ON CONVEX COMPARISON FOR EXTERIOR BERNOULLIPROBLEMS WITH DISCONTINUOUS ANISOTROPYWILLIAM M FELDMAN AND NORBERT POŽÁR

Abstract. We give a new proof of a convex comparison principle for exterior
Bernoulli free boundary problems with discontinuous anisotropy.

1. Introduction
Let K be a closed convex subset of Rd with nonempty interior. We will consider
the convexity properties of the minimal supersolution of the anisotropic Bernoulli
problemin {u > 0} \ K, ∆u = 0|∇u| = Q(nx ) on ∂{u > 0},
(1.1)

 u=1
on K.
Here nx is the inner unit normal to ∂{u > 0} at x and Q : S d−1 → (0, ∞) is only
assumed to be bounded above and below and upper semi-continuous
Q(n) ≥ lim sup Q(n′ ).
n′ →nSuch problems, with discontinuous anisotropy, naturally arise from periodic homogenization scaling limits [5, 6, 9, 10, 14].
In the context of homogenization and other scaling limits, one cannot directly
show the minimality property of the limiting supersolutions. Instead the following
weak subsolution property is natural. Essentially this is a viscosity subsolution
property that only allows to test the free boundary condition with one-dimensional
test functions.
Definition 1.1. We say that φ ∈ C ∞ (U ) is one-dimensional in U if it is of theform φ(x) = f (x · p) in U for some f ∈ C ∞ (R) and p ∈ Rd , |p| = 1.
Definition 1.2. A weak subsolution of (1.1) is a nonnegative function u ∈ C(Rd )that is compactly supported, satisfies u ≤ 1 on K, is harmonic in {u > 0} \ K,and such that, whenever U is an open neighborhood and one-dimensional φ with∆φ < 0 touches u from above in {u > 0} at x ∈ ∂{u > 0} ∩ U with strict orderingu < φ on {u > 0} ∩ ∂U , then|∇φ(x)| ≥ Q(p),where p is from Definition 1.1.
The following supersolution definition is standard.
2020 Mathematics Subject Classification. 35R35, 35E10.
Key words and phrases. Bernoulli problem, comparison principle, discontinuous anisotropy.
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W. M. FELDMAN AND N. POŽÁR

Definition 1.3. A supersolution of (1.1) is a nonnegative function u ∈ C(Rd ) thatis compactly supported, u ≥ 1 on K, is harmonic in {u > 0} \ K, and wheneverφ ∈ C ∞ (U ), U open, ∆φ > 0 and ∇φ 6= 0 in U , touches u from below at x ∈∂{u > 0} ∩ U then
)
(
∇φ
(x) .
|∇φ(x)| ≤ Q|∇φ|We give a new proof of the following theorem.
Theorem 1.4. A supersolution of (1.1) is minimal if and only if it satisfies the
weak subsolution property Definition 1.2. In other words, a supersolution that is alsoa weak subsolution is the unique minimal supersolution. Furthermore the minimal
supersolution has convex superlevel sets.
This result has already been proved in [10] by Smart and the first author. This
comparison principle, for discontinuous anisotropies, was an important component
in the proof of the homogenization scaling limit of minimal supersolutions in thatwork and in [9]. This paper gives a new proof which we feel has valuable simplicity
and intuition. The main new ideas in this paper are in Lemma 3.4 and Corollary
3.7 below. We give a brief explanation about those new ideas here.
The diﬀiculty in using the weak subsolution condition is that, on a facet F (seeSection 2.1 for definitions) with inward normal n of the free boundary of a quasiconvex subsolution u, the weak subsolution condition only guarantees thatmax |∇u| ≥ Q(n).
F

Unfortunately if a supersolution touches u from above on this facet it may nottouch at the point where the maximum in the subsolution condition is saturated.
We would like instead the strong subsolution condition(1.2)

min |∇u| ≥ Q(n).
F

The first idea, exploited in Lemma 3.4, is that the weak and strong subsolution
conditions above are the same at exposed points, facets F which are singletons. By
Straszewicz’s Theorem, Theorem 2.2 quoted below, all extreme points are limits of
exposed points so, along with some additional convexity arguments we can showthat weak subsolutions actually satisfy the strong condition (1.2) on all facets atleast when Q is continuous.
This idea does not directly work in the case of upper semicontinuous Q; if it
did, all facets of a solution would actually be trivial. However it turns out, somewhat unexpectedly, that we can use a simple monotone continuous approximationargument of the upper semicontinuous Q to prove a comparison principle in the
discontinuous case too. This is the content of Corollary 3.7 below.
There have been several works on existence, uniqueness, convexity, and regularity of solutions of Bernoulli-type problem (1.1) with Q(n) ≡ 1 or continuous.
Classical work of Beurling [2], and later Schaeffer [16] and Acker [1], showed the existence/uniqueness of a convex solution in 2-d. Hamilton [11] gave a different proofin 2-d by a Nash-Moser implicit function theorem. Later Henrot and Shahgholian[12, 13] gave proofs based on maximum principle including nonlinear problems withx-dependent boundary conditions (satisfying a concavity assumption). Bianchini[3] extended this to anisotropic interior Bernoulli problems and to Finsler metrics [4]. The present paper is most similar to these works [3, 4, 12, 13]; for us the

CONVEX COMPARISON FOR ANISOTROPIC BERNOULLI PROBLEMS

3

main diﬀiculty is in proving comparison principle with only the weak subsolution
condition.
Acknowledgments. The authors appreciate discussions with Inwon Kim which
helped to motivate the proof idea. W. Feldman was partially supported by the
NSF grant DMS-2009286. N. Požár was partially supported by JSPS KAKENHI
Kiban C Grant No. 23K03212.
2. Convexity properties
2.1. Extreme and exposed points of convex sets. We recall several notions
and results from convex analysis; see for example [15, Sec. 18].
For a convex set X, a face is a convex subset Y ⊂ X such that every line segment
in X with relative interior point1 in Y has both end points in Y . If a face of X isa point, it is called an extreme point of X.
The convex set Y = {x ∈ X : h(x) = maxX h} for some linear function h is aface of X, and it is called an exposed face. If an exposed face X is a point, it iscalled an exposed point of X. Also we call a facet of X to be a nontrivial exposed
face, i.e. not the entire set X.
The following two results about extreme points and exposed points will play a
key role.
Lemma 2.1 ([15, Cor. 18.5.1]). A closed bounded convex set is the convex hull of
its extreme points.
Theorem 2.2 (Straszewicz’s Theorem[15, Th. 18.6]). Let X be a closed convex
set. Every extreme point of X is the limit of a sequence of exposed points of X.
2.2. Convexity properties of a gradient of a harmonic function. Let ∅ 6=
U ⊂⊂ V be two bounded open convex sets. We also assume that V is inner regular.
Let v be the unique solution of∆v = 0 in V \ U ,v = 1 on ∂U,
(2.1)


v = 0 on ∂V.
By [8, Th. 2.1] the super-level sets {v > t} are convex for all t ∈ (0, 1). This implies
convexity of the super-level sets of |∇v| on the facets of ∂V .
Lemma 2.3. Let U , V and v be as above and let Λ be a tangent hyperplane to ∂V .
Recall that V is uniformly inner regular. Set F = ∂V ∩ Λ. Then 1/|∇v| is convex
on F . In particular, its sublevel sets{x ∈ F : 1/|∇v(x)| < 1/τ } = {x ∈ F : |∇v(x)| > τ }are convex for any τ > 0.
Proof. Let n be the unit normal of Λ so that it is also the inner unit normal of ∂V
on F .
Fix ξ0 , ξ1 ∈ F and set ξt := (1 − t)ξ0 + tξ1 . For any θ > 0 we definehθ (t) := inf{h > 0 : v(ξt + hn) > θ},t ∈ [0, 1].

1(1 − t)x + ty for t ∈ (0, 1) are relative interior points of segment x–y.

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W. M. FELDMAN AND N. POŽÁR

By continuity of v we have hθ (t) > 0. Since {v > θ} % V as θ → 0 and {v > 0} is
inner regular at ξt we have hθ (t) < ∞ for suﬀiciently small θ > 0 for all t ∈ [0, 1].
In fact, hθ (t) → 0 as θ → 0. Since {v > θ} is convex by [8], hθ is convex on [0, 1].
Thereforeθv(ξt + hθ (t)n)= lim.
|∇v(ξt )| = ∇v(ξt ) · n = lim
θ→0 hθ (t)θ→0hθ (t)
By inner regularity and Hopf’s lemma |∇v(ξt )| > 0 for all t ∈ [0, 1] and hence wededuce that t 7→ 1/|∇v(ξt )| is convex since t 7→ hθ (t)/θ is convex and pointwise
limits of convex functions are convex.
□
Remark 2.4. It is not in general true that |∇v| is concave on F . Here is a counterexample on an unbounded V .
In d = 2, consider the top half-space V := {x2 > 0} and the potential between
a point charge at (0, 1) and the plane {x2 = 0} = ∂V :v(x) = − log(x21 + (x2 − 1)2 ) + log(x21 + (x2 + 1)2 ),and set U = {v > 1}, a convex set. We have|∇v(x)| =4∂v(x1 , 0) = 2,∂x2
x1 + 1x ∈ ∂V,

which is not convex.
3. Comparison principle
In this section we establish the proof of comparison between a weak subsolution
and a supersolution when at least one has convex support. When the supersolution
has convex support, we are able to directly show comparison with only semicontinuous Q. However, if only the weak subsolution has convex support we need toadditionally assume that Q is continuous. To finally use this result to establish
uniqueness of function that is both a supersolution and a weak subsolution withconvex support when Q is upper semicontinuous, we monotonically approximate Q
by continuous functions.
We recall the notions of sup- and inf-convolutions that we will use to regularize
free boundaries of the solutions. For a set V we define(3.1)

Vr := {x : Br (x) ⊂ V }.

V r := V + Br ,For a continuous u, we then define u as the solution of (2.1) with K r and {u > 0}r
in place of U and V . Similarly replacing U by Kr and V by {u > 0}r leads to ur .
If u is a weak subsolution so is ur , and furthermore {ur > 0} is uniformly inner
regular with radius r. Similarly, if u is supersolution so is ur and {ur > 0} is
uniformly outer regular.
Let us recall a stability property of supersolutions.
rLemma 3.1. Let Q be a upper semicontinuous function. Then supersolutions inthe sense of Definition 1.3 are stable under uniform convergence.
Proof. Let {uk } be a sequence of supersolutions and let u be its uniform limit.
Clearly u is continuous, u ≥ 1 on K and u is harmonic in {u > 0}.
Suppose that φ ∈ C ∞ (U ), ∆φ > 0 and ∇φ 6= 0 in U , touches u from below
at x ∈ ∂{u > 0} ∩ U . We can assume that u − φ has a strict minimum at x.
By uniform convergence, all points of minimum of uk − φ converge to x as k →CONVEX COMPARISON FOR ANISOTROPIC BERNOULLI PROBLEMS

5∞. By maximum principle for harmonic functions the minima are located on∂{uk > 0} ∩ U for suﬀiciently large k. Let xk be a sequence of such minima. For
large k we have ∇φ(xk ) ≤ Q(∇φ(xk )/|∇φ(xk )|). By sending k → ∞ and by theupper-semicontinuity of Q we deduce that this also holds at x. Therefore u is asupersolution.
□
3.1. Outer regular points have a classical normal derivative. First we show
a useful technical lemma: the normal derivative is well defined at outer regular freeboundary points, and for a supersolution it satisfies the supersolution condition.
Lemma 3.2. Suppose that u is a supersolution, and x0 ∈ ∂{u > 0} is an outerregular free boundary point. Then |∇u(x0 )| is well-defined and|∇u(x0 )| ≤ Q(nx0 ).
Proof. Without loss assume that x0 = 0 and outward normal determined by theexterior ball is −ed . By [7, Lemma 11.18] the blow-up sequenceu(rx)→ αxd in non-tangential cones for some α ∈ [0, k∇uk∞ ].
r
On the other hand the blow-up sequence ur (x) is uniformly Lipschitz continuouswith ur (0) = 0 and so every subsequence has a subsequence converging uniformly
on Rn . Every subsequential limit must be zero on xn ≤ 0 by the exterior ballcondition, and must agree with αxn on xn > 0 by the non-tangential limit. Thus
the sequence actually converges locally uniformly to α(xd )+ . By Lemma 3.1 forupper semicontinuous Q, the limit α(xn )+ is also a viscosity supersolution and so
α ≤ Q(ed ).
□ur (x) =3.2. Comparison principle: convex supersolution.
Lemma 3.3. Assume Q is upper semicontinuous. Suppose that v is supersolutionof (1.1) with V := {v > 0} convex, and u is a weak subsolution of (1.1). Thenv ≥ u.
Proof. By translation we can assume that the interior of K contains the origin.
Since v and u are harmonic in their positive sets, it is suﬀicient to show that{u > 0} =: U ⊂ V := {v > 0}.
Let us suppose that this is not the case. Recall the sup and inf-convolutions (3.1).
There exist r > 0 suﬀiciently small and a > 1 such that U r ⊂ aVr , ∂U r ∩ a∂Vr 6= ∅and K r ⊂ aKr by convexity of K. Set ṽ(x) := vr (a−1 x). By maximum principle
for harmonic functions, ur ≤ ṽ. Moreover ṽ is a strict supersolution.
Let us refer to ur , ṽ, U r and aVr as u, v, U , V , respectively, in the following.
Let x0 ∈ ∂U ∩ ∂V . Since U is inner regular and V outer regular, there is a uniquesupporting normal n0 at x0 . Let P0 be the supporting hyperplane to V at x0 . Then
Γ := ∂{u > 0} ∩ P0 is compact and all points of Γ are uniformly inner and outerregular points of ∂V and, by Lemma 3.2,|∇v(x)| ≤ Q(n0 ) − δ for all x ∈ Γ,
for some δ > 0 by the strict supersolution property. By [10, Lemma 3.14]v(x) ≤ (Q(n0 ) − δ + Cr1/3 )((x − x0 ) · n0 )+ on Γ + Br (0)so taking r = cδ 3 we find that
φ(x) = (Q(n0 ) − 12 δ)((x − x0 ) · n0 )+6

W. M. FELDMAN AND N. POŽÁR

touches u from above on Γ with strict ordering on ∂(Γ + Br (0)). This contradicts
the weak subsolution property of u.
□
3.3. Comparison principle: convex subsolution. Next we consider the casewhen a supersolution touches a convex weak subsolution from above. In [10, Lemmas 3.15 and 3.16] this case relied on a geometric argument using convexity allowing
to compare the gradient of the touching supersolution at a sequence of points withnormal νn approaching the touching direction ν0 to the gradient of the subsolution
on its νn facet. In place of this argument we are able to take advantage of wellknown results from convex analysis in the case of continuous Q. The idea is thatthe weak subsolution condition is not weak at exposed points, and we can exploit
the quasi-convexity of the gradient along with Straszewicz’s Theorem, Theorem 2.2,
to derive a strong subsolution condition on exposed faces.
Lemma 3.4. Suppose that v is a weak subsolution of (1.1) with continuous Q,
harmonic in {v > 0} \ K. Further assume that V = {v > 0} is convex anduniformly inner regular, and n is some normal direction with facet Vn = {x ∈ V :n · x = inf x∈V x · n}. Thenmin |∇v| ≥ Q(n).
VnNote that the weak subsolution property just says maxVn |∇v| ≥ Q(n). Actually
this allows us to conclude, in the setting of the Lemma, that Vn must be trivial (asingle point). If we did not assume that Q(n) was continuous we would only obtainmin |∇v| ≥ lim
inf Q(n′ ).
′
Vn

n →n

Proof. Since Vn is the convex hull of its extreme points by Lemma 2.1 and thesuperlevel set {x ∈ Vn : |∇v|(x) ≥ Q(n)} is convex by Lemma 2.3, it is enough toestablish the inequality at the extreme points of Vn .
Let x0 be an extreme point of Vn . By inner regularity, n is the inner unit normalof ∂V at x0 . Clearly x0 is also an extreme point of V . By Theorem 2.2 there is asequence xj → x0 of exposed points of V . At exposed points of V , we have by the
subsolution condition directly|∇v|(xj ) ≥ Q(nxj ).
Since ∂V is uniformly inner and outer regular|∇v|(x0 ) = lim |∇v|(xj ) ≥ lim Q(nxj ) = Q(n),
n→∞

j→∞

where we used the standard gradient regularity in C 1,1 domains, see [10, Lemma
3.13] for proof.
□
Now applying Lemma 3.4 along with a typical dilation argument to create atouching point we will get a comparison principle.
Lemma 3.5. Assume Q is continuous. Suppose that u is supersolution of (1.1),
and v is a weak subsolution of (1.1) with {v > 0} convex. Then v ≤ u.
Proof. By translation we can assume that the interior of K contains the origin.
Since v and u are harmonic in their positive sets, it is suﬀicient to show that
V := {v > 0} ⊂ {u > 0} =: U .

CONVEX COMPARISON FOR ANISOTROPIC BERNOULLI PROBLEMS

7

Let us suppose that this is not the case. Recall the sup and inf-convolutions (3.1).
There exist r > 0 suﬀiciently small and a > 1 such that V r ⊂ aUr , ∂V r ∩ a∂Ur 6= ∅and K r ⊂ aKr by convexity of K. Set ũ(x) := ur (a−1 x). By maximum principle
for harmonic functions, v r ≤ ũ.
At any point x ∈ ∂V r ∩ a∂Ur we have |∇v r | ≥ Q(n) by Lemma 3.4. However,this is a contradiction with the fact that ũ cannot be touched from below by a test
function with a slope larger than Q(n)/a by supersolution property.
□
3.4. Existence and uniqueness of a quasi-convex solution. Now we can use
the comparison principle, in the case of continuous Q, to show uniqueness andquasi-convexity of solutions.
Theorem 3.6 (cf. [10, Th. 3.10]). Assume that K is nonempty convex compact set
with inner regular boundary and assume that Q is continuous. Then there exists a
unique continuous function u with compact support that is both a supersolution anda weak subsolution. Moreover {u > 0} is convex.
Proof. The existence follows by Perron’s method taking the infimum u of supersolutions with convex support. The support of u must be convex since all supersolutions
are bounded from below by a harmonic function with 1 on K with support equal tothe intersection of the convex support of the supersolutions. A standard argument
yields that u is also a supersolution. And u must be a weak subsolution for otherwise we could touch it from above by a one dimensional C ∞ test function thatis a strong supersolution on a neighborhood of the contact set. By shifting thistest function down a smaller supersolution with convex support can be created,leading to a contradiction. Finally, uniqueness follows from the convex comparison□in Lemma 3.3 and Lemma 3.5.
The comparison principle and convexity in the case of upper semicontinuous Q
now follows by a monotone approximation argument.
Corollary 3.7. Suppose Q is upper semicontinuous and let u be the minimal
supersolution of (1.1). Then:(1) {u > 0} is convex(2) Any supersolution is larger than u, and u is larger than any weak subsolution, and therefore any supersolution which is also a weak subsolution is
identical to u.
Proof. Since Q is upper semi-continuous there is a monotone decreasing sequence ofcontinuous Qj : S d−1 → (0, ∞) with Qj & Q. Let uj be the minimal supersolutionof (1.1) corresponding to the equation |∇uj | ≤ Qj (nx ) on ∂{uj > 0}, and u be theminimal supersolution corresponding to |∇u| ≤ Q(n).
Since Qj ≥ Q we have u is a supersolution of |∇u| ≤ Qj (nx ) on ∂{u > 0} sou ≥ uj .
Similarly, since the Qj are monotone decreasing, the uj are a monotone increasing
sequence.
Call u∞ := limj→∞ uj ≤ u. We claim that u∞ is a supersolution of (1.1) which
will mean u∞ ≥ u and hence u∞ = u. The supersolution condition is standard
stability of the viscosity solution property with respect to uniform convergence,
Lemma 3.1.

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By Theorem 3.6 the sets {uj > 0} are convex for all j. Since they converge
monotonically upwards to {u > 0} that set is convex as well.
Now let v be a weak subsolution, since we now know that {u > 0} is convex wecan apply Lemma 3.3 to find v ≤ u. If v is also a supersolution of (1.1) then v ≥ u,by the minimality of u, and so we conclude that v = u.
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Department of Mathematics, University of Utah, Salt Lake City, Utah, USA
Email address: feldman@math.utah.edu
Faculty of Mathematics and Physics, Institute of Science and Engineering, Kanazawa
University, Kakuma, Kanazawa 920-1192, Japan
Email address: npozar@se.kanazawa-u.ac.jp