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From a journal entitled Certain subclass of starlike functions by Gao and Zhou in 2007, they mentioned that " since $ k(z)=\frac{z}{1-zt}$ is convex in open unit disk $E,z:|z|<1$, $k(\bar{z})= \bar{k(z)}$ and $k(z)$ maps real axis to real axis". How to show that it is convex in $E$ ?

The definition of convex function from Univalent function Volume 1 by A.W.Goodman: A set of domain, D in the plane is called convex if for every pair of point $w_{1}$ and $w_{2}$ in the interior of D, the line segment joining $w_{1}$ and $w_{2}$ is also in the interior of D. If a function $f(z)$ maps E onto a convex domain, then $f(z)$ is called a convex function.

Thank you.

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What is your definition of "convex" for a complex function? –  Henning Makholm Dec 14 '11 at 15:21
    
Now, what is $t$ that you've added to the definition of $k$? –  Thomas Andrews Dec 14 '11 at 15:38
    
@HenningMakholm: already put it in my details question –  DRN Dec 14 '11 at 15:50
    
@ThomasAndrews:I just at the beginning stage to understand this, that is all they have been wrote in that journal. –  DRN Dec 14 '11 at 15:56

1 Answer 1

If your original definition, $k(z)=\frac{z}{1-z}$ is what you want, then $k$ is a Moebius transformation which send $1$ to $\infty$. Any Moebius transformation is $1-1$ and onto the Riemann sphere, $\mathbb C \cup \{\infty\}$, and a Moebius transformation such that $k(1)=\infty$ has the property that any circle through $1$ gets mapped to a line, and the interiors and exteriors get mapped to half-planes on either side of that line.

So $k$ maps the interior of the unit ball onto a half-plane, which is necessarily convex.

In general, any Moebius transformation $m(z)$ sends any ball either onto a half-plane, another ball, or the complement of the closure of the ball. So for $m(z)$ to be convex on a ball, it is necessary and sufficient to prove that either $m(z_0)=\infty$ for some $z_0$ on the boundary of the ball, or $m(z)$ is bounded in the ball.

In the case $k_t(z)=\frac{z}{1-tz}$, this function is bounded on the unit ball if $|t|<1$ and has $z_0=\bar{t}$ with $h(z_0)=\infty$. If $|t|>1$, then you can easily show that $k_t$ is unbounded on $E$.

So, $k_t$ is convex on $E$ if and only if $|t|\leq 1$.

For a general Moebius transformation, $m$, and ball, $B$, $m$ is convex on $B$ if and only if $m^{-1}(\infty)\notin B$.

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yes,it is true that $|t|\leq 1$. I am not familiar with ball, $B$ and unit ball, is it same with disk or unit disk? –  DRN Dec 14 '11 at 16:06
    
Yes, ball is just something we use in higher dimensions. Actually, a lot of places where I wrote "unit ball" I meant "ball" more generally, and everywhere I wrote "ball" you can just read it as "disk." –  Thomas Andrews Dec 14 '11 at 16:10
    
Depending on what you definition of disk is, it might differ from the a disk in that a ball does not contain the boundary. So a ball is always a set of the form: $B_{z_0,r}=\{z\in\mathbb C: |z-z_0|<r\}$ –  Thomas Andrews Dec 14 '11 at 16:16
    
Really good information. I am clear with your answers. Thank you so much. –  DRN Dec 14 '11 at 16:20

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