Bloch constant for $\mathbb {T}$-holomorphic mappings

Now we are ready to prove that there is a Bloch constant for the class of $\mathbb {T}$-holomorphic mapping of $\mathbb {C}^{2}$ on the unit Ball. The proof requires the natural ``disc" of $\mathbb {C}_{2}$ called the $\mathbb {C}_{2}$-disc and defined as follows: $D(a_1+a_2i_2,r_1,r_2):=\{z_1+z_2i_2:z_1+z_2i_2=w_1e_1+w_2e_2,\vert w_1-(a_1-a_2i_1)\vert<r_1,\vert w_2-(a_1+a_2i_1)\vert<r_2\}$. Also, we call $D(a_1+a_2i_2,r):=D(a_1+a_2i_2,r,r)$ the $\mathbb {C}_{2}$-disc of radius $r$. This kind of disc is in fact the $\mathbb {C}_{2}$-cartesian product of two discs of $\mathbb {C}_{1}$. Also, we need to remark that: $f^\prime(0)=1$ iff $\mathcal{J}_{f}(0)=I$ (the identity matrix).

Theorem 6   There is a positive constant $d$ such that if $f\in TH({B}^{2})$ with $f^\prime(0)\in\mathbb {C}_{1}\backslash\{0\}$, then f maps some subdomain of ${B}^{2}$ biholomorphically onto a ball of radius $d\cdot\vert f^\prime(0)\vert$. In particular, if $\mathcal{J}_{f}(0)=I$, the radius is $d$.

Proof. We note first that $D:=D(0,1)\subseteq{B}^{2}(0,1)$ [13]. In particular $f \in TH(D)$ and thus by Theorem 5, we can write $f=f_{e1}e_1+f_{e2}e_2$ on $D$ with $f_{ei}$ holomorphic on $D_i$ where $D_i$:=$h_i$(D)={ $w_{i}\in \mathbb {C}_{1}$:$\vert w_{i}\vert<1$} for $i=1,2$. Also, $f^\prime$(0)= $f^\prime_{e1}(0)e_1+f^\prime_{e2}(0)e_2$ [13, Theorem 24.3] and then $f^\prime(0)$ invertible implies $f^\prime_{e1}(0)\neq0$ and $f^\prime_{e2}(0)\neq0$. Then, by the Bloch theorem in one variable, there exists a positive constant $b$ such that $f_{ei}$ maps some subdomain $G_{i}$ of $D_{i}$ biholomorphically onto a disc $B_{i}(c_i,b\cdot \vert f^\prime_{ei}(0)\vert)$. Now, define

$$G:=\{z_{1}+z_{2}i_{2}\in\mathbb {C}_{2} : z_{1}+z_{2}i_{2}=w_{1}e_{1}+w_{2}e_{2}, (w_{1},w_{2})\in G_{1}\times G_{2}\}.$$

In fact, $G=G_1\times_e G_2$ is a domain of $\mathbb {C}_{2}$ and $G\subseteq D(0,1)\subseteq{B}^{2}(0,1)$. Then, $f=f_{e1}e_1+f_{e2}e_2$, with ${f}_{ei}:{G}_{i}\longrightarrow B_{i}(c_i,{b}\cdot \vert{f}^\prime_{ei}(0)\vert)$ biholomorphic for $i=1,2$. Let $c:={c}_{1}e_{1}+{c}_{2}e_{2}$, then

$$f:G\longrightarrow D(c,b\cdot \vert f^\prime_{e1}(0)\vert, b\cdot \vert f^\prime_{e2}(0)\vert)$$

is $\mathbb {T}$-biholomorphic. By a result in [13], $B^2(c,min(\frac{r_1}{\sqrt{2}},\frac{r_2}{\sqrt{2}}))\subseteq D(c,r_1,r_2)$ and so $B^2(c,min(b\cdot \frac{\vert f^\prime_{e1}(0)\vert}{\sqrt{2}},b\cdot \frac{\ver... ...eteq D(c,b\cdot \vert f^\prime_{e1}(0)\vert,b\cdot \vert f^\prime_{e2}(0)\vert)$. Thus, for the domain

$$G^\prime :=f^{-1}(B^2(c,min(b\cdot \frac{\vert f^\prime_{e1... ...} {\sqrt{2}})\subseteq G\subseteq D(0,1)\subseteq B^{2}(0,1),$$

the function $f$ is $\mathbb {T}$-biholomorphic from $G^\prime\subseteq B^2(0,1)$ to a ball of center $c$ and of radius equal to $b\cdot\frac{min(\vert f^\prime_{e1}(0)\vert,\vert f^\prime_{e2}(0)\vert)}{\sqrt{2}}$. Finally, we remark that $f^\prime(0)\in\mathbb {C}_{1}\backslash\{0\}$ implies $f^\prime_{e1}(0)=f^\prime_{e2}(0)$. Then, $d\cdot \vert f^\prime(0)\vert =b\cdot\frac{min(\vert f^\prime_{e1}(0)\vert,\vert f^\prime_{e2}(0)\vert)}{\sqrt{2}}$ if and only if $d=\frac{b}{\sqrt{2}}$, because $\vert f^\prime(0)\vert=\left[\frac{\vert f^\prime_{e1}(0)\vert^2+\vert f^\prime_{e2}(0)\vert^2}{2}\right]^{1/2}.\Box$

We have just seen that there exists a Bloch constant for the class of $\mathbb {T}$-holomorphic mappings with $\mathcal{J}_{f}(0)=I$ 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: $D(0,1)$, of bicomplex numbers.

Theorem 7   Let $f\in$TH($D$) with $f^\prime(0)\in\mathbb {C}_{1}\backslash\{0\}$. Then there is a positive constant $a$ such that $f$ maps some subdomain of $D$ biholomorphically onto a $\mathbb {C}_{2}$-disc of radius $a\cdot\vert f^\prime(0)\vert$. In particular, if $\mathcal{J}_{f}(0)=I$, the radius is $a$.

Proof. The proof is contained in the proof of Theorem 6.$\Box$

On this special domain of $\mathbb {C}_{2}$, 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 $f:U\longrightarrow\mathbb {C}^{2}$ be a $\mathbb {T}$-holomorphic injection with $U$ open, then $f$ is a $\mathbb {T}$-biholomorphic mapping from $U$ to $f(U)$.

Proof. Because $f$ is a holomorphic injection, we know [11] that $f(U)$ is open in $\mathbb {C}^{2}$, that $f$ is a biholomorphic mapping from $U$ to $f(U)$, and $det\mathcal{J}_{f}(z)\neq 0$, $\forall z\in U$. Thus $f^\prime(z)$ will be an invertible number $\forall z\in U$. Now, we want to prove that:

$$\lim_{\stackrel{\scriptstyle w \rightarrow w_{0}} {\scriptsc... ..._{0})}{w-w_{0}} \mbox{ } exists,\mbox{ }\forall w_{0}\in f(U).$$

Let $z=f^{-1}(w)$ and $z_{0}=f^{-1}(w_{0})$. Then for $w-w_{0}$ invertible we obtain: $\frac{f^{-1}(w)-f^{-1}(w_{0})}{w-w_{0}}=\frac{z-z_{0}}{f(z)-f(z_{0})}$. Also, because $f\in TH(U)$ and $f^\prime(z_{0})$ invertible we have: $(f(z)-f(z_0))/(z-z_0)$ is invertible for $z$ near $z_0$ and

$$\lim_{\stackrel{\scriptstyle z \rightarrow z_{0}} {\script... ...\prime(z_{0})} \mbox{ } exists,\mbox{ }\forall z_{0}\in U.$$ (3.1)

Thus, $\forall\epsilon>0$, there exists $\delta_{1}>0$ such that $\left\vert\frac{z-z_{0}}{f(z)-f(z_{0})}-\frac{1}{f^\prime(z_{0})}\right\vert<\frac{\epsilon}{2}$, whenever $\vert z-z_0\vert<\delta_{1}$ and $z-z_0$ is invertible. Choose $\delta>0$ such that $\vert w-w_0\vert<\delta$ implies $\vert z-z_0\vert<\delta_{1}$. Then,

$$\left\vert\frac{z-z_{0}}{f(z)-f(z_{0})}-\frac{1}{f^\prime(z_{0})}\right\vert<\frac{\epsilon}{2},$$ (3.2)

whenever $\vert w-w_0\vert<\delta$ and both $w-w_0$ and $z-z_0$ are invertible. If $\vert w-w_0\vert<\delta$ and $w-w_0$ invertible but $z-z_0$ is not, there always exists $\epsilon^\prime>0$ such that $\vert(z+\epsilon^\prime)-z_0\vert<\delta_1$, both $(z+\epsilon^\prime)-z_0$ and $f(z+\epsilon^\prime)-f(z_0)$ are invertible and $\left\vert\frac{z-z_{0}}{f(z)-f(z_{0})}-\frac{1}{f^\prime(z_{0})}\right\vert\... ...\frac{1}{f^\prime(z_{0})}\right\vert <\frac{\epsilon}{2}+\frac{\epsilon}{2}.$ Thus, $\vert w-w_0\vert<\delta$ and $w-w_0$ invertible implies (3.2). This concludes the proof.$\Box$

It is now possible to describe the exact value of the Bloch constant for the class of $\mathbb {T}$-holomorphic mappings with $\mathcal{J}_{f}(0)=I$ on the unit $\mathbb {C}_{2}$-disc. In fact, it turns out, that it coincides precisely with the classical Bloch constant.

Notation 1   $\alpha:=inf\{\alpha_{f}:f\in TH(D(0,1))\mbox{ with }\mathcal{J}_{f}(0)=I\},$ $\hspace{0.75in}\alpha_{f}:=sup\{a:f(D(0,1))\mbox{ contains a univalent $\mathbb {C}_{2}$-disc of radius $a$}\},$

Theorem 8  

$$\alpha=\beta,$$

where $\beta$ is the Bloch constant of one variable.

Proof. Again, let $f:=f_{e1}e_1+f_{e2}e_2$ on $D$ with $f_{ei}$ holomorphic on $D_i$ where $D_i$:=$h_i$(D)={ $w_{i}\in \mathbb {C}_{1}$:$\vert w_{i}\vert<1$} for i=1, 2. Moreover, suppose $f^\prime_{ei}(0)=1$ for i=1, 2; then, by the definition of the Bloch constant for one variable, $\forall\epsilon>0$ there exists a univalent disc of radius $c_\epsilon$ for $f_{e1}$ and $f_{e2}$ such that $c_\epsilon \geq\beta-\epsilon$. Hence, $\forall f\in TH(D)$ with $\mathcal{J}_{f}(0)=I$, $\alpha_f \geq\beta-\epsilon$ $\forall\epsilon>0$ and thus $\alpha\geq\beta$. Also, we know by [18] that there exists $g\in H(B)$ such that:

$$g^\prime(0)=1\mbox{ and } \beta_{g}=\beta.$$

Let us now define:

$$f(z_1+z_2i_2):=g(z_1-z_2i_1)e_1+g(z_1+z_2i_1)e_2.$$

Then, $f_{1}(z_1,z_2)=\frac{g(z_1-z_2i_1)+g(z_1+z_2i_1)}{2}$ and $f_{2}(z_1,z_2)=\frac{g(z_1-z_2i_1)-g(z_1+z_2i_1)}{2}i_1$ so by Theorem 4 and the remark after Theorem 3, $f \in TH(D)$ with $f^\prime(0)=1$. We want to show that for this $f$, $\alpha_{f}\leq\beta$. If not, i.e. $\alpha_{f}>\beta$, then $f(D(0,1))$ contains a univalent $\mathbb {C}_2$-disc of radius $c^\prime$ such that $c^\prime>\beta$. Thus $f$ maps a subdomain $G\subseteq D(0,1)$ biholomorphically onto a $\mathbb {C}_2$-disc of radius $c^\prime$. But $f\in TH(D(0,1))$, so by Lemma 1:

$$f:G\rightarrow D(0,c^\prime)\mbox{ is $\mathbb {T}$-biholomorphic}.$$

This is a contradiction because Theorem 5 applied to $f$ and $f^{-1}$ forces $g$ to map the subdomain $h_i(G)\in B$ biholomorphically onto a disc of radius $c^\prime$ for $i=1,2.\Box$

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

Definition 5   We say that $f$ has a $\mathbb {T}$-univalent ball if $f$ has a $\mathbb {T}$-biholomorphic univalent ball.

Notation 2   $\delta:=inf\{\delta_{f}:f\in TH({B}^{2}(0,1))\mbox{ with }\mathcal{J}_{f}(0)=I\},$ $\hspace{0.75in}\delta_{f}:=sup\{d:f({B}^{2}(0,1)) \mbox{ contains a univalent ball of radius $d$}\},$ $\hspace{0.75in}\delta^\prime:=inf\{\delta_{f}^\prime:f\in TH({B}^{2}(0,1))\mbox{ with } \mathcal{J}_{f}(0)=I\}.$ $\hspace{0.75in}\delta_{f}^\prime:=sup\{d:f({B}^{2}(0,1))\mbox{ contains a $\mathbb {T}$-univalent ball of radius $d$}\},$

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

Theorem 9  

$$\frac{\beta}{\sqrt{2}}\leq\delta\leq\sqrt{2}\beta,$$

where $\beta$ is the Bloch constant of one variable.

Proof. First we prove that $\frac{\beta}{\sqrt{2}}\leq\delta^\prime\leq\delta$. Suppose $f\in TH(B^{2}(0,1))$ with $\mathcal{J}_{f}(0)=I$. By the proof of Theorem 6, for every $b<\beta$, $f({B}^{2}(0,1))$ contains, in fact, a $\mathbb {T}$-univalent ball of radius $d=\frac{b}{\sqrt{2}}$. In fact, $\forall\epsilon>0$ there is a ${b}_{\epsilon}$ such that ${b}_{\epsilon}\geq\beta-\epsilon$, so setting ${d}_{\epsilon}=\frac{{b}_{\epsilon}}{\sqrt{2}}$, we have ${d}_{\epsilon}\geq\frac{{\beta}-\epsilon}{\sqrt{2}}$ which implies ${\delta}_{f}^\prime\geq \frac{\beta -\epsilon}{\sqrt{2}}$, $\forall\epsilon>0$ and thus

$$\delta_{f}^\prime\geq \frac{\beta}{\sqrt{2}}.$$

Hence $\frac{\beta}{\sqrt{2}}\leq$ inf $\delta_{f}^\prime:=\delta^\prime$. Finally, because a $\mathbb {T}$-univalent ball for $f$ is a univalent ball, we have $\delta_{f}^\prime\leq\delta_{f}$ and then $\delta^\prime\leq\delta$. In fact, by Lemma 1 we have $\delta=\delta^\prime$. The second part of the proof is to prove that $\delta\leq\sqrt{2}\beta$. We will prove this by contradiction. Suppose that there exists $\epsilon>0$ such that $\delta>\sqrt{2}(\beta + \epsilon)$. Then

$$\delta_{f}>\sqrt{2}(\beta + \epsilon),\hspace{0.1in} \forall f\in TH_{N}(B^{2}(0,1)),$$

where $TH_{N}(B^{2}(0,1)):=\{ f\in TH(B^{2}(0,1))$ such that $\mathcal{J}_{f}(0)=I\}$. Hence, $\forall f\in TH_N(B^{2}(0,1))$ there exists, $c_{f}$ such that $c_f>\sqrt{2}(\beta + \epsilon)$ and $f$ maps a subdomain $M$ of $B^{2}(0,1)$ biholomorphically onto $B^{2}(w_f,c_f)$, a ball centered at $w_{f}$ of radius $c_f$. However, by Theorem 8, we know that $\mbox{ }\forall\epsilon>0$ there exists $g\in TH_{N}(D(0,1))$ such that $g(D(0,1))$ cannot contain a univalent $\mathbb {C}_{2}$-disc of radius $r>\beta +\epsilon$. Also $B^{2}(0,\frac{1}{\sqrt{2}}) \subseteq D(0,1)$, so $g\in TH_{N}(B^2(0,\frac{1}{\sqrt{2}}))$. Let us define:

$$g^*(w)=\sqrt{2}g(\frac{w}{\sqrt{2}})\mbox{ on } B^2(0,1).$$

Then $g^*(w)\in TH_{N}(B^2(0,1))$. Hence, there exists $c_{g^*}$ such that $c_{g^*}>\sqrt{2}(\beta + \epsilon)$ and $g^*$ maps a subdomain $W$ of $B^{2}(0,1)$ biholomorphically onto $B^{2}(w_{g^*},c_{g^*})$. Then, $g$ maps the subdomain $\frac{W}{\sqrt{2}}$ biholomorphically onto $B^2(\frac{w_{g^*}}{\sqrt{2}},\frac{c_{g^*}}{\sqrt{2}})$. But $D(\frac{w_{g^*}}{\sqrt{2}},\frac{c_{g^*}}{\sqrt{2}})\subseteq B^{2}(\frac{w_{g^*}}{\sqrt{2}},\frac{c_{g^*}}{\sqrt{2}})$, so $g$ maps, biholomorphically

$$g^{-1}(D(\frac{w_{g^*}}{\sqrt{2}},\frac{c_{g^*}}{\sqrt{2}})) ... ...x{ onto } D(\frac{w_{g^*}}{\sqrt{2}},\frac{c_{g^*}}{\sqrt{2}}).$$

This contradicts the way in which g was chosen because $g^{-1}(D(\frac{w_{g^*}}{\sqrt{2}},\frac{c_{g^*}}{\sqrt{2}}))\subseteq \frac{W}{\sqrt{2}}\subseteq D(0,1)$ with $\frac{c_{g^*}}{\sqrt{2}}>\beta +\epsilon.\Box$