Jtdylktuy

The following is a refinement of Schwarz's Lemma. I am trying to prove for $$|z| leq R$$ that
$$|f(z)|leq M left(frac{M|z|+|a|R}{|a||z|+MR}right)$$ where $$f(z)$$ is analytic in $$|z|leq R$$ such that $$|f(z)| leq M$$ on $z=R$ and $f(0)=a$ where $|a|< M$. I know we want to take a derivative, but still have not come up with a solution.

Consider $$F(z)=Mleft(frac{f(Rz)-a}{overline{a}f(Rz)-M^2}right)$$ If $f$ is analytic in $|z|leq 1$ and $|f(z)|leq 1$ in $|z|leq 1$, then $$|f(z)|leq frac{(1-|a|)|z^2|+|bz|+|a|(1-|a|)}{|a|(1-|a|)|z^2|+|bz|+(1-|a|)}$$ for $|z|leq 1$ where $a=f(0)$ and $b=f '(0)$. Thus $$f(z)=frac{a+frac{b}{1+a}z-z^2}{1-frac{b}{1+a}z-az^2}$$

Consider $F(z)=Mleft(frac{f(Rz)-a}{overline{a}f(Rz)-M^2}right)$,$hspace{0.3cm}$ $|z|leq 1$ $hspace{0.3cm} Longrightarrow|Rz|leq R hspace{0.5cm}$

Let $$hspace{0.2cm} overline{a}f(Rz)-M^2=0 hspace{0.1cm}\ Longleftrightarrow |a||f(Rz)|= M^2 Longleftrightarrow |f(Rz)|=frac{M^2}{|a|}>M$$ $|F(z)| leq Mhspace{0.3cm} $ on $hspace{0.3cm}|z|=R hspace{0.1cm} Longleftrightarrow |f(Rz)|leq M hspace{0.3cm}$ on $hspace{0.3cm}|z|leq 1 $ $$Longrightarrow|overline{a}f(Rz)| leq |a|M <M^2 Longrightarrow overline{a}f(Rz)-M^2neq 0$$ in $|z| leq 1 $. $$$$So, $F(z)$ is analytic in $|z|leq 1$. $$F(0)=Mleft(frac{f(0)-a}{overline{a}f(0)-M^2}right)=0$$ $$ $$Let $f(Rz)=w$, $hspace{0.4cm}w leq Mhspace{0.3cm}$ on $hspace{0.3cm}|z|=1$. $hspace{0.3cm}$ Now suppose $$left|M frac{w-a}{overline{a}w-M^2}right|^2=M^2 frac{|w|^2 + |a|^2-2hspace{0.1cm} Re(overline{a}w)}{M^4+|a|^2 |w|^2-2hspace{0.1cm} Re(M^2 overline{a}w)}leq 1$$ $Longleftrightarrow M^2 |w|^2+M^2|a|^2leq M^4+|a|^2 |w|^2$ $$Longleftrightarrow M^2 |a|^2-|a|^2|w|^2leq M^4-M^2 |w|^2 $$ $$Longleftrightarrow |a|^2leq M^2 Longleftrightarrow |a| leq M $$ $$$$So,$hspace{0.3cm}F(z) leq1 hspace{0.2cm}$ on $hspace{0.2cm}|z|=1.$ $hspace{0.2cm}$ By Schwarz's Lemma, we have $|F(z)|
leq |z|$, $hspace{0.2cm}$so $$Mleft|frac{f(Rz)-a}{overline{a}f(Rz)-M^2}right| leq|z| Longrightarrow left|frac{frac{f(Rz)}{M}-frac{a}{M}}{1-frac{{a}}{M}frac{f(Rz)}{M}}right|hspace{0.3cm} leq |z|$$ $$ Longrightarrow frac{frac{|f(Rz)|}{M}-frac{|a|}{M}}{1-frac{{|a|}}{M}frac{|f(Rz)|}{M}}hspace{0.3cm} leq |z|$$ $$Longrightarrow frac{|f(Rz)|}{M}-frac{|a|}{M} hspace{0.2cm}leq hspace{0.1cm} |z|-frac{|z||a||f(Rz)|}{M^2}$$ $$Longrightarrow frac{|f(Rz)|}{M}+frac{|z||a||f(Rz)|}{M^2} hspace{0.2cm}leq hspace{0.1cm} |z|+frac{|a|}{M}$$ $$Longrightarrow |f(Rz)| left(frac{1}{M}+hspace{0.2cm}frac{|z||a|}{M^2}right)leq hspace{0.1cm} |z|+frac{|a|}{M}$$ $$Longrightarrow |f(Rz)|leq frac{|z|+frac{|a|}{M}}{frac{1}{M}+frac{|z||a|)}{M^2}}hspace{0.3cm}= Mleft( frac{M|z|+|a|}{M+|z||a|}right)$$ Now let $hspace{0.2cm} u=
Rz$, $hspace{0.3cm}|z|leq 1$, $hspace{0.2cm}$then $hspace{0.2cm} z=frac{u}{R}$, $hspace{0.2cm}$and $hspace{0.2cm}|u|leq R$.$hspace{0.3cm}$ So, $$|f(u)|leq M left(frac{Mfrac{|u|}{R}+|a|}{M+|a|frac{|u|}{R}}right)=Mleft( frac{M|u|+|a|R}{|a||u|+MR}right)hspace{0.2cm}$$ Therefore, the result follows replacing $u$ with $z$. $hspace{0.9cm}Box$