suivant: -holomorphy and quasiregularity monter: A Bloch Constant for précédent: Preliminaries

# Bloch constant for -holomorphic mappings

Now we are ready to prove that there is a Bloch constant for the class of -holomorphic mapping of on the unit Ball. The proof requires the natural disc" of called the -disc and defined as follows: . Also, we call the -disc of radius . This kind of disc is in fact the -cartesian product of two discs of . Also, we need to remark that: iff (the identity matrix).

Theorem 6   There is a positive constant such that if with , then f maps some subdomain of biholomorphically onto a ball of radius . In particular, if , the radius is .

Proof. We note first that [13]. In particular and thus by Theorem 5, we can write on with holomorphic on where :=(D)={ :} for . Also, (0)= [13, Theorem 24.3] and then invertible implies and . Then, by the Bloch theorem in one variable, there exists a positive constant such that maps some subdomain of biholomorphically onto a disc . Now, define

In fact, is a domain of and . Then, , with biholomorphic for . Let , then

is -biholomorphic. By a result in [13], and so . Thus, for the domain

the function is -biholomorphic from to a ball of center and of radius equal to . Finally, we remark that implies . Then, if and only if , because

We have just seen that there exists a Bloch constant for the class of -holomorphic mappings with on the unit ball and now we wish to estimate this Bloch constant. For this, we need first to work on the natural unit disc: , of bicomplex numbers.

Theorem 7   Let TH() with . Then there is a positive constant such that maps some subdomain of biholomorphically onto a -disc of radius . In particular, if , the radius is .

Proof. The proof is contained in the proof of Theorem 6.

On this special domain of , it is possible to find the exact value of the Bloch constant. For this, we need to prove the following lemma, which is itself of interest:

Lemma 1   Let be a -holomorphic injection with open, then is a -biholomorphic mapping from to .

Proof. Because is a holomorphic injection, we know [11] that is open in , that is a biholomorphic mapping from to , and , . Thus will be an invertible number . Now, we want to prove that:

Let and . Then for invertible we obtain: . Also, because and invertible we have: is invertible for near and
 (3.1)

Thus, , there exists such that , whenever and is invertible. Choose such that implies . Then,
 (3.2)

whenever and both and are invertible. If and invertible but is not, there always exists such that , both and are invertible and Thus, and invertible implies (3.2). This concludes the proof.

It is now possible to describe the exact value of the Bloch constant for the class of -holomorphic mappings with on the unit -disc. In fact, it turns out, that it coincides precisely with the classical Bloch constant.

Notation 1

Theorem 8

where is the Bloch constant of one variable.

Proof. Again, let on with holomorphic on where :=(D)={ :} for i=1, 2. Moreover, suppose for i=1, 2; then, by the definition of the Bloch constant for one variable, there exists a univalent disc of radius for and such that . Hence, with , and thus . Also, we know by [18] that there exists such that:

Let us now define:

Then, and so by Theorem 4 and the remark after Theorem 3, with . We want to show that for this , . If not, i.e. , then contains a univalent -disc of radius such that . Thus maps a subdomain biholomorphically onto a -disc of radius . But , so by Lemma 1:

This is a contradiction because Theorem 5 applied to and forces to map the subdomain biholomorphically onto a disc of radius for

The followings definitions are used to prove the main result of this article:

Definition 5   We say that has a -univalent ball if has a -biholomorphic univalent ball.

Notation 2

It is now possible to find the following estimates for our Bloch constant on the unit ball:

Theorem 9

where is the Bloch constant of one variable.

Proof. First we prove that . Suppose with . By the proof of Theorem 6, for every , contains, in fact, a -univalent ball of radius . In fact, there is a such that , so setting , we have which implies , and thus

Hence inf . Finally, because a -univalent ball for is a univalent ball, we have and then . In fact, by Lemma 1 we have . The second part of the proof is to prove that . We will prove this by contradiction. Suppose that there exists such that . Then

where such that . Hence, there exists, such that and maps a subdomain of biholomorphically onto , a ball centered at of radius . However, by Theorem 8, we know that there exists such that cannot contain a univalent -disc of radius . Also , so . Let us define:

Then . Hence, there exists such that and maps a subdomain of biholomorphically onto . Then, maps the subdomain biholomorphically onto . But , so maps, biholomorphically

This contradicts the way in which g was chosen because with

suivant: -holomorphy and quasiregularity monter: A Bloch Constant for précédent: Preliminaries
Dominic Rochon
2000-07-26