Blow-Up Sequences and Blow-Up Limits

Abstract

Let D be an open set in $\mathbb {R}^d$ and $u:D\to \mathbb {R}$ be a (non-negative) local minimizer of $\mathcal F_\Lambda$ in D

Let D be an open set in $\mathbb {R}^d$ and $u:D\to \mathbb {R}$ be a (non-negative) local minimizer of $\mathcal F_\Lambda$ in D. Recall that, by Theorem 3.1, we have that u is locally Lipschitz continuous in D. Let x ₀ ∈ ∂ Ω_u ∩ D be a given point on the free boundary. For every r > 0, we define the rescaled function

$$\displaystyle \begin{aligned} u_{x_0,r}(x):=\frac 1r u(x_0+rx). \end{aligned}$$

Let (r _n)_n≥1 be a vanishing sequence of positive numbers. We say that the sequence of functions $u_{x_0,r_n}$ is a blow-up sequence. We notice that $u_{x_0,r_n}$ is not defined on the entire $\mathbb {R}^d$ (since a priori we might have that $D\neq \mathbb {R}^d$), its domain of definition being the set

$$\displaystyle \begin{aligned} \frac 1r(-x_0+D):=\big\{x\in\mathbb{R}^d\ :\ x_0+rx\in D \big\}. \end{aligned}$$

On the other hand, since r _n converges to zero, for every fixed R > 0, there exists m > 0 such that, for every n ≥ m, $u_{x_0,r_n}$ is defined on B _R, that is,

$$\displaystyle \begin{aligned} B_R\subset \frac 1{r_n}(-x_0+D). \end{aligned}$$

Now since,

$$\displaystyle \begin{aligned} \nabla u_{x_0,r_n}(x)=\nabla u(x_0+rx)\quad \text{for every}\quad x\in B_R, \end{aligned}$$

we have that

$$\displaystyle \begin{aligned} \|\nabla u_{x_0,r_n}\|{}_{L^\infty(B_R)}=\|\nabla u\|{}_{L^\infty(B_{{R}/{r_n}}(x_0))}. \end{aligned}$$

Since u is locally Lipschitz continuous and u(x ₀) = 0, we get that the sequence $u_{x_0,r_n}$ is uniformly bounded and equicontinuous on B _R. Thus, by the Theorem of Ascoli-Arzelà, we obtain that there is a subsequence of $u_{x_0,r_n}$ that converges uniformly in the ball B _R. Repeating this argument for every (natural number) R > 0 and extracting a diagonal sequence, we get that there exists a function $u_0:\mathbb {R}^d\to \mathbb {R}$ such that, for every R > 0, the sequence $u_{x_0,r_n}$ converges uniformly to u ₀ in B _R,

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\|u_{x_0,r_n}-u_0\|{}_{L^\infty(B_R)}=0\quad \text{for every}\quad R>0. \end{aligned} $$

(6.1)

FormalPara Definition 6.1 (Blow-Up Limit)

We will say that the function $u_0:\mathbb {R}^d\to \mathbb {R}$ is a blow-up limit of u at x ₀ if (6.1) does hold.

We notice that every blow-up limit u ₀ of a local minimizer u of $\mathcal F_\Lambda$ is non-negative, Lipschitz continuous (in $\mathbb {R}^d$) and vanishes in zero. We also stress that there might be numerous blow-up limits, each one depending on the choice of the (sub-)sequence $u_{x_0,r_n}$. If this is the case, then we simply say that the blow-up limit is not unique. For instance, the function $\phi :B_1\to \mathbb {R}$ defined in polar coordinates as (see Fig. 6.1)

$$\displaystyle \begin{aligned} \phi(\rho,\theta)=\rho \max\{0,\cos{}(\theta+\ln\rho)\} \end{aligned}$$

has infinitely many blow-up limits in zero (but it is not a local minimizer of the functional $\mathcal F_\Lambda$). We will denote the family of all blow-up limits of u at x ₀ by $\mathcal {B}\mathcal {U}_{u}(x_0)$. The classification of all the possible blow-up limits and the uniqueness of the blow-up limit at a given point x ₀ ∈ ∂ Ω_u are both central questions in the free boundary regularity theory, which do not have a complete answer yet.

Fig. 6.1

Example of a (Lipschitz) function with infinitely many blow-up limits in zero

In this chapter we will decompose the free boundary into a regular and singular parts according to the structure of the space of blow-up limits at the points of ∂ Ω_u. The Sects. 6.1, 6.2, and 6.3 are dedicated to the proof of the following result.

FormalPara Proposition 6.2 (Convergence of the Blow-Up Sequences)

Let D be an open subset of $\mathbb {R}^d$ and let $u:D\to \mathbb {R}$ be non-negative, $u\in H^1_{loc}(D)$ and a local minimizer of $\mathcal F_\Lambda$ in D. Let x ₀ ∈ ∂ Ω _u ∩ D and let r _n → 0 be a vanishing sequence of positive real numbers such that the blow-up sequence $u_{x_0,r_n}$ converges locally uniformly to the blow-up limit $u_0:\mathbb {R}^d\to \mathbb {R}$ in the sense of (6.1). Then, there is a subsequence such that, for every R > 0, we have:

• (i) the sequence $u_{x_0,r_n}$ converges to u ₀ strongly in H ¹(B _R);

• (ii) the sequence of characteristic functions converges to in L ¹(B _R), where

  $$\displaystyle \begin{aligned} \Omega_n:=\{u_{x_0,r_n}>0\}\qquad \mathit{\text{and}}\qquad \Omega_0:=\{u_0>0\}\,; \end{aligned}$$

• (iii) the sequence of sets $\overline \Omega _n$ converges locally Hausdorff in B _R to $\overline \Omega _0$;

• (iv) u ₀ is a non-trivial local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^d$.

In particular, Sect. 6.1 is dedicated to the strong convergence of the blow-up sequences (claims (i) and (ii)) and the optimality of the blow-up limits (claim (iv)); the main result of this section (Lemma 6.3) is more general and will also be used in the proof of Theorem 1.9. Section 6.2 is dedicated to the local Hausdorff convergence of the free boundaries (claim (iii)); the results of this section apply both to Theorem 1.2 and Theorem 1.9. In Sect. 6.3, we conclude the proof of Proposition 6.2.

In Sect. 6.4, we define the regular part Reg(∂ Ω_u) and the singular part Sing(∂ Ω_u) of the free boundary. Moreover, we prove that the singular set Sing(∂ Ω_u) has zero (d − 1)-dimensional Hausdorff measure (Proposition 6.12). We notice that this result applies to Theorems 1.2, 1.4, and 1.9, but is interesting only for Theorem 1.2, in which we do not make use of monotonicity formulas. In fact, in Sect. 10, we will obtain better estimates on the dimension of the singular set by means of the Weiss’ monotonicity formula, which we will apply to both Theorem 1.4 and Theorem 1.9.

6.1 Convergence of Local Minimizers

In this section we prove the strong convergence of the blow-up sequences and the minimality of the blow-up limits at every point of the free boundary of a local minimizer. Our result (Lemma 6.3) is more general and applies also to other free boundary problems; for instance, we will use it in the proof of Theorem 1.9.

Lemma 6.3

Let Λ > 0 be a given constant, $B_R\subset \mathbb {R}^d$ and u _n ∈ H ¹(B _R) be a sequence of non-negative functions such that:

• (a) every u _n is a local minimizer of $\mathcal F_\Lambda$ in B _R or, more generally, satisfies

  $$\displaystyle \begin{aligned} \mathcal F_\Lambda(u_n,B_R)\le \mathcal F_\Lambda(u_n+\varphi,B_R)+{\varepsilon}_n\quad \mathit{\text{for every}}\quad \varphi\in H^1_0(B_r)\quad \mathit{\text{and every}}\quad r<R\ , \end{aligned}$$

  where ε _n is a vanishing sequence of positive constants.

• (b) the sequence u _n is uniformly bounded in H ¹(B _R), that is, for some constant C > 0,

  $$\displaystyle \begin{aligned} \|u_n\|{}_{H^1(B_R)}^2=\mathcal F_0(u_n,B_R)+\int_{B_R}u_n^2\,dx\le C\qquad \mathit{\text{for every}}\qquad n\ge 1. \end{aligned}$$

Then, there is a function u _∞∈ H ¹(B _R) such that, up to a subsequence, we have

• (i) u _n converges to u _∞ strongly in H ¹(B _r), for every 0 < r < R;

• (ii) the sequence of characteristic functions converges to strongly in L ¹(B _r) and pointwise almost-everywhere in B _r , for every 0 < r < R;

• (iii) u _∞ is a local minimizer of $\mathcal F_\Lambda$ in B _R.

Proof

The idea of the proof is very similar to the one in Lemma 3.14, but is more involved due to the presence of the measure term. Up to extracting a subsequence, we can suppose that the sequence u _n converges to a function u _∞∈ H ¹(B _R) weakly in H ¹(B _R), strongly in L ²(B _R) and pointwise (Lebesgue) almost-everywhere in B _R. We set for simplicity

$$\displaystyle \begin{gathered} \Omega_n=\{u_n>0\}\quad and \quad \Omega_\infty=\{u_\infty>0\}. \end{gathered} $$

The weak H ¹-convergence implies that for every 0 < r ≤ R

$$\displaystyle \begin{aligned} \|\nabla u_{\infty}\|{}_{L^2(B_r)}\le\liminf_{n\to\infty}\|\nabla u_{n}\|{}_{L^2(B_r)}, \end{aligned} $$

(6.2)

with an equality, if and only if, (up to a subsequence) the convergence is strong in B _r. On the other hand, the pointwise convergence of u _n implies that for almost-every x ∈ B _R

$$\displaystyle \begin{aligned} x\in\Omega_{\infty}\ \Rightarrow\ u_\infty(x)>0\ \Rightarrow\ u_n(x)>0\ \text{ for large }n \ \Rightarrow\ x\in\Omega_{n}\ \text{ for large } \ n. \end{aligned}$$

In particular, this implies that

and so, by the Fatou Lemma, for every 0 < r ≤ R, we have

$$\displaystyle \begin{aligned} |\Omega_{\infty}\cap B_r|\le \liminf_{n\to\infty}|\Omega_{n}\cap B_r|, \end{aligned} $$

(6.3)

with an equality, if and only if, (again, up to a subsequence) converges strongly to in L ¹(B _r). Notice that, up to extracting a subsequence we may assume that the limits in the right-hand sides of (6.3) and (6.2) do exist.

In order to prove (i) and (ii), it is sufficient to prove that, for fixed 0 < r < R, we have

$$\displaystyle \begin{aligned} \|\nabla u_{\infty}\|{}_{L^2(B_r)}=\liminf_{n\to\infty}\|\nabla u_{n}\|{}_{L^2(B_r)}\qquad \text{and}\qquad |\Omega_\infty\cap B_r|=\liminf_{n\to\infty}|\Omega_n\cap B_r|. \end{aligned} $$

(6.4)

Let $\eta :B_R\to \mathbb {R}$ be a function such that

$$\displaystyle \begin{aligned} \eta\in C^\infty(B_R)\,,\quad 0\le\eta\le 1\quad \text{in}\quad B_R\,,\quad \eta=1\quad \text{on}\quad \partial B_R\,,\quad \eta=0\quad \text{on}\quad B_r\,. \end{aligned} $$

(6.5)

Consider the test function $\tilde u_n=\eta u_{n}+(1-\eta )u_\infty$. Since u _n is a local minimizer for $\mathcal F_\Lambda$ in B _R, and since $u_n=\tilde u_n$ on ∂B _R, we have $\mathcal F_\Lambda (u_n,B_R)\le \mathcal F_\Lambda (\tilde u_n,B_R)+{\varepsilon }_n$, that is,

$$\displaystyle \begin{aligned} 0\le \int_{B_{R}}|\nabla \tilde u_n|{}^2\,dx-\int_{B_{R}}|\nabla u_n|{}^2\,dx+ \Lambda|\tilde\Omega_n\cap B_R|-\Lambda|\Omega_n\cap B_{R}|+{\varepsilon}_n, \end{aligned}$$

where we have set $\tilde \Omega _n:=\{\tilde u_n>0\}$. We first estimate

$$\displaystyle \begin{aligned} |\tilde\Omega_n\cap B_R|-| \Omega_n\cap B_{R}| & =|\tilde\Omega_n\cap \{\eta=0\}|-| \Omega_n\cap \{\eta=0\}|\\ & \qquad +|\tilde\Omega_n\cap \{\eta>0\}|-|\Omega_n\cap \{\eta>0\}|\\ & =|\Omega_\infty\cap \{\eta=0\}|-|\Omega_n\cap \{\eta=0\}|\\ & \qquad +|(\Omega_n\cup\Omega_\infty)\cap \{\eta>0\}|-|\Omega_n\cap \{\eta>0\}|\\ & \le|\Omega_\infty\cap \{\eta=0\}|-|\Omega_n\cap \{\eta=0\}|-|\{\eta>0\}|. \end{aligned} $$

By the Fatou Lemma on the set , we have that

and so, we get

$$\displaystyle \begin{aligned} \limsup_{n\to\infty}\Big(|\tilde\Omega_n\cap B_R|-| \Omega_n\cap B_{R}|\Big)\le \limsup_{n\to\infty}\Big(|\Omega_\infty\cap B_r|-|\Omega_n\cap B_r|\Big)-|\{\eta>0\}|. \end{aligned} $$

(6.6)

We next calculate

$$\displaystyle \begin{aligned} |\nabla \tilde u_n|{}^2-|\nabla u_n|{}^2 & =\big|\nabla(\eta u_{n}+(1-\eta)u_\infty)\big|{}^2-|\nabla u_n|{}^2\\ & =\big|(u_{n}-u_\infty)\nabla\eta+\eta\nabla u_n+(1-\eta)\nabla u_\infty\big|{}^2-|\nabla u_n|{}^2. \end{aligned} $$

Now since u _n → u _∞ strongly in L ²(B _R), we have that

By the weak H ¹ convergence of u _n to u _∞ on the set , we have

$$\displaystyle \begin{aligned} {} \limsup_{n\to\infty}\int_{B_{R}}\Big(|\nabla \tilde u_n|{}^2-|\nabla u_n|{}^2\Big)\,dx\le\limsup_{n\to\infty}\int_{B_r}\Big(|\nabla u_\infty|{}^2-|\nabla u_n|{}^2\Big)\,dx + \int_{\{\eta>0\}}|\nabla u_\infty|{}^2\,dx. \end{aligned}$$

This estimate, together with (6.6) and the minimality of u _n, gives

$$\displaystyle \begin{aligned} \liminf_{n\to\infty}\mathcal F_\Lambda(u_n, B_r) & =\liminf_{n\to\infty}\int_{B_r}|\nabla u_n|{}^2\,dx+\Lambda|\Omega_n\cap B_r|\\ & \le \int_{B_r}|\nabla u_\infty|{}^2\,dx+\Lambda|\Omega_\infty\cap B_r|+\int_{\{\eta>0\}}|\nabla u_\infty|{}^2\,dx+\Lambda|\{\eta>0\}|\\ & =\mathcal F_\Lambda(u_\infty, B_r)+\int_{\{\eta>0\}}|\nabla u_\infty|{}^2\,dx+\Lambda|\{\eta>0\}|. \end{aligned} $$

Since η is arbitrary, we finally obtain

$$\displaystyle \begin{aligned} \liminf_{n\to\infty}\mathcal F_\Lambda(u_n, B_r)\le \mathcal F_\Lambda(u_\infty, B_r), \end{aligned}$$

which implies (6.4) and, as a consequence, the claims (i) and (ii).

We now prove (iii). Let 0 < r < R and $\varphi \in H^1_0(B_r)$. We will show that

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(u_\infty, B_r)\le \mathcal F_\Lambda(u_\infty+\varphi, B_r). \end{aligned} $$

(6.7)

In order to prove(6.7), we will use the optimality of u _n and we will pass to the limit. We notice that, for a fixed n ≥ 1, the natural competitor is simply u _n + φ. Unfortunately, we cannot follow this strategy since we do NOT a priori know that

$$\displaystyle \begin{aligned} \displaystyle\lim_{n\to\infty}|\{u_n+\varphi>0\}|=|\{u_\infty+\varphi>0\}|. \end{aligned}$$

Thus, we consider a function $\eta :B_R\to \mathbb {R}$ that satisfies (6.5) and is such that the set $\mathcal N:=\{\eta <1\}$ is a ball strictly contained in B _R. Precisely, we have that the following inclusions do hold:

$$\displaystyle \begin{aligned} \{\varphi\neq0\}\subset B_r\subset\{\eta=0\}\subset \mathcal N=\{\eta<1\}\subset B_R, \end{aligned}$$

the last two inclusions being strict. We define the competitor

$$\displaystyle \begin{aligned} v_n=u_n+\varphi+(1-\eta)(u_\infty-u_n), \end{aligned}$$

and we set for simplicity v _∞ := u _∞ + φ. Now, since φ = 0 on , we have that v _n = v _∞ on the set {η = 0} and (6.7) is equivalent to

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(u_\infty,\mathcal N)\le \mathcal F_\Lambda(v_\infty,\mathcal N). \end{aligned} $$

(6.8)

By the points (i) and (ii), we have that

$$\displaystyle \begin{aligned} {} \mathcal F_\Lambda(u_\infty,\mathcal N)=\lim_{n\to\infty}\mathcal F_\Lambda(u_n,\mathcal N). \end{aligned}$$

The optimality of u _n and the strong H ¹ convergence of u _n to u _∞ in $\mathcal N$ give

$$\displaystyle \begin{aligned} \lim_{n\to\infty} \mathcal F_\Lambda(u_n,\mathcal N)\le\liminf_{n\to\infty}\mathcal F_\Lambda(v_n,\mathcal N)=\int_{\mathcal N}|\nabla v_\infty|{}^2\,dx+\Lambda\,\liminf_{n\to\infty} |\{v_n>0\}\cap \mathcal N|. \end{aligned} $$

(6.9)

Moreover, since

$$\displaystyle \begin{aligned} v_n=v_\infty\quad \text{on the set}\quad \{\eta=0\}, \end{aligned}$$

we have

$$\displaystyle \begin{aligned} |\{v_n>0\}\cap \mathcal N| & = |\{v_n>0\}\cap\{\eta=0\}|+|\{v_n>0\}\cap \{0<\eta<1\}|\\ & \le |\{v_\infty>0\}\cap \mathcal N|+|\{0<\eta<1\}|, \end{aligned} $$

which, together with (6.9) and (6.8), gives

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(u_\infty,\mathcal N)=\lim_{n\to\infty} \mathcal F_\Lambda(u_n,\mathcal N)\le \mathcal F_\Lambda(v_\infty,\mathcal N)+|\{0<\eta<1\}|. \end{aligned} $$

Now, since the set {0 < η < 1} is arbitratry, we get (6.8) and so, the claim (iii). □

6.2 Convergence of the Free Boundary

This section is dedicated to the proof of Proposition 6.2 (iii). In particular, we define the notion of local Hausdorff convergence (see Definition 6.4 below) and we prove several results, which are general and can be used in the context of different free boundary problems.

Definition 6.4 (Local Hausdorff Convergence)

Suppose that X _n is a sequence of closed sets in $\mathbb {R}^d$ and Ω is an open subset of $\mathbb {R}^d$. We say that X _n converges locally Hausdorff in Ω to (the closed set) X, if for every compact set $\mathcal K\subset \Omega$ and every open set $\mathcal U$, such that $\mathcal K\subset \mathcal U\subset \Omega$, we have

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\mathrm{dist}_{\,\mathcal K, \mathcal U}(X_n,X)=0, \end{aligned}$$

where, for any pair of closed subset X, Y  of Ω, we define

$$\displaystyle \begin{aligned} \text{dist}_{\,\mathcal K, \mathcal U}(X,Y):=\max\Big\{\max_{x\in X\cap\mathcal K}\mathrm{dist}\,(x,Y\cap\mathcal U),\max_{y\in Y\cap\mathcal K}\mathrm{dist}\,(y,X\cap\mathcal U)\Big\}. \end{aligned}$$

Lemma 6.5 (Hausdorff Convergence of the Supports)

Let B _R be the ball of radius R in $\mathbb {R}^d$ . Let $u_n:B_{2R}\to \mathbb {R}$ be a sequence of continuous non-negative functions such that:

• (a) u _n converges uniformly in B _2R to the continuous non-negative function $u_0:B_{2R}\to \mathbb {R}$;

• (b) u _n is uniformly non-degenerate, that is, there is a strictly increasing function

  $$\displaystyle \begin{aligned} \underline\omega:[0,+\infty)\to[0,+\infty), \end{aligned}$$

  such that $\underline \omega (0)=0$ and

  $$\displaystyle \begin{aligned} \|u_n\|{}_{L^\infty(B_r(x_0))}\ge \underline\omega(r)\quad \mathit{\text{for every}}\quad x_0\in \overline\Omega_{u_n}\cap B_{\frac{3R}2}\ ,\quad r\in(0,\frac{R}2)\quad \mathit{\text{and}}\quad n\in\mathbb{N}\,. \end{aligned}$$

Then the sequence of closed sets $\overline \Omega _{u_n}$ converges locally Hausdorff in B _R to $\overline \Omega _{u_0}$.

Proof

We first prove the non-degeneracy of u ₀. Suppose that $x\in \overline \Omega _{u_0}\cap \overline B_{R}$ and r ≤ R∕2. Then, there is y ∈ B _r∕2(x) such that u ₀(y) > 0 and so, for n large enough we have that u _n(y) > 0. By the non-degeneracy of u _n, there is a point z _n ∈ B _r∕2(y) such that $u_n(z_n)\ge \underline \omega (\frac {r}2)$. Up to a subsequence z _n converges to some $z\in \overline B_{\frac {r}2}(y)$. By the uniform convergence of u _n we have

$$\displaystyle \begin{aligned} u_0(z)=\lim_{n\to\infty}u_n(z_n)\ge\underline\omega(\frac{r}2), \end{aligned}$$

which proves that

$$\displaystyle \begin{aligned} \|u_0\|{}_{L^\infty(B_r(x))}\ge \underline\omega(\frac{r}2)\quad \text{for every}\quad x\in\overline\Omega_{u_0}\cap \overline B_{R}\quad \text{and every}\quad r\le \frac{R}2\,. \end{aligned}$$

We can now prove the local Hausdorff convergence of $\overline \Omega _{u_n}$ to $\overline \Omega _{u_0}$. Let $\mathcal K\subset B_R$ be a given compact set and $\mathcal U\subset B_R$ be an open set containing $\mathcal K$. Let δ > 0 be the distance from $\mathcal K$ to the boundary of $\mathcal U$. We reason by contradiction. Indeed, suppose that there is ε > 0 such that $\displaystyle \text{dist}_{\,\mathcal K, \mathcal U}\big (\overline \Omega _{u_n},\overline \Omega _{u_0}\big )>{\varepsilon }.$ Then, up to extracting a subsequence, we can assume that one of the following does hold:

• (1) There is a sequence (x _n)_n such that

  $$\displaystyle \begin{aligned} x_n\in \overline\Omega_{u_n}\cap\mathcal K\quad \text{and}\quad \mathrm{dist}\,(x_n,\overline\Omega_{u_0}\cap\mathcal U)\ge{\varepsilon}. \end{aligned}$$

• (2) There is a sequence (x _n)_n such that

  $$\displaystyle \begin{aligned} x_n\in \overline\Omega_{u_0}\cap\mathcal K\quad \text{and}\quad \mathrm{dist}\,(x_n,\overline\Omega_{u_n}\cap\mathcal U)\ge{\varepsilon}. \end{aligned}$$

Moreover, we can assume that 0 < ε ≤ δ.

Suppose that (1) holds. Since $x_n\in \overline \Omega _{u_n}$ we have that there is $y_n\in B_{\frac {\varepsilon }2}(x_n)\subset \mathcal U$ such that $u_n(y_n)> \underline \omega (\frac {{\varepsilon }}2)$. On the other hand, (1) implies that u ₀(y _n) = 0, in contradiction with the uniform convergence of u _n to u ₀.

Suppose that (2) holds. By the non-degeneracy of u ₀ we have that there is $y_n\in B_{\frac {\varepsilon }2}(x_n)\subset \mathcal U$ such that $u_0(y_n)\ge \underline \omega (\frac {{\varepsilon }}4)$. On the other hand u _n(y _n) = 0, in contradiction with the uniform convergence of u _n to u ₀. □

Lemma 6.6 (Hausdorff Convergence of the Zero Level Sets)

Let B _R be the ball of radius R in $\mathbb {R}^d$ . Let $u_n:B_{2R}\to \mathbb {R}$ be a sequence of continuous non-negative functions such that:

• (a) u _n converges uniformly in B _2R to the continuous non-negative function $u_0:B_{2R}\to \mathbb {R}$;

• (b) u _n(0) = 0 and u _n satisfies the following uniform growth condition:

  $$\displaystyle \begin{aligned} u_n(x)\ge \underline\omega\big(\mathrm{dist}\,(x,\{u_n=0\}\cap \overline B_{2R})\big)\quad \mathit{\text{for every}}\quad x\cap \overline B_R\quad \mathit{\text{and every}}\quad n\in\mathbb{N}\,, \end{aligned}$$

  where $\underline \omega :[0,+\infty )\to [0,+\infty )$ is a strictly increasing function such that $\underline \omega (0)=0$.

Then the sequence of closed sets {u _n = 0} converges locally Hausdorff in B _R to {u ₀ = 0}.

Proof

Let $\mathcal K\subset B_R$ be a compact set and let $\mathcal U\subset B_R$ be an open set containing $\mathcal K$. Let δ > 0 be the distance from $\mathcal K$ to the boundary $\partial \mathcal U$. We reason by contradiction and we suppose that there is ε ∈ (0, δ) such that

$$\displaystyle \begin{aligned} \displaystyle\text{dist}_{\,\mathcal K, \mathcal U}\big(\{u_n=0\},\{u_0=0\}\big)\ge{\varepsilon}. \end{aligned}$$

Then, up to a subsequence, we have one of the following possibilities:

• (1) There is a sequence (x _n)_n such that

  $$\displaystyle \begin{aligned} x_n\in \{u_n=0\}\cap\mathcal K\quad \text{and}\quad \text{dist}(x_n,\{u_0=0\}\cap\mathcal U)\ge {\varepsilon}. \end{aligned}$$

• (2) There is a sequence (x _n)_n such that

  $$\displaystyle \begin{aligned} x_n\in \{u_0=0\}\cap\mathcal K\quad \text{and}\quad \text{dist}(x_n,\{u_n=0\}\cap\mathcal U)\ge {\varepsilon}. \end{aligned}$$

Suppose first that (1) holds. Up to extracting a subsequence, we can suppose that x _n converges to $x_0\in \mathcal K$. By the uniform convergence of u _n and the continuity of u ₀, we have

$$\displaystyle \begin{aligned} u_n(x_0)\le u_n(x_n)+|u_0(x_n)-u_n(x_n)|+|u_0(x_0)-u_0(x_n)|+|u_n(x_0)-u_0(x_0)|\to 0. \end{aligned}$$

Passing to the limit as n →∞, we get that u ₀(x ₀) = 0, which is a contradiction since

$$\displaystyle \begin{aligned} \text{dist}\,(x_0,\{u_0=0\}\cap\mathcal U)\ge \lim_{n\to\infty}\text{dist}\,(x_n,\{u_0=0\}\cap\mathcal U)\ge {\varepsilon}. \end{aligned}$$

Suppose now that (2) holds. Now, let y _n be the point in B _2R ∩{u _n = 0} that realizes the distance from x _n to this set. There are two possibilities:

• . In this case, we have |x _n − y _n|≥ δ.

• $y_n\in \mathcal U$. Then, we have $\text{dist}(x_n,\{u_n=0\}\cap \mathcal U)=|x_n-y_n|\ge {\varepsilon }$.

In both cases, we have that |x _n − y _n|≥ ε. By the uniform growth condition (b), we have

$$\displaystyle \begin{aligned} u_n(x_n)\ge \underline\omega(|x_n-y_n|)\ge \underline\omega({\varepsilon}), \end{aligned}$$

which is a contradiction with the uniform convergence of u _n to u ₀. □

Lemma 6.7 (Hausdorff Convergence of the Free Boundaries)

Let B _R be the ball of radius R in $\mathbb {R}^d$ . Let $u_n:B_{2R}\to \mathbb {R}$ be a sequence of continuous non-negative functions and $u_0:B_{2R}\to \mathbb {R}$ be a continuous non-negative function such that:

• (a) the sequence $\overline \Omega _{u_n}$ converges locally Hausdorff in B _R to $\overline \Omega _{u_0}$;

• (b) the sequence {u _n = 0} converges locally Hausdorff in B _R to {u ₀ = 0}.

Then, $\partial \Omega _{u_n}$ converges locally Hausdorff in B _R to $\partial \Omega _{u_0}$.

Proof

Let $\mathcal K\subset B_R$ be a fixed compact set and $\mathcal U\subset B_R$ be a given open set. Let δ > 0 be the distance between $\mathcal K$ and $\partial \mathcal U$. Let ε ∈ (0, δ) be fixed.

Let $x_0\in \partial \Omega _{u_0}\cap \mathcal K$. By the Hausdorff convergence of $\overline \Omega _{u_n}$ and {u _n = 0}, we get that, for n large enough, there are points

$$\displaystyle \begin{aligned} y_n\in \overline\Omega_{u_n}\cap \mathcal U\qquad \text{and}\qquad z_n\in \{u_n=0\}\cap \mathcal U, \end{aligned}$$

such that

$$\displaystyle \begin{aligned} |x_0-y_n|<{\varepsilon}\qquad \text{and}\qquad |x_0-z_n|<{\varepsilon}. \end{aligned}$$

Since u _n is continuous, there is a point w _n on the segment [y _n, z _n] such that $w_n\in \partial \Omega _{w_n}$. Moreover, by construction $w_n\in B_{\varepsilon }(x_0)\subset \mathcal U$. Since x ₀ is arbitrary, we get that

$$\displaystyle \begin{aligned} \max_{x\in \partial\Omega_{u_0}\cap\,\mathcal K}\text{dist}\,(x,\partial\Omega_{u_n}\cap\mathcal U)<{\varepsilon}. \end{aligned}$$

Conversely, let $x_n\in \partial \Omega _{u_n}\cap \mathcal K$ be fixed. Using again the Hausdorff convergence of $\overline \Omega _{u_n}$ and {u _n = 0}, we get that, for n large enough, there are points

$$\displaystyle \begin{aligned} y_0\in \overline\Omega_{u_0}\cap\,\mathcal U\qquad \text{and}\qquad z_0\in \{u_0=0\}\cap\,\mathcal U, \end{aligned}$$

such that

$$\displaystyle \begin{aligned} |x_n-y_0|<{\varepsilon}\qquad \text{and}\qquad |x_n-z_0|<{\varepsilon}. \end{aligned}$$

Now, by the continuity of u ₀, we get that there is a point w ₀ on the segment [y ₀, z ₀] such that $w_0\in \partial \Omega _{w_0}\cap B_{\varepsilon }(x_n)$. Since x _n is arbitrary, we get

$$\displaystyle \begin{aligned} \max_{x\in \partial\Omega_{u_n}\cap\,\mathcal K}\text{dist}\,(x,\partial\Omega_{u_0}\cap\mathcal U)<{\varepsilon}, \end{aligned}$$

which concludes the proof. □

6.3 Proof of Proposition 6.2

By the local Lipschitz continuity of u, we have that for any fixed R > 0, the sequence $u_n=u_{x_0,r_n}$ is uniformly bounded in H ¹(B _R). Thus, applying Lemma 6.3, we get at once the claims (i), (ii) and (iv) of Proposition 6.2. We notice that the fact that the blow-up limit is non-trivial (u ₀ ≡ 0) follows by the non-degeneracy of u, which assures that for every n ≥ N and every R > 0, there is a point $x_n\in \overline B_R$ such that $u_{x_0,r_n}(x_n)\ge \kappa$, where κ is a constant that depends only on Λ and the dimension d. The Hausdorff convergence of the free boundary (Proposition 6.2 (iii)) follows by Lemma 6.5; Lemma 6.6 and finally, by Lemma 6.7. Notice that the non-degeneracy condition of Lemma 6.5 follows by Proposition 4.1, while the uniform growth condition of Lemma 6.6 is a consequence of the following lemma (Lemma 6.8).

Lemma 6.8

Let $u:B_{2R}\to \mathbb {R}$ be a continuous non-negative function such that:

• (1) u(0) = 0;

• (2) u satisfies the following non-degeneracy condition:

  $$\displaystyle \begin{aligned} \|u\|{}_{L^\infty(B_r(x))}\ge \kappa r\quad \mathit{\text{for every}}\quad x\in \overline\Omega_u\cap B_R\quad \mathit{\text{and every}}\quad r\in(0,R), \end{aligned}$$

  for some given constant κ > 0;

• (3) u is harmonic in Ω _u ∩ B _2R.

Then, u satisfies the following growth condition:

$$\displaystyle \begin{aligned} u(x)\ge \frac{\kappa}{2^{d+1}}\mathrm{dist}\,(x,\{u=0\}\cap B_{2R})\quad \mathit{\text{for every}}\quad x\in B_R\,. \end{aligned}$$

Proof

Suppose that x ₀ ∈ Ω_u ∩ B _R and let y ₀ ∈ ∂ Ω_u ∩ B _2R be such that

$$\displaystyle \begin{aligned} r:=|x_0-y_0|=\mathrm{dist}\,(x_0,\{u=0\}\cap B_{2R}). \end{aligned}$$

Then, the non-degeneracy of u implies that there is a point z ₀ ∈ B _r∕2(x ₀) at which

$$\displaystyle \begin{aligned} u(z_0)\ge \kappa \frac{r}2. \end{aligned}$$

Now, since u is harmonic in Ω_u, we get

$$\displaystyle \begin{aligned} \int_{B_{\frac{r}2}(z_0)}u(x)\,dx\ge |B_{\frac{r}2}| u(z_0)\ge \frac{\kappa\,\omega_d\, r^{d+1}}{2^{d+1}}. \end{aligned}$$

Since u is non-negative and harmonic in B _r(x ₀) , we have that

$$\displaystyle \begin{aligned} u(x_0)=\frac{1}{|B_r|}\int_{B_{r}(x_0)}u(x)\,dx\ge \frac{1}{|B_r|} \int_{B_{\frac{r}2}(z_0)}u(x)\,dx\ge \frac{\kappa\,\omega_d\, r^{d+1}}{\omega_d\,r^d\,2^{d+1}}=\frac{\kappa}{2^{d+1}}r, \end{aligned}$$

which concludes the proof. □

As an immediate corollary, we obtain the following well-known result (see for instance [3]), which we give here for the sake of completeness.

Corollary 6.9

Suppose that u is a (non-negative) minimizer of $\mathcal F_\Lambda$ in the ball $B_{2R}\subset \mathbb {R}^d$ such that u(0) = 0. Then, there are constants C ₁ and C ₂ , depending only on Λ and d, such that the following inequality does hold:

$$\displaystyle \begin{aligned} C_1\,\mathrm{dist}\,(x,\{u=0\}\cap B_{2R})\le u(x)\le C_2\,\mathrm{dist}\,(x,\{u=0\}\cap B_{2R})\quad \mathit{\text{for every}}\quad x\in B_R\,. \end{aligned}$$

Proof

The first inequality follows by Lemma 6.8, while the second one is due to the Lipschitz continuity of u (see Theorem 3.1). □

6.4 Regular and Singular Parts of the Free Boundary

In this section, we define the regular and the singular parts of the free boundary.

We notice that we will use exactly the same definition of regular and singular parts in Theorems 1.2, 1.4, 1.9 and 1.10.

Let D be a bounded open set in $\mathbb {R}^d$ and let $u:D\to \mathbb {R}$ be a non-negative continuous function (in particular, one can take u to be a minimizer of $\mathcal F_\Lambda$ in D). Let x ₀ be a fixed point on the free boundary ∂ Ω_u ∩ D, where Ω_u = {u > 0}.

Definition 6.10 (Decomposition of the Free Boundary)

We say that x ₀ is a regular point if there exists a blow-up limit u ₀ of u at x ₀ (see Definition 6.1) of the form

$$\displaystyle \begin{aligned} u_0(x)=\sqrt{\Lambda}\,(x\cdot\nu)_+\quad \text{for every}\quad x\in\mathbb{R}^d, \end{aligned}$$

for some unit vector $\nu \in \mathbb {R}^d$. We will denote the set of all regular points x ₀ ∈ ∂ Ω_u ∩ D by Reg(∂ Ω_u), and we define the singular part of the free boundary as

In Chap. 8, we will prove that Reg(∂ Ω_u) is an open subset of ∂ Ω_u and is a C ^{1, α}-regular surface in $\mathbb {R}^d$. In this section, we will prove that the reduced boundary ∂ ^∗ Ω_u is actually a subset of the regular part Reg(∂ Ω_u) and (as a consequence) that the singular set is small. Precisely, we will show that

$$\displaystyle \begin{aligned} \mathcal{H}^{d-1}\big(Sing(\partial\Omega_u)\big)=0. \end{aligned}$$

We start with the following lemma.

Lemma 6.11

Let D be a bounded open set in $\mathbb {R}^d$ and u be a minimizer of $\mathcal F_\Lambda$ in D. Let x ₀ ∈ ∂ Ω _u ∩ D be a free boundary point, for which there exists a unit vector $\nu \in \mathbb {R}^d$ and a vanishing sequence r _n → 0 such that

(6.10)

where $\Omega _n:=\frac 1{r_n}(-x_0+\Omega _u)$ and $H_\nu :=\{x\in \mathbb {R}^d\ :\ x\cdot \nu >0\}$ . Then, x ₀ ∈ Reg(∂ Ω _u).

Proof

Let u _n be the blow-up sequence

$$\displaystyle \begin{aligned} u_n(x):=u_{x_0,r_n}(x)=\frac 1{r_n}u(x_0+r_nx). \end{aligned}$$

Notice that Ω_n = {u _n > 0}. By Proposition 6.2, we have that, up to a subsequence and for every R > 0, u _n converges locally uniformly in B _R and strongly in H ¹ to a function u ₀, which is a non-negative Lipschitz continuous global minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^d$. Moreover, we have that the sequence of characteristic functions converges in L ¹(B _R) to . In particular, this implies that $\Omega _{u_0}=H_\nu$ almost everywhere. Now, the minimality of u ₀ and the fact that |{u ₀ = 0}∩ H _ν| = 0 implies that u ₀ is harmonic in H _ν. By the maximum principle, we get that $\Omega _{u_0}=H_\nu$. Thus, u ₀ is C ^∞ smooth up to the boundary ∂H _ν (where it vanishes).

We will next prove that

$$\displaystyle \begin{aligned} \nabla u_0=\sqrt{\Lambda}\,\nu\quad \text{on}\quad \partial H_\nu. \end{aligned}$$

Indeed, suppose that this is not the case. Then, there are two possibilities:

• (1) there is a point y ∈ ∂H _ν such that ∇u ₀ = Aν for some $A>\sqrt {\Lambda }$;

• (2) there is a point y ∈ ∂H _ν such that ∇u ₀ = Bν for some $0<B<\sqrt {\Lambda }$.

Suppose that (1) holds. Let h _r,R be the radial solution from Proposition 2.15, where r is large enough and R = R(r) is uniquely determined by r. Recall that:

$$\displaystyle \begin{aligned} r<R\,,\qquad \lim_{r\to\infty}|R-(r+1)|=0\,,\qquad \{h_{r,R}>0\}=B_R\,, \end{aligned}$$

Moreover, the function $\sqrt {\Lambda }\,h_{r,R}$ is a local minimizer of $\mathcal F_\Lambda$ in . Let $y_r\in \mathbb {R}^d$ be such that the ball B _R(y _r) is contained in H _ν and is tangent to ∂H _ν at y. Let r > 0 be fixed and such that

$$\displaystyle \begin{aligned} |\nabla h_{r,R}|<\frac 12+\frac{A}{2\sqrt{\Lambda}}. \end{aligned}$$

Then, there is ε > 0 small enough, for which the function

$$\displaystyle \begin{aligned} \tilde h(x):=\sqrt{\Lambda}\,h_{r,R}(x+{\varepsilon}\nu) \end{aligned}$$

satisfies the following conditions:

• the support of $\tilde h$ is not entirely contained in H _ν, that is, $\{\tilde h>0\}\cap \{u_0=0\}\neq \emptyset$;

• $\tilde h>u_0$ only in a small neighborhood of y, precisely, $\{\tilde h>u_0\}\subset B_{{1}/{2}}(y)$.

Next, we notice that both $\tilde h$ and u ₀ are minimizers of $\mathcal F_\Lambda$ in B := B _1∕2(y). Since, by construction u ₀ ≥ h on ∂B, we get that

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(\tilde h,B)\le \mathcal F_\Lambda(u_0\wedge \tilde h,B)\qquad \text{and}\qquad \mathcal F_\Lambda(u_0,B)\le \mathcal F_\Lambda(u_0\vee \tilde h,B). \end{aligned} $$

(6.11)

On the other hand,

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(\tilde h,B)+ \mathcal F_\Lambda(u_0,B)=\mathcal F_\Lambda(u_0\wedge \tilde h,B)+ \mathcal F_\Lambda(u_0\vee \tilde h,B), \end{aligned}$$

which means that both the inequalities in (6.11) are equalities and that both the functions $\tilde h\wedge u_0$ and $\tilde h\vee u_0$ are minimizers of $\mathcal F_\Lambda$ in B. For instance, this means that $\tilde h\vee u_0$ is harmonic in the set $\{\tilde h>0\}\cap B$, which is impossible since by construction $\tilde h\vee u_0$ is not C ¹ (for instance, the gradient is not continuous on the segment [y, y _r]). Thus, (1) cannot happen. By the same argument, also (2) cannot happen, which means that

$$\displaystyle \begin{aligned} |\nabla u_0|=\sqrt{\Lambda}\quad \text{on}\quad \partial H_\nu. \end{aligned}$$

Now, by the unique continuation principle we have that $u_0(x)=\sqrt {\Lambda }\,(x\cdot \nu )$ on H _ν. Indeed, the function $\tilde u_0$, defined as

is harmonic in the entire space $\mathbb {R}^d$ and so, it should coincide everywhere with the function $x\mapsto \sqrt {\Lambda }\, (x\cdot \nu )$. This concludes the proof. □

Proposition 6.12 (The Singular Set Is Negligible)

Let D be a bounded open set in $\mathbb {R}^d$ and u ∈ H ¹(D) be a minimizer of $\mathcal F_\Lambda$ in D. Then, $\mathcal {H}^{d-1}\big (Sing(\partial \Omega _u)\big )=0.$

Proof

By Proposition 5.3, Ω_u has locally finite perimeter in D. Let ∂ ^∗ Ω_u be the reduced boundary of Ω_u. It is well-known (see for instance [43, Theorem 5.15]) that, for every x ₀ ∈ ∂ ^∗ Ω_u, there is a unit vector $\nu \in \mathbb {R}^d$ such that the property (6.10) does hold. Thus, by Lemma 6.11, we have that ∂ ^∗ Ω_u ⊂ Reg(∂ Ω_u). On the other hand, by the Second Theorem of Federer (see [43]), we have that

(6.12)

Recall that, by Lemma 5.1, there are no points of density 1 and 0 on the free boundary, that is,

$$\displaystyle \begin{aligned} (\partial\Omega_u\cap D) \cap \big(\Omega_u^{(1)}\cup \Omega_u^{(0)}\big)=\emptyset. \end{aligned}$$

Thus, by (6.12)

Now, by the definition of the singular part, we have

which concludes the proof. □