R E S E A R C H
Open Access
Characterizations of some function spaces
associated with Bloch type spaces on the unit
ball of
C
n
Jineng Dai
1and Baobin Wang
2**Correspondence: wbb1818@126.com 2School of Mathematics and Statistics, South Central University for Nationalities, Wuhan, 430074, China
Full list of author information is available at the end of the article
Abstract
Motivated by characterizations of Bloch type spaces, we define some related function spaces and characterize them in this paper. Our results reveal the reason in theory that some equivalent characterizations of the Bloch type spaceBαrequire extra conditions for
α
.MSC: 32A18
Keywords: Bloch type space; unit ball
1 Introduction
LetBnbe the unit ball of the complex Euclidean spaceCnand∂Bnbe the unit sphere. The
class of all holomorphic functions onBnis denoted byH(Bn). For ≤α<∞, letHα∞be
the space of holomorphic functionsf onBnsatisfyingsupz∈Bn( –|z|
)α|f(z)|<∞. When
α= , we writeH∞forH∞.
Forf∈H(Bn), its complex gradient is defined by
∇f(z) =
∂f
∂z
(z), . . . , ∂f ∂zn
(z)
.
As introduced by Timoney in [], the Bloch space on the unit ballBnis the space of all f∈H(Bn) such thatsupz∈BnQf(z) <∞, where
Qf(z) =sup
( –|z|)|∇f(z),w|
( –|z|)|w|+|z,w| : =w∈C
n
.
Let <α<∞. The Bloch type spaceBαconsists of functionsf∈H(B
n) satisfying
sup z∈Bn
–|z|α ∇f(z) <∞.
It is well known thatBα=H∞
α–forα> . Moreover, whenα> /, a functionf ∈Bαif and only if
sup z∈Bn
–|z|α–Qf(z) <∞. ()
But it is not true for the case <α≤/ in the setting of several complex variables. For example (see []), when <α< /,f ∈Bα if and only ifsup
z∈Bn( –|z|
)α–G
f(z) <∞,
where
Gf(z) =sup
( –|z|)|∇f(z),w|
( –|z|)α|w|+|z,w| : =w∈C
n
.
Thus an interesting question arises naturally: what is the property for a holomorphic func-tionf onBnsatisfying () when <α≤/. Below we define the space
Tα=
f ∈H(Bn) :sup z∈Bn
–|z|α–Qf(z) <∞
.
In Section of this paper, we prove thatf ∈T/ if and only if the directional derivatives off in the directions perpendicular to the radial direction are uniformly bounded. In par-ticular,Tαis a trivial space consisting of constants forα< /.
In Holland and Walsh [] gave another characterization for the Bloch space on the unit discD, namely,f belongs to the Bloch space if and only if
sup z,w∈D
z=w
–|z|/ –|w|/|f(z) –f(w)|
|z–w| <∞.
Ren and Tu [] extended the above form to the unit ballBn. Zhao [] generalized these
results as follows.
Theorem A Let <α≤.Letλbe any real number satisfying the following properties: () ≤λ≤αif <α< ;
() <λ< ifα= ;
() α– ≤λ≤if <α≤.
Then a holomorphic function f on Bnis in Bαif and only if
sup z,w∈Bn
z=w
–|z|λ –|w|α–λ|f(z) –f(w)|
|z–w| <∞. ()
Zhao gave some examples showing that the conditions onαandλin Theorem A cannot be improved.
Motivated by Theorem A, we denote the following function space:
Sα,λ=
f ∈H(Bn) : sup z,w∈Bn
z=w
–|z|λ –|w|α–λ|f(z) –f(w)| |z–w| <∞
,
whereαandλare real numbers. Our purpose is to characterize the spaceSα,λforαand
λwithout satisfying those conditions in Theorem A. Ifλ< orλ>α, by the maximum modulus principle, it is easy to see thatSα,λconsists of constants. So we always assume
≤λ≤α. In Section we showSα,λ⊂Bαwhenαandλdo not satisfy the conditions of
spaceBλ+orBα–λ+ in terms of different numbersαandλ. Our results reveal in theory instead of examples that the conditions onαandλin Theorem A cannot be improved.
Throughout this paper, constants are denoted byC, and they are positive finite quantities and not necessarily the same in each occurrence.
2 Characterizations ofTα
Theorem . The following statements are equivalent: (i) f∈T/;
(ii) There exists a constantC> such that
∇f(z),ζ ≤C ()
for allz∈Bnandζ ∈∂Bnwithz,ζ= ;
(iii) For all≤i,j≤n,
sup z∈Bn zi
∂f
∂zj
(z) –zj
∂f
∂zi
(z) <∞. ()
Proof (i)⇔(ii): Forz∈Bnandζ∈∂Bnwithz,ζ= , we have
–|z|–/Qf(z)≥
( –|z|)/|∇f(z),ζ|
( –|z|)|ζ|+|z,ζ| = ∇f(z),ζ . ()
This shows that (i)⇒(ii).
For the converse, if (ii) holds, we have ( –|z|)/|∇f(z),z| ≤Cfor allz∈B
n(see []).
When /≤ |z|< and =w∈Cn, by the projection theorem, there existsζ ∈∂Bnsuch
thatζ,z= and
w=wz+wζ,
wherew=w,z/|z|and|w|=|w||z|+|w|. Thus,
∇f(z),w ≤ w
∇f(z),z + w
∇f(z),ζ
≤C z,w –|z|–/+C|w|.
It follows that
( –|z|)/|∇f(z),w|
( –|z|)|w|+|z,w| ≤C
for all /≤ |z|< and =w∈Cn. On the other hand, for ≤ |z|< /, we have
( –|z|)/|∇f(z),w|
( –|z|)|w|+|z,w| ≤
( –|z|)/|∇f(z),w|
( –|z|)|w| = ∇f(z) ≤C.
(ii)⇔(iii): Without loss of generality, we only need to show that
sup z∈BN z
∂f
∂z (z) –z
∂f
∂z
(z) <∞.
Ifz=z= , there is nothing to prove. If|z|+|z|= , put
ζ =
z
|z|+|z|
, –z
|z|+|z|
, , . . . ,
.
Obviously,ζ∈∂Bnandz,ζ= . Therefore,
z
∂f
∂z (z) –z
∂f
∂z
(z) =|z|+|z| ∇f(z),ζ ≤C.
Conversely, suppose (iii) holds. When |z| ≤/, it is clear that () holds. For |z|=
|z|+· · ·+|zn|> /, there existszi (≤i≤n) such that|zi|> /(√n). We may
as-sume that|z|> /(
√
n). LetV={w∈Cn:z,w= }. ThenVis a subspace ofCnand a
basis ofV is{v, . . . ,vn–}, where
vi= (–zi+, , . . .
i–
,z, , . . . ,
n–(i+)
), i= , . . . ,n– . ()
Therefore, forζ= (ζ, . . . ,ζn)∈∂Bnwithz,ζ= , there exist scalarsk, . . . ,kn–such that ζ is expressed as a linear combination ofv, . . . ,vn–in only one way. That is,
ζ =kv+kv+· · ·+kn–vn–. ()
By () and (), we get
ζ= –kz–kz–· · ·–kn–zn
and
ζi+=kiz, i= , . . . ,n– . ()
Note that|ζ|= and|z|> /(
√
n). Hence, it follows from () that
|ki|=|
ζi+|
|z|
< √n, i= , . . . ,n– . ()
Thus we have
∇f(z),ζ=
n–
i=
ki
∇f(z),vi
= n– i= ki
–zi+ ∂f
∂z (z) +z
∂f
∂zi+ (z)
.
The desired result follows from () and (). This finishes the proof of Theorem ..
From Theorem . and the result of [], we see thatT/⊂B/. Meanwhile, it is evident thatBα⊂T
Example Let
f(z,z) = ( –z)/.
Note that|z|+|z|< forz= (z,z)∈B. Then
z
∂f
∂z (z) –z
∂f
∂z
(z) = |z| | –z|/
<( –|z| )/
( –|z|)/ <
√
.
On the other hand, when <α< / andz= (r, )→(, ) ( <r< ),
–|z|α ∇f(z) = ( –|z| )α
| –z|/
= ( –r )α
( –r)/ → ∞.
Thereforef∈T/by Theorem ., butf∈Bα( <α< /).
Example Let
g(z,z) =zlog( –z).
Then, forz= (z,z)∈B,
–|z|/ ∇g(z) ≤ –|z|/ z –z
+ log( –z)
≤ –|z|/
( –|z|)/ –|z|
+ log( –z)
≤C.
Meanwhile, letz= (r, )→(, ) ( <r< ), we have
z
∂g
∂z (z) –z
∂g
∂z
(z) = –rlog( –r)→ ∞.
Thusg∈T/butg∈B/.
Lemma . Let n> , ≤i,j≤n and i=j.Suppose that f ∈H(Bn),g∈H(Bn)and
lim
|z|→
zif(z) –zjg(z) = .
Then f ≡and g≡.
Proof We may assumei= ,j= without loss of generality. Leth(z) =zf(z) –zg(z). For each fixedζ = (ζ, . . . ,ζn)∈∂Bn, define the slice functionhζ(λ) =h(λζ) on the unit disk D={λ:|λ|< }. Then
lim
|λ|→
λζf(λζ, . . . ,λζn) –λζg(λζ, . . . ,λζn) = .
That is,
lim
|λ|→
Since the function ζf(λζ, . . . ,λζn) –ζg(λζ, . . . ,λζn) is holomorphic on the unit disk |λ|< , by the maximum modulus principle of one complex variable, it follows that
ζf(λζ, . . . ,λζn) –ζg(λζ, . . . ,λζn) =
for allλ∈D. Therefore, for anyζ∈∂Bnandλ∈D,
h(λζ) =λζf(λζ, . . . ,λζn) –λζg(λζ, . . . ,λζn) = .
When =z∈Bn, letλ=|z|andζ=z/|z|. Then we get
h(z) =h(λζ) =zf(z) –zg(z) = .
Hence, for allz∈Bn,
zf(z) =zg(z).
Sincef andg are holomorphic onBn, we can concludef(z) =g(z)≡. The proof is
fin-ished.
An immediate consequence of Lemma . is the following theorem.
Theorem . Let n> andα< /.If f ∈Tα,then f is constant.
Proof By () it follows that |∇f(z),ζ| ≤C( –|z|)/–α for z∈B
n andζ ∈∂Bn with z,ζ= . Sinceα< /, we get lim|z|→|∇f(z),ζ|= . Using a similar argument as in the proof of Theorem ., we havelim|z|→|zj∂∂zfi(z) –zi
∂f
∂zj(z)|= for all ≤i,j≤n. Thus
the desired result follows from Lemma ..
3 Characterizations ofSα,λ
In the section we will characterizeSα,λexplicitly for real numbersαandλin several cases.
For this, we need the following lemma which plays an important role in the proof of The-orem ..
Lemma . Letα> .Letλbe any real number satisfying the following properties: () <λ<α– if <α≤;
() <λ≤α/ifα> . Let
H(x,y) =x
λyα–λ
y–x
y
x dτ τλ+.
Then there exists a constant C > such that H(x,y)≤ C for any x and y satisfying
<x,y≤and x=y.
Proof Lett=x/y. Thent∈(, )∪(,∞), and
H(x,y) =H(ty,y) = t
λyα–
( –t)
y
Lets=τ/y, we have
H(x,y) =t
λyα–(λ+)
( –t)
t ds sλ+ =y
α–(λ+)G(t),
where
G(t) = t
λ
( –t)
t ds sλ+ =
–tλ
λ( –t).
It is evident thatG(t) is continuous on (, )∪(,∞), and
lim t→G(t) =
λ, limt→G(t) = .
For the case (), since <λ< , we get
lim
t→∞G(t) = .
Noticing thatyα–(λ+)≤ fory∈(, ], we conclude thatH(x,y) is bounded in this case. For the case (), we easily see that <λ<α– . Write
H(x,y) =H(x,x/t) = x
α–(λ+)
tα–(λ+)G(t).
It is clear that
lim t→∞
G(t)
tα–(λ+) =tlim→∞
tλ–α
λ .
The above limit is /λforλ=α/ and for <λ<α/. ThereforeH(x,y) is also bounded in the case () sincexα–(λ+)≤ forx∈(, ]. The proof is complete.
Theorem . Letα> .Letλbe any real number satisfying the following properties: () <λ<α– if <α≤;
() <λ≤α/ifα> . Then Sα,λ=Bλ+.
Proof Letf ∈Bλ+. For anyz,w∈B
n, since
f(z) –f(w) =
d[f(tz+ ( –t)w)]
dt dt
=
n
k=
(zk–wk)
∂f
∂zk
tz+ ( –t)wdt,
we get
f(z) –f(w) ≤C|z–w|
(∇f)tz+ ( –t)w dt
≤C|z–w|
dt
≤C|z–w|
dt
( –t|z|– ( –t)|w|)λ+
=C|z–w|
dt
[t( –|z|) + ( –t)( –|w|)]λ+.
If|z|=|w|, noting thatλ+ <α, we get
dt
[t( –|z|) + ( –t)( –|w|)]λ+ = ( –|z|)λ+ ≤
C
( –|z|)λ+
≤ C
( –|z|)λ( –|w|)α–λ.
If|z| =|w|, letτ=t( –|z|) + ( –t)( –|w|). By Lemma ., we have
dt
[t( –|z|) + ( –t)( –|w|)]λ+ =
( –|z|) – ( –|w|)
–|z|
–|w|
dt
τλ+
≤ C
( –|z|)λ( –|w|)α–λ.
Therefore,
f(z) –f(w) ≤ C|z–w| ( –|z|)λ( –|w|)α–λ,
which shows thatf∈Sα,λ.
Conversely, iff ∈Sα,λ, it follows that
sup z∈Bn
z=
–|z|λ|f(z) –f()| |z| <∞.
Thus we get
sup z∈Bn
–|z|λ f(z) <∞,
namely,f∈Hλ∞and sof∈Bλ+. This finishes the proof of the theorem.
Theorem . Letα> .Letλbe any real number satisfying the following properties: () <λ<αif <α≤;
() α/ <λ<αifα> . Then Sα,λ=Bα–λ+.
Proof If <λ<α for <α≤, then <α–λ<α– . Ifα/ <λ<α forα> , then < α–λ<α/. Applying Theorem ., we immediately conclude thatSα,λ=Bα–λ+.
For any pointw∈Bn–{}, we recall that the bi-holomorphic mappingϕwofBn, which
interchanges the points andw, is defined by
ϕw(z) =
w–Pw(z) –swQw(z)
wheresw=
–|w|,P
w(z) =|zw,w|wandQw(z) =z–Pw(z). Whenw= , letϕw(z) = –z. The
pseudo-hyperbolic distance betweenwandzis denoted byρ(w,z) =|ϕw(z)|.
Theorem . Letα≥.Then Sα,λ=H∞forλ= orλ=α.
Proof It suffices to prove forα≥ andλ= . Iff ∈Sα,, that is,
sup z,w∈Bn
z=w
–|w|α|f(z) –f(w)|
|z–w| <∞. ()
Then
sup z∈Bn
z=
|f(z) –f()|
|z| <∞,
which implies thatf ∈H∞.
Conversely, assumef∈H∞. Then we have (see Lemma in [])
f(z) –f(w) ≤Cρ(z,w) ()
for allz,w∈Bn. On the other hand,
w–Pw(z) –swQw(z) =|z–w|+ z,w –|z||w|≤ |z–w|,
and so
ρ(w,z)≤ |z–w|
| –z,w|≤ |z–w|
–|w|. ()
By () and (), we get () sinceα≥. Thusf∈Sα,. The proof is complete.
Remark . We easily see thatλ+ <αin Theorem ., andλ> in Theorem .. Com-bining with Theorem ., we conclude thatSα,λ⊂Bα and the inclusion is strict for real
numbersαandλwhich do not satisfy the conditions of Theorem A.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
BW proposed the problem. JD and BW together finished the proof. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Wuhan University of Technology, Wuhan, 430070, China.2School of Mathematics and Statistics, South Central University for Nationalities, Wuhan, 430074, China.
Acknowledgements
JD was supported by the Natural Science Foundation of China under 11301404 and the Fundamental Research Funds for the Central Universities (WUT: 2015IVA069). BW was supported by the Natural Science Foundation of China under 11201348, the Fund of China Scholarship Council and the special Fund of Basic Scientific Research of Central Colleges of South Central University for Nationalities under CZQ13015.
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