U N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K A L U B L I N – P O L O N I A
VOL. LXXII, NO. 2, 2018 SECTIO A 57–70
PAWEŁ SOBOLEWSKI
Products of Toeplitz and Hankel operators on the Bergman space in the polydisk
Abstract. In this paper we obtain a condition for analytic square integrable functions f, g which guarantees the boundedness of products of the Toeplitz operators TfTg¯ densely defined on the Bergman space in the polydisk. An analogous condition for the products of the Hankel operators HfHg∗ is also given.
1. Introduction. Let D be the open unit disk in the complex plane C. For a fixed positive integer n ≥ 2, the unit polydisk D
nis the Cartesian product of n copies of D. By dA we will denote the Lebesgue volume measure on D
n, normalized so that A(D
n) = 1.
The Bergman space A
2= A
2(D
n) is the space of all analytic functions on D
nsuch that
kf k
2= Z
Dn
|f (z)|
2dA(z) < ∞.
For w = (w
1, w
2, . . . , w
n) ∈ D
nthe reproducing kernel in A
2is the function K
wgiven by
K
w(z) =
n
Y
j=1
1
(1 − ¯ w
jz
j)
2, z ∈ D
n.
If h·, ·i is the inner product in L
2(D
n), then for every function f ∈ A
2we have
hf, K
wi = f (w), w ∈ D
n.
2010 Mathematics Subject Classification. 30H05, 32A36.
Key words and phrases. Toeplitz operator, Bergman space.
In the special case when f = K
w, we obtain kK
wk
2= hK
w, K
wi = K
w(w) =
n
Y
j=1
1
(1 − |w
j|
2)
2, w ∈ D
n. So, the normalized reproducing kernel for A
2is
k
w(z) =
n
Y
j=1
1 − |w
j|
2(1 − ¯ w
jz
j)
2, z ∈ D
n.
Now we quote the definition of the Toeplitz operator. The orthogonal projection P from L
2(D
n) onto A
2is defined by
P (f )(w) = hf, K
wi = Z
Dn
f (z)
n
Y
j=1
1
(1 − ¯ z
jw)
2dA(z), f ∈ L
2(D
n), w ∈ D
n. For a function f ∈ L
∞and h ∈ A
2the Toeplitz operator T
fis given by
T
fh(w) = P (f h)(w), w ∈ D
n.
Similarly, the Hankel operator H
facting on A
2is defined as H
fh = f h − P (f h), h ∈ A
2,
and P is the projection mentioned above. It is clear that H
fh ∈ A
2⊥. Both operators T
fand H
fcan be defined when the symbol f belongs to the space L
2(D
n). In that case the Toeplitz and Hankel operators are densely defined on the Bergman space A
2, that is on H
∞.
Let w
i, i = 1, 2, . . . , n, belong to the unit disk D. For each w
iwe define an automorphism ϕ
wiof D by
ϕ
wi(z
i) = w
i− z
i1 − ¯ w
iz
i, z
i∈ D, i = 1, 2, . . . , n.
Then the map
ϕ
w(z) = (ϕ
w1(z
1), ϕ
w2(z
2), . . . , ϕ
wn(z
n)), z, w ∈ D
nis an automorphism of the polydisk D
n, in fact, ϕ
−1w= ϕ
w. The real Jaco- bian of ϕ
wis equal to
|k
w|
2=
n
Y
j=1
(1 − |w
j|
2)
2|1 − ¯ w
jz
j|
4, thus we have change-of-variable formula
Z
Dn
(h ◦ ϕ
w)(z)dA(z) = Z
Dn
h(z)|k
w(z)|
2dA(z),
whenever such integrals make sense.
2. Problem and results. As we mentioned, the Toeplitz operator may be considered when the index f belongs to the space L
2(D
n). If f ∈ A
2, then by the definition of the Toeplitz operator, we have
T
f¯h(w) = P ( ¯ f h)(w) = Z
Dn
f (z)h(z)
n
Y
j=1
1
(1 − ¯ z
jw)
2dA(z), w ∈ D
n. The main problem in this note is what conditions must be satisfied by functions f, g ∈ A
2to guarantee that the product of the Toeplitz operators T
fT
g¯is bounded on the Bergman space A
2in the polydisk D
n. We provide a sufficient condition for boundedness of such products. Similarly, we give a sufficient condition to ensure that the product of the Hankel operators H
fH
g∗is bounded on the space (A
2)
⊥, where H
∗is the adjoint of H.
For u ∈ L
2(D
n) we denote
˜
u(w) = B[u](w) = Z
Dn
(u ◦ ϕ
w)(z)dA(z), w ∈ D
n.
In [9] Stroethoff and Zheng established the following necessary condition for boundedness of the products T
fT
¯gon the unit disk D.
Theorem 1. Let f and g be in A
2. If T
fT
¯gis bounded, then sup
w∈D
|f | g
2(w)g |g|
2(w) < ∞.
In the same paper the authors also gave a little stronger sufficient condi- tion.
Theorem 2. Let f and g be in A
2. If there is a positive constant ε such that
sup
w∈D
|f |
^2+ε(w) |g|
^2+ε(w) < ∞, then T
fT
¯gis bounded.
There is a conjecture that the necessary condition is also a sufficient condition for boundedness. But in view of a counter-example of Nazarov [6]
for Toeplitz products on the Hardy space, it may not be possible to prove that this necessary condition is also sufficient.
Stroethoff and Zheng [12] showed the analogous results on the Bergman spaces of the polydisk [11], weighted Bergman space of the unit disk [13]
and the unit ball [12]. Next, Miao in [4] gave an interesting way to transfer Theorem 1 and Theorem 2 to the space A
pα, 1 < p < ∞, α > −1, of the unit ball. Recently, Michalska and Sobolewski [5] improved a sufficient condition on boundedness of T
fT
¯gon A
pα.
A similar problem concerns the products of the Hankel operators H
fH
g∗.
Such operators are densely defined on space (A
2)
⊥. The following condition
for the Hankel products on the unit disk was established by Stroethoff and
Zheng in [9].
Theorem 3. Let f and g be in L
2(D, dA). If H
fH
g∗is bounded on (A
2)
⊥, then
sup
w∈D
kf ◦ ϕ
w− P (f ◦ ϕ
w)k
L2kg ◦ ϕ
w− P (g ◦ ϕ
w)k
L2< ∞.
The same authors showed that this necessary condition is, like for T
fT
¯g, very close to being sufficient.
Theorem 4. Let f and g be in L
2(D, dA). If there is a positive constant ε such that
sup
w∈D
kf ◦ ϕ
w− P (f ◦ ϕ
w)k
L2+εkg ◦ ϕ
w− P (g ◦ ϕ
w)k
L2+ε< ∞, then the product H
fH
g∗is bounded on (A
2)
⊥.
Their theorems were extended to the weighted Bergman spaces of the unit ball by Lu and Liu [2] and for the Bergman space of the polydisk by Lu and Shang [3].
In this paper we provide a sufficient condition for the boundedness of the operators T
fT
¯gand H
fH
g∗.
For u ∈ L
1, ε > 0 and w ∈ D
nwe define B
ε[u](w) =
Z
Dn
(u ◦ ϕ
w)(z)
n
Y
i=1
log
1+ε1
1 − |z
i| dA(z),
where ϕ
wis the automorphism of D
nand z = (z
1, z
2, . . . , z
n). The following theorems are the main results in this paper.
Theorem 5. Let f, g ∈ A
2. If there is a positive constant ε > 0 such that sup
w∈Dn
B
ε[|f |
2](w)B
ε[|g|
2](w) < ∞, then the operator T
fT
g¯is bounded on A
2.
Theorem 6. Let f, g ∈ L
2(D
n). If there is a positive constant ε > 0 such that
sup
w∈Dn
(f ◦ ϕ
w− P (f ◦ ϕ
w))
n
Y
j=1
log
(1+ε)/21 1 − |z
j|
L2
×
(g ◦ ϕ
w− P (g ◦ ϕ
w))
n
Y
j=1
log
(1+ε)/21 1 − |z
j|
L2
< ∞,
then the operator H
fH
g∗is bounded on (A
2)
⊥.
After sending this paper for publication we found that Theorem 5 is
contained in a result obtained in [1].
3. Proofs. A very important role in our considerations is played by the for- mula for the inner product in A
2introduced in [11]. Let α = {α
1, α
2, . . . , α
m} be a nonempty subset of {1, 2, . . . , n} with α
1< α
2< . . . < α
m. We define the measure on D
nby
dµ
α(z) = 3
n−m6
m(1 − |z
1|
2)
2(1 − |z
2|
2)
2. . . (1 − |z
n|
2)
2× Y
j∈α
(5 − 2|z
j|)
2dA(z
1)dA(z
2) . . . dA(z
n)
and
dµ
∅(z) = 3
n(1 − |z
1|
2)
2(1 − |z
2|
2)
2. . . (1 − |z
n|
2)
2dA(z
1)dA(z
2) . . . dA(z
n), where m is the cardinality of α. Let us set D
jh = ∂h/∂z
jand
D
αh = D
α1D
α2. . . D
αmh, D
∅h = h.
For f, g ∈ A
2we have
(1)
Z
Dn
f (z)g(z)dA(z) = X
α
Z
Dn
D
αf (z)D
αg(z)dµ
α(z),
where α runs over all subsets of {1, 2, . . . , n}.
We start with some lemmas which we will apply to prove the main theo- rems.
Lemma 1. Let f ∈ A
2, h ∈ H
∞and ε > 0. If α = {α
1, α
2, . . . , α
m} is a subset of {1, 2, . . . , n}, then
|D
αT
fα¯h(w)| ≤ C
n
Y
i=1
1
(1 − |w
i|
2) B
ε[|f |
2](w)
12× Z
Dn
|h(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)
!
12for all w ∈ D
n.
Proof. First we show the inequality for α = ∅.
|T
f¯h(w)| ≤ 2
nZ
Dn
|f (z)|
n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
1+ε21 1 − |ϕ
wi(z
i)|
× |h(z)|
n
Y
i=1
1
|1 − w
iz
i|
n
Y
i=1
log
−1+ε21
1 − |ϕ
wi(z
i)| dA(z)
≤ C Z
Dn n
Y
i=1
1
(1 − |w
i|
2)
2|f (z)|
2n
Y
i=1
(1 − |w
i|
2)
2|1 − w
iz
i|
4n
Y
i=1
log
1+ε1 1 − |ϕ
wi(z
i)|
!
12× Z
Dn
|h(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−(1+ε)1
1 − |ϕ
wi(z
i)| dA(z)
!
12≤ C
n
Y
i=1
1
(1 − |w
i|
2) B
ε[|f |
2](w)
12× Z
Dn
|h(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−(1+ε)1
1 − |ϕ
wi(z
i)| dA(z)
!
12.
In the case α = {1, 2, . . . , n}, we have
|D
αT
f¯h(w)| ≤ 2
nZ
Dn
|f (z)||h(z)|
n
Y
i=1
|z
i|
|1 − w
iz
i|
3dA(z)
≤ Z
Dn
|f (z)|
n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
1+ε21 1 − |ϕ
wi(z
i)|
× |h(z)|
n
Y
i=1
1
|1 − w
iz
i|
n
Y
i=1
log
−1+ε21
1 − |ϕ
wi(z
i)| dA(z).
Following the previous calculations, we obtain the desired inequality. It remains to consider the case when α is a proper subset of {1, 2, . . . , n}.
Then
|D
αT
f¯h(w)| ≤ Z
Dn
|f (z)||h(z)| Y
i∈α
2|z
i|
|1 − w
iz
i|
3Y
i /∈α
1
|1 − w
iz
i|
2dA(z)
≤ C Z
Dn
|f (z)|
n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
1+ε21 1 − |ϕ
wi(z
i)|
× |h(z)|
n
Y
i=1
1
|1 − w
iz
i|
n
Y
i=1
log
−1+ε21
1 − |ϕ
wi(z
i)| dA(z),
where the last inequality follows from
Y
j∈α
2z
j(1 − ¯ w
jz
j)
3Y
j /∈α
1 (1 − ¯ w
jz
j)
2≤ C
n
Y
j=1
1
|1 − ¯ w
jz
j|
3.
Lemma 2. Let ε > 0, u ∈ (A
2)
⊥, f ∈ L
2(D
n), α = {α
1, α
2, . . . , α
m} ⊂ {1, 2, . . . , n}, α
1< α
2< . . . < α
m. Then
|D
αH
f∗u(w)| ≤ C
n
Y
j=1
1 1−|w
j|
2(f ◦ ϕ
w− P (f ◦ ϕ
w))
n
Y
j=1
log
(1+ε)/21 1 − |z
j|
×
Z
Dn
|u(z)|
2n
Y
j=1
1
|1 − ¯ z
jw
j|
2n
Y
j=1
log
−1−ε1
1 − |ϕ
wj(z
j)| dA(z)
1 2
.
Proof. The proof will proceed in three steps as above. Suppose first that α = ∅. Then
hH
f∗u, K
wi =
n
Y
j=1
1
1 − |w
j|
2hH
f∗u, k
wi =
n
Y
j=1
1
1 − |w
j|
2hu, H
fk
wi.
In view of [8, Proposition 1] we may write
H
fk
w= (f − P (f ◦ ϕ
w) ◦ ϕ
w) k
wand
hH
f∗u, K
wi =
n
Y
j=1
1
1 − |w
j|
2hu, (f − P (f ◦ ϕ
w) ◦ ϕ
w) k
wi.
Thus, by H¨ older’s inequality, we obtain
|hu, (f − P (f ◦ ϕ
w) ◦ ϕ
w) k
w(z)i|
= Z
Dn
u(z)
n
Y
j=1
log
−1+ε21
1 − |ϕ
wj(z
j)| (f − P (f ◦ ϕ
w) ◦ ϕ
w) (z)k
w(z)
×
n
Y
j=1
log
1+ε21
1 − |ϕ
wj(z
j)| dA(z)
≤
Z
Dn
| (f −P (f ◦ ϕ
w) ◦ ϕ
w) (z)|
2|k
w(z)|
2n
Y
j=1
log
1+ε1
1−|ϕ
wj(z
j)| dA(z)
1 2
×
Z
Dn
|u(z)|
2n
Y
j=1
log
−1−ε1
1 − |ϕ
wj(z
j)| dA(z)
1 2
.
By the change-of-variable formula z 7→ ϕ
w(z) and using that |1 − ¯ z
jw
j| ≤ 2, we have
|hu, (f − P (f ◦ ϕ
w) ◦ ϕ
w) k
w(z)i|
≤ C
(f ◦ ϕ
w− P (f ◦ ϕ
w))
n
Y
j=1
log
(1+ε)/21 1 − |z
j|
×
Z
Dn
|u(z)|
2n
Y
j=1
1
|1 − ¯ z
jw
j|
2n
Y
j=1
log
−1−ε1
1 − |ϕ
wj(z
j)| dA(z)
1 2
.
This proves the first case. Now, let α = {1, 2, . . . , n}. Then H
f∗u(w) = P ( ¯ f u)(w) =
Z
Dn
f (z)u(z)
n
Y
j=1
1
(1 − w
jz ¯
j)
2dA(z).
Hence
D
αH
f∗u(w) = Z
Dn
f (z)u(z)
n
Y
j=1
2¯ z
j(1 − w
j¯ z
j)
3dA(z).
Let
F
w(z) = P (f ◦ ϕ
w) ◦ ϕ
w(z)
n
Y
j=1
2z
j(1 − ¯ w
jz
j)
3. The function F
wbelongs to ∈ A
2, thus
hu, F
wi = Z
Dn
u(z)P (f ◦ ϕ
w) ◦ ϕ
w(z)
n
Y
j=1
2z
j(1 − ¯ w
jz
j)
3dA(z) ≡ 0.
So,
D
αH
f∗u(w) = D
αH
f∗u(w) − hu, F
wi
= Z
Dn
u(z)(f (z) − P (f ◦ ϕ
w) ◦ ϕ
w(z))
n
Y
j=1
2z
j(1 − ¯ w
jz
j)
3dA(z).
Using H¨ older’s inequality, we get
|D
αH
f∗u(w)|
≤ C
Z
Dn
|u(z)|
2n
Y
j=1
1
|1 − ¯ z
jw
j|
2n
Y
j=1
log
−1−ε1
1 − |ϕ
wj(z
j)| dA(z)
1 2
×
n
Y
j=1
1 1 − |w
j|
2×
Z
Dn
| (f −P (f ◦ ϕ
w) ◦ ϕ
w) (z)|
2|k
w(z)|
2n
Y
j=1
log
1+ε1
1−|ϕ
wj(z
j)| dA(z)
1 2
= C
n
Y
j=1
1 1 − |w
j|
2×
Z
Dn
|u(z)|
2n
Y
j=1
1
|1 − ¯ z
jw
j|
2n
Y
j=1
log
−1−ε1
1 − |ϕ
wj(z
j)| dA(z)
1 2
×
(f ◦ ϕ
w− P (f ◦ ϕ
w))
n
Y
j=1
log
(1+ε)/21 1 − |z
j|
L2
.
Suppose now that α = {α
1, α
2, . . . , α
m} is a nonempty subset of {1, 2, . . . , n}.
Then
D
αH
f∗u(w) = Z
Dn
f (z)u(z) Y
j∈β
2¯ z
j(1 − w
jz ¯
j)
3Y
j /∈β
1
(1 − w
jz ¯
j)
2dA(z).
Putting
F
w(z) = P (f ◦ ϕ
w) ◦ ϕ
w(z) Y
j∈β
2z
j(1 − ¯ w
jz
j)
3Y
j /∈β
1 (1 − ¯ w
jz
j)
2and using the fact that
Y
j∈β
2z
j(1 − ¯ w
jz
j)
3Y
j /∈β
1 (1 − ¯ w
jz
j)
2≤ C
n
Y
j=1
1
|1 − ¯ w
jz
j|
3, we obtain
|D
βH
f∗u(w)|
≤ C Z
Dn
|u(z)|
n
Y
j=1
1
|1− ¯ w
jz
j| |f (z) − P (f ◦ ϕ
w) ◦ ϕ
w(z)|
n
Y
j=1
1
|1 − ¯ w
jz
j|
2dA(z).
Using the same arguments as in the proof of Lemma 1, the stated result
follows.
Now, we give the proofs of the main theorems.
Proof of Theorem 5. Let u, v ∈ H
∞. We show that
|hT
fT
g¯u, vi| ≤ Ckukkvk.
By (1), we get
hT
fT
g¯u, vi = hT
¯gu, T
f¯vi
= Z
Dn
T
g¯u(w)T
f¯v(w)dA(w)
= X
α
Z
Dn
D
αT
¯gu(w)D
αT
f¯v(w)dµ
α(w).
Using Lemma 1, we obtain
|hT
fT
g¯u, υi| ≤ C X
α
Z
Dn
Z
Dn n
Y
i=1
1
(1 − |w
i|
2) B
ε[|f |
2](w)
12× Z
Dn
|u(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)
!
12× Z
Dn n
Y
i=1
1
(1 − |w
i|
2) B
ε[|g|
2](w)
12× Z
Dn
|υ(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)
!
12
dµ
α(z)
≤ C sup
w∈Dn
B
ε[|f |
2](w)B
ε[|g|
2](w)
12X
α
Z
Dn n
Y
i=1
1 (1 − |w
i|
2)
2× Z
Dn
|u(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)
!
12× Z
Dn
|υ(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)
!
12dµ
α(w).
Since
dµ
α(z) = 3
n−m6
mn
Y
j=1
(1 − |z
j|
2)
2Y
j∈α
(5 − 2|z
j|)
2dA(z
1)dA(z
2) . . . dA(z
n)
≤ 3
nn
Y
j=1
(1 − |z
j|
2)
2dA(z
1)dA(z
2) . . . dA(z
n), we get
|hT
fT
¯gu, υi| ≤ C sup
w∈Dn
B
ε[|f |
2](w)B
ε[|g|
2](w)
12× Z
Dn
Z
Dn
|u(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)
!
12× Z
Dn
|v(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)
!
12
dA(w).
Now, applying H¨ older’s inequality and Fubini’s theorem, we have
|hT
fT
g¯u, υi| ≤ C sup
w∈Dn
B
ε[|f |
2](w)B
ε[|g|
2](w)
12× Z
Dn
Z
Dn
|u(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)dA(w)
!
12× Z
Dn
Z
Dn
|υ(z)|
2n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(z)dA(w)
!
12= C sup
w∈Dn
B
ε[|f |
2](w)B
ε[|g|
2](w)
12× Z
Dn
|u(z)|
2Z
Dn n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(w)dA(z)
!
12
× Z
Dn
|υ(z)|
2Z
Dn n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(w)dA(z)
!
12. It remains to prove that the integral
I = Z
Dn n
Y
i=1
1
|1 − w
iz
i|
2n
Y
i=1
log
−1−ε1
1 − |ϕ
wi(z
i)| dA(w)
is convergent independently of z. Indeed, the change-of-variable formula
ζ = ϕ
z(w) and the fact that |ϕ
wi(z
i)| = |ϕ
zi(w
i)| imply
I = Z
Dn n
Y
i=1
|1 − z
iw
i|
2(1 − |z
i|
2)
2n
Y
i=1
log
−1−ε1 1 − |ϕ
zi(w
i)|
n
Y
i=1
(1 − |z
i|
2)
2|1 − z
iw
i|
4dA(w)
= Z
Dn n
Y
i=1
|1 − z
iϕ
zi(ζ
i)|
2(1 − |z
i|
2)
2n
Y
i=1
log
−1−ε1
1 − |ζ
i| dA(ζ)
= Z
Dn n
Y
i=1
(1−|zi|2)2
|1−ziζi|2
(1 − |z
i|
2)
2n
Y
i=1
log
−1−ε1
1 − |ζ
i| dA(ζ)
=
n
Y
i=1
Z
D
1
|1 − z
iζ
i|
2log
−1−ε1
1 − |ζ
i| dA(ζ
i).
We need only to show that I
j=
Z
D
1
|1 − z
jζ
j|
2log
−1−ε1
1 − |ζ
j| dA(ζ
j) ≤ C for j = 1, 2, . . . , n. Let ζ
j= re
iθ.
According to Theorem 1.7 in [14], we have Z
2π0
1
|1 − z
jre
iθ|
2dθ ≤ C
1 − |z|r ≤ C 1 − r . Therefore
I
j≤ C 1 π
Z
10
r
1 − r log
−1−ε1 1 − r dr.
By the change-of-variable formula, I
j≤ C
Z
+∞0
t
−1−ε(1 − e
−t)dt
= C Z
10
t
−1−ε(1 − e
−t)dt + Z
+∞1
t
−1−ε(1 − e
−t)dt
≤ C Z
10
t
−εdt + Z
+∞1
t
−1−εdt.
Clearly, for ε ∈ (0, 1) the integrals I
iare bounded by a constant which is independent of z. Finally, we conclude that
|hT
fT
g¯u, υi| ≤ Ckukkυk,
which proves the theorem.
Proof of Theorem 6. To prove the theorem we need to use Lemma 2 and the method used in the proof of Theorem 5. The details are left to the
reader.
Now, we propose one additional theorem concerning products of Toeplitz and Hankel operators T
fH
g∗. The following result can be proved in much the same way as Theorem 5 and Theorem 6.
Theorem 7. Let f ∈ A
2, g ∈ L
2(D
n). If sup
Dn
B
ε[|f |
2](w)
(g ◦ ϕ
w− P (g ◦ ϕ
w))
n
Y
j=1
log
(1+ε)/21 1 − |z
j|
L2
< ∞,
then the operator T
fH
g∗is bounded on (A
2)
⊥.
It is clear that the above condition also gives the boundedness of H
gT
f¯. The next proposition reveals that Theorem 5 extends Theorem 2.
Proposition 1. Let f, g ∈ A
2and ε > 0. Then for all w ∈ D
n, B
ε[|f |
2]B
ε[|g|
2] ≤ C B[|f |
2+ε]B
ε[|g|
2+ε]
2/(2+ε).
Proof. Let w ∈ D
n. Then by the change-of-variable formula and H¨ older’s inequality we have
B
ε[|f |
2](w) = Z
Dn
|f (z)|
2n
Y
i=1
log
1+ε1 1 − |ϕ
wi(z
i)|
n
Y
j=1
(1 − |w
j|
2)
2|1 − ¯ w
jz
j|
4dA(z)
≤
Z
Dn
|f (z)|
2+ε(z)
n
Y
j=1
(1 − |w
j|
2)
2|1 − ¯ w
jz
j|
4dA(z)
2 2+ε
×
Z
Dn n
Y
j=1
log
(1+ε)(2+ε)ε1
1 − |ϕ
wi(z
i)|
nY
j=1
(1 − |w
j|
2)
2|1 − ¯ w
jz
j|
4dA(z)
ε 2+ε
= {B[|f |
2+ε](w)}
2+ε2
Z
Dn n
Y
j=1
log
(1+ε)(2+ε)ε1
1 − |z
i|
dA(z)
ε 2+ε
.
Since the last integral is convergent, our claim follows. References
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Paweł Sobolewski Institute of Mathematics
Maria Curie-Skłodowska University pl. M. Curie-Skłodowskiej 1 20-031 Lublin
Poland
e-mail: pawel.sobolewski@umcs.eu Received September 20, 2018
