Convexity of the free boundary for an exterior free boundary problem involving the perimeter

We prove that if the given compact set $K$ is convex then a minimizer of the functional $$ I(v)=\int_{B_R} |\nabla v|^p dx+\text{Per}(\{v>0\}),\,1<p<\infty, $$ over the set $\{v\in H^1_0(B_R)|\,\, v\equiv 1\,\,\text{on}\,\, K\subset B_R\}$ has a convex support, and as a result all its level sets are convex as well. We derive the free boundary condition for the minimizers and prove that the free boundary is analytic and the minimizer is unique.

Here we set Per({v > 0}) = +∞ if χ {v>0} / ∈ BV (R n ). This problem is the one-phase exterior analogue of the problem introduced in [ACKS] for a functional with general convex function F (t) in the first term (in [ACKS] they treat the case F (t) = t 2 ).
In [ACKS] (two phase, p = 2) and [Maz] (one phase, 1 < p < ∞) it is proved that the minimizers are Lipschitz continuous. This gives that the free boundary Γ u := ∂{x|u(x) > 0} is an almost minimal surface, thus C 1,1/2 -smooth. Let us recall some facts from the theory of almost minimal surfaces following [T].
As it is shown in [ACKS] and [Maz] (one phase, 1 < p < ∞) the Lipschitz regularity of the minimizer gives that the free boundary is an almost minimal surface with α(r) = r hence the reduced boundary Γ * u := ∂ * {x|u(x) > 0} is C 1, 1 2 regular and the singular set is of Hausdorff dimension n − 8 or less.
Remark 2. Any blow-up at the almost minimal surface is a minimal cone (see [T], p. 85). Thats why all points of the free boundary at which we can find a supporting smooth surface belong to the smooth part Γ * u . In this paper we restrict ourselves to the case F (t) = t p , p > 1, i.e., the functional (2) though we want to mention that the same ideas and methods will work in the general case (1) if we put some additional (rather weak) conditions on the function F . The main result of this paper is the following theorem. Theorem A. If K ⋐ B R is a convex set with non-empty interior and u is a minimizer of (2) over the set {v ∈ H 1 0 (B R )|v ≡ 1 on K} then the minimizer is unique, the set {u > 0} is convex and the free boundary ∂{u > 0} ∩ B R is an analytic surface.
We also derive the free boundary condition in case of general (nonconvex) K and prove that any minimizer u K satisfies the following inclusion Ω u K ⊂ Ω u cov(K) , see the notations below. This means for instance that for large R we will have Ω u K ⋐ B R .
It is noteworthy that convexity results for the so-called Bernoulli free boundary problem has been extensively studied, see [A], [HS1], [HS2] and the references therein.
1.2. Notations. In the sequel we use the following notations: ∂Ω u the free boundary, Γ * u ∂ * Ω u the reduced boundary of Ω u (see [EG]), cov(U) the convex hull of the set U.
1.3. Organization of the paper. In Sections 2 and 3 we develop some technical tools which will be used in the proofs coming after.
In Section 4 we derive the free boundary condition and prove that the reduced free boundary is analytic. An interesting geometric result about the mean curvature of the boundary of the convex hull of a nonconvex domain is proved in Section 5. The main result of the paper, the convexity of the free boundary and the uniqueness of the minimizer is proved in Section 6.
2. An energy estimate for p-harmonic extensions Assume K ⋐ Ω 1 ⊂ Ω 2 , where K, Ω 1 , Ω 2 are open and bounded subsets of R n with non-empty interior, and that u j minimizes the functional in the class of functions {v ∈ H 1 0 (Ω j )| v ≡ 1 on K} (j = 1, 2). Then we say that u 2 is the p-harmonic extension of u 1 from Ω 1 to Ω 2 . The use of word "extension" is a little bit misleading here, since u 1 = u 2 in Ω 1 , but we keep it as an analogy to extension by zero, which would be standard in this situation, because of zero boundary data on the boundary of Ω 1 . We also extend functions in H 1 0 (Ω) by zero and assume that they are defined in all R n .
In this section we prove the following lemma.
From the convexity and monotonicity of t p it follows that Φ is convex in t. So we can write This gives us exactly the following in Ω 1 . Integrating partially in the domain Ω 1 like in (5) we get that Remark 4. Note that the first inequality in (4) does not require smoothness assumptions on ∂Ω i , i = 1, 2.

The Hopf lemma for p-harmonic functions in domains with Liapunov-Dini boundary
Let us present the definition of Liapunov-Dini surface following [W].
Definition 5. A Liapunov-Dini surface S is a closed, bounded (n − 1)−dimensional surface satisfying the following conditions: (a) At every point of S there is a uniquely defined tangent (hyper-)plane, and thus also a normal.
(b) There exits a Dini modulus of continuity ǫ(t) such that if β is the angle between two normals, and r is the distance between their foot points, then the inequality β ≤ ǫ(r) holds.
(c) There is a constant ρ > 0 such that for any point x ∈ S any line parallel to the normal at x meets S ∩ B ρ (x) at most once.
Note that a domain E with Liapunov-Dini boundary satisfies a kind of interior and exterior Dini condition in the following sense: There exists a convex Liapunov-Dini domain K such that for any point x 0 ∈ ∂E there exists a translation and rotation K x 0 of the domain K satisfying The following lemma is proved in [J], see also [MPS].
Lemma 6. Let Ω\K be a convex ring and u be its p-capacitary potential. Then Now we formulate and prove the main result of this section, which might be a known result but we could not find any reference.
Lemma 7. Assume u is a p-harmonic function in the domain U. Further assume y ∈ ∂U, ∂U satisfies the interior and exterior Dini conditions locally near y and u(x) ≥ u(y) for all x ∈ U. Then there exist positive constants r 0 , c, C such that Proof. Let us take the function w to be the minimizer of the Dirichlet integral in {v ∈ H 1 0 (K 2 )|v ≡ 1 on K 1 }, where K 1 and K 2 are convex domains with Liapunov-Dini boundary and K 1 ⋐ K 2 . Thus we have ∆w = 0 on K 2 \K 1 . From the Hopf lemma for harmonic functions (Thm. 2.5, [W]) and the convexity and regularity of the level sets of w (see [L]) we know that ∇w(x) = 0, for any x ∈ K 2 \K 1 . Now we will prove the existence of a smooth, convex function f : . This will mean that the function f (w) is a sub-solution for ∆ p and has non-vanishing gradient, thus it will work as a standard barrier function.
We have is the well known infinity Laplace operator 1 . On the other hand We see that for p ≥ 2 we can take f (t) ≡ t. This follows from the Lemma 6. In case 1 < p < 2 we continue as follows. We have from [W] that the derivatives of w are continuous up to the boundary and do not vanish. Moreover we have bounds for the second derivatives of w near the boundary (formula (2.4.1) in [W]) where ζ ∈ L 1 (0, dist(K 1 , R n \K 2 )/2), and d(x) is the distance function from the boundary of the domain K 2 \K 1 . Coming back to our case there exists a function ζ 1 (t) ∈ L 1 ((0, 1)) ∩ C((0, 1)) such that Thus we can take for instance where the constant c > 0 is chosen to get f (1) = 1. Note that the function 1 − f (w) is a super-solution for ∆ p in K 2 \K 1 and will give us bounds from above.
Remark 8. In case of the C 1,α boundary the existence of the gradient of the function u at the boundary is known (see [Li]) and we can write

The free boundary condition
First let us prove that for convex K the free boundary stays away from the set K.
Lemma 9. Let the set K be convex and u be the minimizer of (1). Then there exists a constant δ depending on n, p and the set K such Proof. Let us take the points y ∈ Γ u and x ∈ K such that dist(y, . Then we have from the isoperimetric inequality and from the minimality condition that where c depends on the dimension. From the convexity of K and the fact that it has a non-empty interior we know that there is a cone C and r K > 0 such that for any point y ∈ ∂K there exists a rotation and translation of the set C r K := C ∩B r K such that 0 → y and C r K is mapped into K. In other words at any point of ∂K we can put a conical set of fixed opening and length r K lying inside K. This gives that H n−1 (∂B r (y) ∩ K) ≥ c K r n−1 , for r < r K and all y ∈ ∂K, c K depends only on K. Let us take the function ζ(x) to be the p-harmonic potential of the convex ring B 1 (0)\(B 1/4 (0) ∩ C).
Step 1: We first exclude the case r 0 = 0. Assume y ∈ Γ u ∩ K. We take as a perturbation of u the function v(x) := max(u(x), ζ r (x)), where ζ r (x) := ζ((x − y)/r) and we can without loss of generality assume that {x|ζ r (x) = 1} ⊂ K for all r < r K . Note that the function ζ r is the p-harmonic extension of the function u from {u < ζ r } ∩ Ω u to {u < ζ r }, which together with Lemma 3 and Remark 4 gives the following The second inequality uses the fact of u being a minimizer. On the other hand using (7) (remember that r 0 is assumed to be 0) we obviously can find constants c 1 , c 2 depending only on K, n and p such that a contradiction. In the second and third inequalities of (9) we used (7) and the fact that cr −1 < |∇ξ r | < Cr −1 in B r (y)\B 3r/4 (y) for some positive constants depending on r K .
Step 2: Now we know that r 0 > 0 and we can use the estimates used by Mazzone (Lemma 3.2, [Maz]) for the terms in (8). If we denote by d ′ (x) := dist(x, ∂B r (y)) we obtain that On the other hand as in (9) Br(y)\Ωu where c depends only on n and K. Summing up we obtain that where all constants depend only on n, p and K.
Lemma 10. The reduced free boundary Γ * is analytic and Γ * ∩ B R satisfies the free boundary condition where κ is the mean curvature. Moreover on Ω u ∩ ∂B R we have pointwise the inequality Proof.
For a vector field η ∈ C 1 0 (N δ ; R n ), sup η ≤ 1 and small enough ǫ consider the bijective map Φ ǫ (x) = x + ǫη(x) and the function u ǫ (y) = u(Φ −1 ǫ (y)). From the minimality of u we have that Let us now calculate the terms above. We are following the book of Ambrosio, Fusco, Pallara ( [AFP], page 360), where all these calculations are carried out in a similar situation. Since On the other hand where div S F (x) = n k=1 ∇ S F k (x), e k is the tangential divergence of F on surface S and ∇ S f is the projection of ∇f (x) on the tangent space T x S (see Definition 7.27 and Theorem 7.31 in [AFP]).

Integrating by parts in
where ν is the normal vector, and Noting that ∇|∇u(x)| p , η = p|∇u| p−2 η, ∇ 2 · u∇u and summing up and letting ǫ go to 0 we obtain that for any η ∈ C 1 0 (N δ ; R n ). If we now rewrite the left hand side in terms of function φ and use the Proposition 7.40 from [AFP] we obtain that weakly in {|x ′ | < δ}. The C α regularity of the right hand side and the theory of quasilinear elliptic equations (see [GT]) give that φ is C 2,α , so the reduced free boundary is C 2,α -smooth as well and the free boundary condition (11) is true pointwise on Γ * .
Step 2: Now since higher regularity of the boundary implies the higher regularity of the function u up to the boundary, we can use the bootstrapping argument and obtain arbitrary smoothness, so the boundary is C ∞ .
Step 3: The analyticity follows from the theory of elliptic coercive systems (see [KNS]). Here we refer to the paper of Argiolas ( [Ar], p. 144), where a similar problem is treated in all details.
Step 4: The inequality (12) is due to the fact that we can carry out the domain variation only in one direction near ∂B R .

A concavity result
From now on we denote by κ(∂U) the interior mean curvature (in viscosity sense) of the C 1,1 part of the boundary of a domain U as follows. Assume 0 ∈ ∂U and the interior normal ν ∂U (0) shows in the direction of the e-axis. We take where S A = {(x, e)|e = Ax, x } and A is the set of all symmetric matrices A such that the set S A (the graph of a quadratic polynomial) locally touches ∂U from inside.
Let us consider the convex hull cov(U) of a (non-convex) set U with C 2 boundary. Note that then cov(U) has a C 1,1 boundary (see [KK]). For notational reasons let us assume U ⊂ R n+1 = {(x, e)|x ∈ R n , e ∈ R}.
The following lemma will be useful and is easy to prove.
Proof. We need to show that 1 κ(∂cov(U)) for all x 1 , x 2 ∈ (y 0 , z 0 ). Without loss of generality we can assume x 1 = (−1, 0, . . . , 0) and x 2 = (1, 0, . . . , 0). Since the supporting planes of cov(U) at x 1 and x 2 coincide we can further assume that the graphs of quadratic polynomials given by positive symmetric matrices A 1 and A 2 locally touch the boundary ∂cov(U) from inside and 0 < 2TrA i − κ(x i ) < ǫ for i = 1, 2. Since x 1 , x 2 lie on the x 1 axis we can assume that for i = 1, 2 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000  Let us now consider the sets which touch the boundary of cov(U) from inside locally at the points x 1 and x 2 respectively. Here x ′ = (x 2 , . . . , x n ). We will now "calculate" the intersection of the convex hull of this two sets with the plane {x|x 1 = 0}. This will locally touch the boundary ∂cov(U) from inside and give us the desired estimate on the mean curvature. The intersection of the convex hull of these two sets with the mentioned plane is We are going to calculate explicitly the expression on the right hand side. So for each x ′ we are looking for the minimum of the following function After differentiation in y ′ and some (simple) calculations we get that the infimum in (14) is attained at the values Substituting now the values of y ′ and z ′ into (14) and using the identity Note that the invertibility of B 1 + B 2 and B −1 1 + B −1 2 follows from the strict positivity of all eigenvalues of B 1 , B 2 . In three dimensions, when matrices B 1 , B 2 are given by positive numbers b 1 , b 2 , this interesting result is illustrated in Figure 1.
The proof now follows from the inequalities bellow: (15) 2 κ(∂cov(U)) Note that ǫ > 0 is arbitrary small and we have the first and the third inequalities in (15) by the construction of B 1 , B 2 and from the properties of the convex hull. The second inequality can be found in [ALL]. The case when κ(∂cov(U))(x) = 0 for some x ∈ (y 0 , z 0 ) follows from (15).

Convexity of the free boundary
In the proof of the key Lemma 14 we will use the following lemma (Lemma 4.1, [LS]). Let K ⊂ U be a compact convex set, U be open and non-convex and cov(U) be the convex hull of U. Further assume that the function u minimizes the functional (2) over the set {v ∈ H 1 0 (cov(U))|v ≡ 1 on K} and that the segment [y 0 , z 0 ] ⊂ ∂cov(U). Then the following lemma is true.
is convex on (y 0 , z 0 ). This is due to the fact (see [L]) that the level sets of a p-harmonic potential in a convex ring are convex.
The following lemma is key to the proof of the main result.
Then ∂cov(Ω u ) is locally a C 1,1 surface and is a solution of the (pointwise) free boundary inequality where κ is the interior mean curvature.
Proof. The C 1,1 regularity of the ∂cov(Ω u ) follows from the fact that at all points of ∂Ω u ∩ ∂cov(Ω u ) we have a supporting plane, thus (Remark 2) ∂Ω u is smooth in the neighborhood of this points, i.e., all singular points of ∂Ω u have positive distance from ∂cov(Ω u ). This means that ∂cov(Ω u ) is as regular as a convex hull of a domain with smooth boundary, that is C 1,1 (see [KK]). We get the desired inequality on ∂cov(Ω u ) ∩ ∂Ω u from the maximum principle and Lemma 10.
Assume now that x 0 ∈ ∂cov(Ω u )\∂Ω u . From the definition of the convex hull it follows that we can always write x 0 = m k=1 α k y k , y k ∈ ∂cov(Ω u ) ∩ ∂Ω u , α k > 0, m k=1 α k = 1, 2 ≤ m ≤ n. We proceed by induction in m. Assume there exist two points y 1 , y 2 ∈ ∂cov(Ω u ) ∩ ∂Ω u such that y 1 , x 0 and y 2 lay on one line.
We need to show that We know that 1 |∇u c (x)| and thus 1 |∇u c (x)| p is convex on [y 1 , y 2 ] (Lemma 13). Since (18) is true at the points y 1 and y 2 the proof follows from the concavity of 1 κ(∂cov(Ωu))(x) on the line segment (y 1 , y 2 ) and its lower semi-continuity (Lemmas 11 and 12).
The induction step m ⇒ m + 1 finishes the proof.
Theorem. If K is convex and u is a minimizer of (2) then Ω u is also convex.
Corollary 16. Using the same method as in the proof of the theorem one can easily prove the uniqueness of the minimizer by a contradiction argument. Note that in the two-phase (interior) case (see [ACKS]) the minimizer is not unique.
Corollary 17. Again by the same method one can prove that the domain Ω u of a minimizer u of the problem with general compact set K (even non-connected) is included in the domain Ωũ of the minimizerũ of the problem with compact set cov(K).
Corollary 18. For large enough R Proof. Due to the previous corollary we need to prove this only for convex K. If y ∈ ∂B R ∩ Γ u K then by convexity the conical set C(y, K) := {x|x ∈ [y, z], z ∈ K} ⊂ Ω u K and where the constant c depends on the set K. This contradicts to the fact that the total energy I(u) should decrease with R. the same. To see this let as fix any unit vector e = (α 1 , . . . , α n ) in R n and note that f α 1 t (α 2 t, . . . , α n t) − f 0 (α 2 t, . . . , α n t) = (f α 1 t (α 2 , . . . , α n ) − f 0 (α 2 , . . . , α n ))t 2 = o(t 2 ), as t → 0+.