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Duality of the Nonreflexive Bergman Space of the Upper Half Plane and Composition Groups

E. O. Gori1 , J. O. Bonyo2,∗

1Department of Pure and Applied Mathematics, Maseno University, P.O. Box 333-40105, Maseno, Kenya
omondierick75@gmail.com1

2Department of Mathematics, Multimedia University of Kenya, P.O. Box 15653-00503, Nairobi, Kenya
jbonyo@mmu.ac.ke

Correspondence: jbonyo@mmu.ac.ke

Abstract. We identify the predual of the nonreflexive Bergman space of the upper half plane withthe little Bloch space of the upper half plane consisting of those functions vanishing at point i . Usingthe duality pairing as well as the composition groups on the nonreflexive Bergman space, we obtainthe groups of composition operators defined on the identified predual. We identify the infinitesimalgenerator of each group and prove the strong continuity property. We then obtain the spectra of thegenerator Γ, determine the resolvents and further obtain the spectra and the norms of the resultingresolvents.

1. Introduction
Let C be the complex plane. The set D := {z ∈ C : |z | < 1} is called the open unit disc. Let

dA denote the area measure on D, normalized so that the area of D is 1. In terms of rectangularand polar coordinates, we have: dA(z) = 1
πdxdy = r

πdrdθ, where z = x + iy = re iθ ∈ D. For
α ∈ R, α > −1, we define a positive Borel measure dmα on D by dmα(z) = (1−|z |2)αdA(z), andthus dmα is a probability measure. Moreover, if α = 0, then dmo = dA. We consider dmα as aweighted measure and a generalization of dA. On the other hand, the set U := {ω ∈ C : =(ω) > 0}denotes the upper half of the complex plane C, with =(ω) being the imaginary part of ω ∈ C. For
α > −1, we define a weighted measure on U by dµα(ω) = (=(ω))αdA(ω), where ω ∈ U. Again itcan easily be seen that α = 0 coincides with the unweighted measure. The function ψ(z) = i(1+z)

1−zis referred to as the Cayley transform and maps the unit disc D conformally onto the upper half-plane U with the inverse ψ−1(ω) = ω−i
ω+i .For an open subset Ω of C, let H(Ω) denote the space of analytic functions on Ω. For 1 ≤ p <∞,

Received: 29 Feb 2024.
Key words and phrases. Duality, Nonreflexive Bergman space, Bloch space, Composition semigroups, Infinitesimalgenerator, Spectrum, Resolvent. 1

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https://doi.org/10.28924/ada/ma.4.14
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Eur. J. Math. Anal. 10.28924/ada/ma.4.14 2

α > −1, the weighted Bergman space of the upper half-plane U is defined by
Lpa(U, µα) :=

{
f ∈ H(U) : ‖f ‖Lpa(U,µα) =

(∫
U
|f (z)|pdµα(z)

) 1
p

<∞

}
.

In particular, Lpa(U, µα) = Lp(U, µα) ∩ H(U), where Lp(U, µα) or simply Lp(µα) denotes theclassical Lebesque spaces with respect to the weighted measure dµα. It is important to note thatthe case α = 0 yields the unweighted Bergman space. Lpa(U, µα) is a Banach space with respectto the norm
‖f ‖Lpa(U,µα) =

(∫
U
|f (ω)|pdµα(ω)

) 1
p

.

For p = 2, L2
a(U, µα) is a Hilbert space. The growth condition for the weighted Bergman spacefunctions on U is given by: For every f ∈ Lpa(U, µα), γ = α+2

p and ω ∈ U, there exists a constant
K such that,

|f (ω)| ≤
K‖f ‖

(=(ω))γ
.

For a detailed account of the theory of Bergman spaces, we refer to [7, 11,13].On the other hand, the Bloch space of the unit disk, denoted by B∞(D), is defined by
B∞(D) := {f ∈ H(D) : ‖f ‖B∞,1(D) = sup

z∈D
(1− |z |2)|f ′(z)| <∞},

with the norm on B∞(D) given by ‖f ‖B∞(D) := |f (0)|+‖f ‖B∞,1(D), while ‖.‖B∞,1(D) is a seminorm.The Bloch space of the upper half plane denoted by B∞(U) is defined by
B∞(U) := {f ∈ H(U) : ‖f ‖B∞,1(U) = sup

ω∈U
=(ω)|f ′(ω)| <∞},

with the norm given by ‖f ‖B∞(U) = |f (i)| + ‖f ‖B∞,1(U). The little Bloch space of the unit diskdenoted by B∞,◦(D) is defined as
B∞,◦(D) := {f ∈ H(D) : lim

|z |→1
(1− |z |2)|f ′(z)| = 0}

but with the same norm as B∞(D), while for the upper half-plane, the little Bloch space is denotedby B∞,◦(U) and is defined by
B∞,◦(U) := {f ∈ H(U) : lim

=(ω)→0
=(ω)|f ′(ω)| = 0}

with the same norm as B∞(U). For a comprehensive theory of Bloch spaces, see [13,14].The duality properties of Bergman spaces are well known in literature. For instance in [13, Theorem4.2.9], it is proved that for 1 < p < ∞, 1
p + 1

q = 1 and α > −1, the dual of the Bergman space
Lpa(D, mα) is given by (Lpa(D, mα))∗ ≈ Lqa(D, mα) under the duality pairing,

〈g, f 〉 =

∫
D
g(z)f (z)dmα (g ∈ Lpa(D, mα), f ∈ Lqa(D, mα)).

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Eur. J. Math. Anal. 10.28924/ada/ma.4.14 3For the non-reflexive Bergman space on the unit disk, L1
a(D, mα), it is shown in [13, Theorems 5.1.4and 5.2.8] that the dual and predual spaces of L1

a(D, mα) are the Bloch space and the little Blochspace respectively. In particular, (L1
a(D, mα))∗ ≈ B∞(D) and (B∞,◦(D))∗ ≈ L1

a(D, mα) under theduality pairings given by respectively,
〈g, f 〉 =

∫
D
g(z)f (z)dmα(z) (g ∈ L1

a(D, mα), f ∈ B∞(D)),

and
〈g, f 〉 =

∫
D
g(z)f (z)dmα(z) (f ∈ L1

a(D, mα), g ∈ B∞,◦(D)).

For the corresponding spaces of the upper half plane, it has been proved and noted that thedual space of the reflexive Bergman space of the upper half plane Lpa(U, µα) is Lqa(U, µα) for
1 < p, q < ∞ with 1

p + 1
q = 1 under a similar pairing as above. See for instance, [2–4] or [11]for details. When p = 1, the space L1

a(U, µα) is non-reflexive, and it’s recently that the dual wasdetermined by Kang [8] as we give in Theorem 2.1 stated in the next section. Apparently, thepredual of L1
a(U, µα) is not explicitly clear from the literature. Generally, there’s no unified andcomprehensive exposition of properties of the analytic spaces of upper half plane U as there are forthe corresponding spaces on the unit disk D. Therefore, the main focus of this paper is to determinethe predual of L1

a(U, µα), that is, identifying the space whose dual is L1
a(U, µα).Let Aut(U) denotes the collection of all automorphisms of U. For ϕt ∈ Aut(U), t ≥ 0, we define acomposition operator on H(U) by Cϕt f := f ◦ϕt . The corresponding group of weighted compositionoperator on H(U) is therefore given by Tt f := Sϕt f = (ϕ′t)

γf ◦ ϕt for some appropriate weight
γ. Motivated by the work of Arvanitidis and Siskakis in [1], the current second author and threeothers in [3] classified all the self - analytic maps of the upper half plane into three distinct groups,namely: the scaling, the translation and the rotation groups. They then studied both the semigroupand spectral properties of the corresponding groups of weighted composition operators. As for theproperties of the adjoint groups on the reflexive weighted Bergman spaces Lpa(U, µα), 1 < p <∞,only the scaling group was considered in [3] and later completed for the other two groups by thesecond author in [4]. In this paper, we therefore determine the groups of composition operators onthe predual of non-reflexive Bergman space of the upper half-plane, L1

a(U, µα) and investigate theadjoint properties of the groups of weighted composition operators on nonreflexive Bergman space
L1
a(U, µα).Let X and Y be Banach spaces over C. The space L(X, Y ) = {T : X → Y such that T is linearand continuous}, endowed with the operator norm ‖T‖ = sup‖x‖≤1 ‖Tx‖, is a Banach space [5].We write L(X,X) = L(X). T is said to be a closed operator if its graph {(x, T x) | x ∈ D(T )}in X × Y is closed. Let T be a closed operator on X . The resolvent set of T , ρ(T ) is given by
ρ(T ) = {λ ∈ C : λI − T is invertible or bijective} and its spectrum σ(T ) = C \ ρ(T ). Therefore
σ(T ) ∪ ρ(T ) = C. The spectral radius of T is defined by r(T ) = sup{|λ| : λ ∈ σ(T )} with the

https://doi.org/10.28924/ada/ma.4.14


Eur. J. Math. Anal. 10.28924/ada/ma.4.14 4relation r(T ) ≤ ‖T‖. The point spectrum σp(T ) = {λ ∈ C : Tx = λx for some 0 6= x ∈ dom(T )}.For λ ∈ ρ(T ), the operator R(λ, T ) := (λI − T )−1 is, by the closed graph theorem a boundedoperator on X and is called the resolvent of T at the point λ or simply the resolvent operator.In fact, ρ(T ) is an open subset of C and R(λ, T ) : ρ(T ) → L(X) is an analytic function. For adetailed theory on spectra, we refer to [5, 6, 9, 12].
2. Predual of Non-reflexive Bergman space of the upper half-plane L1

a(U, µα)

Let B∞(U, i) denote the subspace of the Bloch space B∞(U) consisting of functions vanishingat point i . Therefore B∞(U, i) is defined as
B∞(U, i) := {f ∈ B∞(U) : f (i) = 0}.

Then B∞(U, i) is a closed subspace of B∞(U) and therefore is a Banach space with respect to thenorm ‖f ‖B∞,i := ‖f ‖B∞(U) = ‖f ‖B∞,1(U), see [8]. Similarly, let B∞,◦(U, i) denotes the subspace of
B∞,◦(U) consisting of functions vanishing at i . Therefore

B∞,◦(U, i) := {f ∈ B∞,◦(U) : f (i) = 0},

with the norm ‖f ‖B∞,i := ‖f ‖B∞(U) = ‖f ‖B∞,1(U). Again, B∞,◦(U, i) is a Banach space with respectto the norm given above.The following result due to Kang [8] gives the dual of L1
a(U, µα);

Theorem 2.1. For any α ∈ R, α > −1, we have

(L1
a(U, µα))∗ ≈ B∞(U, i),

under the integral pairing

〈g, f 〉 =

∫
U
g(ω)f (ω)dµα(ω) (g ∈ L1

a(U, µα), f ∈ B∞(U, i)).

With the help of Theorem 2.1 above, we determine the predual space of L1
a(U, µα), that is, a setwhose dual is L1

a(U, µα), but first we state some results.Let C(U) be the algebra of complex valued continuous functions on U = U
⋃
∂U, and C◦(U) bethe subalgebra of C(U) consisting of functions f such that f (ω)→ 0 as =(ω)→ 0.

Proposition 2.2. Let C◦(U) be the subalgebra of C(U) consisting of functions f such that f (ω) −→
0 as Im(ω) −→ 0 and C◦(D) be the subalgebra of C(D) consisting of functions f with f (z) → 0

as |z | → 1−. Then C◦(U) = {g ◦ ψ−1 : g ∈ C◦(D)}.

Proof. Let K ⊂ U be compact. Since Cayley transform ψ : D → U is a continuous bijection, itfollows that K ⊂ U is compact if and only if ψ−1(K) is compact in D. If f ∈ C◦(U) and ε > 0,

https://doi.org/10.28924/ada/ma.4.14


Eur. J. Math. Anal. 10.28924/ada/ma.4.14 5then there exists K compact in U such that supw∈U\K |f (w)| < ε.Now, g = f ◦ ψ is continuous on D with f = g ◦ ψ−1, and
sup

z∈D\ψ−1(K)

|g(z)| = sup
w∈U\K

|f (w)| < ε.

�

Proposition 2.3. Let Cψ be the composition by ψ operator. Then

(1) f ∈ B∞(U) if and only if Cψf ∈ B∞(D). In particular, ‖f ‖B∞,1(U) = 1
2‖Cψf ‖B∞,1(D).

(2) f ∈ B∞,◦(U) if and only if Cψf ∈ B∞,◦(D).
(3) f ∈ L1(U, µα) if and only if Sψf ∈ L1(D, mα). In particular, ‖f ‖L1

a(U,µα) = 1
2α ‖Sψf ‖L1(D,mα).

(4) f ∈ L∞(U, µα) if and only if Cψf ∈ L∞(D, mα).

Proof. For (1), if f ∈ B∞(U), then by definition,
‖f ‖B∞,1(U) = sup

ω∈U
(=(ω))|f ′(ω)| = sup

z∈D

1− |z |2

|1− z |2
∣∣f ′(ψ(z))

∣∣
=

1

2
sup
z∈D

(1− |z |2)|ψ′(z)||f ′(ψ(z))| =
1

2
sup
z∈D

(1− |z |2)|(f ◦ ψ)′(z)|

=
1

2
‖f ◦ ψ‖B∞,1(D).

For (2), we have f ∈ B∞,0(U) is equivalent to
lim

=(ω)→0
(=(ω))|f ′(ω)| = lim

=(ψ(z))→0

1− |z |2

|1− z |2 |f
′(ψ(z))| =

1

2
sup
|z |→1

(1− |z |2)|(f ◦ ψ)′(z)| = 0,

which in turn is equivalent to f ◦ ψ ∈ B∞,0(D), as desired. For f ∈ L1
a(U, µα), we have

‖f ‖L1
a(U,µα) =

∫
U
|f (ω)| dµα(ω)

=

∫
U
|f (ω)|(=(ω))α dA(ω)

=

∫
D
|f (ψ(z))|

(
1− |z |2

|1− z |2

)α
|ψ′(z)|2 dA(z)

= 1
2α

∫
D
|f (ψ(z))||ψ′(z)|α+2 (1− |z |2) dA(z)

= 1
2α

∫
D
|(ψ′(z))γ(f ◦ ψ)(z)| dmα(z)

= 1
2α ‖Sψf ‖L1(D,mα),

which proves (3). Now, f ∈ L∞(U, µα) means that f is essentially bounded which implies that
f ◦ ψ is essentially bounded as well. Since ψ is an invertible mapping from D onto U, it followsthat f ◦ ψ ∈ L∞(D, mα). The converse follows similarly. This completes the proof. �

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Eur. J. Math. Anal. 10.28924/ada/ma.4.14 6

Remark 1. It is easy to verify that Cψ−1 = C−1
ψ . Proposition 2.3 above therefore implies that Cψ

is an is an isometry up to a constant and at the same time invertible on the respective spaces with
the inverse also acting on the appropriate spaces.
More generally, let {V1, V2} = {D,U}, and let LF (Vi , Vj) denote the collection of conformal map-
pings from Vi onto Vj . Then LF (Vi , Vi) = Aut(Vi), and if h ∈ LF (Vi , Vj), then g ∈ Aut(Vj) 7→
h−1 ◦ g ◦ h ∈ Aut(Vi) is an isomorphism from Aut(Vi) onto Aut(Vj). For each g ∈ LF (Vi , Vj), we
define a weighted composition operator Sg : H(Vj) → H(Vi), by

Sgf (z) = (g′(z))γf (g(z)), for all z ∈ Vi . (2.1)
We note that if g ∈ LF (Vi , Vj) and h ∈ LF (Vj , Vi), then it is clear by chain rule that Sh ◦Sg = Sg◦h

and S−1
g = Sg−1 .

Now, using Propositions 2.2 and 2.3 above, we obtain the following result which is the upperhalf-plane analogue of [13, Lemma 5.14].
Proposition 2.4. For t > 0, α > −1, let the integral operator T on H(D) be defined by

T f (z) = (1− |z |2)t
∫
D

f (w)

(1− zw)2+t+α
dmα(w).

Let S be the corresponding integral operator on H(U) defined by
S := Cψ−1TCψ . Then the following properties hold:(a) S = (α+ t + 1)S2,(b) S is a bounded embedding of B∞(U) into L∞(U) and(c) S is an embedding of B∞,◦(U) into C◦(U).

Proof. From [13, Lemma 5.14], we have,
S = Cψ−1TCψ = Cψ−1 (α+ t + 1)T 2Cψ

= (α+ t + 1)Cψ−1T 2Cψ

= (α+ t + 1)S2,

which proves (a).For (b), we have
B∞(U)

Cψ−−→ B∞(D)
T−→ L∞(D)

Cψ−1

−−−→ L∞(U).

Now, Cψ is an isometry of B∞(U) onto B∞(D) up to constant, T is a bounded embedding of
B∞(D) into L∞(D) [13, Lemma 5.14], Cψ−1 is also an isometry of L∞(D) onto L∞(U), it thereforefollows that S = Cψ−1TCψ is a bounded embedding of B∞(U) into L∞(U).For (c), we have

B∞,◦(U)
Cψ−−→ B∞,◦(D)

T−→ C◦(D)
Cψ−1

−−−→ C◦(U).

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Eur. J. Math. Anal. 10.28924/ada/ma.4.14 7

Cψ is a bijection of B∞,◦(U) into B∞,◦(D), T is an embedding of B∞,◦(D) into C◦(D) [13, Lemma5.14], while on the other hand, Cψ−1 is also a bijection of C◦(D) into C◦(U). Therefore S =

Cψ−1TCψ is an embedding of B∞,◦(U) into C◦(U), which completes the proof. �

We now establish the predual space of L1
a(U, µα) as we give in the following theorem:

Theorem 2.5. For any α > −1, we have;

(B∞,◦(U, i))∗ ≈ L1
a(U, µα),

under the pairing

〈g, f 〉 =

∫
U
g(ω)f (ω)dµα(ω),

where g ∈ B∞,◦(U, i) and f ∈ L1
a(U, µα). Here, B∞,◦(U, i) is equipped with the same norm as

B∞(U, i).

Proof. If f ∈ L1
a(U, µα), then by Theorem 2.1 above, g 7−→ ∫

U g(ω)f (ω)dµα(ω) defines a boundedlinear functional on B∞,◦(U, i). Conversely, if F is a bounded linear functional on B∞,◦(U, i), wewant to show that there exists a function f ∈ L1
a(U, µα) such that F (g) =

∫
U g(ω)f (ω)dµα(ω) for

g in a dense set of B∞,◦(U, i).Now we fix any positive parameter t and consider the embedding S of B∞,◦(U, i) into C◦(U)as given by Proposition 2.4. The space X = S(B∞,◦(U, i)) is a closed subspace of C◦(U) and
F ◦ S−1 : X → C is a bounded linear functional on X since F and S−1 are both bounded. Bythe Hahn-Banach extension theorem, F ◦ S−1 extends to a bounded linear functional on C◦(U).By the Riesz representation theorem, there exists a finite weighted measure µα on U such that
‖µα‖ = ‖F ◦S−1‖ and F ◦S−1(h) =

∫
U h(z)dµα(z), h ∈ C◦(U). In particular, if g is a polynomial(polynomials are dense in B∞,◦(U, i)), then F (g) = F ◦S−1◦S(g) =

∫
U Sg(z)dµα(z). By Fubini’stheorem, we have F (g) =

∫
U g(ω)f (ω)dµα(ω), where f = Cψ−1TCψ which is bounded since T isbounded. �

3. Groups of weighted composition operators on predual of L1
a(U, µα)

As remarked in the section 1, the automorphisms of the upper half plane U were identified andclassified into three distinct groups according to the location of their fixed points in [3], namely:the scaling, the translation and the rotation groups. Since the corresponding groups of compositionoperators for the rotation group are defined on the analytic spaces of the unit disk, we shall onlyconsider groups of composition operators associated with the scaling and the translation groups inthis paper.

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Eur. J. Math. Anal. 10.28924/ada/ma.4.14 83.1. Scaling group. The automorphisms of this group are of the form ϕt(z) = k tz , where z ∈ Uand k, t ∈ R with k 6= 0. As noted in [3] and without loss of generality, we consider the analyticself maps ϕt : U → U of the form ϕt(z) = e−tz for z ∈ U. The corresponding group of weightedcomposition operators on Lpa(U, µα) is given by Tt f (z) = e−tγf (e−tz), for all f ∈ Lpa(U, µα),where γ=α+2
p and 1 ≤ p <∞. For p = 1, (Tt)t≥0 is defined on L1

a(U, µα) with γ = α+ 2.Following Theorem 2.5, the predual of L1
a(U, µα) is given by the duality relation

(B∞,◦(U, i))∗ ≈ L1
a(U, µα) (3.1)

under the integral pairing
〈g, f 〉 =

∫
U
g(w)f (w)dµα(w), (3.2)

where g ∈ B∞,◦(U, i) and f ∈ L1
a(U, µα).Using the duality pairing above, we obtain the corresponding group of weighted composition op-erators on B∞,◦(U, i) as below:Let g ∈ B∞,◦(U, i) and f ∈ L1

a(U, µα), then,
〈g, Tt f 〉 =

∫
U
g(z)e−tγf (e−tz)dµα(z)

=

∫
U
g(z)e−tγf (e−tz)(=(z))αdA(z).

By change of variables, let ω = e−tz , then z = etω, dA(ω) = e−2tdA(z) and =(z) = et Im(ω).Then,
〈g, Tt f 〉 =

∫
U
g(etω)e−tγf (ω)eαt(=(ω))αe2tdA(ω)

=

∫
U
g(etω)e−tγetγf (ω)dµα(ω)

=

∫
U
g(etω)f (ω)dµα(ω) = 〈T ∗t g, f 〉. (3.3)

where T ∗t g(ω) = g(etω).Now, St := T ∗t is defined on B∞,◦(U, i). But we see that Stg(i) = g(et i) 6= 0 and therefore Stgdoes not vanish at i . This means that St does not map B∞,◦(U, i) onto itself, and there (St)t≥0 isnot a good semigroup. We therefore propose two remedies to correct the defect. The first one is toapply a correction factor by writing Stg(ω) = g(etω)− g(et i) for all g ∈ B∞,◦(U, i). This meansthat Stg(i) = 0 and therefore St maps B∞,◦(U, i) onto itself, as desired.The second remedy is to redefined St to act on B∞,◦(U) instead of B∞,◦(U, i). This simply meansthat the domain of St has been enlarged and therefore St is well defined on B∞,◦(U, i).
We shall now carry out a complete study of both the semigroup and spectral properties of thisgroup on B∞,◦(U).

https://doi.org/10.28924/ada/ma.4.14


Eur. J. Math. Anal. 10.28924/ada/ma.4.14 93.1.1. Semigroup properties. In this section, we investigate the semigroup properties and determinethe infinitesimal generator Γ of (St)t≥0 on B∞,◦(U) where, Stg(w) := g(etw). We begin byproving the strong continuity property.
Theorem 3.1. Let Stg(w) := g(etw) be a semigroup of composition operators defined on B∞,◦(U).
Then, (St)t∈R is a strongly continuous group of isometries on B∞,◦(U).

Proof. It is clear from the definition that (St)t∈R is a group. To prove that (St)t∈R is an isometryon B∞,◦(U), we have;
‖Stg‖B∞,◦(U) = sup

ω∈U
=(ω)|Stg′(ω)|

= sup
ω∈U
=(ω)et |g′(etω)|.

Now by change of variables, let z = etω then ω = e−tz , and =(ω) = e−t=(z). Therefore,
‖Stg‖B∞,◦(U) = sup

z∈U
e−t=(z)et |g′(z)|

= sup
z∈U
=(z)|g′(z)|

= ‖g‖B∞,◦(U), as desired.
For strongly continuity, we first take note that St = Cϕ−t since Stg(ω) = g(ϕt(ω)). Thenby Proposition 2.3, it is easy to see that Cψ−t is strongly continuous on B∞,◦(U) if and only if
(Cψ−1◦ϕ−t◦ψ)t∈R is strongly continuous on B∞,◦(D). Now by simple computation of ψ−1◦ϕ−t◦ψ(z),we obtain;

ψ−1 ◦ ϕ−t ◦ ψ(z) =
z − 1−et

1+et

1− 1−et
1+et z

=
z − at

1− atz
,

where at = 1−et
1+et . As t → 0, at → 0. Let ha(z) = z−at

1−atz = ψ−1 ◦ ϕ−t ◦ ψ(z), then for strongcontinuity, it therefore suffices to show that ‖Cha f − f ‖B∞,◦(D) → 0 as a→ 0 (at → 0). Using thedensity of polynomials in B∞,◦(D), let f (z) = zn. Then Chazn − zn = (ha(z))n − zn, n ≥ 1, and
(Cha f − f )′(z) = n[(ha(z))n−1h′a(z)− zn−1].

But ha(z) = z−at
1−atz , and hence h′a(z) = 1−atat

(1−atz)2 . Therefore,
(Cha f − f )′(z) = n

[
(ha(z))n−1(1− atat)

(1− atz)2
− zn−1

]
= n

[
( z−at1−atz )n−1(1− atat)

(1− atz)2
− zn−1

]
= n

[
(z − at)n−1(1− atat)− zn−1((1− atz)n+1)

(1− atz)n+1

]
.

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Eur. J. Math. Anal. 10.28924/ada/ma.4.14 10Now,
lim
a→0
‖Cha f − f ‖B∞,◦(D)

= lim
a→0

(
sup
z∈D

(1− |z |2)|(Cha f − f )′|(z)

)
= lim

t→0

(
sup
z∈D

(1− |z |2)

∣∣∣∣n [(zn−1)(1)− zn−1(1)

(1)n+1

]∣∣∣∣)
= lim

t→0

(
sup
z∈D

(1− |z |2)
∣∣n[zn−1 − zn−1]

∣∣)
= 0.

Hence, (St)t∈R is strongly continuous on B∞,◦(U), as claimed. �

Theorem 3.2. The infinitesimal generator Γ of (St)t≥0 on B∞,◦(U) is given by Γg(ω)=ωg′(ω)

with the domain D(Γ) = {g ∈ B∞,◦(U) : ωg′(ω) ∈ B∞,◦(U)}.

Proof. By definition, the infinitesimal generator denoted by Γ of (St)t≥0 is given by;
Γg(ω) = lim

t→0+

g(etω)− g(ω)

t
=

∂

∂t
g(etω)

∣∣∣∣
t=0

= ωg′(ω).

It therefore follows that D(Γ) ⊆ {g ∈ B∞,◦(U) : ωg′(ω) ∈ B∞,◦(U)}. To prove the reverseinclusion, we let g ∈ B∞,◦(U) be such that ωg′(ω) ∈ B∞,◦(U). Then for ω ∈ U, we have;
Stg(ω)− g(ω) =

∫ t

0

∂

∂s
g(esω) ds

=

∫ t

0

esωg′(esω) ds

=

∫ t

0

SsG(ω) ds where G(ω) = ωg′(ω).

Thus,
lim
t→0+

Stg − g
t

= lim
t→0+

1
t

∫ t

0

SsG(ω) ds

and strong continuity of (Ss)t≥0 implies that 1
t

∫ t
0 ‖SsG − G‖ds → 0 as t → 0+. Hence D(Γ) ⊇

{g ∈ B∞,◦(U) : ωg′(ω) ∈ B∞,◦(U)}, which completes the proof. �

3.1.2. Spectral properties. Now for the spectral properties, we obtain the spectra of the generator
Γ, determine the resolvents and further obtain the spectra and the norms of the resulting resolvents.
Theorem 3.3. Let Γ be the infinitesimal generator of (St)t∈R on B∞,◦(U). Then σp(Γ) = ∅ and
σ(Γ) = iR. In particular, Γ is an unbounded operator on B∞,◦(U).

Before we prove this theorem, we first give the following Lemma:
Lemma 3.4. If ν ∈ C and c ∈ R, we have

(1) g(ω) = cων /∈ B∞,0(U) for any c

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Eur. J. Math. Anal. 10.28924/ada/ma.4.14 11

(2) f (ω) = (w − i)ν ∈ B∞,0(U) if and only if <(ν) < 0.

In particular, g(ω) /∈ B∞,0(U) for any c and f (ω) ∈ B∞,0(U) if and only if <(ν) < 0.

Proof. From Proposition 2.3, we know that g ∈ B∞,◦(U) if and only if g ◦ ψ ∈ B∞,◦(D). Then for
z ∈ D,

(g ◦ ψ)(z) = g(ψ(z)) = c(ψ(z))ν = c(
i(1 + z)

1− z )ν

= ci(1 + z)ν(1− z)−ν .

Now g ◦ ψ ∈ H(D) if and only if <(ν) > 0 and <(−ν) > 0 which is not possible, and therefore
g ◦ ψ /∈ H(D). Hence g /∈ B∞,0(U). This proves (1).

For (2), following [3, Lemma 3.2], for any ν ∈ C, (w − i)ν ∈ H(U) if and only if <(ν) < 0 since
γ = 0 in this case.The particular cases follow immediately since B∞,0(U, i) ⊆ B∞,0(U) and g(i) 6= 0 for (1), while
f (i) = 0 for (2). �

Proof of Theorem 3.3. To obtain the point spectrum of Γ, let λ be an eigenvalue of Γ and g be thecorresponding eigenvector. Then Γg(ω) = λg(ω) is equivalent to ωg′(ω) = λg(ω) which yields
ωg′(ω)
ω = λg(ω)

ω by dividing both sides by ω. By integrating both sides, we obtain g(ω) = cωλ,which is not in B∞,◦(U) for any c . Therefore σp(Γ) = ∅.Since each St is an invertible isometry, its spectrum satisfies σ(St) ⊆ ∂D. Therefore the spectralmapping theorem for strongly continuous groups [10, Theorem 2.3] implies that etσ(Γ) ⊆ σ(St) ⊆
∂D. Now let λ ∈ σ(Γ), then |etλ| = 1 which further implies that <(λ) = 0. Thus λ ∈ iR andtherefore σ(Γ) ⊆ iR.We now need to show the reverse inclusion, that is, iR ⊆ σ(Γ). Fix λ ∈ iR and assume λ /∈ σ(Γ)which implies that the resolvent operator R(λ,Γ) : B∞,◦(U) → B∞,◦(U) is bounded. Considerthe function h(w) = (w − i)−(λ+1). Then <(−(λ + 1)) = −1 < 0 and following Lemma 3.4, itis immediate that h ∈ B∞,◦(U). The image function f = R(λ,Γ)h is equivalent to (λ − Γ)f = hwhich yields a differential equation

f ′(ω)−
λ

ω
f (ω) = −

h(ω)

ω
,

whose general solution is
f (ω) = (ω − i)−λ + cωλwhich does not belong to B∞,◦(U) for any c , by Lemma 3.4. Thus h /∈ R(λ − Γ) and so σ(Γ) =

iR. �

Theorem 3.5. Let Γ be the infinitesimal generator of (St)t∈R. Then the following hold;

(1) For λ ∈ ρ(Γ), and h ∈ B∞,◦(U) then,

https://doi.org/10.28924/ada/ma.4.14


Eur. J. Math. Anal. 10.28924/ada/ma.4.14 12(i) R(λ,Γ)h(ω) = ωλ
∫∞
ω

1
zλ+1 h(z) dz , if <(λ) > 0.(ii) R(λ,Γ)h(ω) = −ωλ

∫ ω
0

1
zλ+1 h(z) dz, if <(λ) < 0.

(2) σ(R(λ,Γ)) =
{
ω : |ω − 1

2<(λ) | = 1
2<(λ)

}
.

(3) r(R(λ,Γ)) = ‖R(λ,Γ)‖ = 1
|<(λ)| .

Proof. To prove (1), we take note the resolvent set is given as ρ(Γ) = {λ ∈ C : <(λ) 6= 0}. Wetherefore consider the following cases:
Case 1: If <(λ) > 0, then the resolvent operator is given by the Laplace transform: For every
h ∈ B∞,◦(U), we have R(λ,Γ)h =

∫∞
0 e−λtSthdt with convergence in norm. Therefore, R(λ,Γ)h =∫∞

0 e−λth(etω)dt. By change of variables, let z = etω, then ω = e−tz , dzdt = ωet then dt = dz
ωet =

dz
z . Therefore when t = 0⇒ z = ω and t =∞⇒ z =∞, and so;

R(λ,Γ)h(ω) =

∫ ∞
ω

e−λth(z)
dz

z
=

∫ ∞
ω

( z
ω

)−λ 1

z
h(z)dz

= ωλ
∫ ∞
ω

1

zλ+1
h(z)dz.

Case 2: If <(λ) < 0, then R(λ,Γ)h = −R(−λ,−Γ)h = −
∫∞

0 eλth(e−tω)dt. Then again bychange of variables, let z = e−tω, then et = ω
z , dzdt = −ωe−t and dt = −dz

ωe−t = −dzz . Therefore
t = 0⇒ z = w and t =∞⇒ z = 0 and so;

R(λ,Γ)h(w) = −
∫ 0

ω

eλth(z).−
dz

z
= −

∫ ω

0

(ω
z

)λ
h(z).

dz

z

= −ωλ
∫ ω

0

(
1

z

)λ
.
1

z
h(z)dz

= −ωλ
∫ ω

0

1

zλ+1
h(z)dz.

To prove (2), we use the spectral mapping theorem for the resolvents which asserts that σ(R(λ,Γ)) ={
1

λ−µ : µ ∈ σ(Γ)
}
\ {0} for λ ∈ ρ(Γ). Therefore,
σ(R(λ,Γ)) =

{
1

λ− i r : r ∈ R
}
\ {0}

=

{
1

<(λ) + i(Im(λ)− r)
: r ∈ R

}
\ {0}.

Rationalizing the denominator and simplifying we get
σ(R(λ,Γ)) =

{
(<(λ)−i(=(λ)−r))

(<(λ))2+(=(λ)−r)2 : r ∈ R
}
.Now by letting w = (<(λ)−i(=(λ)−r))

(<(λ))2+(=(λ)−r)2 , subtracting 1
2<(λ) and finding the magnitude of both sideswe get, ∣∣∣∣w − 1

2<(λ)

∣∣∣∣2 =
1

(2<(λ))2
,

https://doi.org/10.28924/ada/ma.4.14


Eur. J. Math. Anal. 10.28924/ada/ma.4.14 13and so ∣∣∣∣w − 1

2<(λ)

∣∣∣∣ =
1

2<(λ)
.

Therefore, σ(R(λ,Γ)) =
{
w : |w − 1

2<(λ) | = 1
2<(λ)

}
. For part (3), the spectral radius r(R(λ,Γ))is given by;

r(R(λ,Γ)) = sup{|w | : w ∈ σ(R(λ,Γ))}

= sup

{
|w | :

∣∣∣∣w − 1

2<(λ)

∣∣∣∣ =
1

2<(λ)

}
=

1

|<(λ)| .Finally to determine ‖R(λ,Γ)‖, we use the Hille Yosida theorem as well as the fact that the spectralradius is always bounded by the norm.Therefore,
1

|<(λ)| = r(R(λ,Γ)) ≤ ‖R(λ,Γ)‖ ≤
1

|<(λ)| .Thus, r(R(λ,Γ)) = ‖R(λ,Γ)‖ = 1
|<(λ)| , as desired. �

3.2. Translation Group. In this group the automorphisms are of the form ϕt(z) = z + kt , where
z ∈ U and k, t ∈ R with k 6= 0. As noted earlier in subsection 3.1, without loss of generality welet k = 1 and consider the self analytic maps ϕt : U → U given by ϕt(z) = z + t for z ∈ U.Then the corresponding group of composition operators defined on L1

a(U, µα) is therefore given by
Tt f (z) = f (z + t), for all f ∈ Lpa(U, µα).Now using the duality relation given by equation (3.1) and its sesquilinear pairing given by equation(3.2), we have:Let g ∈ B∞,◦(U, i) and f ∈ L1

a(U, µα), then
〈g, Tt f 〉 =

∫
U
g(z)f (z + t)dµα(z)

=

∫
U
g(z)f (z + t)(=(z))αdA(z).

Now by a change of variables, let ω = z + t , then z = ω − t and dA(ω) = dA(z). Therefore,
〈g, Tt f 〉 =

∫
U
g(ω − t)f (ω)(=(ω))αdA(ω)

=

∫
U
g(ω − t)f (ω)dµα(ω)

= 〈T ∗t g, f 〉. (3.4)
Now, we define St := T ∗t on B∞,◦(U, i). But again we see that just as in the case of the scalinggroup, Stg(i) = g(i − t) and therefore Stg(i) does not vanish at point i . This means that St doesnot map B∞,◦(U, i) onto itself. We can therefore apply similar remedies proposed in subsection3.1 above. In the next sections, we study the semigroup properties of (St)t≥0 on B∞,◦(U).

https://doi.org/10.28924/ada/ma.4.14


Eur. J. Math. Anal. 10.28924/ada/ma.4.14 143.2.1. Semigroup properties. In this section, we begin by proving the strong continuity property.
Theorem 3.6. Let Stg(ω) := g(ω−t) be a semigroup of composition operators defined on B∞,◦(U).
Then, (St)t∈R is a strongly continuous group of isometries on B∞,◦(U).

Proof. It is clear from the definition that (St)t∈R is a group. To prove that (St)t∈R is an isometry,then by the definition of isometry, we have;
‖Stg‖B∞,◦(U) = sup

ω∈U
=(ω)|(Stg)′(ω)|

= sup
ω∈U
=(ω)|g′(ω − t)|.

By change of variables, let z = ω − t then ω = z + t and =(ω) = =(z). Hence,
‖Stg‖B∞,◦(U) = sup

z∈U
=(z)|g′(z)|

= ‖g‖B∞,◦(U), as desired.
For strongly continuity property, we argue as we did in the previous section. We note that St =

Cϕ−t which is strongly continuous on B∞,◦(U) if and only if (Cψ−1◦ϕ−t◦ψ)t∈R is strongly continuouson B∞,◦(D), which consists of functions vanishing at point 0.We compute ψ−1 ◦ ϕ−t ◦ ψ(z). Let at = t
2i+t and bt = 2i−t

2i+t , then a straight forward calculationyields
ψ−1 ◦ ϕ−t ◦ ψ(z) =

z − at
bt + atz

= ha(z),

where we have let ha(z) = z−at
bt+atz

. Clearly, t → 0 as at → 0 and bt → 1. It therefore suffices toshow that ‖Cha f −f ‖B∞,◦(D) → 0 as t → 0. Using density of polynomials in B∞,◦(D), we let f (z) =

zn. Then Chazn−zn = (ha(z))n−zn, n ≥ 1. Therefore (Cha f −f )′(z) = n[(ha(z))n−1h′a(z)−zn−1].But ha(z) = z−at
bt+atz

⇒ h′a(z) = (bt+atz)(1)−(z−at)(at)
(bt+atz)2 . Therefore by substituting,

(Cha f − f )′(z) = n[(ha(z))n−1h′a(z)− zn−1]

= n

[(
z − at
bt + atz

)n−1 (bt + atz)− (z − at)(at)

(bt + atz)2
− zn−1

]
= n

[
(z − at)n−1(bt + atz)− (z − at)(at)

(bt + atz)n+1
− zn−1

]
.

Now,
lim
t→0+

‖Cha f − f ‖B∞,◦(D)
= lim

t→0+

(
sup
z∈D

(1− |z |2)|(Cha f − f )′|(z)

)
= lim

t→0+

(
sup
z∈D

(1− |z |2)∣∣∣∣n [(z − at)n−1(bt + atz)− (z − at)(at)

(bt + atz)n+1
− zn−1

]∣∣∣∣)

https://doi.org/10.28924/ada/ma.4.14


Eur. J. Math. Anal. 10.28924/ada/ma.4.14 15

= lim
t→0+

(
sup
z∈D

(1− |z |2)

∣∣∣∣n[zn−1 − 0− zn−1]

1

∣∣∣∣)
= 0.

Hence (St)∈R is strongly continuous, as claimed. �

Theorem 3.7. The infinitesimal generator Γ of (St)t≥0 on B∞,◦(U) is given by Γg(ω) = −g′(ω)

with the domain D(Γ) = {g ∈ B∞,◦(U) : g′ ∈ B∞,◦(U)}.

Proof. By definition, the infinitesimal generator Γ on B∞,◦(U) is given by;
Γg(ω) = lim

t→0+

g(ω − t)− g(ω)

t
=

∂

∂t
g(ω − t)

∣∣∣∣
t=0

= −g′(ω).

Therefore D(Γ) ⊂ {g ∈ B∞,◦(U) : −g′ ∈ B∞,◦(U)}. Conversely, let g ∈ B∞,◦(U) be such that
−g′ ∈ B∞,◦(U). Thus we have;

Stg − g
t

= 1
t

∫ t

0

∂

∂s
Ssg ds

and for every ω ∈ U, ∂
∂sSsg(ω) = −g′(ω − s) = Ssg

′(ω). Thus,∥∥∥∥Ssg − gt
− f ′

∥∥∥∥ ≤ 1
t

∫ t

0

∥∥Ts f ′ − f ′∥∥ ds → 0 as t → 0

by strong continuity, and therefore D(Γ) ⊇ {g ∈ B∞,◦(U) : −g′ ∈ B∞,◦(U)}, which completes theproof. �

acknowledgement
This work was completed during the period when the second author was visiting the AristotleUniversity of Thessaloniki, Greece. He would like to sincerely that the Simon’s Foundation forfunding his visit. He would also wish to thank his host Prof. Aristomenis G. Siskakis and thedepartment of Mathematics for the unmatched hospitality

References
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Eur. J. Math. Anal. 10.28924/ada/ma.4.14 16

[7] P. Duren, A. Schuster, Bergman spaces, Mathematical Surveys and Monographs 100, Amer. Math. Soc., Providence,RI, 2004.[8] S. H. Kang, Some duality of weighted Bergman spaces of the half-plane, Bull. Korean Math. Soc. 42 (2005), 385-396.[9] K. B. Laursen, M. M. Neumann, An introduction to local spectral theory, Clarendon Press, Oxford, 2000.[10] A. Pazy, Semigroups of linear operators and applications to partial differential equations, Applied MathematicalSciences 40, Springer, New York, 1983.[11] M. M. Peloso, Classical spaces of Holomorphic functions, Technical report, Universìt di Milano, 2014.[12] W. Rudin, Functional Analysis, McGraw-Hill, Inc. New York (1991).[13] K. Zhu, Operator Theory in function spaces, Marcel Dekker Inc. New York, Basel (1990).[14] K. Zhu, Bloch type spaces of analytic functions, Rocky Mountain J. Math. 23 (1993), 1143–1177.

https://doi.org/10.28924/ada/ma.4.14

	1. Introduction
	2. Predual of Non-reflexive Bergman space of the upper half-plane L1a(U,) 
	3. Groups of weighted composition operators on predual of L1a(U,)
	3.1. Scaling group
	3.2. Translation Group

	acknowledgement
	References

