A Study on Some Properties of F-Composition Operators on the Hardy Space H2

Lemma 1.1:

¹¹ Suppose that $\phi$ and $\psi$ be two h s-maps of $\mathbb{U}$ . Then we get:

Lemma 1.2:

¹¹ Suppose that $\phi$ and $\psi$ be two h s-maps of $\mathbb{U}$ then, we obtain

Abood¹¹ in 2021, introduced F-induced composition mapping acting on the Hardy space ${\mathbb{H}}^2$ as follows:

Definition 1.3:

¹¹ Let F be a holomorphic function on $\mathbb{U}$ of the form $F\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ which is absolutely convergent in an open unit disk. If $\Vert C_{\phi }\Vert <1$ , then $F_{\phi }≔F\left(C_{\phi }\right)={\sum }_{n=0}^{\infty }{a}_n{C}_{\phi }^n$ is a bounded linear operator on ${\mathbb{H}}^2$ and is called the F-induced composition mapping of ${\mathbb{H}}^2$ induced by F and $\phi$ . As we know, the holomorphic functions are a fundamental concept in the analysis of complex functions. For more details, see.^17–19

This paper includes four sections. In section two, we introduce some basic axioms for F-induced composition mapping and present many of the results obtained. In section three, we studied the conditions and descriptions of each $\phi$ and F such that the F-induced composition mapping is an invertible operator and presented some important results. Finally, a discussion and conclusions are presented.

Theorem 2.1:

¹¹

Note that if $F$ admits a power series representation of nonnegative, whose coefficients are real numbers, then

Reminder, we know that $x\perp y$ means $x$ orthogonal to $y$ .

Corollary 2.2:

Consider $\phi$ as an inner self-map of $\mathbb{U}$ such that $C_{\phi }^{n}f\perp C_{\phi }^{m}f$ for each $n,m\in \mathbb{N}$ , $n\ne m,f\in {\mathbb{H}}^2$ , then:

Proof:

Since $C_{\phi }^{n}f\perp C_{\phi }^{m}f$ for each $n,m\in \mathbb{N},n\ne m,$ then for each $f\in {\mathbb{H}}^2$ we have:

(from Lemma 1.1 (1), $C_{\phi }^n=C_{{\phi }_n})$

Thus, $\Vert F_{\phi }\Vert =\sqrt{\sum _{n=0}^{\infty }{\left|a_n\right|}^2\frac{1+\left|\phi \left(0\right)\right|}{1-\left|\phi \left(0\right)\right|}}$ . □

Recall that for each $\alpha \in \mathbb{U}$ ,

Thus, for each $f\in {\mathbb{H}}^2$ , $f\left(z\right)=\sum _{n=0}^{\infty }\widehat{f}\left(n\right)z^n$ .

The adjoint of the operator induced by composition on a family of functions ${\left\{k_{\alpha }\right\}}_{\alpha \in \wedge }$ in ${\mathbb{H}}^2$ have been discussed in^3,21 and²² by, $C_{\phi }^{\ast }{k}_{\alpha }=k_{\phi \left(\alpha \right)}$ .

In the following proposition we will compute the adjoint of F-induced composition mapping on ${\left\{k_{\alpha }\right\}}_{\alpha \in \wedge }$ .

Proposition 2.3:

For each $\alpha \in \mathbb{U}$ , the adjoint of the F-induced composition mapping is

Proof:

For all $f\in {\mathbb{H}}^2$ ,

Thus, $F_{\phi }^{\ast }{k}_{\alpha }={\sum }_{n=0}^{\infty }\overline{a_n}{k}_{{\phi }_n\left(\alpha \right)}.$ □

Proposition 2.4:

Let $F\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ then $F_{\phi }^{\ast }={\sum }_{n=0}^{\infty }\overline{a_n}{\left(C_{\phi }^{\ast }\right)}^n.$

To learn more about multiplying two infinite series, known as the Cauchy product, see.¹⁵

Proposition 2.5:

Assume that we have two h s-maps $\psi$ and $\phi$ of $\mathbb{U}$ , then:

Proof:

□

We now discuss the relationship between F-induced composition mapping and some classes of operators. Recall that operator $T\in B\left(H\right)$ is called a self-adjoint if $T=T^{\ast }$ and normal operator if $T^{\ast }T=T{T}^{\ast }$ . Moreover, $T$ is called unitary if $T^{\ast }T=T{T}^{\ast }=I$ . By^19–20 and.^23–27

Remark 2.6:

Assume that $F\left(z\right)=\sum _{n=0}^{\infty }{a}_n{z}^n$ , then $F_{\phi }$ is a self-adjoint operator if and only if F has real coefficients and $C_{\phi }$ is a self-adjoint operator.

Proof:

Note that,

Thus, one can easily observe that $F_{\phi }=F_{\phi }^{\ast }$ if and only if the sequence of coefficients ${\left\{a_n\right\}}_{n\in N}$ is real and ${\left(C_{\phi }^{\ast }\right)}^n=C_{\phi }^n$ for each $n\in N$ , implying that $C_{\phi }$ is a self-adjoint operator. □

From Remark 2.6 and Lemma 1.2 (3) the following consequences.

Corollary 2.7:

$F_{\phi }$ is a self-adjoint operator if and only if ${\left\{a_n\right\}}_{n\in N}$ are real coefficients and $\phi \left(z\right)=\mathit{\gamma z}$ , $\gamma \in ℝ$ , $\left|\gamma \right|\le 1$ .

Proposition 2.8:

Assume that $F\left(z\right)=\sum _{n=0}^{\infty }{a}_n{z}^n$ with nonzero coefficients, then $F_{\phi }$ is a normal operator on ${\mathbb{H}}^2$ if and only if $C_{\phi }$ is a normal operator on ${\mathbb{H}}^2$ .

Proof:

Note that, by,²⁸ because the following series are absolutely convergent in $\mathbb{U}$ , we have

Hence, it is clear that $F_{\phi }^{\ast }{F}_{\phi }=F_{\phi }{F}_{\phi }^{\ast }$ if and onl $y$ if ${\left(C_{\phi }^{\ast }\right)}^n{C}_{\phi }^m=C_{\phi }^m{\left(C_{\phi }^{\ast }\right)}^n$ for each $n,m\in \mathbb{N}$ , that is, $C_{\phi }$ is a normal operator. □

We can obtain the next result by Proposition 2.8 and Lemma 1.1 (1).

Corollary 2.9:

$F_{\phi }$ is a normal operator for ${\mathbb{H}}^2$ if and only if $\phi \left(z\right)=\mathit{\gamma z}$ , $\gamma \in ℂ$ , $\left|\gamma \right|\le 1$ .

Proposition 2.10:

Assume that $F\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ with nonzero coefficients. Then $F_{\phi }$ is a unitary operator on ${\mathbb{H}}^2$ if and onl $y$ if $C_{\phi }$ is a unitary operator on ${\mathbb{H}}^2$ and the sequence ${\left\{a_n\right\}}_{n\in N}$ of module one.

Proof:

Note that,

Therefore, it is clear that $F_{\phi }{F}_{\phi }^{\ast }={F_{\phi }}^{\ast }{F}_{\phi }=I$ if and only if $C_{\phi }^m{\left(C_{\phi }^{\ast }\right)}^n={\left(C_{\phi }^{\ast }\right)}^n{C}_{\phi }^m=I$ , and $\overline{a_n}{a}_m=1$ for each $n,m\in \mathbb{N}$ . Thus, $F_{\phi }$ is a unitary operator and $\left|a_n\right|=1$ for each $n\in \mathbb{N}$ . □

From Proposition 2.10 and Lemma 1.1. (2), we obtain the following correlation.

Corollary 2.11:

$F_{\phi }$ is the unitary operator on ${\mathbb{H}}^2$ if and only if $\phi \left(z\right)=\mathit{\gamma z}$ , $\gamma \in ℂ$ , $\left|\gamma \right|=1$ , and the sequence ${\left\{a_n\right\}}_{n\in N}$ of module one.

Proposition 2.12:

Suppose that $F\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ with nonzero coefficients, the $n{F}_{\phi }$ is an isometric operator on ${\mathbb{H}}^2$ if and only if $C_{\phi }$ is an isometric operator and the sequence ${\left\{a_n\right\}}_{n\in N}$ of module one.

Proof:

Note that, $F_{\phi }^{\ast }{F}_{\phi }=\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }\overline{a_n}{a}_m{\left(C_{\phi }^{\ast }\right)}^n{C}_{\phi }^m=I$ .

It follows that, $F_{\phi }^{\ast }{F}_{\phi }=I$ if and only if ${\left(C_{\phi }^{\ast }\right)}^n{C}_{\phi }^m=I$ and $\overline{a_n}{a}_m=1$ for each $n,m\in \mathbb{N}$ . Hence, $F_{\phi }$ is isometric if and only if $C_{\phi }$ is isometric and the sequence ${\left\{a_n\right\}}_{n\in N}$ of modules one. □

Theorem 3.1:

If $F\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ with nonzero coefficients and $\phi$ is non-constant, then $F_{\phi }$ is a one-to-one operator on ${\mathbb{H}}^2$ .

Proof:

Assume that for some $f,g\in {\mathbb{H}}^2,F_{\phi }f=F_{\phi }g$ . We to show that $f\equiv g$ on $\mathbb{U}$ . Note that, for every $z\in \mathbb{U}$ , we have ${\sum }_{n=0}^{\infty }{a}_n{C}_{\phi }^{n}f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{C}_{\phi }^{n}g\left(z\right)$ .

Therefore, ${\sum }_{n=0}^{\infty }{a}_n(C_{\phi }^{n}f\left(z\right)-C_{\phi }^{n}g\left(z\right))=0$ . Because all the series are absolutely convergent in $\mathbb{U}$ , and uniformly converge on every compact set in $\mathbb{U}$ , then $C_{\phi }^n\left(f\left(z\right)\right)=C_{\phi }^n\left(g\left(z\right)\right)$ for each $z\in \mathbb{U}$ , $n\in \mathbb{N}$ , because $a_n\ne 0$ for each $n\in \mathbb{N}$ .

Therefore $f\left({\phi }_n\left(z\right)\right)=g\left({\phi }_n\left(z\right)\right)$ for all $z\in \mathbb{U}$ . This implies that $f\equiv g$ on ${\phi }_n\left(\mathbb{U}\right)$ . Because $\phi$ is non-constant, ${\phi }_n$ is non-constant for all $n\in \mathbb{N}$ . However ${\phi }_n$ is a holomorphic self-map of $\mathbb{U}$ for all $n\in \mathbb{N}$ , and ${\phi }_n$ is an open map for all $n\in \mathbb{N}$ ,.³¹ Therefore, $f\equiv g$ on the non-empty open subset of $\mathbb{U}$ . Thus, by Taylors theorem, we have $f\equiv g$ on $\mathbb{U}$ . Thus, $F_{\phi }$ is one-to-one. □

Theorem 3.2:

I $f$ $\mathcal{R}\left(F_{\phi }\right)$ is dense in ${\mathbb{H}}^2$ , then ${\phi }_n$ is one to one fo $r$ all $n\in \mathbb{N}$ .

Proof:

Assume the converse; that is ${\phi }_n$ is not one to one, then there exists $\alpha ,\beta \in \mathbb{U},\alpha \ne \beta$ such that ${\phi }_n\left(\alpha \right)={\phi }_n\left(\beta \right)$ . Hence,

$F_{\phi }^{\ast }(k_{\alpha }-k_{\beta })=F_{\phi }^{\ast }{k}_{\alpha }-F_{\phi }^{\ast }{k}_{\beta }=\sum _{n=0}^{\infty }\overline{a_n}{k}_{{\phi }_n\left(\alpha \right)}-k_{{\phi }_n\left(\beta \right)}=0,(\text{since}{\phi }_{n\left(\alpha \right)}={\phi }_{n\left(\beta \right)}).$

Thus, $F_{\phi }^{\ast }(k_{\alpha }-k_{\beta })=0$ . Therefore, $k_{\alpha }-k_{\beta }\in \mathit{ker}{F}_{\phi }^{\ast }$ . However $k_{\alpha }\ne k_{\beta }$ , since $\alpha \ne \beta$ , it follows that $\mathit{ker}{F}_{\phi }^{\ast }\ne 0$ . However it is well-known that ${\mathbb{H}}^2=\mathit{ker}{F}_{\phi }^{\ast }⨁\overline{\mathcal{R}\left(F_{\phi }\right)}$ , then ${\mathbb{H}}^2\ne \overline{\mathcal{R}\left(F_{\phi }\right)}$ . Hence, $\mathcal{R}\left(F_{\phi }\right)$ is not dense in ${\mathbb{H}}^2$ , which is contradictory. Therefore, ${\phi }_n$ is one-to-one for all $n\in \mathbb{N}$ . □

Theorem 3.3:

If $F\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ with nonzero coefficients, and $F_{\phi }$ is onto, then ${\phi }_n$ is onto for all $n\in \mathbb{N}$ .

Proof:

Assume the converse; that is ${\phi }_n$ is not onto for all $n\in \mathbb{N}$ . This implies that ${\phi }_n\left(\mathbb{U}\right)\ne \mathbb{U}$ for all $n\in \mathbb{N}$ . This implies that there exists, $y\in \mathbb{U}$ and $y\notin {\phi }_n\left(\mathbb{U}\right)$ for all $n\in \mathbb{N}$ . Now, consider the functions $f$ and $g$ , for every $z\in \mathbb{U}$

Since $y\in \mathbb{U}$ , then $f$ is not holomorphic on $\mathbb{U}$ , thus $f\notin {\mathbb{H}}^2$ .

On the other hand, clearly $g$ is holomorphic on $\mathbb{U}$ , since $y\notin {\phi }_n\left(\mathbb{U}\right)$ . We claim $y\notin \mathcal{R}\left(F_{\phi }\right)$ . Suppose otherwise, that there exists $h\in {\mathbb{H}}^2$ such that $F_{\phi }h=g$ . Hence, for every $z\in \mathbb{U}/\left\{y\right\}$ $F_{\phi }h\left(z\right)=g\left(z\right)$ , ${\sum }_{m=0}^{\infty }{a}_m{C}_{\phi }^{m}h\left(z\right)=g\left(z\right)$ . It follows that,

Hence, ${\sum }_{m=0}^{\infty }{a}_{m}h\left({\phi }_m\left(z\right)\right)=a_{n}f\left({\phi }_n\left(z\right)\right).$

Thus, $f\equiv g$ on a non-empty open subset of $\mathbb{U}/\left\{y\right\}$ and thus by Taylor theorem $h\equiv f$ on all $\mathbb{U}/\left\{y\right\}$ . But we have $h\in {\mathbb{H}}^2$ , this implies that $f\in {\mathbb{H}}^2$ , which is a contradiction, since $f\notin {\mathbb{H}}^2$ . Thus, $g\notin \mathcal{R}\left(F_{\phi }\right)$ , so $F_{\phi }$ is not onto, which is a contradiction, it follows that ${\phi }_n$ is onto for all $n\in \mathbb{N}$ . □

Our goal is to describe the inevitability of F-induced composition mapping in the Hardy space ${\mathbb{H}}^2$ . From Theorems 3.2 and 3.3, a straightforward can be obtained as follows:

Theorem 3.4:

Let $F$ have nonzero coefficients. If $F_{\phi }$ is invertible, ${\phi }_n$ is invertible for all $n\in \mathbb{N}$ .

Before discuss the inverse of Theorem 3.4 we need the next Lemma.

Lemma 3.5:

If $F$ is a nonzero series such t $\mathit{hat}F\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ . Subsequently, $F_{\phi }{F}_{\psi }=F_{\psi }{F}_{\phi }=I$ if and only if $a_0^2=1$ and $a_n=0$ for all $n\ge 1$ .

Proof:

Note that,

If and only if $a_0^2=1$ and $a_n=0$ for all $n\ge 1$ . □

Corollary 3.6:

Consider $\phi$ to be invertible and $F$ to be a nonzero series $F\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ , then $F_{\phi }$ is invertible such that $F_{\phi }=\mathit{\lambda I}$ where $\left|\lambda \right|=1$ such that $F_{\phi }^{-1}=F_{{\phi }^{-1}}$ .

Proof:

Because $\phi$ is invertible, then ${\phi }^{-1}$ exists. Applying Lemma 3.5 by taking $\psi ={\phi }^{-1}$ , we have $F_{\phi }{F}_{{\phi }^{-1}}=F_{{\phi }^{-1}}{F}_{\phi }=I$ if and only if $a_n=0$ for all $n\ge 1$ , $a_0^2=1$ .

It follows that $F_{\phi }=\mathit{\lambda I}$ , $\left|\lambda \right|=1$ and $F_{\phi }^{-1}=F_{{\phi }^{-1}}$ . □