¹¹ Suppose that φ and ψ be two h s-maps of U then, we obtain 1.

C φ is normal if and only if φ ( z ) = γz , γ ∈ ℂ , | γ | ≤ 1 .

Abood ¹¹ in 2021, introduced F-induced composition mapping acting on the Hardy space H 2 as follows:

¹¹ Let F be a holomorphic function on U of the form F ( z ) = ∑ n = 0 ∞ a n z n which is absolutely convergent in an open unit disk. If ‖ C φ ‖ < 1 , then F φ ≔ F ( C φ ) = ∑ n = 0 ∞ a n C φ n is a bounded linear operator on H 2 and is called the F-induced composition mapping of H 2 induced by F and φ . 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 φ 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.

Consider φ as an inner self-map of U such that C φ n f ⊥ C φ m f for e ach n , m ∈ ℕ , n ≠ m , f ∈ H 2 , then: ‖ F φ ‖ = ∑ n = 0 ∞ | a n | 2 1 + | φ ( 0 ) | 1 − | φ ( 0 ) | .

Since C φ n f ⊥ C φ m f for each n , m ∈ ℕ , n ≠ m , then for e ach f ∈ H 2 we have: ‖ F φ f ‖ 2 = ‖ ∑ n = 0 ∞ a n C φ n f ‖ 2 = ∑ n = 0 ∞ | a n | 2 ‖ C φ n f ‖ 2

(from Lemma 1.1 (1), C φ n = C φ n ) = ∑ n = 0 ∞ | a n | 2 ‖ C φ n f ‖ 2 = ∑ n = 0 ∞ | a n | 2 1 + | φ n ( 0 ) | 1 − | φ n ( 0 ) | ‖ f ‖ ( because φ n is the inner function ) .

Thus, ‖ F φ ‖ = ∑ n = 0 ∞ | a n | 2 1 + | φ ( 0 ) | 1 − | φ ( 0 ) | . □

Recall that for each α ∈ U , k α ( z ) = 1 1 − α ¯ z = ∑ n = 0 ∞ α ¯ n z n .

Thus, for each f ∈ H 2 , f ( z ) = ∑ n = 0 ∞ f ̂ ( n ) z n . < f , k α > = ∑ n = 0 ∞ f ̂ ( n ) α n = f ( α ) .

The adjoint of the operator induced by composition on a family of functions { k α } α ∈ ∧ in H 2 have been discussed in ^{3, 21} and ²² by, C φ ∗ k α = k φ ( α ) .

In the following proposition we will compute the adjoint of F-induced composition mapping on { k α } α ∈ ∧ .

For each α ∈ U , the adjoint of the F-induced composition mapping is F φ ∗ k α = ∑ n = 0 ∞ a n ¯ k φ n ( α ) ( z ) .

For all f ∈ H 2 , < f , F φ ∗ k α > = < F φ f , k α > = < ∑ n = 0 ∞ a n C φ n ( f ) , k α > = ∑ n = 0 ∞ a n < C φ n ( f ) , k α > = ∑ n = 0 ∞ a n < C φ n ( f ) , k α > = ∑ n = 0 ∞ a n < f ∘ φ n , k α > ( but < f , k α > = f ( α ) ) = ∑ n = 0 ∞ a n ( f ∘ φ n ( α ) ) = ∑ n = 0 ∞ a n f ( φ n ( α ) ) = ∑ n = 0 ∞ a n < f , k φ n ( α ) > = < f , ∑ n = 0 ∞ a n ¯ k φ n ( α ) >

Thus, F φ ∗ k α = ∑ n = 0 ∞ a n ¯ k φ n ( α ) . □

Let F ( z ) = ∑ n = 0 ∞ a n z n then F φ ∗ = ∑ n = 0 ∞ a n ¯ ( C φ ∗ ) n .

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

Assume that we have two h s-maps ψ and φ of U , then: F φ F ψ = ∑ n = 0 ∞ ∑ k = 0 n a k a n − k C φ k C ψ n − k .

F φ F ψ = ∑ n = 0 ∞ a n C φ n ∑ m = 0 ∞ a m C ψ m = ∑ n = 0 ∞ ∑ k = 0 n a k a n − k C φ k C ψ n − k .

□

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

Assume that F ( z ) = ∑ n = 0 ∞ a n z n , then F φ is a self-adjoint operator if and only if F has real coefficients and C φ is a self-adjoint operator.

Note that, F φ ∗ = ∑ n = 0 ∞ a n ¯ ( C φ ∗ ) n , F φ = ∑ n = 0 ∞ a n C φ n .

Thus, one can easily observe that F φ = F φ ∗ if and only if the sequence of coefficients { a n } n ∈ N is real and ( C φ ∗ ) n = C φ n for each n ∈ N , implying that C φ is a self-adjoint operator. □

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

F φ is a self-adjoint operator if and only if { a n } n ∈ N are real coefficients and φ ( z ) = γz , γ ∈ ℝ , | γ | ≤ 1 .

Assume that F ( z ) = ∑ n = 0 ∞ a n z n with nonzero coefficients, then F φ is a normal operator on H 2 if and only if C φ is a normal operator on H 2 .

Note that, by, ²⁸ because the following series are absolutely convergent in U , we have F φ ∗ F φ = ∑ n = 0 ∞ a n ¯ ( C φ ∗ ) n ∑ m = 0 ∞ a m C φ m = ∑ n = 0 ∞ ∑ m = 0 ∞ a n ¯ a m ( C φ ∗ ) n C φ m . F φ F φ ∗ = ∑ m = 0 ∞ a m C φ m ∑ n = 0 ∞ a n ¯ ( C φ ∗ ) n = ∑ m = 0 ∞ ∑ n = 0 ∞ a m a n ¯ C φ m ( C φ ∗ ) n .

Hence, it is clear that F φ ∗ F φ = F φ F φ ∗ if and onl y if ( C φ ∗ ) n C φ m = C φ m ( C φ ∗ ) n for each n , m ∈ ℕ , that is, C φ is a normal operator. □

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

F φ is a normal operator for H 2 if and only if φ ( z ) = γz , γ ∈ ℂ , | γ | ≤ 1 .

Assume that F ( z ) = ∑ n = 0 ∞ a n z n with nonzero coefficients. Then F φ is a unitary operator on H 2 if and onl y if C φ is a unitary operator on H 2 and the sequence { a n } n ∈ N of module one.

Note that, F φ F φ ∗ = ∑ m = 0 ∞ a m C φ m ∑ n = 0 ∞ a n ¯ ( C φ ∗ ) n = ∑ m = 0 ∞ ∑ n = 0 ∞ a m a n ¯ C φ m ( C φ ∗ ) n = I . F φ ∗ F φ = ∑ n = 0 ∞ a n ¯ ( C φ ∗ ) n ∑ m = 0 ∞ a m C φ m = ∑ n = 0 ∞ ∑ m = 0 ∞ a n ¯ a m ( C φ ∗ ) n C φ m = I .

Therefore, it is clear that F φ F φ ∗ = F φ ∗ F φ = I if and only if C φ m ( C φ ∗ ) n = ( C φ ∗ ) n C φ m = I , and a n ¯ a m = 1 for each n , m ∈ ℕ . Thus, F φ is a unitary operator and | a n | = 1 for each n ∈ ℕ . □

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

F φ is the unitary operator on H 2 if and only if φ ( z ) = γz , γ ∈ ℂ , | γ | = 1 , and the sequence { a n } n ∈ N of module one.

Suppose that F ( z ) = ∑ n = 0 ∞ a n z n with nonzero coefficients, the n F φ is an isometric operator on H 2 if and only if C φ is an isometric operator and the sequence { a n } n ∈ N of module one.

Note that, F φ ∗ F φ = ∑ n = 0 ∞ ∑ m = 0 ∞ a n ¯ a m ( C φ ∗ ) n C φ m = I .

It follows that, F φ ∗ F φ = I if and only if ( C φ ∗ ) n C φ m = I and a n ¯ a m = 1 for each n , m ∈ ℕ . Hence, F φ is isometric if and only if C φ is isometric and the sequence { a n } n ∈ N of modules one. □

Assume that for some f , g ∈ H 2 , F φ f = F φ g . We to show that f ≡ g on U . Note that, for every z ∈ U , we have ∑ n = 0 ∞ a n C φ n f ( z ) = ∑ n = 0 ∞ a n C φ n g ( z ) .

Therefore, ∑ n = 0 ∞ a n ( C φ n f ( z ) − C φ n g ( z ) ) = 0 . Because all the series are absolutely convergent in U , and uniformly converge on every compact set in U , then C φ n ( f ( z ) ) = C φ n ( g ( z ) ) for each z ∈ U , n ∈ ℕ , because a n ≠ 0 for each n ∈ ℕ .

Therefore f ( φ n ( z ) ) = g ( φ n ( z ) ) for all z ∈ U . This implies that f ≡ g on φ n ( U ) . Because φ is non-constant, φ n is non-constant for all n ∈ ℕ . However φ n is a holomorphic self-map of U for all n ∈ ℕ , and φ n is an open map for all n ∈ ℕ ,. ³¹ Therefore, f ≡ g on the non-empty open subset of U . Thus, by Taylors theorem, we have f ≡ g on U . Thus, F φ is one-to-one. □

I f R ( F φ ) is dense in H 2 , then φ n is one to one fo r all n ∈ ℕ .

Assume the converse; that is φ n is not one to one, then there exists α , β ∈ U , α ≠ β such that φ n ( α ) = φ n ( β ) . Hence,

F φ ∗ ( k α − k β ) = F φ ∗ k α − F φ ∗ k β = ∑ n = 0 ∞ a n ¯ k φ n ( α ) − k φ n ( β ) = 0 , ( since φ n ( α ) = φ n ( β ) ) .

Thus, F φ ∗ ( k α − k β ) = 0 . Therefore, k α − k β ∈ ker F φ ∗ . However k α ≠ k β , since α ≠ β , it follows that ker F φ ∗ ≠ 0 . However it is well-known that H 2 = ker F φ ∗ ⨁ R ( F φ ) ¯ , then H 2 ≠ R ( F φ ) ¯ . Hence, R ( F φ ) is not dense in H 2 , which is contradictory. Therefore, φ n is one-to-one for all n ∈ ℕ . □

If F ( z ) = ∑ n = 0 ∞ a n z n with nonzero coefficients, and F φ is onto, then φ n is onto for all n ∈ ℕ .

Assume the converse; that is φ n is not onto for all n ∈ ℕ . This implies that φ n ( U ) ≠ U for all n ∈ ℕ . This implies that there exists, y ∈ U and y ∉ φ n ( U ) for all n ∈ ℕ . Now, consider the functions f and g , for every z ∈ U f ( z ) = 1 z − y and g ( z ) = a n φ n ( z ) − y .

Since y ∈ U , then f is not holomorphic on U , thus f ∉ H 2 .

On the other hand, clearly g is holomorphic on U , since y ∉ φ n ( U ) . We claim y ∉ R ( F φ ) . Suppose otherwise, that there exists h ∈ H 2 such that F φ h = g . Hence, for every z ∈ U / { y } F φ h ( z ) = g ( z ) , ∑ m = 0 ∞ a m C φ m h ( z ) = g ( z ) . It follows that, ∑ m = 0 ∞ a m C φ m h ( z ) = g ( z ) .

Hence, ∑ m = 0 ∞ a m h ( φ m ( z ) ) = a n f ( φ n ( z ) ) .

Thus, f ≡ g on a non-empty open subset of U / { y } and thus by Taylor theorem h ≡ f on all U / { y } . But we have h ∈ H 2 , this implies that f ∈ H 2 , which is a contradiction, since f ∉ H 2 . Thus, g ∉ R ( F φ ) , so F φ is not onto, which is a contradiction, it follows that φ n is onto for all n ∈ ℕ . □

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

Let F have nonzero coefficients. If F φ is invertible, φ n is invertible for all n ∈ ℕ .

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

If F is a nonzero series such t hat F ( z ) = ∑ n = 0 ∞ a n z n . Subsequently, F φ F ψ = F ψ F φ = I if and only if a 0 2 = 1 and a n = 0 for all n ≥ 1 .

Note that, F φ F ψ = ∑ n = 0 ∞ a n C φ n ∑ m = 0 ∞ a m C ψ m = ∑ n = 0 ∞ ∑ m = 0 ∞ a n a m C φ n C ψ m = I , and F ψ F φ = ∑ m = 0 ∞ a m C ψ m ∑ n = 0 ∞ a n C φ n = ∑ m = 0 ∞ ∑ n = 0 ∞ a m a n C ψ m C φ n = I .

If and only if a 0 2 = 1 and a n = 0 for all n ≥ 1 . □

Consider φ to be invertible and F to be a nonzero series F ( z ) = ∑ n = 0 ∞ a n z n , then F φ is invertible such that F φ = λI where | λ | = 1 such that F φ − 1 = F φ − 1 .

Because φ is invertible, then φ − 1 exists. Applying Lemma 3.5 by taking ψ = φ − 1 , we have F φ F φ − 1 = F φ − 1 F φ = I if and only if a n = 0 for all n ≥ 1 , a 0 2 = 1 .

It follows that F φ = λI , | λ | = 1 and F φ − 1 = F φ − 1 . □