cn Department of Mathematics, Tianjin University, Tianjin 300072, People ’s Republic of China Abstract This paper characterizes the boundedness and compactness of the weighted differenti
Trang 1R E S E A R C H Open Access
Weighted differentiation composition operators from weighted bergman space to nth weighted space on the unit disk
Liang Zhang and Hong-Gang Zeng*
* Correspondence: zhgng@tju.edu.
cn
Department of Mathematics,
Tianjin University, Tianjin 300072,
People ’s Republic of China
Abstract This paper characterizes the boundedness and compactness of the weighted differentiation composition operator from weighted Bergman space to nth weighted space on the unit disk of≤
2000 Mathematics Subject Classification: Primary: 47B38; Secondary: 32A37, 32H02, 47G10, 47B33
Keywords: weighted differentiation composition operators, weighted Bergman space, nth weighted space, boundedness, compactness
1 Introduction LetDbe the open unit disk in the complex space≤, dA the Lebesegue measure onD normalized so that A(D) = 1 LetH(D)be the space of all analytic functions onD Let a >-1, p >0 f is said to belong to the weighted Bergman space, denoted by
A p α (= A p α D)), if f ∈ H(D)and
f p
= (α + 1)
D
| f (z) | p
(1− | z |2
) dA < ∞.
When 0 < p <1, it is complete metric space; when p≥1, it is a Banach space
weighted space on the unit disk, denoted byω (n)
μ (D), consists of all f ∈ H(D)such that
b (n)
μ (D)
f
= sup
z∈Dμ (z) | f (n) (z) | < ∞.
For n = 0, the space becomes the weighted-type spaceH∞μ(D); for n = 1, the Bloch-type spaceB μ(D); and for n = 2, the Zygmud-typeZμ(D) For more details about these spaces, we recommend the readers to ([1,2])
The expressionb (n)
μ (D)(f )defines a semi-norm on the nth weighted space ω (n)
μ (D), while the natural norm is given by
f ω (n)
μ (D)=
n−1
j=0
f ( j ) (0) + b
ω (n)
μ(D)
f
© 2011 Zhang and Zeng; licensee Springer This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in
Trang 2With this norm, ω (n)
μ (D)becomes a Banach space The little nth weighted space, denoted byω (n) μ,0(D), is a closed subspace ofω (n)
μ (D), consisting of those f for which
lim
|z|→1 μ (z) | f (n) (z) | = 0.
Let be a non-constant analytic self-map ofD, u(z) ∈ H(D), and m Î N The weighted differentiation composition operatorD m
ϕ,uis defined by
D m ϕ,u f (z) = u (z) f (m) (ϕ (z)) ,
forz∈D, f ∈ H(D) If m = 1, u (z) = ’ (z), thenD m
ϕ,u = DC ϕ; if let m = 1, u (z) = 1, then D m ϕ,u = C ϕ D
Recently, there have been some interests in studying some particular cases of opera-tors, such as DC , C DandD m
ϕ,u, between different function spaces From those
stu-dies, they gave some sufficient and necessary conditions for these operators to be
bounded and compact Concerning these results, we also recommend the interested
readers to ([3-9])
In this paper, we characterize the boundedness and compactness of the operatorD m
ϕ,u
from A p
αto nth weighted space For the case ofD m ϕ,u : A p α → ω (n)
μ , we have the following
results:
Theorem 1 Assume that p >0, a >-1, n, m Î N, μ is a weight onD, is a non-constant analytic self-map ofD, and u ∈ H(D) Then,
(1a)D m ϕ,u : A p α → ω (n)
μ is bounded if and only if for each kÎ {0, 1, , n}
sup
z∈D
μ(z)n l=k C l
n u (n−l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
(1b) D m ϕ,u : A p α → ω (n)
μ is compact if and only ifD m ϕ,u : A p α → ω (n)
μ is bounded and for
each kÎ {0, 1, , n}
lim
|ϕ(z)|→1
μ(z)n l=k C l n u (n−l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
1−ϕ(z)2 k+m+ α+2
p
= 0
(2)
For the case ofD m ϕ,u : A p α → ω (n)
μ,0, our main results are the following:
Theorem 2 Assume that p > 0, a > -1, n, m Î N, μ is a weight onD, is a non-constant analytic self-map ofD, andu ∈ H(D) Then,
(2a) D m ϕ,u : A p α → ω (n)
μ,0is bounded if and only if D m ϕ,u : A p α → ω (n)
each k Î {0, 1, , n},
lim
|z|→1 μ(z)
n
l=k
C l n u (n−l) (z) B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
(2b) D m ϕ,u : A p α → ω (n)
μ,0is compact if and only if D m ϕ,u : A p α → ω (n)
μ,0is bounded and for
each k Î {0, 1, , n},
Trang 3|z|→1
μ(z)n l=k C l n u (n−l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
1−ϕ(z)2 k+m+ α+2p = 0
(4)
The organization of this paper is as follows: we give some lemmas in Section 2, and then prove Theorem 1 in Section 3 and Theorem 2 in Section 4, respectively
Throughout this paper, we will use the symbol C to denote a finite positive number, and it may differ from one occurrence to the other
2 Some Lemmas
Lemma 1 is a direct consequence of the well-known estimate in ([10], Proposition
1.4.10) Hence, we omit its proof
Lemma 1 Assume that p > 0, a > - 1, n Î N, n > 0, andw∈D Then the function
g w,n (z) =
1− |w|2n
(1 − wz) n+2+p α
belongs to A p
α Moreover,supw∈D g w,n A p α < ∞.
The next lemma comes from ([11])
Lemma 2 Assume that p > 0, a > -1, n Î N, and z∈D Then, there is a positive constant C independent of f such that
f (n) (z) ≤C f A
p α
1− |z|2n+2+α
p
Lemma 3 Let p > 0, a > -1, m Î N,a = m + 1 + α+2 p and
D n+1=
· · ·
n−1
j=0
a + j n−1
j=0
a + j + 1
· · · n−1
j=0
a + j + n
.
Then, D n+1= n
j=1 j! Proof With a = m + 1 + α+2 p and replacing n by n + 1 in ([12], Lemma 2.3), the lemma easily follows □
The next lemma can be found in ([7], Lemma 4)
Lemma 4 Assume n Î N, g,u ∈ H(D)and is an analytic self-map ofD Then,
u(z)g( ϕ(z))(n)
k=0 g (k)
ϕ(z) n l=k C l
n u (n−l) (z) B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
,
where
B l,k(ϕ(z), ϕ(z), , ϕ (l−k+1)) =
k1 , ,k l
l!
k1! k l!
l
j=1
ϕ (j) (z) j!
k j
and the sum in (5) is overall non-negative integers k1, , klsatisfying k1+ k2+ + k1=
k and k1+ 2k2+ + lkl= l
By a proof in a standard way ([1], Proposition 3.11), we can get the next lemma
Trang 4Lemma 5 Supposeu ∈ H(D), p > 0,a > - 1, n, m Î N and is a non-constant analytic
ϕ,u : A p α → ω (n)
D m
ϕ,u : A p α → ω (n)
μis bounded and for any bounded sequence{fk}k ÎNin A p
αwhich converges
to zero uniformly on compact subsets ofDas k® ∞, we haveD m
ϕ,u f k→ 0inω (n)
μ as k® ∞
Lemma 6 Suppose n Î N and μ is a radial weight such that lim|z|®1μ (z) = 0 A closed set K in ω (n)
μ,0is compact if and only if it is bounded and satisfies
lim
|z|→1 supf ∈K μ (z)f (n) (z) = 0.
Proof The proof of this Lemma is followed by standard arguments similar to those outlined in ([13]) We omit the details □
3 The Proof of Theorem 1
(1a) Boundedness ofD m ϕ,u
We will prove the sufficiency first Suppose that the conditions in (1) hold Then, for any f ∈ A p
α, from Lemma 2 and Lemma 4, we obtain
μ(z)
D m ϕ,u f(n)
(z)
=μ(z)
n
k=0
f (k+m)
ϕ(z)
n
l=k
C l n u (n −l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
≤ μ(z)
n
k=0
f (k+m)
ϕ(z)
n
l=k
C l n u (n −l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
≤ C f A p
α
n
k=0
μ(z)
l=k n C l n u (n −l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
1−ϕ (z)2 k+m+ α+2p
(6)
For z = 0 and every dÎ {0, 1, , n - 1},
D m ϕ,u f(d)
(0)
=
d
k=0
f (k+m)(ϕ(0))
d
l=k
C l d u (d −l) (0)B l,k
ϕ(0),ϕ(0), ϕ (l −k+1)(0)
≤ C f A p
α
d
k=0
d
l=k
C l
d u (d−l) (0)B l,k
ϕ(0),ϕ(0), , ϕ (l −k+1)(0)
1− |ϕ (0)|2k+m+ α+2p
(7)
From (1), (6), and (7), we know that D m
ϕ,u : A p α → ω (n)
μ is bounded.
Conversely, suppose that D m ϕ,u : A p α → ω (n)
μ is bounded Then, there exists a constant
C such that D m
ϕ,u f ω (n)
μ ≤ C f A p
α, for all f ∈ A p
α.
For a fixedw∈D, and constants c1, c2, , cn + 1, set
g w (z) =
n+1
j=1
c j
j +2+p α
j +2+p α+ 1
.j +2+p α + m− 1
1− |w|2j
(1 − wz) j+2+p α
Trang 5Applying Lemma 1 and triangle inequality, it is easy to get that g w ∈ A p
αfor every
w∈D Moreover, we have that
sup
w∈D g w A
p
Now we show that for each sÎ {m, m + 1, , m + n}, there are constants c1, c2, , cn+1, such that,
g (s) w (w) = w
s
1− | w |2s+ α+2p , g w (t) (w) = 0, t ∈ {m, , m + n}\{s} (10)
Indeed, by differentiating function gw for each sÎ {m, m + 1, , m + n}, the system in (10) becomes
c1+ c2 +· · · + c n+1= 0
(m + 1 + α+2 p )c1+ (m + 2 + α+2 p )c2 +· · · + (m + n + 1 + α+2
p )c n+1= 0
(m + 1 + α+2 p ) (s + α+2
p )c1 +· · · + (m + n + 1 + α+2
p ) .n + s + α+2 p
c n+1= 1
(m + 1 + α+2 p ) (m + n + α+2
p )c1 +· · · + (m + n + 1 + α+2
p ) .m + 2n + α+2 p
c n+1= 0
(11)
By Lemma 3, the determinant of system (11) is different from zero, which implies the statement For each k Î {0, 1, 2, , n}, we choose the corresponding family of
func-tions that satisfy (10) with s = m + k and denote it by gw,k For each fixed k Î {0, 1, ,
n}, the boundedness of the operatorD m
ϕ,u : A p α → ω (n)
implies that for each (w) ≠ 0,
μ(w) ϕ (w) m+kn
l=k C l
n u (n−l) (w)B l,k
ϕ(w), ϕ(w), , ϕ (l−k+1) (w)
1− | ϕ(w) |2k+m+ α+2p
≤ C sup
w ∈D D
m ϕ,u (g ϕ(w),k) ω (n)
μ ≤ C D m ϕ,u A p
α →ω (n)
μ
(12)
From (12), it follows that for each kÎ {0, 1, , n},
sup
|ϕ(z)|>12
μ(z)n l=k C l
n u (n−l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
(1− | ϕ(z) |2)k+m+ α
+2
p
≤ C D m ϕ,u A p
α →ω (n)
μ
(13)
Now we use the test functions
h k (z) = z k+m, k = 0, 1, , n.
For each kÎ N, it is easy to get that
h k ∈ A p
α, h kA p
By applying Lemma 4 to the h0(z) = zm, we get
(D m ϕ,u h0)(n) (z)
= h (m)0 (ϕ(z))
n
l=0
C l n u (n −l) (z)B l,0
ϕ(z), ϕ(z), , ϕ (l+1) (z)
= m!
n
C l n u (n −l) (z) B l,0
ϕ(z), ϕ(z), , ϕ (l+1) (z)
,
Trang 6which along with boundedness of the operatorD m
ϕ,u : A p α → ω (n)
μ implies that
m! sup
z∈D μ(z)
n
l=0
C l n u (n −l) (z)B l,0
ϕ(z), ϕ(z), , ϕ (l+1) (z)
≤ C D m ϕ,u A p
α →ω (n)
μ (15) Assume now that we have proved the following inequalities
sup
z∈D μ(z)
n
l=j
C l n u (n −l) (z)B l,j
ϕ(z), ϕ(z), , ϕ (l −j+1) (z)
≤C D m ϕ,u
A p α →ω (n)
μ (16)
for jÎ {0, 1, , k - 1}, k ≤ n
Apply Lemma 4 to the hk(z) = zm+k, and knowing that z(s)≡ 0 for s >m + k and the boundedness of the operatorD m
ϕ,u : A p α → ω (n)
μ , we get
(D m ϕ,u h k)(n) (z)
≥
k−1
j=0
(m + k) (k − j + 1)(ϕ(z)) (k−j)
n
l=j
C l n u (n −l) (z)B l,j
ϕ(z), , ϕ (l −j+1) (z)
−
(m + k)!
n
l=k
C l n u (n −l) (z)B l,k
ϕ(z), , ϕ (l −k+1) (z)
(17)
Using hypothesis (16), we can know that
sup
z∈D μ(z)
n
l=k
C l n u (n −l) (z) B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
≤ C D m ϕ,u A p
α →ω (n)
μ (18)
for each k Î {0, 1, , n} Then, for each k Î {0, 1, , n},
sup
|ϕ(z)|≤12
μ(z)n l=k C l n u (n −l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
1− | ϕ(z) |2k+m+ α+2p
≤ C sup
z ∈D μ(z)
n
l=k
C l n u (n −l) (z)B l,k
ϕ(z), ϕ(z), ϕ (l −k+1) (z)
≤ C D m ϕ,u A p
α →ω (n) μ
(19)
From (13) and (19), we know that (1) holds
(1b) Compactness ofD m
ϕ,u.
ϕ,u : A p α → ω (n)
μ is bounded and (2) holds Then, by (1a), (1) holds Let
f i
i∈Nbe a sequence in A p α, such that,supi∈N f i A p α ≤ Mand ficonverges to 0 uniformly
on compact subsets ofDas i ® ∞ By the assumption, for any ε > 0, there is a δ Î (0,
1), such that, for each kÎ {0, 1, , n} and δ < | (z) | < 1,
μ(z)n l=k C l
n u (n −l) (z)B
l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
Trang 7
From Lemma 2 and (20), we have
sup
i∈N μ(z)
n
k=0
f i (k+m)
ϕ(z)
n
l=k
C l n u (n −l) (z)B l,k
ϕ(z), ϕ (l −k+1) (z)
≤ C sup
i∈N f i A
p α
n
k=0
sup
z∈D
μ(z)
n
l=k
C l
n u (n −l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
1− | ϕ(z) |2k+m+ α+2p
≤ CM (n + 1) ε.
(21)
If | (z) | ≤ r, then by Cauchy’s estimate and (19), we have
supμ(z)
n
k=0
f i (k+m)(ϕ(z))
n
l=k
C l n u (n −l) (z)B l,k
ϕ(z), ϕ (l −k+1) (z)
≤ C
n
k=0
sup
|ϕ(z)|≤r
f (m+k)
For j = 0, 1, , n - 1, we have
(D m
Applying (21), (22), and (23), we know that D m
ϕ,u f i ω (n)
μ → 0, (i → ∞) From Lemma
5,D m
ϕ,u : A p α → ω (n)
μ is compact.
Conversely, suppose that D m ϕ,u : A p α → ω (n)
μ is compact, then D m ϕ,u : A p α → ω (n)
μ is
bounded Let (zi)i ÎN be a sequence inD such that | (zi)| ® 1, i ® ∞ If such a
sequence does not exist, then the condition in (2) is easily satisfied Now, assume that
when || ||∞= 1 and (2) does not hold, then there is kÎ {0, 1, , n} and δ > 0 such that
μ(z i)n
l=k C l
n u (n−l) (z i )B l,k
ϕ(z
i),ϕ(z
i), ϕ (l−k+1) (z
i)
1− | ϕ(zi)|2k+m+ α+2
p
≥ δ.
Let g i (z) = g ϕ(z i ),k (z), i∈N, k Î 0, 1, , n be as in Theorem 1 Then,sup
i∈N g i A
p
α ≤ M
Lemma 5, we have that for kÎ {0, 1, , n}
lim
m ϕ,u g ϕ(z i ),k ω (n)
On the other hand, from (12), we obtain
D m ϕ,u g ϕ(z i ),k ω (n)
μ
≥ μ(z i)| ϕ(z i)|k+mn
l=k C l
n u (n−l) (z i )B l,k
ϕ(z i),ϕ(z i), ϕ (l −k+1) (z i)
1− | ϕ(z i)|2k+m+ α+2p
> δ
3
(25)
for large enough i From (24) and (25), this is a contradiction So, (2) holds
Now the proof of Theorem 1 is completed
Trang 84 The Proof of Theorem 2
(2a) Boundedness ofD m
ϕ,u.
First, suppose thatD m
ϕ,u : A p α → ω (n)
μ is bounded and (3) holds For each polynomial p
(z), we obtain |p(m+k)(z)|≤ Cp, zÎ D, Cpis s constant depending on p
And
μ(z)(D m
ϕ,u p) (n) (z)
=μ(z)
n
k=0
p (k+m)(ϕ(z))
n
l=k
C l n u (n−l) (z)B l,k
ϕ(z), , ϕ (l−k+1) (z)
≤ C p n
k=0
| μ(z) |
n
l=k
C l n u (n −l) (z)B l,k
ϕ(z), , ϕ (l −k+1) (z)
→ 0, |z| → 1.
(26)
From (26), we have that for each polynomial p(z), D m
ϕ,u p ∈ ω (n)
μ,0 Since the set of
poly-nomials is dense in A p α, we have that for each f ∈ A p
α, there is a sequence of
polyno-mials (p k)k∈N, such that f − p kA p
D m ϕ,u : A p α → ω (n)
μ, we have that
D m ϕ,u f − D m
ϕ,u p k ω (n)
μ ≤ D m ϕ,u A p
α →ω (n)
μ f − p k A p
α → 0, k → ∞.
Then, D m
ϕ,u f ∈ ω (n)
μ,0, from which the boundedness ofD m ϕ,u : A p α → ω (n)
μ,0follows.
ϕ,u : A p α → ω (n)
D m
ϕ,u : A p α → ω (n)
μ is bounded Then, taking the test functions hk (z) = zm+k for each k
Î {0, 1, , n}, we obtain D m ϕ,u z m+k ∈ ω (n)
μ,0 By the proof of Theorem 1, for each kÎ {0, 1, , n},
lim
|z|→1 μ (z)
n
l=k
C l n u (n−l) (z)B l,k
ϕ(z), ϕ(z), ϕ (l−k+1) (z)
= 0
(2a) is completed
(2b) Compactness ofD m
ϕ,u.
First, assume that D m
ϕ,u : A p α → ω (n)
μ,0is compact, so it is bounded and (3) holds.
Hence, if || ||∞< 1,
μ(z)n l=k C l
n u (n −l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
1−ϕ (z)2 k+m+ α + 2
p
≤ μ(z)n
l=k C l n u (n −l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
1− ϕ 2
∞
k+m+ α+2p
→ 0, (|z| → 1)
(27)
Trang 9If || ||∞= 1, sinceD m
ϕ,u : A p α → ω (n)
μ is compact too and (2) holds, then for allε > 0, there is an r Î (0, 1), such that when r < | (z) | < 1, for k Î {0, 1, , n},
μ(z)n l=k C l
n u (n −l) (z)B
l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
By (3), we know there is aδ Î (0, 1), such that δ < |z| < 1, for k Î {0, 1, , n},
μ(z)
n
l=k
C l n u (n−l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
< ε(1 − r2)k+m+ α
+2
p
(29)
Then, whenδ < |z| < 1 and r < | (z) | < 1 for k Î {0, 1, , n}, we get
μ(z)n l=k C l
n u (n −l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l −k+1) (z)
(1−ϕ(z)2
)k+m+ α
+2
p
In addition, when | (z) | ≤ r and δ < |z| < 1, we have
μ(z)n l=k C l
n u (n−l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
1− | ϕ(z) |2k+m+ α+2p
< μ (z)n
l=k C l
n u (n−l) (z)B l,k
ϕ(z), ϕ(z), , ϕ (l−k+1) (z)
(1− r2)k+m+ α
+2
p
< ε.
(31)
Combining (30) and (31), we know (4) holds
ϕ,u : A p α → ω (n)
μ,0is bounded and (4) holds Taking the
supre-mum in (6) for all f in the unit ball of A p
α, and using the condition (4), we have
lim|z|→1supfAp
α≤1 μ(z)(D m
ϕ,u f ) (n) (z) = 0, from which by Lemma 6, the compactness
ofD m
ϕ,u : A p α → ω (n)
μ,0follows.
Now the proof of Theorem 2 is finished
Acknowledgements
We are grateful to the referee(s) for many helpful comments on the manuscript Hong-Gang Zeng is supported in
part by the National Natural Science Foundation of China (Grand Nos 10971153, 10671141).
Authors ’ contributions
LZ found the question and drafted the manuscript HGZ joined in the discussion about the question and revised the
paper All authors read and approved the final manuscript
Competing interests
The authors declare that they have no competing interests.
Received: 29 January 2011 Accepted: 21 September 2011 Published: 21 September 2011
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doi:10.1186/1029-242X-2011-65 Cite this article as: Zhang and Zeng: Weighted differentiation composition operators from weighted bergman space to nth weighted space on the unit disk Journal of Inequalities and Applications 2011 2011:65.
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... as: Zhang and Zeng: Weighted differentiation composition operators from weighted bergman space to nth weighted space on the unit disk Journal of Inequalities and Applications 2011 2011:65....doi:10.1016/0022-247X(72)90081-9
12 Stevi ć, S: composition operators from the weighted Bergman space to the nth weighted- type space on the unit disk.
Hindawi Publ Corp...
7 Stevi ć, S: Weighted differentiation composition operators from H ∞ and Bloch spaces to nth weighted- type spaces on< /small>
the unit disk Appl Math Comput 216(12),