Some questions about complex harmonic functions

Abstract

In this paper, we propose composition products in the class of complex harmonic functions so that the composition of two such functions is again a complex harmonic function. From here, we begin the study of the iterations of the functions of this class showing briefly its potential to be a topic of future research. In parallel, we define and study composition operators on a Hardy type space denoted by $HH^{2}(\mathbb {D})$ of complex harmonic functions also introduced for us in the present work. The symbols of these composition operators have of form $\chi +\overline{\pi }$ where $\chi ,\pi$ are analytic functions from $\mathbb {D}$ into $\mathbb {D}$. We also analyze the space of bounded linear operators on $HH^{2}(\mathbb {D})$.

1 Introduction

Unlike what happens in the class of analytical functions which is closed under the customary composition, the usual composition product of two harmonic functions is not in general a harmonic function. This fact causes that some problems which are studied for a long time in the space of analytical functions do not make sense or are difficult to translate and treat on the set of complex harmonic functions with the tools of the complex variable. We give two typical examples: the theory of linear composition operator whose symbols are complex harmonic functions and the corresponding theory of iterations for complex harmonic functions.

In this note, we will introduce two composition products in the space complex harmonic function in order to overcome these difficulties. From these products, in this work we begin the study of the problems before mentioned in this class of complex harmonic functions.

A continuous function $f=u+iv$ defined in a domain $D\subset \mathbb {C}$ is harmonic in D if u and v are real harmonic functions in D which are not necessarily conjugate. If D is simply connected domain we can write $f=h+\overline{g}$, where h and g are analytic functions on D and $\overline{g}$ denotes the function $z\longrightarrow \overline{g(z)}$. In this paper, we should only consider complex harmonic functions $f=h+\overline{g}$ for which $g\ne 0$. From now on, all plane domain D must be considered simply connected.

It is clear that we can consider f as a function of two variable z and $\overline{z}$, in this case $\partial _{z}f(z)=\partial _{z}h(z)$ and $\partial _{\overline{z}}f(z)=\overline{\partial _{z}g(z)}=\partial _{\overline{z}}\overline{g}(\overline{z})$. Let us denote by H(D) the set of all complex harmonic functions defined on D. Obviously $f\in H(D)$ if and only if $\partial ^{2}_{z\overline{z}}f=0$. Observe also that if for $f=h+\overline{g}\in H(D)$ one has $g=0$, then f is analytic, and u and v satisfy the Cauchy-Riemann equations. We can denote by A(D) the set of all analytic functions on D. For $f=h+\overline{g}\in H(D)$ we say that h is its analytic part and g will be called the co-analytic part. In the opinion of the authors the basic reference for the study of complex harmonic functions is [7]. The reader can also consult the articles [5, 9, 11] on the matter.

The theory of iterations of a function f of the complex variable z studies the sequence the ‘iterates’ $\{f_{n}(z)\}$ defined by

$$\begin{aligned} f_{0}(z)=z,\qquad f_{1}(z)=f(z),\qquad f_{n+1}(z)=f_{1}(f_{n}(z)),\qquad n=0,1,2,\ldots . \end{aligned}$$

In the general theory of iterations, the fixed points of f play a fundamental role both in local theories and others which deal with the behavior of the sequence $\{f_{n}(z)\}$ and the solutions of some functional equations in the neighbourhood of fixed points. In this work, we begin the theory of iteration of complex harmonic functions. The classical theory on the dynamic of complex functions can be found in the references: [2, 4, 12, 16].

In another direction, for two given functions f and g, the composition of them $f\circ g$ has played a main role in several areas of the mathematics. In particular, in the functional analysis and the operator theory led to the study of composition operators. Being it a research topic since the late 60 s of the last century (see [15] for more detail). On the other hand, an essential objective of the theory of linear composition operators is to obtain information of the operators from the properties of the functional class on which the operators act.

We recall that the composition operators on Hardy spaces are fundamental in the Berkson-Porta representation theorem of semigroups on $H^{p}(\mathbb {D})$ (see [3]). Also, it is well known the fundamental role of the composition operators and the conjugation techniques initially arisen in the theory of iterations of analytical functions A(D), research which we wish to undertake in a forthcoming paper to the class H(D). This program also aims to insert, in this context, the theory of integrable systems.

Recently, the theory of complex harmonic functions has found applications in fluid mechanics. For instance, using complex harmonic maps A. Aleman and A. Constantin provided an approach towards obtaining explicit solutions to the incompressible two-dimensional Euler equations [1]. We recommend also to see the paper [6], where the authors obtain a complete solution to the problem of classifying all two-dimensional ideal fluid flows with harmonic Lagrangian labelling maps.

2 Dynamic of complex harmonic functions and beyond

2.1 Two composition products for complex harmonic functions

We introduce a first composition law between harmonic functions for which the function $z+\overline{z}\in H(\mathbb {C})$ should be the identity function.

Definition 1

Let $f_{1}=h_{1}+\overline{g_{1}}$ and $f_{2}=h_{2}+\overline{g_{2}}$ be two functions of $H(D_{1})$ and $H(D_{2})$ respectively, such that $h_{2}(D_{2})\subset D_{1}$ and $g_{2}(D_{2})\subset D_{1}$. We define a harmonic composition product of $f_{1}$ with $f_{2}$, which is denoted by $f_{1}\circleddash f_{2}$, in the following form

$$\begin{aligned} f_{1}\circleddash f_{2}(z)=h_{1}\circ h_{2}(z)+\overline{g_{1}\circ g_{2}(z)},\qquad \text {for all} \qquad z\in D_{2}, \end{aligned}$$

(1)

where “$\circ$” denotes the usual composition between analytic functions. The composition defined by (1) will be called direct harmonic composition.

Let $H_{s}(D)$ be the subset of H(D) consists of whose functions $f=h+\overline{g}$ for which $h:D\longrightarrow D$ and $g:D\longrightarrow D$. For example, if $I_{0}(z)=z+\overline{z}$, then $I_0\in H_{s}(D)$. Although for a $f\in H_{s}(D)$ given, we can not guarantee in general that $f(D)\subset D$, we call to the elements of $H_{s}(D)$ harmonic $\circleddash$-automorphisms of D with respect to the composition $\circleddash$.

Our first result is the following

Proposition 2

$(H_{s}(D),\circleddash )$ is a monoid and the function $I_0$ is its identity.

Proof

The proof follows taking into account that the usual composition for analytic functions is an associative operation. On other hand, it is trivial to verify that $I_0$ is the identity with respect to $\circleddash$. $\square$

It is known that the parts of a complex harmonic function may not be univalent even though the function itself is. In the following, that class of functions will be excluded from our analysis. This leads to the following definition.

Definition 3

In this paper, a univalent harmonic function should be a function $f=h+\overline{g}\in H_{s}(D)$ for which both h and g are univalent analytic functions (see [8, 10]). Then $f^{-1}=h^{-1}+\overline{g^{-1}}\in H_{s}(D)$ and $f\circleddash f^{-1}=I_0=f^{-1}\circleddash f$.

We introduce a second composition product between harmonic maps.

Definition 4

Let $f_{1}=h_{1}+\overline{g_{1}}$ and $f_{2}=h_{2}+\overline{g_{2}}$ be two functions of $H(D_{1})$ and $H(D_{2})$ respectively, such that $h_{2}(D_{2})\subset D_{1}$ and $g_{2}(D_{2})\subset D_{1}$. We define another harmonic composition $f_{1}\circledcirc f_{2}$ between the two functions $f_{1}$ and $f_{2}$ in the form

$$\begin{aligned} f_{1}\circledcirc f_{2}(z) =h_{1}\circ g_{2}(z)+\overline{g_{1}\circ h_{2}(z)}, \qquad \text {for all} \qquad z\in D_{2}, \end{aligned}$$

(2)

where, as before, “$\circ$” denotes the usual composition of analytic functions. Next, this composition is called the crossed harmonic composition. Here, the function $I_0$ is the identity.

Notice that the composition $\circledcirc$ is not associative.

Example 1

A Möbius harmonic transformation is a harmonic function of the form

$$\begin{aligned} R(z)=\frac{az+b}{cz+d}+\overline{\frac{lz+n}{sz+t}}=T_{A}(z)+\overline{T_{B}(z)}, \end{aligned}$$

where

$$\begin{aligned} A=\left( \begin{array}{cc} a &{} b \\ c &{} d \\ \end{array} \right) , \quad B=\left( \begin{array}{cc} l &{} n \\ s &{} t \\ \end{array} \right) \end{aligned}$$

are matrices of $GL_{2}(\mathbb {C})$. Let $R_{1}(z)=T_{A_{1}}(z)+\overline{T_{B_{1}}(z)}$ and $R_{2}(z)=T_{A_{2}}(z)+\overline{T_{B_{2}}(z)}$ be two Möbius harmonic transformations, then

$$\begin{aligned} R_{1}(z)\circleddash R_{2}(z)=T_{A_{1}}(z)\circ T_{A_{2}}(z)+\overline{T_{B_{1}}(z)\circ T_{B_{2}}(z)}=T_{A_{1}A_{2}}(z)+\overline{T_{B_{1}B_{2}}(z)}, \end{aligned}$$

(3)

and

$$\begin{aligned} R_{1}(z)\circledcirc R_{2}(z)=T_{A_{1}}(z)\circ T_{B_{2}}(z)+\overline{T_{B_{1}}(z)\circ T_{A_{2}}(z)}=T_{A_{1}B_{2}}(z)+\overline{T_{B_{1}A_{2}}(z)}. \end{aligned}$$

(4)

In what follows, if $f=h+\overline{g}$ then the function $\overline{f}=g+\overline{h}$ will be called the conjugated complex harmonic function of f. Let $f=h+\overline{g}\in H_{s}(D)$ be given, we define $f^{n,\circleddash }=f\circleddash f^{n-1,\circleddash }$ and $f^{n,\circledcirc }=f^{n-1,\circledcirc }\circledcirc f$; for $n\ge 2$, where $f^{1,\circleddash }=f^{1,\circledcirc }=f$ and $f^{0,\circleddash }(z)=z+\overline{z}$. Then

$$\begin{aligned}{} & {} f^{k,\circleddash }(z)=h^{k}(z)+\overline{g^{k}(z)},\nonumber \\{} & {} f^{k,\circledcirc }(z)=h(g^{k-1}(z))+\overline{g(h^{k-1}(z))} =(h+\overline{g})\circledcirc f^{k-1,\circleddash }(z)=f\circledcirc f^{k-1,\circleddash }(z),\nonumber \\ \end{aligned}$$

(5)

for all $z\in D$ and $1\le k$. We represent $h^{k}=h\circ h^{k-1}$ and the same for $g^{k}$, for all k. Also, here and throughout the section $\overline{D}=\{\overline{z}\;:\;z\in D\}$.

Suppose that $f=h+\overline{g}\in H_{s}(D)$, we say that the sequence $\{f^{n,\circleddash }(z)\}$ is convergent in the point $z\in D$, if by definition, there exits a complex number $\mu +\overline{\omega }$ such that $f^{n,\circleddash }(z)\longrightarrow \mu +\overline{\omega }$.

Example 2

Assume that $f=h+\overline{g}\in H_{s}(D)$ is such that $\{h^{n}(z)\}$ converges to a finite point $\mu$ and $\{g^{n}(z)\}$ also converges to a finite point $\omega$ for some $z\in D$, clearly $\mu , \omega \in cl(D)$, where cl(D) stands for the closure of D. It is well known that $\mu$ and $\omega$ should be fixed points of h and g, respectively. Then, the sequence $\{f^{n,\circleddash }(z)\}$ converges to $\mu +\overline{\omega }$, while $\{f^{n,\circledcirc }(z)\}$ is convergent to $h(\omega )+\overline{g(\mu )}$.

We give the following useful general result

Proposition 5

Let us suppose that $f=h+\overline{g}\in H_{s}(D)$, then we have

• if $z\in D$ and the sequence $\{f^{n,\circleddash }(z)\}$ is convergent to $\mu +\overline{\omega }$, then $\mu +\overline{\omega }=h(\mu )+\overline{g(\omega )}$,

• if the sequence $\{f^{n,\circleddash }(z)\}$ is convergent to $\mu +\overline{\omega }$ in $z\in D$, then $\{f^{n,\circledcirc }(z)\}$ converges to $h(\omega )+\overline{g(\mu )}$ in z. Moreover, if f is univalent then $\{f^{n,\circledcirc }(z)\}$ converges in $z\in D$ if and only if $\{f^{n, \circleddash }(z)\}$ converges in this point.

Proof

From the first equality of (5) it follows that $f^{n+1,\circleddash }(z)=f\circleddash f^{n,\circleddash }(z)$ for all $n\in \mathbb {N}$, then passing to the limit on both sides, we obtain $\mu +\overline{\omega }=f\circleddash (\mu +\overline{\omega })=h(\mu )+\overline{g(\omega )}$ which proves the first statement. Also, for the second equality of (5) one has $f^{n,\circledcirc }(z)=f\circledcirc f^{n-1,\circleddash }(z)$ for all $n\ge 1$ and for our assumptions $\{f^{n,\circleddash }(z)\}$ is convergent to $\mu +\overline{\omega }$ which implies that $\{f^{n,\circledcirc }(z)\}$ converges to $(h+\overline{g})\circledcirc (\mu +\overline{\omega })= h(\omega )+\overline{g(\mu )}$. Finally, observe that if $f=h+\overline{g}$ is univalent, then further we have $(h^{-1}+\overline{g^{-1}})\circleddash f^{n,\circledcirc }(z)=(\overline{f})^{n,\circleddash }(z)$. It shows that $\{f^{n,\circledcirc }(z)\}$ converges if and only if $\{f^{n, \circleddash }(z)\}$ converges. $\square$

Definition 6

The complex $\zeta$ is said to be a finite $\mathfrak {h}$-fixed point for the complex harmonic function $f=h+\overline{g}\in H_{s}(D)$ if $\zeta =\mu +\overline{\omega }$ and satisfies the equation $\mu +\overline{\omega }=h(\mu )+\overline{g(\omega )}$.

Suppose that $f=h+\overline{g}\in H_{s}(D)$ such that $h(\mu )=\mu$ and $g(\omega )=\omega$, then $\mu +\overline{\omega }=h(\mu )+\overline{g(\omega )}$. Thus $\zeta =\mu +\overline{\omega }$ is a $\mathfrak {h}$-fixed point of f. These types of points will be called induced $\mathfrak {h}$-fixed points and they constitute probably isolated $\mathfrak {h}$-fixed points. In particular, if $g(0)=0$ then all the usual fixed points of h are $\mathfrak {h}$-fixed points of $f=h+\overline{g}$. In the same way, if $h(0)=0$ then each fixed point of g is $\mathfrak {h}$-fixed point of $f=h+\overline{g}$. It shows that 0 is a $\mathfrak {h}$-fixed point of $f=h+\overline{g}$ whenever $h(0)=g(0)=0$.

In the case $D=\mathbb {C}$ is reasonable to include $\zeta =\infty$ in the analysis of the $\mathfrak {h}$-fixed points of a complex harmonic functions. We say that

• 1. $\mu +\overline{\infty }$ is an infinite $\mathfrak {h}$-fixed point of $f=h+\overline{g}$ if $h(\mu )=\mu$ and $g(\infty )=\infty$,

• 2. $\infty +\overline{\omega }$ is an infinite $\mathfrak {h}$-fixed point of $f=h+\overline{g}$ if $h(\infty )=\infty$ and $g(\omega )=\omega$,

• 3. $\infty +\overline{\infty }$ is an infinite $\mathfrak {h}$-fixed point of $f=h+\overline{g}$ if $h(\infty )=\infty$ and $g(\infty )=\infty$.

For example, for a Möbius harmonic function $R(z)=T_{A}(z)+\overline{T_{B}(z)}$, we can have up to 4 $\mathfrak {h}$-fixed points. There are several possibilities which could be showed and analyzed from the point of view of the convergence of the sequence $\{R^{n,\circleddash }(z)\}$.

• The Möbius functions $T_{A}(z)$ and $T_{A}(z)$ have a single fixed point:

  • Both Möbius functions $T_{A}(z)$ and $T_{B}(z)$ have $\infty$ as their only fixed point, that is, $\infty +\overline{\infty }$ is the unique $\mathfrak {h}$-fixed point. Then, $R(z)=z+\overline{z}$ or $R(z)=R_{\beta }(z)=z+\beta +\overline{z}$ where $\beta \ne 0$. Hence, $R^{n,\circleddash }_{\beta }(z)\longrightarrow \infty +\overline{z}=\infty$ for all $z\in \mathbb {C}$,

  • The sets of fixed points of $T_{A}(z)$ and $T_{B}(z)$ are $FP_{A}=\{\mu \}$ and $FP_{B}=\{\infty \}$ respectively. In this case, one has $R^{n,\circleddash }(z)\longrightarrow \mu +\overline{z}$ or $R^{n,\circleddash }(z)\longrightarrow \mu +\overline{\infty }=\infty$ for all $z\in \mathbb {C}$,

  • $FP_{A}=\{\infty \}$ and $FP_{B}=\{\omega \}$, then $R^{n,\circleddash }(z)\longrightarrow z+\overline{\omega }$ or $R^{n,\circleddash }(z)\longrightarrow \infty +\overline{\omega }=\infty$, for all $z\in \mathbb {C}$,

  • For $FP_{A}=\{\mu \}$ and $FP_{B}=\{\omega \}$, we find that $R^{n,\circleddash }(z)\longrightarrow \mu +\overline{\omega }$, for all $z\in \mathbb {C}$.

• In the remaining cases, $R^{n,\circleddash }(z)$ converges for all $z\in \mathbb {C}$ if and only if $\infty$ does not belong to $FP_{A}\cup FP_{B}$ and there are $\mu \in FP_{A}$ and $\omega \in FP_{B}$ such that $h^{n}(z)\longrightarrow \mu$ and $g^{n}(z)\longrightarrow \omega$.

For the remainder of this section, we will only work with the composition product $\circleddash$.

Theorem 7

Let $\mu +\overline{\omega }\in D$ be a $\mathfrak {h}$-fixed point of $f=h+\overline{g}\in H(D)$. Assume that there exist a neighborhood $V\subset D$ of $\mu +\overline{\omega }$ and two positive constants $\rho _{h}<1$ and $\rho _{g}<1$ such that $h(V)\subset V$, $g(V)\subset V$ and the following inequalities

$$\begin{aligned} |h(z)-h(\mu )|<\rho _{h}|z-h(\mu )|, \end{aligned}$$

(6)

$$\begin{aligned} |g(z)-g(\omega )|<\rho _{g}|z-g(\omega )|, \end{aligned}$$

(7)

hold for all $z\in V$. Then sequence $\{f^{n,\circleddash }\}_{n\ge 1}$ converges uniformly to $\mu +\overline{\omega }$ in V. In this case, we say that $\mu +\overline{\omega }$ is an attracting $\mathfrak {h}$-fixed point.

Proof

From (6) and (7) follow that for all n with $2\le n$ we obtain

$$\begin{aligned} |h^{n}(z)-h(\mu )|<(\rho _{h})^{n}|z-h(\mu )|,\,\,\,\,\,\hbox {for all } z\in V, \end{aligned}$$

(8)

and

$$\begin{aligned} |g^{n}(z)-g(\omega )|<(\rho _{g})^{n}|z-g(\omega )|,\,\,\,\,\,\,\,\,\hbox {for any }z\in V. \end{aligned}$$

(9)

Then using (8) and (9), we have

$$\begin{aligned} |f^{n,\circleddash }(z)-(\mu +\overline{\omega })|&=|(h^{n}(z)-h(\mu ))+\overline{(g^{n}(z)-g(\omega ))}| \\&<|h^{n}(z)-h(\mu )|+|g^{n}(z)-g(\omega )| \\&<(\rho _{h})^{n}|z-h(\mu )|+(\rho _{g})^{n}|z-g(\omega )|, \end{aligned}$$

consequently, the sequence of harmonic iterates $\{f^{n,\circleddash }\}$ converges uniformly to $\mu +\overline{\omega }$ in V. $\square$

Example 3

Assume now that $U=\{z\in \mathbb {C}: |z|<1\}$ and let us choose arbitrary $\alpha$ and $\beta$ in U. Then $z_{0}=0$ is a attracting $\mathfrak {h}$-fixed point of $f(z)=\alpha z+\beta \overline{z}$.

Now, we present a generalization of a result due to Koenigs (see [4], page 31). Let $\mu +\overline{\omega }$ be a $\mathfrak {h}$-fixed point of $f=h+\overline{g}\in H_{s}(D)$ then $\lambda =\partial _{z} h(\mu )$ and $\theta =\partial _{z}g(\omega )$ are called the multipliers of f.

Theorem 8

Assume that $f=h+\overline{g}\in H_{s}(U)$ has a fixed point in $z_{0}=0$ (this means that $h(0)=0=g(0)$) with multipliers $\lambda$ and $\theta$ satisfying $0<|\lambda | <1$, $0<|\theta | <1$. Then there exists a neighborhood $V\subset U$ of zero and $\varphi =\varphi _{h}+\overline{\varphi _{g}} \in H(V)$ such that

$$\begin{aligned} \varphi \circleddash f=\lambda \varphi _{h}+\overline{\theta \varphi _{g}}=(\lambda z+\overline{\theta z})\circleddash \varphi , \end{aligned}$$

(10)

and $\varphi (0)=0$.

Proof

By a well known Theorem due to Koenigs (1884) there are two neighborhoods of zero $V_{h}\subset U$ and $V_{g}\subset U$ and two functions $\varphi _{h}$ and $\varphi _{g}$ which are analytic in $V_{h}$ and $V_{g}$ respectively, such that, $\varphi _{h}\circ h(z)=\lambda \varphi _{h}(z)$ for all $z\in V_{h}$ and $\varphi _{g}\circ g(z)=\theta \varphi _{g}(z)$ for any $z\in V_{g}$, where $\lambda =\partial _{z} h(0)$ and $\theta =\partial _{z}g(0)$. Observe that necessarily must be $\varphi _{h}(0)=0=\varphi _{g}(0)$. Let us define $V=V_{h}\cap V_{g}$ then $\varphi (z)=\varphi _{h}(z)+\overline{\varphi _{g}(z)}\in H(V)$ and satisfies (10). $\square$

If f satisfies the hypothesis of the previous Theorem we must add that $z_{0}=0$ is also an attracting $\mathfrak {h}$-fixed point since evidently the conditions (6) and (7) hold.

Proposition 9

Under the hypothesis of the Theorem 8 the sequence of iterates $\{f^{n,\circleddash }\}$ converges uniformly to $z_{0}=0$ on some neighborhood $V_{0}$.

Proof

It is clear that $z_{0}=0$ is an ordinary attracting fixed point of h and g. Hence, the iterates $h^{n}$ and $g^{n}$ constitute two sequences which converge uniformly to 0 on neighborhoods $V_{1}$ and $V_{2}$ respectively. Since, $f^{n,\circleddash }=h^{n}+\overline{g^{n}}$ it shows that the sequence if iterates $\{f^{n,\circleddash }\}$ converges uniformly to $z_{0}=0$ in $V_{0}=V_{1}\cap V_{2}$. $\square$

Let us assume now that $z_{0}=0$ is a $\mathfrak {h}$-fixed point of $f=h+\overline{g}\in H_{s}(U)$ such that $\lambda =0$ and $0<\theta <1$. We recall that $h(0)=0=g(0)$, and consider h(z) of the form

$$\begin{aligned} h(z)=a_{p}z^{p}+\cdots ,\,\,\,\,a_{p}\ne 0,\,\,p\ge 2, \end{aligned}$$

(11)

then by a well known result of Boettcher (1904) (the result can be found in [4], page 33) there exists a neighborhood of zero $V_{h}$ and an analytic function $\varphi _{h}$ over it such that

$$\begin{aligned} \varphi _{h}\circ h(z)=(\varphi (z))^{p},\,\,\,\,\hbox {for any }z\in V_{h}. \end{aligned}$$

(12)

Hence, we can enunciate the following result

Theorem 10

Assume that $f=h+\overline{g}\in H_{s}(U)$ has a $\mathfrak {h}$-fixed point in $z_{0}=0$ (that is, $h(0)=0=g(0)$) with multipliers $\lambda$ and $\theta$ satisfying $\lambda =0$, $0<|\theta | <1$. Suppose we are given h by (11), then there exists a neighborhood V of zero and $\varphi =\varphi _{h}+\overline{\varphi _{g}} \in H(V)$ such that

$$\begin{aligned} \varphi \circleddash f=(\varphi _{h})^{p}+\overline{\theta \varphi _{g}}=(z^{p}+\overline{\theta z})\circleddash \varphi , \end{aligned}$$

(13)

where $\varphi _{h}(0)=0$ or $\varphi _{h}(0)$ is a $(p-1)$ th root of unity.

Analogous results can be proved in the cases when $0<|\lambda |<1, \theta =0$ and $\lambda =0=\theta$.

2.2 Classification of Möbius harmonic functions

The classification of the usual Möbius functions is a classic subject with many applications, see [13]. Here, we propose to introduce the topic in the class of Möbius harmonic functions.

We directly start with a definition

Definition 11

The normal form of a Möbius harmonic functions $R(z)=\frac{az+b}{cz+d}+\overline{\frac{lz+n}{sz+t}}=T_{A}(z)+\overline{T_{B}(z)}$, is the Möbius harmonic functions

$$\begin{aligned} R_{n}(z)=T_{J_{A}}(z)+\overline{T_{J_{B}}(z)}, \end{aligned}$$

(14)

where $J_{A}$ and $J_{B}$ are the Jordan normal forms of the matrices A and B respectively.

Note that $R_{n}(z)$ can be obtained by means of the following conjugation

$$\begin{aligned} R_{n}(z)&=\left( T_{N}(z)+\overline{T_{M}(z)}\right) \,\ominus \,\left( T_{A}(z)+\overline{T_{B}(z)}\right) \,\ominus \,\left( T_{N^{-1}}(z)+\overline{T_{M-1}(z)}\right) \\ {}&=\left( T_{NAN^{-1}}(z)+\overline{T_{MBM^{-1}}(z)}\right) , \end{aligned}$$

where the matrices M and N are selected such that $NAN^{-1}=J_{A}$ and $MBM^{-1}=J_{B}$. Observe that $J_{A}$ and $J_{B}$ could be in one of the following ways

$$\begin{aligned} \left( \begin{array}{cc} \lambda &{} 1 \\ 0 &{} \lambda \\ \end{array} \right) ,\,\,\,\,\,\,\,\,\,\hbox {or}\,\,\,\,\,\,\,\,\,\left( \begin{array}{cc} \lambda &{} 0 \\ 0 &{} \mu \\ \end{array} \right) ,\,\,\,\,\,\,\hbox {with}\,\,\,\,\,\,\lambda \ne \mu . \end{aligned}$$

Therefore, $T_{J_{A}}$ and $T_{J_{B}}$ are essentially of the form :

$$\begin{aligned} z\longrightarrow z+\lambda ^{-1},\,\,\,\,\,\,\,\hbox {or}\,\,\,\,\,z\longrightarrow cz, \end{aligned}$$

where $c\ne 1$.

Remark 12

Note that

$$\begin{aligned} R^{m,\ominus }(z)&=\left( T_{N^{-1}}(z)+\overline{T_{M^{-1}}(z)}\right) \ominus \,\left( T_{J_{A}}(z)+\overline{T_{J_{B}}(z)}\right) ^{m,\ominus } \,\ominus \left( T_{N}(z)+\overline{T_{M}(z)}\right) \nonumber \\&=\left( T_{N^{-1}}(z)+\overline{T_{M^{-1}}(z)}\right) \ominus \, R_{n}^{m{,\ominus }}(z)\,\ominus \left( T_{N}(z)+\overline{T_{M}(z)}\right) \nonumber \\&=\left( T_{N^{-1}}(z)+\overline{T_{M^{-1}}(z)}\right) \ominus \,\left( T_{(J_{A})^{m}}(z)+\overline{T_{(J_{B})^{m}}(z)}\right) \,\ominus \left( T_{N}(z)+\overline{T_{M}(z)}\right) . \nonumber \\ \end{aligned}$$

(15)

We say that $R(z)=T_{A}(z)+\overline{T_{B}(z)}$ is strongly parabolic if

$$\begin{aligned} J_{A}=\left( \begin{array}{cc} \lambda _{a} &{} 1 \\ 0 &{} \lambda _{a} \\ \end{array} \right) ,\,\,\,\,\,\,\,\,\,\hbox {and}\,\,\,\,\,\,\,\,\,J_{B}=\left( \begin{array}{cc} \lambda _{b} &{} 1 \\ 0 &{} \lambda _{b} \\ \end{array} \right) , \end{aligned}$$

(16)

that is $T_{J_{A}}(z)=z+\lambda _{a}^{-1}$ and $T_{J_{B}}(z)=z+\lambda _{b}^{-1}$. Thus, $R_{n}(z)=z+\overline{z}+\lambda _{a}^{-1}+(\overline{\lambda _{b}})^{-1}$, so

$$\begin{aligned} z=-\overline{\left( \lambda _{a}^{-1}+ (\overline{\lambda _{b}})^{-1} \right) }, \end{aligned}$$

is the unique fixed point of $R_{n}(z)$. The Möbius harmonic function R(z) is said to be strongly non-parabolic if $R_{n}(z)=c_{a}z+\overline{c_{b}}\,\overline{z}$, thus $z=0$ is the fixed point of the normal form. Last, R(z) is called of mixed type if

$$\begin{aligned} T_{J_{A}}(z)=z+\lambda _{a}^{-1},\,\,\,\,\,\,\,\, \hbox {and}\,\,\,\,\,\,\,\,\,\,\, T_{J_{B}}(z)=c_{b}z, \end{aligned}$$

or

$$\begin{aligned} T_{J_{A}}(z)=c_{a}z,\,\,\,\,\,\,\,\, \hbox {and}\,\,\,\,\,\,\,\,\,\,\, T_{J_{B}}(z)=z+\lambda _{b}^{-1}, \end{aligned}$$

that is $R_{n}(z)$ has one of the following forms

$$\begin{aligned} R_{n}(z)=z+\lambda _{a}^{-1}+\overline{c_{b}}\,\overline{z},\,\,\,\,\,\,\, \hbox {or}\,\,\,\,\,\,\,\,R_{n}(z)=c_{a}z+\overline{z}+\lambda _{b}^{-1}, \end{aligned}$$

in the first case, $z=-\left( \overline{\lambda _{a}}c_{b}\right) ^{-1}$ is the fixed point of $R_{n}(z)$. The second harmonic function has a fixed point if and only if $c_{a}$ and $\lambda _{b}^{-1}$ are reals, being it equal to $x=\frac{-\lambda _{b}^{-1}}{c_{a}}$.

3 Composition operators on the space $HH^{2}(\mathbb {D})$

The main object of this section is to investigate the theory of composition operators in the framework of the complex harmonic functions. The section is organized as follows, first of all we introduce an analogue of Hardy space in the class of complex harmonic functions and give some results for the linear operators defined in this space, in particular we develop with some depth the theory of linear composition operators whose symbols are complex harmonic functions. We conclude the section showing the relationship of these composition operators with the corresponding complex harmonic reproducing kernels.

3.1 The Hardy type space $HH^{2}(\mathbb {D})$ and its corresponding space $\mathcal {B}(HH^{2}(\mathbb {D}))$

We recall that the separable Hilbert space $H^{2}(\mathbb {D})$ consists of all analytic functions having power series representations with square-summable complex coefficients, in other words

$$\begin{aligned} H^{2}(\mathbb {D})=\left\{ f:\mathbb {D}\rightarrow \mathbb {C}: f(z)=\sum _{n=0}^{\infty }a_{n}z^{n},\,\,\sum _{n=0}^{\infty }|a_{n}|^{2}< \infty \right\} . \end{aligned}$$

Any function of $H^{2}(\mathbb {D})$ is analytic in the open unit disc $\mathbb {D}$. The inner product on $H^{2}$ is defined as follows $\langle f, g\rangle =\sum _{n=0}^{\infty }a_{n}\overline{b_{n}}$; for $f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}$ and $g(z)=\sum _{n=0}^{\infty }b_{n}z^{n}$. Thus, the norm of the vector $f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}$ is $\Vert f\Vert =\left( \sum _{n=0}^{\infty }|a_{n}|^{2}\right) ^{\frac{1}{2}}$. Each analytic function $\phi$ mapping the unit disc into itself defines a composition operator $C_{\phi }: H^{2}(\mathbb {D})\longrightarrow H^{2}(\mathbb {D})$ of the form $(C_{\phi }f)(z)=f(\phi (z))$ for all $f\in H^{2}(\mathbb {D})$ and $C_{\phi }\in \mathcal {B}(H^{2}(\mathbb {D}))$ where $\mathcal {B}(H^{2}(\mathbb {D}))$ is the space of all bounded linear operators of $H^{2}(\mathbb {D})$ into $H^{2}(\mathbb {D})$. The function $\phi$ is called the symbol of $C_{\phi }$. Moreover, we recall that

$$\begin{aligned} \Vert C_{\phi }\Vert \le \sqrt{\frac{1+|\phi (0)|}{1-|\phi (0)}}. \end{aligned}$$

In particular, if $\phi (0)=0$, then $\Vert C_{\phi }\Vert =1$. See for instance [14, 15].

We use $HH^{2}(\mathbb {D})$ to denote the space of harmonic complex functions $h+\overline{g}$ where $h,g\in H^{2}(\mathbb {D})$. It is clear that $HH^{2}(\mathbb {D})$ is a vector spaces. Moreover, it is possible to introduced in $HH^{2}(\mathbb {D})$ an inner product, more exactly one has

Lemma 13

$HH^{2}(\mathbb {D})$ is a complex Hilbert space with respect to the following inner product

$$\begin{aligned} (a+\overline{b},c+\overline{d})_{HH^{2}(\mathbb {D})} =(a,c)_{H^{2}(\mathbb {D})}+\overline{(b,d)_{H^{2}(\mathbb {D})}}= (a,c)_{H^{2}(\mathbb {D})}+(d,b)_{H^{2}(\mathbb {D})},\nonumber \\ \end{aligned}$$

(17)

for all $a+\overline{b}, c+\overline{d}\in HH^{2}(\mathbb {D})$.

Proof

Indeed,

$$\begin{aligned} (a+\overline{b},a+\overline{b})_{HH^{2}(\mathbb {D})}=\parallel a \parallel ^{2}_{H^{2}(\mathbb {D})}+\parallel b \parallel ^{2}_{H^{2}(\mathbb {D})}\ge 0, \end{aligned}$$

and $(a+\overline{b},a+\overline{b})_{HH^{2}(\mathbb {D})}=0$ if and only if $\parallel a \parallel ^{2}_{H^{2}(\mathbb {D})}=0$ and $\parallel b \parallel ^{2}_{H^{2}(\mathbb {D})}=0$, that is if $a+\overline{b}=0$. Now

$$\begin{aligned}&(\alpha _{1}(a_{1}+\overline{b_{1}})+\alpha _{2}(a_{2}+\overline{b_{2}}),c+\overline{d})_{HH^{2}(\mathbb {D})}\\&\quad = ((\alpha _{1}a_{1}+\alpha _{2}a_{2})+\overline{(\overline{\alpha _{1}}b_{1}+\overline{\alpha _{2}}b_{2})},c+\overline{d})_{HH^{2}(\mathbb {D})} \\&\quad =((\alpha _{1}a_{1}+\alpha _{2}a_{2}),c)_{H^{2}(\mathbb {D})}+\overline{((\overline{\alpha _{1}}b_{1}+\overline{\alpha _{2}}b_{2}),d)_{H^{2}(\mathbb {D})}} \\ {}&\quad =\alpha _{1}\left[ (a_{1},c)_{H^{2}(\mathbb {D})}+\overline{(b_{1},d)_{H^{2}(\mathbb {D})}}\right] + \alpha _{2}\left[ (a_{2},c)_{H^{2}(\mathbb {D})}+\overline{(b_{2},d)_{H^{2}(\mathbb {D})}}\right] \\&\quad =\alpha _{1}((a_{1}+\overline{b_{1}}),(c+\overline{d}))_{HH^{2}(\mathbb {D})}+\alpha _{1}((a_{2}+\overline{b_{2}}),(c+\overline{d}))_{HH^{2}(\mathbb {D})}, \end{aligned}$$

for all $\alpha _{1},\alpha _{2}\in \mathbb {C}$ and any $(a_{1}+\overline{b_{1}}),(a_{2}+\overline{b_{2}}),(c+\overline{d})\in HH^{2}(\mathbb {D})$.

Also,

$$\begin{aligned} (a+\overline{b},c+\overline{d})_{HH^{2}(\mathbb {D})}{} & {} =(a,c)_{H^{2}(\mathbb {D})}+(d,b)_{H^{2}(\mathbb {D})} =\overline{(c,a)_{H^{2}(\mathbb {D})}} +\overline{(b,d)_{H^{2}(\mathbb {D})}}\\{} & {} =\overline{(c+\overline{d},a+\overline{b})_{HH^{2}(\mathbb {D})}}\,\,. \end{aligned}$$

Finally, if $\{a_{n}+\overline{b_{n}}\}_{n\ge 0}$ is a sequence of Cauchy in $HH^{2}(\mathbb {D})$, then $\{a_{n}\}_{n\ge 0}$ and $\{b_{n}\}_{n\ge 0}$ are of Cauchy in $H^{2}(\mathbb {D})$. It shows that the sequence $\{a_{n}+\overline{b_{n}}\}_{n\ge 0}$ is convergent, thus $HH^{2}(\mathbb {D})$ is complete. $\square$

Lemma 14

The functions $\{e_{n}, f_{n}\}_{n\ge 0}$, where $e_{n}=z^{n}$ and $f_{n}=\overline{z^{n}}$ for all $n\in \mathbb {N}$ form an orthonormal basis of $HH^{2}(\mathbb {D})$.

Proof

In fact

$$\begin{aligned} (e_{n},f_{m})_{HH^{2}(\mathbb {D})}=(z^{n},\overline{z^{m}})_{HH^{2}(\mathbb {D})}=(z^{n},0)_{H^{2}(\mathbb {D})}+\overline{(0,z^{m})_{H^{2}(\mathbb {D})}}=0, \end{aligned}$$

if $n,m\in \mathbb {N}\backslash \{0\}$. Now, $(e_{n},e_{m})_{HH^{2}(\mathbb {D})}=(z^{n},z^{m})_{H^{2}(\mathbb {D})}=\delta _{nm}$ and $(f_{n},f_{m})_{HH^{2}(\mathbb {D})}=\overline{(z^{n},z^{m})_{H^{2}(\mathbb {D})}}=\delta _{nm}$. On the other hand, being $a+\overline{b}=\sum a_{n}z^{n}+\overline{\sum b_{n}z^{n}}$ then $(a+\overline{b}, e_{n}(z))_{HH^{2}(\mathbb {D})}=(a,z^{n})_{H^{2}(\mathbb {D})}+\overline{(b,0)_{H^{2}(\mathbb {D})}}=a_{n}$, and $(a+\overline{b}, f_{n}(z))_{HH^{2}(\mathbb {D})}=(a,0)_{H^{2}(\mathbb {D})}+\overline{(b,z^{n})_{H^{2}(\mathbb {D})}}=\overline{b_{n}}$. Hence, if $a+\overline{b} \perp e_{n}(z)$ and $a+\overline{b} \perp f_{n}(z)$ for all $n\in \mathbb {N}$, it follows that $a=b=0$. $\square$

From now on, when there is no possibility of confusion, we will not use sub-indices to indicate to what space corresponds the internal product or the respective norm.

Let $H(\mathbb {D})$ be the space of complex harmonic functions defined on $\mathbb {D}$. Two compositions defined in previous Sect. 2 maintain the linearity on the left. More general composition product of $H(\mathbb {D})\times H(\mathbb {D})$ into $H(\mathbb {D})$ keeping the linearity on the left is the following

$$\begin{aligned} (a+\overline{b})\propto _{(\alpha ,\beta ,\gamma ,\delta )} (c+\overline{d})=\alpha (a\circ c)+\beta (\overline{b\circ d})+\gamma (a\circ d)+\delta (\overline{b\circ c}), \end{aligned}$$

(18)

where $\alpha ,\beta ,\gamma ,\delta \in \mathbb {C}$ are arbitrary but fixed. Note that $\propto _{(1,1,0,0)}=\circleddash$ and $\propto _{(1,0,0,1)}=\circledcirc$

Let us denote $AH(\mathbb {D})=\{\varphi +\overline{\pi }: \varphi ,\pi \in A(\mathbb {D},\mathbb {D})\}$, where $A(\mathbb {D},\mathbb {D})$ is the space of all analytic self-mappings on $\mathbb {D}$. Also, let us consider $\varphi +\overline{\pi }\in AH(\mathbb {D})$ and the composition operator $C^{(\alpha ,\beta ,\gamma ,\delta )}_{\varphi +\overline{\pi }}:\,\,HH^{2}(\mathbb {D})\longrightarrow HH^{2}(\mathbb {D})$ defined by $C^{(\alpha ,\beta ,\gamma ,\delta )}_{\varphi +\overline{\pi }}(a+\overline{b})=(a+\overline{b})\propto _{(\alpha ,\beta ,\gamma ,\delta )} (\varphi +\overline{\pi })=\alpha (a\circ \varphi )+\beta (\overline{b\circ \pi })+ \gamma (a\circ \pi )+\delta (\overline{b\circ \varphi })$, if $a+\overline{b}\in HH^{2}(\mathbb {D})$.

Remark 15

The action of $C^{(\alpha ,\beta ,\gamma ,\delta )}_{\varphi +\overline{\pi }}$ can be written in the form

$$\begin{aligned} C^{(\alpha ,\beta ,\gamma ,\delta )}_{\varphi +\overline{\pi }}(a+\overline{b})&=\alpha C_{\varphi }a+\beta \overline{C_{\pi }b}+\gamma C_{\pi }a +\delta \overline{C_{\varphi }b}\nonumber \\&= \left[ \alpha C_{\varphi }a+\gamma C_{\pi }a \right] +\left[ \,\, \overline{\overline{\beta } C_{\pi }b+\overline{\delta } C_{\varphi }b}\,\,\right] \nonumber \\ {}&=A^{(\alpha ,\beta ,\gamma ,\delta )}a+ \overline{B^{(\alpha ,\beta ,\gamma ,\delta )}b}, \end{aligned}$$

(19)

where $A^{(\alpha ,\beta ,\gamma ,\delta )}=\alpha C_{\varphi }+\gamma C_{\pi }$ and $B^{(\alpha ,\beta ,\gamma ,\delta )}=\overline{\beta }C_{\pi }+\overline{\delta }C_{\varphi }$.

The next result establishes that the given operator are bounded.

Lemma 16

The operators $A^{(\alpha ,\beta ,\gamma ,\delta )}$ and $B^{(\alpha ,\beta ,\gamma ,\delta )}$ defined in the previous remark are bounded linear operators on $H^{2}(\mathbb {D})$.

Proof

The linearity of $A^{(\alpha ,\beta ,\gamma ,\delta )}$ and $B^{(\alpha ,\beta ,\gamma ,\delta )}$ follows from the fact that both are linear combinations of usual bounded linear composition operators whose symbols are analytic functions mapping the unit disk into itself. For the same reason $A^{(\alpha ,\beta ,\gamma ,\delta )}$ and $B^{(\alpha ,\beta ,\gamma ,\delta )}$ are bounded operators, moreover

$$\begin{aligned} \Vert A^{(\alpha ,\beta ,\gamma ,\delta )}\Vert \le |\alpha |\Vert C_{\varphi }\Vert +|\gamma |\Vert C_{\pi }\Vert \le |\alpha |\sqrt{\frac{1+\varphi (0)}{1-\varphi (0)}} +|\gamma |\sqrt{\frac{1+\pi (0)}{1-\pi (0)}}, \end{aligned}$$

(20)

and

$$\begin{aligned} \Vert B^{(\alpha ,\beta ,\gamma ,\delta )}\Vert \le |\delta |\Vert C_{\varphi }\Vert +|\beta |\Vert C_{\pi }\Vert \le |\delta |\sqrt{\frac{1+\varphi (0)}{1-\varphi (0)}} +|\beta |\sqrt{\frac{1+\pi (0)}{1-\pi (0)}}, \end{aligned}$$

(21)

$\square$

The previous result leads to our next definition

Definition 17

Suppose that $A,B\in \mathcal {B}(H^{2}(\mathbb {D}))$ and define $A+\overline{B}: HH^{2}(\mathbb {D})\longrightarrow HH^{2}(\mathbb {D})$ of the following form

$$\begin{aligned} (A+\overline{B})(a+\overline{b})=Aa+\overline{Bb}, \end{aligned}$$

for all $a+\overline{b}\in HH^{2}(\mathbb {D})$.

Then, the following result holds

Theorem 18

For all $A,B\in \mathcal {B}(H^{2}(\mathbb {D}))$, we have $A+\overline{B}\in \mathcal {B}(HH^{2}(\mathbb {D}))$.

Proof

It is clear that $A+\overline{B}$ is defined in all $HH^{2}(\mathbb {D})$. First, we will prove that $A+\overline{B}$ is linear

$$\begin{aligned} (A+\overline{B})(\alpha _{1}(a_{1}+\overline{b_{1}})+\alpha _{2}(a_{2}+\overline{b_{2}}))&=(A+\overline{B})\left( (\alpha _{1}a_{1}+\alpha _{2}a_{2}) +\overline{(\overline{\alpha _{1}}b_{1}+\overline{\alpha _{2}}b_{2})}\right) \\ {}&=A(\alpha _{1}a_{1}+\alpha _{2}a_{2})+\overline{B(\overline{\alpha _{1}}b_{1}+\overline{\alpha _{2}}b_{2})} \\&=(\alpha _{1}Aa_{1}+\alpha _{2}Aa_{2})+\overline{(\overline{\alpha _{1}}Bb_{1}+\overline{\alpha _{2}}Bb_{2})}\\&=\alpha _{1}(Aa_{1}+\overline{Bb_{1}})+\alpha _{2}(Aa_{2}+\overline{Bb_{2}}) \\&=\alpha _{1}(A+\overline{B})(a_{1}+\overline{b_{1}})+\alpha _{2}(A+\overline{B})(a_{2}+\overline{b_{2}}), \end{aligned}$$

thus $A+\overline{B}$ is linear. It remains to be shown that $A+\overline{B}$ is bounded. In fact

$$\begin{aligned} \Vert (A+\overline{B})(a+\overline{b})\Vert ^{2}&=\Vert Aa\Vert ^{2}+\Vert Bb\Vert ^{2}\le \Vert A\Vert ^{2}\Vert a\Vert ^{2}+\Vert B\Vert ^{2}\Vert b\Vert ^{2} \\ {}&\le \max \{\Vert A\Vert ^{2},\Vert B\Vert ^{2}\} (\Vert a\Vert ^{2}+\Vert b\Vert ^{2})\\&= \max \{\Vert A\Vert ^{2},\Vert B\Vert ^{2}\}\,\Vert (a+\overline{b})\Vert ^{2}, \end{aligned}$$

thus

$$\begin{aligned} \Vert A+\overline{B}\Vert \le \sqrt{\max \{\Vert A\Vert ^{2},\Vert B\Vert ^{2}\}}. \end{aligned}$$

(22)

$\square$

Notice that from the proof of the theorem follows that $\Vert A\Vert ,\Vert B\Vert \le \Vert A+\overline{B}\Vert$.

Corollary 19

We have $C^{(\alpha ,\beta ,\gamma ,\delta )}_{\varphi +\overline{\pi }}\in \mathcal {B}(HH^{2}(\mathbb {D}))$.

Theorem 20

All operator $L\in \mathcal {B}(HH^{2}(\mathbb {D}))$ such that $L\left( H^{2}(\mathbb {D})\right) \subset H^{2}(\mathbb {D})$ and $L\left( \overline{H^{2}(\mathbb {D})}\right) \subset \overline{H^{2}(\mathbb {D})}$ is of the form $L=A_{L}+\overline{B_{L}}$ where this is understood in the sense of the Definition 17, and $A_{L},B_{L}\in \mathcal {B}(H^{2}(\mathbb {D}))$.

Proof

From the linearity of L for every $a+\overline{b}$, we have that $L(a+\overline{b})=La+L\overline{b}$, moreover $La \in H^{2}(\mathbb {D})$ and $L\overline{b}\in \overline{H^{2}(\mathbb {D})}$. Define now $A_{L}a=La$ and $B_{L}b=\overline{L\overline{b}}$ then $A_{L}$ and $B_{L}$ are linear. Also, it is evident that $D(A_{L})=D(B_{L})=H^{2}(\mathbb {D})$ and $L(a+\overline{b})=A_{L}a+\overline{B_{L}b}=(A_{L}+\overline{B_{L}})(a+\overline{b})$ in the sense of the definition 17. Next, we will see that both operators $A_{L}$ and $B_{L}$ are bounded. Indeed,

$$\begin{aligned} \Vert Aa\Vert ^{2}_{H^{2}(\mathbb {D})}+\Vert Bb\Vert ^{2}_{H^{2}(\mathbb {D})}&=\Vert Aa+\overline{Bb}\Vert ^{2}_{HH^{2}(\mathbb {D})}=\Vert (A+\overline{B}) (a+\overline{b})\Vert ^{2}_{HH^{2}(\mathbb {D})} \\ {}&=\Vert L(a+\overline{b})\Vert ^{2}_{HH^{2}(\mathbb {D})}\le \Vert L\Vert (\Vert a\Vert ^{2}_{H^{2}(\mathbb {D})}+\Vert b\Vert ^{2}_{H^{2}(\mathbb {D})}), \end{aligned}$$

it shows that $\Vert A\Vert _{\mathcal {B}(H^{2}(\mathbb {D}))},\Vert B\Vert _{\mathcal {B}(H^{2}(\mathbb {D}))}\le \Vert L\Vert _{\mathcal {B}(HH^{2}(\mathbb {D}))}$. $\square$

The following lemma will be needed to establish some results of the next subsection.

Lemma 21

Suppose that $L=A+\overline{B}\in \mathcal {B}(HH^{2}(\mathbb {D}))$. Then, $L^{*}=A^{*}+\overline{B^{*}}$ and moreover $L^{*}\in \mathcal {B}(HH^{2}(\mathbb {D}))$.

Proof

For any $f+\overline{g},l+\overline{m}\in HH^{2}(\mathbb {D})$ one has

$$\begin{aligned} \left( (A+\overline{B})(f+\overline{g}) ,l+\overline{m} \right)&=\left( Af+\overline{Bg},l+\overline{m} \right) \\ {}&=(Af,l)+\overline{(Bg,m)}=(f,A^{*}l)+\overline{(g,B^{*}m)} \\ {}&=(f+\overline{g}, A^{*}l+\overline{B^{*}m})=(f+\overline{g},\left( A^{*}+\overline{B^{*}}\,\right) (l+\overline{m})), \end{aligned}$$

now, that $A^{*}+\overline{B^{*}}$ is bounded follows from Theorem 18. $\square$

Consider now $\mathcal {E}=\mathcal {B}(H^{2}(\mathbb {D}))+\overline{\mathcal {B}(H^{2}(\mathbb {D}))}\subset \mathcal {B}(HH^{2}(\mathbb {D}))$. The space $\mathcal {E}$ may be regarded a vector space if one defines $(A+\overline{B})+(C+\overline{D})=(A+C)+\overline{(B+D)}$ and $\lambda (A+\overline{B})=\lambda A+\overline{\lambda B}$ where $\lambda \in \mathbb {C}$. Observe that if $A+\overline{B}\in \mathcal {E}$ then $(A+\overline{B})^{*}=A^{*}+\overline{B^{*}}\in \mathcal {E}$. The composition of operators of this type leads to the equality $(A+\overline{B})(C+\overline{D}) =AC+\overline{BD}\in \mathcal {E}$. Next, we intend to show that $\mathcal {E}$ is a closed subspace of $\mathcal {B}(HH^{2}(\mathbb {D}))$, suppose that $A_{n}+\overline{B}_{n}\longrightarrow L$ in $\mathcal {B}(HH^{2}(\mathbb {D}))$, this means that $\{A_{n}+\overline{B}_{n}\}$ is a Cauchy sequence then so are $\{A_{n}\}$ and $\{B_{n}\}$. Taking into account that $\mathcal {B}(H^{2}(\mathbb {D}))$ is complete it follows that there are $A, B\in \mathcal {B}(H^{2}(\mathbb {D}))$ such that $A_{n}\longrightarrow A$ and $B_{n}\longrightarrow B$ which implies that $A_{n}+\overline{B}_{n}\longrightarrow A+\overline{B}$, thus $L=A+\overline{B}$.

Proposition 22

For all $A+\overline{B} \in \mathcal {B}(HH^{2}(\mathbb {D}))$ we obtain

• 1. $e^{A+\overline{B}}:=\sum _{n=0}^{\infty }\frac{(A+\overline{B})^{n}}{n!}\in \mathcal {B}(HH^{2}(\mathbb {D}))$.

• 2. The following equality holds

  $$\begin{aligned} e^{A+\overline{B}}=e^{A}+\overline{e^{B}}, \end{aligned}$$

  (23)

Proof

It is trivial to see that

$$\begin{aligned} \Vert e^{A+\overline{B}}\Vert \le e^{\max \{\Vert A\Vert ,\Vert B\Vert \}}, \end{aligned}$$

thus $e^{A+\overline{B}}$ is a bounded linear operator. On the other hand,

$$\begin{aligned} e^{A+\overline{B}}{} & {} =\sum _{n=0}^{\infty }\frac{(A+\overline{B})^{n}}{n!}=\sum _{n=0}^{\infty }\frac{A^{n}+\overline{B^{n}}}{n!}=\sum _{n=0}^{\infty } \left( \frac{A^{n}}{n!}+\overline{\frac{B^{n}}{n!}} \right) \\{} & {} =\sum _{n=0}^{\infty } \frac{A^{n}}{n!}+\overline{\sum _{n=0}^{\infty } \frac{B^{n}}{n!}}=e^{A}+\overline{e^{B}}. \end{aligned}$$

$\square$

3.2 Reproducing kernel and some composition operators

In this part of our work, we will characterize the composition operators introduced above, in terms of complex harmonic reproduction kernels.

Definition 23

We shall say that $k(\lambda ,z)+\overline{i(\lambda ,z)}$ is a reproducing kernel in $HH^{2}(\mathbb {D})$ if

$$\begin{aligned} (a(z)+\overline{b(z)},k(\lambda ,z)+\overline{i(\lambda ,z)})=a(\lambda )+\overline{b(\lambda )}, \end{aligned}$$

for all $a+\overline{b}\in HH^{2}(\mathbb {D})$ and for all $\lambda \in \mathbb {D}$. It is clear that in this case both $k(\lambda ,z)$ and $i(\lambda ,z)$ must be reproducing kernels in $H^{2}(\mathbb {D})$.

Below, by a simple composition operator, we mean an operator of the form $C^{(1,1,0,0)}_{\varphi +\overline{\pi }}=C_{\varphi +\overline{\pi }}=C_{\varphi }+\overline{C_{\pi }}$. We recall that $\varphi , \pi : \mathbb {D}\longrightarrow \mathbb {D}$ are analytic functions.

Proposition 24

For all reproducing kernel $k(\lambda ,z)+\overline{i(\lambda ,z)}$ we have

$$\begin{aligned} (C_{\varphi +\overline{\pi }})^{*}(k(\lambda ,z)+\overline{i(\lambda ,z)})=k(\varphi (\lambda ),z)+\overline{i(\pi (\lambda ),z)}. \end{aligned}$$

(24)

Proof

For every $a+\overline{b}\in HH^{2}(\mathbb {D})$ we obtain

$$\begin{aligned}&\left( a+\overline{b},(C_{\varphi +\overline{\pi }})^{*}\left( k(\lambda ,z)+\overline{i(\lambda ,z)}\right) \right) _{HH^{2}(\mathbb {D})}\\ {}&\quad = \left( C_{\varphi +\overline{\pi }}(a+\overline{b}), k(\lambda ,z)+\overline{i(\lambda ,z)} \right) _{HH^{2}(\mathbb {D})} \\ {}&\quad =\left( C_{\varphi }a+\overline{C_{\pi } b}, k(\lambda ,z)+\overline{i(\lambda ,z)} \right) _{HH^{2}(\mathbb {D})} \\ {}&\quad =(C_{\varphi }a, k(\lambda ,z))_{H^{2}(\mathbb {D})}+\overline{(C_{\pi }b, i(\lambda ,z))_{H^{2}(\mathbb {D})}} \\ {}&\quad =((a\circ \varphi )(z), k(\lambda ,z))_{H^{2}(\mathbb {D})}+\overline{((b\circ \pi )(z), i(\lambda ,z))_{H^{2}(\mathbb {D})}} \\&\quad =a(\varphi (\lambda ))+\overline{b(\pi (\lambda ))}=a\circ \varphi (\lambda )+\overline{b\circ \pi (\lambda )}. \end{aligned}$$

But also,

$$\begin{aligned} \left( a+\overline{b},k(\varphi (\lambda ),z)+\overline{i(\pi (\lambda ),z)} \right) _{H^{2}(\mathbb {D})}&=(a, k(\varphi (\lambda ),z))_{H^{2}(\mathbb {D})}+ \overline{(b, i(\varphi (\lambda ),z))_{H^{2}(\mathbb {D})}}\\ {}&=a(\varphi (\lambda ))+\overline{b(\pi (\lambda ))}=a\circ \varphi (\lambda )+\overline{b\circ \pi (\lambda )}, \end{aligned}$$

and since $a+\overline{b}$ is arbitrary, (24) holds. $\square$

Theorem 25

The operator $L=A+\overline{B}$ where $A,B\in \mathcal {B}(H^{2}(\mathbb {D}))$ is a simple composition operator on $HH^{2}(\mathbb {D})$ if and only if $A^{*}$ and $B^{*}$ map the space of reproducing kernels of $H^{2}(\mathbb {D})$ in itself.

Proof

We already have seen that (see (24)) $C^{*}_{\varphi +\overline{\pi }}(k(\lambda ,z)+\overline{i(\lambda ,z)})=k(\varphi (\lambda ),z)+\overline{i(\pi (\lambda ),z)}$. Moreover, from the lemma 21, we conclude that $C^{*}_{\varphi +\overline{\pi }}=C^{*}_{\varphi }+\overline{C^{*}_{\pi }}$ hence $C^{*}_{\varphi }k(\lambda ,z)=k(\varphi (\lambda ),z)$ and $C^{*}_{\pi }i(\lambda ,z)=i(\pi (\lambda ),z)$.

Reciprocally, suppose $(A^{*}+\overline{B^{*}})(k(\lambda ,z)+\overline{i(\lambda ,z)})=A^{*}k(\lambda ,z)+\overline{B^{*}i(\lambda ,z)} =k(\lambda ^{'},z)+\overline{i(\lambda ^{''},z)}$, where $k(\lambda ^{'},z)$ and $i(\lambda ^{''},z)$ are reproducing kernels in $H^{2}(\mathbb {D})$. It is clear that $A^{*}k(\lambda ,z)=k(\lambda ^{'},z)$ and $B^{*}i(\lambda ,z)=i(\lambda ^{''},z)$. Define $\varphi , \pi :\mathbb {D}\longrightarrow \mathbb {D}$ in the following form $\lambda ^{'}=\varphi (\lambda )$ and $\lambda ^{''}=\pi (\lambda )$, Observe that It is possible because $\lambda ^{'}, \lambda ^{''}\in \mathbb {D}$. Then, $\forall f \in H^{2}(\mathbb {D})$

$$\begin{aligned} (Af)(\lambda ){} & {} =(Af,k(\lambda ,z))=(f,A^{*}k(\lambda ,z))=(f,k(\lambda ^{'},z))=(f,k(\varphi (\lambda ),z))\\{} & {} =f(\varphi (\lambda )), \end{aligned}$$

and $\forall g \in H^{2}(\mathbb {D})$

$$\begin{aligned} (Bg)(\lambda ){} & {} =(Bg,i(\lambda ,z))=(g,B^{*}i(\lambda ,z))=(g,i(\lambda ^{''},z))=(g,k(\pi (\lambda ),z))\\{} & {} =g(\pi (\lambda )). \end{aligned}$$

Taking into account that $A, B: H^{2}(\mathbb {D})\longrightarrow H^{2}(\mathbb {D})$ if $f=g=z$ then $(Az)(\lambda )= \varphi (\lambda )\in H^{2}(\mathbb {D})$ and $(Bz)(\lambda )=\pi (\lambda )\in H^{2}(\mathbb {D})$. Since $f,g \in H^{2}(\mathbb {D})$ are arbitrary, we obtain $A=C_{\varphi }$ and $B=C_{\pi }$. It follows that $L=C_{\varphi }+\overline{C_{\pi }}$. $\square$

Theorem 26

$L=A+\overline{B}\in \mathcal {E}$ is a simple composition operator in $HH^{2}(\mathbb {D})$ if and only if $Le_{n}=Ae_{n}=(Ae_{1})^{n}=(L e_{1})^{n}$ and $Lf_{n}=\overline{Be_{n}}=\overline{(Be_{1})^{n}} =\left( \,\overline{Be_{1}}\, \right) ^{n}=(L f_{1})^{n}$ for all $n\in \mathbb {N}$, that is, if and only if $Ae_{n}=(Ae_{1})^{n}$ and $Be_{n}=(Be_{1})^{n}$ for any $n\in \mathbb {N}$.

Proof

Suppose first that $L=C_{\varphi }+\overline{C_{\pi }}$ then $Le_{n}=(C_{\varphi }+\overline{C_{\pi }})e_{n}=C_{\varphi }z^{n} =\varphi ^{n}(z)=(C_{\varphi }z)^{n}=(Le_{1})^{n}$ for all $n\in \mathbb {N}$. In the same way, $\forall n\in \mathbb {N}$ we have $Lf_{n}=(C_{\varphi }+\overline{C_{\pi }})f_{n}=\overline{C_{\pi }z^{n}}=\overline{\pi ^{n}(z)}=\left( \overline{\pi (z)}\right) ^{n} =\left( \,\overline{C_{\pi }z}\,\right) ^{n}=\left( Lf_{1} \right) ^{n}$.

Conversely, assume that we have $L=A+\overline{B}$ where $A,B\in \mathcal {B}(H^{2}(\mathbb {D}))$ such that $Az^{n}=(Az)^{n}$ and $Bz^{n}=(Bz)^{n}$. We must show that L is a simple composition operator. Define $\varphi (z)=Az$ and $\pi (z)=Bz$, then $\varphi ,\pi \in H^{2}(\mathbb {D})$. Now, $Le_{n}=Lz^{n}=Az^{n}=(Az)^{n}=\varphi ^{n}(z)$ and $Lf_{n}=L\overline{z^{n}}=\overline{Bz^{n}}=\overline{(Bz)^{n}}=\overline{\pi ^{n}(z)}$. It implies that if $l+\overline{m}=\Sigma \, l_{n}e_{n}+\Sigma \, \overline{m_{n}e_{n}}$ hence $L(l+\overline{m})=(A+\overline{B})(l+\overline{m})=\Sigma \, l_{n}Ae_{n}+ \Sigma \, \overline{m_{n}Be_{n}}=\Sigma \, l_{n}\varphi ^{n}(z)+ \Sigma \, \overline{m_{n}\pi ^{n}(z)}=(C_{\varphi }+\overline{C_{\pi }})(l+\overline{m})$ and since $l+\overline{m}\in HH^{2}(\mathbb {D})$ is arbitrary, in order to see that $L=A+\overline{B}$ is a simple composition operator is sufficient to prove that both functions $\varphi , \pi$ map $\mathbb {D}$ into $\mathbb {D}$. For this purpose it is sufficient to point out that $(\varphi (z))^{n}=Ae_{n}$ and $(\pi (z))^{n}=Be_{n}$, hence $\Vert (\varphi (z))^{n}\Vert \le \Vert A\Vert$ and $\Vert (\pi (z))^{n}\Vert \le \Vert B\Vert$ (see [14] page 169). $\square$

For $l+\overline{m},p+\overline{q}\in HH^{2}(\mathbb {D})$, we define the product $(l+\overline{m})(p+\overline{q})=lp+\overline{mq}$.

Theorem 27

An operator $L=A+\overline{B}$ where $A,B\in \mathcal {B}(H^{2}(\mathbb {D}))$ is a simple composition operator if and only if

$$\begin{aligned} L\left( (l+\overline{m})(p+\overline{q}) \right) = \left( L(l+\overline{m})L(p+\overline{q}) \right) , \end{aligned}$$

(25)

Proof

Note that if $L=A+\overline{B}$ then $L\left( (l+\overline{m})(p+\overline{q}) \right) =L(lp+\overline{mq})=A(lp)+\overline{B(mq)}$. Hence, if $L=C_{\varphi }+\overline{C_{\pi }}$ where $\varphi ,\pi : \mathbb {D}\longrightarrow \mathbb {D}$ are analytic functions, then

$$\begin{aligned} (C_{\varphi }+\overline{C_{\pi }})\left( (l+\overline{m})(p+\overline{q}) \right)&=C_{\varphi }(lp)+\overline{C_{\pi }(mq)} =(C_{\varphi }l)\,(C_{\varphi }p)+\overline{(C_{\pi }m)\,(C_{\pi }q)} \\&=(C_{\varphi }l+\overline{C_{\pi }m}\,)(C_{\varphi }p+\overline{C_{\pi }q}\,)\\&=\left( (C_{\varphi }+\overline{C_{\pi }}\,)(l+\overline{m})\right) \left( (C_{\varphi }+\overline{C_{\pi }}\,)(p+\overline{q})\right) . \end{aligned}$$

On the other hand, suppose that $L=A+\overline{B}$ satisfies (25), then $Le_{n}=Lz^{n}=\underbrace{Lz\cdots Lz}_{n-factors}=(Le_{1})^{n}$ and $Lf_{n}=L\overline{z^{n}}=L(\overline{z})^{n}=\underbrace{L\overline{z}\cdots L\overline{z}}_{n-factors}=(Lf_{1})^{n}$. So, from the previous lemma follows that L is a simple composition operator. $\square$

Remark 28

If $L=A+\overline{B}\in B(HH^{2}(\mathbb {D}))$ it has been pointed out above that $L^{*}=A^{*}+\overline{B^{*}}$. Then

$$\begin{aligned} LL^{*}(l+\overline{m})=(A+\overline{B})(A^{*}+\overline{B^{*}}) (l+\overline{m})=(A+\overline{B})(A^{*}l+\overline{B^{*}m})= (AA^{*}l+\overline{BB^{*}m}), \end{aligned}$$

on the other hand, $L^{*}L(l+\overline{m})=A^{*}Al+\overline{B^{*}Bm}$. It shows that $(LL^{*}-L^{*}L)e_{n}=(AA^{*}-A^{*}A)e_{n}$ and $(LL^{*}-L^{*}L)f_{n}=\overline{(BB^{*}-B^{*}B)e_{n}}$. Hence, L is a normal operator if and only if A and B are normal operators. In the particular case in which $L=C_{\varphi +\overline{\pi }}=C_{\varphi }+\overline{C_{\pi }}$, it will be a normal operator if and only if $C_{\varphi }$ and $C_{\pi }$ are normal operators, that is, if and only if $\varphi (z)=\lambda z$ and $\pi (z)=\mu z$ where $|\lambda |\le 1$ and $|\mu |\le 1$.