164 lines
4.1 KiB
Markdown
164 lines
4.1 KiB
Markdown
# Math416 Lecture 16
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## Answer checking for exam
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### Q1
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Cauchy riemann equations:
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$$
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\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
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$$
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Liouville's Theorem:
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Any non-constant entire function is unbounded.
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So $\cos(z)$ is unbounded in $\mathbb{C}$.
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$$
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\text{Log}(-e^2) = \ln|-e^2| + i\arg(-e^2) = -2 + \pi i
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$$
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At any point $z_0\in \mathbb{C}\setminus\{0\}$, there is an open set $z_0\in U\subset \mathbb{C}$ and a branch of logarithm defined on $U$.
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### Q2
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Power series expansion
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### Q3
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limit superior
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### Q4
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Bound integral
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### Q5
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$f_n$ converges pointwise to $f$ on $U$ if $\forall z\in U$, $\forall \epsilon > 0$, $\exists N$ s.t. $\forall n\geq N$, $|f_n(z)-f(z)| < \epsilon$.
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$f_n$ converges uniformly to $f$ on $U$ if $\forall \epsilon > 0$, $\exists N$ s.t. $\forall n\geq N$, $\forall z\in U$, $|f_n(z)-f(z)| < \epsilon$.
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Show for $|z|<1$, $f_n(z)=z^n$ converges pointwise to $0$ but not uniformly to $0$.
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(a) pointwise convergence:
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$|z^n| = |z|^n < \epsilon$ if $n > \frac{\ln\epsilon}{\ln|z|}$.
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(b) uniform convergence:
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No matter how small $\epsilon$ is, there is always a $z$ s.t. $|z^n| > \epsilon$ for all $n$.
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## Continue from last lecture
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### Schwarz's Lemma
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Let $f$ be an holomorphic function that maps the unit disk $D(0,1)$ to itself and $f(0)=0$. Then $|f(z)|\leq |z|$ for all $z\in D(0,1)$
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#### Schwarz-Pick's Lemma
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(see exercise 7.17.2)
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Let $f$ be an holomorphic function that maps the unit disk $D(0,1)$ to itself. Then $\forall z,w\in D(0,1)$,
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$$
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\left|\frac{f(z)-f(w)}{1-\overline{f(w)}f(z)}\right|\leq \left|\frac{z-w}{1-\overline{w}z}\right|
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$$
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> Recall the Möbius map
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>
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> $$\phi_\alpha(z) = \frac{z-\alpha}{1-\overline{\alpha}z}$$
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>
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> is a homeomorphism of the unit disk.
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>
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> So we can use the Möbius to restate the Schwarz-Pick's Lemma as:
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>
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> $$|\phi_{f(w)}(f(z))|\leq |\phi_w(z)|$$
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Suppose we defined $g=\phi_{f(w)}\circ f\circ \phi_{-w}$, then $g$ is a holomorphic function that maps the unit disk to itself and $g(0)=0$.
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By Schwarz's Lemma, let $z\in D(0,1)$, $|g(z)|\leq |z|$.
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$$
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|\phi_{f(w)}(f(\phi_{-w}(z)))|\leq |z|
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$$
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Let $\zeta=\phi_{-w}(z)$, then $\zeta=\frac{z+w}{1+\overline{w}z}\in D(0,1)$, so $|\zeta|=\phi_w(z)$.
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#### Extension of Schwarz-Pick's Lemma in hyperbolic metric
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Suppose we defined the distance on $\mathbb{C}$ as $d(z,w)=|\frac{z-w}{1-\overline{w}z}|$.
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We claim that this is a metric on $\mathbb{C}$. $\forall z,w,v\in \mathbb{C}$:
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(a) $d(z,w)=0$ if and only if $z=w$ and $d(z,w)> 0$ otherwise.
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(b) $d(z,w)=d(w,z)$.
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(c) $d(z,w)\leq d(z,v)+d(v,w)$.
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We call this metric the Pseudo hyperbolic metric.
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> Hyperbolic metric:
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>
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> $$ \text{Hypdist}(z,w)=\tanh^{-1}(d(z,w))$$
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>
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> Where $d(z,w)=|\frac{z-w}{1-\overline{w}z}|$
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So we can restate the Schwarz-Pick's Lemma as:
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$$
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d(f(z),f(w))\leq d(z,w)
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$$
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And in hyperbolic metric, it becomes:
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$$
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\text{Hypdist}(f(z),f(w))\leq \text{Hypdist}(z,w)
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$$
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Suppose the equality holds for Schwarz-Pick's Lemma, then $|g(z)|=\tau z$ where $|\tau|=1$.
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Computation ignored here.
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Then $f$ is a Möbius map that is automorphism of the unit disk.
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### Existence of harmonic conjugate
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Suppose $f=u+iv$ is holomorphic on a domain $U\subset \mathbb{C}$. Then $u=\text{Re}(f)$ is harmonic on $U$. That is $\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$.
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#### Theorem 7.18
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Let $u$ be a real harmonic function on a convex domain $G\subset \mathbb{C}$. Then there exists $g\in O(G)$ such that $\text{Re}(g)=u$. Moreover, $g$ is unique up to an additive imaginary constant.
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Proof:
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Existence next time.
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Uniqueness:
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Suppose $g,h\in O(G)$ s.t. $\text{Re}(g)=\text{Re}(h)=u$.
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$\text{Re}g=u=\text{Re}h$ on $G$.
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If we can show that $(g-h)'=0$ on $G$, then we win.
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Let $g=u+iv$, $h=u+iw$.
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By the Cauchy-Riemann equations,
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$$
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\begin{aligned}
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\frac{\partial}{\partial x}(g-h)&=\frac{\partial}{\partial x}i(v-w)\\
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&=i\left(\frac{\partial u}{\partial y}-\frac{\partial u}{\partial y}\right)\\
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&=0
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\end{aligned}
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$$
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Suppose $G=\mathbb{C}\setminus\{0\}$, then $u=\ln|z|=\frac{1}{2}\ln(x^2+y^2)$, which is harmonic.
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Continue next time.
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