RIEMANN-STIELTJES OPERATORS FROM F(p, q,s) SPACES
TO α-BLOCH SPACES ON THE UNIT BALL
SONGXIAO LI
Received 5 December 2005; Accepted 19 April 2006
Let H(B) denote the space of all holomorphic functions on the unit ball B Cn . We
1
investigate the following integral operators: Tg ( f )(z) = 0 f (tz)g(tz)(dt/t), Lg ( f )(z) =
1
n
j =1 z j (∂h/∂z j )(z)
0 f (tz)g(tz)(dt/t), f H(B), z B, where g H(B), and h(z) =
is the radial derivative of h. The operator Tg can be considered as an extension of the
Cesàro operator on the unit disk. The boundedness of two classes of Riemann-Stieltjes
operators from general function space F(p, q,s), which includes Hardy space, Bergman
space, Q p space, BMOA space, and Bloch space, to α-Bloch space Ꮾα in the unit ball is
discussed in this paper.
Copyright © 2006 Songxiao Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let z = (z1 ,...,zn ) and w = (w1 ,...,wn ) be points in the complex vector space Cn and
z,w = z1 w̄1 + + zn w̄n .
(1.1)
Let dv stand for the normalized Lebesgue measure on Cn . For a holomorphic function f
we denote
f =
∂f
∂f
,...,
.
∂z1
∂zn
(1.2)
Let H(B) denote the class of all holomorphic functions on the unit ball. Let f (z) =
for the radial derivative of f H(B) [21]. It is easy to see that,
j =1 z j (∂ f /∂z j )(z) stand
if f H(B), f (z) = α aα zα , where α is a multiindex, then
n
f (z) = αaα zα .
α
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 27874, Pages 1–14
DOI 10.1155/JIA/2006/27874
(1.3)
2
Riemann-Stieltjes operators from F(p, q,s) to Ꮾα
The α-Bloch space Ꮾα (B) = Ꮾα , α > 0, is the space of all f
H(B) such that
α bα ( f ) = sup 1 z2 f (z) < .
(1.4)
z B
On Ꮾα the norm is introduced by
f Ꮾ
α
= f (0) + bα ( f ).
(1.5)
With this norm Ꮾα is a Banach space. If α = 1, we denote Ꮾα simply by Ꮾ.
For a,z B, a = 0, let ϕa denote the Möbius transformation of B taking 0 to a defined
by
ϕa (z) =
a Pa (z) 1 z2 Qa (z)
,
1 z,a
(1.6)
where Pa (z) is the projection of z onto the one dimensional subspace of Cn spanned by a
and Qa (z) = z Pa (z) which satisfies (see [21])
ϕa Æ ϕa = id,
ϕa (0) = a,
2
1 a2 1 z2
1 ϕa (z) =
.
1 z,a2
ϕa (a) = 0,
(1.7)
Let 0 < p, s < , n 1 < q < . A function f
F(p, q,s)(B) (see [19, 20]) if
p
p
f F(p,q,s)
= f (0) + sup
B
a B
H(B) is said to belong to F(p, q,s) =
f (z) p 1 z2 q g s (z,a)dv(z) < ,
(1.8)
where g(z,a) = log ϕa (z)1 is Green’s function for B with logarithmic singularity at a.
We call F(p, q,s) general function space because we can get many function spaces, such
as BMOA space, Q p space (see [9]), Bergman space, Hardy space, Bloch space, if we take
special parameters of p, q,s in the unit disk setting, see [20]. If q + s 1, then F(p, q,s)
is the space of constant functions.
For an analytic function f (z) on the unit disk D with Taylor expansion f (z) =
n
n=0 an z , the Cesàro operator acting on f is
Ꮿ f (z) =
n =0
1 ak z n .
n + 1 k =0
n
(1.9)
The integral form of Ꮿ is
Ꮿ( f )(z) =
1
z
z
0
f (ζ)
1
1ζ
dζ =
1
z
z
0
f (ζ) ln
1
1ζ
dζ,
(1.10)
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3
taking simply as a path of the segment joining 0 and z, we have
Ꮿ( f )(z) =
1
0
f (tz) ln
1ζ 1
ζ =tz
dt.
(1.11)
The following operator:
zᏯ( f )(z) =
z
0
f (ζ)
dζ,
1ζ
(1.12)
is closely related to the previous operator and on many spaces the boundedness of these
two operators is equivalent. It is well known that Cesàro operator acts as a bounded linear
operator on various analytic function spaces (see [4, 8, 11–13, 16] and the references
therein).
Suppose that g H(D), the operator
Jg f (z) =
z
0
f (ξ)dg(ξ) =
1
0
f (tz)zg (tz)dt =
z
0
f (ξ)g (ξ)dξ,
z D,
(1.13)
where f H(D), was introduced in [10] where Pommerenke showed that Jg is a bounded
operator on the Hardy space H 2 (D) if and only if g BMOA. The operator Jg acting on
various function spaces have been studied recently in [1–3, 14, 17, 18].
Another operator was recently defined in [18], as follows:
Ig f (z) =
z
0
f (ξ)g(ξ)dξ.
(1.14)
The above operators Jg , Ig can be naturally extended to the unit ball. Suppose that
g : B C1 is a holomorphic map of the unit ball, for a holomorphic function f , define
Tg f (z) =
1
0
f (tz)
dg(tz)
=
dt
1
0
f (tz)g(tz)
dt
,
t
z B.
(1.15)
This operator is called Riemann-Stieltjes operator (or extended-Cesàro operator). It was
introduced in [5], and studied in [5–7, 15, 17].
Here, we extend operator Ig for the case of holomorphic functions on the unit ball as
follows:
Lg f (z) =
1
0
f (tz)g(tz) dtt ,
z B.
(1.16)
To the best of our knowledge operator Lg on the unit ball is introduced in the present
paper for the first time.
The purpose of this paper is to study the boundedness of the two Riemann-Stieltjes
operators Tg , Lg from F(p, q,s) to α-Bloch space. The corollaries of our results generalized
the former results and some results are new even in the unit disk setting.
4
Riemann-Stieltjes operators from F(p, q,s) to Ꮾα
In this paper, constants are denoted by C, they are positive and may differ from one
occurrence to the other. a b means that there is a positive constant C such that a Cb.
Moreover, if both a b and b a hold, then one says that a b.
2. Tg ,Lg : F(p, q,s) Ꮾα
In order to prove our results, we need some auxiliary results which are incorporated in
the following lemmas. The first one is an analogy of the following one-dimensional result:
z
0
f (ζ)g (ζ)dζ
z
= f (z)g (z),
0
f (ζ)g(ζ)dζ
= f (z)g(z).
(2.1)
Lemma 2.1 [5]. For every f ,g H(B), it holds that
Tg ( f ) (z) = f (z)g(z),
Lg ( f ) (z) = f (z)g(z).
Proof. Assume that the holomorphic function f g has the expansion
Tg ( f ) (z) = 1
0 α
aα (tz)
α dt
t
=
aα
α
α
z
α
=
α aα z
aα z α ,
(2.2)
α.
Then
(2.3)
α
which is what we wanted to prove. The proof of the second formula is similar and will be
omitted.
The following lemma can be found in [19].
Lemma 2.2. For 0 < p, s < ,
Ꮾ(n+1+q)/ p and
n 1 < q < , q + s > 1, if f F(p, q,s), then f f Ꮾ
(n+1+q)/ p
C f F(p,q,s) .
(2.4)
The following lemma can be found in [15].
Lemma 2.3. If f
Ꮾα , then
⎧
⎪
⎪
⎪ f (0) + f Ꮾα ,
⎪
⎪
⎪
⎪
⎨
1
f (z) C f (0) + f Ꮾα log
1
z2 ,
⎪
⎪
⎪
⎪
f Ꮾα ,
⎪
⎪
⎪ f (0) + ⎩
2 α1
1 z
0 < α < 1;
α = 1,
(2.5)
α > 1,
for some C independent of f .
2.1. Case p < n + 1 + q. In this section we consider the case p < n + 1 + q. Our first result
is the following theorem.
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5
Theorem 2.4. Let g be a holomorphic function on B, 0 < p, s < , n 1 < q < , q + s >
1, n + 1 + q pα, p < n + 1 + q. Then Tg : F(p, q,s) Ꮾα is bounded if and only if
sup 1 z2
z B
α+1(n+1+q)/ p g(z) < .
(2.6)
Moreover, the following relationship:
Tg F(p,q,s)Ꮾα
sup 1 z2
z B
α+1(q+n+1)/ p g(z)
(2.7)
holds.
Proof. For f ,g H(B), note that Tg f (0) = 0, by Lemmas 2.1, 2.2, and 2.3,
Tg f Ꮾα
α = sup 1 z2 Tg f (z)
z B
α = sup 1 z2 f (z)g(z)
z B
C f Ꮾ
(n+1+q)/ p
sup 1 z2
z B
C f F(p,q,s) sup 1 z2
z B
(2.8)
α+1(n+1+q)/ p g(z)
α+1(n+1+q)/ p g(z).
Therefore (2.6) implies that Tg : F(p, q,s) Ꮾα is bounded.
Conversely, suppose Tg : F(p, q,s) Ꮾα is bounded. For w B, let
1 w2
fw (z) = (n+1+q)/ p .
1 z,w
(2.9)
It is easy to see that
fw (w) = 1 w2
1
(n+1+q)/ p1 ,
w2
(n+1+q)/ p .
1 w2
fw (w) (2.10)
If w = 0 then fw 1 obviously belongs to F(p, q,s). From [19] we know that fw F(p, q,s), moreover there is a positive constant K such that supw B fw F(p,q,s) K. Therefore
α α 1 z2 fw (z)g(z) = 1 z2 Tg fw (z) Tg fw Ꮾα K Tg F(p,q,s)Ꮾα ,
(2.11)
for every z,w B.
From this and (2.10), we get
1 w2
α+1(n+1+q)/ p g(w) = 1 w 2 α fw (w)g(w) K Tg F(p,q,s)Ꮾα ,
(2.12)
from which (2.6) follows. From the above proof, we see that (2.7) holds.
6
Riemann-Stieltjes operators from F(p, q,s) to Ꮾα
Theorem 2.5. Let g be a holomorphic function on B, 0 < p, s < , n 1 < q < , q + s >
1, n + 1 + q pα, p < n + 1 + q. Then Lg : F(p, q,s) Ꮾα is bounded if and only if
sup 1 z2
z B
α(n+1+q)/ p g(z) < .
(2.13)
Moreover, the following relationship:
L g F(p,q,s)Ꮾα
sup 1 z2
α(n+1+q)/ p g(z)
(2.14)
z B
holds.
Proof. Assume that (2.13) holds. Let f (z) F(p, q,s)
sup 1 z2
z B
Ꮾ(n+1+q)/ p , then
(n+1+q)/ p f (z) < .
(2.15)
Therefore by Lemmas 2.1 and 2.2 we have
L g f Ꮾα
α = sup 1 z2 Lg f (z)
z B
α = sup 1 z2 f (z)g(z)
z B
sup 1 z2
z B
C f Ꮾ
(n+1+q)/ p f (z) sup 1 z2 α(n+1+q)/ p g(z)
z B
(n+1+q)/ p
sup 1 z2
(2.16)
α(n+1+q)/ p g(z)
z B
C f F(p,q,s) sup 1 z2
α(n+1+q)/ p g(z).
z B
Here we used the fact Lg f (0) = 0. It follows that Lg is bounded.
Conversely, if Lg : F(p, q,s) Ꮾα is bounded. Let β(z,w) denote the Bergman metric
between two points z and w in B. It is well known that
1 + ϕz (w)
1
.
β(z,w) = log
2
1 ϕz (w)
(2.17)
For a B and r > 0 the set
D(a,r) = z B : β(a,z) < r ,
a B,
(2.18)
is a Bergman metric ball at a with radius r. It is well known that (see [21])
1 a2
n+1
1 a,z2(n+1)
1
n+1
1 z2
1
n+1
1 a2
1
D(a,r)
(2.19)
Songxiao Li
7
when z D(a,r). For w B, let fw (z) be defined by (2.9), then by (2.10) and (2.19) we
have
2(n+1+q)/ p g(w)2 w 4
1 w2
fw (w)g(w)2 C
n+1
1 w2
C
D(w,r)
C
n+1
1 w2
D(w,r)
D(w,r)
fw (z)2 g(z)2 dv(z)
fw (z)2 g(z)2 1 z2 2α 1
2α dv(z)
1 z2
(2.20)
dv(z)
2 2α fw (z)2 g(z)2
2α+n+1 sup 1 z
2
1 z z D(w,r)
2
C
2α Lg fw Ꮾα ,
2
1 w
that is,
1 w2
2α2(n+1+q)/ p g(w)2 w 4 C Lg fw 2 α
Ꮾ
CK 2 Lg 2F(p,q,s)Ꮾ .
α
(2.21)
Taking supremum in the last inequality over the set 1/2 w < 1 and noticing that by
the maximum modulus principle there is a positive constant C independent of g H(B)
such that
sup 1 w2
w1/2
α(q+n+1)/ p g(w) C
sup w4 1 w2
1/2w<1
α(q+n+1)/ p g(w).
(2.22)
Therefore
sup 1 w2
α(q+n+1)/ p g(w) < C Lg z B
F(p,q,s)Ꮾα ,
(2.23)
the result follows.
Remark 2.6. Note that if α < (q + n + 1)/ p in Theorem 2.5, then the condition (2.13) is
equivalent to g 0.
Corollary 2.7. Let g be a holomorphic function on B, α > 0. Then the operator Tg : A2 Ꮾα is bounded if and only if
sup 1 z2
z B
α(n+1)/2 g(z) < .
(2.24)
α(n+1)/21 g(z) < .
(2.25)
Lg : A2 Ꮾα is bounded if and only if
sup 1 z2
z B
8
Riemann-Stieltjes operators from F(p, q,s) to Ꮾα
Tg : H 2 Ꮾα is bounded if and only if
sup 1 z2
z B
αn/2 g(z) < .
(2.26)
αn/21 g(z) < .
(2.27)
Lg : H 2 Ꮾα is bounded if and only if
sup 1 z2
z B
2.2. Case p > n + 1 + q
Theorem 2.8. Let g be a holomorphic function on B, 0 < p, s < , n 1 < q < , q + s >
1, α 0, n + 1 + q pα, p > n + 1 + q. Then Tg : F(p, q,s) Ꮾα is bounded if and only if
g Ꮾα . Moreover, the following relationship:
Tg F(p,q,s)Ꮾα
sup 1 z2 α g(z)
(2.28)
z B
holds.
Proof. Since f
F(p, q,s)
Tg f Ꮾ(n+1+q)/ p , by Lemmas 2.1, 2.2, and 2.3,
Ꮾα
α = sup 1 z2 f (z)g(z)
z B
(2.29)
C f F(p,q,s) sup 1 z2 α g(z).
z B
Therefore g Ꮾα implies that Tg : F(p, q,s) Ꮾα is bounded.
Conversely, suppose Tg : F(p, q,s) Ꮾα is bounded. For w B, let
1 w2
(p+n+1+q)/ p
fw (z) = 2(n+1+q)/ p
1 z,w
1 w2
(n+1+q)/ p + 1.
1 z,w
(2.30)
Then it is easy to see that
fw (w) = 1,
C 1 w2
fw (z) ,
1 z,w (n+1+q+p)/ p
w2
(n+1+q)/ p .
1 w2
fw (w) (2.31)
Songxiao Li
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By [19], we know that fw F(p, q,s), moreover there exists a constant L such that
supz B fw F(p,q,s) L. Hence
g(w)2 = fw (w)g(w)2 C
n+1
1 w2
C
D(w,r)
C
n+1
1 w2
D(w,r)
D(w,r)
fw (z)2 g(z)2 dv(z)
fw (z)2 g(z)2 1 z2 2α 1
2α dv(z)
1 z2
2 2
dv(z)
2 2α fw (z) g(z)
2α+n+1 sup 1 z
1 z2
z D(w,r)
2
C
2α Tg fw Ꮾα ,
1 w2
(2.32)
that is,
α 1 w2 g(w) C Tg fw Ꮾα CLTg F(p,q,s)Ꮾα
for every w B. The result follows.
(2.33)
Theorem 2.9. Let g be a holomorphic function on B, 0 < p, s < , n 1 < q < , q + s >
1, α 0, n + 1 + q pα, p > n + 1 + q. Then Lg : F(p, q,s) Ꮾα is bounded if and only if
sup 1 z2
z B
α(n+1+q)/ p g(z) < .
(2.34)
Moreover, the following relationship:
L g F(p,q,s)Ꮾα
sup 1 z2
α(n+1+q)/ p g(z)
(2.35)
z B
holds.
Proof. Suppose (2.34) holds. Let f (z) F(p, q,s)
sup 1 z2
z B
Hence
L g f Ꮾα
Ꮾ(n+1+q)/ p , then
(n+1+q)/ p f (z) < .
(2.36)
α = sup 1 z2 f (z)g(z)
z B
sup 1 z2
z B
(n+1+q)/ p f (z) sup 1 z2 α(n+1+q)/ p g(z)
z B
C f Ꮾ(n+1+q)/ p sup 1 z2 α(n+1+q)/ p g(z)
z B
C f F(p,q,s) sup 1 z2
z B
It follows that Lg is bounded.
α(n+1+q)/ p g(z).
(2.37)
Riemann-Stieltjes operators from F(p, q,s) to Ꮾα
10
Conversely, if Lg : F(p, q,s) Ꮾα is bounded, for w B, let fw (z) be defined by (2.30).
Then by (2.31),
2(n+1+q)/ p g(w)2 w 4
1 w2
fw (w)g(w)2 C
n+1
1 w2
C
D(w,r)
C
n+1
1 w2
D(w,r)
D(w,r)
fw (z)2 g(z)2 dv(z)
fw (z)2 g(z)2 1 z2 2α 1
2α dv(z)
1 z2
(2.38)
2 2
dv(z)
2 2α fw (z) g(z)
2α+n+1 sup 1 z
1 z2
z D(w,r)
2
C
2α Lg fw Ꮾα ,
1 w2
that is,
1 w2
2α2(n+1+q)/ p g(w)2 w 4 C Lg fw 2 α
Ꮾ
CLLg 2F(p,q,s)Ꮾ .
α
(2.39)
Similarly to the proof of Theorem 2.5, we get the desired result.
2.3. Case p = n + 1 + q
Theorem 2.10. Let g be a holomorphic function on B, 0 < p, s < , n 1 < q < , q + s >
1, s > n, α 1, p = n + 1 + q. Then Tg : F(p, q,s) Ꮾα is bounded if and only if
sup 1 z2
α
log
z B
1
1 z
2
g(z) < .
(2.40)
Moreover the following relationship:
Tg F(p,q,s)Ꮾα
sup 1 z2
α
log
z B
1
1 z
2
g(z)
(2.41)
holds.
Proof. Since f
F(p, q,s)
Tg f Ꮾα
Ꮾ, by Lemmas 2.1, 2.2, and 2.3,
α = sup 1 z2 Tg f (z)
z B
α = sup 1 z2 f (z)g(z)
(2.42)
z B
C f F(p,q,s) sup 1 z2
z B
α
log
1
1 z 2
g(z).
Therefore (2.40) implies that Tg is a bounded operator from F(p, q,s) to Ꮾα .
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11
Conversely, suppose Tg is a bounded operator from F(p, q,s) to Ꮾα . For w B, let
fw (z) = log
1
.
1 z,w
(2.43)
Then by [19] we see that fw F(p, q,s) and
fw (w) = log
w2 .
1 w2
fw (w) 1
,
1 w2
Moreover there is a positive constant M such that supw
log
1
1 w2
2
C
n+1
1 w2
D(w,r)
f F(p,q,s) M. Hence
g(w)2
2
= fw (w)g(w) C
B
(2.44)
C
n+1
1 w2
D(w,r)
D(w,r)
fw (z)2 g(z)2 dv(z)
fw (z)2 g(z)2 1 z2 2α 1
2α dv(z)
1 z2
(2.45)
2 2
dv(z)
2 2α fw (z) g(z)
2α+n+1 sup 1 z
2
1 z
z D(w,r)
2
C
2α Tg fw Ꮾα ,
2
1 w
that is,
1 w2
α
log
1
g(w) C Tg fw α CM Tg Ꮾ
F(p,q,s)Ꮾα .
1 w2
(2.46)
The result follows.
Theorem 2.11. Let g be a holomorphic function on B, 0 < p, s < , n 1 < q < , q + s >
1, s > n, α 1, p = n + 1 + q. Then Lg : F(p, q,s) Ꮾα is bounded if and only if
sup 1 z2
z B
α1 g(z) < .
(2.47)
Moreover the following relationship:
L g holds.
F(p,q,s)Ꮾα
sup 1 z2
z B
α1 g(z)
(2.48)
Riemann-Stieltjes operators from F(p, q,s) to Ꮾα
12
Proof. Suppose (2.47) holds. Let f (z) F(p, q,s)
. By Lemmas 2.1 and 2.2 we have
L g f Ꮾα
Ꮾ, then supz B (1 z2 ) f (z) <
α = sup 1 z2 f (z)g(z)
z B
C f Ꮾ sup 1 z2
z B
g(z)
(2.49)
C f F(p,q,s) sup 1 z2 α1 g(z).
z B
It follows that Lg is bounded.
Conversely, if Lg : F(p, q,s) Ꮾα is bounded. For w B, let fw (z) be defined by (2.43),
then by (2.44) we have
2 g(w)2 w 4
1 w2
fw (w)g(w)2 C
n+1
1 w2
C
D(w,r)
C
n+1
1 w2
D(w,r)
D(w,r)
fw (z)2 g(z)2 dv(z)
fw (z)2 g(z)2 1 z2 2α 1
2α dv(z)
1 z2
(2.50)
2 2
dv(z)
2 2α fw (z) g(z)
2α+n+1 sup 1 z
1 z2
z D(w,r)
2
C
2α Lg fw Ꮾα ,
1 w2
that is,
1 w2
2α2 g(w)2 w 4 C Lg fw 2 α .
Ꮾ
(2.51)
Similarly to the proof of Theorem 2.5, we get the desired result.
Similarly to the proof of Theorems 2.10 and 2.11, we can obtain the following results.
We omit the details.
Corollary 2.12. Let g be a holomorphic function on B, 0 < p < , and α 1. Then Tg :
Q p Ꮾα is bounded if and only if
sup 1 z2
α
log
z B
1 g(z) < .
1 z2
(2.52)
α1 g(z) < .
(2.53)
Lg : Q p Ꮾα is bounded if and only if
sup 1 z2
z B
Corollary 2.13. Let g be a holomorphic function on B. Then Lg : Ꮾ Ꮾ is bounded if and
only if g H .
Especially, we have the following known result (see [6, 15, 17]).
Songxiao Li
13
Corollary 2.14. Let g be a holomorphic function on B. Then Tg : Ꮾ Ꮾ is bounded if
and only if
sup 1 z2 log
z B
1 g(z) < .
1 z2
(2.54)
Acknowledgment
This author is supported partly by the NNSF of China (no. 10371051, 10371069).
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Songxiao Li: Department of Mathematics, JiaYing University, 514015, Meizhou,
GuangDong, China
Current address: Department of Mathematics, Shantou University, 515063, Shantou,
GuangDong, China
E-mail addresses: lsx@mail.zjxu.edu.cn; jyulsx@163.com