Isometric multiplication operators and weighted composition operators of BMOA

Abstract

Let u be an analytic function in the unit disk $\mathbb{D}$ and φ be an analytic self-map of $\mathbb{D}$. We give characterizations of the symbols u and φ for which the multiplication operator $M_{u}$ and the weighted composition operator $M_{u,\varphi }$ are isometries of BMOA.

1 Introduction

In order to state our results, we first introduce some notations and definitions. Let $\mathbb{D}$ denote the open unit disk in the complex plane $\mathbb{C}$, that is, $\mathbb{D}=\{z:|z|<1,z\in \mathbb{C}\}$. Its boundary is the unit circle $\partial \mathbb{D}=\{z:|z|=1,z\in \mathbb{C}\}$. A lot of analytic function spaces over $\mathbb{D}$ have been introduced in last half-century, such as Hardy, Bergman, Besov, Bloch, and Dirichlet spaces.

Let $H^{2}$ denote the classical analytic Hardy space on the unit disk $\mathbb{D}$. We will consider the multiplication operator $M_{u}: f\rightarrow uf$ of BMOA space, which consists analytic functions $f\in H^{2}$ whose boundary functions have bounded mean oscillation on $\partial \mathbb{D}$. In this manuscript we suppose that u is analytic in $\mathbb{D}$ and continuous on $\partial \mathbb{D}$. The norm of the BMOA space is defined as

$$\begin{aligned} \Vert f \Vert _{\mathrm{BMOA}}= \bigl\vert f(0) \bigr\vert + \Vert f \Vert _{*}, \end{aligned}$$

where

$$\begin{aligned} \Vert f \Vert _{*}= \sup_{a\in \mathbb{D}} \bigl\Vert f( \sigma _{a})-f(a) \bigr\Vert _{2}. \end{aligned}$$

Here $\|\cdot \|_{2}$ is the norm of $H^{2}(\mathbb{D})$ and $\sigma _{a}$ is the conformal automorphism of $\mathbb{D}$, $\sigma _{a}(z)=\frac{a-z}{1-\bar{a}z}$ for $a,z\in \mathbb{D}$. There are some equivalent norms of the BMOA space. For example, for a measurable complex-valued function f defined on $\partial \mathbb{D}$, define

$$\begin{aligned} f_{I}=\frac{1}{ \vert I \vert } \int _{I}f\bigl(e^{it}\bigr)\,dt, \end{aligned}$$

where $I\subset \partial \mathbb{D}$ is an arc and $|I|$ is the length of I, normalized so that $|I|\leq 1$. The function f has bounded mean oscillation if the seminorm

$$\begin{aligned} \Vert f \Vert '_{*}= \sup_{I} \frac{1}{ \vert I \vert } \int _{I} \bigl\vert f\bigl(e^{it} \bigr)-f_{I} \bigr\vert \,dt< \infty, \end{aligned}$$

as I ranges over all arcs in $\partial \mathbb{D}$, then $\mathrm{BMOA}=\{f\in H^{1}:\|f\|'_{*}<\infty \}$, 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 $\mathbb{D}$, such as Hardy, Bergman, and Bloch spaces. In [4], the author studied the invertible and Fredholm multiplication operators on Bergman spaces. In [5], the authors studied the compact multiplication operators with BMO symbols. In [6], the authors studied the $C^{*}$-algebra in the commutant of the multiplication operators with the symbol on the finite-dimensional Bergman spaces. In [7], the authors studied the commutants and reducing subspaces of multiplication operators on the 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 the linear isometries of spaces of continuous functions and some spaces of analytic functions, such as Hardy and Bergman spaces, in the unit disk $\mathbb{D}$ have the forms of weighted composition operators. There are some studies of the properties of isometric and weighted composition operators on spaces of analytic functions on $\mathbb{D}$ and other domains; these can be seen in [14–17]. In [18], the authors studied isometric pointwise multipliers of weighted Bergman, Bloch, and Dirichlet-type spaces in the unit disk. They obtained that the isometric pointwise multipliers of these spaces are unimodular constants.

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

2 Isometric multiplication operators

The Hardy space $H^{2}(\mathbb{D})$ consists of all analytic functions on $\mathbb{D}$ whose power series have square summable coefficients. For every $f\in H^{2}$, the radial limit $f(\zeta )=\lim_{r\rightarrow 1^{-}}f(r\zeta )$ exists almost everywhere on $\partial \mathbb{D,}$ and a standard norm on $H^{2}$ is given by

$$\begin{aligned} \Vert f \Vert _{2}=\biggl( \int _{\partial \mathbb{D}} \bigl\vert f(\zeta ) \bigr\vert ^{2} \,dm(\zeta )\biggr)^{ \frac{1}{2}}. \end{aligned}$$

Here dm is the Lebesgue measure on $\partial \mathbb{D}$ normalized so that $m(\partial \mathbb{D})=1$. Also, the norm on $H^{2}$ can be represented as

$$\begin{aligned} \Vert f \Vert _{2}=\sup_{0< r< 1}\biggl( \frac{1}{2\pi } \int _{0}^{2\pi } \bigl\vert f\bigl(re^{i \theta } \bigr) \bigr\vert ^{2}\,d\theta \biggr)^{\frac{1}{2}}. \end{aligned}$$

Let X and Y be metric spaces with metrics $d_{X}$ and $d_{Y}$. A map $T: X\rightarrow Y$ is called an isometry if for any $x_{1},x_{2}\in X$ one has

$$\begin{aligned} d_{Y}(Tx_{1},Tx_{2})=d_{X}(x_{1},x_{2}). \end{aligned}$$

Given two normed vector spaces X and Y, a linear isometry is a linear map $T: X\rightarrow Y$ that preserves the norms,

$$\begin{aligned} \Vert Tx \Vert _{Y}= \Vert x \Vert _{X}, \end{aligned}$$

for all $x\in X$.

If $M_{u}$ is an isometric operator of a Banach space $X\neq 0$, then it is also a bounded operator of X. Hence, if $M_{u}$ is an isometry of BMOA then u must be an $H^{\infty }$ function with $H^{\infty }$-norm smaller than or equal to 1. This can be seen in [18–20].

We will consider the question when the multiplication operator $M_{u}$ is an isometry of BMOA space. We start with the following proposition.

Proposition 1

If $M_{u}$ is an isometry of BMOA space, then $u\in \mathrm{BMOA}$ and $\|u\|_{\mathrm{BMOA}}=1$.

Proof

Suppose $M_{u}$ is an isometry of BMOA, then

$$\begin{aligned} \Vert M_{u}f \Vert _{\mathrm{BMOA}}= \Vert uf \Vert _{\mathrm{BMOA}}= \Vert f \Vert _{\mathrm{BMOA}} \end{aligned}$$

for every $f\in \mathrm{BMOA}$. That is,

$$\begin{aligned} \bigl\vert u(0)f(0) \bigr\vert +\sup_{a\in \mathbb{D}} \bigl\Vert u( \sigma _{a})f(\sigma _{a})-u(a)f(a) \bigr\Vert _{2}= \bigl\vert f(0) \bigr\vert +\sup_{a\in \mathbb{D}} \bigl\Vert f(\sigma _{a})-f(a) \bigr\Vert _{2} . \end{aligned}$$

(1)

Letting $f=1$ in (1), we get

$$\begin{aligned} \Vert u \Vert _{\mathrm{BMOA}}= \bigl\vert u(0) \bigr\vert +\sup _{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})-u(a) \bigr\Vert _{2}= \Vert 1 \Vert _{\mathrm{BMOA}}=1 . \end{aligned}$$

(2)

 □

It is easy to see that $\|u\|_{\mathrm{BMOA}}=1$ is a necessary but not sufficient condition for $M_{u}$ to be an isometry of BMOA space. The following is an example.

Example 1

Let $u=z$, then $\|z\|_{\mathrm{BMOA}}=1$, but $M_{z}$ is not an isometry of BMOA, since

$$\begin{aligned} \bigl\Vert \sigma _{a}(z) \bigr\Vert _{\mathrm{BMOA}}= \bigl\vert \sigma _{a}(0) \bigr\vert +\sup_{b\in \mathbb{D}} \bigl\Vert \sigma _{a}\bigl(\sigma _{b}(z)\bigr)-\sigma _{a}(b) \bigr\Vert _{2}=1+ \vert a \vert \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \bigl\Vert M_{z} \sigma _{a}(z) \bigr\Vert _{\mathrm{BMOA}}&= \bigl\vert 0\cdot \sigma _{a}(0) \bigr\vert +\sup_{b\in \mathbb{D}} \bigl\Vert \sigma _{b}(z)\sigma _{a}\bigl(\sigma _{b}(z)\bigr)-b \cdot \sigma _{a}(b) \bigr\Vert _{2} \\ &=\sup_{b\in \mathbb{D}}\sqrt{1- \vert b \vert ^{2} \bigl\vert \sigma _{a}(b) \bigr\vert ^{2}}= 1 \end{aligned} \end{aligned}$$

for $a\neq 0$.

Now we describe the isometric multiplication operators of BMOA space in the following.

Proposition 2

Suppose $M_{u}$ is an isometry of BMOA, then the function u has one of the following properties:

• (1) The function $u(z)$ is a constant function of modulus 1;

• (2) There exists a real number $\lambda \in [0,1]$ such that $u(z)=\frac{c}{1-\lambda z}$. Here, $c=u(0)$, $\lambda =1-|u(0)|$.

Proof

Suppose $M_{u}$ is an isometry of BMOA, let $f(z)=z$ in (1), then we can get

$$\begin{aligned} \Vert M_{u}z \Vert _{\mathrm{BMOA}}= \Vert uz \Vert _{\mathrm{BMOA}}=\sup_{a\in \mathbb{D}} \bigl\Vert u( \sigma _{a})\sigma _{a}-u(a)a \bigr\Vert _{2}= \Vert z \Vert _{\mathrm{BMOA}}. \end{aligned}$$

Moreover,

$$\begin{aligned} \Vert z \Vert _{\mathrm{BMOA}}=\sup_{a\in \mathbb{D}} \Vert \sigma _{a}-a \Vert _{2}= \sup_{a\in \mathbb{D}}\sqrt{1- \vert a \vert ^{2}}=1. \end{aligned}$$

Hence,

$$\begin{aligned} \sup_{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})\sigma _{a}-u(a)a \bigr\Vert _{2}=1 . \end{aligned}$$

(3)

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

$$\begin{aligned} \begin{aligned} \Vert 1+z \Vert _{\mathrm{BMOA}}&=1+\sup _{a\in \mathbb{D}} \bigl\Vert 1+ \sigma _{a}-(1+a) \bigr\Vert _{2} \\ &=1+\sup_{a\in \mathbb{D}} \Vert \sigma _{a}-a \Vert _{2}=1+1=2. \end{aligned} \end{aligned}$$

By (1) again, we obtain

$$\begin{aligned} \begin{aligned}& \bigl\Vert M_{u}(1+z) \bigr\Vert _{\mathrm{BMOA}}\\ &\quad= \Vert u+uz \Vert _{\mathrm{BMOA}} \\ &\quad= \bigl\vert u(0) \bigr\vert +\sup_{a\in \mathbb{D}} \bigl\Vert u( \sigma _{a})+u(\sigma _{a}) \sigma _{a}- \bigl(u(a)+u(a)a\bigr) \bigr\Vert _{2} \\ &\quad= \bigl\vert u(0) \bigr\vert +\sup_{a\in \mathbb{D}} \bigl\Vert u( \sigma _{a})-u(a)+u( \sigma _{a})\sigma _{a}-u(a)a \bigr\Vert _{2}. \end{aligned} \end{aligned}$$

Since $\|M_{u}(1+z)\|_{\mathrm{BMOA}} = \|1+z\|_{\mathrm{BMOA}}$, we get

$$\begin{aligned} \bigl\vert u(0) \bigr\vert +\sup_{a\in \mathbb{D}} \bigl\Vert u( \sigma _{a})-u(a)+u(\sigma _{a}) \sigma _{a}-u(a)a \bigr\Vert _{2}=2 . \end{aligned}$$

(4)

For simplicity, we denote

$$\begin{aligned} F_{a}(z)=u\bigl(\sigma _{a}(z)\bigr)-u(a), G_{a}(z)= u\bigl(\sigma _{a}(z)\bigr) \sigma _{a}(z)-u(a)a. \end{aligned}$$

Now from (2)–(4), we have

$$\begin{aligned} \sup_{a\in \mathbb{D}} \Vert F_{a}+G_{a} \Vert _{2}=2- \bigl\vert u(0) \bigr\vert =\sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}+\sup_{a\in \mathbb{D}} \Vert G_{a} \Vert _{2}. \end{aligned}$$

(5)

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

$$\begin{aligned} \begin{aligned} &\sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}^{2}+\sup_{a \in \mathbb{D}} \Vert G_{a} \Vert _{2}^{2}+2\sup _{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}\sup _{a\in \mathbb{D}} \Vert G_{a} \Vert _{2} \\ &\quad=\sup_{a\in \mathbb{D}}\bigl( \Vert F_{a} \Vert _{2}^{2}+ \Vert G_{a} \Vert _{2}^{2}+ \langle F_{a},G_{a}\rangle + \langle G_{a},F_{a}\rangle \bigr) \\ &\quad\leq \sup_{a\in \mathbb{D}}\bigl( \Vert F_{a} \Vert _{2}^{2}+ \Vert G_{a} \Vert _{2}^{2}\bigr)+ \sup_{a\in \mathbb{D}}\bigl(\langle F_{a},G_{a}\rangle +\langle G_{a},F_{a} \rangle \bigr) \\ &\quad\leq \sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}^{2}+\sup_{a \in \mathbb{D}} \Vert G_{a} \Vert _{2}^{2}+\sup_{a\in \mathbb{D}} \bigl( \langle F_{a},G_{a}\rangle +\langle G_{a},F_{a}\rangle \bigr) \\ &\quad\leq \sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}^{2}+\sup_{a \in \mathbb{D}} \Vert G_{a} \Vert _{2}^{2}+2\sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2} \Vert G_{a} \Vert _{2} \\ &\quad\leq \sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}^{2}+\sup_{a \in \mathbb{D}} \Vert G_{a} \Vert _{2}^{2}+2\sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}\sup_{a\in \mathbb{D}} \Vert G_{a} \Vert _{2}. \end{aligned} \end{aligned}$$

So we obtain

$$\begin{aligned} \sup_{a\in \mathbb{D}}\bigl( \Vert F_{a} \Vert _{2}^{2}+ \Vert G_{a} \Vert _{2}^{2}\bigr)= \sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}^{2}+\sup_{a\in \mathbb{D}} \Vert G_{a} \Vert _{2}^{2} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \sup_{a\in \mathbb{D}}\bigl(\langle F_{a},G_{a}\rangle +\langle G_{a},F_{a} \rangle \bigr)&=2\sup_{a\in \mathbb{D}} \Vert F_{a} \Vert _{2} \Vert G_{a} \Vert _{2}\\ &=2 \sup _{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}\sup _{a\in \mathbb{D}} \Vert G_{a} \Vert _{2}. \end{aligned} \end{aligned}$$

(6)

Consider two cases. In the first, there exists $a_{0}\in \mathbb{D}$ such that

$$\begin{aligned} \langle F_{a_{0}},G_{a_{0}}\rangle = \Vert F_{a_{0}} \Vert _{2} \Vert G_{a_{0}} \Vert _{2}= \sup _{a\in \mathbb{D}} \Vert F_{a} \Vert _{2}\sup _{a\in \mathbb{D}} \Vert G_{a} \Vert _{2}. \end{aligned}$$

It follows that there exist $\lambda \geq 0$ and $a_{0}\in \mathbb{D}$ such that $F_{a_{0}}=\lambda G_{a_{0}}$. If $F_{a_{0}}=\lambda G_{a_{0}}$, we can obtain that $\lambda =1-|u(0)|$ from (2) and (3). There are two cases:

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

• If $\lambda \neq 0$,then

  $$\begin{aligned} u\bigl(\sigma _{a_{0}}(z)\bigr)-u(a_{0})=\lambda \bigl[u\bigl( \sigma _{a_{0}}(z)\bigr)\sigma _{a_{0}}(z)-u(a_{0})a_{0} \bigr] \end{aligned}$$

  (7)

  for every $z\in \mathbb{D}$. Let $z=a_{0}$ in (7), then $(1-\lambda a_{0})u(a_{0})=u(0)$. It is easy to see that $u(a_{0})\neq 0$.

  • If $a_{0}=0$, we have $u(z)=\frac{u(0)}{1-\lambda z}=\frac{c}{1-\lambda z}$, where $c=u(0)$ is a constant;

  • If $a_{0}\neq 0$, then λ can be written as $\lambda =\frac{u(a_{0})-u(0)}{u(a_{0})a_{0}}$. Letting $z=\sigma _{a_{0}}(z)$ in (7), we obtain

    $$\begin{aligned} u(z)-u(a_{0})=\lambda \bigl[u(z)z-u(a_{0})a_{0} \bigr], \end{aligned}$$

    that is,

    $$\begin{aligned} u(z)=u(a_{0})\frac{1-\lambda a_{0}}{1-\lambda z}= \frac{u(a_{0})u(0)a_{0}}{u(a_{0})a_{0}-(u(a_{0})-u(0))z}. \vadjust{\goodbreak} \end{aligned}$$

    (8)

If $z=0$ in (8), then

$$\begin{aligned} u(0)= u(a_{0}) (1-\lambda a_{0}). \end{aligned}$$

This shows that $u(z)=\frac{u(0)}{1-\lambda z}$.

Now consider the second case and suppose that there exists a sequence $\{a_{n}\}$ in $\mathbb{D}$ converging to $\xi _{0}\in \partial \mathbb{D}$ such that

$$\begin{aligned} \lim_{n\rightarrow \infty } \langle F_{a_{n}},G_{a_{n}}\rangle = \lim_{n \rightarrow \infty } \Vert F_{a_{n}} \Vert _{2} \Vert G_{a_{n}} \Vert _{2} =1- \bigl\vert u(0) \bigr\vert . \end{aligned}$$

Since

$$\begin{aligned} \begin{aligned} \langle F_{a},G_{a}\rangle &= \bigl\langle u(\sigma _{a})-u(a),u(\sigma _{a}) \sigma _{a}-u(a)a\bigr\rangle \\ &= \int _{\partial \mathbb{D}} \bigl\vert u\bigl(\sigma _{a}(\xi ) \bigr) \bigr\vert ^{2} \overline{\sigma _{a}(\xi )}\,dm(\xi )- \bigl\vert u(a) \bigr\vert ^{2}\bar{a}, \end{aligned} \end{aligned}$$

we obtain

$$\begin{aligned} \lim_{n\rightarrow \infty } \langle F_{a_{n}},G_{a_{n}}\rangle =0=1- \bigl\vert u(0) \bigr\vert . \end{aligned}$$

We get $|u(0)|=1$ from the above equation, so $u(z)$ is a constant function of modulus one. □

It is a question whether $M_{u}$ is an isometry of BMOA space when u has the form in Proposition 2. We have the following result.

Theorem 3

Let $u(z)=\frac{u(0)}{1-\lambda z}$, where $\lambda =1-|u(0)|$. Then $M_{u}$ is an isometry of BMOA space only if $|u(0)|=1$. Therefore, $M_{u}$ is an isometry of BMOA space if and only if $u(z)$ is a constant function of modulus 1.

Proof

Let $u(z)=\frac{u(0)}{1-\lambda z}$, then

$$\begin{aligned} \Vert M_{u} 1 \Vert _{\mathrm{BMOA}}= \bigl\vert u(0) \bigr\vert +\sup_{a\in \mathbb{D}} \biggl\Vert \frac{1}{1-\lambda \sigma _{a}(z)}- \frac{1}{1-\lambda a} \biggr\Vert _{2}= \Vert z \Vert _{\mathrm{BMOA}}=1. \end{aligned}$$

By calculation, we have

$$\begin{aligned} \bigl\vert u(0) \bigr\vert +\sup_{a\in \mathbb{D}} \biggl\Vert \frac{1}{1-\lambda \sigma _{a}(z)}-\frac{1}{1-\lambda a} \biggr\Vert _{2}= \bigl\vert u(0) \bigr\vert \biggl(1+ \sqrt{\frac{\lambda ^{2}}{1-\lambda ^{2}}}\sup_{a\in \mathbb{D}} \sqrt{ \frac{1- \vert a \vert ^{2}}{ \vert 1-\lambda a \vert ^{2}}}\biggr). \end{aligned}$$

We want to get the extremum value of the function $f(a)= \frac{1-|a|^{2}}{|1-\lambda a|^{2}}$ for $a\in \mathbb{D}$. Suppose $a=x+iy$, then

$$\begin{aligned} f(a)=\frac{1-x^{2}-y^{2}}{(1-\lambda x)^{2}+(\lambda y)^{2}}. \end{aligned}$$

Letting $f_{x}=0, f_{y}=0$, by calculation and analysis, we obtain that $x=\lambda, y=0$, which means that $a=\lambda$. Let $\lambda =1-|u(0)|$ in $|u(0)|(1+\frac{\lambda }{1-\lambda ^{2}})=1$. Through calculation, we can get $|u(0)|=1$, which means $u(z)$ is a constant function of modulus 1.

Therefore, we get that $M_{u}$ is an isometry of BMOA space if and only if u is a constant function of modulus 1 from Proposition 2. □

3 Isometric weighted composition operators

In [14], the authors characterized the isometric composition operators of BMOA, they got the following results:

Lemma 4

([14, Proposition 2.1])

A composition operator $C_{\varphi }$ is an isometry of BMOA if and only if $\varphi (0)=0$ and $\|C_{\varphi }f\|_{*}=\|f\|_{*}$ for all $f\in \mathrm{BMOA}$.

Lemma 5

([14, Theorem 3.1])

The following conditions are equivalent:

• $\|C_{\varphi }f\|_{*}=\|f\|_{*}$ for all $f\in \mathrm{BMOA}$.

• $\|\sigma _{w}\circ \varphi \|_{*}=1$ for all $w\in \mathbb{D}$.

• The map φ satisfies the following property: for every $w\in \mathbb{D}$, there is a sequence $(a_{n})$ in $\mathbb{D}$ such that $\varphi (a_{n})\rightarrow w$ and $\|\varphi _{a_{n}}\|_{2}\rightarrow 1$, as $n\rightarrow \infty$.

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

In this section we will study the properties of isometric weighted composition operator of BMOA space. We assume that $\varphi (z)$ is an analytic self-map of $\mathbb{D}$ and continuous in $\overline{\mathbb{D}}$.

Lemma 6

If $M_{u,\varphi }$ is an isometry of BMOA space, then $\|u\|_{\mathrm{BMOA}}=1$.

Proof

If $M_{u,\varphi }$ is an isometry, then $\|M_{u,\varphi }f\|_{\mathrm{BMOA}}=\|f\|_{\mathrm{BMOA}}$ for every $f\in \mathrm{BMOA}$. Letting $f=1$ in $\|M_{u,\varphi }f\|_{\mathrm{BMOA}}=\|f\|_{\mathrm{BMOA}}$, we can get the result. □

Proposition 7

If $M_{u,\varphi }$ is an isometry of BMOA space, then the functions $u(z)$ and $\varphi (z)$ have one of the following properties:

• The function $u(z)$ is a constant function of modulus 1 and $C_{\varphi }$ is an isometry of BMOA;

• There exist a real number $\lambda \in [0,1]$ and a constant c such that $u(z)=\frac{c}{1-\lambda \varphi (z)}$, where $c=u(0)(1-\lambda \varphi (0))$ and $1-|u(0)|=\lambda (1-|u(0)\varphi (0)|)$.

Proof

Suppose $M_{u,\varphi }$ is an isometry of BMOA space, then $\|M_{u,\varphi }f\|_{\mathrm{BMOA}}=\|f\|_{\mathrm{BMOA}}$ for every $f\in \mathrm{BMOA}$, that is,

$$\begin{aligned} \begin{aligned} & \bigl\vert u(0)f\bigl(\varphi (0)\bigr) \bigr\vert +\sup_{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})f \bigl( \varphi (\sigma _{a})\bigr)-u(a)f\bigl(\varphi (a)\bigr) \bigr\Vert _{2} \\ &\quad= \bigl\vert f(0) \bigr\vert +\sup_{a\in \mathbb{D}} \bigl\Vert f( \sigma _{a})-f(a) \bigr\Vert _{2} . \end{aligned} \end{aligned}$$

(9)

Letting $f=1$ in (9), we get

$$\begin{aligned} \begin{aligned} \bigl\vert u(0) \bigr\vert +\sup _{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})-u(a) \bigr\Vert _{2}=1. \end{aligned} \end{aligned}$$

(10)

Letting $f=z$ in (9), we obtain

$$\begin{aligned} \begin{aligned} \bigl\vert u(0)\varphi (0) \bigr\vert +\sup _{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a}) \varphi (\sigma _{a})-u(a)\varphi (a) \bigr\Vert _{2}=1 . \end{aligned} \end{aligned}$$

(11)

Letting $f=1+z$ in (9) yields

$$\begin{aligned} \bigl\vert u(0)+u(0)\varphi (0) \bigr\vert +\sup_{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})-u(a)+ u(\sigma _{a})\varphi ( \sigma _{a})-u(a)\varphi (a) \bigr\Vert _{2}=2. \end{aligned}$$

From (9)–(11), we have

$$\begin{aligned} \begin{aligned} & \bigl\vert u(0)+u(0)\varphi (0) \bigr\vert + \sup_{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})-u(a)+ u(\sigma _{a})\varphi (\sigma _{a})-u(a)\varphi (a) \bigr\Vert _{2} \\ &\quad= \bigl\vert u(0) \bigr\vert + \bigl\vert u(0)\varphi (0) \bigr\vert + \sup_{a\in \mathbb{D}} \bigl\Vert u( \sigma _{a})-u(a) \bigr\Vert _{2} +\sup_{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a}) \varphi (\sigma _{a})-u(a)\varphi (a) \bigr\Vert _{2}. \end{aligned} \end{aligned}$$

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

$$\begin{aligned} \bigl\vert u(0) \bigr\vert \bigl\vert 1+\varphi (0) \bigr\vert = \bigl\vert u(0) \bigr\vert \bigl(1+ \bigl\vert \varphi (0) \bigr\vert \bigr) \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} &\sup_{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})-u(a)+ u(\sigma _{a}) \varphi (\sigma _{a})-u(a)\varphi (a) \bigr\Vert _{2} \\ &\quad=\sup_{a\in \mathbb{D}}\bigl( \bigl\Vert u(\sigma _{a})-u(a) \bigr\Vert _{2}+ \bigl\Vert u( \sigma _{a})\varphi ( \sigma _{a})-u(a)\varphi (a) \bigr\Vert _{2}\bigr) \\ &\quad=\sup_{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})-u(a) \bigr\Vert _{2}+ \sup_{a\in \mathbb{D}} \bigl\Vert u(\sigma _{a})\varphi (\sigma _{a})-u(a) \varphi (a) \bigr\Vert _{2}. \end{aligned} \end{aligned}$$

In the same way as in the proof of Proposition 2, we can get the existence of $0\leq \lambda \leq 1$ and $a_{0}\in \mathbb{D}$ such that

$$\begin{aligned} u\bigl(\sigma _{a_{0}}(z)\bigr)-u(a_{0})=\lambda \bigl(u\bigl( \sigma _{a_{0}}(z)\bigr)\varphi \bigl( \sigma _{a_{0}}(z) \bigr)-u(a_{0})\varphi (a_{0})\bigr), \end{aligned}$$

or that of a sequence $\{a_{n}\}$ in $\mathbb{D}$ converging to $\xi _{0}\in \partial \mathbb{D}$ such that

$$\begin{aligned} \begin{aligned} &\lim_{n\rightarrow \infty }\bigl\langle u(\sigma _{a_{n}})-u(a_{n}), u(\sigma _{a_{n}})\varphi (\sigma _{a_{n}})-u(a_{n})\varphi (a_{n}) \bigr\rangle \\ &\quad=\lim_{n\rightarrow \infty } \bigl\Vert u(\sigma _{a_{n}})-u(a_{n}) \bigr\Vert _{2} \bigl\Vert u(\sigma _{a_{n}})\varphi (\sigma _{a_{n}})-u(a_{n}) \varphi (a_{n}) \bigr\Vert _{2} \\ &\quad=\bigl(1- \bigl\vert u(0) \bigr\vert \bigr) \bigl(1- \bigl\vert u(0) \varphi (0) \bigr\vert \bigr). \end{aligned} \end{aligned}$$

First, we assume $u(0)\neq 0$. Letting $z=\sigma _{a_{0}}(z)$ in $u(\sigma _{a_{0}}(z))-u(a_{0})=\lambda (u(\sigma _{a_{0}}(z)) \varphi (\sigma _{a_{0}}(z))-u(a_{0})\varphi (a_{0}))$, we have

$$\begin{aligned} u(z) \bigl(1-\lambda \varphi (z)\bigr)=u(a_{0}) \bigl(1-\lambda \varphi (a_{0})\bigr). \end{aligned}$$

• If $\lambda =0$, then $u(z)$ is a constant function of modulus one and $C_{\varphi }$ is an isometry on BMOA. The weighted composition operator $M_{u,\varphi }$ is constructed by an isometric multiplication operator and an isometric composition operator;

• If $\lambda \neq 0$, then $u(z)(1-\lambda \varphi (z))=u(a_{0})(1-\lambda \varphi (a_{0}))$, that is,

  $$\begin{aligned} u(z)=u(a_{0})\frac{1-\lambda \varphi (a_{0})}{1-\lambda \varphi (z)}. \end{aligned}$$

  Letting $z=0$, we have $u(a_{0})(1-\lambda \varphi (a_{0}))=u(0)(1-\lambda \varphi (0))$. That is,

  $$\begin{aligned} u(z)=\frac{c}{1-\lambda \varphi (z)}. \end{aligned}$$

  Here $c=u(0)(1-\lambda \varphi (0))$.

In the other hand, if

$$\begin{aligned} \lim_{n\rightarrow \infty }\bigl\langle u(\sigma _{a_{n}})-u(a_{n}), u(\sigma _{a_{n}})\varphi (\sigma _{a_{n}})-u(a_{n}) \varphi (a_{n}) \bigr\rangle =\bigl(1- \bigl\vert u(0) \bigr\vert \bigr) \bigl(1- \bigl\vert u(0)\varphi (0) \bigr\vert \bigr), \end{aligned}$$

then we can obtain that $|u(0)|=1$ or $|u(0)\varphi (0)|=1$.

• If $|u(0)|=1$, then $u(z)$ is a constant function of modulus one and the composition operator $C_{\varphi }$ is an isometry;

• If $|u(0)|\neq 1$ and $|u(0)\varphi (0)|=1$, we get a contradiction since $|u(0)|$ is less than 1 by (10) and $\varphi (z)$ is an analytic self-map of $\mathbb{D}$;

• If $|u(0)|=1$ and $|u(0)\varphi (0)|=1$, we get a contraction since $\varphi (z)$ is an analytic self-map of $\mathbb{D}$.

Secondly, let $u(0)=0$. Arguing as we did in the case $u(0)\neq 0$, we have:

• If $u(\sigma _{a_{0}}(z))-u(a_{0})=\lambda (u(\sigma _{a_{0}}(z)) \varphi (\sigma _{a_{0}}(z))-u(a_{0})\varphi (a_{0}))$, then u and φ have properties similar to those in the case $u(0)\neq 0$;

• If $\lim_{n\rightarrow \infty }\langle u(\sigma _{a_{n}})-u(a_{n}), u(\sigma _{a_{n}})\varphi (\sigma _{a_{n}})-u(a_{n})\varphi (a_{n}) \rangle =(1-|u(0)|)(1-|u(0)\varphi (0)|)$, we get a contraction since the left-hand side of the equation is 0, while the right-hand side is not.

 □

In the following we consider whether $M_{u,\varphi }$ is an isometry of BMOA space when $u(z)$ has the form $\frac{c}{1-\lambda \varphi (z)}$.

Theorem 8

Let $u(z)=\frac{c}{1-\lambda \varphi (z)}$, where $c=u(0)(1-\lambda \varphi (0))$, then $M_{u,\varphi }$ is an isometry of BMOA only if $u(z)$ is a constant function of modulus 1. Therefore, $M_{u,\varphi }$ is an isometry of BMOA if and only if both $M_{u}$ and $C_{\varphi }$ are isometries of BMOA.

Proof

If $u(z)=\frac{c}{1-\lambda \varphi (z)}$, let $f(z)=1-\lambda z$, then

$$\begin{aligned} M_{u,\varphi }f(z)=c= u(0) \bigl(1-\lambda \varphi (0)\bigr). \end{aligned}$$

Since $\| M_{u,\varphi }f \|_{\mathrm{BMOA}}=\|f\|_{\mathrm{BMOA}}$ and $\|1-\lambda z\|_{\mathrm{BMOA}}=1+\lambda$, we get

$$\begin{aligned} \bigl\vert u(0) \bigl(1-\lambda \varphi (0)\bigr) \bigr\vert =1+\lambda. \end{aligned}$$

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

Therefore, we obtain that if $M_{u,\varphi }$ is an isometry of BMOA, then $M_{u}$ is an isometry of BMOA. Then $M_{u,\varphi }$ is an isometry of BMOA if and only if both $M_{u}$ and $C_{\varphi }$ are isometries of BMOA. □