isometric multiplication operators and weighted

Geng Journal of Inequalities and Applications (2021) 2021:155 https://doi.org/10.1186/s13660-021-02689-6

R E S E A R C H Open Access

Isometric multiplication operators andweighted composition operators of BMOALigang Geng1*

*Correspondence:[email protected] of Mathematics andStatistics, Chongqing Technologyand Business University, 19 XuefuAvenue, Nan’an District, Chongqing,China

AbstractLet u be an analytic function in the unit disk D and ϕ be an analytic self-map of D. Wegive characterizations of the symbols u and ϕ for which the multiplication operatorMu and the weighted composition operatorMu,ϕ are isometries of BMOA.

MSC: 46B04; 46E15; 47B33; 47B38

Keywords: Isometric; Multiplication Operator; Weighted Composition Operator;BMOA

1 IntroductionIn order to state our results, we first introduce some notations and definitions. Let D de-note the open unit disk in the complex plane C, that is, D = {z : |z| < 1, z ∈ C}. Its bound-ary is the unit circle ∂D = {z : |z| = 1, z ∈ C}. A lot of analytic function spaces over D havebeen introduced in last half-century, such as Hardy, Bergman, Besov, Bloch, and Dirichletspaces.

Let H2 denote the classical analytic Hardy space on the unit disk D. We will consider themultiplication operator Mu : f → uf of BMOA space, which consists analytic functions f ∈H2 whose boundary functions have bounded mean oscillation on ∂D. In this manuscriptwe suppose that u is analytic in D and continuous on ∂D. The norm of the BMOA spaceis defined as

‖f ‖BMOA =∣∣f (0)

∣∣ + ‖f ‖∗,

where

‖f ‖∗ = supa∈D

∥∥f (σa) – f (a)

∥∥

2.

Here ‖ · ‖2 is the norm of H2(D) and σa is the conformal automorphism of D, σa(z) = a–z1–az

for a, z ∈ D. There are some equivalent norms of the BMOA space. For example, for ameasurable complex-valued function f defined on ∂D, define

fI =1|I|

∫

If(

eit)dt,

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where I ⊂ ∂D is an arc and |I| is the length of I , normalized so that |I| ≤ 1. The functionf has bounded mean oscillation if the seminorm

‖f ‖′∗ = sup

I

1|I|

∫

I

∣∣f

(

eit) – fI∣∣dt < ∞,

as I ranges over all arcs in ∂D, then BMOA = {f ∈ H1 : ‖f ‖′∗ < ∞}, this can be seen in [1–3].Properties of multiplication operators have been studied on various classical Banach

spaces of analytic functions in the unit disk D, such as Hardy, Bergman, and Blochspaces. In [4], the author studied the invertible and Fredholm multiplication operatorson Bergman spaces. In [5], the authors studied the compact multiplication operators withBMO symbols. In [6], the authors studied the C∗-algebra in the commutant of the multi-plication operators with the symbol on the finite-dimensional Bergman spaces. In [7], theauthors studied the commutants and reducing subspaces of multiplication operators onthe Bergman spaces.

Properties of linear isometries of those spaces have received a great deal of attention.Readers can see parts of research in this direction in [8–13]. An interesting fact is that thelinear isometries of spaces of continuous functions and some spaces of analytic functions,such as Hardy and Bergman spaces, in the unit disk D have the forms of weighted com-position operators. There are some studies of the properties of isometric and weightedcomposition operators on spaces of analytic functions on D and other domains; these canbe seen in [14–17]. In [18], the authors studied isometric pointwise multipliers of weightedBergman, Bloch, and Dirichlet-type spaces in the unit disk. They obtained that the isomet-ric pointwise multipliers of these spaces are unimodular constants.

In this work we will study the properties of the symbols of isometric multiplication andweighted composition operators of BMOA space.

2 Isometric multiplication operatorsThe Hardy space H2(D) consists of all analytic functions on D whose power series havesquare summable coefficients. For every f ∈ H2, the radial limit f (ζ ) = limr→1– f (rζ ) existsalmost everywhere on ∂D, and a standard norm on H2 is given by

‖f ‖2 =(∫

∂D

∣∣f (ζ )

∣∣2 dm(ζ )

) 12

.

Here dm is the Lebesgue measure on ∂D normalized so that m(∂D) = 1. Also, the normon H2 can be represented as

‖f ‖2 = sup0<r<1

(1

2π

∫ 2π

0

∣∣f

(

reiθ )∣∣2 dθ

) 12

.

Let X and Y be metric spaces with metrics dX and dY . A map T : X → Y is called anisometry if for any x1, x2 ∈ X one has

dY (Tx1, Tx2) = dX(x1, x2).


Given two normed vector spaces X and Y , a linear isometry is a linear map T : X → Ythat preserves the norms,

‖Tx‖Y = ‖x‖X ,

for all x ∈ X.If Mu is an isometric operator of a Banach space X = 0, then it is also a bounded operator

of X. Hence, if Mu is an isometry of BMOA then u must be an H∞ function with H∞-normsmaller than or equal to 1. This can be seen in [18–20].

We will consider the question when the multiplication operator Mu is an isometry ofBMOA space. We start with the following proposition.

Proposition 1 If Mu is an isometry of BMOA space, then u ∈ BMOA and ‖u‖BMOA = 1.

Proof Suppose Mu is an isometry of BMOA, then

‖Muf ‖BMOA = ‖uf ‖BMOA = ‖f ‖BMOA

for every f ∈ BMOA. That is,

∣∣u(0)f (0)

∣∣ + sup

a∈D

∥∥u(σa)f (σa) – u(a)f (a)

∥∥

2 =∣∣f (0)

∣∣ + sup

a∈D

∥∥f (σa) – f (a)

∥∥

2. (1)

Letting f = 1 in (1), we get

‖u‖BMOA =∣∣u(0)

∣∣ + sup

a∈D

∥∥u(σa) – u(a)

∥∥

2 = ‖1‖BMOA = 1. (2)�

It is easy to see that ‖u‖BMOA = 1 is a necessary but not sufficient condition for Mu to bean isometry of BMOA space. The following is an example.

Example 1 Let u = z, then ‖z‖BMOA = 1, but Mz is not an isometry of BMOA, since

∥∥σa(z)

∥∥

BMOA =∣∣σa(0)

∣∣ + sup

b∈D

∥∥σa

(

σb(z))

– σa(b)∥∥

2 = 1 + |a|

and∥∥Mzσa(z)

∥∥

BMOA =∣∣0 · σa(0)

∣∣ + sup

b∈D

∥∥σb(z)σa

(

σb(z))

– b · σa(b)∥∥

2

= supb∈D

√

1 – |b|2∣∣σa(b)∣∣2 = 1

for a = 0.

Now we describe the isometric multiplication operators of BMOA space in the follow-ing.

Proposition 2 Suppose Mu is an isometry of BMOA, then the function u has one of thefollowing properties:


(1) The function u(z) is a constant function of modulus 1;(2) There exists a real number λ ∈ [0, 1] such that u(z) = c

1–λz . Here, c = u(0),λ = 1 – |u(0)|.

Proof Suppose Mu is an isometry of BMOA, let f (z) = z in (1), then we can get

‖Muz‖BMOA = ‖uz‖BMOA = supa∈D

∥∥u(σa)σa – u(a)a

∥∥

2 = ‖z‖BMOA.

Moreover,

‖z‖BMOA = supa∈D

‖σa – a‖2 = supa∈D

√

1 – |a|2 = 1.

Hence,

supa∈D

∥∥u(σa)σa – u(a)a

∥∥

2 = 1. (3)

Let f (z) = 1 + z in (1), then

‖1 + z‖BMOA = 1 + supa∈D

∥∥1 + σa – (1 + a)

∥∥

2

= 1 + supa∈D

‖σa – a‖2 = 1 + 1 = 2.

By (1) again, we obtain

∥∥Mu(1 + z)

∥∥

BMOA

= ‖u + uz‖BMOA

=∣∣u(0)

∣∣ + sup

a∈D

∥∥u(σa) + u(σa)σa –

(

u(a) + u(a)a)∥∥

2

=∣∣u(0)

∣∣ + sup

a∈D

∥∥u(σa) – u(a) + u(σa)σa – u(a)a

∥∥

2.

Since ‖Mu(1 + z)‖BMOA = ‖1 + z‖BMOA, we get

∣∣u(0)

∣∣ + sup

a∈D

∥∥u(σa) – u(a) + u(σa)σa – u(a)a

∥∥

2 = 2. (4)

For simplicity, we denote

Fa(z) = u(

σa(z))

– u(a), Ga(z) = u(

σa(z))

σa(z) – u(a)a.

Now from (2)–(4), we have

supa∈D

‖Fa + Ga‖2 = 2 –∣∣u(0)

∣∣ = sup

a∈D‖Fa‖2 + sup

a∈D‖Ga‖2. (5)


Due to the properties of supremum, triangle inequality of norm, and (5), we have

supa∈D

‖Fa‖22 + sup

a∈D‖Ga‖2

2 + 2 supa∈D

‖Fa‖2 supa∈D

‖Ga‖2

= supa∈D

(‖Fa‖22 + ‖Ga‖2

2 + 〈Fa, Ga〉 + 〈Ga, Fa〉)

≤ supa∈D

(‖Fa‖22 + ‖Ga‖2

2)

+ supa∈D

(〈Fa, Ga〉 + 〈Ga, Fa〉)

≤ supa∈D

‖Fa‖22 + sup

a∈D‖Ga‖2

2 + supa∈D

(〈Fa, Ga〉 + 〈Ga, Fa〉)

≤ supa∈D

‖Fa‖22 + sup

a∈D‖Ga‖2

2 + 2 supa∈D

‖Fa‖2‖Ga‖2

≤ supa∈D

‖Fa‖22 + sup

a∈D‖Ga‖2

2 + 2 supa∈D

‖Fa‖2 supa∈D

‖Ga‖2.

So we obtain

supa∈D

(‖Fa‖22 + ‖Ga‖2

2)

= supa∈D

‖Fa‖22 + sup

a∈D‖Ga‖2

2

and

supa∈D

(〈Fa, Ga〉 + 〈Ga, Fa〉)

= 2 supa∈D

‖Fa‖2‖Ga‖2

= 2 supa∈D

‖Fa‖2 supa∈D

‖Ga‖2.(6)

Consider two cases. In the first, there exists a0 ∈ D such that

〈Fa0 , Ga0〉 = ‖Fa0‖2‖Ga0‖2 = supa∈D

‖Fa‖2 supa∈D

‖Ga‖2.

It follows that there exist λ ≥ 0 and a0 ∈ D such that Fa0 = λGa0 . If Fa0 = λGa0 , we canobtain that λ = 1 – |u(0)| from (2) and (3). There are two cases:

• If λ = 0, then u(z) is a constant function of modulus one;• If λ = 0,then

u(

σa0 (z))

– u(a0) = λ[

u(

σa0 (z))

σa0 (z) – u(a0)a0]

(7)

for every z ∈D. Let z = a0 in (7), then (1 – λa0)u(a0) = u(0). It is easy to see thatu(a0) = 0.– If a0 = 0, we have u(z) = u(0)

1–λz = c1–λz , where c = u(0) is a constant;

– If a0 = 0, then λ can be written as λ = u(a0)–u(0)u(a0)a0

. Letting z = σa0 (z) in (7), we obtain

u(z) – u(a0) = λ[

u(z)z – u(a0)a0]

,

that is,

u(z) = u(a0)1 – λa0

1 – λz=

u(a0)u(0)a0

u(a0)a0 – (u(a0) – u(0))z. (8)


If z = 0 in (8), then

u(0) = u(a0)(1 – λa0).

This shows that u(z) = u(0)1–λz .

Now consider the second case and suppose that there exists a sequence {an} in D con-verging to ξ0 ∈ ∂D such that

limn→∞〈Fan , Gan〉 = lim

n→∞‖Fan‖2‖Gan‖2 = 1 –∣∣u(0)

∣∣.

Since

〈Fa, Ga〉 =⟨

u(σa) – u(a), u(σa)σa – u(a)a⟩

=∫

∂D

∣∣u

(

σa(ξ ))∣∣2σa(ξ ) dm(ξ ) –

∣∣u(a)

∣∣2a,

we obtain

limn→∞〈Fan , Gan〉 = 0 = 1 –

∣∣u(0)

∣∣.

We get |u(0)| = 1 from the above equation, so u(z) is a constant function of modulusone. �

It is a question whether Mu is an isometry of BMOA space when u has the form inProposition 2. We have the following result.

Theorem 3 Let u(z) = u(0)1–λz , where λ = 1 – |u(0)|. Then Mu is an isometry of BMOA space

only if |u(0)| = 1. Therefore, Mu is an isometry of BMOA space if and only if u(z) is a con-stant function of modulus 1.

Proof Let u(z) = u(0)1–λz , then

‖Mu1‖BMOA =∣∣u(0)

∣∣ + sup

a∈D

∥∥∥∥

11 – λσa(z)

–1

1 – λa

∥∥∥∥

2= ‖z‖BMOA = 1.

By calculation, we have

∣∣u(0)

∣∣ + sup

a∈D

∥∥∥∥

11 – λσa(z)

–1

1 – λa

∥∥∥∥

2=

∣∣u(0)

∣∣

(

1 +√

λ2

1 – λ2 supa∈D

√

1 – |a|2|1 – λa|2

)

.

We want to get the extremum value of the function f (a) = 1–|a|2|1–λa|2 for a ∈ D. Suppose a =

x + iy, then

f (a) =1 – x2 – y2

(1 – λx)2 + (λy)2 .

Letting fx = 0, fy = 0, by calculation and analysis, we obtain that x = λ, y = 0, which meansthat a = λ. Let λ = 1 – |u(0)| in |u(0)|(1 + λ

1–λ2 ) = 1. Through calculation, we can get |u(0)| =1, which means u(z) is a constant function of modulus 1.


Therefore, we get that Mu is an isometry of BMOA space if and only if u is a constantfunction of modulus 1 from Proposition 2. �

3 Isometric weighted composition operatorsIn [14], the authors characterized the isometric composition operators of BMOA, they gotthe following results:

Lemma 4 ([14, Proposition 2.1]) A composition operator Cϕ is an isometry of BMOA ifand only if ϕ(0) = 0 and ‖Cϕ f ‖∗ = ‖f ‖∗ for all f ∈ BMOA.

Lemma 5 ([14, Theorem 3.1]) The following conditions are equivalent:• ‖Cϕ f ‖∗ = ‖f ‖∗ for all f ∈ BMOA.• ‖σw ◦ ϕ‖∗ = 1 for all w ∈D.• The map ϕ satisfies the following property: for every w ∈D, there is a sequence (an) inD such that ϕ(an) → w and ‖ϕan‖2 → 1, as n → ∞.

A weighted composition operator can be considered as a composition of multiplica-tion and composition operators. We want to know the relationship between isometricweighted composition operator, composition operator, and multiplication operator.

In this section we will study the properties of isometric weighted composition operatorof BMOA space. We assume that ϕ(z) is an analytic self-map of D and continuous in D.

Lemma 6 If Mu,ϕ is an isometry of BMOA space, then ‖u‖BMOA = 1.

Proof If Mu,ϕ is an isometry, then ‖Mu,ϕ f ‖BMOA = ‖f ‖BMOA for every f ∈ BMOA. Lettingf = 1 in ‖Mu,ϕ f ‖BMOA = ‖f ‖BMOA, we can get the result. �

Proposition 7 If Mu,ϕ is an isometry of BMOA space, then the functions u(z) and ϕ(z) haveone of the following properties:

• The function u(z) is a constant function of modulus 1 and Cϕ is an isometry of BMOA;• There exist a real number λ ∈ [0, 1] and a constant c such that u(z) = c

1–λϕ(z) , wherec = u(0)(1 – λϕ(0)) and 1 – |u(0)| = λ(1 – |u(0)ϕ(0)|).

Proof Suppose Mu,ϕ is an isometry of BMOA space, then ‖Mu,ϕ f ‖BMOA = ‖f ‖BMOA forevery f ∈ BMOA, that is,

∣∣u(0)f

(

ϕ(0))∣∣ + sup

a∈D

∥∥u(σa)f

(

ϕ(σa))

– u(a)f(

ϕ(a))∥∥

2

=∣∣f (0)

∣∣ + sup

a∈D

∥∥f (σa) – f (a)

∥∥

2.(9)

Letting f = 1 in (9), we get

∣∣u(0)

∣∣ + sup

a∈D

∥∥u(σa) – u(a)

∥∥

2 = 1. (10)

Letting f = z in (9), we obtain

∣∣u(0)ϕ(0)

∣∣ + sup

a∈D

∥∥u(σa)ϕ(σa) – u(a)ϕ(a)

∥∥

2 = 1. (11)


Letting f = 1 + z in (9) yields

∣∣u(0) + u(0)ϕ(0)

∣∣ + sup

a∈D

∥∥u(σa) – u(a) + u(σa)ϕ(σa) – u(a)ϕ(a)

∥∥

2 = 2.

From (9)–(11), we have

∣∣u(0) + u(0)ϕ(0)

∣∣ + sup

a∈D


∥∥

2

=∣∣u(0)

∣∣ +

∣∣u(0)ϕ(0)

∣∣ + sup

a∈D

∥∥u(σa) – u(a)

∥∥

2 + supa∈D


∥∥

2.

Due to the properties of supremum and triangle inequality, we can get

∣∣u(0)

∣∣∣∣1 + ϕ(0)

∣∣ =

∣∣u(0)

∣∣(

1 +∣∣ϕ(0)

∣∣)

and

supa∈D


∥∥

2

= supa∈D

(∥∥u(σa) – u(a)

∥∥

2 +∥∥u(σa)ϕ(σa) – u(a)ϕ(a)

∥∥

2

)

= supa∈D

∥∥u(σa) – u(a)

∥∥

2 + supa∈D


∥∥

2.

In the same way as in the proof of Proposition 2, we can get the existence of 0 ≤ λ ≤ 1 anda0 ∈D such that

u(

σa0 (z))

– u(a0) = λ(

u(

σa0 (z))

ϕ(

σa0 (z))

– u(a0)ϕ(a0))

,

or that of a sequence {an} in D converging to ξ0 ∈ ∂D such that

limn→∞

⟨

u(σan ) – u(an), u(σan )ϕ(σan ) – u(an)ϕ(an)⟩

= limn→∞

∥∥u(σan ) – u(an)

∥∥

2

∥∥u(σan )ϕ(σan ) – u(an)ϕ(an)

∥∥

2

=(

1 –∣∣u(0)

∣∣)(

1 –∣∣u(0)ϕ(0)

∣∣)

.

First, we assume u(0) = 0. Letting z = σa0 (z) in u(σa0 (z)) – u(a0) = λ(u(σa0 (z))ϕ(σa0 (z)) –u(a0)ϕ(a0)), we have

u(z)(

1 – λϕ(z))

= u(a0)(

1 – λϕ(a0))

.

• If λ = 0, then u(z) is a constant function of modulus one and Cϕ is an isometry onBMOA. The weighted composition operator Mu,ϕ is constructed by an isometricmultiplication operator and an isometric composition operator;

• If λ = 0, then u(z)(1 – λϕ(z)) = u(a0)(1 – λϕ(a0)), that is,

u(z) = u(a0)1 – λϕ(a0)1 – λϕ(z)

.


Letting z = 0, we have u(a0)(1 – λϕ(a0)) = u(0)(1 – λϕ(0)). That is,

u(z) =c

1 – λϕ(z).

Here c = u(0)(1 – λϕ(0)).In the other hand, if

limn→∞

⟨

u(σan ) – u(an), u(σan )ϕ(σan ) – u(an)ϕ(an)⟩

=(

1 –∣∣u(0)

∣∣)(

1 –∣∣u(0)ϕ(0)

∣∣)

,

then we can obtain that |u(0)| = 1 or |u(0)ϕ(0)| = 1.• If |u(0)| = 1, then u(z) is a constant function of modulus one and the composition

operator Cϕ is an isometry;• If |u(0)| = 1 and |u(0)ϕ(0)| = 1, we get a contradiction since |u(0)| is less than 1 by (10)

and ϕ(z) is an analytic self-map of D;• If |u(0)| = 1 and |u(0)ϕ(0)| = 1, we get a contraction since ϕ(z) is an analytic self-map

of D.Secondly, let u(0) = 0. Arguing as we did in the case u(0) = 0, we have:• If u(σa0 (z)) – u(a0) = λ(u(σa0 (z))ϕ(σa0 (z)) – u(a0)ϕ(a0)), then u and ϕ have properties

similar to those in the case u(0) = 0;• If limn→∞〈u(σan ) – u(an), u(σan )ϕ(σan ) – u(an)ϕ(an)〉 = (1 – |u(0)|)(1 – |u(0)ϕ(0)|), we

get a contraction since the left-hand side of the equation is 0, while the right-hand sideis not. �

In the following we consider whether Mu,ϕ is an isometry of BMOA space when u(z) hasthe form c

1–λϕ(z) .

Theorem 8 Let u(z) = c1–λϕ(z) , where c = u(0)(1 –λϕ(0)), then Mu,ϕ is an isometry of BMOA

only if u(z) is a constant function of modulus 1. Therefore, Mu,ϕ is an isometry of BMOA ifand only if both Mu and Cϕ are isometries of BMOA.

Proof If u(z) = c1–λϕ(z) , let f (z) = 1 – λz, then

Mu,ϕ f (z) = c = u(0)(

1 – λϕ(0))

.

Since ‖Mu,ϕ f ‖BMOA = ‖f ‖BMOA and ‖1 – λz‖BMOA = 1 + λ, we get

∣∣u(0)

(

1 – λϕ(0))∣∣ = 1 + λ.

By calculation and analysis, we have |u(0)| = 1, which means that u(z) is a constant functionof modulus 1.

Therefore, we obtain that if Mu,ϕ is an isometry of BMOA, then Mu is an isometry ofBMOA. Then Mu,ϕ is an isometry of BMOA if and only if both Mu and Cϕ are isometriesof BMOA. �

AcknowledgementsThe author would like to thank the editor and referees for their careful reading and valuable suggestions to make thearticle reader-friendly.


FundingThe author is partially supported by the National Natural Science Foundation of China (11701056, 11271388, 11426046);Natural Science Foundation of Chongqing (cstc2018jcyjAx0595); Chongqing Key Laboratory of Social Economy andApplied Statistics.

Competing interestsThe author declares that he has no competing interests.

Authors’ contributionsLigang Geng performed the research and wrote the paper. Author read and approved the final manuscript.

Publisher’s NoteSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 24 November 2020 Accepted: 31 August 2021

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