$\mathbb {T}$-holomorphy and quasiregularity

As we saw in the introduction, there is a Bloch theorem for K-quasiregular mappings. Then, to justify our Bloch theorem on the unit ball, we need to prove that the new class of mappings is not totally included in the class of $K$-quasiregular mappings. First, with Definition 1 it is easy to show the following characterization:

Remark 1   If $f\in TH(B^{2})$ then $f$ is $K$-quasiregular iff

$$\left\vert\frac{\partial f_{1}}{\partial z_{1}} \right\ver... ...2}}{\partial z_{1}}\right)^{2}\right\vert\mbox{ on $B^{2}$.}$$

The following examples will clearly show that a $\mathbb {T}$-holomorphic mapping is not necessarily quasiregular.

Example 1   Let $f(w)=u\cdot w$, where $u$ and $w$ are in $\mathbb {C}_{2}$. If $u\not\equiv 0$ is noninvertible then, f is not quasiregular. If $u$ is invertible, then $f$ is $K$-quasiregular for

$$K=\left(\frac{\vert u_1\vert^{2}+\vert u_2\vert^{2}}{\vert{u_1}^{2}+{u_2}^{2}\vert}\right)^{1/2},$$

where $u=u_1+u_2i_2.$ In particular, if $u\in\mathbb {C}_{1}$, then $f$ is conformal. However, we note in this example, that $f:\mathbb {C}^{2} \longrightarrow\mathbb {C}^{2}$ is a linear transformation which is nondegenerate if and only if $u$ is invertible.

In fact, any injective holomorphic mapping of the closed ball $\bar{B}^{2}$ into $\mathbb {C}^{2}$ is $K$-quasiregular for some $K$, but it is interesting to estimate $K$. In the last example, it is important to specify that $f$ is clearly an entire $\mathbb {T}$-holomorphic mapping and that the multiplication is the multiplication between bicomplex numbers. The next examples will show that there exist some nontrivial mappings which are simultaneously quasiregular and $\mathbb {T}$-holomorphic with $\mathcal{J}_{f}(0)=I$.

Example 2   If $f(w)=e^{w}$ then $f$ is $K$-quasiregular on $B^{2}$ iff $K\geq\sqrt{cosh(2)}$. Moreover, because $(e^{w})^\prime\vert _{w=0}=e^{0}=1$, we have $\mathcal{J}_{e^{w}}(0)=I$.

Proof. Let $f_{e1}$ and $f_{e2}$ be holomorphic functions on $\mathbb {C}_1$. Then we know that

$$f(z_{1}+z_{2}{i}_{2})=f_{e1}(z_{1}-z_{2}{i}_{1})e_{1}+f_{e2}(z_{1}+z_{2}{i}_{1})e_{2}$$

is $\mathbb {T}$-holomorphic with $f^\prime(z_{1}+z_{2}{i}_{2})=f^\prime_{e1}(z_{1}-z_{2}{i}_{1})e_{1}+ f^\prime_{e2}(z_{1}+z_{2}{i}_{1})e_{2}$. First, we seek conditions on $f^\prime_{e1}$ and $f^\prime_{e2}$ such that $f$ is $K$-quasiregular. By Remark 1 it is easy to show that $f$ is $K$-quasiregular on $B^{2}$ iff

$$\vert f^\prime_{e1}(z_{1}-z_{2}i_{1})+f^\prime_{e2}(z_{1}+z_{... ...(z_{1}-z_{2}i_{1})-f^\prime_{e2}(z_{1}+z_{2}i_{1})\vert^{2}\leq$$

$$4K^{2}\vert f^\prime_{e1}(z_{1}-z_{2}i_{1})f^\prime_{e2}(z_{1}+z_{2}i_{1})\vert$$

on $B^{2}$, that is

$$\vert f^\prime_{e1}(z_{1}-z_{2}i_{1})\vert^{2}+\vert f^\pri... ...{e1}(z_{1}-z_{2}i_{1})f^\prime_{e2}(z_{1}+z_{2}i_{1})\vert.$$ (4.3)

Now, $e^{z_{1}+z_{2}i_{2}}=e^{z_{1}-z_{2}i_{1}}e_{1}+e^{z_{1}+z_{2}i_{1}}e_{2}$, so in this case $f^\prime_{e1}(z_{1}-z_{2}i_{1})=e^{z_{1}-z_{2}i_{1}}$ and $f^\prime_{e2}(z_{1}-z_{2}i_{1})=e^{z_{1}+z_{2}i_{1}}$. Hence (4.1) becomes

$$\vert e^{-2z_{2}{i}_{1}}\vert+\vert e^{2z_{2}{i}_{1}}\vert\leq2K^{2}.$$

Let $z_{2}=x+yi_{1}$. Then, because $z_{1}+z_{2}i_{2}\in B^{2}$, we have that $\vert y\vert<1$. Moreover, $$\sup_{\{\vert y\vert<1\}}\{e^{2y}+e^{-2y}\}=e^{2}+e^{-2}$$ and then $K\geq \sqrt{\frac{e^{2}+e^{-2}}{2}}$, i.e. $K\geq\sqrt{cosh(2)}$.$\Box$

In Example 2, because $f^\prime(0)=1$, we know by Theorem 6 that $\delta_f\geq d$. Since the mapping $f$ is $K$-quasiregular, we already know that $\delta_f\geq {c}_K$ by Theorem 2. However, Theorems 2 and 6 merely assert the existence of the constants $c_K$ and $d$ without giving any estimates for these constants. From Theorem 9, on the other hand, we have an interesting lower estimate $\delta_f\geq\delta\geq\beta/\sqrt{2}$, by invoking lower estimates on the classical Bloch constant $\beta$ [6]. One can also give lower estimates for the Bloch constant for $K$-quasiregular holomorphic mappings [7].

The next example is a mapping $f$ for which $\delta_f\geq d$ by Theorem 6, but for which it is impossible to invoke Theorem 2.

Example 3   If $f(w)=w+\frac{w^{2}}{2}$, then $f$ is an entire $\mathbb {T}$-holomorphic (normalized) mapping, but for all $K$ is not $K$-quasiregular.

Proof. The function $f$ is normalized because $f^\prime(w)=1+w$ and then $f^\prime(0)=1$. Also, $w_{0}=-1/2+1/2j$ is in $B^{2}$ with $f^\prime(w_{0})=1/2+1/2j$ which is noninvertible. Hence $f$ cannot satisfy the criteria of Remark 1 at $w_{0}$ and then for all $K$, $f$ cannot be $K$-quasiregular. Actually, quasiregular holomorphic mappings in $\mathbb {C}^{2}$ are necessarily locally injective hence locally quasiconformal, but we wished to avoid invoking this rather deep theorem (see [8]).$\Box$

Now, we show that our class of mappings includes some $K$-quasiregular mappings for arbitrary values of K. Then, Theorems 6 and 9 give geometric information about $K$-quasiregular mappings for a subclass including different values of $K$.

Example 4   If $f(z_{1}+z_{2}i_{2})=\left(\frac{e^{n(z_{1}-z_{2}i_{1})}}{n}\right)e_{1}+(z_{1}+z_{2}i_{1})e_{2}$, then $f$ is an entire $\mathbb {T}$-holomorphic (normalized) mapping which is $K$-quasiregular in $B^{2}$ with $K$ becoming necessarily bigger as $n$ increases.

Proof. First, we see that $f$ is normalized because $f^\prime_{e1}(0)=e^{n0}=1$ and $f^\prime_{e2}(z_{1}+z_{2}i_{1})=1,$ i.e. $f^\prime(0)=1e_{1}+1e_{2}=1$. Also, by (4.1) $f$ is $K$-quasiregular on $B^{2}$ iff $\vert e^{n(z_{1}-z_{2}i_{1})}\vert^{2}+1 \leq 2K^{2}\vert e^{n(z_{1}-z_{2}i_{1})}\vert$ on $B^{2}$. Then, we must have

$$K\displaystyle{\geq\sqrt{\frac{\frac{1}{\vert e^{n(z_{1}-z_{2... ...}+\vert e^{n(z_{1}-z_{2}i_{1})}\vert}{2}} \mbox{ on $B^{2}$.}}$$

However, $${\sup_{\{w\in B^{2}\}}\sqrt{\frac{\frac{1}{\vert e^{n(z_{1}-z_{2}i... ...e{\sqrt{\frac{\frac{1}{e^{n(a^\prime+d^\prime)}}+e^{n(a^\prime+d^\prime)}}{2}}}$$, where $a^\prime+d^\prime$ can be taken to be positive. Finally we see that the last expression goes to infinity as $n\rightarrow\infty$.$\Box$

Finally, we give an example of a mapping which is $\mathbb {T}$-holomorphic and biholomorphic without being quasiregular on the unit ball of $\mathbb {C}^{2}$. In fact, we show that the class of $\mathbb {T}$-biholomorphic mappings cannot be totaly included in the class of quasiconformal mappings on the unit ball.

Example 5   If $f(z_{1}+z_{2}i_{2})=\left(-\frac{(z_{1}-z_{2}i_{1})^{2}}{2\sqrt{2}}+(z_{1}-z_{2}i_{1})\right)e_{1}+(z_{1}+z_{2}i_{1})e_{2}$ then $f$ is an entire $\mathbb {T}$-holomorphic (normalized) mapping which is biholomorphic but not quasiregular on the unit ball of $\mathbb {C}^{2}$.

Proof. Because $f^\prime_{ei}$ is nonzero on $h_{i}(D(0,\sqrt{2}))$ for $i=1,2$ and $B^2(0,1)\subseteq D(0,\sqrt{2})$, then $f$ is $\mathbb {T}$-biholomorphic on $B^2$. However, $f$ cannot be K-quasiregular on $B^2$ because the relationship (4.1) will fail for all $K$ as $z_{1}-z_{2}i_{1}\rightarrow \sqrt{2}$ with $z_1-z_2i_1\in B^2$.$\Box$