161 lines
4.5 KiB
Markdown
161 lines
4.5 KiB
Markdown
# Math416 Lecture 23
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## Chapter 9: Generalized Cauchy Theorem
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### Separation lemma
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Let $\Omega$ be an open subset in $\mathbb{C}$, let $K\subset \Omega$ be compact. Then There exists a simple contour $\Gamma$ such that
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$$
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K\subset \text{int}(\Gamma)\subset \Omega
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$$
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#### Corollary 9.9 for separation lemma
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Let $\Gamma$ be the contour constructed in the separation lemma. Let $f\in O(\Omega)$ be holomorphic on $\Omega$. Then $\forall z_0\in K$ such that
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$$
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f(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{z-z_0}dz
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$$
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Proof:
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Suppose $h\in O(G)$, then $\int_{\partial S} h(z)dz=0$, by Cauchy's theorem for square, followed from the triangle case.
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So $\int_{\Gamma} h(z)dz=0=\sum_{j=1}^n \int_{\partial S_j} h(z)dz$
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Fix $z_0\in K$,
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$$
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g(z_0)=\begin{cases}
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\frac{f(z)-f(z_0)}{z-z_0} & z\neq z_0 \\
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f'(z_0) & z=z_0
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\end{cases}
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$$
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So $\int_{\Gamma} g(z)dz=0$
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Thus
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$$
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\begin{aligned}
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\int_{\Gamma}\frac{f(z)}{z-z_0}dz-\int_{\Gamma}\frac{f(z_0)}{z-z_0}dz&=0 \\
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\int_{\Gamma}\frac{f(z)}{z-z_0}dz&=f(z_0)\int_{\Gamma}\frac{1}{z-z_0}dz \\
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&=f(z_0)\cdot 2\pi i
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\end{aligned}
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$$
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QED
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#### Theorem 9.10 Cauchy's Theorem
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Let $\Omega$ be an open subset in $\mathbb{C}$, let $\Gamma$ be a contour with $int(\Gamma)\subset \Omega$. Let $f\in O(\Omega)$ be holomorphic on $\Omega$. Then
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$$
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\int_{\Gamma} f(z)dz=0
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$$
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Proof:
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Let $K\subset \mathbb{C}\setminus \text{ext}(\Gamma)$.
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By separation lemma, $\exists \Gamma_1$ s.t. $K\subset \text{int}(\Gamma_1)\subset \Omega$.
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Notice that Separation lemma ensured that $w\neq z$ for all $w\in \Gamma_1, z\in \Gamma$.
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By Corollary 9.9, $\forall z\in K, f(z)=\frac{1}{2\pi i}\int_{\Gamma_1}\frac{f(w)}{w-z}dw$
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$$
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\int_{\Gamma} f(z)dz=\frac{1}{2\pi i}\int_{\Gamma}\left[\int_{\Gamma_1}\frac{f(w)}{w-z}dw\right]dz
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$$
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By Fubini's theorem (In graduate course for analysis),
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$$
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\begin{aligned}
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\int_{\Gamma} f(z)dz&=\frac{1}{2\pi i}\int_{\Gamma_1}\left[\int_{\Gamma}\frac{f(w)}{w-z}dz\right]dw \\
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&=\frac{1}{2\pi i}\int_{\Gamma_1}f(w)\left[\int_{\Gamma}\frac{1}{w-z}dz\right]dw \\
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&=\frac{1}{2\pi i}\int_{\Gamma_1}f(w)\cdot 2\pi i \ \text{ind}_{\Gamma}(w)dw \\
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&=0
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\end{aligned}
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$$
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Since the winding number for $\Gamma$ on $w\in \Gamma_1$ is 0. ($w$ is outside of $\Gamma$)
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QED
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### Homotopy
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Suppose $\gamma_0, \gamma_1$ are two curves from
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$[0,1]$ to $\Omega$ with same end points $P,Q$.
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A homotopy is a continuous function of curves $\gamma_t, 0\leq t\leq 1$, deforming $\gamma_0$ to $\gamma_1$, keeping the end points fixed.
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Formally, if $H:[0,1]\times [0,1]\to \Omega$ is a continuous function satsifying
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1. $H(s,0)=\gamma_0(s)$, $\forall s\in [0,1]$
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2. $H(s,1)=\gamma_1(s)$, $\forall s\in [0,1]$
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3. $H(0,t)=P$, $\forall t\in [0,1]$
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4. $H(1,t)=Q$, $\forall t\in [0,1]$
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Then we say $H$ is a homotopy between $\gamma_0$ and $\gamma_1$. (If $\gamma_0$ and $\gamma_1$ are closed curves, $Q=P$)
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#### Lemma 9.12 Technical Lemma
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Let $\phi:[0,1]\times [0,1]\to \mathbb{C}\setminus \{0\}$ is continuous. Then there exists a continuous map $\psi:[0,1]\times [0,1]\to \mathbb{C}$ such that $e^\phi=\psi$. Moreover, $\psi$ is unique up to an additive constant in $2\pi i\mathbb{Z}$.
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Proof:
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Let $\phi_t(s)=\phi(s,t)$, $0\leq t\leq 1$.
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Then $\exists \psi_{00}$ such that $e^{\psi_{00}(s)}=\phi(0,t)$.
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$\exists \psi_{t}(s)$ such that $e^{\psi_{t}(s)}=\phi_t(s)$.
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We want to show $\psi_t(s)$ is continuous in $t$.
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Since $\exists \epsilon>0$ such that $\phi(s,t)$ is at least $\epsilon$ away from $0$ for all $s\in [0,1]$ and $t\in [0,1]$.
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Moreover, $\phi(s,t)$ is uniformly continuous.
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So $\exists \delta>0$ such that $|\phi(s,t)-\phi(s,t_0)|<\epsilon$ if $|t-t_0|<\delta$.
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Therefore,
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$$
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\begin{aligned}
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\left|\frac{\phi(s,t)}{\phi(s,t_0)}-1\right|&<\frac{\epsilon}{\phi(s,t_0)}
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&<1
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\end{aligned}
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$$
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So $\text {Re} \frac{\phi(s,t)}{\phi(s,t_0)}>0$.
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Therefore, $\text{Log} \frac{\phi(s,t)}{\phi(s,t_0)}=\chi(s,t)$ is continuous on $s\in [0,1], t\in [t_0-\delta, t_0+\delta]$.
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So $e^{\chi(s,t)}=\frac{\phi(s,t)}{\phi(s,t_0)}$, $\chi(s,t_0)=0,\forall s\in [0,1]$
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Define $\tilde{\psi}(s,t)=\chi(s,t)+\chi(s,t_0)$. So this function is continuous.
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And $e^{\tilde{\psi}(s,t)}=e^{\chi(s,t)+\chi(s,t_0)}=e^{\chi(s,t)}\cdot e^{\chi(s,t_0)}=\phi(s,t)$.
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$$
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\begin{aligned}
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\tilde{\psi}(0,t_0)&=\chi(0,t_0)+\psi(0,t_0) \\
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&=0+\psi_{00}(t_0) \\
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&=\psi_{00}(t_0)
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\end{aligned}
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$$
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$\tilde{\psi}(s,0)$ and $\psi(t,0)$ on $t\in[t_0-\delta, t_0+\delta]$ are both logs of the same function, and agree to each other on $t_0$.
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Therefore, $\tilde{\psi}(s,0)=\psi(s,0)+\text{const}$
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QED
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#### Theorem 9.13 Cauchy's Theorem for Homotopic Curves
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