125 lines
3.5 KiB
Markdown
125 lines
3.5 KiB
Markdown
# Math416 Lecture 15
|
|
|
|
## Review on Cauchy Integrals
|
|
|
|
The cauchy integral of function $\phi$ (may not be holomorphic) on curve $\Gamma$ (may not be closed) is
|
|
|
|
$$
|
|
\int_{\Gamma}\frac{\phi(\zeta)}{\zeta-z}d\zeta
|
|
$$
|
|
|
|
The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain $D$, then the integral of $f$ over any closed curve $\gamma$ in $D$ is 0.
|
|
|
|
$$
|
|
\int_{\gamma}f(z)dz = 0
|
|
$$
|
|
|
|
The Cauchy integral formula states that if $f$ is holomorphic on a simply connected domain $D$, then $f$ over any closed curve $\gamma$ in $D$ is
|
|
|
|
$$
|
|
f(z) = \frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}d\zeta
|
|
$$
|
|
|
|
## Continue on Cauchy Integrals (Chapter 7)
|
|
|
|
### Convergence of functions
|
|
|
|
#### Theorem 7.15 Weierstrass Convergence Theorem
|
|
|
|
Limit of a sequence of holomorphic functions is holomorphic.
|
|
|
|
Let $G$ be an open subset of $\mathbb{C}$ and let $\left(f_n\right)_{n\in\mathbb{N}}$ be a sequence of holomorphic functions on $G$ that converges locally uniformly to $f$ on $G$. Then $f$ is holomorphic on $G$.
|
|
|
|
Proof:
|
|
|
|
Let $z_0\in G$. There exists a neighborhood $\overline{B_r(z_0)}\subset G$ of $z_0$ such that $\left(f_n\right)_{n\in\mathbb{N}}$ converges uniformly on $\overline{B_r(z_0)}$.
|
|
|
|
Let $C_r=\partial B_r(z_0)$.
|
|
|
|
By Cauchy integral formula, we have
|
|
|
|
$$
|
|
f_n(z_0) = \frac{1}{2\pi i}\int_{C_r}\frac{f_n(\zeta)}{\zeta-z_0}d\zeta
|
|
$$
|
|
|
|
$\forall z\in B_r(z_0)$, we have $\frac{f(w)}{w-\zeta}$ converges uniformly on $C_r$.
|
|
|
|
So $\lim_{n\to\infty}f_n(z_0) = f(z_0) = \frac{1}{2\pi i}\int_{C_r}\frac{f(w)}{w-z_0}dw$
|
|
|
|
So $f$ is holomorphic on $G$.
|
|
|
|
QED
|
|
|
|
#### Theorem 7.16 Maximum Modulus Principle
|
|
|
|
If $f$ is a non-constant holomorphic function on a domain $D$ (open and connected subset of $\mathbb{C}$), then $|f|$ does not attain a local maximum value on $D$.
|
|
|
|
Proof:
|
|
|
|
Assume at some point $z_0\in D$, $|f(z_0)|$ is a local maximum. $\exists r>0$ such that $\forall z\in \overline{B_r(z_0)}$, $|f(z)|\leq |f(z_0)|$.
|
|
|
|
If $f(z_0)=0$, then $f(z)$ is identically 0 on $B_r(z_0)$. (by identity theorem)
|
|
|
|
Else, we can assume that without loss of generality that $f(z_0)>0$. By mean value theorem,
|
|
|
|
$$
|
|
f(z_0) = \frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})d\theta
|
|
$$
|
|
|
|
So $f(z_0) =$
|
|
|
|
/* TRACK LOST */
|
|
|
|
#### Corollary 7.16.1 Minimum Modulus Principle
|
|
|
|
If $f$ is a non-constant holomorphic function on a domain $D$ (open and connected subset of $\mathbb{C}$), and $f$ is non zero on $D$, then $\frac{1}{f}$ does not attain a local minimum value on $D$.
|
|
|
|
Proof:
|
|
|
|
Let $g(z) = \frac{1}{f(z)}$. $g$ is holomorphic on $D$.
|
|
|
|
QED
|
|
|
|
#### Theorem 7.17 Schwarz Lemma
|
|
|
|
Let $f$ be a holomorphic map of the open unit disk $D$ into itself, and $f(0)=0$. Then $\forall z\in D$, $|f(z)|\leq |z|$ and $|f'(0)|\leq 1$.
|
|
|
|
And the equality holds if and only if $f$ is a rotation, that is, $f(z)=e^{i\theta}z$ for some $\theta\in\mathbb{R}$.
|
|
|
|
Proof:
|
|
|
|
Let
|
|
|
|
$$
|
|
g(z) = \begin{cases}
|
|
\frac{f(z)}{z} & z\neq 0 \\
|
|
f'(0) & z=0
|
|
\end{cases}
|
|
$$
|
|
|
|
We claim that $g$ is holomorphic on $D$.
|
|
|
|
For $z\neq 0$, $g(z)$ is holomorphic since $f$ is holomorphic on $D$.
|
|
|
|
For $z=0$, $g(z)$ is holomorphic since $f$'s power series expansion has $c_0=f(0)=0$. $g'(0)=f'(0)=c_1+c_2z+c_3z^2+\cdots$.
|
|
|
|
So $g$ is (analytic) thus holomorphic on $D$.
|
|
|
|
On the boundary of $D$, $|g(z)|\leq\frac{1}{r} \cdot 1$. By maximum modulus principle, $|g(z)|\leq 1$ on $D$.
|
|
|
|
So $|f(z)|\leq |z|$ on $D$.
|
|
|
|
And $|f'(0)|\leq 1$.
|
|
|
|
QED
|
|
|
|
#### Schwarz-Pick Lemma
|
|
|
|
Let $f$ be a holomorphic map of the open unit disk $D$ into itself, then for any $z,w\in D$,
|
|
|
|
$$
|
|
\frac{|f(z)-f(w)|}{|1-\overline{f(w)}f(z)|}\leq\frac{|z-w|}{|1-\overline{w}z|}=\rho(z,w)
|
|
$$
|
|
|
|
Prove after spring break.
|