Dimension of the Singular Set

Abstract

In this chapter, we prove Theorem 1.4. As in the original work of Weiss (see [52]), we will use the so-called Federer’s dimension reduction principle, which first appeared in [32].

In this chapter, we prove Theorem 1.4. As in the original work of Weiss (see [52]), we will use the so-called Federer’s dimension reduction principle, which first appeared in [32].

This chapter is organized as follows.

• In Sect. 10.1 we give the definitions of the Hausdorff measure and Hausdorff dimension; we also state and prove the main properties of the Hausdorff measure, which we will need for the proof of Theorem 1.4.

• In Sect. 10.2 we give a general result for the convergence of the singular sets of a sequence of functions.

• In Sect. 10.3 we study the structure of the singular set of the one-homogeneous global minimizers of $\mathcal F_\Lambda$.

• Finally, in Sect. 10.4, we use the results of the previous subsections (Lemmas 10.7 and 10.12) to prove Theorem 1.4.

10.1 Hausdorff Measure and Hausdorff Dimension

In this section we define the notions of Hausdorff measure and Hausdorff dimension and we also give their main properties. For more details, we refer to the book [31].

We recall that, for every s > 0, δ ∈ (0, +∞] and every set $E\subset \mathbb {R}^d$,

$$\displaystyle \begin{aligned} \begin{array}{ll} \mathcal{H}^{s}_\delta(E):= & {\displaystyle\frac{\omega_{s}}{2^s}}\,\inf\Big\{{\displaystyle\sum_{j=1}^\infty\left(\text{diam}\,U_j\right)^{s}}\ :\ \text{for every family of sets } \{U_j\}_{j=1}^\infty\\ & \qquad \qquad \displaystyle \text{such that}\ E\subset\bigcup_{j=1}^\infty U_j\ \text{and}\ \text{diam}\,U_j\le\delta,\ \text{for every}\ j\ge 1\Big\}, \end{array} \end{aligned} $$

(10.1)

where, for any s ∈ (0, +∞), the constant ω _s is defined as

$$\displaystyle \begin{aligned} \omega_s:=\frac{\pi^{\frac{s}2}}{\Gamma(\frac{s}2+1)}\qquad \text{where}\qquad \Gamma(s):=\int_0^{+\infty} x^{s-1}e^x\,dx. \end{aligned}$$

Definition 10.1 (Hausdorff Measure)

For any s ≥ 0, $\mathcal {H}^s(E)$ denotes the s-dimensional Hausdorff measure of a set $E\subset \mathbb {R}^d$ and is defined as:

$$\displaystyle \begin{aligned} \mathcal H^{s}(E):=\lim_{\delta\to0_+}\mathcal{H}^{s}_\delta(E)=\sup_{\delta>0}\mathcal{H}^{s}_\delta(E). \end{aligned}$$

Remark 10.2

The constant in (10.1) is chosen in such a way that we have

$$\displaystyle \begin{aligned} \mathcal{H}^d(B_r)=|B_r|=\omega_dr^d\qquad \text{and}\qquad \mathcal{H}^{d-1}(\partial B_r)=d\omega_dr^{d-1}. \end{aligned}$$

Definition 10.3

The Hausdorff dimension of a set $E\subset \mathbb {R}^d$ is defined as

$$\displaystyle \begin{aligned} \dim_{\mathcal H} E:=\inf\big\{s>0\ :\ \mathcal{H}^s(E)=0\big\}. \end{aligned}$$

The following elementary properties of the Hausdorff measure are an immediate consequence of the definitions of $\mathcal H^s$, $\mathcal H^s_\delta$ and $\mathcal H^s_\infty$.

Proposition 10.4 (Properties of the Hausdorff Measure)

• (i) For every s > 0 and δ ∈ (0, ∞], the set functionals $\mathcal H^s$ and $\mathcal H^s_\delta$ are translation invariant and increasing with respect to the set inclusion. Moreover, we have

  $$\displaystyle \begin{aligned} \mathcal{H}^s(rE)=r^s\mathcal{H}^s(E)\quad \mathit{\text{and}}\quad \mathcal{H}^s_\infty(rE)=r^s\mathcal{H}^s_\infty(E)\quad \mathit{\text{for any}}\quad E\subset\mathbb{R}^d\quad \mathit{\text{and}}\quad r>0. \end{aligned}$$

• (ii) The function $\delta \mapsto \mathcal {H}^s_\delta$ is non-decreasing in δ. In particular, we have

  $$\displaystyle \begin{aligned} \mathcal{H}^s(E)\le \mathcal{H}^s_\delta(E)\le \mathcal{H}^s_\infty(E)\quad \mathit{\text{for any}}\quad E\subset\mathbb{R}^d\quad \mathit{\text{and any}}\quad \delta>0. \end{aligned}$$

• (iii) Given s > 0 and $E\subset \mathbb {R}^{d}$ , we have that

  $$\displaystyle \begin{aligned} \mathcal{H}^s(E)=0\quad \mathit{\text{if and only if}}\quad \mathcal{H}^{s}_\infty(E)=0. \end{aligned}$$

• (iv) Given a sequence of sets $E_j\subset \mathbb {R}^d$ , s > 0 and δ ∈ (0, +∞] we have that

  $$\displaystyle \begin{aligned} \mathcal{H}^s_\delta(E)\le \sum_{j=1}^\infty \mathcal{H}^s_\delta(E_j)\qquad \mathit{\text{where}}\qquad E=\bigcup_{j=1}^\infty E_j. \end{aligned}$$

  In particular, $\mathcal {H}^s(E)=0$ if and only if $\mathcal {H}^{s}(E_j)=0$ , for every j ≥ 1.

Lemma 10.5 (Existence of Points of Positive Density)

Let s > 0 and let $K\subset \mathbb {R}^{d}$ be a given set. If $\mathcal {H}^s(K)>0$ , then there is a point x ₀ ∈ K such that

$$\displaystyle \begin{aligned} \limsup_{r\to0}\frac{\mathcal{H}^{s}\big(K\cap B_r(x_0)\big)}{r^s}>0. \end{aligned} $$

(10.2)

Proof

Suppose that (10.2) does not hold. Then, we have

$$\displaystyle \begin{aligned} \limsup_{r\to0}\frac{\mathcal{H}^{s}\big(K\cap B_r(x_0)\big)}{r^s}=0. \end{aligned} $$

(10.3)

Let K _δ,ε ⊂ K be the set

$$\displaystyle \begin{aligned} K_{\delta,{\varepsilon}}=\big\{x\in K\ :\ \mathcal{H}^s(K\cap B_r(x))\le {\varepsilon} r^s\quad \text{for every}\quad r\le \delta\big\}. \end{aligned}$$

By (10.3), we have that

$$\displaystyle \begin{aligned} \bigcup_{\delta>0}K_{\delta,{\varepsilon}}=\bigcup_{n=1}^\infty K_{\delta,\frac1n}=K\qquad \text{for every fixed}\qquad {\varepsilon}>0. \end{aligned} $$

(10.4)

Let now δ and ε be fixed and let {U _i}_i≥1 be a family of sets of diameter diam U _i ≤ δ such that K _δ,ε ⊂⋃ iU _i. Then, the subadditivity of $\mathcal {H}^s_\delta$ gives that

$$\displaystyle \begin{aligned} \mathcal{H}^s_\delta(K_{\delta,{\varepsilon}})& \le \sum_{i=1}^\infty \mathcal{H}^s_\delta(U_i\cap K_{\delta,{\varepsilon}})\le \sum_{i=1}^\infty \mathcal{H}^s(U_i\cap K_{\delta,{\varepsilon}})\\ & \le \sum_{i=1}^\infty \mathcal{H}^s(U_i\cap K)\le \sum_{i=1}^\infty {\varepsilon} \big(\text{diam}\, U_i\big)^s, \end{aligned} $$

where the last inequality holds since the set U _i ∩ K is contained in a ball of radius

$$\displaystyle \begin{aligned} r_i=\text{diam}\, U_i\le \delta. \end{aligned}$$

Taking the infimum over all coverings C _i with sets of diameter less than or equal to δ, we get that

$$\displaystyle \begin{aligned} \mathcal{H}^s_\delta(K_{\delta,{\varepsilon}})\le {\varepsilon}\, \frac{2^s}{\omega_s}\,\mathcal{H}^s_\delta(K_{\delta,{\varepsilon}}), \end{aligned}$$

and so, for ε small enough, $\mathcal {H}^s_\delta (K_{\delta ,{\varepsilon }})=0$, which implies that $\mathcal {H}^s(K_{\delta ,{\varepsilon }})=0$. Finally, (10.4) and the subadditivity of $\mathcal {H}^s$ imply that $\mathcal {H}^s(K)=0$, which is a contradiction. □

Lemma 10.6 (Dimension Reduction: Lemma I)

Let s > 0. Let $E\subset \mathbb {R}^{d-1}$ be a given set and let $\tilde E=E\times \mathbb {R}\subset \mathbb {R}^d$ . If $\mathcal {H}^s(E)=0$ , then also $\mathcal {H}^{s+1}(\tilde E)=0$.

Proof

We will prove that $\mathcal {H}^{s+1}(E\times [0,T])=0$ for every T > 0. In fact, this implies that $\mathcal {H}^{s+1}(E\times [-T,T])=0$ and since

$$\displaystyle \begin{aligned} \displaystyle\tilde E=\bigcup_{T>0}E\times [-T,T],\end{aligned}$$

we get $\mathcal {H}^{s+1}(\tilde E)=0$.

Since $\mathcal {H}^s(E)=0$, for every ε > 0, there is a family of balls $B^{\prime }_{r_i}(x_i)\subset \mathbb {R}^{d-1}$ such that

$$\displaystyle \begin{aligned} E\subset \bigcup_{i\ge 1} B^{\prime}_{r_i}(x_i)\qquad \text{and}\qquad \sum_{i=1}^\infty r_i^s\le{\varepsilon}\,. \end{aligned}$$

Let now T be fixed. For every $i\in \mathbb {N}$, we consider the point $x_{i,k}\in \mathbb {R}^d$ of coordinates x _i,k = (x _i, kr _i), for k = 0, 1, …, K _i, where $K_i:=\left [{T}/{r_i}\right ]+1$ and the family of balls $B_{2r_i}(x_{i,k})$. Notice that

$$\displaystyle \begin{aligned} x'\times[0,T]\subset \bigcup_{k} B_{2r_i}(x_{i,k})\qquad \text{for every}\qquad x'\in B_{r_i}^{\prime}(x_i)\subset\mathbb{R}^{d-1}. \end{aligned}$$

Thus, the family of balls $\{B_{2r_i}(x_{i,k})\}_{i,k}$ is a covering of E × [0, T]. We now estimate,

$$\displaystyle \begin{aligned} \mathcal{H}^{s+1}_\infty\big(E\times[0,T]\big)& \le \sum_{i=1}^\infty\sum_{k=1}^{K_i} \left(2r_i\right)^{s+1}= 2^{s+1} \sum_{i=1}^\infty\sum_{k=1}^{K_i} r_i^{s+1}\\ & = 2^{s+1} \sum_{i=1}^\infty (K_i+1) r_i^{s+1}\le 2^{s+1} \sum_{i=1}^\infty\frac{2T}{r_i} r_i^{s+1}, \end{aligned} $$

where the last inequality follows by the fact that, for T large enough,

$$\displaystyle \begin{aligned} K_i+1\le \frac{T}{r_i}+2\le \frac{2T}{r_i}. \end{aligned}$$

Thus, we get

$$\displaystyle \begin{aligned} \mathcal{H}^{s+1}_\infty\big(E\times[0,T]\big)\le 2^{s+2}T \sum_{i=1}^\infty r_i^{s}\le 2^{s+2}T {\varepsilon}, \end{aligned}$$

which concludes the proof. □

10.2 Convergence of the Singular Sets

In this section we will prove a general result (Lemma 10.7) for the convergence of the singular sets, which applies both to minimizers of $\mathcal F_\Lambda$ (Theorem 1.4) and to measure-constrained minimizers (Theorem 1.9). Recall that, if $D\subset \mathbb {R}^d$ is an open set, $u:D\to \mathbb {R}$ a given (continuous and non-negative) function, then for every ball B _r(x ₀) ⊂ D, we define

$$\displaystyle \begin{aligned} u_{x_0,r}:B_1\to\mathbb{R},\qquad u_{x_0,r}(x)=\frac 1ru(x_0+rx). \end{aligned}$$

We say that a boundary point x ₀ ∈ ∂ Ω_u ∩ D is regular (and we write x ₀ ∈ Reg(∂ Ω_u)), if there is a sequence r _n → 0 such that

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\|u_{x_0,r_n}-h_\nu\|{}_{L^\infty(B_1)}=0\,, \end{aligned}$$

where for simplicity we set

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

and we recall that

$$\displaystyle \begin{aligned} \|u_{x_0,r_n}-h_\nu\big\|{}_{L^\infty(B_1)}=\|u(x)-h_\nu(x - x_0)\|{}_{L_x^\infty(B_r(x_0))}. \end{aligned}$$

We say that a point x ₀ is singular if it is not regular, that is, if

Lemma 10.7 (Convergence of the Singular Sets)

Suppose that $D\subset \mathbb {R}^d$ is a bounded open set. Let $u_n:D\to \mathbb {R}$ be a sequence of continuous non-negative functions satisfying the following conditions:

• (a) Uniform ε -regularity. There are constants ε > 0 and R > 0 such that the following holds:

  if $n\in \mathbb {N}$, $x_0\in \partial \Omega _{u_n}\cap D$ and r ∈ (0, R) are such that B _r(x ₀) ⊂ D and

  $$\displaystyle \begin{aligned} \|u_n-h_\nu(\cdot-x_0)\|{}_{L^\infty(B_r(x_0))}\le{\varepsilon} r \quad \mathit{\text{for some}}\quad \nu\in\partial B_1, \end{aligned}$$

  then $\partial \Omega _{u_n}=\mathit{\text{Reg}}\,(\partial \Omega _{u_n})$ in B _r∕2(x ₀).

• (b) Uniform non-degeneracy. There are constants κ > 0 and r ₀ > 0 such that the following holds: if $n\in \mathbb {N}$, $x_0\in \partial \Omega _{u_n}\cap D$ and r ∈ (0, r ₀) are such that B _r(x ₀) ⊂ D, then

  $$\displaystyle \begin{aligned} \|u_n\|{}_{L^\infty(B_r(x_0))}\ge \kappa\,r\,. \end{aligned}$$

• (c) Uniform convergence. The sequence u _n converges locally uniformly in D to a (continuous and non-negative) function $u_0:D\to \mathbb {R}$.

Then, for every compact set K ⊂ D, the following claim does hold:

$$\displaystyle \begin{aligned} \begin{array}{c} \mathit{\text{ For every open set }} \ U\subset D \mathit{\text{ containing }\ } \ \mathrm{Sing}\,(\partial\Omega_{u_0})\cap K,\\ \mathit{\text{ there exists }} \ n_0\in \mathbb{N} \ \mathit{\text{ such that: }}\\ \mathit{\text{Sing}}\,(\partial\Omega_{u_n})\cap K\subset U\quad \mathit{\text{for every}}\quad n\ge n_0. \end{array} \end{aligned} $$

(10.5)

In particular, for every s > 0,

$$\displaystyle \begin{aligned} \mathcal{H}^s_\infty\big(\mathit{\text{Sing}}\,(\partial\Omega_{u_0})\cap K\big)\ge\limsup_{n\to\infty}\mathcal{H}^s_\infty\big(\mathrm{Sing}\,(\partial\Omega_{u_n})\cap K\big). \end{aligned} $$

(10.6)

Proof

The semicontinuity of the Hausdorff measure (10.6) follows by (10.5) and the definition of $\mathcal {H}^s_\infty$. Thus, it is sufficient to prove (10.5). Arguing by contradiction, we suppose that there are a compact set K ⊂ D and an open set U ⊂ D such that

$$\displaystyle \begin{aligned} \mathrm{Sing}\,(\partial\Omega_{u_0})\cap K\subset U, \end{aligned}$$

but (up to extracting a subsequence of u _n) there is a sequence

Up to extracting a further sequence we may assume that there is a point x ₀ such that

We claim that $x_0\in \partial \Omega _{u_0}$. Indeed, the uniform convergence of u _n implies that u ₀(x ₀) = 0. On the other hand, the non-degeneracy hypothesis (b) implies that, for every r > 0 small enough,

$$\displaystyle \begin{aligned} \|u_0\|{}_{L^\infty(B_r(x_0))}& \ge \liminf_{n\to\infty}\Big(\|u_n\|{}_{L^\infty(B_r(x_0))}-\|u_n-u_0\|{}_{L^\infty(B_r(x_0))}\Big)\\ & \ge \liminf_{n\to\infty}\|u_n\|{}_{L^\infty(B_{\frac{r}2}(x_n))}\ge \kappa \frac{r}{2}, \end{aligned} $$

which gives that $x_0\in \partial \Omega _{u_0}$.

Now, we notice that, since U contains $Sing(\partial \Omega _{u_0})\cap K$ and x ₀∉U, we have that

$$\displaystyle \begin{aligned} x_0\in Reg(\partial\Omega_{u_0}). \end{aligned}$$

By definition of $Reg(\partial \Omega _{u_0})$, there is a sequence r _n → 0 and a unit vector ν ∈ ∂B ₁ such that

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\frac 1{r_n}\|u_0-h_\nu(\cdot-x_0)\|{}_{L^\infty(B_{r_n}(x_0))}=0. \end{aligned}$$

In particular, there exists r ∈ (0, R) such that B _r(x ₀) ⊂ D and

$$\displaystyle \begin{aligned} \|u_0(x)-h_\nu(\cdot-x_0)\|{}_{L^\infty(B_r(x_0))}\le\frac{{\varepsilon}}{3} r \,. \end{aligned}$$

By the continuity of u ₀ and h _ν, we get that, for n large enough,

$$\displaystyle \begin{aligned} \|u_0-h_\nu(\cdot-x_n)\|{}_{L^\infty(B_r(x_n))}\le\frac{2{\varepsilon}}{3} r\,. \end{aligned}$$

Since, u _n converges to u ₀ locally uniformly in D, we get that, for n large enough,

$$\displaystyle \begin{aligned} \|u_n-h_\nu(\cdot-x_n)\|{}_{L^\infty(B_r(x_n))}\le{\varepsilon} r\,. \end{aligned}$$

Thus, (a) implies that $x_n\in \mathrm {Reg}\,(\partial \Omega _{u_n})$, in contradiction with the initial assumption. □

10.3 Dimension Reduction

In this section, we study the singularities of the global one-homogeneous minimizers of $\mathcal F_\Lambda$. In particular, we prove Theorem 1.4 in the case when u is one-homogeneous. This (significant) simplification is essential for the proof of Theorem 1.4 since we already know that the blow-up limits of a local minimizer are global one-homogeneous minimizers and we will prove (see Lemma 10.7) that the dimension of the singular set of a blow-up limit does not decrease if we choose the free boundary point to have non-zero Hausdorff density (see Lemma 10.5).

Remark 10.8 (The Singular Set of a One-Homogeneous Function Is a Cone)

Suppose that $z:\mathbb {R}^d\to \mathbb {R}$ is a non-negative one-homogeneous local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^d$. Then, for any singular free boundary point , we have that $\{tx_0\, :\, t\in \mathbb {R}\}\subset \mathrm {Sing}\,(\partial \Omega _z).$ This claim follows by the fact that Reg (∂ Ω_u) is a cone. and that

Lemma 10.9 (Blow-Up Limits of One-Homogeneous Functions)

Let $z:\mathbb {R}^d\,{\to }\,\mathbb {R}$ be a one-homogeneous locally Lipschitz continuous function. Let 0 ≠ x ₀ ∈ ∂ Ω _z . Let r _n → 0 and $z_{r_n,x_0}$ be a a blow-up sequence converging locally uniformly to a function $z_0:\mathbb {R}^d\to \mathbb {R}$ . Then z ₀ is invariant in the direction x ₀ , that is,

$$\displaystyle \begin{aligned} z_0(x+tx_0)=z_0(x)\quad \mathit{\text{for every}}\quad x\in\mathbb{R}\quad \mathit{\text{and every}}\quad t\in\mathbb{R}. \end{aligned}$$

Proof

Let $t\in \mathbb {R}$ be fixed. Then, we have

$$\displaystyle \begin{aligned} z_0(x+tx_0) & =\lim_{n\to\infty}z_{r_n,x_0}(x+tx_0)=\lim_{n\to\infty}\frac 1{r_n}z\big(x_0+r_n(x+tx_0)\big)\\ & =\lim_{n\to\infty}\frac{1+tr_n}{r_n}z\big(x_0+\frac{r_n}{1+tr_n}x\big)=\lim_{n\to\infty}\frac{1}{r_n}z\big(x_0+r_nx\big)=z_0(x), \end{aligned} $$

where the third inequality follows by the homogeneity of z and the fourth inequality follows by the Lipschitz continuity of z. Precisely, setting $L=\|\nabla z\|{ }_{L^\infty (B_1(x_0))}$, we have

$$\displaystyle \begin{aligned} \Big|\frac{1+tr_n}{r_n}z& \big(x_0+\frac{r_n}{1+tr_n}x\big)-\frac{1}{r_n}z\big(x_0+r_nx\big)\Big|\\ & \le t\left|z\right|\big(x_0+\frac{r_n}{1+tr_n}x\big)+\frac 1{r_n}\Big|z\big(x_0+\frac{r_n}{1+tr_n}x\big)-z\big(x_0+r_nx\big)\Big|\\ & \le t\frac{r_nL|x|}{1+tr_n}+\frac 1{r_n} \frac{tr_n^2L|x|}{1+tr_n}, \end{aligned} $$

which converges to zero as n →∞. □

Lemma 10.10 (Translation Invariant Global Minimizers)

Let $u:\mathbb {R}^{d-1}\to \mathbb {R}$ be a non-negative function, $u\in H^1_{loc}(\mathbb {R}^{d-1})$ and let $\tilde u:\mathbb {R}^d\to \mathbb {R}$ be the function defined by

$$\displaystyle \begin{aligned} \tilde u(x)=u(x')\quad \mathit{\text{for every}}\quad x=(x',x_d)\in\mathbb{R}^d. \end{aligned}$$

Then, u a local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^{d-1}$ if and only if $\tilde u$ a local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^{d}$.

Proof

Suppose first that $\tilde u$ is not a local minimizer of $\mathcal F_\Lambda$. Then, there is a function $\tilde v:\mathbb {R}^d\to \mathbb {R}$ such that $\tilde u=\tilde v$ outside the cylinder $\mathcal C_R:=B_R^{\prime }\times (-R,R)\subset \mathbb {R}^{d-1}\times \mathbb {R}$ and such that $\mathcal F_\Lambda (\tilde u,\mathcal C_R)>\mathcal F_\Lambda (\tilde v,\mathcal C_R)$.

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(u,B_R{\prime}) & =\int_{B_R^{\prime}}|\nabla_{x'}\, u|{}^2\,dx'+\Lambda\,\big|B_R^{\prime}\cap \{u>0\}\big|\\ & =\frac{1}{2R}\left(\int_{\mathcal C_R}|\nabla \tilde u|{}^2\,dx+\Lambda\,\big|\mathcal C_R\cap \{\tilde u>0\}\big|\right)=\frac{1}{2R}\mathcal F_\Lambda(\tilde u,\mathcal C_R)\\ & >\frac{1}{2R}\mathcal F_\Lambda(\tilde v,\mathcal C_R)=\frac{1}{2R}\left(\int_{\mathcal C_R}|\nabla \tilde v|{}^2\,dx+\Lambda\,\big|\mathcal C_R\cap \{\tilde v>0\}\big|\right)\\ & \ge \frac{1}{2R}\int_{-R}^R\left(\int_{B_R^{\prime}}|\nabla_{x'}\, \tilde v(x',x_d)|{}^2\,dx'+\Lambda\,\big|B_R^{\prime}\cap \{\tilde v(\cdot,x_d)>0\}\big|\right)\,dx_d\\ & \ge \int_{B_R^{\prime}}|\nabla_{x'}\, \tilde v(x',t)|{}^2\,dx'+\Lambda\,\big|B_R^{\prime}\cap \{\tilde v(\cdot,t)>0\}\big|\,, \end{aligned} $$

for some t ∈ (−R, R), which exists due to the mean-value theorem. Thus, also u is not a local minimizer of $\mathcal F_\Lambda$.

Conversely, suppose that u is not a local minimizer of $\mathcal F_\Lambda$. Then, there is a function $v:\mathbb {R}^{d-1}\to \mathbb {R}$ such that u = v outside a ball $B_R^{\prime }\subset \mathbb {R}^{d-1}$ and $\mathcal F_\Lambda (u,B_R^{\prime })>\mathcal F_\Lambda (v,B_R^{\prime })$. We now define the function

$$\displaystyle \begin{aligned} \tilde v(x',x_d)=v(x')\phi_t(x_d), \end{aligned}$$

where for any t > 0, we define the function $\phi _t:\mathbb {R}\to [0,1]$ as

$$\displaystyle \begin{aligned} \phi_t(x_d):=\begin{cases} 1\quad \text{if}\quad |x_d|\le t,\\ 0\quad \text{if}\quad |x_d|\ge t+1,\\ x_d+t+1\quad \text{if}\quad -t-1\le x_d\le -t,\\ x_d-t\quad \text{if}\quad t\le x_d\le t+1. \end{cases}\end{aligned}$$

Then,

$$\displaystyle \begin{aligned} \big|\mathcal C_{R,t+1}\cap \{\tilde v>0\}\big|=2(t+1)\big|B_{R}^{\prime}\cap \{v>0\}\big|, \end{aligned}$$

where $\mathcal C_{R,t}:=B_R^{\prime }\times (-t,t)$. Thus, we have

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(\tilde v,\mathcal C_{R,t+1}) & =\int_{\mathcal C_{R,t+1}}|\nabla \tilde v|{}^2\,dx+\Lambda \big|\mathcal C_{R,t+1}\cap \{\tilde v>0\}\big|\\ & \le 2t\mathcal F_\Lambda(v, B_{R}^{\prime})+2\int_{B_R^{\prime}}v^2\,dx'+2\big|B_{R}^{\prime}\cap \{v>0\}\big|. \end{aligned} $$

Choosing t large enough, we have that

$$\displaystyle \begin{aligned} 2t\mathcal F_\Lambda(v, B_{R}^{\prime})+2\int_{B_R^{\prime}}v^2\,dx'+2\big|B_{R}^{\prime}\cap \{v>0\}\big|\le 2t\mathcal F_\Lambda(u, B_{R}^{\prime}). \end{aligned}$$

Since,

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(\tilde u,\mathcal C_{R,t+1})=2(t+1)\mathcal F_\Lambda(u, B_{R}^{\prime}), \end{aligned}$$

we get that

$$\displaystyle \begin{aligned} \mathcal F_\Lambda(\tilde v,\mathcal C_{R,t+1})< \mathcal F_\Lambda(\tilde u,\mathcal C_{R,t+1}), \end{aligned}$$

which concludes the proof. □

Lemma 10.11 (Singular One-Homogeneous Global Minimizers in $\mathbb {R}^{d^\ast }$)

Let $z:\mathbb {R}^{d^\ast }\to \mathbb {R}$ be a non-negative one-homogeneous local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^{d^\ast }$ . Then, one of the following does hold:

• (1) $z(x)=\sqrt {\Lambda }\,(x\cdot \nu )$ for some $\nu \in \mathbb {R}^{d^\ast }$ (in this case Sing (∂ Ω _z) = ∅);

• (2) Sing (∂ Ω _z) = {0}.

In other words,

In particular, this means that $\dim _{\mathcal H}\mathrm {Sing}\,(\partial \Omega _z)=0$.

Proof

Suppose that there is a point such that x ₀ ∈Sing (∂ Ω_z). Then, by Remark 10.8 we have that tx ₀ ∈Sing (∂ Ω_z) for every $t\in \mathbb {R}$. In particular, we can suppose that |x ₀| = 1 and, without loss of generality, we set x ₀ = e _d. Let now z ₀ be a blow-up limit of z at x ₀. Then, z ₀ is a one-homogeneous local minimizer of $\mathcal F_\Lambda$. Moreover, by Lemma 10.9 we have that z ₀(x ^′, t) = z ₀(x ^′, 0) for every $x'\in \mathbb {R}^{d-1}$. Now, Lemma 10.10 implies that the function $z_0^{\prime }:=z_0(\cdot ,0):\mathbb {R}^{d-1}\to \mathbb {R}$ is still a local minimizer of $\mathcal F_\Lambda$. Moreover, the origin $0'\in \mathbb {R}^{d-1}$ is a singular point for $\partial \Omega _{z_0^{\prime }}$ in contradiction with the definition of d ^∗. □

Lemma 10.12 (Dimension Reduction: Lemma II)

Suppose that d ≥ d ^∗ and that $z:\mathbb {R}^{d}\to \mathbb {R}$ is a non-negative one-homogeneous local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^{d}$ . Then,

$$\displaystyle \begin{aligned} \mathcal{H}^{d-d^\ast+s}\big(\mathrm{Sing}\,(\partial\Omega_z)\big)=0\qquad \mathit{\text{for every}}\qquad s>0\,. \end{aligned}$$

Proof

Let s > 0 be fixed. The claim in the case d = d ^∗ follows by Lemma 10.11. We will prove the claim by induction. Indeed, suppose that the claim holds in dimension d − 1, with d − 1 ≥ d ^∗, and let $z:\mathbb {R}^{d}\to \mathbb {R}$ be a non-negative one-homogeneous local minimizer. If such that $\mathcal {H}^{d-d^\ast +s}\big (\mathrm {Sing}\,(\partial \Omega _z)\big )>0$, then, by Lemma 10.5, there is a point x ₀ ∈Sing (∂ Ω_z), a constant ε > 0 and a sequence r _n → 0 such that

$$\displaystyle \begin{aligned} \mathcal{H}^{d-d^\ast+s}\big(\mathrm{Sing}\,(\partial\Omega_z)\cap B_{r_n}(x_0)\big)\ge {\varepsilon} r_n^{d-d^\ast+s}\qquad \text{for every}\qquad n\in\mathbb{N}, \end{aligned}$$

which can be re-written as

$$\displaystyle \begin{aligned} \mathcal{H}^{d-d^\ast+s}\big(\mathrm{Sing}\,(\partial\Omega_{z_n})\cap B_{1}\big)\ge {\varepsilon}\qquad \text{for every}\qquad n\in\mathbb{N}, \end{aligned} $$

(10.7)

where we have set $z_n(x):=\frac 1{r_n}z(x_0+r_nx)$.

Without loss of generality, we can assume that x ₀ = e _d. Now, up to a subsequence, z _n converges to a blow-up limit z ₀ of z. By Lemma 10.9 and Lemma 10.10, we have that:

• (1) z ₀(x′, x _d) = z ₀(x′, 0) for every $x'\in \mathbb {R}^{d-1}$ and every $x_d\in \mathbb {R}$;

• (2) $z_0^{\prime }:=z_0(\cdot ,0):\mathbb {R}^{d-1}\to \mathbb {R}$ is one-homogeneous local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^{d-1}$.

By hypothesis, we have that

$$\displaystyle \begin{aligned} \mathcal{H}^{d-1-d^\ast+s}\big(\mathrm{Sing}\,(\partial\Omega_{z_0^{\prime}})\big)=0. \end{aligned}$$

The translation invariance of z ₀ now implies that

$$\displaystyle \begin{aligned} \mathrm{Sing}\,(\partial\Omega_{z_0})=\mathrm{Sing}\,(\partial\Omega_{z_0^{\prime}})\times\mathbb{R}\,, \end{aligned}$$

so, Lemma 10.6 gives

$$\displaystyle \begin{aligned} \mathcal{H}^{d-d^\ast+s}\big(\mathrm{Sing}\,(\partial\Omega_{z_0})\big)=0, \end{aligned}$$

which is a contradiction with (10.6) of Lemma 10.7 and (10.7). □

10.4 Proof of Theorem 1.4

In this section, we will give an estimate on the dimension of the singular set. The result is more general and applies to different situations, for instance to almost-minimizers and measure-constrained minimizers.

Proposition 10.13 (Dimension of the Singular Set)

Let $D\subset \mathbb {R}^d$ be a bounded open set and $u:D\to \mathbb {R}$ a continuous non-negative function. Let the regular and singular sets Reg(∂ Ω _u) and Sing(∂ Ω _u) of the free boundary ∂ Ω _u ∩ D be defined as in the beginning of Sect. 10.2 . Suppose that u satisfies the following hypotheses:

• (a) ε -regularity. There are constants ε > 0 and R > 0 such that the following holds:

  If x ₀ ∈ ∂ Ω _u ∩ D and r ∈ (0, R) are such that B _r(x ₀) ⊂ D and

  $$\displaystyle \begin{aligned} \|u(x)-\sqrt{\Lambda}\, ((x-x_0)\cdot\nu)_+\|{}_{L^\infty_x(B_r(x_0))}\le{\varepsilon} r \quad \mathit{\text{for some}}\quad \nu\in\partial B_1, \end{aligned} $$

  (10.8)

  then ∂ Ω _u = Reg (∂ Ω _u) in B _r∕2(x ₀).

• (b) Non-degeneracy. There are constants κ > 0 and r ₀ > 0 such that the following holds: if $n\in \mathbb {N}$ , x ₀ ∈ ∂ Ω _u ∩ D and r ∈ (0, r ₀) are such that B _r(x ₀) ⊂ D, then

  $$\displaystyle \begin{aligned} \|u\|{}_{L^\infty(B_r(x_0))}\ge \kappa\,r\,. \end{aligned}$$

• (c) Convergence of the blow-up sequences. Every blow-up sequence

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

  with x ₀ ∈ ∂ Ω _u ∩ D and r _n → 0, admits a subsequence that converges locally uniformly to a blow-up limit $u_0:\mathbb {R}^d\to \mathbb {R}$.

• (d) Homogeneity and minimality of the blow-up limits. Every blow-up limit of u is a one-homogeneous global minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^d$.

Then,

• (i) if d < d ^∗ , then Sing (∂ Ω _u) is empty;

• (ii) if d = d ^∗ , then Sing (∂ Ω _u) is locally finite;

• (iii) if d > d ^∗ , then $\dim _{\mathcal {H}}\,\mathrm {Sing}\,(\partial \Omega _u)\le d-d^\ast$.

Proof

Suppose first that d < d ^∗. Let x ₀ ∈ ∂ Ω_u ∩ D and let r _n → 0 be a infinitesimal sequence such that $u_{r_n,x_0}$ converges locally uniformly to a blow-up limit u ₀ (such a sequence exists by the hypothesis (b)). By (c), u ₀ is a one-homogeneous local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^d$. By definition of d ^∗, we get that $\mathrm {Sing}\,(\partial \Omega _{u_0})=\emptyset$. This means that every blow-up limit of u ₀ is of the form $\sqrt {\Lambda }\,(x\cdot \nu )_+$, for some ν ∈ ∂B ₁. In particular, it holds for every blow-up limit in zero. Since u ₀ is one-homogeneous, the blow-up of u ₀ in zero is u ₀ itself and so,

$$\displaystyle \begin{aligned} u_0(x)=\sqrt{\Lambda}\,(x\cdot\nu)_+\qquad \text{for some}\qquad \nu\in\partial B_1. \end{aligned}$$

Thus, for n large enough, we get that

$$\displaystyle \begin{aligned} \|u_{r_n,x_0}(x)-\sqrt{\Lambda}\, (x\cdot\nu)_+\|{}_{L^\infty_x(B_1)}\le {\varepsilon}, \end{aligned}$$

which, by the definition of $u_{r_n,x_0}$ gives precisely (10.8). Thus, by (a), we get that x ₀ is a regular point, x ₀ ∈Reg (∂ Ω_u). Since x ₀ is arbitrary, we conclude that Sing (∂ Ω_u) = ∅.

Let now d = d ^∗. Suppose by contradiction that there is a sequence of points x _n ∈Sing (∂ Ω_u) converging to a point x ₀ ∈ D ∩Sing (∂ Ω_u). Let r _n := |x _n − x ₀|. Up to extracting a subsequence, we can assume that the blow-up sequence $u_n:= u_{r_n,x_0}$ converges to a blow-up limit $u_0:\mathbb {R}^d\to \mathbb {R}$. By (c), u ₀ is a one-homogeneous local minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^d$. On the other hand, notice that for every n > 0 the point $\xi _n=\frac {x_n-x_0}{r_n}\in \partial B_1$ is a singular point for u _n. Up to extracting a subsequence, we may assume that ξ _n converges to a point ξ ₀ ∈ ∂B ₁. By Lemma 10.7, we get that $\xi _0 \in \mathrm {Sing}\,(\partial \Omega _{u_0})$, in contradiction with Lemma 10.11.

Finally, we consider the case d > d ^∗. Let s > 0 be fixed. We will prove that $\mathcal {H}^{d-d^\ast +s} \big (\mathrm {Sing}\,(\partial \Omega _{u})\big )=0$. Suppose that this is not the case and $\mathcal {H}^{d-d^\ast +s}\big (\mathrm {Sing}\,(\partial \Omega _{u})\big )>0$. By Lemma 10.5 we have that there is a point x ₀ ∈Sing (∂ Ω_u) and a sequence r _n → 0 such that

$$\displaystyle \begin{aligned} \mathcal{H}^{d-d^\ast+s}\big(\mathrm{Sing}\,(\partial \Omega_{u})\cap B_{r_n}(x_0)\big)\ge {\varepsilon} r_n^{d-d^\ast+s}. \end{aligned}$$

Taking, $u_n=u_{r_n,x_0}$, we get that

$$\displaystyle \begin{aligned} \mathcal{H}^{d-d^\ast+s}\big(\mathrm{Sing}\,(\partial \Omega_{u_n})\cap B_{1}\big)\ge {\varepsilon}. \end{aligned}$$

Using (b), we can suppose that, up to extracting a subsequence, u _n converges to a blow-up limit u ₀. By (c), u ₀ is a one-homogeneous minimizer of $\mathcal F_\Lambda$ in $\mathbb {R}^d$. Now, Lemma 10.7, we get that $\mathcal {H}^{d-d^\ast +s}\big (\mathrm {Sing}\,(\partial \Omega _{u_0})\cap B_{1}\big )\ge {\varepsilon }$, which is in contradiction with Lemma 10.12. □