Radial Growth and Hardy-Littlewood-Type Theorems on Hyperbolic Harmonic Functions

Filomat 29:2 (2015), 361–370 Published by Faculty of Sciences and Mathematics, DOI 10.2298/FIL1502361C University of Nis,ˇ Serbia Available at: http://www.pmf.ni.ac.rs/filomat

Shaolin Chena, Zhenhua Sub

In celebration of Matti Vuorinen’s 65-th birthday aDepartment of Mathematics and Computational Science, Hengyang Normal University, Hengyang, Hunan 421008, People’s Republic of China. bDepartment of Mathematics, Huaihua University, Huaihua, Hunan 418008, People’s Republic of China.

Abstract. In this paper, we ﬁrst show that a result of Girela et al. on analytic functions can be extended to hyperbolic-harmonic functions, and then we establish Hardy-Littlewood-type theorems on hyperbolic harmonic functions.

1. Introduction and main results

For n 2, let Rn denote the usual real vector space of dimension n. Sometimes it is convenient to ≥ n identify each point x = (x ,..., xn) R with an n 1 column matrix so that 1 ∈ ×    x1    x  .  =  .  .   xn

n For a = (a ,..., an) and x R , we deﬁne the Euclidean inner product , by 1 ∈ h· ·i

x, a = x a + + xnan h i 1 1 ··· so that the Euclidean length of x is deﬁned by

1/2 2 2 1/2 x = x, x = ( x + + xn ) . | | h i | 1| ··· | | n Denote a ball in R with center x0 and radius r by

Bn(x , r) = x Rn : x x < r . 0 { ∈ | − 0| }

2010 Mathematics Subject Classiﬁcation. Primary 31C05, 46E15; Secondary 30J60, 30C62 Keywords. Hardy-Littlewood type theorem; majorant; hyperbolic harmonic function. Received: 9 November 2014; Accepted: 12 January 2015 Communicated by Miodrag Mateljevic´ This research was partly supported by the National Natural Science Foundation of China (No. 11401184 and No. 11326081), the Hunan Province Natural Science Foundation of China (No. 2015JJ3025), the Excellent Doctoral Dissertation of Special Foundation of Hunan Province (higher education 2050205), the Construct Program of the Key Discipline in Hunan Province. Email addresses: [email protected]. (Shaolin Chen), [email protected] (Zhenhua Su) Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 362

In particular, Bn denotes the unit ball Bn(0, 1). Set D = B2, the open unit disk in the complex plane C. Let Ω be a proper subdomain of Rn. A function f 2(Ω) is called hyperbolic harmonic (brieﬂy, h- harmonic, in the following) function in Ω if it satisﬁes the∈hyperbolic C Laplace’s equation

2 2 2 ∆hu = (1 x ) ∆u + 2(n 2)(1 x ) u, x = 0, − | | − − | | h∇ i where ∆ denotes the ordinary Laplacian operator and denotes the gradient. Recall that hyperbolic harmonic functions are solutions of the Laplace-Beltrami equation∇ with respect to the Poincar´emetric

Xn ds2 = (1 x 2)2 dx2 − | | k k=1 in the unit ball Bn. Obviously, when n = 2, all h-harmonic functions are harmonic functions. We refer to [2, 3, 6, 14, 18, 28, 29] for more details of h-harmonic functions. It turns out that if ψ (∂Bn), then the Dirichlet problem ∈ C   n ∆h f = 0 in B   f = ψ on ∂Bn

n has an unique solution in (B ) and can be represented by C Z f (x) = Ph[ψ](x) = Ph(x, ζ)ψ(ζ)dσ(ζ), (1) ∂Bn

n where dσ is the unique normalized surface measure on ∂B and Ph(x, ζ) is the hyperbolic Poisson kernel deﬁned by !n 1 2 − 1 x n n Ph(x, ζ) = − | | (x B , ζ ∂B ). x ζ 2 ∈ ∈ | − | Throughout this paper, we use C to denote the various positive constants, whose value may change from one occurrence to the next. A continuous increasing function ω : [0, ) [0, ) with ω(0) = 0 is called a majorant if ω(t)/t is non-increasing for t > 0. Given a subset Ω of∞Rn,→ a function∞ f : Ω Rm (m 1) is said to belong to the → ≥ Lipschitz space Λω(Ω) if there is a positive constant C such that

f (x) f (y) Cω( x y ) for all x, y Ω. (2) | − | ≤ | − | ∈

For δ0 > 0, let Z δ ω(t) dt C ω(δ), 0 < δ < δ0 (3) 0 t ≤ · and Z ∞ ω(t) dt C δ 2 ω(δ), 0 < δ < δ0, (4) δ t ≤ · where ω is a majorant. A majorant ω is said to be regular if it satisﬁes the conditions (3) and (4) (see [12, 13, 26]). n Let Ω be a proper subdomain of R . We use dΩ(x) to denote the Euclidean distance from x to the boundary ∂Ω of Ω. In particular, we always use d(x) to denote the Euclidean distance from x to the boundary of Bn. Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 363

n A proper subdomain G of R is said to be Λω-extension if Λω(G) = locΛω(G), where locΛω(G) denotes the set of all functions f : G Rm (m 1) satisfying (2) with a fixed positive constant C, whenever x G →1 ≥ n ∈ and y G such that x y < dG(x). Obviously, B is a Λ -extension domain. ∈ | − | 2 ω In [22], the author proved that G is a Λω-extension domain if and only if each pair of points x, y G can be joined by a rectifiable curve γ G satisfying ∈ ⊂ Z ω(d (τ)) G ds(τ) Cω( x y ) (5) γ dG(τ) ≤ | − | with some fixed positive constant C = C(G, ω), where ds stands for the arc length measure on γ. Furthermore, the author also proved that Λω-extension domains exist only for majorants ω satisfying (3). See [13, 15, 22] for more details on Λω-extension domains. For p (0, ], the Hardy class Hp(Bn) consists of those functions f : Bn R such that f is measurable, ∈ ∞ → Mp(r, f ) exists for all r (0, 1) and f p < , where ∈ k k ∞   sup Mp(r, f ), if p (0, ), Z !1/p  ∈ ∞ f = 0

Theorem 1.1. ([16, Theorem 1(a)]) Let p (2, ). For r (0, 1), if f is an analytic function in D such that ∈ ∞ ∈ 1 Mp(r, f 0) = O as r 1, 1 r → − then for all β > 1/2,

1 β Mp(r, f ) = O log as r 1. (6) 1 r → − In [16, P464, Equation (26)], Girela and Pelaez´ asked whether β in (6) can be substituted by 1/2. This problem was aﬃrmatively settled by Girela, Pavlovic and Pelaez´ in [17] (see [17, Theorem 1.1]). We show that Theorem 1.6 can be extended to h-harmonic functions in Bn with β = 1/2. On the related topics, see [5, 7, 8, 10, 27].

Theorem 1.2. Let p [2, ) and ω be a majorant. For r (0, 1), if f is h-harmonic from Bn into R such that ∈ ∞ ∈ 1 Mp(r, f ) Cω , ∇ ≤ 1 r − then

1 " n 2 2 Z 1 # 2 2 rp(p 1)(1 + r) − C ω(1) 1 Mp(r, f ) f (0) + − ω dt . ≤ | | n 1 1 rt − 0 − Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 364

Especially, if n = 2 and ω(t) = t, then

1 ! 1 2 Mp(r, f ) = O log as r 1 (7) 1 r → − and the estimate of (7) is sharp. Krantz [21] proved a Hardy-Littlewood-type theorem for harmonic functions in Bn with respect to the majorant ω(t) = ω (t) = tα (0 < α 1) as follows. For the extended discussion on this topic, see [4, 9]. α ≤ Theorem 1.3. ([21, Theorem 15.8]) Let u be a harmonic function in Bn and 0 < α 1. Then u satisﬁes ≤ ωα d(x) u(x) C for any x Bn |∇ | ≤ d(x) ∈ if and only if u(x) u(y) Cω ( x y ) for any x, y Bn. | − | ≤ α | − | ∈ We generalize Theorem 1.3 to the following form.

Theorem 1.4. Let ω be a majorant satisfying (3), Ω be a Λω-extension domain and f be a h-harmonic function from Ω into R. Then f Λω(Ω) if and only if ∈ ω dΩ(x) f (x) C for any x Ω. |∇ | ≤ dΩ(x) ∈ n B Let ω be a majorant and D be a bounded set of R . We use Λω(D) to denote all the bounded continuous functions f in D with the norm ( ) f (x) f (y) f ω,D = sup | − | < . k k x,y D,x,y ω( x y ) ∞ ∈ | − | Taking another majorant ω0, we deﬁne the operator norm

n Ph[ f ] ω0,B Ph ω ω0 = sup k k . k k → B n f Bn f Λ (∂B ), f Bn ,0 ω,∂ ∈ ω k kω,∂ k k For each a ∂Bn, we deﬁne ∈ n ηa (ζ) = ω( ζ a ) for ζ ∂B . ,ω | − | ∈ We refer to [1] for the similar deﬁnitions of harmonic functions. B n Proposition 1.5. Let ω be a majorant. Then ηa Λ (B ). ,ω ∈ ω The following result is the classical Hardy-Littlewood Theorem. Theorem 1.6. ([11, Theorem 5.1]) Let f be an analytic function in D and continuous in D. Then for some 0 < α 1, ≤ iθ iθ f (e 1 ) f (e 2 ) Cω ( θ θ ) for any 0 θ , θ < 2π | − | ≤ α | 1 − 2| ≤ 1 2 if and only if ωα d(z) f 0(z) C for any z D. | | ≤ d(z) ∈ The following result is a Hardy-Littlewood-type theorem for hyperbolic functions. For the extensive discussions on this topics, see [1, 13, 23–25]. Theorem 1.7. Let ω be a majorant. Then for each a ∂Bn and any x Bn, there is a constant C 1 such that ∈ ∈ ≥ Ph[ηa,ω](x) Cω( x a ), if and only if Ph ω ω < . ≤ | − | k k → ∞ We will prove Theorems 1.2, 1.4 and 1.7 in section 2. Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 365

2. Proofs of the main results

Lemma 2.1. ([28, Lemma 3.2]) If f 2(Bn), then 1 n 2 n 2 ∈ C d R r − (1 r ) − R a f r d − f x d x and ( ) dr ∂Bn ( ζ) σ(ζ) = n Bn (r) ∆h ( ) τ( ) R R R b f r d f 1 x r f x d x where ( ) ∂Bn ( ζ) σ(ζ) = (0) + Bn (r) ( , )∆h ( ) τ( ), R | |

Z r 1 (1 s2)n 2 1 x r − ds ( , ) = −n 1 , | | n x s − | |

dVN Bn dτ = (1 x 2)n and dVN denotes the normalized Lebesgue volume measure on . −| |

Proof of Theorem 1.2

Let f be h-harmonic in Bn. By elementary calculations, we see that

p 2 2 p 2 2 ∆h( f ) = p(p 1)(1 x ) f − f . (8) | | − − | | | | |∇ | For r [0, 1), the Holder’s¨ inequality yields ∈ Z p 2 2 2 p 2 f (rζ) − f (rζ) dσ(ζ) Mp(r, f ) Mp− (r, f ). (9) ∂Bn | | |∇ | ≤ ∇ ·

By (8), (9) and Lemma 2.1, we obtain

Z p p p Mp(r, f ) = f (0) + 1( x , r)∆h( f (x) )dτ(x) | | Bn(0,r) | | | | Z p p 2 2 2 2 = f (0) + p(p 1) f (x) − f (x) 1( x , r)(1 x ) dτ(x) | | − Bn(0,r) | | |∇ | | | − | | Z r n 1 Z ρ − 1(ρ, r) f p np p f p 2 f 2d d = (0) + ( 1) 2 n 2 (ρζ) − (ρζ) σ(ζ) ρ | | − (1 ρ ) n | | |∇ | 0 − − ∂B Z r n 11 p ρ − (ρ, r) 2 p 2 f (0) + np(p 1) M (ρ, f )M − (ρ, f )dρ. (10) ≤ | | − (1 ρ2)n 2 p ∇ p 0 − −

By computations, we obtain

Z r 1 (1 s2)n 2 1 r − ds (ρ, ) = −n 1 n ρ s − Z r 1 s2 n 2ds n 1 (1 ) − ≤ nρ − ρ − Z r (1 + r)n 2 − s n 2ds n 1 (1 ) − ≤ nρ − ρ − n 2 n 1 (1 + r) (1 ρ) − − − . (11) ≤ n(n 1) ρn 1 − − Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 366

Applying (10) and (11), we get

Z r n 1 ρ − 1(ρ, r) M2(r, f ) f (0) 2 + np(p 1) M2(ρ, f )dρ p ≤ | | − (1 ρ2)n 2 p ∇ 0 − − n 2 Z r p(p 1)(1 + r) − f (0) 2 + − (1 ρ)M2(ρ, f )dρ ≤ | | n 1 − p ∇ − 0 n 2 Z 1 rp(p 1)(1 + r) − = f (0) 2 + − (1 rt)M2(tr, f )dt | | n 1 − p ∇ − 0 n 2 2 Z 1 rp(p 1)(1 + r) − C 1 f (0) 2 + − (1 rt)ω2 dt ≤ | | n 1 − 1 rt − 0 − n 2 2 Z 1 rp(p 1)(1 + r) − C 1 1 = f (0) 2 + − ω (1 tr) ω dt | | n 1 1 rt − 1 rt − 0 − − n 2 2 Z 1 rp(p 1)(1 + r) − C ω(1) 1 f (0) 2 + − ω dt, ≤ | | n 1 1 rt − 0 − which gives 1 " n 2 2 Z 1 # 2 2 rp(p 1)(1 + r) − C ω(1) 1 Mp(r, f ) f (0) + − ω dt . ≤ | | n 1 1 rt − 0 − In particular, if n = 2 and ω(t) = t, then the estimate of (7) is sharp. The proof of the sharpness part follows from [16, Theorem 1(b)]. The proof of this theorem is complete.

Lemma 2.2. Let ω be a majorant. Then (1) ω is subadditive, that is, if t, s > 0, then ω(s + t) ω(s) + ω(t); (2) for t > 0, if λ 1, then ω(λt) λω(t). ≤ ≥ ≤ Proof. We ﬁrst prove (1). Since ω(t)/t is nonincreasing for t > 0, we see that for s, t > 0,

ω(s) ω(t) ω(t + s) ω(s) + ω(t) ω(s + t) = s + t (s + t) − s t − t + s ! ! ω(s) ω(s + t) ω(t) ω(s + t) = s + t s − s + t t − s + t 0. ≥ (2) easily follows from the monotonicity of ω(t)/t for t > 0. The proof of this lemma is complete.

Proof of Theorem 1.4

We first prove the sufficiency. Since Ω is a Λω-extension domain, we see that for any x, y Ω, by using (5), there is a rectifiable curve γ Ω joining x to y such that ∈ ⊂ Z f (x) f (y) f (ζ) ds(ζ) | − | ≤ γ |∇ | Z ω dΩ(ζ) C ds(ζ) ≤ γ dΩ(ζ) Cω( x y ). ≤ | − | Now we come to prove the necessity. Let x = (x1, , xn) Ω and r = dΩ(x)/2. Then by Lemma 2.2, for n ··· ∈ all y = (y1, , yn) B (x, r), ··· ∈ f (x) f (y) Cω( x y ) 2Cω(d (x)). | − | ≤ | − | ≤ Ω Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 367

For all y Bn(x, r), using (1), we get ∈ Z f (y) = Ph(y, ζ) f (rζ + x)dσ(ζ), ∂Bn where !n 1 2 2 − r y x n Ph(y, ζ) = − | − | and ζ = (ζ , , ζn) ∂B . y x rζ 2 1 ··· ∈ | − − | By elementary calculations, for each k 1, 2, , n , we have ∈ { ··· } h 2 2 2 i n 2 (y x ) y x rζ + (r y x )(y rζ x ) ∂Ph(y, ζ) n−1 k k k k k n y − − | − − | − | − | − − = 2( 1) Ph( , ζ) 4 . ∂yk − − y x rζ | − − | Then for all y Bn(x, r/2), ∈

∂Ph(y, ζ)

∂yk n 2 h i 2 2 − 2 2 2 r y x yk xk y x rζ + (r y x ) yk rζk xk 2(n 1) − | − | | − || − − | − | − | | − − | ≤ − y x rζ 2n ! | − − | r2n 4 9r3 3r3 2(n 1) − + ≤ − y x rζ 2n 8 2 | − − | 2n 1 21(n 1) r − − 2n ≤ 4 r 2 2(n 1) 2 − 21(n 1) = · − , r which implies that

  1 Xn 2  2  f (y) =  f (y) |∇ |  yk  k=1 n Z 2 1 n X ∂ o 2 = Ph(y, ζ)( f (rζ + x) f (x))dσ(ζ) Bn ∂yk − k=1 ∂ n Z X ∂ Ph(y, ζ)( f (rζ + x) f (x))dσ(ζ) ≤ Bn ∂yk − k=1 ∂ Xn Z ∂ Ph(y, ζ) f (rζ + x) f (x) dσ(ζ) ≤ Bn ∂yk − k=1 ∂ Z

√n Ph(y, ζ) f (rζ + x) f (x) dσ(ζ) ≤ ∂Bn ∇ − 2(n 1) Z 2 − 21n(n 1) · − f (rζ + x) f (x) dσ(ζ) ≤ r ∂Bn − 2(n 1) 2 − 21n(n 1)Cω(r) · − ≤ r ω dΩ(x) 2n 1 2 − 21n(n 1)C . ≤ · − dΩ(x) The proof of this theorem is complete. Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 368

Proof of Proposition 1.5 n Let ζ1, ζ2 ∂B . Without loss of generality, we may assume ζ1 a < ζ2 a . Then by Lemma 2.2, we have ∈ | − | | − |

ηa (ζ ) ηa (ζ ) = ω( ζ a ) ω( ζ a ) | ,ω 2 − ,ω 1 | | 2 − | − | 1 − | = ω( ζ a ζ a + ζ a ) ω( ζ a ) | 2 − | − | 1 − | | 1 − | − | 1 − | ω( ζ a ζ a ) + ω( ζ a ) ω( ζ a ) ≤ | 2 − | − | 1 − | | 1 − | − | 1 − | ω( ζ ζ ), ≤ | 2 − 1| which gives ηa,ω(ζ2) ηa,ω(ζ1) | − | 1. ω( ζ ζ ) ≤ | 2 − 1| n Hence ηa Bn 1 < . On the other hand, for each a ∂B , k ,ωkω,∂ ≤ ∞ ∈ n ηa (ζ) = ω( ζ a ) ω(2) < for ζ ∂B . ,ω | − | ≤ ∞ ∈ B n Then ηa is bounded and ηa Λ (B ). The proof of this Proposition is complete. ,ω ,ω ∈ ω Proof of Theorem 1.7 B n n We ﬁrst prove the necessity. Let f Λω(∂B ). We only need to prove that for any x, y B , there is a positive constant C such that ∈ ∈

Ph[ f ](x) Ph[ f ](x) C f w,∂Bn ω( x y ). − ≤ k k | − | Without loss of generality, we assume that

0 < d(y) d(x). (12) ≤ Let x , y ∂Bn such that d(x) = x x and d(y) = y y , respectively. For ζ ∂Bn, let F(ζ) = f (ζ) f (x ). 0 0 ∈ | − 0| | − 0| ∈ − 0 We divide the proof into two cases. Case 1. x y d(x)/2. For z Bn(x, d(x)/2), using Lemma 2.2, we see that | − | ≤ ∈ d(x) 3d(x) z x z x + x x + d(x) = , | − 0| ≤ | − | | − 0| ≤ 2 2 which yields that

Ph[F](z) f Bn Ph[ηx ](z) | | ≤ k kw,∂ | 0,ω | C f w,∂Bn ω( x y ) ≤ k k | − |! 3d(x) C f Bn ω ≤ k kw,∂ 2 3C f Bn ω(d(x)). ≤ 2 k kw,∂ Using arguments similar to those in the necessity’s proof of Theorem 1.4, for z Bn(x, d(x)/2), there is a positive constant C such that ∈

ω(d(x)) Ph[ f ](z) = Ph[F](z) C f Bn , |∇ | |∇ | ≤ k kw,∂ d(x) Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 369 which gives that Z

Ph[ f ](x) Ph[ f ](y) Ph[ f ](z) dz − ≤ [x,y] |∇ || | ω(d(x)) C f Bn x y ≤ k kw,∂ d(x) | − | C f Bn ω( x y ), ≤ k kw,∂ | − | where [x, y] denotes the segment from x to y. Case 2. x y > d(x)/2. We use the similar approach as in the proof of [1, Theorem 1.1] to prove this case. By (12),| we− know| that x y > d(x)/2 d(y)/2. By elementary calculations, we see that there is positive constant C such that | − | ≥

Ph[ f ](y) f (y0) C f w,∂Bn ω( x y ). (14) − ≤ k k | − |

By using x0 y0 x y + d(x) + d(y) < 5 x y and Lemma 2.2, we know that there is positive constant C such that | − | ≤ | − | | − |

f (x0) f (y0) C f w,∂Bn ω( x0 y0 ) 5C f w,∂Bn ω( x y ). (15) − ≤ k k | − | ≤ k k | − | By (13), (14) and (15), we conclude that

Ph[ f ](x) Ph[ f ](y) Ph[ f ](x) f (x0) Ph[ f ](y) f (y0) + f (x0) f (y0) − ≤ − − − −

Ph[ f ](x) f (x0) + Ph[ f ](y) f (y0) + f (x0) f (y0) ≤ − − − 7C f Bn ω( x y ). ≤ k kw,∂ | − | Now we come to prove the suﬃciency. By Proposition 1.5, we have

Ph[ηa,ω] ω ,Bn Ph ω ω ηa,ω ω,∂Bn < , k k 0 ≤ k k → k k ∞ which implies that, for x, y Bn, there is a positive constant C such that ∈

Ph[ηa,ω](x) Ph[ηa,ω](y) Cω( x y ). − ≤ | − | Let x ∂Bn such that d(x) = x x . By letting y tends to x yields that 0 ∈ | 0 − | 0

Ph[ηa,ω](x) ηa,ω(x0) Cω( x0 x ), − ≤ | − | which gives

Ph[ηa ](x) ηa (x ) + Cω( x x ) ω( x a ) + Cω( x a ). (16) ,ω ≤ ,ω 0 | 0 − | ≤ | 0 − | | − | Applying (16) and the inequality x a x x + x a 2 x a , we conclude that | 0 − | ≤ | 0 − | | − | ≤ | − |

Ph[ηa ](x) Cω( x a ) + ω( x a ) (C + 2)ω( x a ). ,ω ≤ | − | | 0 − | ≤ | − | The proof of this theorem is complete. Sh. Chen, Z. Su / Filomat 29:2 (2015), 361–370 370

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