




































©2023 Ada Academica https://adac.eeEur. J. Math. Anal. 3 (2023) 23doi: 10.28924/ada/ma.3.23
Strong Continuity of Composition Semigroups on the Generalized Bloch Spaces of the Upper

Half Plane

K. A. Wandera1, J. O. Bonyo2,∗ , D. O. Ambogo1
1Department of Pure and Applied Mathematics, Maseno University, P.O. Box 333-40105, Maseno - Kenya

kwandera1@gmail.com, ambogos@maseno.ac.ke
2Department of Mathematics, Multimedia University of Kenya, P.O. Box 15653-00503, Nairobi - Kenya

jbonyo@mmu.ac.ke
∗Correspondence Author

Abstract. We investigate strong continuity of composition semigroups on the generalized Blochspaces of the upper half plane. These composition semigroups are induced by automorphisms ofthe upper half plane as classified into three distinct groups in [3].

1. Introduction
Consider H(Ω) as the Fréchet space of analytic functions f : Ω→ C endowed with the topologyof uniform convergence on compact subsets of Ω. A function f ∈ H(D) is in the Bloch space of theunit disc B(D) if

‖f ‖B1(D) := sup
z∈D

(1− |z |2)|f ′(z)| <∞

and in the little Bloch space of the unit disc B0(D) if
lim
|z |−→1

(1− |z |2)|f ′(z)| = 0.

For f ∈ B(D), we define the norm on B(D) by
‖f ‖B(D) := |f (0)|+ ‖f ‖B1(D),where ‖.‖B1(D) is a seminorm on B(D).Bloch space of the upper half plane B(U) is a set of analytic functions f ∈ H(U) such that

‖f ‖B1(U) := sup
ω∈U
=(ω)|f ′(ω)| <∞.

For f ∈ B(U), we define the norm on B(U) by
‖f ‖B(U) := |f (i)|+ ‖f ‖B1(U),

Received: 5 Jun 2023.
Key words and phrases. composition semigroup; analytic functions; self analytic maps; Bloch spaces; unit disc; upperhalf plane; strong continuity; infinitesimal generator. 1

https://adac.ee
https://doi.org/10.28924/ada/ma.3.23
https://orcid.org/0000-0002-6442-4211


where ‖.‖B1(U) is a seminorm on B(U).Let α > 0 be a real number, we define the generalized Bloch space of the unit disc, Bα(D) asthe space of all functions f ∈ H(D) such that
‖f ‖Bα1 (D) := sup

z∈D

(
1− |z |2

)α |f ′(z)| <∞.

For f ∈ Bα(D), we define the norm on Bα(D) by
‖f ‖Bα(D) := |f (0)|+ ‖f ‖Bα1 (D). (1)

We also define the corresponding generalized little Bloch space of the unit disc as the space of allfunctions f ∈ H(D) for which
lim
|z |→1

(
1− |z |2

)α |f ′(z)| = 0,

with the same norm given by (1). Here, Bα(D) and Bα◦ (D) are both Banach spaces with respectto the norm ‖.‖Bα(D). The generalized little Bloch space of the unit disc, Bα◦ (D) is the closure ofthe set of polynomials in the norm topology of Bα(D). For more details see [17, 18]. The space
B(D) has been studied by many authors because of its intrinsic interest since its introduction[1, 4, 8, 10, 13, 14, 18]. In [17], the generalized Bloch spaces of the open unit disc, Bα(D) aredefined and proved to be Banach spaces with respect to their norm. Zhu [17] further establishedgeneralized little Bloch spaces of the unit disc Bα◦ (D), as closed, separable subspaces of Bα(D).There is scanty literature on the properties of the generalized Bloch spaces of the upper half plane
Bα(U), including whether they are Banach spaces. Composition semigroups on Bloch spaces ofthe unit disc have been studied in literature, see for instance [2,11,12] and references therein. Onstrong continuity of composition semigroups, Siskakis [12] proved that no nontrivial compositionsemigroups are strongly continuous on the Bloch space of the unit disc B(D). The correspondingstudy of composition semigroups defined on the Bloch spaces of the upper half plane has not yetbeen exhausted. Moreover, existing works on the half plane, see [7,13], have neither exhausted theinvestigation of properties of these semigroups nor considered these generalizations. In this papertherefore, we investigate the properties of the generalized Bloch spaces of the upper half plane asBanach spaces and extend the study of semigroups of composition operators to the setting of thegeneralized Bloch spaces of the upper half plane.

2. Preliminaries and Definitions
Let C be the complex plane. The set D := {z ∈ C : |z | < 1} is called the open unit disc.On the other hand, the set U := {ω ∈ C : =(ω) > 0} denotes the upper half of the complexplane C, where =(ω) is the imaginary part of ω ∈ C. The function ψ(z)= i(1+z)

1−z is referred to asthe Cayley transform and maps the unit disc D conformally onto the upper half-plane U, with theinverse ψ−1(ω) = ω−i
ω+i mapping the upper half plane U, onto the unit disc, D. We refer to [16] for2



details. Let α > 0 be a real number. A function f ∈ H(U) belongs to the generalized Bloch spaceof the upper half plane, Bα(U) if
‖f ‖Bα1 (U) := sup

ω∈U
= (ω)α |f ′(ω)| <∞

with the norm given by
‖f ‖Bα(U) := |f (i)|+ ‖f ‖Bα1 (U).The corresponding generalized little Bloch space of the upper half plane, Bα0 (U)is defined as

Bα◦ (U) := {f ∈ H(U) : lim
=(ω)−→0

= (ω)α |f ′(ω)| = 0}

having the same norm as Bα(U). There is little literature on the properties of the generalized Blochspaces of the upper half plane as Banach spaces. Let X be a Banach space. A one-parameterfamily (Tt)t≥0 is a semigroup of bounded linear operators on X , if(i) To = I (Identity operator on X), and(ii) Tt+s = Tt ◦ Ts for every t, s,≥ 0 (Semigroup property).A semigroup (Tt)t≥0 of bounded linear operators on X is strongly continuous if
lim
t→0+

‖Ttx − x‖ = 0 for all x ∈ X.
The infinitesimal generator denoted by Γ of (Tt)t≥0 is defined by

Γx := lim
t→0+

Ttx − x
t

=
∂

∂t
(Ttx)

∣∣∣∣
t=0

for each x ∈ dom(Γ),

where dom(Γ) denotes the domain of Γ given by
dom(Γ) =

{
x ∈ X : lim

t→0+
Ttx − x

t
exists} .

We define a group of bounded linear operators as
(Tt)t∈R =

Tt , t ≥ 0,

T−t , t ≥ 0.

if both (Tt)t≥0 and (T−t)t≥0 are semigroups on X . For more details see [5,6,9]. Suppose ϕ : Ω→ Ωis a self analytic map. The composition operator induced by ϕ on H(Ω) is defined as
Cϕ(f ) = f o ϕ,

for all f ∈ H(Ω). On the other hand, given t ≥ 0 we define a semigroup as a family (ϕt)t≥0 ofself analytic maps on Ω satisfying the following properties(i) ϕ0(z) = z (Identity map on Ω).(ii) ϕt+s = ϕt ◦ ϕs ,∀ t, s ≥ 0 (Semigroup property).(iii) ϕt → ϕ0 uniformly on compact subsets of Ω as t → 0.3



Composition semigroup induced by ϕt on H(Ω) is defined as
Cϕt (f ) = f o ϕt , for all f ∈ H(Ω).

3. Generalized Bloch Spaces of the Upper Half Plane
In this section, we study properties of the generalized Bloch spaces as Banach spaces. We alsorelate functions in the generalized Bloch space of the upper half plane U to their counterparts inthe unit disc D. Following [17,18], it’s well known that Bα(D) and Bα0 (D) are Banach spaces withrespect to the norm ‖.‖Bα(D). Moreover the set of analytic polynomials C[z ] :=

{ ∞∑
n=0

an z
n : z ∈ C

}
is dense in Bα0 (D). These results are not explicitly clear from the literature in the setting of theupper half plane U.In the following theorem, we establish the completeness of Bα(U) with respect to the norm ‖.‖Bα(U).
Theorem 3.1. Bα(U) is a Banach space with respect to the norm ‖.‖Bα(U)

Proof. It’s clear that (Bα(U), ‖.‖Bα(U)
) is a normed space. Now we prove that the space Bα(U)is complete in ‖.‖Bα(U). Let (fk)k denote a Cauchy sequence in Bα(U). For ε > 0, there exists

N ∈ N such that ‖fk − fl‖Bα(U) < ε, ∀ k, l > N. Hence by the definition of the norm, we have forall ∀ k, l > N,

|fk(i)− fl(i)|+ sup
ω∈U
= (ω)α |f ′k(ω)− f ′l (ω)| < ε,

which means that
|fk(i)− fl(i)| < ε and (=(ω))α |f ′k(ω)− f ′l (ω)| < ε,

for ω ∈ U. So, (fk(i))k∈N is Cauchy in C. By the completeness of C, (fk(i))k converges to a limit,say u0. Similarly, (f ′k(ω)
)
k∈N is Cauchy in C and therefore converges to a limit, say g.Since |f ′k(ω)− f ′l (ω)| < ε
=(ω)α and f ′k(ω)→ g uniformly on compact subsets of U, then g ∈ H(U).Now, take f such that f ′(ω) = g(ω)∀ω ∈ U and f (i) = u0.Thus, ∀ ε > 0, ∃N such that ∀k, l > N ,

= (ω)α |f ′k(ω)− f ′l (ω)| < ε, ∀ω ∈ U.

Taking limits as l →∞, then ∀k > N ,
= (ω)α |f ′k(ω)− f ′(ω)| < ε, ∀ω ∈ U.

It follows that
‖fk − f ‖Bα(U) = |fk(i)− f (i)|+ sup

ω∈U
= (ω)α |f ′k(ω)− f ′(ω)| < ε

4



and so ‖fk − f ‖Bα(U) → 0 as k →∞.Now, it remains to show that f ∈ Bα(U). We have
= (ω)α |f ′(ω)| = = (ω)α |f ′(ω)− f ′k(ω) + f ′k(ω)|

≤ = (ω)α |f ′(ω)− f ′kω|+ = (ω)α |f ′k(ω)|

< ε+ = (ω)α |f ′k(ω)| <∞

since (fk)k ⊂ Bα(U).Now, taking supremum over all ω ∈ U in the above equation, we have that
sup
ω∈U
= (ω)α |f ′(ω)| <∞

which implies that f ∈ Bα(U), as desired. �

As an immediate consequence, we have
Corollary 3.2. B(U) is a Banach space with respect to the norm ‖ . ‖B(U)

Proof. Follows immediately by taking α = 1 in Theorem 3.1. �

Under the norm ‖ . ‖Bα(U), the space Bα0 (U) also becomes a Banach space as in the followingtheorem,
Theorem 3.3. Bα0 (U) is a Banach space with respect to the norm ‖ . ‖Bα(U).

Proof. Following Theorem 3.1, we need to show that every sequence in Bα0 (U) convergent in Bα(U)has its limit in Bα0 (U).Let (fn) ⊂ Bα0 (U) and g ∈ Bα(U) be such that fn → g as n → ∞. We need to prove that
g ∈ Bα0 (U). Since fn, g are holomorphic on compact subsets of U, and fn → g, we have f ′n → g′uniformly. Now that fn ⊂ Bα0 (U), we have

lim
=(ω)→0

(=(ω))α |f ′n(ω)| = 0,∀ n. (2)
Since limn→∞ f

′
n = g′, we have

lim
=(ω)→0

(=(ω))α |g′(ω)| = lim
=(ω)→0

(=(ω))α | lim
n→∞

f ′n(ω)|

which is equivalent to
lim

=(ω)→0
(=(ω))α |g′(ω)| = lim

n→∞

(
lim

=(ω)→0
(=(ω))α |f ′n(ω)|

)
.

Following equation (2), we see that
lim

=(ω)→0
(=(ω))α |g′(ω)| = 0.

So, g ∈ Bα0 (U), completing the proof. �5



As a consequence, we have the following,
Corollary 3.4. B0(U) is a Banach space with respect to the norm ‖.‖B(U)

Proof. Follows immediately by taking α = 1 in Theorem 3.3. �

In the next results, we generate a relationship between functions in the generalized Bloch spaceof the upper half plane U and their counterparts in the unit disc D
Proposition 3.5. Let f ∈ Bα(U) and ψ be the Cayley transform, then f ∈ Bα(U) if and only if
f ◦ ψ ∈ Bα(D)

Proof. It suffices to prove that ‖f ‖Bα1 (U) <∞ if and only if ‖f ◦ ψ‖Bα1 (D) <∞. Let f be a functionin Bα(U). Then by definition,
‖f ‖Bα1 (U) = supω∈U=(ω)α|f ′(ω)| <∞.

Now, by changing variables, let ω = ψ(z), where ψ is the Cayley transform. Then
=(ω) =

ω − ω
2i

=
ψ(z)− ψ(z)

2i
.

Using ψ(z) = i(1+z)
1−z and ψ(z) = −i(1+z)

1−z , we have
=(ω) =

i(1+z)
1−z −

−i(1+z)
1−z

2i

=
i(1 + z)(1− z) + i(1 + z)(1− z)

2i(1− z)(1− z)

=
i(2− 2zz)

2i(1− z)(1− z)

=
1− |z |2

|1− z |2 .We get the absolute of ψ′(z) = 2i
(1−z)2 as
|ψ′(z)| =

2

|1− z |2 . (3)
Now, by definition we have

‖f ‖Bα1 (U) = sup
z∈D

(
1− |z |2

|1− z |2

)α
|f ′(ψ(z))|.

From equation (3), we have |1− z |2 = 2
|ψ′(z)| , therefore

‖f ‖Bα1 (U) =
1

2α
sup
z∈D

(1− |z |2)α|ψ′(z)|α|f ′(ψ(z))|.
6



Since, (f ◦ ψ)′(z) = f ′(ψ(z))ψ′(z), we have
|ψ′(z)|α|f ′(ψ(z))| = |ψ′(z)(f ◦ ψ)′(z)||ψ′(z)α−1| and hence

‖f ‖Bα1 (U) =
1

2α
sup
z∈D

(1− |z |2)α|ψ′(z)(f ◦ ψ)′(z)||ψ′(z)α−1|

=
1

2α
|ψ′(z)α−1|‖f ◦ ψ‖Bα1 (D),which is finite if and only if ‖f ◦ ψ‖Bα1 (D) is finite. This completes the proof. �

An immediate consequence is the following,
Corollary 3.6. Let f ∈ B(U) and ψ be the Cayley transform, then

‖f ‖B1(U) =
1

2
‖f ◦ ψ‖B1(D) (4)

In particular, a function f ∈ B(U) if and only if f ◦ ψ ∈ B(D).

Proof. This follows immediately from Proposition 3.5 by taking α = 1. �

4. Composition Semigroups on the Generalized Little Bloch Space of the Upper Half Plane
In [3], the non trivial automorphisms of the upper half plane U were classified according to thelocation of their fixed points into three distinct classes namely; scaling, translation and rotationgroups. In this section, we determine composition semigroups induced by these automorphismgroups of the upper half plane U, on the generalized Bloch space of the upper half plane Bα(U). Wethen employ the theory of linear operators on Banach spaces to investigate the semigroup propertiesof the induced composition semigroup. For any given semigroup ϕt , the induced operator semigroup

Cϕt is known to be strongly continuous on the little Bloch space. On the other hand, no non trivialcomposition semigroup is strongly continuous on the Bloch space, see [11]. Therefore, we shalldetermine the composition semigroup induced by these automorphism groups on the generalizedlittle Bloch space of the upper half plane, Bα0 (U). Further, we show that composition semigroupsinduced by scaling and translation groups are strongly continuous on Bα0 (U). We also establishstrong continuity of composition semigroups induced by rotation group on Bα0 (D). The infinitesimalgenerator is identified and its domain stated.
4.1. 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], the semigroup properties of the induced compositionoperators will differ significantly depending on whether 0 < k < 1 or k > 1. Thus for 0 < k < 1,we consider without loss of generality, the analytic self maps ϕt : U −→ U of the form

ϕt(z) = e−tz, z ∈ U. (5)
The composition semigroup induced by equation (5) on Bα0 (U) is given by

Cϕt f (z) = (f ◦ ϕt) (z) = f
(
e−tz

)
7



It can be easily proved that (Cϕt )t∈R is a group on Bα0 (U).In what follows, we prove that the composition semigroup given by equation (4.1) fails to be anisometry on Bα0 (U).

Proposition 4.1. The operator Cϕt fails to be an isometry on Bα0 (U).

Proof. By the definition of the norm, we have for all f ∈ Bα0 (U)

‖Cϕt f ‖Bα(U) = |Cϕt f (i)|+ sup
ω∈U
=(ω)α| (Cϕt f )′ (ω)|

= |f (e−t i)|+ sup
ω∈U
=(ω)α|e−t f ′(e−tω)|.

Now by change of variables:Let z = e−tω, then ω = etz , and =(ω) = et=(z). Therefore,
‖Cϕt f ‖Bα(U) = |f (e−t i)|+ sup

z∈U
etα=(z)α|e−t f ′(z)|

= |f (e−t i)|+ e(α−1)t sup
z∈U
=(z)α|f ′(z)|

6= |f (i)|+ sup
z∈U
=(z)α|f ′(z)| = ‖f ‖Bα(U),

which completes the proof. �

Next, we prove that the operator Cϕt given by (4.1) is strongly continuous on Bα0 (U).
Theorem 4.2. (Cϕt )t∈R is strongly continuous on Bα0 (U).

Proof. To prove strong continuity of (Cϕt )t∈R, it suffices to show that ‖Cϕt f −f ‖Bα(U) → 0 as t → 0.That is, | (Cϕt f − f ) (i)|+‖Cϕt f −f ‖Bα1 (U) → 0 as t → 0. This is equivalent to | (Cϕt f − f ) (i)| → 0and ‖Cϕt f − f ‖Bα1 (U) → 0, as t → 0. For the former, we have
| (Cϕt f − f ) (i)| = |Cϕt f (i)− f (i)| (6)

= |f (ϕt(i))− f (i)|

= |f (e−t i)− f (i)| → 0 as t → 0,

as desired. We now prove that ‖Cϕt f − f ‖Bα1 (U) → 0 as t → 0. Recall that ψ : D→ U, ϕt : U→

U andψ−1 : U → D. We can therefore have D ψ−→ U ϕt−→ U ψ−1−−→ D. Now, let Xt = ψ−1 ◦ ϕt ◦ ψ :

D→ D. If (ϕt)t≥0 is an automorphism of the upper half plane U, then (Xt)t≥0 is an automorphismof the unit disc D. Since Xt = ψ−1 ◦ ϕt ◦ ψ, it follows that ‖Cϕt f − f ‖Bα1 (U) → 0 as t → 0 if andonly if ‖CXt f ∗ − f ∗‖Bα(D) → 0 as t → 0 8



Cayley transform is given by ψ(z) = i(1+z)
1−z . We therefore have

ψ−1 ◦ ϕ−t ◦ ψ(z) = ψ−1 (ϕt (ψ(z))) .

= ψ−1
(
ϕt

(
i(1 + z)

1− z

))
= ψ−1

(
e−t

(
i(1 + z)

1− z

))
.

Substituting ψ−1(z) = z−i
z+i , we obtain

ψ−1 ◦ ϕ−t ◦ ψ(z) =
e−t( i(1+z)1−z )− i

e−t( i(1+z)1−z ) + i
.

Simplifying the fraction, we have
ψ−1 ◦ ϕ−t ◦ ψ(z) =

z + e−tz − 1 + e−t

−z + e−tz + 1 + e−t
.

Now, by factorizing z and dividing both the numerator and denominator by (1 + e−t), we obtain
ψ−1 ◦ ϕ−t ◦ ψ(z) =

z − (1−e
−t)

(1+e−t)

1− (1−e
−t)

1+e−t z
.

Let bt = 1−e−t
1+e−t , and substitute to obtain

ψ−1 ◦ ϕ−t ◦ ψ(z) =
z − bt

1− btz
:= Xt(z).

Further, we apply density of polynomials in Bα0 (D) to prove that for f ∗ ∈ Bα0 (D), we have ‖CX t f ∗−
f ∗‖Bα1 (D) → 0 as t → 0.By the definition of the norm, we have

lim
t→0+

‖CX t f ∗ − f ∗‖Bα(D) = lim
t→0+

|(CX t f ∗ − f ∗)(0)|+ sup
z∈D

(
1− |z |2

)α |(CX t f ∗ − f ∗)′(z)|.

Let f ∗(z) = zn and z ∈ D.
We need to show that ‖ (CXt f

∗ − f ∗) ‖Bα1 (D) → 0, as t → 0.Since
CXtz

n − zn = (Xt(z))n − zn, n ≥ 1,

differentiating (Xt(z))n − zn with respect to z , we obtain
(CXt f

∗ − f ∗)′(z) = n(Xt(z))n−1X ′t(z)− nzn−1

= n[(Xt(z))n−1X ′t(z)− zn−1].

Substituting for
Xt(z) =

z − bt
1− btz9



and
X ′t(z) =

(1− btz)1− (z − bt)(−bt)
(1− btz)2

=
(1− b2t )

(1− btz)2
,

we obtain
(CXt f

∗ − f ∗)′(z) = n

[(
z − bt

1− btz

)n−1 (1− b2t )

(1− btz)2
− zn−1

]
= n

[
(z − bt)n−1(1− b2t )

(1− btz)n−1(1− btz)2
− zn−1

]
= n

[
(z − bt)n−1(1− b2t )− zn−1(1− btz)n+1

(1− btz)n+1

]
.

It therefore follows that limt→0+ ‖CX t f ∗ − f ∗‖Bα1 (D) is equivalent to
lim
t→0+

(
(sup
z∈D

(
1− |z |2

)α ∣∣∣∣n [(z − bt)n−1(1− b2t )− zn−1(1− btz)n+1

(1− btz)n+1

]∣∣∣∣) .
Now, let bt → 0 as t → 0, we obtain

lim
t→0+

‖CX t f ∗ − f ∗‖Bα1 (D) = sup
z∈D

(1− |z |2)α
∣∣n[zn−1 − zn−1]

∣∣
= 0.

Since limt→0+ ‖(CXt f ∗ − f ∗‖Bα1 (D) = 0, it follows that
lim
t→0+

(
‖Cϕt f − f ‖Bα1 (U)

)
= 0.

Therefore ‖Cϕt f − f ‖Bα(U) = |ϕt f (i))− f (i)|+ ‖Cϕt f − f ‖Bα1 (U) → 0 as t→ 0, as desired. �

In the next proposition, we compute the infinitesimal generator and determine the domain of thecomposition semigroup in equation (4.1).
Proposition 4.3. The infinitesimal generator Γ of (Cϕt )t≥0 on Bα0 (U) is given by Γf (z) = −zf ′(z)

with the domain dom (Γ) = {f ∈ Bα0 (U) : zf ′(z) ∈ Bα0 (U)}.

Proof. Using the definition of the infinitesimal generator Γ of (Cϕt )t≥0, for f ∈ Bα0 (U) we have
Γf (z) = lim

t→0+
Cϕt f (z)− f (z)

t

= lim
t→0+

f
(
e−tz

)
− f (z)

t

=
∂

∂t
f (e−tz)

∣∣∣∣
t=0

= −zf ′(z).10



This implies that Γf (z) = −zf ′(z) and therefore dom(Γ) ⊆ {f ∈ Bα0 (U) : zf ′ ∈ Bα0 (U)}. To provereverse inclusion, we let f ∈ Bα0 (U) be such that zf ′ ∈ Bα0 (U). Then for z ∈ U,
Cϕt f (z)− f (z)

t
=

1

t

∫ t

0

∂

∂s
(Cϕs f (z))ds

=
1

t

∫ t

0

(−e−szf ′(e−sz))ds

=
1

t

∫ t

0

CϕsF (z)ds, whereF (z) = −zf ′(z).

Since F (z) is a function in Bα0 (U), it remains to show that the limit of F (z) exist in Bα0 (U). Thus
lim
t→0+

Cϕs f (z)− f (z)

t
= lim
t→0+

1

t

∫ t

0

CϕsF (z)ds.

By strong continuity of (Cϕs )s≥0 we have
1

t

∫ t

0

‖CϕsF − F‖ds → 0 as t → 0+.

Hence
{f ∈ Bα0 (U) : zf ′ ∈ Bα0 (U)} ⊆ dom(Γ).This completes the proof. �

4.2. Translation group. In this case the automorphisms are of the form ϕt(z) = z + kt , where
z ∈ U and k, t ∈ R with k 6= 0. As noted in [3], we can consider the self analytic maps of U of theform

ϕt(z) = z + t. (7)The composition semigroup induced by translation group on Bα0 (U) is given by
Cϕt f (z) = f (z + t). (8)

The proof of our results given in equation (8) as a group on Bα0 (U) is basic, we therefore omit thedetails.We shall now prove that the composition semigroup in equation (8), fails to be an isometry on
Bα0 (U).

Proposition 4.4. The operator Cϕt fails to be an isometry on Bα0 (U).

Proof. By norm definition, we have
‖Cϕt f ‖Bα(U) = |Cϕt f (i)|+ sup

z∈U
=(z)α| (Cϕt f )′ (z)|

= |f (i + t)|+ sup
z∈U
=(z)α|f ′(z + t)|.

Now by change of variables: Let z + t = ω then z = ω − t , and =(z) = =(ω). Therefore,
‖Cϕt f ‖Bα(U) = |f (i + t)|+ sup

ω∈U
=(ω)α|f ′(ω)|. (9)

11



The right hand side of equation (9) is not equal to the norm ‖f ‖Bα(U) for any t > 0. This impliesthat (8) is not an isometry on Bα0 (U). This completes the proof. �

In the following results, we investigate the strong continuity of the composition semigroup inequation (8) on Bα0 (U).
Proposition 4.5. The operator Cϕt is strongly continuous on Bα0 (U).

Proof. We need to show that ‖Cϕt f − f ‖Bα(U) → 0 as t → 0. This approach is similar to (7). Weomit the details. We compute the automorphism of the unit disc D, denoted by Xt as follows
Xt(z) = ψ−1 (ϕt (ψ(z)))

= ψ−1
(
ϕt

(
i(1 + z)

1− z

))
= ψ−1

(
i(1 + z)

1− z + t

)
.

Since the inverse of Cayley transform is given by ψ−1 = z−i
z+i , we substitute to obtain

Xt =

i(1+z)
1−z − t − i
i(1+z)
1−z − t + i

=

i(1+z)
1−z − (t + i)

i(1+z)
1−z + (i − t)

.

We simplify further by multiplying both the numerator and denominator by (1− z) to obtain
Xt(z) =

i(1 + z) + (t − i)(1− z))

i(1 + z) + (t + i)(1− z)

=
(2i − t)z − t
(2i + t)− tz .By dividing both the numerator and denominator by 2i − t , we get

Xt =
z + t

2i−t
2i+t
2i−t −

t
2i−t z.Letting kt = t

2i−t and mt = 2i+t
2i−t . We have

Xt =
z + kt
mt − ktz

.

Next, we apply density of polynomials in Bα0 (D) to prove that for f ∗ ∈ Bα0 (D), we have ‖CX t f ∗ −
f ∗‖Bα1 (D) → 0 as t → 0.

lim
t→0+

‖CX t f ∗ − f ∗‖Bα1 (D) = lim
t→0+

(
sup
z∈D

(
1− |z |2

)α |(CX t f ∗ − f ∗)′(z)|
)
.

Using density of polynomials in Bα0 (D), let f ∗(z) = zn and z ∈ D be such that
CX tz

n − zn = (Xt(z))n − zn, n ≥ 1. (10)12



Now, differentiating (Xt(z))n − zn with respect to z , we get
(CX t f

∗ − f ∗)′(z) = n(Xt(z))n−1X ′t(z)− nzn−1

= n[(Xt(z))n−1X ′t(z)− zn−1]. (11)
We also differentiate Xt = z+kt

mt−ktz by quotient rule to obtain
X ′t(z) =

(mt − ktz)1− (z + kt)(−kt)
(mt − ktz)2

=
mt + k2t

(mt − ktz)2
.

Substituting for Xt = z+kt
mt+ktz

and X ′t(z) =
mt−k2t
(mt−ktz)2 in equation (11) we have

(CX t f
∗ − f ∗)′(z) = n[(Xt(z))n−1X ′t(z)− zn−1]

= n

[
(z + kt)

n−1(mt − ktz2)− zn−1(mt − ktz)n+1

(mt − ktz)n+1

]
.

It therefore follows that as t → 0, we have
‖CX t f ∗ − f ∗‖Bα(D) = (|(Xt(0))n − 0|)

+ (sup
z∈D

(
1− |z |2

)α ∣∣n[(Xt(z))n−1X ′t(z)− zn−1]
∣∣ = 0.

Therefore ‖Cϕt f − f ‖Bα(U) = |ϕt f (i)) − f (i)| + ‖Cϕt f − f ‖Bα1 (U) → 0 as t → 0, as desired. Thiscompletes the proof. �

In the next theorem, we obtain the infinitesimal generator of the strongly continuous compositionsemigroup given in equation (8).
Theorem 4.6. The infinitesimal generator Γ of (Cϕt )t≥0 on Bα0 (U) is given by Γf (z)=f ′(z) with
the domain dom(Γ) = {f ∈ Bα0 (U) : f ′(z) ∈ Bα0 (U)}.

Proof. Using the definition of the infinitesimal generator Γ, for f ∈ Bα0 (U), we have;
Γf (z) = lim

t→0+
f (z + t)− f (z)

t

=
∂

∂t
f (z + t)

∣∣∣∣
t=0

= f ′(z).

This means that dom(Γ) ⊂ {f ∈ Bα0 (U) : f ′(z) ∈ Bα0 (U)}.It remains to prove the reverse inclusion. Let f ∈ Bα0 (U) be such that f ′(z) ∈ Bα0 (U).Then for z ∈ U, we have;
Cϕt f (z)− f (z) =

∫ t

0

∂

∂s
f (z + s)ds

=

∫ t

0

f ′(z)ds.

13



Letting F (z) = f ′(z), we obtain
Cϕt f (z)− f (z) =

∫ t

0

F (z)ds.

This implies that F (z) = f ′(z) is a function of Bα0 (U). It remains to show that the limit of F (z)exists in Bα0 (U). Since
Cϕt f (z)− f (z)

t
=

1

t

∫ t

0

F (z)ds,

we now take limits as t → 0+ and invoke strong continuity of (Cϕs )s≥0 to obtain
lim
t→0+

1

t

∫ t

0

‖CϕsFds − F‖ = 0.

Hence dom(Γ) ⊇ {f ∈ Bα0 (U) : f ′(z) ∈ Bα0 (U)} which completes the proof. �

5. Rotation Group
The induced composition semigroups for rotation group are defined on the analytic spaces of theunit disk. We shall therefore generate composition semigroups induced by rotation group on thegeneralized little Bloch space of the disc. The results obtained can then be mapped onto the upperhalf plane by use of Cayley transform. In this case, the self analytic maps of D are of the form

ϕt(z) = e iktz . We consider the composition semigroup induced by the rotation group on Bα0 (D)given by
Cϕt f (z) = (f ◦ ϕt) (z)

= f
(
e itz

)
, (12)

for all f ∈ Bα0 (D).It can be easily shown that (Cϕt )t≥0 and (Cϕ−t)t≥0 are semigroups on Bα0 (D) thus (Cϕt )t∈Rdefines a group on Bα0 (D).Moreover, this group is an isometry, as we prove in the next proposition.
Proposition 5.1. The operator Cϕt given by (12) is an isometry on Bα0 (D).

Proof. We shall prove that for each t ∈ R, the group (Cϕt )t∈R is an isometry on Bα0 (D). It sufficesto prove that
‖Cϕt f ‖Bα(D) = ‖f ‖Bα(D).It follows from the definition that

‖Cϕt f ‖Bα(D) = |Cϕt f (0)|+ sup
z∈D

(
1− |z |2

)α |(Cϕt f )′(z)|

= |(e it)f (0)|+ sup
z∈D

(
1− |z |2

)α |e it f ′(e itz)|

= |f (0)|+ sup
z∈D

(
1− |z |2

)α |f ′(e itz)|.
14



Now, let ω = e itz so that z = e−itω. Then;
‖Cϕt f ‖Bα(D) = |f (0)|+ sup

ω∈D

(
1− |e−itω|2

)α |f ′(ω)|)

= |f (0)|+ sup
ω∈D

(1− |ω|2)α|f ′(ω)|

= ‖f ‖Bα(D).

�

Theorem 5.2. The operator Cϕt given by (12) is strongly continuous on Bα0 (D).

Proof. Since polynomials are dense in Bα0 (D), it suffices to show that (Cϕt )t∈R is strongly contin-uous on Bα0 (D) that is, for a polynomial (zn)n≥0 where z ∈ D we obtain
lim
t→0+

‖Cϕtzn − zn‖Bα(D) = 0.

Clearly,
lim
t→0+

‖Cϕtzn − zn‖Bα(D) = lim
t→0+

|Cϕt f (0)− f (0)|+
(

sup
z∈D

(1− |z |2)α|(Cϕtzn − zn)′|)
)
.

But
Cϕtz

n − zn = (e int − 1)zn.So its derivative is given by
(Cϕtz

n − zn)′ = n(e int − 1)zn−1,

implying that
lim
t→0+

‖Cϕtzn − zn‖Bα(D) = lim
t→0+

|e it f (0)− f (0)|+
(

sup
z∈D

(1− |z |2)α|nzn−1||(e int − 1)|)
)
.

Hence,
lim
t→0+

‖Cϕtzn − zn‖Bα(D) = 0 as desired .
�

Proposition 5.3. The infinitesimal generator Γ of (Cϕt ) is given by Γf (z) = iz f ′(z) with the
domain dom(Γ) = {f ∈ Bα0 (D) : zf ′(z) ∈ Bα0 (D)}.

Proof. We obtain the infinitesimal generator as follows
Γf (z) = lim

t→0+
Cϕt(z)− f (z)

t

=
∂

∂t
f (e itz)

∣∣∣∣
t=0

= iz f ′(z).

15



It therefore follows that dom(Γ) ⊆ {f ∈ Bα0 (D)} : zf ′(z) ∈ Bα0 (D)}. On the other hand, let
f ∈ Bα0 (D)} be such that zf ′(z) ∈ Bα0 (D)}, then for z ∈ D we have by the Fundamental theoremof Calculus,

Cϕt f (z)− f (z) =

∫ t

0

∂

∂s
(Cϕs f (z))ds

=

∫ t

0

ie iszf ′(e isz)ds

=

∫ t

0

CϕsF (z)ds,

where F (z) = iz f ′(z) is a function in Bα0 (D). Thus
lim
t→0+

Cϕt f − f
t

= lim
t→0+

1

t

∫ t

0

CϕsFds

and strong continuity of (Cϕs )s≥0 implies that ‖1t ∫ t0 CϕsFds −F‖ ≤ 1
t

∫ t
0 ‖CϕsF −F‖ds → 0+ as

t → 0+. Thus dom(Γ) ⊇ {f ∈ Bα0 (D) : zf ′(z) ∈ Bα0 (D)}, as desired. �

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17

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https://doi.org/10.1007/0-387-27539-8

	1. Introduction
	2.  Preliminaries and Definitions
	3. Generalized Bloch Spaces of the Upper Half Plane
	4. Composition Semigroups on the Generalized Little Bloch Space of the Upper Half Plane
	4.1. Scaling group
	4.2. Translation group

	5. Rotation Group
	References

