Sharp theorems on multipliers and distances in harmonic function spaces in higher dimension
aa r X i v : . [ m a t h . C V ] J un SHARP THEOREMS ON MULTIPLIERS AND DISTANCES INHARMONIC FUNCTION SPACES IN HIGHER DIMENSION
MILOˇS ARSENOVI´C † AND ROMI F. SHAMOYAN
Abstract.
We present new sharp results concerning multipliers and distanceestimates in various spaces of harmonic functions in the unit ball of R n . Introduction and preliminaries
The aim of this paper is twofold. One is to describe spaces of multipliers betweencertain spaces of harmonic functions on the unit ball. We note that so far there areno results in this direction in the multidimensional case, where the use of sphericalharmonics is a natural substitute for power series expansion. In fact, even the caseof the unit disc has not been extensively studied in this context. We refer the readerto [8], where multipliers between harmonic Bergman type classes were considered,and to [4] and [5] for the case of harmonic Hardy classes. Most of our results arepresent in these papers in the special case of the unit disc.The other topic we investigate is distance estimates in spaces of harmonic func-tions on the unit ball. This line of investigation can be considered as a continuationof papers [1], [6] and [7].Let B be the open unit ball in R n , S = ∂ B is the unit sphere in R n , for x ∈ R n we have x = rx ′ , where r = | x | = qP nj =1 x j and x ′ ∈ S . Normalized Lebesguemeasure on B is denoted by dx = dx . . . dx n = r n − drdx ′ so that R B dx = 1. Wedenote the space of all harmonic functions in an open set Ω by h (Ω).We consider harmonic weighted Bergman spaces A pα ( B ) on B defined for α > − < p ≤ ∞ by A pα ( B ) = ( f ∈ h ( B ) : k f k p,α = (cid:18)Z B | f ( rx ′ ) | p (1 − r ) α r n − drdx ′ (cid:19) /p < ∞ ) ,A ∞ α ( B ) = (cid:26) f ∈ h ( B ) : k f k ∞ ,α = sup x ∈ B | f ( x ) | (1 − | x | ) α < ∞ (cid:27) . It is easy to show that the spaces A pα = A pα ( B ) are Banach spaces for 1 ≤ p ≤ ∞ and complete metric spaces for 0 < p < < p < ∞ , 0 ≤ r < f ∈ h ( B ) we set M p ( f, r ) = (cid:18)Z S | f ( rx ′ ) | p dx ′ (cid:19) /p , † Supported by Ministry of Science, Serbia, project M144010. Mathematics Subject Classification 2010 Primary 42B15, Secondary 42B30. Key words andPhrases: Multipliers, harmonic functions, Bergman spaces, mixed norm spaces, distance estimates. with the usual modification to cover the case p = ∞ . Weighted Hardy spaces aredefined, for α ≥ < p ≤ ∞ , by H pα ( B ) = H pα = { f ∈ h ( B ) : k f k p,α = sup r< M p ( f, r )(1 − r ) α < ∞} . For α = 0 the space H pα is denoted simply by H p .For 0 < p ≤ ∞ , 0 < q ≤ ∞ and α > k f k p,q ; α defined by(1) k f k p,q ; α = (cid:18)Z M q ( f, r ) p (1 − r ) αp − r n − dr (cid:19) /p , f ∈ h ( B ) , again with the usual interpretation for p = ∞ , and the corresponding spaces B p,qα ( B ) = B p,qα = { f ∈ h ( B ) : k f k p,q ; α < ∞} . It is not hard to show that these spaces are complete metric spaces and that formin( p, q ) ≥ A ∞ α = H ∞ α for α ≥ B ∞ ,qα = H qα for 0 < q ≤ ∞ , α >
0. We alsohave, for 0 < p ≤ p ≤ ∞ , B p , α ⊂ B p , α , see [3].Next we need certain facts on spherical harmonics and Poisson kernel, see [8] fora detailed exposition. Let Y ( k ) j be the spherical harmonics of order k , j ≤ ≤ d k ,on S . Next, Z ( k ) x ′ ( y ′ ) = d k X j =1 Y ( k ) j ( x ′ ) Y ( k ) j ( y ′ )are zonal harmonics of order k . Note that the spherical harmonics Y ( k ) j , ( k ≥ ≤ j ≤ d k ) form an orthonormal basis of L ( S , dx ′ ). Every f ∈ h ( B ) has anexpansion f ( x ) = f ( rx ′ ) = ∞ X k =0 r k b k · Y k ( x ′ ) , where b k = ( b k , . . . , b d k k ), Y k = ( Y ( k )1 , . . . , Y ( k ) d k ) and b k · Y k is interpreted in thescalar product sense: b k · Y k = P d k j =1 b jk Y ( k ) j . We often write, to stress dependenceon a function f ∈ h ( B ), b k = b k ( f ) and b jk = b jk ( f ), in fact we have linear functionals b jk , k ≥ , ≤ j ≤ d k on the space h ( B ).We denote the Poisson kernel for the unit ball by P ( x, y ′ ), it is given by P ( x, y ′ ) = P y ′ ( x ) = ∞ X k =0 r k d k X j =1 Y ( k ) j ( y ′ ) Y ( k ) j ( x ′ )= 1 nω n − | x | | x − y ′ | n , x = rx ′ ∈ B , y ′ ∈ S , where ω n is the volume of the unit ball in R n . We are going to use also a Bergmankernel for A pβ spaces, this is the following function(2) Q β ( x, y ) = 2 ∞ X k =0 Γ( β + 1 + k + n/ β + 1)Γ( k + n/ r k ρ k Z ( k ) x ′ ( y ′ ) , x = rx ′ , y = ρy ′ ∈ B . For details on this kernel we refer to [3], where the following theorem can be found.
ILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN 3
Theorem 1 ([3]) . Let p ≥ and β ≥ . Then for every f ∈ A pβ and x ∈ B we have f ( x ) = Z Z S n − Q β ( x, y ) f ( ρy ′ )(1 − ρ ) β ρ n − dρdy ′ , y = ρy ′ . This theorem is a cornerstone for our approach to distance problems in the caseof the unit ball. The following lemma from [3] gives estimates for this kernel.
Lemma 1.
1. Let β > . Then, for x = rx ′ , y = ρy ′ ∈ B we have | Q β ( x, y ) | ≤ C (1 − rρ ) −{ β } | ρx − y ′ | n +[ β ] + C (1 − rρ ) β . If, moreover, β > is an integer, then we have | Q β ( x, y ) | ≤ C | ρx − y ′ | n + β .
2. Let β > − . Then Z S n − | Q β ( rx ′ , y ) | dx ′ ≤ C (1 − rρ ) β , | y | = ρ, ≤ r < .
3. Let β > n − , , ≤ r < and y ′ ∈ S n − . Then Z S n − dx ′ | rx ′ − y ′ | β ≤ C (1 − r ) β − n +1 . Lemma 2 ([3]) . Let α > − and λ > α + 1 . Then Z (1 − r ) α (1 − rρ ) λ dr ≤ C (1 − ρ ) α +1 − λ , ≤ ρ < . Lemma 3.
Let G ( r ) , ≤ r < , be a positive increasing function. Then, for α > − , β > − , γ ≥ and < q ≤ we have (3) (cid:18)Z G ( r ) (1 − r ) β (1 − ρr ) γ r α dr (cid:19) q ≤ C Z G ( r ) q (1 − r ) βq + q − (1 − ρr ) qγ r α dr, ≤ ρ < . A special case of the above lemma appears in [2], for reader’s convenience weproduce a proof.
MILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN
Proof.
We use a subdivision of I = [0 ,
1) into subintervals I k = [ r k , r k +1 ), k ≥ r k = 1 − − k . Since 1 − ρr k ≍ − ρr k +1 , 0 ≤ ρ <
1, we have J = (cid:18)Z G ( r ) (1 − r ) β (1 − ρr ) γ r α dr (cid:19) q = X k ≥ Z I k G ( r ) (1 − r ) β (1 − ρr ) γ r α dr q ≤ X k ≥ (cid:18)Z I k G ( r ) (1 − r ) β (1 − ρr ) γ r α dr (cid:19) q ≤ X k ≥ − kqβ G q ( r k +1 ) (cid:18)Z I k r α dr (1 − ρr ) γ (cid:19) q ≤ C X k ≥ − kqβ G q ( r k +1 )2 − kq (1 − ρr k +1 ) − qγ ≤ C X k ≥ − kqβ G q ( r k +1 )2 − kq (1 − ρr k ) − qγ ≤ C X k ≥ G q ( r k +1 ) Z I k +1 (1 − r ) βq + q − r α dr (1 − ρr ) qγ ≤ C Z G ( r ) q (1 − r ) βq + q − (1 − ρr ) qγ r α dr. (cid:3) Lemma 4.
For δ > − , γ > n + δ and integer β > we have Z B | Q β ( x, y ) | γn + β (1 − | y | ) δ dy ≤ C (1 − | x | ) δ − γ + n , x ∈ B . Proof.
Using Lemma 1 and Lemma 2 we obtain: Z B | Q β ( x, y ) | γn + β (1 − | y | ) δ dy ≤ C Z B (1 − | y | ) δ | ρrx ′ − y ′ | γ dy ≤ C Z (1 − ρ ) δ Z S dy ′ | ρrx ′ − y ′ | γ dy ′ dρ ≤ C Z (1 − ρ ) δ (1 − rρ ) n − γ − dρ ≤ C (1 − r ) n + δ − γ . (cid:3) Lemma 5 ([3]) . For real s, t such that s > − and t + n > we have Z (1 − r ) s r t + n − dr = 12 Γ( s + 1)Γ( n/ t )Γ( s + 1 + n/ t ) . We set R n +1+ = { ( x, t ) : x ∈ R n , t > } ⊂ R n +1 . We usually denote the points in R n +1+ by z = ( x, t ) or w = ( y, s ) where x, y ∈ R n and s, t > < p < ∞ and α > − A pα ( R n +1+ ) = ˜ A pα = ( f ∈ h ( R n +1+ ) : Z R n +1+ | f ( x, t ) | p t α dxdt < ∞ ) . Also, for p = ∞ and α >
0, we set˜ A ∞ α ( R n +1+ ) = ˜ A ∞ α = f ∈ h ( R n +1+ ) : sup ( x,t ) ∈ R n +1+ | f ( x, t ) | t α < ∞ . ILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN 5
These spaces have natural (quasi)-norms, for 1 ≤ p ≤ ∞ they are Banach spacesand for 0 < p ≤ R n +1+ by P ( x, t ), i.e. P ( x, t ) = c n t ( | x | + t ) n +12 , x ∈ R n , t > . For an integer m ≥ Q m ( z, w ), where z = ( x, t ) ∈ R n +1+ and w = ( y, s ) ∈ R n +1+ , by Q m ( z, w ) = ( − m +1 m ! ∂ m +1 ∂t m +1 P ( x − y, t + s ) . The terminology is justified by the following result from [3].
Theorem 2.
Let < p < ∞ and α > − . If < p ≤ and m ≥ α + n +1 p − ( n + 1) or ≤ p < ∞ and m > α +1 p − , then (4) f ( z ) = Z R n +1+ f ( w ) Q m ( z, w ) s m dyds, f ∈ ˜ A pα , z ∈ R n +1+ . The following elementary estimate of this kernel is contained in [3]:(5) | Q m ( z, w ) | ≤ C (cid:2) | x − y | + ( s + t ) (cid:3) − n + m +12 , z = ( x, t ) , w = ( y, s ) ∈ R n +1+ . Multipliers on spaces of harmonic functions
In this section we present our results on multipliers between spaces of harmonicfunctions on the unit ball. The following definitions are needed to formulate thesetheorems.
Definition 1.
For a double indexed sequence of complex numbers c = { c jk : k ≥ , ≤ j ≤ d k } and a harmonic function f ( rx ′ ) = P ∞ k =0 P d k j =1 r k b jk ( f ) Y ( k ) j ( x ′ ) we define ( c ∗ f )( rx ′ ) = ∞ X k =0 d k X j =1 r k c jk b jk ( f ) Y ( k ) j ( x ′ ) , rx ′ ∈ B , if the series converges in B . Similarly we define convolution of f, g ∈ h ( B ) by ( f ∗ g )( rx ′ ) = ∞ X k =0 d k X j =1 r k b jk ( f ) b jk ( f ) Y ( k ) j ( x ′ ) , rx ′ ∈ B , it is easily seen that f ∗ g is defined and harmonic in B . Definition 2.
For t > and a harmonic function f ( x ) = P ∞ k =0 b k ( f ) Y k ( x ′ ) on theunit ball we define a fractional derivative of order t of f by the following formula: (Λ t f )( x ) = ∞ X k =0 r k Γ( k + n/ t )Γ( k + n/ t ) b k ( f ) · Y k ( x ′ ) , x = rx ′ ∈ B . Clearly, for f ∈ h ( B ) and t > t h is also harmonic in B . Definition 3.
Let X and Y be subspaces of h ( B ) . We say that a double indexedsequence c is a multiplier from X to Y if c ∗ f ∈ Y for every f ∈ X . The vectorspace of all multipliers from X to Y is denoted by M H ( X, Y ) . MILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN
Clearly every multiplier c ∈ M H ( X, Y ) induces a linear map M c : X → Y . If,in addition, X and Y are (quasi)-normed spaces such that all functionals b jk arecontinuous on both spaces X and Y , then the map M c : X → Y is continuous, as iseasily seen using the Closed Graph Theorem. We note that this holds for all spaceswe consider in this paper : A pα , B p,qα and H pα .The first part of the lemma below appeared, in dimension two, in [5]. Lemma 6.
Let f, g ∈ h ( B ) have expansions f ( rx ′ ) = ∞ X k =0 r k d k X j =1 b jk Y ( k ) j ( x ′ ) , g ( rx ′ ) = ∞ X l =0 r k d k X i =1 c il Y ( l ) i ( x ′ ) . Then we have Z S ( g ∗ P y ′ )( rx ′ ) f ( ρx ′ ) dx ′ = ∞ X k =0 r k ρ k d k X j =1 b jk c jk Y ( k ) j ( y ′ ) , y ′ ∈ S , ≤ r, ρ < . Moreover, for every m > − , y ′ ∈ S and ≤ r, ρ < we have Z S ( g ∗ P y ′ )( rx ′ ) f ( ρx ′ ) dx ′ = 2 Z Z S Λ m +1 ( g ∗ P y ′ )( rRx ′ ) f ( ρRx ′ )(1 − R ) m R n − dx ′ dR. Proof.
The first assertion of this lemma easily follows from the orthogonalityrelations for spherical harmonics Y ( k ) j . Using Lemma 5 and orthogonality relationswe have I = 2 Z Z S Λ m +1 ( g ∗ P y ′ )( rRx ′ ) f ( ρRx ′ )(1 − R ) m R n − dx ′ dR = 2 Z ∞ X k =0 r k ρ k R k + n − (1 − R ) m Γ( k + n/ m + 1)Γ( k + n/ m + 1) d k X j =1 b jk c jk Y ( k ) j dR = ∞ X k =0 r k ρ k d k X j =1 b jk c jk Y ( k ) j ( y ′ ) , which proves the second assertion. (cid:3) We note that ( g ∗ P y ′ )( rx ′ ) = ( g ∗ P x ′ )( ry ′ ) and Λ t ( g ∗ P y ′ )( x ) = (Λ t g ∗ P y ′ )( x ),these easy to prove formulae are often used in our proofs.In this section f m,y stands for the harmonic function f m,y ( x ) = Q m ( x, y ), y ∈ B .We often write f y instead of f m,y . Let us collect some norm estimates of f y . Lemma 7.
For < p ≤ ∞ and m > we have M ∞ ( f m,y , r ) ≤ C (1 − | y | r ) − n − m , m ∈ N . (6) M ( f m,y , r ) ≤ C (1 − | y | r ) − − m . (7) k f m,y k B p, α ≤ C (1 − | y | ) α − − m , m > α − , α > . (8) k f m,y k B p, ∞ α ≤ C (1 − | y | ) α − n − m , m ∈ N , m > α − n, α > . (9) k f m,y k A α ≤ C (1 − | y | ) α − m , m > α > − . (10) k f m,y k H α ≤ C (1 − | y | ) α − − m m > α − , α ≥ . (11) Proof.
Using Lemma 1 we obtain M ∞ ( f m,y , r ) = max x ′ ∈ S | Q m ( y, rx ′ ) | ≤ max x ′ ∈ S C | ρrx ′ − y ′ | n + m = C (1 − r | y | ) − n − m , ILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN 7 which gives (6). The estimate (7) follows from Lemma 1. The estimates (8), forfinite p , and (10) follow from Lemma 2 and (7). Similarly, for finite p (9) followsfrom (6) and Lemma 2. Next, using (7), k f m,y k H α ≤ C sup ≤ r< (1 − r ) α (1 − rρ ) − m − , ρ = | y | . The function φ ( r ) = (1 − r ) α (1 − rρ ) − m − attains its maximum on [0 ,
1] at r = 1 − (1 − ρ ) αρ (1 + m − α ) , as is readily seen by a simple calculus, and this suffices to establish (11) and there-fore (8) for p = ∞ . Finally, (9) directly follows from Lemma 1. (cid:3) In this section we are looking for sufficient and/or necessary condition for adouble indexed sequence c to be in M H ( X, Y ), for certain spaces X and Y ofharmonic functions. We associate to such a sequence c a harmonic function(12) g c ( x ) = g ( x ) = X k ≥ r k d k X j =1 c jk Y ( k ) j ( x ′ ) , x = rx ′ ∈ B , and express our conditions in terms of g c . Our main results give conditions in termsof fractional derivatives of g c , however it is possible to obtain some results on thebasis of the following formula, contained in Lemma 6:(13) ( c ∗ f )( r x ′ ) = Z S ( g ∗ P y ′ )( rx ′ ) f ( ry ′ ) dy ′ . Using continuous form of Minkowski’s inequality, or more generally Young’s in-equality, this formula immediately gives the following proposition.
Proposition 1.
Let c = { c jk : k ≥ , ≤ j ≤ d k } be a double indexed sequence andlet g ( x ) = P k ≥ r k P d k j =1 c jk Y ( k ) j ( x ′ ) be the corresponding harmonic function. If Z S | ( g ∗ P y ′ )( rx ′ ) | p dx ′ ≤ C, y ′ ∈ S , ≤ r < , then c ∈ M H ( H , H p ) .More generally, if /q + 1 /p = 1 + 1 /r , where ≤ p, q, r ≤ ∞ , α + γ = β , α, β, γ ≥ and g ∈ H pγ , then c ∈ M H ( H qα , H rβ ) . The first part of the following lemma, which gives necessary conditions for c tobe a multiplier, is based on [2]. Lemma 8.
Let < p, q ≤ ∞ , ≤ s ≤ ∞ and m > α − . Assume a double indexedsequence c = { c jk : k ≥ , ≤ j ≤ d k } is a multiplier from B p, α to B q,sβ and g = g c is defined in (12) . Then the following condition is satisfied: (14) N s ( g ) = sup ≤ ρ< sup y ′ ∈ S (1 − ρ ) m +1 − α + β (cid:18)Z S | Λ m +1 ( g ∗ P x ′ )( ρy ′ ) | s dx ′ (cid:19) /s < ∞ , where the case s = ∞ requires usual modification.Also, let < p ≤ ∞ , ≤ s ≤ ∞ and m > α − . If a double indexed sequence c = { c jk : k ≥ , ≤ j ≤ d k } is a multiplier from B p, α to H sβ , then the abovefunction g satisfies condition (14). MILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN
Proof.
Let c ∈ M H ( B p, α , B q,sβ ), and assume both p and q are finite, the infinitecases require only small modifications. We have k M c f k B q,sβ ≤ C k f k B p, α for f in B p, α . Set h y = M c f y , then we have(15) h y ( x ) = X k ≥ r k ρ k k X j =1 Γ( k + n/ m + 1)Γ( k + n/ m + 1) c jk Y ( k ) j ( y ′ ) Y ( k ) j ( x ′ ) , x = rx ′ ∈ B , moreover(16) k h y k B q,sβ ≤ C k f y k B p, α . This estimate and Lemma 8 give(17) k h y k B q,sβ ≤ C (1 − | y | ) α − m − , y ∈ B . Note that h y ( x ) = Λ m +1 ( g ∗ P y ′ )( ρx ), using monotonicity of M s ( h y , r ) we obtain: I y ′ ( ρ ) = (cid:18)Z S | Λ m +1 ( g ∗ P x ′ )( ρ y ′ ) | s dx ′ (cid:19) /s = (cid:18)Z ρ (1 − r ) βq − r n − dr (cid:19) − /q × Z ρ (1 − r ) βq − r n − (cid:18)Z S | Λ m +1 ( g ∗ P y ′ )( ρ x ′ ) | s dx ′ (cid:19) q/s dr ! /q ≤ C (1 − ρ ) − β (cid:18)Z ρ (1 − r ) βq − r n − M qs ( h y , r ) dr (cid:19) /q ≤ C (1 − ρ ) − β k h y k B q,sβ . (18)Combining (18) and (17) we obtain (cid:18)Z S | Λ m +1 ( g ∗ P x ′ )( ρ y ′ ) | s dx ′ (cid:19) /s ≤ C (1 − ρ ) α − β − m − , which is equivalent to (14). The case s = ∞ is treated similarly.Next we consider c ∈ M H ( B p, α , H sβ ), assuming 0 < p ≤ ∞ . Set h y = M c h y = g ∗ f y . We have, by Lemma 7, k f y k B p, α ≤ C (1 − | y | ) α − m − , y ∈ B , and, by continuity of M c , k h y k H sβ ≤ C k f y k B p, α . Therefore k h y k H sβ ≤ C (1 − | y | ) α − m − , y ∈ B . Setting y = ρy ′ we have I y ′ ( ρ ) = (cid:18)Z S | Λ m +1 ( g ∗ P x ′ )( ρ y ′ ) | s dx ′ (cid:19) /s = (cid:18)Z S | Λ m +1 ( g ∗ P y )( ρx ′ ) | s dx ′ (cid:19) /s = M s ( h y , ρ ) ≤ (1 − | y | ) − β k h y k H sβ . The last two estimates yield (cid:18)Z S | Λ m +1 ( g ∗ P x ′ )( ρ y ′ ) | s dx ′ (cid:19) /s ≤ C (1 − | y | ) α − β − m − , | y | = ρ which is equivalent to (14). (cid:3) One of main results of this paper is a characterization of the space M H ( B p, α , B q, β )for 0 < p ≤ q ≤ ∞ . The following theorem treats the case p >
1, while Theorem 6below covers the case 0 < p ≤ ILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN 9
Theorem 3.
Let < p ≤ q ≤ ∞ , and m > α − . Then for a double indexedsequence c = { c jk : k ≥ , ≤ j ≤ d k } the following conditions are equivalent:1. c ∈ M H ( B p, α , B q, β ) .2. The function g ( x ) = P k ≥ r k P d k j =1 c jk Y ( k ) j ( x ′ ) is harmonic in B and satisfiesthe following condition (19) N ( g ) < ∞ . Proof.
Since necessity of (19) is contained in Lemma 8 we prove sufficiency ofcondition (19). We assume p and q are finite, the remaining cases can be treated ina similar manner. Take f ∈ B p, α and set h = M c f . Applying the operator Λ m +1 to both sides of equation (13) we obtain(20) Λ m +1 h ( rx ) = Z S Λ m +1 ( g ∗ P y ′ )( x ) f ( ry ′ ) dy ′ . Now we estimate the L norm of the above function on | x | = r : M (Λ m +1 h, r ) ≤ Z S M (Λ m +1 ( g ∗ P y ′ ) , r ) | f ( ry ′ ) | dy ′ ≤ M ( f, r ) sup y ′ ∈ S Z S | Λ m +1 ( g ∗ P y ′ )( rx ′ ) | dx ′ ≤ M ( f, r ) N ( g )(1 − r ) α − β − m − . (21)Since, Z M p ( h, r )(1 − r ) βp − r n − dr ≤ C Z (1 − r ) p ( m +1) M p (Λ m +1 h, r )(1 − r ) βp − r n − dr, see [3], we have k h k pB p, β ≤ C Z (1 − r ) p ( m +1) M p (Λ m +1 h, r )(1 − r ) βp − r n − dr ≤ CN p ( g ) Z M p ( f, r )(1 − r ) αp − r n − dr = CN p ( g ) k f k pB p, α , and therefore k h k B p, β ≤ k f k B p, α . Since k h k B q, β ≤ C k h k B p, β the proof is complete. (cid:3) Next we consider multipliers from B p, α to H sβ , in the case 0 < p ≤ Theorem 4.
Let β ≥ , < p ≤ , s ≥ and m > α − . Then, for a doubleindexed sequence c = { c jk : k ≥ , ≤ j ≤ d k } the following two conditions areequivalent:1. c ∈ M H ( B p, α , H sβ ) .2. The function g ( x ) = P k ≥ r k P d k j =1 c jk Y ( k ) j ( x ′ ) is harmonic in B and satisfiesthe following condition: (22) N s ( g ) < ∞ . Proof.
The necessity of condition (22) is contained in Lemma 8. Now we turnto the sufficiency of (22). We chose f ∈ B p, α and set h = c ∗ f . Then, by Lemma 6:(23) h ( r x ′ ) = 2 Z Z S Λ m +1 ( g ∗ P ξ )( rRx ′ ) f ( rRξ )(1 − R ) m R n − dξdR and this allows us to obtain the following estimate: M s ( h, r ) ≤ Z (1 − R ) m R n − (cid:13)(cid:13)(cid:13)(cid:13)Z S Λ m +1 ( g ∗ P ξ )( rRx ′ ) f ( rRξ ) dξ (cid:13)(cid:13)(cid:13)(cid:13) L s ( S ,dx ′ ) dR ≤ Z (1 − R ) m R n − M ( f, rR ) sup ξ ∈ S k Λ m +1 ( g ∗ P ξ )( rRx ′ ) k L s dR ≤ CN s ( g ) Z (1 − R ) m M ( f, rR )(1 − rR ) α − β − m − dR ≤ CN s ( g ) Z M ( f, rR )(1 − rR ) α − β − dR ≤ CN s ( g ) Z M ( f, rR ) (1 − R ) α (1 − rR ) β +1 dR. Note that M ( f, rR ) is increasing in 0 ≤ R <
1, therefore we can combine Lemma3 and the above estimate to obtain, for 1 / ≤ r < M ps ( h, r ) ≤ CN ps ( g ) Z M p ( f, rR ) (1 − R ) αp + p − (1 − rR ) pβ + p dR ≤ CN ps ( g )(1 − r ) − pβ Z M ( f, R )(1 − R ) αp − dR ≤ CN ps ( g )(1 − r ) − pβ k f k pB p, α Therefore M s ( h, r ) ≤ CN s ( g )(1 − r ) − β k f k B p, α , which completes the proof of theTheorem. (cid:3) The omitted case p = ∞ is treated in our next theorem, which gives a charac-terization of the space M H ( H α , H pβ ). Theorem 5.
Let α ≥ , β > , ≤ p ≤ ∞ and m > α − . Then for adouble indexed sequence c = { c jk : k ≥ , ≤ j ≤ d k } the following conditions areequivalent:1. c ∈ M H ( H α , H pβ ) .2. The function g ( x ) = P k ≥ r k P d k j =1 c jk Y ( k ) j ( x ′ ) is harmonic in B and satisfiesthe following condition: (24) N p ( g ) < ∞ . In the case p = ∞ condition (24) is interpreted in the usual manner. Proof.
Let us assume c ∈ M H ( H α , H pβ ) and set h y = M c f y for y ∈ B . Then wehave, by continuity of M c and by Lemma 7: k h y k H pβ ≤ C k f y k H α ≤ C (1 − | y | ) α − m − . ILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN 11
On the other hand,(25) k h y k H pβ ≥ (1 − ρ ) β M p ( h y , ρ ) ≥ (1 − ρ ) β (cid:18)Z S | Λ m +1 ( g ∗ P x ′ )( ρ y ) | p dx ′ (cid:19) /p , and the above estimates imply (24). Now we prove sufficiency of the condition(24). Choose f ∈ H α and set h = c ∗ f . We apply continuous form of Minkowski’sinequality to (20) to obtain M p (Λ m +1 h, r ) ≤ M ( f, r ) sup y ′ ∈ S M p (Λ m +1 ( g ∗ P y ′ ) , r ) ≤ N p ( g )(1 − r ) α − β − m − M ( f, r ) . Therefore sup r< (1 − r ) m +1+ β M p (Λ m +1 h, r ) ≤ C k f k H α and it follows, see [3], thatsup r< (1 − r ) β M p ( h, r ) ≤ C k f k H α as required. The case p = ∞ is treated the sameway. (cid:3) Since H ∞ β = A ∞ β the p = ∞ case of this theorem gives a complete description ofthe space M H ( H α , A ∞ β ). The next proposition gives necessary conditions for c tobe in M H ( X, A ∞ β ) for some spaces X . Proposition 2.
Let m ∈ N and m > α . Let us consider, for a double indexedsequence c = { c jk : k ≥ , ≤ j ≤ d k } the following conditions:1. c ∈ M H ( A α , A ∞ β ) .2. c ∈ M H ( B p, α , A ∞ β ) .3. The function g ( x ) = P k ≥ r k P d k j =1 c jk Y ( k ) j ( x ′ ) is harmonic in B and satisfiesthe following condition: (26) M t ( g ) = sup ≤ ρ< sup x ′ ,y ′ ∈ S (1 − ρ ) t | Λ m +1 ( g ∗ P x ′ )( ρy ′ ) | < ∞ . Then we have: ⇒ with t = m + β − α and ⇒ with t = m + 1 + β − α .Proof. Let X be one of the spaces A α , B p, α . As in the previous theorems, wechoose a multiplier c from X to A ∞ β and note that k c ∗ f k A ∞ β ≤ C k f k X . We applythis to f y , y = ρy ′ ∈ B to obtain, with h y = c ∗ f y , the estimate k h y k A ∞ β ≤ C k f y k X . Next, k h y k A ∞ β ≥ (1 − ρ ) β M ∞ ( h y , ρ ) = (1 − ρ ) β sup x ′ ∈ S | h y ( ρx ′ ) | = (1 − ρ ) β sup x ′ ∈ S | Λ m +1 ( g ∗ P x ′ )( ρ y ′ ) | . and both implications follow from Lemma 7. (cid:3) The next theorem complements Theorem 3, in less general form it appeared in[2] and for completness of exposition we present, with permission of the authors, aproof.
Theorem 6.
Let < p ≤ , m > α − and p ≤ q ≤ ∞ . Then for a double indexedsequence c = { c jk : k ≥ , ≤ j ≤ d k } the following conditions are equivalent:1. c ∈ M H ( B p, α , B q, β ) .
2. The function g ( x ) = P k ≥ r k P d k j =1 c jk Y ( k ) j ( x ′ ) is harmonic in B and satisfiesthe following condition (27) N ( g ) < ∞ . Proof.
Necessity of condition (27) follows from Lemma 8. Now we prove suffi-ciency of condition (27). Let f ∈ B p, α ( B ) and set h = c ∗ f . Then, using Lemma 6,we have: Z S | h ( rρx ′ ) | dx ′ ≤ Z Z S Z S | Λ m +1 ( g ∗ P x ′ )( rRξ ) || f ( ρRξ ) | (1 − R ) m R n − dξdx ′ dR ≤ C Z sup ξ ∈ S Z S | Λ m +1 ( g ∗ P x ′ )( rRξ ) | dx ′ ! Z S | f ( ρRξ ) | dξ (1 − R ) m R n − dR, and letting ρ → Z S | h ( rx ′ ) | dx ′ ≤ C Z sup ξ ∈ S Z S | Λ m +1 ( g ∗ P x ′ )( rRξ ) | dx ′ ! Z S | f ( Rξ ) | dξ (1 − R ) m R n − dR. For each fixed ξ ∈ S the function ψ ξ ( R ) = Z S | Λ m +1 ( g ∗ P x ′ )( rRξ ) | dx ′ = Z S | Λ m +1 ( g ∗ P ξ )( rRx ′ ) | dx ′ is increasing for 0 ≤ R <
1, due to subharmonicity of u ξ ( x ) = | Λ m +1 ( g ∗ P ξ )( rx ) | .Therefore the function G r ( R ) = sup ξ ∈ S Z S | Λ m +1 ( g ∗ P x ′ )( rRξ ) | dx ′ ! Z S | f ( Rξ ) | dξ, ≤ R < (cid:18)Z S | h ( rx ′ | dx ′ (cid:19) p ≤ C (cid:18)Z G r ( R )(1 − R ) m R n − dR (cid:19) p ≤ C Z G r ( R ) p (1 − R ) mp + p − R n − dR. Since, for 0 ≤ r < G r ( R ) ≤ N ( g ) M ( f, R )(1 − rR ) α − β − m − we have, usingLemma 2 k h k pB p, β = Z (cid:18)Z S | h ( rx ′ ) | dx ′ (cid:19) p (1 − r ) pβ − r n − dr ≤ CN ( g ) p Z M ( f, R ) p (1 − R ) mp + p − R n − Z (1 − r ) pβ − r n − dr (1 − rR ) p ( m +1+ β − α ) dR ≤ CN ( g ) p Z M ( f, R ) p (1 − R ) pα − dR = C k f k pB p, α and we proved k h k B p, β ≤ C k f k B p, α This, together with inequality k h k B q, β ≤ C k h k B p, β finishes the proof. (cid:3) ILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN 13 Estimates for distances in harmonic function spaces in the unitball and related problems in R n +1+ In this section we investigate distance problems both in the case of the unit balland in the case of the upper half space.
Lemma 9.
Let < p < ∞ and α > − . Then there is a C = C p,α,n such that forevery f ∈ A pα ( B ) we have | f ( x ) | ≤ C (1 − | x | ) − α + np k f k A pα , x ∈ B . Proof.
We use subharmonic behavior of | f | p to obtain | f ( x ) | p ≤ C (1 − | x | ) n Z B ( x, −| x | ) | f ( y ) | p dy ≤ C (1 − | x | ) − α (1 − | x | ) n Z B ( x, −| x | ) | f ( y ) | p (1 − | y | ) α dy ≤ C (1 − | x | ) − α − n k f k pA pα . (cid:3) This lemma shows that A pα is continuously embedded in A ∞ α + np and motivates thedistance problem that is investigated in Theorem 7. Lemma 10.
Let < p < ∞ and α > − . Then there is C = C p,α,n such that forevery f ∈ ˜ A pα and every ( x, t ) ∈ R n +1+ we have (28) | f ( x, t ) | ≤ Cy − α + n +1 p k f k ˜ A pα . The above lemma states that ˜ A pα is continuously embedded in ˜ A ∞ α + n +1 p , its proofis analogous to that of Lemma 9.For ǫ > t > f ∈ h ( B ) we set U ǫ,t ( f ) = U ǫ,t = { x ∈ B : | f ( x ) | (1 − | x | ) t ≥ ǫ } . Theorem 7.
Let p > , α > − , t = α + np and β > max( α + np − , αp ) , β ∈ N . Set,for f ∈ A ∞ α + np ( B ) : t ( f ) = dist A ∞ α + np ( f, A pα ) ,t ( f ) = inf ( ǫ > Z B Z U ǫ,t | Q β ( x, y ) | (1 − | y | ) β − t dy ! p (1 − | x | ) α dx < ∞ ) . Then t ( f ) ≍ t ( f ) .Proof. We begin with inequality t ( f ) ≥ t ( f ). Assume t ( f ) < t ( f ). Thenthere are 0 < ǫ < ǫ and f ∈ A pα such that k f − f k A ∞ t ≤ ǫ and Z B Z U ǫ,t ( f ) | Q β ( x, y ) | (1 − | y | ) β − t dy ! p (1 − | x | ) α dx = + ∞ . Since (1 − | x | ) t | f ( x ) | ≥ (1 − | x | ) t | f ( x ) | − (1 − | x | ) t | f ( x ) − f ( x ) | for every x ∈ B weconclude that (1 − | x | ) t | f ( x ) | ≥ (1 − | x | ) t | f ( x ) | ≥ (1 − | x | ) t | f ( x ) | − ǫ and therefore( ǫ − ǫ ) χ U ǫ,t ( f ) ( x )(1 − | x | ) − t ≤ | f ( x ) | , x ∈ B . Hence + ∞ = Z B Z U ǫ,t ( f ) | Q β ( x, y ) | (1 − | y | ) β − t dy ! p (1 − | x | ) α dx = Z B (cid:18)Z B χ U ǫ,t ( f ) ( y )(1 − | y | ) t | Q β ( x, y ) | (1 − | y | ) β dy (cid:19) p (1 − | x | ) α dx ≤ C ǫ,ǫ Z B (cid:18)Z B | f ( y ) || Q β ( x, y ) | (1 − | y | ) β dy (cid:19) p (1 − | x | ) α dx = M, and we are going to prove that M is finite, arriving at a contradiction. Let q bethe exponent conjugate to p . We have, using Lemma 4, I ( x ) = (cid:18)Z B | f ( y ) | (1 − | y | ) β | Q β ( x, y ) | dy (cid:19) p = (cid:18)Z B | f ( y ) | (1 − | y | ) β | Q β ( x, y ) | n + β ( np + β − ǫ ) | Q β ( x, y ) | n + β ( nq + ǫ ) dy (cid:19) p ≤ Z B | f ( y ) | p (1 − | y | ) pβ | Q β ( x, y ) | n + pβ − pǫn + β dy (cid:18)Z B | Q β ( x, y ) | n + qǫn + β dy (cid:19) p/q ≤ C (1 − | x | ) − pǫ Z B | f ( y ) | p (1 − | y | ) pβ | Q β ( x, y ) | n + pβ − pǫn + β dy for every ǫ >
0. Choosing ǫ > α − pǫ > − M ≤ C Z B | f ( y ) | p (1 − | y | ) pβ Z B (1 − | x | ) α − pǫ | Q β ( x, y ) | n + pβ − pǫn + β dxdy ≤ C Z B | f ( y ) | p (1 − | y | ) α dy < ∞ . In order to prove the remaining estimate t ( f ) ≤ Ct ( f ) we fix ǫ > t ( f ) is finite and use Theorem 1, with β > max( t − , f ( x ) = Z B \ U ǫ,t ( f ) Q β ( x, y ) f ( y )(1 − | y | ) β dy + Z U ǫ,t ( f ) Q β ( x, y ) f ( y )(1 − | y | ) β dy = f ( x ) + f ( x ) . Since, by Lemma 4, | f ( x ) | ≤ β R B | Q β ( x, y ) | (1 − | w | ) β − t dy ≤ C (1 − | x | ) − t we have k f k A ∞ t ≤ Cǫ . Thus it remains to show that f ∈ A pα and this follows from k f k pA pα ≤ k f k pA ∞ t Z B Z U ǫ,t ( f ) | Q β ( x, y ) | (1 − | y | ) β − t dy ! p (1 − | x | ) α dx < ∞ . (cid:3) The above theorem has a counterpart in the R n +1+ setting. As a preparation forthis result we need the following analogue of Lemma 4. Lemma 11.
For δ > − , γ > n + 1 + δ and m ∈ N we have Z R n +1+ | Q m ( z, w ) | γn + m +1 s δ dyds ≤ Ct δ − γ + n +1 , t > . ILOˇS ARSENOVI´C AND ROMI F. SHAMOYAN 15
Proof.
Using Fubini’s theorem and estimate (5) we obtain I ( t ) = Z R n +1+ | Q m ( z, w ) | γn + m +1 s δ dyds ≤ C Z ∞ s δ (cid:18)Z R n dy [ | y | + ( s + t ) ] γ (cid:19) ds = C Z ∞ s δ ( s + t ) n − γ ds = Ct δ − γ + n +1 . (cid:3) For ǫ > λ > f ∈ h ( R n +1+ ) we set: V ǫ,λ ( f ) = { ( x, t ) ∈ R n +1+ : | f ( x, t ) | t λ ≥ ǫ } . Theorem 8.
Let p > , α > − , λ = α + n +1 p , m ∈ N and m > max( α + n +1 p − , αp ) .Set, for f ∈ ˜ A ∞ α + n +1 p ( R n +1+ ) : s ( f ) = dist ˜ A ∞ α + n +1 p ( f, ˜ A pα ) ,s ( f ) = inf ( ǫ > Z R n +1+ Z V ǫ,λ Q m ( z, w ) s β − t dyds ! p t α dxdt < ∞ ) . Then s ( f ) ≍ s ( f ) . The proof of this theorem closely parallels the proof of the previous one, in fact,the role of Lemma 4 is taken by Lemma 11 and the role of Theorem 1 is taken byTheorem 2. We leave details to the reader.
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Faculty of mathematics, University of Belgrade, Studentski Trg 16, 11000 Belgrade,Serbia
E-mail address : [email protected]
Bryansk University, Bryansk Russia