ON LIPSCHITZ CONTINUITY OF HARMONIC QUASIREGULAR MAPS ON THE UNIT BALL IN R

Annales Academiæ Scientiarum Fennicæ Mathematica

Volumen 33, 2008, 315–318

ON LIPSCHITZ CONTINUITY OF HARMONIC QUASIREGULAR MAPS ON THE UNIT BALL IN R

Miloš Arsenović, Vesna Kojić and Miodrag Mateljević

University of Belgrade, Faculty of Mathematics Studentski Trg 16, Belgrade, Serbia; [email protected]

University of Belgrade, Faculty of Organizational Sciences Jove Ilića 154, Belgrade, Serbia; [email protected]

University of Belgrade, Faculty of Mathematics Studentski Trg 16, Belgrade, Serbia; [email protected]

Abstract. We show that Lipschitz continuity ofφ: Sⁿ⁻¹→Rⁿ implies Lipschitz continuity of its harmonic extensionu=P[φ] :Bⁿ →Rⁿ, provideduis a quasiregular map.

1. Introduction and notations

It is known, even for n = 2, that Lipschitz continuity of φ: T → C, where T = {z ∈ C: |z| = 1}, does not imply Lipschitz continuity of u = P[φ]. In fact u = P[φ] is Lipschitz continuous iff the Hilbert transform of _dθ^dφ(e^iθ) (which is defined almost everywhere and bounded since φ is Lipschitz) is also in L^∞(T) (see [7]).

Here, for any n≥2, P[φ](x) =

Z

Sⁿ⁻¹

P(x, ξ)φ(ξ)dσ(ξ), x∈Bⁿ,

where P(x, ξ) = ^1−|x|_|x−ξ|n² is the Poisson kernel for the unit ballBⁿ ={x∈ Rⁿ: |x| <

1},dσis the normalized surface measure on the unit sphereSⁿ⁻¹andφ:Sⁿ⁻¹ →Rⁿ is a continuous mapping.

The situation is different forC^α (or Hölder) continuousφ: Sⁿ⁻¹ →Rⁿ,0< α <

1, i.e., for φ satisfying |φ(ξ)−φ(η)| ≤C|ξ−η|^α. In that case Hölder continuity of φ implies Hölder continuity of its harmonic extension u=P[φ], (see [2, 5]). In the case n= 2 it is a classical result, following from Privalov’s theorem (see [7]).

Our aim is to show that Lipschitz continuity is preserved by harmonic extension, if the extension is quasiregular. The analogous statement is, as noted, true for Hölder continuity without assumption of quasiregularity.

2000 Mathematics Subject Classification: Primary 31B25; Secondary 30C65, 31B05.

Key words: Quasiregular mappings, harmonic mappings, Lipschitz spaces.

316 Miloš Arsenović, Vesna Kojić and Miodrag Mateljević

2. Result

Theorem 1. Assume φ: Sⁿ⁻¹ →Rⁿ satisfies the Lipschitz condition

|φ(ξ)−φ(η)| ≤L|ξ−η|, ξ, η ∈Sⁿ⁻¹, and assumeu=P[φ] : Bⁿ→Rⁿ isK-quasiregular. Then

|u(x)−u(y)| ≤C⁰|x−y|, x, y ∈Bⁿ, whereC⁰ depends on L, K and n only.

Kalaj obtained a related result, but under additional assumption of C^1,α regu- larity ofφ, (see [3]).

The main part of the proof is the estimate of the tangential derivatives of u, and in that part quasiregularity plays no role. We choose x₀ =rξ₀ ∈Bⁿ, r = |x|, ξ₀ ∈ Sⁿ⁻¹. Let T =T_x0rSⁿ⁻¹ be the n−1-dimensional tangent plane at x₀ to the sphererSⁿ⁻¹. We want to prove that

(1) kD(u|_T)(x₀)k ≤C(n)L.

Without loss of generality we can assume ξ₀ = e_n and x₀ = re_n. By a simple calculation

∂

∂x_jP(x, ξ) = −2x_j

|x−ξ|ⁿ −n(1− |x|²) x_j−ξ_j

|x−ξ|ⁿ⁺². Hence, for1≤j < n we have

∂

∂x_jP(x₀, ξ) =n(1− |x₀|²) ξ_j

|x₀−ξ|ⁿ⁺².

It is important to note that this kernel is odd in ξ (with respect to reflection (ξ₁, . . . ξ_j, . . . , ξ_n) 7→ (ξ₁, . . . ,−ξ_j, . . . , ξ_n)), a typical fact for kernels obtained by differentiation. This observation and differentiation under integral sign gives, for any 1≤j < n,

∂u

∂x_j(x₀) =n(1−r²) Z

Sⁿ⁻¹

ξ_j

|x₀−ξ|ⁿ⁺²φ(ξ)dσ(ξ)

=n(1−r²) Z

Sⁿ⁻¹

ξ_j

|x₀−ξ|ⁿ⁺²(φ(ξ)−φ(ξ₀))dσ(ξ).

Using the elementary inequality|ξ_j| ≤ |ξ−ξ₀|, (1≤j < n,ξ ∈Sⁿ⁻¹) and Lipschitz continuity ofφ we get

¯¯

¯¯∂u

∂x_j(x₀)

¯¯

¯¯≤Ln(1−r²) Z

Sⁿ⁻¹

|ξ_j||ξ−ξ₀|

|x₀ −ξ|ⁿ⁺²dσ(ξ)

≤Ln(1−r²) Z

Sⁿ⁻¹

|ξ−ξ₀|²

|x0 −ξ|ⁿ⁺²dσ(ξ).

In order to estimate the last integral, we split Sⁿ⁻¹ into two subsets E = {ξ ∈ Sⁿ⁻¹: |ξ−ξ₀| ≤1−r}andF ={ξ∈Sⁿ⁻¹: |ξ−ξ₀|>1−r}. Since|ξ−x₀| ≥1−|x₀|

On Lipschitz continuity of harmonic quasiregular maps on the unit ball inRⁿ 317

for all ξ∈Sⁿ⁻¹ we have Z

E

|ξ−ξ₀|²

|x₀−ξ|ⁿ⁺² dσ(ξ)≤(1−r²)⁻ⁿ⁻² Z

E

|ξ−ξ₀|²dσ(ξ)

≤(1−r²)⁻ⁿ⁻² Z _1−r

0

ρ²ρⁿ⁻²dρ

≤ 2

n+ 1(1−r)⁻¹. On the other hand, |ξ−ξ₀| ≤C_n|ξ−x₀| for every ξ ∈F, so

Z

F

|ξ−ξ₀|²

|x0−ξ|ⁿ⁺² dσ(ξ)≤C_nⁿ⁺² Z

F

|ξ−ξ₀|⁻ⁿdσ(ξ)

≤C_n⁰ Z ₂

1−r

ρ⁻ⁿρⁿ⁻²dρ

≤C_n⁰(1−r)⁻¹. Combining these two estimates we get

¯¯

¯¯∂u

∂x_j(x₀)

¯¯

¯¯≤LC(n)

for 1 ≤ j < n. Due to rotational symmetry, the same estimate holds for every derivative in any tangential direction. This establishes estimate (1). Finally, K- quasiregularity gives

kDu(x)k ≤LKC(n).

Now the mean value theorem gives Lipschitz continuity ofu.

We conclude by noting that, for each n≥2, there is a Lipschitz continuous map φ: Sⁿ⁻¹ →Rⁿsuch thatu=P[φ]is not Lipschitz continuous. We first briefly recall a well known example in the plane. Letf(z) = P_∞

n=1zⁿ/n² and letu(z) = Ref(z).

Then

u(z) = X∞

n=1

rⁿcosnθ

n² and zf⁰(z) = X∞

n=1

zⁿ

n =−log(1−z).

So,zf⁰(z) =−log|1−z| −iv(z), where−π/2< v(z)< π/2. Henced_θu(z) = v(z)is a bounded harmonic function while its harmonic conjugate rd_ru(z) = Rezf⁰(z) is not. This implies thatu(e^iθ)is Lipschitz continuous on the unit circle, whileu(z) is not Lipschitz continuous in the disc. However,uhas the following weaker property:

(2) |u(z⁰)−u(z⁰⁰)| ≤C µ

|z⁰−z⁰⁰|+||z⁰| − |z⁰⁰||log 1

||z⁰| − |z⁰⁰||

¶ . This gives a counterexample in any dimension n≥2.

Example 1. Set φ(x₁, x₂, . . . , x_n) = (u(x₁ +ix₂), x₂, . . . , x_n), x ∈ Sⁿ⁻¹. Then φ is a Lipschitz continuous map onSⁿ⁻¹, while its harmonic extensionU =P[φ] is not Lipschitz continuous on the unit ball.

318 Miloš Arsenović, Vesna Kojić and Miodrag Mateljević

It is clear that U(x) = P[φ](x) = (u(x₁ +ix₂), x₂, . . . , x_n), x ∈ Bⁿ, is not Lipschitz continuous sinceu(x1+ix2)is not Lipschitz continuous on the disc. Prov- ing Lipschitz continuity of φ on Sⁿ⁻¹ reduces to checking Lipschitz continuity of u(x1 +ix2) = u(x1, x2) on Sⁿ⁻¹. Choose x⁰ = (z⁰, w⁰) and x⁰⁰ = (z⁰⁰, w⁰⁰) on Sⁿ⁻¹ where z⁰ = (x⁰₁, x⁰₂), w⁰ = (x⁰₃, . . . , x⁰_n), z⁰⁰ = (x⁰⁰₁, x⁰⁰₂) and w⁰⁰ = (x⁰⁰₃, . . . , x⁰⁰_n). Then, using (2),

|U(x⁰)−U(x⁰⁰)|=|u(z⁰)−u(z⁰⁰)| ≤C(d+δlog1 δ) whered=|z⁰−z⁰⁰|and δ=||z⁰| − |z⁰⁰||. On the other hand, we have

|x⁰−x⁰⁰|² =d²+|w⁰−w⁰⁰|² ≥d² + (|w⁰| − |w⁰⁰|)². But

(|w⁰| − |w⁰⁰|)² = (p

1− |z⁰|² −p

1− |z⁰⁰|²)² ≥2δ−δ². Therefore, for small δ > 0, |x⁰ −x⁰⁰| ≥ √

δ. Since δlog ¹_δ = o(√

δ) and, obviously,

|x⁰−x⁰⁰| ≥ d, we have proven Lipschitz continuity of u(x₁, x₂) on the unit sphere Sⁿ⁻¹.

References

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[2] Dyakonov, K. M.: Equivalent norms on Lipschitz-type spaces of holomorphic functions. - Acta Math. 178:2, 143–167, 1997.

[3] Kalaj, D.: On harmonic quasiconformal mappings. - Ann. Acad. Sci. Fenn. Math. 33, 2008 (to appear).

[4] Mateljević, M.: Distortion of harmonic functions and harmonic quasiconformal quasi- isometry. - Rev. Roumaine Math. Pures Appl. 51:5-6, 2006, 711–722.

[5] Nolder, C. A., andD. M. Oberlin: Moduli of continuity and a Hardy–Littlewood theorem.

- In: Complex analysis, Joensuu 1987, Lecture Notes in Math. 1351, Springer, Berlin, 1988, 265–272.

[6] Rickman, S.: Quasiregular mappings. - Springer-Verlag, Berlin, 1993.

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Received 15 July 2007