in D.

Proof

Let Ω := {x ∈ D  :  u(x) > v(x)}. We will prove that Ω = ∅. We first claim that Ω is strictly contained in D, that is

$$\displaystyle \begin{aligned} \partial\Omega\cap\partial D=\emptyset. \end{aligned}$$

Suppose that this is not the case. Then, there is a sequence x _n ∈ Ω converging to some x ₀ ∈ ∂D. Since u and v are continuous, we get that

$$\displaystyle \begin{aligned} v(x_0)-u(x_0)=0. \end{aligned}$$

On the other hand, for every $n\in \mathbb {N}$, we have

$$\displaystyle \begin{aligned} u(x_n)>v(x_n)\ge 0, \end{aligned}$$

which gives that x _n ∈ Ω_u. Then, x _n ∈ Ω_u and thus, x ₀ ∈ ∂ Ω_u. This is a contradiction with the assumption (b).

We will next prove that

$$\displaystyle \begin{aligned} \Omega_u\cap \partial\{u>v\}=\Omega_v\cap \partial\{u>v\}=\emptyset. \end{aligned}$$

We consider the competitors

$$\displaystyle \begin{aligned} u\vee v=\max\{u,v\}\qquad \text{and}\qquad u\wedge v=\min\{u,v\}. \end{aligned}$$

Since

$$\displaystyle \begin{aligned} u\vee v=v\quad \text{on}\quad \partial D\qquad \text{and}\qquad u\wedge v=u\quad \text{on}\quad \partial D, \end{aligned}$$

the minimality of u and v implies that

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(v,D)\le \mathcal F_\Lambda(u\vee v,D)\qquad \text{and}\qquad \mathcal F_\Lambda(u,D)\le \mathcal F_\Lambda(u\wedge v,D). \end{aligned} $$

(12.39)

On the other hand, we have

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(u\vee v,D)+\mathcal F_\Lambda(u\wedge v,D)=\mathcal F_\Lambda(u,D)+\mathcal F_\Lambda(v,D). \end{aligned}$$

Thus, both inequalities in (12.39) are in fact equalities and so u ∧ v is a minimizer of $\mathcal F_\Lambda$ in D. Suppose that

$$\displaystyle \begin{aligned} x_0\in\Omega_u\cap\partial\Omega. \end{aligned}$$

Then, u(x ₀) = v(x ₀) > 0 and by the continuity of u and v, there is a ball B _r(x ₀) such that

$$\displaystyle \begin{aligned} B_r(x_0)\subset \Omega_u\qquad \text{and}\qquad B_r(x_0)\subset \Omega_v. \end{aligned}$$

Thus, both the functions u and u ∧ v are positive and harmonic in B _r(x ₀). Thus, the strong maximum principle implies that u = u ∧ v in B _r(x ₀). This is contradiction with the assumption that x ₀ ∈ ∂{u > v}.

We are now in position to prove that Ω = ∅. Indeed, suppose that this is not the case. Then, for every x ₀ ∈ ∂ Ω_u, we have that u(x ₀) = 0. Thus, we consider the function

Then, $\widetilde u=u$ on ∂D and $\widetilde u\in H^1(D)$ (this follows, from instance from the facts that u is Lipschitz continuous on the compact subsets of D and that $\overline \Omega \subset D$). Thus, $\widetilde u$ is an admissible competitor for u and we have

$$\displaystyle \begin{aligned} 0\ge \mathcal F_\Lambda(u,D)- \mathcal F_\Lambda(\widetilde u,D)=\int_\Omega|\nabla u|{}^2\,dx+\Lambda|\Omega\cap\Omega_u|. \end{aligned}$$

In particular,

$$\displaystyle \begin{aligned} |\Omega|=|\{u>v\}|=|\{u>v\}\cap\{u>0\}|=|\Omega\cap\Omega_u|=0, \end{aligned}$$

and so, Ω = ∅, which concludes the proof. □