Regularity of the Free Boundary for Measure Constrained Minimizers

Abstract

Let D be a connected bounded open set in $\mathbb {R}^d$ and let v ∈ H ¹(D) be a given non-negative function. This chapter is dedicated to the problem

$$\displaystyle \begin{aligned} \min\Big\{\mathcal F_0(u,D)\ :\ u\in H^1(D),\ u-v\in H^1_0(D),\ |\Omega_u\cap D|=m\Big\}, \end{aligned} $$

(11.1)

where m ∈ (0, |D|) is a fixed constant and we recall that

$$\displaystyle \begin{aligned} \mathcal F_0(u,D)=\int_D|\nabla u|{}^2\,dx. \end{aligned}$$

In this chapter, we give the main steps of the proof of Theorem 1.9.

Let D be a connected bounded open set in $\mathbb {R}^d$ and let v ∈ H ¹(D) be a given non-negative function. This chapter is dedicated to the problem

$$\displaystyle \begin{aligned} \min\Big\{\mathcal F_0(u,D)\ :\ u\in H^1(D),\ u-v\in H^1_0(D),\ |\Omega_u\cap D|=m\Big\}, \end{aligned} $$

(11.1)

where m ∈ (0, |D|) is a fixed constant and we recall that

$$\displaystyle \begin{aligned} \mathcal F_0(u,D)=\int_D|\nabla u|{}^2\,dx. \end{aligned}$$

In this chapter, we give the main steps of the proof of Theorem 1.9.

• Section 11.1. Existence of minimizers.

  In this section, we prove that (11.1) admits a solution and that every solution is a non-negative subharmonic function (see Proposition 11.1).

• Section 11.2. Euler-Lagrange equations.

  In this section, we prove that if u is a solution to (11.1), then there exists a Lagrange multiplier Λ ≥ 0 such that the first variation of $\mathcal F_\Lambda$ vanishes, that is,

  $$\displaystyle \begin{aligned} \delta \mathcal F_\Lambda(u,D)[\xi]=0\quad \text{for every}\quad \xi\in C^\infty_c(D;\mathbb{R}^d). \end{aligned}$$

• Section 11.3. Strict positivity of the Lagrange multiplier.

  In this section we prove that Λ > 0.

• Section 11.4. Convergence of the Lagrange multipliers.

  In this section, we prove a technical lemma, that we will use several times in the next section. Roughly speaing, we show that if u _n is a sequence of solutions converging to a solution u, then also the sequence of Lagrange multipliers converge to the Lagrange multipliers of u.

• Section 11.5. Almost optimality of u at small scales.

  In this section, we show that if u is a solution to (11.1), then it minimizes $\mathcal F_\Lambda$ in every ball B _r, up to an error that depends on the radius r and vanishes as r → 0. At this point, the regularity of the free boundary (Theorem 1.9) follows by the same arguments that we used for Theorem 1.2 and Theorem 1.4, the necessary modifications being pointed out in the sketch of the proof given in the introduction.

11.1 Existence of Minimizers

In this section we prove that there is a solution to the problem (11.1). This follows by a standard argument which can be divided in two steps. We will first show that there is a solution u to the auxiliary problem

$$\displaystyle \begin{aligned} \min\Big\{\mathcal F_0(u,D)\ :\ u\in H^1(D),\ u-v\in H^1_0(D),\ |\Omega_u^+\cap D|\le m\Big\}, \end{aligned} $$

(11.2)

where for simplicity we set

$$\displaystyle \begin{aligned} \Omega_u^+=\Omega_{u_+}=\{u>0\}. \end{aligned}$$

Then we will prove that the constraint is saturated, that is, every solution u of (11.2) is such that $|\Omega _u|=|\Omega _u^+|=m$. We give the details in the following proposition.

Proposition 11.1 (Existence of Minimizers)

Let D be a connected bounded open set in $\mathbb {R}^d$ , v ∈ H ¹(D) be a non-negative function and 0 < m < |D|. Then,

• (i) there is a solution to the variational problem (11.1);

• (ii) the function u ∈ H ¹(D) is a solution to (11.1) if and only if it is a solution to (11.2);

• (iii) every solution (to (11.1) and (11.2)) is a non-negative subharmonic function in D and, in particular, is defined at every point of D.

Proof

We will proceed in several steps.

Step 1. There is a solution to the auxiliary problem(11.2). This follows by a direct argument. Indeed, let u _n be a minimizing sequence for (11.2), that is, $u_n-v\in H^1_0(D)$, $|\Omega _u^+\cap D|\le m$ and

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\mathcal F_0(u_n,D)=\inf\Big\{\mathcal F_0(u,D)\ :\ u\in H^1(D),\ u-v\in H^1_0(D),\ |\Omega_u^+\cap D|\le m\Big\}. \end{aligned}$$

Then, up to a subsequence, u _n converges weakly in H ¹(D), strongly in L ²(D) and pointwise a.e. in D to a function u _∞∈ H ¹(D) such that $u_\infty -v\in H^1_0(D)$. Then, we have

$$\displaystyle \begin{aligned} \mathcal F_0(u_\infty,D)\le \liminf_{n\to\infty}\mathcal F_0(u_n,D), \end{aligned}$$

and, by the pointwise convergence of u _n to u _∞,

which means that u _∞ is a solution to (11.2).

Step 2. Every solution u to(11.2) is non-negative. Indeed, this follows simply by the fact that if u = u ₊ − u ₋ is a solution to (11.2), then the function u ₊ still satisfies the constraints $u_+-v\in H^1_0(\Omega )$ and $|\Omega _{u_+}|\le m$, and we have

$$\displaystyle \begin{aligned} \mathcal F_0(u,D)=\mathcal F_0(u_+,D)+\mathcal F_0(u_-,D)\le \mathcal F_0(u_+,D), \end{aligned}$$

with an equality if and only if u ₋≡ 0.

Step 3. Every solution u to(11.2) is subharmonic. This follows by the fact that

$$\displaystyle \begin{aligned} \mathcal F_0(u,D)\le \mathcal F_0(\varphi,D), \end{aligned}$$

for every φ ≤ u with the same boundary values as u. In particular, this means that u is defined pointwise. In fact, we simply consider the representative of u defined as

Step 4. Every solution u to(11.2) satisfies the condition | Ω_u| = m. Indeed, suppose that this is not the case. Let r ₀ > 0 be such that $|B_{r_0}|\le m-|\Omega _u|$. Take x ₀ ∈ D and $r<\min \big \{r_0,\text{dist}(x_0,\partial D)\big \}$. Let h be the harmonic extension of u in B _r(x ₀), that is, h is a solution of the PDE

$$\displaystyle \begin{aligned} \Delta h=0\quad \text{in}\quad B_r(x_0),\qquad h=u\quad \text{on}\quad \partial B_r(x_0). \end{aligned}$$

Then, let $\widetilde u$ be the competitor defined as

Then, $|\Omega _{\widetilde u}|\le m$ and so, the optimality of u gives

$$\displaystyle \begin{aligned} 0\ge \mathcal F_0(u,D)-\mathcal F_0(\widetilde u,D)=\int_{B_r(x_0)}|\nabla h|{}^2\,dx-\int_{B_r(x_0)}|\nabla u|{}^2\,dx=\int_{B_r(x_0)}|\nabla (u-h)|{}^2\,dx, \end{aligned}$$

which means that h = u in B _r(x ₀). In particular, we get that the set {u > 0} is open: if u(x ₀) > 0, then for some r small enough, but then u > 0 in B _r(x ₀) because it coincides with its (non-zero) harmonic extension. On the other hand {u > 0} is also closed. Indeed, if there is a sequence of points x _n converging to x ₀ such that u(x _n) > 0, then the harmonic extension of u in B _r(x ₀) is non-zero, so it is strictly positive, and so, u(x ₀) > 0. Since D is connected, this means that {u > 0} = D, which is a contradiction with the fact that | Ω_u|≤ m < |D|.

Step 5. u ∈ H ¹(D) is a solution to(11.2) if and only if it is a solution to(11.1). This is a trivial consequence of Step 4. □

In the rest of this section, without loss of generality, we will only consider functions u ∈ H ¹(D), which are non-negative and satisfy the following optimality condition:

$$\displaystyle \begin{aligned} \mathcal F_0(u,D)\le \mathcal F_0(v,D)\quad \text{for every}\quad v\in H^1(D)\quad \text{such that}\quad \begin{cases}v-u\in H^1_0(D),\\ |\Omega_u|=|\Omega_v|. \end{cases} \end{aligned} $$

(11.3)

11.2 Euler-Lagrange Equation

In this section, we will prove the existence of a Lagrange multiplier for functions satisfying (11.3). We will follow step-by-step the proof from [46]. Our main result is the following.

Proposition 11.2 (Euler-Lagrange Equation)

Let $D\subset \mathbb {R}^d$ be a connected bounded open set and let the non-negative function $u:D\to \mathbb {R}$ be a solution of (11.3). Then, there is a constant Λ _u > 0 such that

$$\displaystyle \begin{aligned} \delta \mathcal F_0(u,D)[\xi] + \Lambda_u \int_{\Omega_u}\!\!\mathit{\text{div}}\, \xi\, dx=0\qquad \mathit{\text{for every}}\qquad \xi\in C^\infty_c(D;\mathbb{R}^d). \end{aligned} $$

(11.4)

We start with the following lemma.

Lemma 11.3 (Variation of the Measure)

Let D be a connected open set in $\mathbb {R}^d$ and let Ω ⊂ D be a Lebesgue measurable set such that 0 < | Ω| < |D|. Then, there is a smooth vector field $\xi \in C^\infty _c(D;\mathbb {R}^d)$ such that

$$\displaystyle \begin{aligned} \displaystyle \int_{\Omega} \mathit{\text{div}}\,\xi\,dx =1. \end{aligned}$$

Proof

Assume, by contradiction, that we have

$$\displaystyle \begin{aligned} \int_{\Omega}\text{div}\, \xi\, dx=0\qquad \text{for every}\qquad \xi\in C^\infty_c (D;\mathbb{R}^d). \end{aligned} $$

(11.5)

In particular, for every ball B _ρ(x ₀) ⊂ D, we may choose ξ to be the vector field

$$\displaystyle \begin{aligned} \xi(x)=(x-x_0)\phi_{\varepsilon}(x), \end{aligned}$$

where

$$\displaystyle \begin{aligned} 0\le \phi_{\varepsilon}\le 1\quad \text{and}\quad \left\vert \nabla\phi_{\varepsilon}\right\vert \leq \frac{1+{\varepsilon}\rho}{{\varepsilon}\rho}\quad \text{in}\quad B_\rho(x_0), \end{aligned}$$

$$\displaystyle \begin{aligned} \phi_{\varepsilon}=1\quad \text{in}\quad B_{\rho(1-{\varepsilon})}(x_0)\qquad \text{and}\qquad \phi_{\varepsilon}=0\quad \text{on}\quad \partial B_\rho(x_0). \end{aligned}$$

By (11.5), we have

$$\displaystyle \begin{aligned} 0=\int_{\Omega}\text{div}\,\xi\,dx=\int_{\Omega}\big(d\phi_{\varepsilon}(x)+(x-x_0)\cdot\nabla\phi_{\varepsilon}(x)\big)\, dx. \end{aligned}$$

Passing to the limit as ε → 0, we obtain

$$\displaystyle \begin{aligned} d|\Omega\cap B_\rho(x_0)|-\rho\,\mathcal{H}^{d-1}\big(\Omega\cap\partial B_\rho(x_0)\big)=0. \end{aligned}$$

In particular, we get that

$$\displaystyle \begin{aligned} \frac{\partial}{\partial\rho}\left(\frac{|\Omega\cap B_\rho(x_0)|}{\rho^d}\right)=0, \end{aligned}$$

which means that the function ρ↦ρ ^−d| Ω∩ B _ρ(x ₀)| is constant. In particular, if x ₀ ∈ B _r is a point of zero Lebesgue density for Ω, then Ω has zero Lebesgue measure in a neighborhood of x ₀. Precisely, setting we have that

$$\displaystyle \begin{aligned} \text{If}\quad x_0\in\Omega^{(0)}\ ,\quad \text{then}\quad |\Omega\cap B_{r(x_0)}(x_0)|=0\,. \end{aligned} $$

(11.6)

Now, notice that (11.6) is both an open and a closed subset of D. Since, by hypothesis, D is connected, we have that Ω⁽⁰⁾ = D or Ω⁽⁰⁾ = ∅, which concludes the proof. □

Proof of Proposition 11.2

Let $\xi \in C^\infty _c(D;\mathbb {R}^d)$. Using the notations from Lemma 9.5, for any (small enough) $t\in \mathbb {R}$, we set

$$\displaystyle \begin{aligned} \Psi_t=Id+t\xi\,,\qquad \Phi_t=\Psi_t^{-1}\qquad \text{and}\qquad u_t:=u\circ\Psi_t\,. \end{aligned}$$

By Lemma 9.5, we have that

$$\displaystyle \begin{aligned} |\Omega_{u_t}| & = |\Omega_u|+t\int_{\Omega_u} \text{div}\,\xi\,dx + o(t)\qquad \text{and}\\ \mathcal F_0(u_t,D) & =\mathcal F_0(u,D)+t\,\delta\mathcal F_0(u,D)[\xi]+o(t). \end{aligned} $$

Now, let the vector field $\xi _0\in C^\infty _c(D;\mathbb {R}^d)$ be such that

$$\displaystyle \begin{aligned} \int_{\Omega_u} \text{div}\,\xi_0\,dx=1\,. \end{aligned}$$

We are now going to prove that (11.4) holds with

$$\displaystyle \begin{aligned} \Lambda_u:=-\delta\mathcal F_0(u,D)[\xi_0]. \end{aligned}$$

We fix $\xi \in C^\infty _c(D;\mathbb {R}^d)$ and we consider two cases.

Case 1. $\displaystyle \int _{\Omega _u} \text{div}\,{\xi }\,dx = 0.$

Let ξ ₁ := ξ − ηξ ₀ , where η > 0 is a real constant. Then, we have

$$\displaystyle \begin{aligned} \displaystyle\int_{\Omega_u} \text{div}\,{\xi_1}\,dx = -\eta. \end{aligned}$$

Setting u _t = u ∘ Φ_t, where Φ_t := (Id + tξ ₁)⁻¹, we have that, for t > 0 small enough,

$$\displaystyle \begin{aligned} u_t \in H^1_0(D)\qquad \text{and}\qquad |\Omega_{u_t}| \leq |\Omega_u|. \end{aligned}$$

By Proposition 11.1 (see Step 5 of the proof), we have that

$$\displaystyle \begin{aligned} \mathcal F_0(u,D)\le\mathcal F_0(u_t,D). \end{aligned}$$

Taking the derivative at t = 0, we obtain

$$\displaystyle \begin{aligned} \delta \mathcal F_0(u,D)[\xi_1] \ge 0,\, \end{aligned}$$

which can be re-written as

$$\displaystyle \begin{aligned} \delta \mathcal F_0(u,D)[\xi] \geq \eta\, \delta \mathcal F_0(u,D)[\xi_0]. \end{aligned}$$

Since η is arbitrary, we can deduce that

$$\displaystyle \begin{aligned} \delta \mathcal F_0(u,D)[\xi] \geq 0. \end{aligned}$$

Finally, repeating the same argument for − ξ instead of ξ, we obtain that

$$\displaystyle \begin{aligned} \delta \mathcal F_0(u,D)[\xi] = 0, \end{aligned}$$

which concludes the proof of (11.4) in this case.

Case 2. $\displaystyle \int _{\Omega _u} \text{div}\,{\xi }\,dx \neq 0$.

Let $\displaystyle \xi _2 := \xi - \xi _0 \int _{\Omega _u} \text{div}\,{\xi }\,dx$. Then $\displaystyle \int _{\Omega _u} \text{div}\,{\xi _2}\,dx = 0$ and, by Case 1, we obtain

$$\displaystyle \begin{aligned} 0 & =\delta \mathcal F_0(u,D)[\xi_2] = \delta \mathcal F_0(u,D)[\xi] - \delta \mathcal F_0(u,D)[\xi_0] \int_{\Omega_u} \text{div}\,{\xi}\,dx\\ & =\delta \mathcal F_0(u,D)[\xi] + \Lambda_u\int_{\Omega_u} \text{div}\,{\xi}\,dx, \end{aligned} $$

which concludes the proof of (11.4).

It only remains to prove that Λ_u ≥ 0. Indeed, let u _t = u ∘ Φ_t, where Φ_t = (Id − tξ ₀)⁻¹. For t > 0 small enough, we have that $|\Omega _{u_t}|\le |\Omega _u|$. We reason as in Case 1. By Proposition 11.1, we get that $\mathcal F_0(u,D)\le \mathcal F_0(u_t,D)$. Then, taking the derivative at t = 0, we deduce

$$\displaystyle \begin{aligned} \Lambda_u:=\delta\mathcal F_0(u,D)[-\xi_0]\ge 0. \end{aligned}$$

The strict positivity of Λ_u is more involved and follows by Proposition 11.4, which we prove in the next subsection. □

11.3 Strict Positivity of the Lagrange Multiplier

In this section we prove that the Lagrange multiplier from Proposition 11.2 is strictly positive. Precisely, we will show that a function, which is critical for the functional $\mathcal F_0$ has a monotone Almgren frequency function N(r). Now, the monotonicity of the frequency function implies that u cannot decay too fast around the free boundary points. On the other hand, if u is harmonic in Ω_u, then we can use a Caccioppoli inequality to show that if the Lebesgue density of Ω_u is too small, then the decay of u on the balls of radius r should be very fast. The combination of these two estimates implies that the Lebesgue density of Ω_u should be bounded from below at every point of D. In particular, there cannot be points of zero Lebesgue density for Ω_u in D. The results from this subsection come directly from [46], but this unique-continuation argument goes back to the work of Garofalo and Lin [34]. The main result of this subsection is the following.

Proposition 11.4

Let D be a connected open set in $\mathbb {R}^d$ . Suppose that u ∈ H ¹(D) is a non-identically-zero function such that

• (a) u is a solution of the equation

  $$\displaystyle \begin{aligned} {} \Delta u=0\quad \mathit{\text{in}}\quad \Omega_u=\{u\neq 0\}; \end{aligned}$$

• (b) u satisfies the extremality condition

  $$\displaystyle \begin{aligned} \delta \mathcal F_0(u)[\xi]=0\quad \mathit{\text{for every}}\quad \xi\in C^\infty_c(D;\mathbb{R}^d), \end{aligned}$$

  where $\delta \mathcal F_0(u)[\xi ]$ is the first variation of $\mathcal F_0$ in the direction ξ and is given by

  $$\displaystyle \begin{aligned} \delta \mathcal F_0(u)[\xi] := \int_D \Big[2\nabla u\cdot D\xi(\nabla u) -|\nabla u|{}^2\mathit{\text{div}}\, \xi\Big]\,dx. \end{aligned} $$

  (11.7)

Then, .

Remark 11.5

It is sufficient to prove Proposition 11.4 in the case u ≥ 0. In fact, if $u:D\to \mathbb {R}$ satisfies the hypotheses (a) and (b) above, then the function $|u|:D\to \mathbb {R}$ satisfies the same hypotheses.

In the proof of Proposition 11.4 we will use the following Faber-Krahn-type inequality, which was first proved in [10] (we report here the original proof).

Lemma 11.6 (A Faber-Krahn Inequality, [10])

There is a dimensional constant C _d > 0 such that for every ball $B_r\subset \mathbb {R}^d$ and every function v ∈ H ¹(B _r) satisfying

$$\displaystyle \begin{aligned} \frac{|\Omega_v\cap B_r|}{|B_r|}\le\frac 12\,, \end{aligned}$$

we have the inequality

$$\displaystyle \begin{aligned} \int_{B_r}v^2\,dx\le C_d\,r^2\left(\frac{|\Omega_v\cap B_r|}{|B_r|}\right)^{\frac2d}\int_{B_r}|\nabla v|{}^2\,dx\,, \end{aligned} $$

(11.8)

where we recall that Ω _v = {v ≠ 0}.

Proof

We first notice that:

• We can assume that v is non-negative in B _r. In fact, for every v ∈ H ¹(B _r), we have that |v|∈ H ¹(B _r) and the following identities do hold:

  $$\displaystyle \begin{aligned} \Omega_v=\Omega_{|v|}\,,\qquad v^2=|v|{}^2\qquad \text{and}\qquad |\nabla v|{}^2=|\nabla |v||{}^2. \end{aligned}$$

• We can assume that r = 1. Indeed, setting v _r(x) = v(rx), we have that

  $$\displaystyle \begin{aligned} |\Omega_v\cap B_r|=r^d|\Omega_{v_r}\cap B_1|\ ,\qquad \int_{B_r}v^2\,dx={r^d}\int_{B_1}v_r^2\,dx\,, \end{aligned}$$
  $$\displaystyle \begin{aligned} \int_{B_r}|\nabla v|{}^2\,dx={r^{d-2}}\int_{B_1}|\nabla v_r|{}^2\,dx\,. \end{aligned}$$

We now proceed with the proof of (11.8) in the case r = 1 and v ≥ 0 on B ₁.

Step 1. We claim that there is a dimensional constant C _iso > 0 such that

$$\displaystyle \begin{aligned} |\Omega|{}^{\frac{d-1}{d}}\le C_{iso}\, Per(\Omega;B_r)\quad \text{for every}\quad \Omega\subset B_r\quad \text{with}\quad |\Omega|\le \frac 12|B_r|, \end{aligned} $$

(11.9)

where Per( Ω;B _r) is the relative perimeter in the sense of De Giorgi. The claim follows by a standard compactness argument.

Step 2. Let $n\in \mathbb {N}$ and let D _n = {x ⋅ ν ₁ > 0}∩{x ⋅ ν ₂ > 0} be the unbounded domain formed by the intersection of the two hyperplanes {x ⋅ ν ₁ = 0} and {x ⋅ ν ₂ = 0} forming (an interior) angle 2π∕n. We claim that, for every Ω ⊂ B ₁ satisfying $|\Omega |\le \frac {\omega _d}{2}$, there is a radius ρ > 0 such that

$$\displaystyle \begin{aligned} |B_\rho\cap D_n|=|\Omega|\qquad \text{and}\qquad Per(B_\rho\cap D_n;D_n)\le Per(\Omega;B_1). \end{aligned} $$

(11.10)

Indeed, for every Ω, there is a unique ρ > 0 such that |B _ρ ∩ D _n| = | Ω|. We set Ω^∗ := B _ρ ∩ D _n. Then, we have

$$\displaystyle \begin{aligned} \begin{aligned} |\Omega^\ast|{}^{\frac{(d-1)}d} & =n^{-\frac{(d-1)}d}|B_\rho|{}^{\frac{(d-1)}d}=\frac{n^{-\frac{(d-1)}d}}{d\omega_d^{\frac1d}} \,Per(B_\rho)=\frac{n^{\frac1d}}{d\omega_d^{\frac1d}}\, Per(\Omega^\ast;D_n). \end{aligned}\end{aligned}$$

Now, the isoperimetric inequality (11.9) implies

$$\displaystyle \begin{aligned} Per(\Omega;D)\ge C_{iso}^{-1}|\Omega|{}^{\frac{(d-1)}d}=C_{iso}^{-1}|\Omega^\ast|{}^{\frac{(d-1)}d}=C_{iso}^{-1}\, \frac{n^{\frac1d}}{d\omega_d^{\frac1d}} \, Per(\Omega^\ast; D_n). \end{aligned}$$

Taking n large enough, such that $n^{\frac 1d} \ge C_{iso}\, d\omega _d^{\frac 1d}$, we get P( Ω;D) ≥ Per( Ω^∗;D _n), which proves (11.10).

Step 3. For every non-negative function v ∈ H ¹(B ₁) we define the symmetrized function v _∗∈ H ¹(D _n) obtained through the symmetrization of each level of v, that is,

$$\displaystyle \begin{aligned} \{v_\ast>t\}=\{v>t\}^\ast\qquad \text{for every}\qquad t\ge 0\,. \end{aligned}$$

We claim that

$$\displaystyle \begin{aligned} \int_{D_n}v_\ast^2\,dx=\int_{B_1} v^2\,dx\qquad \text{and}\qquad \int_{D_n}|\nabla v_\ast|{}^2\,dx\le \int_{B_1} |\nabla v|{}^2\,dx\,. \end{aligned} $$

(11.11)

The first part of (11.11) follows by the area formula

$$\displaystyle \begin{aligned} \int_{D_n}v_\ast^2\,dx=\int_0^\infty t|\{v_\ast>t\}|\,dt=\int_0^\infty t|\{v>t\}|\,dt=\int_{B_1} v^2\,dx, \end{aligned}$$

while for the second part we will use the co-area formula. Indeed, setting

$$\displaystyle \begin{aligned} f(t):=|\{v>t\}|=|\{v^\ast>t\}|, \end{aligned}$$

we have

$$\displaystyle \begin{aligned} \begin{aligned} \int_{B_1}|\nabla v|{}^2\,dx & =\int_{0}^{+\infty}\Big(\int_{\{v=t\}}|\nabla v|\,d\mathcal{H}^{d-1}\Big)\,dt\\ & \ge \int_0^{+\infty}\Big(\int_{\{v=t\}}|\nabla v|{}^{-1}\,d\mathcal{H}^{d-1}\Big)^{-1}\Big(\mathcal{H}^{d-1}\big(\{v=t\}\cap D\big)\Big)^2\,dt\\ & =\int_0^{+\infty}|f^{\prime}(t)|{}^{-1}\,\Big(\mathcal{H}^{d-1}\big(\{v=t\}\cap B_1\big)\Big)^2\,dt\\ & \ge\int_0^{+\infty}|f^{\prime}(t)|{}^{-1}\,\Big(\mathcal{H}^{d-1}\big(\{ v_\ast=t\}\cap D_n\big)\Big)^2\,dt\\ & = \int_0^{+\infty}\Big(\int_{\{ v_\ast=t\}}|\nabla v_\ast|{}^{-1}\,d\mathcal{H}^{d-1}\Big)^{-1}\Big(\mathcal{H}^{d-1}\big(\{v_\ast=t\}\cap D_n\big)\Big)^2\,dt\\ & =\int_0^{+\infty}\Big(\int_{\{v_\ast=t\}}|\nabla v_\ast|\,d\mathcal{H}^{1}\Big)\,dt=\int_{D_n}|\nabla v_\ast|{}^2\,dx, \end{aligned}\end{aligned}$$

where the first inequality follows by Cauchy-Schwartz and the second one is a consequence of (11.10).

Step 4. We claim that there is a constant C _d,n > 0, depending only on d and n, such that

$$\displaystyle \begin{aligned} \int_{D_n}v_\ast^2\,dx\le C_{d,n}\, \big|\{v_\ast>0\}\big|{}^{\frac2d}\int_{D_n}|\nabla v_\ast|{}^2\,dx\,. \end{aligned} $$

(11.12)

Let $\tilde v_\ast :\mathbb {R}^d\to \mathbb {R}$ be the radially decreasing function defined by

$$\displaystyle \begin{aligned} \tilde v_\ast(x)=v_\ast(y), \end{aligned}$$

where y ∈ D _n is any point such that |x| = |y|. By the classical Faber-Krahn inequality in $\mathbb {R}^d$, there is a dimensional constant C _d such that

$$\displaystyle \begin{aligned} {} \int_{\mathbb{R}^d}\tilde v_\ast^2\,dx\le C_{d}\, \big|\{\tilde v_\ast>0\}\big|{}^{\frac2d}\int_{\mathbb{R}^d}|\nabla \tilde v_\ast|{}^2\,dx\,. \end{aligned}$$

which gives that

$$\displaystyle \begin{aligned} \qquad \int_{D_n}v_\ast^2\,dx\le C_d\, n^{\frac2d}\big|\{v_\ast>0\}\big|{}^{\frac2d}\int_{D_n}|\nabla v_\ast|{}^2\,dx,\qquad \end{aligned}$$

which is precisely (11.12). This, together with (11.11), concludes the proof. □

In the next lemma, we prove that the Almgren frequency function is monotone. This is a classical result, which was first proved by Almgren [2].

Lemma 11.7 (Almgren Monotonicity Formula)

Let u ∈ H ¹(B _R). For r ∈ (0, R], we define

$$\displaystyle \begin{aligned} H(r):=\int_{\partial B_r}u^2\,d\mathcal{H}^{d-1}\qquad \mathit{\text{and}}\qquad D(r):=\int_{B_r}|\nabla u|{}^2\,dx\,, \end{aligned}$$

and, if H(r) ≠ 0, we define the Almgren frequency function

$$\displaystyle \begin{aligned} N(r):=\frac{rD(r)}{H(r)}. \end{aligned}$$

Suppose that u is a solution of the equation

$$\displaystyle \begin{aligned} {} \Delta u=0\quad \mathit{\text{in}}\quad \Omega_u=\{u\neq 0\}; \end{aligned}$$

and satisfies the extremality condition

$$\displaystyle \begin{aligned} \delta \mathcal F_0(u)[\xi]=0\quad \mathit{\text{for every}}\quad \xi\in C^\infty_c(B_R;\mathbb{R}^d), \end{aligned}$$

where $\delta \mathcal F_0(u)[\xi ]$ is given by (11.7). If, moreover, H > 0 on the interval (a, b) ⊂ (0, R), then the frequency function N is non-decreasing on (a, b).

Proof

We first calculate the derivative of H

$$\displaystyle \begin{aligned} H^{\prime}(r) & =\frac{d-1}{r}H(r)+r^{d-1}\frac{\partial}{\partial r}\int_{\partial B_1}u^2(rx)\,d\mathcal{H}^{d-1}(x)\\ & =\frac{d-1}{r}H(r)+2\int_{\partial B_r}u\frac{\partial u}{\partial n}d\mathcal{H}^{d-1}=\frac{d-1}{r}H(r)+2\int_{B_r}|\nabla u|{}^2\,dx, \end{aligned} $$

which we rewrite as

$$\displaystyle \begin{aligned} H^{\prime}(r)=\frac{d-1}{r}H(r)+2 D(r). \end{aligned} $$

(11.13)

Next, we notice that the extremality condition $\delta \mathcal F_0(u)=0$ gives that the following equipartition of the energy does hold:

$$\displaystyle \begin{aligned} 0 & =-(d-2)\int_{B_r}|\nabla u|{}^2\,dx+r\int_{\partial B_r}|\nabla u|{}^2\, d\mathcal{H}^{d-1}-2r\int_{\partial B_r}\left(\frac{\partial u}{\partial n}\right)^2\, d\mathcal{H}^{d-1}, \end{aligned} $$

which can be rewritten as

$$\displaystyle \begin{aligned} -(d-2)D(r)+rD^{\prime}(r)= 2r\int_{\partial B_r}\left(\frac{\partial u}{\partial n}\right)^2\,d\mathcal{H}^{d-1}. \end{aligned} $$

We now compute the derivative of N.

$$\displaystyle \begin{aligned} N^{\prime}(r) & =\frac{D(r)H(r)+rD^{\prime}(r)H(r)-rD(r)H^{\prime}(r)}{H^2(r)}\notag\\ & = \frac{D(r)H(r)+rD^{\prime}(r)H(r)-r D(r)\left(\frac{d-1}{r}H(r)+2D(r)\right)}{H^2(r)}\notag\\ & =\frac{-(d-2)D(r)H(r)+rD^{\prime}(r)H(r)-2rD^2(r)}{H^2(r)}\notag\\ & =\frac{2r}{H^2(r)}\left(H(r)\int_{\partial B_r}\left(\frac{\partial u}{\partial n}\right)^2\,d\mathcal{H}^{d-1}-D^2(r)\right).{} \end{aligned} $$

(11.14)

Notice that, since u is harmonic in Ω_u, we have

$$\displaystyle \begin{aligned} D(r) & =\int_{B_r}|\nabla u|{}^2\,dx=\int_{\partial B_r}u\frac{\partial u}{\partial n}\,d\mathcal{H}^{d-1}, \end{aligned} $$

and so, by the Cauchy-Schwarz inequality and (11.14) we obtain N ^′(r) ≥ 0. □

Remark 11.8 (The Derivative of $\ln H$)

Notice that, by (11.13), we get that

$$\displaystyle \begin{aligned} \frac{d}{dr}\left[\log\left(\frac{H(r)}{r^{d-1}}\right)\right]=2\frac{N(r)}r\,. \end{aligned} $$

(11.15)

We are now in position to prove Proposition 11.4.

Proof of Proposition 11.4

Let x ₀ = 0 ∈ D. We set H(r), D(r) and N(r) to be as in Lemma 11.7 and Remark 11.8. Let r ₀ > 0 be such that $B_{r_0}(x_0)\subset D$ and H(r ₀) > 0. Since u ∈ H ¹(D), there is some ε > 0 such that H > 0 on the interval (r ₀ − ε, r ₀). Then, the function r↦N(r) is non-decreasing in r and so

$$\displaystyle \begin{aligned} N(r)\le N(r_0)\quad \text{for every}\quad r_0-{\varepsilon}<r\le r_0. \end{aligned}$$

By (11.15), we have

$$\displaystyle \begin{aligned} \frac{d}{dr}\left[\log\left(\frac{H(r)}{r^{d-1}}\right)\right]=2\frac{N(r)}r\le \frac{2N(r_0)}{r}, \end{aligned} $$

(11.16)

and integrating we get

$$\displaystyle \begin{aligned} \log\left(\frac{H(r_0)}{r_0^{d-1}}\right)-\log\left(\frac{H(r)}{r^{d-1}}\right)\le \log\Big(\frac{r_0}{r}\Big)\,2 N(r_0)\qquad \text{for every}\qquad r_0-{\varepsilon}<r\le r_0. \end{aligned}$$

This means that, for every ε > 0, H is bounded from below by a positive constant on the interval [r ₀ − ε, r ₀]. In particular, H > 0 on (0, r ₀]. Thus, we can take ε = r ₀.

Let now, r ∈ (0, r ₀∕2]. Integrating the inequality (11.16) from r to 2r, we get

$$\displaystyle \begin{aligned} \log\left(\frac{H(2r)}{H(r)}\right)\le (d-1)\log 2+2\log 2\,N(r_0).\end{aligned}$$

This implies that

$$\displaystyle \begin{aligned} {} H(2r)\le 2^{d-1}4^{N(r_0)}H(r)\qquad \text{for every}\qquad 0<r\le \frac{r_0}2. \end{aligned}$$

Integrating once more in r we get

$$\displaystyle \begin{aligned} \int_{B_{2r}}u^2\,dx\le 2^{d-1}4^{N(r_0)}\int_{B_{r}}u^2\,dx\qquad \text{for every}\qquad 0<r\le \frac{r_0}2. \end{aligned} $$

(11.17)

We next prove a Caccioppoli inequality for u in the ball B _2r. Indeed, let $\phi \in C^\infty _c(\mathbb {R}^d)$ be a cut-off function such that

Since, u is harmonic in Ω_u, the following Caccioppoli inequality does hold:

$$\displaystyle \begin{aligned} \int_{B_r}|\nabla u|{}^2\,dx& \le \int_{B_{2r}}|\nabla (u\phi)|{}^2\,dx=\int_{B_{2r}}\left(u^2|\nabla\phi|{}^2+\nabla u\cdot\nabla (u\phi^2)\right)\,dx\notag\\ & \!=\! \int_{B_{2r}}u^2|\nabla\phi|{}^2\,dx\!-\!\int_{B_{2r}} u\phi^2\Delta u\,dx\!=\! \int_{B_{2r}}u^2|\nabla\phi|{}^2\,dx \!\le\! \frac 4{r^2}\int_{B_{2r}}u^2\,dx.\notag \end{aligned} $$

On the other hand, by Lemma 11.6, there is a dimensional constant C _d > 0 such that:

$$\displaystyle \begin{aligned} \int_{B_r}u^2\,dx\le C_d \, r^2\bigg(\frac{|\Omega_u\cap B_r|}{|B_r|}\bigg)^{\frac2{d}}\int_{B_r}|\nabla u|{}^2\,dx\qquad \text{whenever}\qquad \frac{|\Omega_u\cap B_r|}{|B_r|}\le \frac 12\,. \end{aligned}$$

This, together with the Caccioppoli and the doubling inequality (11.17), gives that

$$\displaystyle \begin{aligned} \int_{B_r}u^2\,dx\le C_d\bigg(\frac{|\Omega_u\cap B_r|}{|B_r|}\bigg)^{\frac2{d}} 4^{N(r_0)}\int_{B_r}u^2\,dx. \end{aligned}$$

Since, $\displaystyle \int _{B_r}u^2\,dx>0$, we get that there is a dimensional constant C _d such that

$$\displaystyle \begin{aligned} \min\left\{\frac 12,\frac{1}{C_d 2^{N(r_0)d}}\right\}\le \frac{|\Omega_u\cap B_r|}{|B_r|}\qquad \text{for every}\qquad 0<r\le\frac{r_0}{2}. \end{aligned}$$

In particular, we have a lower density bound for Ω_u at every point of D, which implies that and concludes the proof. □

11.4 Convergence of the Lagrange Multipliers

In this section we prove that the Lagrange multipliers, associated to the solutions of variational problems with measure constraint in a fixed connected open set $D\subset \mathbb {R}^d$, are continuous with respect to variations of the constraint. This fact will be used several times in the proof of the optimality of the blow-up limits. In the next Lemma, which comes directly from [46], we will use the notation

$$\displaystyle \begin{aligned} {} \delta {\mathrm{Vol}}\,(\Omega)[\xi]:=\int_{\Omega}\text{div}\,\xi\,dx, \end{aligned}$$

for every Lebesgue measurable set $\Omega \subset \mathbb {R}^d$ and every vector field $\xi \in C^\infty _c(\mathbb {R}^d;\mathbb {R}^d)$.

Lemma 11.9 (Convergence of the Lagrange Multipliers)

Let D be a connected bounded open set in $\mathbb {R}^d$ and let $u\in H^1_0(D)$ be a non-negative function for which (11.3) does hold. Let Λ _u be the Lagrange multiplier from (11.4) in D.

Let B ⊂ D be a connected bounded open set such that 0 < m := | Ω _u ∩ B| < |B|.

Let $\left (m_n\right )_{n\geq 1}$ be a sequence such that lim_n→∞ m _n = m and let u _n ∈ H ¹(B) be a solution (which exists due to Proposition 11.1) to the problem

$$\displaystyle \begin{aligned} \min\Big\{\mathcal F_0(v,B)\, :\ v\in H^1(B),\ v-u\in H^1_0(B),\ |\Omega_v|=m_n\Big\}. \end{aligned} $$

(11.18)

Then, we have:

• (i) for every n, there is a Lagrange multiplier $\Lambda _{u_n}>0$ for which

  $$\displaystyle \begin{aligned} \delta \mathcal F_0(u_n,B)[\xi] + \Lambda_{u_n}\delta{\mathrm{Vol}}\,(\Omega_{u_n})[\xi]=0\qquad \mathit{\text{for every}}\qquad \xi\in C^\infty_c(B;\mathbb{R}^d), \end{aligned} $$

  (11.19)

• (ii) for every n, there is a vector field $\xi _n\in C^\infty _c(B;\mathbb {R}^d)$ such that

  $$\displaystyle \begin{aligned} \delta\mathcal F_0(u_n,B)[\xi_n]+\Lambda_{u_n}=0\qquad \mathit{\text{and}}\qquad \delta{\mathrm{Vol}}\,(\Omega_{u_n})[\xi_n]=1\,. \end{aligned} $$

  (11.20)

• (iii) u _n converges strongly in $H^1_0(D)$ and pointwise almost everywhere to a function u _∞ , which is a solution to the problem

  $$\displaystyle \begin{aligned} \min\Big\{\mathcal F_0(v,B)\, :\ v\in H^1(B),\ v-u\in H^1_0(B),\ |\Omega_v| = m\Big\}; \end{aligned} $$

  (11.21)

• (iv) the sequence of characteristic functions converges to pointwise almost everywhere and strongly in L ²(D);

• (v) $\displaystyle \lim _{n\to \infty }\Lambda _{u_n}=\Lambda _{u_\infty }$ , where $\Lambda _{u_\infty }>0$ is the Lagrange multiplier of u _∞ in B, that is,

  $$\displaystyle \begin{aligned} \delta \mathcal F_0(u_\infty,B)[\xi] + \Lambda_{u_\infty}\delta{\mathrm{Vol}}\,(\Omega_{u_\infty})[\xi]=0\qquad \mathit{\text{for every}}\qquad \xi\in C^\infty_c(B;\mathbb{R}^d), \end{aligned} $$

  (11.22)

• (vi) Suppose that B ≠ D and that there is a connected component $\mathcal C$ of such that

  $$\displaystyle \begin{aligned} 0<|\Omega_u\cap \mathcal C|< |\mathcal C|, \end{aligned}$$

  then $\Lambda _{u_\infty }=\Lambda _u$.

Proof

The existence of a solution u _n follows from Proposition 11.1. The existence of a Lagrange multiplier $\Lambda _{u_n}$ and a vector field $\xi _n\in C^\infty _c(B;\mathbb {R}^d)$ with the properties (11.20) follows by Proposition 11.2. Let $\xi _0\in C^\infty _c(B;\mathbb {R}^d)$ be a vector field such that

$$\displaystyle \begin{aligned} {} \delta\mathcal F_0(u,B)[\xi_0]+\Lambda_{u}=0\qquad \text{and}\qquad \delta{\mathrm{Vol}}\,(\Omega_{u})[\xi_0]=1\,. \end{aligned}$$

Setting u _t := u ∘ (Id + tξ ₀)⁻¹, we get that, for t small enough, $u_t-u\in H^1_0(D)$. Moreover, to every n large enough, we can associate a unique $t_n\in \mathbb {R}$ such that

$$\displaystyle \begin{aligned} u-u_{t_n}\in H^1_0(B)\qquad \text{and}\qquad |\Omega_{u_n}|=m_n=|\Omega_{u_{t_n}}|. \end{aligned}$$

Thus, we can use $u_{t_n}$ as a test function in (11.18). Thus, there is a constant C depending only on u and ξ ₀ such that, for every n large enough (say n ≥ n ₀ for some $n_0\in \mathbb {N}$), we have

$$\displaystyle \begin{aligned} \mathcal F_0(u_n,B)\le \mathcal F_0(u_{t_n},B)\le C. \end{aligned}$$

Then the sequence $\left (u_n\right )_{n\geq 1}$ is uniformly bounded in H ¹(B) and so, up to a subsequence, u _n converges weakly in H ¹, strongly in L ² and pointwise almost everywhere to a function u _∞∈ H ¹(B) such that $u_\infty -u\in H^1_0(B)$. In particular, the pointwise convergence of u _n to u _∞ implies that

Thus, we get that

$$\displaystyle \begin{aligned} |\Omega_{u_\infty}|\le \liminf m_n=m, \end{aligned}$$

and so, the minimality of u implies that

$$\displaystyle \begin{aligned} \mathcal F_0(u,B)\le \mathcal F_0(u_\infty,B). \end{aligned}$$

On the other hand, the weak H ¹ convergence of u _n gives that

$$\displaystyle \begin{aligned} \mathcal F_0(u_\infty,B)\le \liminf_{n\to\infty} \mathcal F_0(u_n,B)\le \liminf_{n\to\infty} \mathcal F_0(u_{t_n},B)=\mathcal F_0(u,B), \end{aligned}$$

so, we get $\mathcal F_0(u_\infty ,B)=\mathcal F_0(u,B)$. Thus, u _∞ is a solution to (11.21) and $|\Omega _{u_\infty }|=m$. Moreover, using again the optimality of u _n and the fact that $u_{t_n}$ converges to u, we obtain

$$\displaystyle \begin{aligned} \limsup_{n\to\infty} \mathcal F_0(u_n,B) \!\le\! \limsup_{n\to\infty} \mathcal F_0(u_{t_n},B) \!=\! \mathcal F_0(u,B) \le \mathcal F_0(u_\infty,B) \!\le\! \liminf_{n\to\infty} \mathcal F_0(u_n,B), \end{aligned}$$

which gives that

$$\displaystyle \begin{aligned} \lim_{n\to\infty} \mathcal F_0(u_n,B)= \mathcal F_0(u_\infty,B). \end{aligned}$$

Since u _n converges strongly in L ²(B) and weakly in H ¹(B) to u _∞, we get that the convergence of u _n to u _∞ is strong in H ¹(B).

We next prove (iv). We will first show that convergence strongly in L ²(B) to .

Indeed, we first notice that, up to a subsequence, there is h ∈ L ²(B) such that weakly in L ²(B). On the other hand, the pointwise convergence of u _n to u _∞ implies that

Thus, for any non-negative function φ ∈ L ²(B), the Fatou Lemma implies that

which yields . In particular,

$$\displaystyle \begin{aligned} \| h\|{}_{L^2(B)}^2\ge |\Omega_{u_\infty}|=m. \end{aligned}$$

On the other hand, the weak L ² convergence of to h gives that

As a consequence,

which implies that converges to h strongly in L ²(B). Now, since

we get that , from which we conclude that converges to strongly in L ²(B), and so, up to a subsequence converges to pointwise almost everywhere.

We now prove (v). We first notice that u and u _∞ are both solutions of (11.21). By Proposition 11.2, there is a Lagrange multiplier $\Lambda _\infty :=\Lambda _{u_\infty }>0$ such that (11.22) does hold. Moreover, by (iii) and (iv), we get that, for every fixed $\xi \in C^\infty _c(B;\mathbb {R}^d)$,

$$\displaystyle \begin{aligned} \delta \mathcal F_0(u_\infty,B)[\xi] & =\lim_{n\to\infty}\delta \mathcal F_0(u_n,B)[\xi]\\ \delta\text{Vol}\,(\Omega_{u_\infty})[\xi] & =\lim_{n\to\infty}\delta\text{Vol}\,(\Omega_{u_n})[\xi]. \end{aligned} $$

Now, choosing any $\xi \in C^\infty _c(B;\mathbb {R}^d)$ such that

$$\displaystyle \begin{aligned} \displaystyle\int_{\Omega_{u_\infty}}\!\text{div}\,\xi\,dx\neq0,\end{aligned}$$

and using (11.22) and (11.19) we get that $\Lambda _{u_n}$ converges to Λ_∞.

We prove the last claim (vi). Indeed, since

$$\displaystyle \begin{aligned} \mathcal F_0(u,B)=\mathcal F_0(u_\infty,B)\qquad \text{and}\qquad |\Omega_{u_\infty}|=|\Omega_u\cap B|=m, \end{aligned}$$

we get that the function

is in H ¹(D) and is a solution to the problem

$$\displaystyle \begin{aligned} \min\Big\{\mathcal F_0(v,D)\ :\ v\in H^1(D),\ v-u\in H^1_0(D),\ |\Omega_v|=|\Omega_u|\Big\}. \end{aligned}$$

In particular, $\widetilde u$ is a critical point of $\mathcal F_{\Lambda _\infty }$ in the entire D, that is,

$$\displaystyle \begin{aligned} {} \delta \mathcal F_0(\widetilde u,D)[\xi] + \Lambda_\infty \delta\text{Vol}\,(\Omega_{\widetilde u})[\xi]=0\qquad \text{for every}\qquad \xi\in C^\infty_c(D;\mathbb{R}^d). \end{aligned}$$

On the other hand, in the connected component $\mathcal C$, we have that $\widetilde u =u$ and so, there is a vector field $\xi ^{\prime }_0\in C^\infty _c(\mathcal C;\mathbb {R}^d)$ such that $\delta \text{Vol}\,(\Omega _{u})[\xi _0^{\prime }]=\delta \text{Vol}\,(\Omega _{\widetilde u})[\xi _0^{\prime }]=1$. This implies that

$$\displaystyle \begin{aligned} \Lambda_\infty & =\Lambda_\infty\delta\text{Vol}\,(\Omega_{\widetilde u})[\xi_0^{\prime}]=-\delta\mathcal F_0(\widetilde u,D)[\xi_0^{\prime}]\\ & =-\delta\mathcal F_0(u,D)[\xi_0^{\prime}]=\Lambda_u\delta\text{Vol}\,(\Omega_{u})[\xi_0^{\prime}]=\Lambda_u\,, \end{aligned} $$

which concludes the proof. □

11.5 Almost Optimality of u at Small Scales

Let $D\subset \mathbb {R}^d$ be a connected bounded open set and $u:D\to \mathbb {R}$ be a non-negative function satisfying (11.3). In this section, we will prove the following result, which is analogous to the results of Briançon [5], Briançon-Lamboley [6], and the more recent [46], which are all dedicated to different (and technically more involved) free boundary problems arising in Shape Optimization.

Proposition 11.10

Let D be a connected bounded open set in $\mathbb {R}^d$ and let u ∈ H ¹(D) be a non-negative function satisfying (11.3). Let Λ > 0 be the corresponding Lagrange multiplier, that is, Λ is such that $\delta \mathcal F_\Lambda (u,D)=0$ . Let B ⊂ D be a ball such that:

• 0 < | Ω _u ∩ B| < |B|;

• is connected:

• .

Then, for every ε > 0, there exists r > 0 such that u satisfies the following optimality conditions in every B _r(x ₀) ⊂ B:

$$\displaystyle \begin{aligned} \!\!\!\!\!\!\! \mathcal F_{\Lambda+{\varepsilon}}(u,D)\le \mathcal F_{\Lambda+{\varepsilon}}(v,D)\ \,\mathit{\text{for every}}\ \, v\in H^1(D)\ \,\mathit{\text{such that}}\ \, \begin{cases}v-u\in H^1_0(B_r(x_0)),\\ |\Omega_u|\le|\Omega_v|. \end{cases} \end{aligned} $$

(11.23)

$$\displaystyle \begin{aligned} \!\!\!\!\!\!\!\mathcal F_{\Lambda-{\varepsilon}}(u,D)\le \mathcal F_{\Lambda-{\varepsilon}}(v,D)\ \,\mathit{\text{for every}}\ \, v\in H^1(D)\ \,\mathit{\text{such that}}\ \, \begin{cases}v-u\in H^1_0(B_r(x_0)),\\ |\Omega_u|\ge|\Omega_v|. \end{cases} \end{aligned} $$

(11.24)

Remark 11.11

An immediate consequence of the inwards (11.24) and the outwards (11.23) optimality is that u satisfies the following almost-minimality condition:

$$\displaystyle \begin{aligned} {} & \mathcal F_{\Lambda}(u,D)\le \mathcal F_{\Lambda}(v,D)+{\varepsilon} |B_r|\ \,\text{for every}\ \, v\in H^1(D)\\ & \quad \text{such that}\ \, v-u\in H^1_0(B_r(x_0)). \end{aligned} $$

In order to prove Proposition 11.10 we will use the contradiction argument of Briançon [5]. The proof presented here follows step-by-step the exposition from [46] and uses only the existence of a Lagrange multiplier, variations with respect to smooth vector fields and elementary variational arguments. Roughly speaking, the main idea is to replace the localization condition $u-v\in H^1_0(B_r)$ in (11.24) and (11.23) by a bound on the measure of Ω_v, | Ω_v|≤| Ω_u| + δ, for which the passages to the limit are somehow easier. Proposition 11.10 is a direct consequence of Proposition 11.16

Remark 11.12

We notice that we work in the ball B ⊂ D only because of the fact that we will use several times the convergence of the Lagrange multipliers associated to solutions of auxiliary problems. Indeed, in order to assure the convergence of these Lagrange multipliers to Λ (the Lagrange multiplier of the solution u), we need to work strictly inside the domain D (see Lemma 11.9, claim (vi)).

Definition 11.13 (Upper and Lower Lagrange Multipliers)

We fix u, D and B to be as in Proposition 11.10. We set m := | Ω_u ∩ B|. For any constant δ > 0, we define the upper Lagrange multiplier μ ₊(δ) as follows:

$$\displaystyle \begin{aligned} \mu_+(\delta) = \inf \big\{ \mu\geq 0 \quad \text{for which (11.25) does hold }\big\}\,,\quad \text{where} \end{aligned}$$

$$\displaystyle \begin{aligned} \!\!\!\!\!\!\mathcal F_\mu(u,B)\le \mathcal F_\mu(v,B) \quad \text{for every}\quad v\in H^1(B)\quad \text{such that}\quad \begin{cases} u-v\in H^1_0(B),\\ m\leq |\Omega_v| \leq m+\delta\,. \end{cases} \end{aligned} $$

(11.25)

Analogously, we define the lower Lagrange multiplier μ ₋(δ):

$$\displaystyle \begin{aligned} \mu_-(\delta) = \sup \big\{ \mu\ge 0 \quad \text{for which (11.26) does hold }\big\}\,,\quad \text{where} \end{aligned}$$

$$\displaystyle \begin{aligned} \!\!\!\!\!\!\mathcal F_\mu(u,B)\le \mathcal F_\mu(v,B) \quad \text{for every}\quad v\in H^1(B)\quad \text{such that}\quad \begin{cases} u-v\in H^1_0(B),\\ m-\delta\leq |\Omega_v| \leq m\,. \end{cases} \end{aligned} $$

(11.26)

Lemma 11.14

Suppose that D is a connected bounded open set in $\mathbb {R}^d$ and that u ∈ H ¹(D) is a given non-negative function such that:

• (a) u ≠ 0 and ;

• (b) u is stationary for $\mathcal F_\Lambda$ , that is,

  $$\displaystyle \begin{aligned} \delta\mathcal F_\Lambda(u,D)[\xi]=0\qquad \mathit{\text{for every}}\qquad \xi\in C^\infty_c(D;\mathbb{R}^d). \end{aligned}$$

Then, we have the following claims:

• (i) Suppose that there are δ > 0 and μ > 0 such that u satisfies (11.25). Then, Λ ≤ μ.

• (ii) Suppose that there are δ > 0 and μ > 0 such that u satisfies (11.26). Then, Λ ≥ μ.

Proof

Let us first prove claim (i). By Lemma 11.3 and the hypothesis (a), we get that there is a vector field $\xi \in C^\infty _c(D;\mathbb {R}^d)$ such that

$$\displaystyle \begin{aligned} \delta{\mathrm{Vol}}\,(\Omega_u)[\xi]=\frac{d}{dt}\Big\vert_{t=0}|\Omega_{u_t}|=1, \end{aligned}$$

where u _t = u ∘ (Id + tξ)⁻¹. Since for t small enough, we have that $|\Omega _u|\le |\Omega _{u_t}|\le |\Omega _{u}|+\delta$, the minimality of u gives that

$$\displaystyle \begin{aligned} \mathcal F_\mu(u,D)\le \mathcal F_\mu(u_t,D). \end{aligned} $$

Thus, taking the derivative at t = 0, we get that

$$\displaystyle \begin{aligned} 0\le \frac{d}{dt}\Big\vert_{t=0}\mathcal F_\mu(u_t,D)=\frac{d}{dt}\Big\vert_{t=0}\mathcal F_\Lambda(u_t,D)+(\mu-\Lambda)\frac{d}{dt}\Big\vert_{t=0}|\Omega_{u_t}|=\mu-\Lambda, \end{aligned}$$

which proves (i). The proof of (ii) is analogous. □

As an immediate corollary, we obtain the following lemma.

Lemma 11.15 (μ ₋≤ Λ_u ≤ μ ₊)

Let D be a connected bounded open set in $\mathbb {R}^d$ and u ∈ H ¹(D) be a non-negative function such that (11.3) holds. Let m = | Ω _u| and Λ _u > 0 be the Lagrange multiplier of u in D, that is,

$$\displaystyle \begin{aligned} \delta \mathcal F_0(u,D)[\xi]+\Lambda_u\delta\mathit{\text{Vol }}(\Omega_u)[\xi]\qquad \mathit{\text{for every}}\qquad \xi\in C^\infty_c(D;\mathbb{R}^d). \end{aligned}$$

Let B, μ ₊(δ) and μ ₋(δ) be as in Definition 11.13 . Then, for every δ > 0, we have

$$\displaystyle \begin{aligned} \mu_-(\delta)\le \Lambda_u\le \mu_+(\delta). \end{aligned}$$

Notice that we still might have μ ₋(δ) = 0 and μ ₊(δ) = +∞. In Proposition 11.16 below we will prove that this cannot occur.

Proposition 11.16 (Convergence of the Upper and the Lower Lagrange Multipliers)

Let D be a connected bounded open set in $\mathbb {R}^d$ . Let u ∈ H ¹(D) be a non-negative function satisfying the minimality condition (11.3) in D and let Λ _u > 0 be the Lagrange multiplier of u in D, given by Proposition 11.2 . Let B ⊂ D be as in Proposition 11.10 . Then, we have

$$\displaystyle \begin{aligned} \lim_{\delta\to 0} \mu_+(\delta) = \lim_{\delta\to0} \mu_-(\delta) = \Lambda_u. \end{aligned}$$

Proof

We will work only in the ball B. The presence of the larger domain D is only necessary to assure the convergence of the Lagrange multipliers (see Lemma 11.9) for the different auxiliary problems that we will use below. We will proceed in three steps.

Step 1 We will first prove that μ ₊(δ) < +∞, for every δ > 0. This is equivalent to prove that there is some μ > 0, for which the minimality condition (11.25) is satisfied. Assume, by contradiction, that for every n > 0, there exists some function u _n ∈ H ¹(B) such that

$$\displaystyle \begin{aligned} \mathcal F_n(u_n,B)<\mathcal F_n(u,B)\,,\qquad u_n-u\in H^1_0(B)\qquad \text{and}\qquad m\le |\Omega_{u_n}|\le m+\delta. \end{aligned}$$

Thus, if v _n is a solution of the auxiliary problem

$$\displaystyle \begin{aligned} \min \big\{\mathcal F_0(v,B)+n\big(|\Omega_{v}| - m\big)_+ \ : \ v\in H^1(B)\,,\ v-u\in H^1_0(B),\ |\Omega_v| \leq m+\delta \big\}, \end{aligned} $$

(11.27)

then, we have that

$$\displaystyle \begin{aligned} \mathcal F_0(v_n,B)\le \mathcal F_0(v_n,B)+n\big(|\Omega_{v_n}| - m\big)_+& \le \mathcal F_n(u_n,B)+n\big(|\Omega_{u_n}| - m\big)_+{}\\ & < \mathcal F_0(u,B)+n\big(|\Omega_{u}| - m\big)_+=\mathcal F_0(u,B).\notag \end{aligned} $$

(11.28)

Thus, by Proposition 11.1 (Step 5 of the proof), we obtain $|\Omega _{v_n}|>m.$ Thus, we may assume

$$\displaystyle \begin{aligned} m<|\Omega_{v_n}|\le m+\delta\qquad \text{for every}\qquad n\in\mathbb{N}\,. \end{aligned}$$

Using again (11.28), we obtain

$$\displaystyle \begin{aligned} \mathcal F_0(v_n,B)+n\big(|\Omega_{v_n}|-m\big)<\mathcal F_0(u,B), \end{aligned}$$

which, in particular, implies that

$$\displaystyle \begin{aligned} |\Omega_{v_n}|-m\le \frac 1n\mathcal F_0(u,B)\qquad \text{and}\qquad \lim_{n\to\infty}|\Omega_{v_n}|= m. \end{aligned}$$

Now, notice that, setting $m_n:=|\Omega _{v_n}|$, we have that v _n is a solution of

$$\displaystyle \begin{aligned} \min \big\{\mathcal F_0(v,B)\ : \ v\in H^1(B)\,,\ v-u\in H^1_0(B),\ |\Omega_v|=m_n \big\}. \end{aligned}$$

In particular, there is a Lagrange multiplier $\Lambda _{v_n}$ such that

$$\displaystyle \begin{aligned} \delta\mathcal F_0(v_n,B)[\xi]+\Lambda_{v_n}\delta\text{Vol}\,(\Omega_{v_n})[\xi]=0\quad \text{for every}\quad \xi\in C^\infty_c(B;\mathbb{R}^d)\,, \end{aligned}$$

and a vector field $\xi _n\in C^\infty _c(B;\mathbb {R}^d)$ such that

$$\displaystyle \begin{aligned} \delta\text{Vol}\,(\Omega_{v_n})[\xi_n]=1. \end{aligned}$$

We set $v_n^t = v_n\circ (Id+t\xi _n)^{-1}$. Choosing t > 0 small enough and $n\in \mathbb {N}$ big enough, we get

$$\displaystyle \begin{aligned} v_n^t -u\in H^1_0(B)\qquad \text{and}\qquad m<|\Omega_{v_n^t}|<m+\frac 1n\mathcal F_0(u,B)\le m+\delta. \end{aligned}$$

Then, by (11.27), we have

$$\displaystyle \begin{aligned} \mathcal F_0(v_n,B) + & n\big(|\Omega_{v_n}|-m\big) \le \mathcal F_0(v_n^t,B) + n\big(|\Omega_{v_n^t}|-m\big)\\ & =\mathcal F_0(v_n,B) + t\,\delta\mathcal F_0(v_n,B)[\xi_n] + n\big(|\Omega_{v_n}| + t\,\delta\text{Vol}\,(\Omega_{v_n})[\xi_n]-m\big)+o(t)\\ & =\mathcal F_0(v_n,B) - t\Lambda_{v_n} + n\big(|\Omega_{v_n}| +t -m\big) +o(t), \end{aligned} $$

which implies $n \leq \Lambda _{v_n}$. On the other hand, Lemma 11.9 implies that

$$\displaystyle \begin{aligned} \displaystyle\lim_{n\rightarrow\infty} \Lambda_{v_n}= \Lambda_u<\infty\,, \end{aligned}$$

which is a contradiction. This concludes the proof of Step 1.

Step 2 In this step, we prove that lim_δ→0 μ ₊(δ) = Λ_u.

Let δ _n be an infinitesimal decreasing sequence. We will prove that lim_n→∞ μ ₊(δ _n) = Λ_u.

Fix ε ∈ (0, Λ_u) and set α _n to be

$$\displaystyle \begin{aligned} 0<\alpha_n:=\mu_+(\delta_n) - \varepsilon < \mu_+(\delta_n). \end{aligned}$$

We will show that, for n big enough, α _n ≤ Λ_u. Let u _n ∈ H ¹(B) be solution to the auxiliary problem

$$\displaystyle \begin{aligned} \min \Big\{ \mathcal F_0(v,B) \!+\! \alpha_n\big(|\Omega_v|-m\big)_+ \ : \ v \in H^1(B),\ v\!-\!u\in H^1_0(D), \ |\Omega_v| \le m\!+\!\delta_n \Big\}. \end{aligned} $$

(11.29)

We consider two cases:

Case 1 (of Step 2). Suppose that $|\Omega _{u_n}|\le m$. Then, the optimality of u gives

$$\displaystyle \begin{aligned} \mathcal F_0(u,B) \le \mathcal F_0(u_n,B). \end{aligned}$$

On the other hand, the optimality of u _n gives

$$\displaystyle \begin{aligned} \mathcal F_0(u,B) +\alpha_n \big(|\Omega_{u}|-m\big) & =\mathcal F_0(u,B) +\alpha_n \big(|\Omega_{u}|-m\big)_+\le \mathcal F_0(u_n,B) + \alpha_n\big(|\Omega_{u_n}|-m\big)_+\\ & \le \mathcal F_0(v,B) +\alpha_n \big(|\Omega_{v}|-m\big)_+ = \mathcal F_0(v,B) +\alpha_n \big(|\Omega_{v}|-m\big), \end{aligned} $$

for every v ∈ H ¹(B) such that $u-v\in H^1_0(B)$ and m ≤| Ω_v|≤ m + δ _n, which contradicts the definition of μ ₊(δ _n).

Case 2 (of Step 2). Suppose that $m<|\Omega _{u_n}|\le m+\delta _n.$ Notice that, setting $m_n:=|\Omega _{u_n}|$, the solution u _n to (11.29) is also a solution to the problem

$$\displaystyle \begin{aligned} \min \Big\{ \mathcal F_0(v,B)\ : \ v \in H^1(B)\,,\ v-u\in H^1_0(B)\,, \ |\Omega_v| =m_n \Big\}. \end{aligned}$$

By Proposition 11.2, there is a Lagrange multiplier $\Lambda _{u_n}\geq 0$ such that

$$\displaystyle \begin{aligned} \delta\mathcal F_0(u_n,B)[\xi]+\Lambda_{u_n}\delta\text{Vol}\,(\Omega_{u_n})[\xi]=0\quad \text{for every}\quad \xi\in C^\infty_c(B_r;\mathbb{R}^d)\,, \end{aligned}$$

and a vector field $\xi _n\in C^\infty _c(B_r;\mathbb {R}^d)$ such that $\delta \text{Vol}\,(\Omega _{u_n})[\xi _n]=1.$

We set $u_n^t:=u_n\circ (Id+t\xi _n)^{-1}$. By the minimality of u _n, for t < 0 small enough, we have

$$\displaystyle \begin{aligned} \mathcal F_0(u_n,B) + & \alpha_n\big(|\Omega_{u_n}|-m\big) \leq \mathcal F_0(u_n^t,B) + \alpha_n\big(|\Omega_{u_n^t}|-m\big)\\ & = \mathcal F_0(u_n,B) +t\,\delta \mathcal F_0(u_n,B)[\xi_n]+\alpha_n\big(|\Omega_{u_n}|+t\,\delta\text{Vol}\,(\Omega_{u_n})[\xi_n]-m\big)+o(t)\\ & =\mathcal F_0(u_n,B) -t\Lambda_{u_n} + \alpha_n\big(|\Omega_{u_n}|+t-m\big) + o(t), \end{aligned} $$

from which we deduce that $\Lambda _{u_n}\geq \alpha _n$. Now, by Lemma 11.9 we get that

$$\displaystyle \begin{aligned} \lim_{n\to\infty}\mu_+(\delta_n) = {\varepsilon}+\lim_{n\to\infty}\alpha_n \le {\varepsilon}+\lim_{n\to\infty}\Lambda_{u_n} = {\varepsilon}+\Lambda_u. \end{aligned}$$

Since Λ_u ≤ μ ₊(δ _n) and ε > 0 is arbitrary, we get the claim of Step 2.

Step 3 In this last step we will prove that lim_δ→0 μ ₋(δ) = Λ_u.

It is sufficient to show that, for aby decreasing infinitesimal sequence δ _n → 0, we have

$$\displaystyle \begin{aligned} \displaystyle\Lambda_u =\lim_{n\to\infty}\mu_-(\delta_n), \end{aligned}$$

Precisely, we will show that for any fixed constant ε > 0, we have Λ_u − ε ≤lim_n→∞ μ ₋(δ _n).

Let β _n := μ ₋(δ _n) + ε and u _n be a solution of the problem

$$\displaystyle \begin{aligned} {} \min \Big\{\mathcal F_0(v,B) + \beta_n\big(|\Omega_{v}|-(m-\delta_n)\big)_+ \ :\ v\in H^1(B),\ v-u\in H^1_0(B),\ |\Omega_v| \leq m \Big\}. \end{aligned}$$

We consider three cases:

Case 1 (of Step 3). Suppose that $|\Omega _{u_n}|=m.$

By the minimality of u, we have that $\mathcal F_0(u,B)\le \mathcal F_0(u_n,B)$. Now, the minimality of u _n, gives that, for every v ∈ H ¹(B) such that $v-u\in H^1_0(B)$ and m − δ _n ≤| Ω_v|≤ m, we have

$$\displaystyle \begin{aligned} \mathcal F_0(u,B) + \beta_n|\Omega_u| \le \mathcal F_0(u_n,B) + \beta_n|\Omega_{u_n}|\le \mathcal F_0(v,B) + \beta_n|\Omega_v|, \end{aligned} $$

which contradicts the definition of μ ₋(δ _n).

Case 2 (of Step 3). Suppose that $|\Omega _{u_n}|<m-\delta _n.$

Then we have

$$\displaystyle \begin{aligned} \mathcal F_0(u_n,B) \le \mathcal F_0(u_n+t\varphi,B), \end{aligned}$$

for every $\varphi \in C^\infty _c(B)$ with sufficiently small compact support. This implies that u _n is harmonic in B. By the strong maximum principle, we obtain that either u _n ≡ 0 or u _n > 0 in B, which is impossible for n large enough.

Case 3 (of Step 3). Suppose that $m-\delta _n\le |\Omega _{u_n}|<m$.

We set $m_n:=|\Omega _{u_n}|$. Then, u _n is a solution to the problem

$$\displaystyle \begin{aligned} \min \Big\{ \mathcal F_0(v,B)\ : \ v \in H^1(B),\ v-u\in H^1_0(B),\ |\Omega_v| =m_n \Big\}. \end{aligned}$$

By Proposition 11.2, there is a Lagrange multiplier $\Lambda _{u_n}\geq 0$ such that

$$\displaystyle \begin{aligned} \delta\mathcal F_0(u_n,B)[\xi]+\Lambda_{u_n}\delta\text{Vol}\,(\Omega_{u_n})[\xi]=0\quad \text{for every}\quad \xi\in C^\infty_c(B;\mathbb{R}^d)\,, \end{aligned}$$

and a vector field $\xi _n\in C^\infty _c(B;\mathbb {R}^d)$ such that $\delta \text{Vol}\,(\Omega _{u_n})[\xi _n]=1.$

We set $u_n^t:=u_n\circ (Id+t\xi _n)^{-1}$. Let t > 0 be small enough. Then $u_n^t$ is such that

$$\displaystyle \begin{aligned} u_n^t-v\in H^1_0(B)\qquad \text{and}\qquad |\Omega_{u_n}|=m_n\le |\Omega_{u_n^t}|=m_n+t+o(t)<m\,. \end{aligned}$$

Thus, by the minimality of u _n, we get

$$\displaystyle \begin{aligned} \mathcal F_0(u_n,B) + & \beta_n\big(|\Omega_{u_n}|-(m-\delta_n)\big)\le \mathcal F_0(u_n^t,B) + \beta_n\big(|\Omega_{u_n^t}|-(m-\delta_n)\big)\\ & \le \mathcal F_0(u_n^t,B) + t\,\delta\mathcal F_0(u_n,B)[\xi_n] \\ & \quad + \beta_n\big(|\Omega_{u_n}|+t\,\delta\text{Vol}\,(\Omega_{u_n})[\xi_n]-(m-\delta_n)\big)+o(t)\\ & = \mathcal F_0(u_n,B) -\Lambda_{u_n}t + \beta_n\big(|\Omega_{u_n}|+t-(m-\delta_n)\big) + o(t), \end{aligned} $$

which implies that

$$\displaystyle \begin{aligned} \Lambda_{u_n} \le\beta_n =\mu_-(\delta_n) + {\varepsilon}. \end{aligned}$$

Finally, by Lemma 11.9, we get

$$\displaystyle \begin{aligned} \Lambda_u=\lim_{n\to\infty}\Lambda_{u_n} \le \lim_{n\to\infty}\mu_-(\delta_n) + \varepsilon, \end{aligned}$$

which concludes the proof. □