Complex Anal. Oper. Theory DOI 10.1007/s11785-014-0387-x

Complex Analysis and Operator Theory

Quaternionic F( p, q, s) Function Spaces Augusto G. Miss P. · Lino F. Reséndis O. · Luis M. Tovar S.

Received: 14 May 2013 / Accepted: 7 May 2014 © Springer Basel 2014

Abstract In this paper we develop the necessary tools to generalize the function spaces F( p, q, s) and F0 ( p, q, s) introduced by Zhao in (Zhao, Ann Acad Sci Fenn Math Diss No 105, 1996) to the case of the monogenic functions defined in the unit ball of R3 . Obtaining at the same time the generalization of Q p monogenic spaces presented in (Gürlebeck et al., On Q p −spaces of quaternion-valued functions, 1999) and B q, p monogenic spaces in (El-Sayed Ahmed et al., Complex Var Elliptic Equ 51(2):119–136, 2006). Keywords

F( p, q, s) spaces · Quaternionic Q p and Bloch spaces

Mathematics Subject Classification

30G35 · 46E15

A. G. Miss P. Depto. de Ciencias Básicas, Unidad Profesional Interdisciplinaria en Ingeniería y Tecnologías Avanzadas del IPN, Av. Instituto Politécnico Nacional # 2580, Col. La Laguna Ticomán, 07340 Deleg. Gustavo A. Madero, D.F., Mexico e-mail: [email protected] L. F. Reséndis O. Universidad Autónoma Metropolitana, Unidad Azcapotzalco, C.B.I., Apartado Postal 16-306, Area de Análisis Matemático y sus Aplicaciones, C.P. 02200 Mexico 16, D.F., Mexico e-mail: [email protected] L. M. Tovar S. (B) Escuela Superior de Física y Matemáticas del IPN, Edif. 9, Unidad ALM, Zacatenco del IPN, C.P. 07338 Mexico, D.F., Mexico e-mail: [email protected]

A. G. Miss P. et al.

1 Introduction In [21], Zhao introduced the so called F( p, q, s) spaces. These spaces consist of analytic functions on the unit open complex disk D = {z ∈ C : |z| < 1}, such that  sup

a∈D

| f  (z)| p (1 − |z|2 )q g s (z, a) d x d y < ∞

D

where 0 < p < ∞, −2 < q < ∞, 0 < s < ∞ and g is the Green’s function of the 1 − az unit disk given by g(z, a) = ln | |. These spaces are the generalization of the a−z Qs = F(2, 0, s) spaces introduced by R. Aulaskari and Lappan in [1] for 1 ≤ s < ∞ and for 0 < s < 1 by Aulaskari et al. in [2]. In particular Q1 = H 1 (D) ∩ B M O A(∂D), where H 1 is the Hardy space and B M O A(∂D) is the space of bounded mean oscillation analytic functions on the boundary of the unit disk ∂D. The family F( p, q, s) is quite general, includes, among others, Dirichlet type spaces F( p, q, 0) and α-Bloch spaces. There are several approaches to higher dimensions of Qs spaces. In [9] Güerlebeck et al presented the generalization of Qs spaces to the case of monogenic functions defined in the unit ball of R3 . More results about this subject appear in [4–6,10] and [7]. For generalizations to holomorphic functions defined in the unit ball of Cn , we refer to [15–17]. In [8], El-Sayed et al introduced the Fϕ ( p, q, s) and Fg ( p, q, s) quaternionic spaces and proved its equivalence only for some particular values of p, q and s. We have to remark that in this paper we generalize the quoted El-Sayed’s results. Besides we give several generalizations of important results appearing in [7] and [9] and the methods presented here are simpler. Although we are dealing with monogenic functions, it is remarkable that we use—mainly—real analysis methods to prove a good number of results. Thus Theorem 5.1 present the characterization of Bα (and Bα,0 ) Bloch spaces in terms of Fϕ ( p, q, s) spaces (respectively Fϕ,0 ( p, q, s)). These results generalize Theorem 4.1 in [9], with different techniques. Likewise in Theorems 4.1, 4.2, 3.1 and 3.2 we prove with simpler techniques generalizations of Lemma 5.1, Theorem 4.1 and Proposition 5.1 in [9], Theorem 2.2 in [7] and Theorem 3 in [18]. Finally it is remarkable the family of examples obtained in Example 3.2, by using Example 3.1 and Lemma 2.4. Let H be the skew field of real quaternions, that is, each element a ∈ H can be written in the form a := a0 e0 + a1 e1 + a2 e2 + a3 e3 ,

ak ∈ R, k = 0, 1, 2, 3

where e0 = 1, e1 , e2 , e3 are the basis elements of H, with the multiplication rules e12 = e22 = e32 = −e0 , e1 e2 = −e2 e1 = e3 , e2 e3 = −e3 e2 = e1 , e3 e1 = −e1 e3 = e2 .

Quaternionic F( p, q, s) Function Spaces

The product is extended by linearity. The quaternionic conjugation defined by a = a0 − a1 e1 − a2 e2 − a3 e3 permits to define the norm |a| of a ∈ H by |a|2 = aa = aa = a02 + a12 + a22 + a32 . Therefore, if a ∈ H \ {0}, the quaternion 1 a |a|2

a −1 :=

is the multiplicative inverse of a. Also, the norm satisfies |ab| = |a||b| for each a, b ∈ H. Let  ⊂ R3 be a domain and consider the space C k (, H), k = 0 , 1, 2, . . . of H valued functions f :  → H such that its components f i belong to C k (, R), i.e. the functions are Cl0,2 -valued and the argument is a paravector in Cl0,2 . The Cauchy–Riemann operator is defined by D = e0

∂ ∂ ∂ + e1 + e2 . ∂ x0 ∂ x1 ∂ x2

D is a right linear operator with respect to scalars from H. Its conjugated operator is given by D = e0

∂ ∂ ∂ − e1 − e2 . ∂ x0 ∂ x1 ∂ x2

The solutions of D f = 0 are called (left) hyperholomorphic functions, generalizing the ring of classical analytic functions. In particular D D = D D =  is the Laplacian in R3 and then all hyperholomorphic functions are harmonic. Let z = (z 1 , z 2 ) be two regular variables (see [11]) given by z 1 = x1 e0 − x0 e1 ,

z 2 = x2 e0 − x0 e2

and a multiindex ν = (ν1 , ν2 ), |ν|. Let ν1 elements of the set a1 , . . . , a|ν| be equal to z 1 and ν2 elements be equal to z 2 . Define the following polynomial z˜ ν =

1  ai1 . . . ai|ν| |ν| i∈S|ν|

where i = (i 1 , . . . , i |ν| ) and S|ν| is the symmetric group. Each polynomial z˜ ν is left and right monogenic and they are H−linearly independent. It was shown in [13] that the general form of the Taylor series of left monogenic functions h in a neighborhood of the origen is given by h(x) =

∞   n=0 |ν|=n

z˜ ν cν

cν ∈ H.

(1.1)

A. G. Miss P. et al.

By Theorem 2.1 from [10] ⎛ ⎞   ∞  1    Dh(x) ≤ n⎝ |cν |⎠ |x|n−1 . 2 

(1.2)

|ν|=n

n=1

We identify each point x ∈ R3 with a quaternion of the form x = x0 e0 + x1 e1 + x2 e2 . Let r > 0 and a ∈ R3 . We denote by B(a, r ) the ball with center a and radius r , B(r ) := B(0, r ), and B := B(0, 1), is the unit ball with boundary S = ∂ B. If 0 < r < 1 we define the annulus A(r ) := B \ B(r ). We denote by M. the class of hyperholomorphic (or monogenic) functions on B. Given a ∈ B, the Möbius transform ϕa : B → B is defined by ϕa (x) = (a − x) (1 − ax)−1 which maps the unit ball onto itself. Let g (x, a) =

1 −1 |ϕa (x)|

be a multiple scalar of the fundamental solution of the Laplacian in R3 applied to the Möbius transform ϕa , i.e. g(x, a) is the modified Green’s function in quaternion sense. These definitions are used to generalize the Qs type spaces (see [9]). More precisely, we have the following definitions, (see [8]). Let 0 < p < ∞, −2 < q < ∞, 0 < s < 3 and f ∈ M. Define I p,q,s f : B → [0, ∞) as 

q p  I p,q,s f (a) =  D f (x) 1 − |x|2 g s (x, a) d x. B

Let 0 < p < ∞, −2 < q < ∞, 0 < s < ∞ and f ∈ M. Define J p,q,s f : B → [0, ∞) by  J p,q,s f (a) =

q    D f (x) p 1 − |x|2 (1 − |ϕa (x)|2 )s d x.

B

The set Fg ( p, q, s) is defined by  Fg ( p, q, s) =

f ∈ M : sup I p,q,s f (a) < ∞ a∈B

and Fg,0 ( p, q, s) is defined by  Fg,0 ( p, q, s) =

f ∈M :

lim I p,q,s f (a) = 0 .

|a|→1−

Quaternionic F( p, q, s) Function Spaces

In the same way we define the classes Fϕ ( p, q, s) and Fϕ,0 ( p, q, s) as 

Fϕ ( p, q, s) = f ∈ M : sup J p,q,s f (a) < ∞ , a∈B  Fϕ,0 ( p, q, s) = f ∈ M : lim J p,q,s f (a) = 0 . |a|→1−

In [9], for 0 < s < 3, the Qs spaces of quaternion valued functions given by Qs :=

⎧ ⎨ ⎩

 f ∈ M : sup a∈B

|D f (x)|2 g s (x, a) d x

⎫ ⎬ ⎭

= Fg (2, 0, s)

B

were introduced. This paper started the theory of Qs spaces in odd dimension. In [6], the B p spaces of quaternion valued functions defined by

B p :=

⎧ ⎨ ⎩

 f ∈ M : sup a∈B

 = Fϕ

|D f (x)| p (1 − |x|2 )

3p 2 −3

⎫ ⎬ (1 − |ϕa |2 )3 d x < ∞ ⎭

B

 3p p, − 3, 3 2

were presented. After, in [7] the B q, p spaces of quaternionic functions were studied, where ⎧ ⎫  ⎨ ⎬ 3p B q, p = f ∈ M : sup |D f (x)| p (1 − |x|2 ) 2 −q (1 − |ϕa |2 )q d x < ∞ ⎩ ⎭ a∈B B   3p − q, q . = Fϕ p, 2 The class Fg ( p, q, s) is a H-right module (also is left). This can be easily seen by the inequality (x + y) p ≤ 2 p (x p + y p ). From the definition of the spaces it follows immediately Lemma 1.1 Let 0 < p < ∞, −1 < q < ∞ and 0 < s < 3. Then Fg ( p, q, s) ⊂ Fϕ ( p, q, s)

and

Fg,0 ( p, q, s) ⊂ Fϕ,0 ( p, q, s).

For 0 < p < ∞, −1 < q < ∞, define the D p,q weighted Dirichlet space, as the set of monogenic functions f : B → H satisfying  |D f (x)| p (1 − |x|2 )q d x < ∞. B

A. G. Miss P. et al.

From the definition of Fg ( p, q, s) and Fϕ ( p, q, s) spaces the following results become immediate with a = 0. Lemma 1.2 Let 0 < p < ∞ and −1 < q < ∞. If 0 < s < 3 then Fg ( p, q, s) ⊂ D p,q+s . If 0 < s < ∞ then Fϕ ( p, q, s) ⊂ D p,q+s . Let α > 0. We define the α-Bloch space Bα as the set of monogenic functions f : B → H such that sup (1 − |x|2 )α |D f (x)| < ∞ a∈B

and the little α-Bloch space Bα,0 as the set of monogenic functions f : B → H such that lim (1 − |x|2 )α |D f (x)| = 0.

|a|→1−

2 Preliminaries Given a ∈ B, the Möbius transform ϕa : B → B satisfies 1 1 − |a|2 1 − |ϕa (x)|2 = = |J ϕa (x)| 3 , 2 2 |1 − ax| 1 − |x|

(2.1)

where J ϕ denotes the Jacobian of the function ϕ. For 0 < R < 1 the pseudohyperbolic ball D (a, R) is defined by D(a, R) = {x ∈ B : |ϕa (x)| < R} . This is an euclidean ball, with center and radius given respectively by c=

1 − R2 a, 1 − R 2 |a|2

r=

1 − |a|2 R. 1 − R 2 |a|2

(2.2)

The next result is a consequence of Cauchy–Schwartz inequality. Theorem 2.1 Let  ⊂ Rm be a domain, f :  → Rn be an integrable function on . Then         f  ≤ | f |.     



Corollary 2.1 Let  ⊂ Rm be a domain and f :  → Rn with f = ( f 1 , . . . , f n ) and 1 ≤ p < ∞. If each coordinate function f i :  → R is subharmonic, then | f | p is subharmonic on .

Quaternionic F( p, q, s) Function Spaces

Proof Let Sm−1 be the unit sphere in Rm . By hypothesis, for each i = 1, . . . , n and a∈  f i (a) ≤ f i (a + r ζ ) dσ (ζ ). Sm−1

where dσ (ζ ) denotes the normalized surface element in Sm−1 . Thus by Theorem 2.1  ⎛   n   ⎜  | f (a)| ≤  ⎝ 

i=1

=

⎞2 ⎟ f i (a + r ζ ) dσ (ζ )⎠ ≤

Sm−1

   n   f i2 (a + r ζ ) dσ (ζ ) Sm−1

i=1

| f (a + r ζ )| dσ (ζ ). Sm−1

Since x → x p is a convex function for 1 ≤ p < ∞, by Jensen’s inequality we complete the proof.

 The following result was proved in [10]. Lemma 2.1 Let 0 < p ≤ 2, 0 < r < 1, |a| < 1 and S be the unit sphere. Then there exists C > 0 such that  dσ (ζ ) C C ≤ ≤ . (2.3) p 2p (1 − |a|) p (1 − |a| r ) |1 − ar ζ | S

Now if a = 0  S

dσ (ζ ) 4π = , |1 − aζ |4 (1 − |a|2 )2

 S

dσ (ζ ) 2π 1 + |a| = ln . |1 − aζ |2 |a| 1 − |a|

In particular for 0 < p ≤ 1, there exists C = C( p) > 0 such that  dσ (ζ ) 1 . ≤ C ln |1 − aζ |2 p 1 − |a|

(2.4)

(2.5)

S

Proposition 2.1 Let 0 < R < 1 and h : B → R be a continuous function. If 0 < s < 3, then  sup h(x)g s (x, a) d x < ∞, a∈B

B(R)



h(x)g s (x, a) d x = 0.

lim

|a|→1− B(R)

A. G. Miss P. et al.

Now if 0 < s < ∞ then  h(x)(1 − |ϕa (x)|2 )s d x < ∞,

sup a∈B

B(R)



h(x)(1 − |ϕa (x)|2 )s d x = 0.

lim

|a|→1− B(R)

Proof Let 0 < R < 1 and 0 < s < 3. Since h is bounded on B(R), the first two claims are a consequence of the following estimations: by the change of variable x = ϕa (w) and (2.1) 

 g (x, a) d x = s

B(R)

B(R)



≤ B( 14 )

(1 − |ϕa (x)|)s dx = |ϕa (x)|s

 D(a,R)

(1 − |w|)s (1 − |a|2 )3 dw |w|s |1 − aw|6

(1 − |w|)s (1 − |a|2 )3 dw |w|s |1 − aw|6 

+ D(a,R)\B( 41 ) 2 34

≤ (1 − |a| )

6

(1 − |w|)s (1 − |a|2 )3 dw |w|s |1 − aw|6 

36 B( 14 )

(1 − |w|)s dw + 4s |w|s

 B

(1 − |a|2 )3 dw. |1 − aw|6

Now define  = B(R) × A( 1+R 2 ). Again by (2.1) 



 (1 − |ϕa (x)|)s 1 g (x, a) d x = d x ≤ max (1 − |ϕa (x)|2 )s d x (x,a)∈ |ϕa (x)|s |ϕa (x)|s B(R) B(R)  1 ≤ (1 − |a|2 )s max (1 − |x|2 )s d x. (x,a)∈ |ϕa (x)|s |1 − ax|s s

B(R)

B(R)



The other proofs are similar. Proposition 2.2 Let −2 < q < ∞. (i) If 0 < s < ∞ with −1 < q + s then  a∈B

 (1−|x|2 )q (1−|ϕa (x)|2 )s d x < ∞;

sup B

(1−|x|2 )q (1−|ϕa (x)|2 )s d x = 0.

lim

|a|→1− B

Quaternionic F( p, q, s) Function Spaces

(ii) If 0 < s < 3 with −1 < q + s then  (1 − |x| ) g (x, a) d x < ∞;

sup a∈B

 2 q s

(1 − |x|2 )q g s (x, a) d x = 0.

lim

|a|→1−

B

B 

Proof (i) Since (1 − |ϕa (x)|2 )s ≤ (1 − |ϕa (x)|2 )s for 0 < s < s  < ∞, it is enough to see the case 0 < s < 1. By (2.1) and (2.5) we have 

 (1 − |x|2 )q (1 − |ϕa (x)|2 )s d x =

(1 − |x|2 )q

B

B



= (1 − |a|2 )s B

(1 − |x|2 )s (1 − |a|2 )s dx |1 − ax|2s (1 − |x|2 )q+s dx |1 − ax|2s

1 = (1−|a| )

 (1−r )

2 s

2 q+s 2

r

0

≤ C(1 − |a|2 )s ln

S

1 1 − |a|

dσ (ζ ) dr |1−ar ζ |2s

1 (1 − r 2 )q+s r 2 dr. 0

(2.6) The previous estimation proves (i). (ii) Let 0 < R < 1 be fixed. We now estimate 

 (1 − |x| ) g (x, a) d x =

(1 − |x|2 )q

2 q s

B

B

≤

(1 − |ϕa (x)|)s dx |ϕa (x)|s



1 Rs

(1 − |x|2 )q (1 − |ϕa (x)|2 )s d x B\D(a,R)



(1 − |x|2 )q

+ D(a,R)

(1 − |ϕa (x)|)s dx |ϕa (x)|s

We use (2.6) to estimate the first integral in the sum. The second summand is estimated as follows: by the change of variable x = ϕa (w) and (2.1)  (1 − |x|2 )q D(a,R)

(1 − |ϕa (x)|)s dx |ϕa (x)|s



=

(1 − |ϕa (w)|2 )q B(R)

(1 − |w|)s (1 − |a|2 )3 dw |w|s |1 − aw|6

A. G. Miss P. et al.

 = B(R)

(1 − |a|2 )q (1 − |w|2 )q (1 − |w|)s (1 − |a|2 )3 dw |1 − aw|2q |w|s |1 − aw|6

(1 − |a|2 )q+3 ≤ (1 − R)2q+6

 B(R)

(1 − |w|2 )q+s dw. |w|s

Since −2 < q < ∞ and 0 < s < 3 we get the result.



From [19,20] and Corollary 2.1 we have: Lemma 2.2 Let 1 ≤ p < ∞, a ∈ B and f : B → H be a hyperholomorphic function. Let ψ f,a : B → H given by ψ f,a (x) =

1 − xa D f (ϕa (x)). |1 − ax|3

(2.7)

Then ψ f,a is a hyperholomorphic function and |ψ f,a | p is a subharmonic function. The next result was proved in [14]. Lemma n ∈ N∪{0}, an ≥ 0, In = {k ∈ N : 2n ≤ k < 2n+1 },  2.3 Let α > 0, p > 0, ∞ tn = k∈In ak and f (x) = n=1 an x n . Then there exists a constant C > 0 depending only on p and α such that  ∞ ∞  1  −nα p p 2 tn ≤ (1 − x)α−1 f (x) p d x ≤ C 2−nα tn . C 1

n=0

(2.8)

n=0

0

Two magnitudes A > 0 and B > 0 are similar, denoted by A ≈ B, if there exist two non negative real constants C1 and C2 such that, C1 A ≤ B ≤ C2 A. Proposition 2.3 Let 0 < s < 3, 0 < τ ≤ 1, −1 < q + s < ∞ and h : [0, 1) → R be a positive and increasing function. Then τ

τ h(r )(1 − r )

r dr ≈

h(r )(1 − r )q+s r 2−s dr

q+s 2

0

0

where the right integral is supposed finite. Proof As 0 < s < 3, 0 < r < 1 and r 2 ≤ r 2−s . Then τ

τ h(r )(1 − r )

r dr ≤

h(r )(1 − r )q+s r 2−s dr

q+s 2

0

0

Quaternionic F( p, q, s) Function Spaces

To obtain the other inequality we will use that h is an increasing function. Let t ∈ ( τ4 , τ2 ) then τ

t h(r )(1 − r )

q+s 2−s

r

τ

dr =

0

h(r )(1 − r )

q+s 2−s

r

dr + t

0

t ≤ h(t)

h(r )(1 − r )q+s r 2−s dr

τ (1 − r )q+s r 2−s dr +

h(r )(1 − r )q+s r 2−s dr.

(2.9)

t

0

Let λ = λ(τ ) > 1 given by



 τ 4s 0 (1 − r )q+s r 2−s dr λ(τ ) := max 1, s ,  τ . q+s r 2 dr τ τ (1 − r ) 2

We have the following decomposition τ

t h(r )λ(1 − r )

r dr =

0

τ h(r )λ(1 − r )

q+s 2

r dr +

h(r )(1 − r )q+s r 2−s dr

q+s 2

t

0

τ +

h(r )[λ(1 − r )q+s r 2 − (1 − r )q+s r 2−s ] dr. t

Since t < τ2 , then t

τ (1 − r )q+s r 2−s dr < t

0

as

τ 4

[λ(1 − r )q+s r 2 − (1 − r )q+s r 2−s ] dr,

< t then t

τ (1 − r )

q+s 2−s

h(t)

r

dr < h(t)

[λ(1 − r )q+s r 2 − (1 − r )q+s r 2−s ] dr t

0

τ ≤

h(r )[λ(1 − r )q+s r 2 − (1 − r )q+s r 2−s ] dr t

Finally by (2.9) we have proved τ

τ h(r )(1 − r )

q+s 2−s

0

r

dr ≤ λ

h(r )(1 − r )q+s r 2 dr. 0



A. G. Miss P. et al.

Lemma 2.4 Let f : B → H such that D f (x) = 0 = D f (x) for all x ∈ B. Then ˜ ˜ f (x) = h(z) + l(z)e 2 where h˜ and l˜ are two holomorphic complex functions of the complex variable z = x1 + x2 e1 . Proof Let f = f 0 + f 1 e1 + f 2 e2 + f 3 e3 . If D f (x) = 0 = D f (x) for all x ∈ B, by the definition of the Cauchy–Riemann operator and its conjugate and comparing the coefficients of e0 , e1 , e2 , e3 , we get the following set of equations ∂ f0 ∂ f1 ∂ f2 (x) = 0 and (x) = − (x) ∂ x0 ∂ x1 ∂ x2 ∂ f1 ∂ f0 ∂ f3 (x) = 0 and (x) = − (x) ∂ x0 ∂ x1 ∂ x2 ∂ f2 ∂ f3 ∂ f0 (x) = 0 and (x) = (x) ∂ x0 ∂ x1 ∂ x2 ∂ f3 ∂ f2 ∂ f1 (x) = 0 and (x) = (x). ∂ x0 ∂ x1 ∂ x2 ˜ ˜ = f 2 (z) + Thus f does not depend on x0 . Define h(z) = f 3 (z) + f 0 (z)e1 and l(z) f 1 (z)e1 , where z = x1 + x2 e1 . These functions are holomorphic by the previous relations. 

˜ + l(z)e ˜ We say that f, g ∈ M are equivalents (∼) if f (x) − g(x) = h(z) 2 where h˜ and l˜ are two holomorphic complex functions of the complex variable z = x1 + x2 e1 . If we consider M with this equivalence relation then for 1 ≤ p < ∞, by Minkowski’s inequality   f  = sup I p,q,s f (a) = sup p

a∈B

a∈B

q    D f (x) p 1 − |x|2 g s (x, a) d x

B

defines a norm in Fg ( p, q, s). 3 Properties of Quaternionic Spaces In this section we present several basic properties and some examples of different quaternionic spaces. Proposition 3.1 Let 1 ≤ p < ∞ and −2 < q < ∞. (i) If 0 < s < 3 and q + s ≤ −1 then Fg ( p, q, s) consists only of constant functions. (ii) If 0 < s < ∞ and q + s ≤ −1 then Fϕ ( p, q, s) consists only of constant functions.

Quaternionic F( p, q, s) Function Spaces

Proof (i) Let f ∈ Fg ( p, q, s) be a non constant function. Then there exist x0 ∈ B and 0 < R < 1 such that |D f (x)| > 0 for all x ∈ B(x0 , R). Thus by subharmonicity of |D f | p we have  ∞>

|D f (x)| p (1 − |x|2 )q B

(1 − |x|)s dx |x|s

 |D f (x)| p (1 − |x|2 )q

≥ A(|x0 |−R)

1 ≥

 (1 − r 2 )q (1 − r )s r 2−s

|x0 |

|D f (r ζ )| p dσ (ζ ) dr S

 ≥

(1 − |x|)s dx |x|s

1 |D f (|x0 |ζ )| p dσ (ζ )

(1 − r 2 )q (1 − r )s r 2−s dr = ∞

|x0 |

S

as q + s ≤ −1, we get a contradiction; therefore f is constant. The proof of (ii) is similar. 

From now on, we will suppose −1 < q + s < ∞. Example 3.1 Let −2 < q < ∞, 0 < s ≤ 3 with −1 < q + s < ∞. The spaces Fg ( p, q, s), Fg,0 ( p, q, s), Fϕ ( p, q, s) and Fϕ,0 ( p, q, s) are not empty. Let f : B → H be a monogenic function such that |D f (x)| < M for all x ∈ B. Therefore   |D f (x)| p (1 − |x|2 )q g s (x, a) d x ≤ M (1 − |x|2 )q g s (x, a) d x B

B

and now apply Proposition 2.2. Thus f belongs to the quoted spaces. This condition is satisfied, for example, if f ∈ M ∩ C 1 (B). Other examples can be constructed by choosing constants cν ∈ H such that ∞ 

⎛ n⎝

n=1



⎞ |cν |⎠ = M < ∞

|ν|=n

then f (x) =

∞  

z˜ ν cν ,

cν ∈ H

n=0 |ν|=n

is a monogenic function on B and by (1.2) satisfies |D f (x)| < 2M for all x ∈ B. Now we will present a family of examples using the Lemma 2.4 and the previous example.

A. G. Miss P. et al.

Example 3.2 Let f be defined as in the previous example and consider ˜ ˜ f 1 (x) = f (x) + h(z) + l(z)e 2 where h˜ and l˜ are two holomorphic complex functions of the complex variable z = x1 + x2 e1 , (see Lemma 2.4). Then f 1 belongs to the spaces quoted in Example 3.1. Observe that we can repeat this procedure for each example about Fg ( p, q, s) or Fϕ ( p, q, s) spaces. The following result improves Theorem 3.1 in [18]. Theorem 3.1 Let 0 < p < ∞, −2 < q < ∞. If 0 < s < 3 the function a → I p,q,s f (a) is continuous. (For 0 < s < ∞ the function a → J p,q,s f (a) is also continuous). Proof Let a ∈ B be fixed. We take r such that 0 < Define M = max

(1 − |b|2 )s |ϕb (x)|s |1 − bx|2s

1 + |a| < r < 1 and ε > 0. 2

for all (x, b) ∈ A ((1 + r )/2) × B(r ).

By Lemma 1.2 and the absolute continuity of the integral, there exists such that 

1+r
ε for all R ≤ R  < 1. M

|D f (x)| p (1 − |x|2 )q+s d x < A(R  )

Then for all b ∈ B(r ) and R < R  < 1 we have  |D f (x)| A(R  )

p

(1 − |ϕb (x)|) (1−|x|2 )q |ϕb (x)|s

s

 dx ≤ M

|D f (x)| p (1−|x|2 )q+s d x < ε. A(R  )

Let {an } ⊂ B(r ) such that limn→∞ an = a. To finish the proof observe that is enough to prove 

 lim

n→∞ B(R  )

|D f (x)| p (1 − |x|2 )q g s (x, an ) d x = B(R  )

|D f (x)| p (1−|x|2 )q g s (x, a) d x.

Quaternionic F( p, q, s) Function Spaces

In fact, by the change of variable w = ϕan (x) we get  |D f (x)| p (1 − |x|2 )q g s (x, an ) d x B(R  )

 |D f (ϕan (w))| p (1 − |ϕan (w)|2 )q

= D(an ,R  )



|D f (ϕan (w))| p

= B

(1 − |w|)s (1 − |a|2 )3 dw |w|s |1 − aw|6

(1 − |an |2 )q+3 (1 − |w|)s (1 − |w|2 )q ) χ dw. D(a ,R n |1 − an w|2q+6 |w|s

Observe that lim |D f (ϕan (w))| p

n→∞

2 q+3 (1 − |an |2 )q+3 p (1−|a| )  ) = |D f (ϕa (w))| χ χ D(a,R  ) D(a ,R n |1 − an w|2q+6 |1−aw|2q+6

and since 0 < q + 3, |D f (ϕan (w))| p

(1 − |an |2 )q 1 χ D(an ,r ) ≤ max |D f (x)| p . 2q+6 |1 − an w| (1 − R  )2q+6 x∈B(R  )

The result follows applying the Lebesgue dominated theorem, with the finite measure s (1−|w|2 )q dμ(w) = (1−|w|)|w| dw. s 

(In a similar way we can prove that J p,q,s f is continuous). Theorem 3.2 Let 0 < p < ∞, −2 < q < ∞. If 0 < s < 3 then Fg,0 ( p, q, s) ⊂ Fg ( p, q, s) and if 0 < s < ∞ then Fϕ,0 ( p, q, s) ⊂ Fϕ ( p, q, s) Proof Let f ∈ Fg,0 ( p, q, s). Thus, by Theorem 3.1 we extend continuously the definition of I p,q,s f to B by setting I p,q,s f (a) = 0 when |a| = 1. Then sup I p,q,s f (a) = max I p,q,s f (a) < ∞. a∈B

a∈B

Moreover, there is b ∈ B, such that maxa∈B I p,q,s f (a) = I p,q,s f (b). The other proof is similar. 

4 Characterizations of Fϕ ( p, q, s) Spaces We prove in this section that Fg ( p, q, s) = Fϕ ( p, q, s) and Fg,0 ( p, q, s) = Fϕ,0 ( p, q, s). Likewise we give a characterization of these spaces using Carleson measures on boxes. The next results generalize Lemma 5.1, Theorem 5.1, Proposition 5.1 from [9], and Theorem 2.2 from [7].

A. G. Miss P. et al.

Theorem 4.1 Let 1 ≤ p < ∞ and 0 < s < 3. Let f : B → H be a monogenic function. Then f belongs to the Dirichlet space D p,q+s if and only if  |D f (x)| p (1 − |x|2 )q g s (x, 0) d x < ∞. B

Proof We will prove that 

 |D f (x)| p (1 − |x|2 )q+s d x ≈

B

|D f (x)| p (1 − |x|2 )q g s (x, 0) d x; B

using spherical coordinates this is equivalent to 1 

1  |D f (r ζ )| p (1−r 2 )q+s r 2 dσ (ζ ) dr ≈

0 S

|D f (r ζ )| p (1 − r 2 )q (1 − r )s r 2−s dσ (ζ ) dr. 0 S

Since 1 ≤ 1 + r ≤ 2 and |D f | p is subharmonic the result follows if we apply the previous Proposition 2.3 with τ = 1 and  h(r ) =

|D f (r ζ )| p dσ (ζ ). S



The previous result motivates the following theorem. Theorem 4.2 Let 1 ≤ p < ∞, −2 < q < ∞ and 0 < s < 3. Then Fg ( p, q, s) = Fϕ ( p, q, s)

and

Fg,0 ( p, q, s) = Fϕ,0 ( p, q, s).

and

Fϕ,0 ( p, q, s) ⊂ Fg,0 ( p, q, s).

Proof By Lemma 1.1 we prove only Fϕ ( p, q, s) ⊂ Fg ( p, q, s)

By the change of variable x = ϕa (w) and (2.7), we have  I p,q,s f (a) =

q    D f (x) p 1 − |x|2 g s (x, a) d x

B

 =

|ψ f,a (w)| p (1 − |ϕa (w)|2 )q B

(1 − |w|)s (1 − |a|2 )3 dw |w|s |1 − aw|6−2 p

Quaternionic F( p, q, s) Function Spaces

while  J p,q,s f (a) =

q    D f (x) p 1 − |x|2 (1 − |ϕa (x)|2 )s d x

B

 =

|ψ f,a (w)| p (1 − |ϕa (w)|2 )q (1 − |w|2 )s B

Since

1 2

≤ |w| < 1 implies 1 <

1 ≤ 2s then |w|s

 |ψ f,a (w)| p (1 − |ϕa (w)|2 )q B\B( 21 )

(1 − |w|)s (1 − |a|2 )3 dw |w|s |1 − aw|6−2 p



≈

(1 − |a|2 )3 dw |1 − aw|6−2 p

|ψ f,a (w)| p (1 − |ϕa (w)|2 )q (1 − |w|)s B\B( 21 )

(1 − |a|2 )3 dw. |1 − aw|6−2 p

1 1 1 ≤ 22|q− p+3| , thus, by implies 2|q− p+3| ≤ 2 2 |1 − aw|2(q− p+3) (2.1) and changing to spherical coordinates, it is enough to prove

Since 0 ≤ |w| ≤

1

1

2 

2  |ψ f,a (r ζ )| p (1−r 2 )q (1 − r )s r 2−s dσ (ζ ) dr ≈

0 S

|ψ f,a (r ζ )| p (1−r 2 )q+s r 2 dσ (ζ ) dr 0 S

We have 1 ≤ 1 + r ≤ 2 and by (2.7), |ψ f,a | p is subharmonic. Now the result follows 1 if we apply Proposition 2.3 with τ = and 2  h(r ) = |ψ f,a (r ζ )| p dσ (ζ ). S



After the previous result, from now on we will consider only the spaces Fϕ ( p, q, s) and Fϕ0 ( p, q, s). Remark 4.1 In [8], Theorem 4.1 gives a special case of the previous theorem. The following proposition gives more examples.

∞  n n+1 ν Proposition 4.1 Let  In = {k ∈ N : 2 ≤ k < 2 }, f (x) = n=0 |ν|=n z˜ cν , cν ∈ H and an = |ν|=n |cν |. For 0 < p < ∞ , −1 < q + s < ∞, suppose that ∞  n=0

⎛ 2n( p−q−s)+ p−n ⎝

 k∈In

⎞p ak+1 ⎠ < ∞

A. G. Miss P. et al.

then: (i) If 0 < s ≤ 1, f ∈ Fϕ,0 ( p, q, s), (ii) If 1 < s ≤ 2, f ∈ Fϕ ( p, q, s). Proof By (1.2) and changing to spherical coordinates we have p   1   D f (x) (1 − |x|2 )q (1 − |ϕa (x)|2 )s d x 2  B p   ∞ (1 − |a|2 )s (1 − |x|2 )s n−1 = nan |x| (1 − |x|2 )q dx |1 − ax|2s B

1 = 0

n=1

∞ 

p nan r

 (1 − r 2 )q+s r 2 (1 − |a|2 )s

n−1

n=1

S

dσ (ζ ) dr. |1 − ar ζ |2s

(i) If 0 < s ≤ 1, by (2.5) and Lemma 2.3, there exist C > 0 and C  > 0 such that p   1   D f (x) (1 − |x|)q (1 − |ϕa (x)|2 )s d x 2  B

1 ≤ C(1 − |a| ) ln 1 − |a| 2 s

≤ C2

p+q+s

1  ∞ 0

p nan r

n−1

(1 − r 2 )q+s r 2 dr

n=1

1 p a (1 − |a| ) ln 1 − |a| 1

1 (1 − r )q+s dr

2 s

0

p 1  ∞ n (n + 1)an+1r (1 − r )q+s dr

1 1 − |a| n=1 0 ⎛ ⎞p⎞ ⎛ ∞   1 p ⎝a +C  = C2 p+q+s (1−|a|2 )s ln 2−n(q+s+1) ⎝ (k + 1)ak+1 ⎠ ⎠ 1 1−|a| n=0 k∈In ⎛ ⎞p⎞ ⎛ ∞   1 p ⎝a + C  ≤ C2 p+q+s (1 − |a|2 )s ln 2n( p−q−s)+ p−n ⎝ ak+1 ⎠ ⎠ 1 1 − |a| +C2 p+q+s (1 − |a|2 )s ln

n=0

k∈In

and thus f ∈ F0 ( p, q, s). By (2.3) we get (ii).



Proposition 4.2 The next inclusions are true (i) If 0 < p  < p < ∞ then Fϕ ( p, q, s) ⊂ Fϕ,0 ( p  , q, s)

for − 2 < q < ∞, 0 < s < ∞.

Quaternionic F( p, q, s) Function Spaces

(ii) If −2 < q  < q < ∞ then Fϕ ( p, q  , s) ⊂ Fϕ ( p, q, s),

for 0 < p < ∞, 0 < s < ∞.

(iii) If −2 < q < ∞ and 0 < p < ∞ then Fϕ ( p, q, s  ) ⊂ Fϕ ( p, q, s),

for 0 < s  < s < ∞.

Proof We prove only (i). Let f ∈ Fg ( p, q, s) and 0 < p  < p < ∞. By Hölder’s inequality and with the measure dμ(x) = (1 − |x|2 )q (1 − |ϕa (x)|2 )s d x we have 

⎛ ⎞ p p  p− p p p |D f (x)| dμ(x) ≤ ⎝ |D f (x)| dμ(x)⎠ (μ(B)) p .

B

B



By Proposition 2.2 we get the result. Corollary 4.1 Let 0 < p < ∞, −2 < q < ∞ and 0 < s < ∞. Then: (i) Fϕ ( p, q, s) ⊂ (ii) Fϕ ( p, q  , s) ⊂ (iii) Fϕ ( p, q, s  ) ⊂

Fϕ,0 ( p  , q, s); q 
0< p  < p

Now, we give a characterization of Fϕ ( p, q, s) in terms of Carleson measures. We assume definitions and results of [12]. Let a ∈ B. Define the Carleson tube by    a   < 1 − |a| . E(a) = x ∈ B : x − |a|  

For 0 < s < ∞, a positive Borel measure μ on B is a bounded s-Carleson measure if sup a∈B

μ(E(a)) <∞ (1 − |a|)s

and μ is a s-compact Carleson measure if lim

|a|→1−

μ(E(a)) = 0. (1 − |a|)s

Theorem 4.3 (Theorems 3.1 and 3.2, [12]) Let 0 < s < ∞ and 0 < τ < ∞. A positive Borel measure μ on B is a bounded s-Carleson measure if and only if   sup a∈B

B

s

(1 − |a|2 )τ (1 + |a|2 |x|2 − 2(x, a))

1+τ 2

dμ(x) < ∞

A. G. Miss P. et al.

and μ is a compact s-Carleson measure if and only if   lim

|a|→1−

s

(1 − |a|2 )τ (1 + |a|2 |x|2 − 2(x, a))

B

1+τ 2

dμ(x) = 0.

From this result we obtain the next characterization for Fϕ ( p, q, s) spaces in terms of Carleson measures. Theorem 4.4 Let 0 < p < ∞, −2 < q < ∞ and 0 < s < ∞. Then (i) f ∈ Fϕ ( p, q, s) if and only if dμ(x) = |D f (x)| p (1−|x|2 )q+s dv(x) is a bounded p-Carleson measure. (ii) f ∈ Fϕ,0 ( p, q, s) if and only if dμ(x) = |D f (x)| p (1 − |x|2 )q+s dv(x) is a compact s-Carleson measure. Proof Consider Theorem 4.3, with τ = 1, the identity (2.1) and the fact that |1 − ax|2 = (1 − ax)(1 − xa) = 1 + |a|2 |x|2 − 2 Re (xa). 

So the result follows from the definition of the spaces. For additional information on this topic see [3]. 5 The Quaternionic Bloch and Dirichlet Spaces

In this section we characterize the Bloch space by some special family of Fϕ ( p, q, s) spaces. Similar results can be proved by using the function g s (z, a) as weight, instead of the weight (1 − |ϕa (z)|2 )s . Proposition 5.1 Let 1 ≤ p < ∞, −2 < q < ∞ and 0 < s < ∞. If f ∈ Fϕ ( p, q, s) (respectively f ∈ Fϕ,0 ( p, q, s)) then f ∈ B q+3 (respectively f ∈ B q+3 ,0 ). p

p

Proof Let 0 < R < 1 be fixed and a ∈ B. By the change of variable x = ϕa (w) and Lemma 2.2 we have for 0 < s < ∞ fixed  J p,q,s f (a) ≥

q    D f (x) p 1 − |x|2 (1 − |ϕa (x)|2 )s d x

D(a,R)

 ≥ (1 − R 2 )s B(R)



= (1 − R 2 )s B(R)

·

q    D f (ϕa (w)) p 1 − |ϕa (w)|2



1 − |a|2 |1 − aw|2

 p  1 − wa    |1 − aw|2 p D f (ϕ (w)) a  |1 − aw|3 



q

q

3 1 − |a|2 1 − |w|2 1 − |a|2 |1 − aw|2q

|1 − aw|6

dw

3 dw

Quaternionic F( p, q, s) Function Spaces

 = (1 − R 2 )s (1 − |a|2 )q+3 B(R)



≥

(1 − R 2 )s 1 − |a|2 22(q− p+3)

≥

(1 − R 2 )s 1 − |a|2

q+3

p  

q  1 − wa    1 − |w|2 dw D f (ϕ (w)) a  |1 − aw|3 

B(r )

R 

22(q− p+3)

≥

q+3



q  p 1 − |w|2  1 − wa     |1 − aw|3 D f (ϕa (w)) |1 − aw|2q+6−2 p dw

(1 − R 2 )s 1 − |a|2 22(q− p+3)

q+3

q   ψ f,a (ρζ ) p 1 − ρ 2 ρ 2 dσ (ζ )dρ

0 S

   D f (a) p

q+3 1 − |a|2 p  = C(R) 2(q− p+3)  D f (a) . 2

R

1 − ρ2

q

ρ 2 dρ

0

Now the proposition follows from this estimation, where we have used the subhar

monicity of |ψ f,a | p . Remark 5.1 This result was proved in [8, Proposition 3.3], however the proof presented here is self-contained. We have the following reciprocal result of Proposition 5.1. Proposition 5.2 Let 0 < p < ∞, −2 < q < ∞ and 2 < s < ∞. If f ∈ B q+3 (respectively f ∈ B q+3 ,0 ), then f ∈ Fϕ ( p, q, s), (respectively f ∈ Fϕ,0 ( p, q, s)).

p

p

Proof Let f ∈ B q+3 be a non constant function. Then, there exists 0 < M < ∞ such p

that

q+3   p  D f (x) ≤ M 1 − |x|2 for all x ∈ B, and so by the change of variable x = ϕa (w)  J p,q,s f (a) ≤ B

Mp (1 − |x|2 )q (1 − |ϕa (x)|2 )s d x (1 − |x|2 )q+3 

= Mp B

2 3 1 2 s (1 − |a| ) (1 − |w| ) dw (1 − |ϕa (w)|2 )3 |1 − aw|6

 =M

(1 − |w|2 )s−3 dw

p B

but the last integral is finite since 2 < s < ∞ and so f ∈ Fϕ ( p, q, s).

(5.1)

A. G. Miss P. et al.

We suppose now that f ∈ B q+3 ,0 . Then there exists 0 < R < 1 such that for all p

R < |x| < 1

q+3   p  D f (x) ≤ 1 − |x|2 !

1

εp 2 s−3 dw B (1 − |w| )

"1 . p

By Proposition 2.1 is enough to estimate 

q    D f (x) p 1 − |x|2 (1 − |ϕa (x)|2 )s d x A(R)

≤

ε 2 )s−3 dw (1 − |w| B

ε < 2 )s−3 dw (1 − |w| B

 A(R)

1 (1 − |x|2 )q (1 − |ϕa (x)|2 )s d x (1 − |x|2 )q+3



(1 − |w|2 )s−3 dw < ε B



so this concludes the proof. Combining Propositions 5.1 and 5.2 we have the following theorem:

Theorem 5.1 Let 1 ≤ p < ∞, −2 < q < ∞. The following conditions are equivalent: (i) f ∈ B q+3 (respectively f ∈ B q+3 ,0 ). p

p

(ii) f ∈ Fϕ ( p, q, s) (respectively f ∈ Fϕ,0 ( p, q, s)) for all s > 2. (iii) f ∈ Fϕ ( p, q, s) (respectively f ∈ Fϕ,0 ( p, q, s)) for some s > 2. Remark 5.2 Observe that although the previous result is formally similar to Theorem 3.5 in [8], our techniques enable generalize the result to the little spaces Fϕ,0 ( p, q, s) and B q+3 ,0 . p

Proposition 5.3 Let 0 < p < ∞, −2 < q < ∞, 0 < s < 2 and

q +1 < α < p

q +s+1 . If f ∈ Bα (respectively f ∈ Bα,0 ) then f ∈ Fϕ ( p, q, s) (respectively p f ∈ Fϕ,0 ( p, q, s)). Proof Let f ∈ Bα be a non constant function. Then, there exists 0 < M < ∞ such that

α   1 − |x|2  D f (x) ≤ M for all x ∈ B, and so by the change of variable x = ϕa (w) and by Lemma 2.1 we have  J p,q,s f (a) ≤ B

Mp (1 − |x|2 )q (1 − |ϕa (x)|2 )s d x (1 − |x|2 )αp

Quaternionic F( p, q, s) Function Spaces

 = Mp

(1 − |ϕa (w)|2 )q−αp (1 − |w|2 )s B



≤ M p (1 − |a|2 )q+3−αp B

(1 − |a|2 )3 dw |1 − aw|6

(1 − |w|2 )q+s−αp dw |1 − aw|6+2q−2αp

1 ≤ M (1 − |a| ) p

 (1 − r 2 )q+s−αp r 2−s

2 q−αp+3 0

S

dσ (ζ ) dr |1 − ar ζ |2(q−αp+3)

1 ≈ M p 2q−αp+3 λ

(1 − r )q− pα+s r 2−s dr, 0

and the last integral is finite. For the little spaces imitate the proof of Proposition 5.2. 

Now we prove some results about Dirichlet spaces D p,q . Theorem 5.2 Let 1 ≤ p < ∞, −1 < q < ∞. Then D p,q ⊂ B q+3 ,0 . p

Proof Let 0 < R < 1 be fixed. Imitating the proof of Proposition 5.1 we obtain  |D f (x)| p (1 − |x|2 )q d x ≥ C(R)(1 − |a|2 )q+3 |D f (a)| p .

(5.2)

D(a,R)

Since  |D f (x)| p (1 − |x|2 )q d x < ∞ B

then given ε > 0, there exists 0 < R˜ < 1 such that  |D f (x)| p (1 − |x|2 )q d x < ε. ˜ A( R)

˜ for all a ∈ B with By (2.2) there exists R˜ < R  < 1 such that D(a, R) ⊂ A( R)  

R < |a| < 1. From (5.2) we get our result. Theorem 5.3 Let 1 ≤ p < ∞, −1 < q < ∞. Then D p,q ⊂

# 0
Fϕ,0 ( p, q, s).

A. G. Miss P. et al.

Proof Let f ∈ D p,q and ε > 0. By Theorem 5.2, there exists 0 < R  < 1 such that (1 − |x|2 )q+3 |D f (x)| p < ε for all

0 < R  < |x| < 1.

(5.3)

˜ R  be as in the previous theorem, where R is fixed and we can choose Let R, R,  ˜ R < R < R  < 1. Then we have 

q    D f (x) p 1 − |x|2 (1 − |ϕa (x)|2 )s d x

J p,q,s f (a) = ˜ B( R)



q    D f (x) p 1 − |x|2 (1 − |ϕa (x)|2 )s d x.

+ ˜ A( R)

By Proposition 2.1 the first integral goes to 0 when |a| → 1− . We consider R  < |a| < 1 and the second integral is divided in two integrals. Thus 

q    D f (x) p 1 − |x|2 (1 − |ϕa (x)|2 )s d x

˜ A( R)\D(a,R)



≤ (1 − R 2 )s

q    D f (x) p 1 − |x|2 d x < ε.

˜ A( R)\D(a,R)

Finally by (5.3), the change of variable ϕa (w) = x and (2.1) we have 

q    D f (x) p 1 − |x|2 (1 − |ϕa (x)|2 )s d x

D(a,R)



≤ D(a,R)



≤ε B(R)



q ε 2 |x| 1 − (1 − |ϕa (x)|2 )s d x (1 − |x|2 )q+3 2 3 1 2 s (1 − |a| ) (1 − |w| ) dw (1 − |ϕa (w)|2 )3 |1 − aw|6



=ε

(1 − |w|)s−3 dw B(R)

so we finish the proof.



Remark 5.3 It is remarkable if we compare this result with Theorem 4 in [2], where they proved only that D ⊂ ∩0< p<∞ Q p . The following proposition proves that the Dirichlet spaces are not empty.

Quaternionic F( p, q, s) Function Spaces n n+1 Proposition 5.4 Let  In = {k ∈ N : 2 ≤ k < 2 }, f (x) = cν ∈ H and an = |ν|=n |cν |. Let 0 < p < ∞ , −1 < q < ∞. If ∞ 

⎛ 2n( p−q)+ p−n ⎝

n=0



∞  n=0

|ν|=n

z˜ ν cν ,

⎞p ak+1 ⎠ < ∞,

k∈In

then f ∈ D p,q . Proof By (1.2) and Lemma 2.3 we have 

1 | D f (x)| p (1 − |x|)q d x 2

B

≤

  ∞ B

p nan |x|

(1 − |x| ) d x = 4π

n−1

2 q

n=1

1  ∞ 0

p

(1 − r )q r 2 dr + 4 · (2q + 1)π 0

= C  + 4C · (2q + 1)π

∞  n=0

≤ C  + 4C · (2q + 1)π

∞  n=0

⎛ 2−n(q+1) ⎝

(1 − r 2 )q r 2 dr

p  1  ∞ (n + 1)an+1 r n (1 − r )q dr 0



nan r

n=1

1

≤ 4 · (2q + 1)πa1

p n−1

n=1

⎞p

(k + 1)ak+1 ⎠

k∈In

⎛

2n( p−q)+ p−n ⎝



⎞p ak+1 ⎠ < ∞.

k∈In



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