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%TCIDATA{Created=Tue Jan 02 13:42:18 2001}
%TCIDATA{LastRevised=Thu Apr 19 16:20:54 2001}
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\newtheorem{theorem}{Teorema}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{proof}[theorem]{Bukti}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Akibat}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definisi}
\newtheorem{example}[theorem]{Contoh}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lema}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposisi}
\newtheorem{remark}[theorem]{Catatan}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{Proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}

\title{FUNGSI UNIVALEN}
\author{Endang Cahya}
\maketitle

\bigskip 

\section{Konstruksi  Fungsi   $\mu $ \ Regular}

\bigskip {\LARGE Hasil J.L. Schiff dan W.J. Walker 1990}

\begin{theorem}
{\LARGE Misalkan }$f(z)${\LARGE \ adalah fungsi }$\mu ${\LARGE \ regular
pada disk }$\left| z\right| \leq a.${\LARGE \ Maka untuk }$\left| z\right|
<a,$%
\begin{equation*}
{\LARGE f(z)=}\sum_{n=0}^{\infty }{\LARGE a}_{n}{\LARGE z}^{n}{\LARGE \Psi }%
_{n}\left( \mu r\right) {\LARGE +}\sum_{n=0}^{\infty }\overline{a_{n}}\frac{%
\mu }{2(n+1)}\overline{z^{n+1}}{\LARGE \Psi }_{n+1}\left( \mu r\right) 
\end{equation*}
{\LARGE dimana} 
\begin{equation*}
{\LARGE a}_{n}{\LARGE =}\frac{1}{2\pi a^{n}\Psi _{n}\left( \mu a\right) }%
\int_{0}^{2\pi }{\LARGE e}^{-in\theta }{\LARGE f(ae}^{i\theta }{\LARGE %
)d\theta }
\end{equation*}
\end{theorem}

\begin{proof}
{\LARGE Karena }$f${\LARGE \ panharmonik pada disk, maka }$f${\LARGE \ dapat
diekspansi dalam bentuk } 
\begin{equation*}
{\LARGE f(re}^{i\theta }{\LARGE )=}\sum_{n=0}^{\infty }{\LARGE a}_{n}{\LARGE %
z}^{n}{\LARGE \Psi }_{n}\left( \mu r\right) {\LARGE +}\sum_{n=1}^{\infty }%
{\LARGE a}_{-n}\overline{z^{n}}{\LARGE \Psi }_{n}\left( \mu r\right) 
\end{equation*}
{\LARGE dengan } 
\begin{equation*}
{\LARGE a}_{n}{\LARGE =}\frac{1}{2\pi a^{n}\Psi _{n}\left( \mu a\right) }%
\int_{0}^{2\pi }{\LARGE e}^{-in\theta }{\LARGE f(ae}^{i\theta }{\LARGE %
)d\theta .}
\end{equation*}
{\LARGE Selanjutnya untuk menunjukkan } 
\begin{equation*}
{\LARGE a}_{-n-1}{\LARGE =}\frac{\mu }{2(n+1)}\overline{a_{n}}{\LARGE %
,n=0,1,2,...}
\end{equation*}
{\LARGE cukup menyamakan koefisien }$z^{n+1}${\LARGE \ pada kedua sisi
persamaan }$Lf=\mu \overline{f}.$
\end{proof}

\bigskip

{\LARGE Ide Teorema di atas}

{\LARGE Misalkan u(r,}$\theta ${\LARGE ) panharmonik pada disk x}$%
^{2}+y^{2}\leq a^{2},${\LARGE \ maka untuk 0}$\leq r<a,$%
\begin{equation*}
{\LARGE u(r,\theta )=}\sum_{n=-\infty }^{\infty }{\LARGE c}_{n}{\LARGE I}%
_{\left| n\right| }\left( \mu r\right) {\LARGE e}^{in\theta }
\end{equation*}

{\LARGE dimana} 
\begin{equation*}
{\LARGE c}_{n}{\LARGE =}\frac{1}{2\pi I_{\left| n\right| }\left( \mu
a\right) }\int_{0}^{2\pi }{\LARGE e}^{-in\theta }{\LARGE u(a,\theta )d\theta
.}
\end{equation*}

{\LARGE dan} 
\begin{equation*}
{\LARGE I}_{n}\left( x\right) {\LARGE =}\frac{1}{n!}\left( \frac{x}{2}%
\right) ^{n}{\LARGE \Psi }_{n}{\LARGE (x)}
\end{equation*}

\begin{equation*}
{\LARGE \Psi }_{n}{\LARGE (x)=1+}\frac{\left( \frac{x}{2}\right) ^{2}}{1(n+1)%
}{\LARGE +}\frac{\left( \frac{x}{2}\right) ^{4}}{1.2(n+1)\left( n+2\right) }%
{\LARGE +...}
\end{equation*}

{\LARGE Sekarang perhatikan penulisan} 
\begin{eqnarray*}
{\LARGE u(r,\theta )} &=&\sum_{n=-\infty }^{\infty }{\LARGE c}_{n}{\LARGE I}%
_{\left| n\right| }\left( \mu r\right) {\LARGE e}^{in\theta } \\
&=&\sum_{n=-\infty }^{\infty }{\LARGE c}_{n}\frac{1}{n!}\left( \frac{\mu r}{2%
}\right) ^{n}{\LARGE \Psi }_{\left| n\right| }\left( \mu r\right) {\LARGE e}%
^{in\theta } \\
&=&\sum_{n=-\infty }^{\infty }{\LARGE c}_{n}\frac{1}{n!}\left( \frac{\mu }{2}%
\right) ^{n}{\LARGE \Psi }_{\left| n\right| }\left( \mu r\right) {\LARGE r}%
^{n}{\LARGE e}^{in\theta } \\
&=&\sum_{n=-\infty }^{\infty }{\LARGE a}_{n}\Psi _{\left| n\right| }\left(
\mu r\right) z^{n} \\
&=&\sum_{n=0}^{\infty }{\LARGE a}_{n}\Psi _{\left| n\right| }\left( \mu
r\right) z^{n}+\sum_{n=1}^{\infty }{\LARGE a}_{-n}\Psi _{\left| n\right|
}\left( \mu r\right) \overline{z^{n}}.
\end{eqnarray*}

{\LARGE dan penulisan} 
\begin{eqnarray*}
{\LARGE a}_{n} &=&\frac{1}{n!}\left( \frac{\mu }{2}\right) ^{n}{\LARGE c}_{n}
\\
&=&\frac{1}{n!}\left( \frac{\mu }{2}\right) ^{n}\frac{1}{2\pi I_{\left|
n\right| }\left( \mu a\right) }\int_{0}^{2\pi }{\LARGE e}^{-in\theta }%
{\LARGE u(a,\theta )d\theta } \\
&=&\frac{1}{2\pi a^{n}\Psi \left( \mu a\right) }\int_{0}^{2\pi }{\LARGE e}%
^{-in\theta }{\LARGE u(a,\theta )d\theta }
\end{eqnarray*}

{\LARGE Selanjutnya untuk menunjukkan} $a_{-n-1}=\frac{\mu }{2(n+1)}%
\overline{a_{n}},$ {\LARGE dengan n = 0,1,2,3,....dapat digunakan} $Lf=\mu 
\overline{f}.$

\begin{eqnarray*}
{\LARGE Lf} &=&{\LARGE e}^{i\theta }\left( \frac{\partial }{\partial r}+%
\frac{1}{r}i\frac{\partial }{\partial \theta }\right) {\LARGE (}%
\sum_{n=0}^{\infty }{\LARGE a}_{n}{\LARGE z}^{n}{\LARGE \psi }_{n}{\LARGE %
(\mu r)+}\sum_{n=1}^{\infty }{\LARGE a}_{-n}\overline{z^{n}}{\LARGE \psi }%
_{n}{\LARGE (\mu r))} \\
&=&\sum_{n=0}^{\infty }\left( a_{n}e^{i\theta (n+1)}r^{n+1}.\frac{\mu ^{2}}{%
2(n+1)}\psi _{n+1}(\mu r)\right) {\LARGE +} \\
&&\sum_{n=1}^{\infty }\left( a_{-n}r^{n-1}e^{i\theta (1-n)}\left( 2n\psi
_{n}(\mu r)+\frac{r^{2}\mu ^{2}}{2(n+1)}\psi _{n+1}(\mu r)\right) \right)  \\
&=&\sum_{n=0}^{\infty }\left( a_{n}z^{n+1}.\frac{\mu ^{2}}{2(n+1)}\psi
_{n+1}(\mu r)\right) +\sum_{n=1}^{\infty }(a_{-n}\overline{z^{n-1}}(2n\psi
_{n}(\mu r)+\frac{r^{2}\mu ^{2}}{2(n+1)}\psi _{n+1}(\mu r))) \\
&& \\
&=&\sum_{n=0}^{\infty }\left( a_{n}z^{n+1}.\frac{\mu ^{2}}{2(n+1)}\psi
_{n+1}(\mu r)\right) {\Large +}\sum_{n=1}^{\infty }(a_{-n}\overline{z^{n-1}}%
2n(\psi _{n}(\mu r)+\frac{\left( \frac{\mu r}{2}\right) ^{2}}{n(n+1)}\psi
_{n+1}(\mu r)))
\end{eqnarray*}
{\LARGE dan}
\begin{eqnarray*}
&& \\
{\LARGE \mu }\overline{f\text{ }} &=&\sum_{n=0}^{\infty }{\LARGE \mu }%
\overline{a_{n}}\overline{z^{n}}{\LARGE \psi }_{n}{\LARGE (\mu r)+}%
\sum_{n=1}^{\infty }{\LARGE \mu }\overline{a_{-n}}{\LARGE z}^{n}{\LARGE \psi 
}_{n}{\LARGE (\mu r)} \\
&=&\sum_{n=0}^{\infty }{\LARGE \mu }\overline{a_{-n-1}}{\LARGE z}^{n+1}%
{\LARGE \psi }_{n+1}{\LARGE (\mu r)+}\sum_{n=1}^{\infty }{\LARGE \mu }%
\overline{a_{n-1}}\overline{z^{n-1}}{\LARGE \psi }_{n-1}{\LARGE (\mu r).}
\end{eqnarray*}
{\LARGE Dengan menyamakan koefisien dari }$z^{n+1}${\LARGE \ diperoleh
hubungan}

\begin{equation*}
{\LARGE a}_{n}\frac{\mu ^{2}}{2(n+1)}{\LARGE \psi }_{n+1}{\LARGE (\mu r)=\mu 
}\overline{a_{-n-1}}{\LARGE \psi }_{n+1}{\LARGE (\mu r)}
\end{equation*}
{\LARGE atau }
\begin{equation*}
{\LARGE a}_{n}\frac{\mu }{2(n+1)}{\LARGE =}\overline{a_{-n-1}}
\end{equation*}
{\LARGE atau }
\begin{equation*}
\overline{a_{n}}\frac{\mu }{2(n+1)}{\LARGE =a}_{-n-1}.
\end{equation*}

\bigskip 

\begin{theorem}
{\LARGE Misal }$f(z)=\sum_{n=0}^{\infty }a_{n}z^{n}\Psi _{n}\left( \mu
r\right) +\sum_{n=0}^{\infty }\overline{a_{n}}\frac{\mu }{2(n+1)}\overline{%
z^{n+1}}\Psi _{n+1}\left( \mu r\right) ${\LARGE \ konvergen seragam pada
disk }$\left| z\right| <r<a,${\LARGE \ maka }$f${\LARGE \ fungsi }$\mu $%
{\LARGE \ regular pada }$\left| z\right| <a.$
\end{theorem}

\begin{proof}
{\LARGE Perhatikan bahwa} 
\begin{equation*}
\mu \overline{f(z)}=\sum_{n=0}^{\infty }\mu \overline{a_{n}}\overline{z^{n}}%
\Psi _{n}\left( \mu r\right) +\sum_{n=0}^{\infty }a_{n}\frac{\mu ^{2}}{2(n+1)%
}z^{n+1}\Psi _{n+1}\left( \mu r\right) 
\end{equation*}
{\LARGE dan} 
\begin{eqnarray*}
Lf(z) &=&\sum_{n=0}^{\infty }a_{n}L\left( z^{n}\Psi _{n}\left( \mu r\right)
\right) +\sum_{n=0}^{\infty }\overline{a_{n}}\frac{\mu }{2(n+1)}L\left( 
\overline{z^{n+1}}\Psi _{n+1}\left( \mu r\right) \right)  \\
&=&\sum_{n=0}^{\infty }a_{n}\frac{\mu ^{2}}{2(n+1)}z^{n+1}\Psi _{n+1}\left(
\mu r\right) + \\
&&\sum_{n=0}^{\infty }\overline{a_{n}}\frac{\mu }{2(n+1)}[2(n+1)\overline{%
z^{n}}\Psi _{n+1}\left( \mu r\right) +\overline{z^{n+1}}\frac{\mu ^{2}z}{%
2(n+2)}\Psi _{n+2}\left( \mu r\right) ] \\
&=&\sum_{n=0}^{\infty }\mu \overline{a_{n}}\overline{z^{n}}\Psi _{n}\left(
\mu r\right) +\sum_{n=0}^{\infty }a_{n}\frac{\mu ^{2}}{2(n+1)}z^{n+1}\Psi
_{n+1}\left( \mu r\right) 
\end{eqnarray*}
{\LARGE karena} 
\begin{equation*}
\Psi _{n}(x)=\Psi _{n+1}(x)+\frac{\left( \frac{x}{2}\right) ^{2}}{\left(
n+1\right) \left( n+2\right) }\Psi _{n+2}(x)
\end{equation*}
\end{proof}

\bigskip 

\begin{theorem}
{\LARGE Jika }$f(z)=\sum_{n=0}^{\infty }a_{n}z^{n}${\LARGE \ analitik pada }$%
\left| z\right| <a${\LARGE \ maka fungsi} 
\begin{equation*}
{\LARGE F}_{f}{\LARGE (z)=}\sum_{n=0}^{\infty }{\LARGE a}_{n}{\LARGE z}^{n}%
{\LARGE \Psi }_{n}\left( \mu r\right) {\LARGE +}\sum_{n=0}^{\infty }%
\overline{a_{n}}\frac{\mu }{2(n+1)}\overline{z^{n+1}}{\LARGE \Psi }%
_{n+1}\left( \mu r\right) 
\end{equation*}
{\LARGE adalah }$\mu ${\LARGE \ Regular pada }$\left| z\right| <a.$
\end{theorem}

\bigskip \bigskip {\LARGE Selanjutnya, fungsi }$\mu ${\LARGE \ regular }$f$%
{\LARGE \ ditulis dalam bentuk} 

\begin{equation*}
{\LARGE f(z)=}\sum_{n=0}^{\infty }{\LARGE a}_{n}{\LARGE z}^{n}{\LARGE \Psi }%
_{n}\left( \mu r\right) {\LARGE +}\sum_{n=1}^{\infty }{\LARGE a}_{-n}%
\overline{z^{n}}{\LARGE \Psi }_{n}\left( \mu r\right) \text{, }{\LARGE dimana%
}
\end{equation*}
\begin{equation*}
{\LARGE a}_{-n-1}{\LARGE =}\frac{\mu }{2(n+1)}\overline{a_{n}}{\LARGE %
,n=0,1,2,...}
\end{equation*}

\bigskip 

\bigskip 

{\LARGE Akibat ini mengatakan bahwa ada pemetaan dari ruang fungsi analitik
ke dalam ruang fungsi }$\mu ${\LARGE \ regular. Selanjutnya, misalkan }$A(D)$%
{\LARGE \ ruang fungsi analitik pada disk satuan }$D,${\LARGE \ dan }$A_{\mu
}(D)${\LARGE \ ruang fungsi }$\mu ${\LARGE \ regular pada disk satuan }$D.$%
{\LARGE \ Definisikan pemetaan} 
\begin{equation*}
{\LARGE T:A(D)\rightarrow A}_{\mu }{\LARGE (D)}
\end{equation*}
\begin{equation*}
{\LARGE :f}\text{ \ \ \ \ \ \ }{\LARGE \mapsto F}_{f}
\end{equation*}

\bigskip 

{\LARGE Dapat diperiksa bahwa }$T${\LARGE \ merupakan isomorpisma ruang
vektor ke dalam }$A_{\mu }(D)${\LARGE \ atas real. Catat, jika }$\mu =0$%
{\LARGE \ maka }$T${\LARGE \ identitas. Selanjutnya perhatikan kelas fungsi }%
$S\subset A(D),${\LARGE \ dan notasikan }$S_{\mu }=T(S)\subset A_{\mu }.$%
{\LARGE \ (catatan, fungsi di dalam  }$S_{\mu }${\LARGE \ tidak perlu
univalen)}

{\LARGE \bigskip }

\begin{theorem}
{\LARGE Jika }$f(z)=\sum_{n=0}^{\infty }a_{n}z^{n}\Psi _{n}\left( \mu
r\right) +\sum_{n=1}^{\infty }a_{-n}\overline{z^{n}}\Psi _{n}\left( \mu
r\right) \in S_{\mu },${\LARGE \ maka }
\begin{equation*}
\left| a_{n}-a_{-n}\right| {\LARGE \leq n+}\frac{\mu (n-1)}{2n}.
\end{equation*}
\end{theorem}

\begin{proof}
{\LARGE Dengan menggunakan ketaksamaan Bieberbach-de Branges untuk }$a_{n}$%
{\LARGE \ dan }$a_{-n}=\frac{\mu }{2n}\overline{a_{n-1}}${\LARGE , }$%
n=1,2,3,...${\LARGE \ Maka} 
\begin{eqnarray*}
\left| a_{n}-a_{-n}\right| {\LARGE } &\leq &{\LARGE }\left| a_{n}\right| 
{\LARGE +}\left| a_{-n}\right|  \\
&=&\left| a_{n}\right| +\frac{\mu }{2n}\overline{\left| a_{n-1}\right| } \\
&=&n+\frac{\mu }{2n}\left| a_{n-1}\right|  \\
&=&n+\frac{\mu (n-1)}{2n}.
\end{eqnarray*}
{\LARGE Untuk tanda ''='', dipenuhi oleh fungsi} 
\begin{equation*}
\widetilde{k(z)}=\frac{z}{\left( 1+z\right) ^{2}}=\sum \left( -1\right)
^{n+1}nz^{n}\in S
\end{equation*}
{\LARGE sehingga }$T\left( \widetilde{k}\right) \in S_{\mu }$ {\LARGE dengan}
$a_{n}=\left( -1\right) ^{n+1}n,$ $a_{-n}=\frac{\mu }{2n}\overline{a_{n-1}}=%
\frac{\mu }{2n}\left( -1\right) ^{n}(n-1)$
\end{proof}

\section{\bigskip \bigskip Estimasi  Koefisien \ Fungsi  Analitik \ di  S}

{\LARGE Hasil Bieberbach yang telah dibuktikan Louis de Brange: Untuk
sembarang }$f\in S${\LARGE \ maka }$\left| a_{n}\right| \leq n.$

{\LARGE Goluzin (1949) dan Jenkin (1959), memberikan estimasi : Untuk
sembarang }$f\in S${\LARGE \ , maka} 
\begin{equation*}
-1\leq \left| a_{3}\right| -\left| a_{2}\right| \leq 1.029...
\end{equation*}

{\LARGE Akhirnya Grinspan (1976) memodifikasi hasil Hayman (1963), Milin
(1968), dan Hina (1968) menghasilkan}

{\LARGE estimasi yang cukup baik untuk sembarang }$n,${\LARGE \ yaitu }
\begin{equation*}
-2.97\leq \left| a_{n+1}\right| -\left| a_{n}\right| \leq 3.61,\text{ }%
n=1,2,3,...
\end{equation*}

\section{Kaitan dengan Koefisien Fungsi $\ \mu $ \ Regular}

\begin{corollary}
{\LARGE Jika }$f(z)=\sum_{n=0}^{\infty }a_{n}z^{n}\Psi _{n}\left( \mu
r\right) +\sum_{n=1}^{\infty }a_{-n}\overline{z^{n}}\Psi _{n}\left( \mu
r\right) \in S_{\mu },${\LARGE \ maka} 
\begin{equation*}
-\frac{\mu }{2}\leq 4\left| a_{-4}\right| -3\left| a_{-3}\right| \leq
0.50145\mu .
\end{equation*}
\end{corollary}

\begin{proof}
{\LARGE Dengan menggunakan }$a_{-n}=\frac{\mu }{2n}\overline{a_{n-1}},$%
{\LARGE \ kita peroleh} 
\begin{equation*}
4\left| a_{-4}\right| -3\left| a_{-3}\right| =4\left| \frac{\mu }{8}%
\overline{a_{3}}\right| -3\left| \frac{\mu }{6}\overline{a_{2}}\right| =%
\frac{\mu }{2}\left[ \left| a_{3}\right| -\left| a_{2}\right| \right] ,
\end{equation*}
{\LARGE selanjutnya gunakan hasil Goluzin dan Jenkin.}
\end{proof}

\bigskip 

\begin{corollary}
{\LARGE Jika }$f(z)=\sum_{n=0}^{\infty }a_{n}z^{n}\Psi _{n}\left( \mu
r\right) +\sum_{n=1}^{\infty }a_{-n}\overline{z^{n}}\Psi _{n}\left( \mu
r\right) \in S_{\mu },${\LARGE \ maka} 
\begin{equation*}
-1.4985\mu \leq (n+1)\left| a_{-\left( _{n+1}\right) }\right| -n\left|
a_{-n}\right| \leq 1.805\mu ,\text{ }n=1,2,3,..
\end{equation*}

\begin{proof}
{\LARGE Dengan menggunakan }$a_{-n}=\frac{\mu }{2n}\overline{a_{n-1}},$%
{\LARGE \ kita peroleh} 
\begin{equation*}
(n+1)\left| a_{-\left( n+1\right) }\right| -n\left| a_{-n}\right| =\frac{\mu 
}{2}\left[ \left| \overline{a_{n}}\right| -\left| \overline{a_{n-1}}\right| %
\right] =\frac{\mu }{2}\left[ \left| a_{n}\right| -\left| a_{n-1}\right| %
\right] ,
\end{equation*}
{\LARGE selanjutnya gunakan hasil Grinspan.}
\end{proof}
\end{corollary}

\bigskip 

\end{document}
