A counterexample to the monotone increasing behavior of an Alt-Caffarelli-Friedman formula in the Heisenberg group

In this paper we provide a counterexample about the existence of an increasing monotonicity behavior of a function introduced in \cite{FeFo}, companion of the celebrated Alt-Caffarelli-Friedman monotonicity formula, in the noncommutative framework.


Introduction
In this paper we continue the research about the existence of a monotonicity formula in the Heisenberg group started in [11], see also [13] and [12].More precisely, we prove that there exists a function u such that, denoting u + := sup{u, 0} and u − := sup{−u, 0} defined in a neighborhood of 0 ∈ H 1 , if ∆ H 1 u ± ≥ 0 and u(0) = 0, then ξ = (x, y, t) ∈ H 1 , is not monotone increasing in a possibly small right neighborhood of 0, where H 1 is the first Heisenberg group.
In order to better understand the profile of this result, we recall that in [1] and [5] the celebrated monotonicity formula, in the Euclidean setting, was applied to prove regularity results about viscosity solutions of two phase problems like where A ⊂ R n is an open set and u ∈ C(A) is a viscosity solution, see [4] or [3] for such definition.
In particular, in [1] the authors proved that for every solution u ∈ H 1 (A) of (1.2) and for every P 0 ∈ F(u), the function is monotone increasing in a right neighborhood of 0. Such tool has been widely employed for proving regularity results of the solutions of (1.2).About this, we recall [8] for an overview concerning recent results on twophase problems in the Euclidean framework, see also [3].As a consequence, in order to study a two-phase problem in a noncommutative group, like the Heisenberg one, it would be useful to have the companion monotonicity formula to (1.3) in this framework.Indeed, in [9] the two-phase problem analogous to (1.2) has been settled down, i.e. (1.4) where Ω ⊂ H n and ∆ H n , ∇ H n denote the Kohn-Laplace operator and the horizontal gradient respectively.We refer to Section 3 for the main notation and definitions about this noncommutative structure.
In [11] we proved that if an intrinsic monotone increasing formula exists in H 1 , then it should be like (1.1).More precisely, in that paper, supposing β > 0, we denoted by and we looked for β such that J H n β,u is monotone increasing.Here, Q := 2n + 2 is the homogeneous dimension of H n .For the sake of simplicity, we considered the simplest case only, given by n = 1.Hence, we worked on only.
We did not manage to conclude that J H 1 u is monotone increasing for all the admissible functions, because following the strategy described in [3], we needed of some sharp results in geometric measure theory that, in the Heisenberg group, are not known yet, see [11] for the details.
As a consequence, in order to deepen that research, we decided to follow another strategy already available in the Euclidean case R n .More precisely, for selecting the right exponent β, and possibly deducing the increasing monotone behavior as well of J u in R n , it is useful to study the behavior of the function when u is harmonic.We recall in Section 2 this last approach already used in [18] to introduce the Alt-Caffarelli-Friedman monotone formula.Hence, we decided to follow that parallel proof adapting it to the Heisenberg group.
Nevertheless, on contrary to what we supposed, we discover that there exists at least a function u such that ∆ H 1 u = 0 and is strictly monotone decreasing in a small right neighborhood of 0, differently from what it happens in the Euclidean case.
Hence, starting from this result, we obtain that J H 1 u is strictly monotone decreasing for a careful choice of u.More precisely, J H 1 u cannot be monotone increasing for all the admissible functions, since the behavior of J H 1 u depends on the choice of u itself.In particular, if u = x we have that which satisfies ∆ H 1 u = 0, then J H 1 u is strictly monotone decreasing.This fact depends on the lack of orthogonality of the intrinsic harmonic polynomials in the Heisenberg group.
Our main result, whose proof is contained in Section 6, is the following one.
In Section 5 we show the explicit computation of the fact that I H 1 u is strictly monotone decreasing, obtaining the main tool for proving Theorem 1.1.In Section 7 we provide an extension of our argument and an application exhibiting a genuine nontrivial example of solution to a two-phase free boundary problem in the Heisenberg group.We conclude this introduction pointing out that in Section 4 we study the behavior of harmonic polynomials in order to justify the way we obtained the specific counterexample we need to.In fact, independently to our application, we think that the result contains therein may be interesting and useful by itself, especially for further applications.For instance in case we want to determine other polynomials with special properties with respect to ∆ H 1 .

Preliminary facts: the Euclidean case
In this section, we provide a preliminary fact inspired by the argument contained in [18] that here below we recall in detail.
More precisely, given a harmonic function u in an open set A ⊂ R n , we consider the following function depending on the radius r of the Euclidean ball B r (0) ⊂ A : Here |P | denotes the usual Euclidean norm of P ∈ A ⊂ R n , as well as |∇u(P )| is the usual Euclidean norm of the Euclidean gradient of the function u.
We want to rewrite (2.1) in a more convenient way, for understanding its behavior when r → 0. To this end, we write down the Taylor expansion of u at 0 In particular, in our case, u is harmonic.Hence we can group together the terms with the same homogeneity to achieve where P k (x) is a homogeneous polynomial of degree k.On the other hand, being u harmonic, each P k (x) has to be harmonic as well.As a consequence, we have that u is a sum (possibly infinite) of homogeneous harmonic polynomials.
Now, if we substitute (2.2) into I u , see (2.1), we have to compute the gradient of the homogeneous polynomials P k .For this purpose, we collect this calculation in the following technical lemma.
The proof is a straightforward computation.Therefore, we are ready to rewrite I u using (2.2) and Lemma 2.1 as well.Hence, from (2.2) we have We apply the classical coarea formula, see [10], and performing the change of variables in spherical coordinates to get which yields (2.4) Now, exploiting Lemma 2.1, we point out that ∇P k has each component which is a homogeneous polynomial of degree k − 1.Thus, it holds which entails, according to (2.4), , where and and having used the fact that To recap, we have showed that Hence, we immediately obtain the following result.
Proposition 2.2.Let I u be as in (2.1).Then I u is monotone increasing with respect to r and lim r→0 I u (r) = a 1 , with a 1 defined in (2.5).
Proof.The proof follows by two straightforward computations, using (2.7).Precisely, about the first one, letting r tend to 0 in (2.7), we obtain Concerning the second one, if we compute the first derivative of I u directly from (2.7), we get because the a k 's are nonnegative by virtue of (2.5), and hence I u is indeed monotone increasing.
Let us discuss that lim r→0 I u (r) = a 1 .By definition, we have We then recall that P 1 is the homogeneous polynomial of degree 1 coming from the Taylor expansion of u at 0. As a consequence, according to Lemma 2.1, ∇P 1 is a constant.Specifically, it is the gradient of u at 0. So, it follows This equality tells us that the limit of I u with r tending to 0 is strictly positive depending on whether the gradient of u vanishes in 0 or not.
To conclude this argument in the Euclidean case, we show a proof of (2.6).Precisely, we recall the following result, see [17].
Lemma 2.3.Let P h and P k be two harmonic homogeneous polynomials in B 1 of degree h and k respectively, with h = k.Then, it holds Proof.The proof is a direct consequence of the divergence theorem in the Euclidean setting and Lemma 2.1.Indeed, since P h and P k are both harmonic in B 1 , we achieve from the divergence theorem 0 = because x is the outward normal to ∂B 1 .Now, we can exploit Lemma 2.1 to get an expression of ∇P k , x for any k.Specifically, we have Then, exploiting (2.10), we obtain from (2.9) which gives It is now an immediate consequence to achieve (2.6), simply observing that, according to Lemma 2.1, each component of the gradient of a homogeneous polynomial is still a homogeneous polynomial.

The Heisenberg setting
Following the idea introduced in Section 2, we face the same problem in the framework of H 1 .For the sake of simplicity, we restrict ourselves to the H 1 case, nevertheless the argument holds in H n as well.
We recall here that H n denotes the set R 2n+1 , n ∈ N, n ≥ 1, endowed with the noncommutative inner law in such a way that for every where •, • denotes the usual inner product in R n .
We use the same symbols to denote the vector fields associated with the previous vectors, so that for i = 1, . . ., n, we have: The commutator between the vector fields is otherwise is 0. The intrinsic gradient of a real valued smooth function u in a point P is Now, there exists a unique metric on HH n P = span{X 1 (P ), . . ., X n (P ), Y 1 (P ), . . ., Y n (P )} which makes orthonormal the set of vectors {X 1 , . . ., X n , Y 1 , . . ., Y n }.Thus, for every P ∈ H n and for every U, W ∈ HH n P , Since we mainly work on H 1 , that is to the case in which n = 1, we simply continue introducing the remaining notation in H 1 .In particular, we define a norm associated with the metric on the space span{X, Y }, as it follows: For example, the norm of the intrinsic gradient of a smooth function If ∇ H 1 u(P ) = 0, instead, we say that the point P is characteristic for the smooth surface {u = u(P )}.In particular, for every point M ∈ {u = u(P )}, which is not characteristic, it is well defined the intrinsic normal to the surface {u = u(P )} as follows: The Kohn-Laplace operator is where it results so that ∆ H 1 is a degenerate elliptic operator because the smallest eigenvalue associated with the matrix At this point, we introduce in the Heisenberg group In particular, for every positive number r, the gauge ball of radius r centered at 0 is It is worth to say that this structure is endowed by suitable properties, like the left invariance with respect to the inner law.More precisely, for every point P ∈ H 1 and for every P ∈ H 1 it results where meas 3 denotes the usual Lebesgue measure in R 3 .Moreover, if u is a C 1 function in H 1 and for every In addition, a dilation semigroup is defined as follows: for every r > 0 and for every P ≡ (x, y, t) ∈ H 1 , let δ r (P ) := (rx, ry, r 2 t). (3.4) As a consequence, denoting u r (S) := u(δ r (S)), it results that and The details of all previous properties can be found in [19], or in other handbooks like [7] or [2].Moreover, the following representation theorem holds, see [6], [14] and [16] for further developments.
where n E = n E (x) is the Euclidean unit outward normal to ∂E.
We also have the following intrinsic divergence theorem: where ν H 1 (x) is the intrinsic horizontal unit outward normal to ∂E, given by the (normalized) projection of n E (x) on the fiber HH 1 x of the horizontal fiber bundle HH 1 .

Some properties of the orthogonal polynomials in the Heisenberg group
In this section we develop our approach.Let us take the companion functional to (2.1) in H n , namely, for every P ∈ H n , where Q is 4 in H 1 , and thus I H 1 u becomes On the other hand, recalling the translation invariance properties (3.2), (3.3), we obtain Hence, we can suppose to work, without restrictions, in the standard case in which our function u is defined in a neighborhood of 0 ∈ H 1 .
Following the parallelism with the Euclidean case, we assume that the function At this point, we want to rewrite (4.2), supposing that r is small.Specifically, we have the following equality.Proposition 4.1.Let I H 1 u be defined in (4.2).It holds ) 2) holds in the same way for u, even if k is not necessarily the height of α, repeating the same argument used to obtain (2.3), we have from (4.2) where P k is homogeneous of degree k in H 1 , see Remark 4.2.
We now want to exploit the coarea formula and the change of variables in spherical coordinates in H 1 to further rewrite I H 1 u .Before applying them, we stress that x 2 + y 2 .(4.5)We then achieve, from (4.4), which implies This fact then entails that ξ β is homogeneous of degree |β| + β 3 .Hence, a homogeneous polynomial P k of degree k in H 1 has the form Properties of polynomials in the Heisenberg group have been investigated in [15], where it has been pointed out how the harmonic polynomials in the Heisenberg group have done.
Therefore, to understand the behavior of I H 1 u it is useful to compute the gradient of homogeneous polynomials.Lemma 4.3.Let P k be as in (4.7).Then, it holds Proof.The proof is a straightforward computation.Given a smooth function u : H 1 → R, we denote by ∇ H 1 u as Consequently, we obtain So, it remains to calculate Xξ β and Y ξ β .We get by (4.8) which immediately yields the desired equality.
We are now ready to find the correspondent expression of I H 1 u to (2.7).Precisely, we get its following rewriting.Proposition 4.4.Let I H 1 u be as in (4.2).Then, we have where with and Proof.The proof follows by a rewriting of (4.3).About the first term in (4.3), using Lemma 4.3 and (4.6), we have which yields where we denote Q k (σ) by we first notice that a priori we do not know if the gradients of two H 1harmonic homogeneous polynomials of different degrees are orthogonal on the Koranyi unit sphere ∂B H 1 1 (0), in the sense of satisfying Thus, we go back to the proof of the correspondent result in the Euclidean case, Lemma 2.3.We recall that Lemma 2.3 is a direct consequence of the harmonicity property of the polynomials and the divergence theorem, where in particular we exploit the fact that the outward normal to ∂B 1 is exactly x.In the case of H 1 , we can use the same ingredients, but with the difference that the outward normal to ∂B H 1 1 (0) is if σ is not a characteristic point for ∂B H 1 1 (0), see Section 3.This difference could yield in turn that the correspondent result to (2.8) in H 1 is not true.To investigate this, we compute (∇ First of all, we write the explicit expression of ν H 1 (σ), that is, according to (4.5), (4.14) and the fact that |σ| Hence we obtain, again by virtue of Lemma 4.3, We then note that, differently from Lemma 2.3, (∇ , so we can not expect to achieve the same conclusion (2.7) in the case of H 1 .Specifically, if we repeat the same considerations done in the proof of Lemma 2.3, we reach the equality ) whereas in (4.3) we have Consequently, we can try to compare and with h = k, to understand if the term (4.17) can be zero.Let us compute explicitly again.First, about (4.18), we recall (4.7) and (4.15) and we get Concerning (4.19), we obtain, repeating some of the steps exploited to have (4.13), (∇ with T h,k (σ) defined as At We find which gives where So, the different structure in H 1 with respect to the Euclidean one yields a different expression for the functional I H 1 u compared to I u .However, we can deduce the correspondent statement to Proposition 2.2.Proposition 4.5.Let I H 1 u be as in (4.2).Then defined as in (4.10), and the monotonicity behavior of I H 1 u around r = 0 depends on the sign of a H 1 2,1 , if this term is different from 0. Proof.The proof directly follows by Proposition 4.4.First, letting r tend to 0 in (4.9), we get Next, for the condition on the monotonicity behavior of I H 1 u , from (4.9) we have This expression then entails lim using that a H 1 2,1 = a H 1 1,2 by (4.10).Hence, the monotonicity behavior of I H 1 u around r = 0 depends on the sign of a Let us discuss now, as in the Euclidean case, the condition Proposition 4.6.Let a H 1 1 be as in (4.10).It holds Proof.According to (4.10), we have In particular, recalling (4.11), Q 1 is the square norm of the H 1 -gradient of the homogeneous polynomial of degree 1, P 1 , coming from the Taylor expansion of u at 0. Therefore, exploiting Lemma 4.3 and repeating the considerations done in the Euclidean case, we obtain the thesis.
Remark 4.7.The equality above shows us that the limit of I H 1 u with r tending to 0 is strictly positive depending on whether the H 1 -gradient of u vanishes at 0 or not, in other words whether 0 is a characteristic point of {u = u(0)} or not.Indeed, we have since it is the integral of a strictly positive function on ∂B H 1 1 (0).
Moreover, we want to rewrite a H 1 2,1 to deal with a more explicit tool.Specifically, we have the following.Proposition 4.8.Let a H 1 2,1 be as in (4.10).Then, we have Proof.By virtue of (4.12), it holds In particular, we remark that we can not have, at the same time, |γ|+γ 3 = 1 and γ 3 = 0, because if γ 3 = 0, |γ| + γ 3 ≥ 2. Therefore, the terms in (4.23) all satisfy γ 3 = 0 and the only possibilities for γ are γ = (1, 0, 0) and γ = (0, 1, 0).So, we can simplify (4.23) and we reach Now, we first focus on In parallel, to obtain β 3 = 0, the only possible choice is β = (0, 0, 1), otherwise again |β| + β 3 ≥ 3. To recap, in (4.25) we have the terms Next, let us look at and since the structure is very similar to (4.25), we can repeat the same considerations with β 2 instead of β 1 and we have Consequently, according to (4.26) and (4.27), (4.24) becomes Recalling (4.10), we then achieve the desired expression of a H 1 2,1 .Remark 4.9.Looking into (4.22)carefully, we point out that each term is, up to a constant, of the form As a consequence, we get a H 1 2,1 = 0.This fact forces us to revisit the condition for the monotonicity behavior of I H 1 u around r = 0. Indeed, a H 1 2,1 = 0 tells us, from the proof of Proposition 4.5, that lim r→0 (I H 1 u ) ′ (r) = 0, which does not yield a sign of (I H 1 u ) ′ around r = 0.

Let us then analyze (I H 1
u ) ′ more.To this end, looking at the proof of Proposition 4.5, we recall that we have Hence, for small radii the terms with r to power 1 are the ones which establish the sign of (I H 1 u ) ′ .Let us focus on these terms.Precisely, we obtain the next result.Proposition 4.10.Let a H 1 2 and a H 1 3,1 be as in (4.10).We have that a H 1 2 is nonnegative, whereas a H 1 3,1 depends on the function u defining I H 1 u .
Proof.First, we rewrite (4.29) to obtain where we have exploited again that a H 1 3,1 = a H 1 1,3 by (4.10).We want to study at this point a H 1 2 + 2a H 1 3,1 .Looking into (4.10),we note that a H 1 2 involves the polynomial Q 2 , see (4.11).This polynomial is made up of monomials which are either positive or with at least one of the variables with odd exponent.Since (4.28) is true in the same way for monomials with at least one of the variables with odd exponent, just the positive monomials give a contribution in a H 1 2 .Consequently, a H 1 2 is nonnegative.In parallel, if we look at the definition of a H 1 3,1 , again in (4.10), the situation is similar, but with the difference that we have monomials with all the variables with even exponent where the coefficients are in the form b β b γ .Thus the sign of such monomials depends on the sign of b β b γ .In other words, in principle, the sign of a H 1 3,1 is related to the specific function u which defines I H 1 u .

Explicit computation of the counterexample
In this section we explicitly check that fixing the polynomial Proof.First, recalling that u defining I H 1 u has to be H 1 -harmonic, we show that ∆ H 1 u = 0. Directly computing, we get which implies ∆ H 1 u = −12x + 12x = 0. Now, we focus on the behavior of I H 1 u .Substituting (5.1) in (4.4), it holds Next, repeating the same steps exploited to achieve (4.3), we have which yields with Explicitly calculating the derivative of I H 1 u as in (5.2) and letting r → 0, we reach the thesis since a H 1 3,1 is positive by definition.

Nonexistence of an Alt-Caffarelli-Friedman type monotonicity formula in H 1
In this section, we show that an Alt-Caffarelli-Friedman type monotonicity formula in H 1 does not hold, at least considering the function (1.1).In fact, we can exploit the counterexample to the increasing monotonicity of I H 1 u provided in Lemma 5.1.
Proof of Theorem 1.1.We first note that J H 1 u (r) = I H 1 u + (r)I H 1 u − (r).(6.1)So, since I H 1 u + and I H 1 u − are nonnegative, we reach the desired result if we prove that they are both monotone decreasing.We claim that I H 1 u + (r) = I H 1 u − (r).Before proving this, we show that the claim implies the monotone decreasing behavior of I H 1 u + (r) and I H 1 u − (r).Indeed, by the obvious fact that I H 1 u (r) = I H 1 u + (r) + I H 1 u − (r), we deduce I H 1 u (r) = 2I H 1 u + (r) = 2I H 1 u − (r), which immediately gives the decreasing monotonicity of I H 1 u + and I H 1 u − from Lemma 5.1.As a byproduct of this remark, in this way, we prove that J H 1 u (r) is monotone decreasing, because it is the product of two positive monotone decreasing functions, see (6.1).

A further generalization with application
Let us consider now the following two phase continuous function , where, as usual, u(x, y, t) = x − 3yt − 2x 3 .Then, we first note that Since the zero level of u α 1 ,α 2 is the same of u, we remark that Hence, performing the same change of variables introduced in (6.2), we obtain On the other hand, in this case, keeping in mind (7.2) it results As a consequence of Lemma 5.1, from (7.3) follows that are monotone decreasing in a right neighborhood of 0. Hence, we conclude from (7.1) that J H 1 uα 1 ,α 2 is monotone decreasing because it is the product of the two positive monotone decreasing functions I H 1 α 2 u − and I H 1 α 1 u + .We conclude the paper with an application coming from the two phase function u α 1 ,α 2 .
The construction of non-trivial solutions of two-phase free boundary problems is not easy even in the Euclidean setting.

( 7 . 5 )
this point, if we compare (4.20) and (4.21), we see that the two terms are not the same.Therefore, we can not expect to have (4.17)equal to 0 in (4.3) and the correspondent expression to (2.7) in H 1 .Let us look then at what we achieve substituting (4.13) and (4.21) in (4.3).
.33) Now, the terms in (4.32) and (4.33) with σ x σ y and σ t give null contribution in a H 1 3,1 , for (4.28) extended to monomials with at least one of the variables with odd exponent.Consequently, in a H 1 3,1 it is sufficient to consider 3b