updates

content/Math401/Extending_thesis/Math401_S4.md

@@ -87,7 +87,52 @@ $$
 P_s(z)=\{v\in \mathbb{C}^d: |v_k-z_k|<s, k=1,2,\cdots,d\}
 $$
 
-If $z\in U$, we cha choose $s$ small enough such that $P_s(z)\subset U$.
+If $z\in U$, we cha choose $s$ small enough such that $\overline{P_s(z)}\subset U$ so that we can claim that $F(z)=(\pi s^2)^{-d}\int_{P_s(z)}F(v)d\mu(v)$ is well-defined.
+
+If $d=1$. Then by Taylor series at $v=z$, since $F$ is analytic in $U$ we have
+
+$$
+F(v)=F(z)+\sum_{k=1}^{\infty}a_n(v-z)^n
+$$
+
+Since the series converges uniformly to $F$ on the compact set $\overline{P_s(z)}$, we can interchange the integral and the sum.
+
+Using polar coordinates with origin at $z$, $(v-z)^n=r^n e^{in\theta}$ where $r=|v-z|, \theta=\arg(v-z)$.
+
+For $n\geq 1$, the integral over $P_s(z)$ (open disk) is zero (by Cauchy's theorem).
+
+So,
+
+$$
+\begin{aligned}
+F(z)&=(\pi s^2)^{-1}\int_{P_s(z)}F(z)+\sum_{k=1}^{\infty}a_n(v-z)^n d\mu(v)\\
+&=(\pi s^2)^{-1}F(z)+(\pi s^2)^{-1}\sum_{k=1}^{\infty}a_n\int_{P_s(z)}r^n e^{in\theta} d\mu(v)\\
+&=(\pi s^2)^{-1}F(z)
+\end{aligned}
+$$
+
+For $d>1$, we can use the same argument to show that
+
+Let $\mathbb{I}_{P_s(z)}(v)=\begin{cases}1 & v\in P_s(z) \\0 & v\notin P_s(z)\end{cases}$ be the indicator function of $P_s(z)$.
+
+$$
+\begin{aligned}
+F(z)&=(\pi s^2)^{-d}\int_{U}\mathbb{I}_{P_s(z)}(v)\frac{1}{\alpha(v)}F(v)\alpha(v) d\mu(v)\\
+&=(\pi s^2)^{-d}\langle \mathbb{I}_{P_s(z)}\frac{1}{\alpha},F\rangle_{L^2(U,\alpha)}
+\end{aligned}
+$$
+
+By definition of inner product.
+
+So $\|F(z)\|^2\leq (\pi s^2)^{-2d}\|\mathbb{I}_{P_s(z)}\frac{1}{\alpha}\|^2_{L^2(U,\alpha)} \|F\|^2_{L^2(U,\alpha)}$.
+
+All the terms are bounded and finite.
+
+For part 2, we need to show that $\forall z\in U$, we can find a neighborhood $V$ of $z$ and a constant $d_z$ such that
+
+$$
+|F(z)|^2\leq d_z \|F\|^2_{L^2(U,\alpha)}
+$$
 
 </details>
 

content/Math416/Math416_L1.md

@@ -81,7 +81,8 @@ $$
 |z_1+z_2|\leq |z_1|+|z_2|
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Geometrically, the triangle inequality states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
 
@@ -97,6 +98,8 @@ $$
 \end{aligned}
 $$
 
+</details>
+
 Suppose $2(|z_1||z_2|-|z_1z_2|)=0$, and $\overline{z_1}z_2$ is a non-negative real number $c$, then $|z_1||z_2|=|z_1z_2|$...
 
 > What is the use of this?
@@ -113,7 +116,8 @@ $$
 
 The sum of the squares of the lengths of the diagonals of a parallelogram equals the sum of the squares of the lengths of the sides.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $z_1,z_2$ be two complex numbers representing the two sides of the parallelogram, then the sum of the squares of the lengths of the diagonals of the parallelogram is $|z_1-z_2|^2+|z_1+z_2|^2$, and the sum of the squares of the lengths of the sides is $2|z_1|^2+2|z_2|^2$.
 
@@ -125,7 +129,7 @@ $$
 \end{aligned}
 $$
 
-QED
+</details>
 
 #### Definition 1.9
 
@@ -143,12 +147,15 @@ $$
 z^n=r^n\text{cis}(n\theta)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 For $n=0$, $z^0=1=1\text{cis}(0)$.
 
 For $n=-1$, $z^{-1}=\frac{1}{z}=\frac{1}{r}\text{cis}(-\theta)=\frac{1}{r}(cos(-\theta)+i\sin(-\theta))$.
 
+</details>
+
 Application:
 
 $$

content/Math416/Math416_L18.md

@@ -26,21 +26,27 @@ $$
 \int_{C(z_0,r)} f(z) dz = \sum_{n=-\infty}^{\infty} c_n \int_{C(z_0,r)} (z-z_0)^n dz
 $$
 
-> $$
-\int_{C(z_0,r)} (z-z_0)^n dz = \begin{cases}
-    2\pi i, & n=-1 \\
-    0, & n\neq -1
-\end{cases}$$
-> Proof:
-> $\gamma(t)=z_0+re^{it}, t\in[0,2\pi]$
-> $$\begin{aligned}
+<details>
+<summary>Additional Proof</summary>
+
+$$
+\int_{C(z_0,r)} (z-z_0)^n dz = \begin{cases} 2\pi i, & n=-1 \\0, & n\neq -1\end{cases}
+$$
+
+Proof:
+
+$\gamma(t)=z_0+re^{it}, t\in[0,2\pi]$
+$$
+\begin{aligned}
 \int_{C(z_0,r)} (z-z_0)^n dz &= \int_0^{2\pi} (z_0+re^{it}-z_0)^n ire^{it} dt \\
 &= ir^{n+1} \int_0^{2\pi} e^{i(n+1)t} dt \\
 &= \begin{cases}
     2\pi i, & n=-1 \\
     \int_0^{2\pi} e^{i(n+1)t} dt = \frac{1}{i(n+1)}e^{i(n+1)t}\Big|_0^{2\pi} = 0, & n\neq -1
 \end{cases}
-\end{aligned}$$
+\end{aligned}
+$$
+</details>
 
 So,
 

content/Math416/Math416_L4.md

@@ -44,7 +44,8 @@ $$
 
 > Looks like the chain rule.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 We want to show that
 
@@ -87,7 +88,7 @@ $$
 \end{aligned}
 $$
 
-QED
+</details>
 
 #### Definition 2.12 (Conformal function)
 
@@ -111,7 +112,8 @@ Suppose $f$ is real differentiable, let $a=\frac{\partial f}{\partial z}(z_0)$,
 
 Let $\gamma(t_0)=z_0$. Then $(f\circ \gamma)'(t_0)=a\gamma'(t_0)+b\overline{\gamma'(t_0)}$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $f=u+iv$, $u,v$ are real differentiable.
 
@@ -144,7 +146,7 @@ $$
 \end{aligned}
 $$
 
-QED
+</details>
 
 #### Theorem of differentiability
 
@@ -152,7 +154,8 @@ Let $f:G\to \mathbb{C}$ be a function defined on an open set $G\subset \mathbb{C
 
 Then, $f$ is conformal at every point $z_0\in G$ if and only if $f$ is holomorphic at $z_0$ and $f'(z_0)\neq 0$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 We prove the equivalence in two parts.
 
@@ -193,7 +196,7 @@ $$
 $$
 for any differentiable curve $\gamma$ through $z_0$, then the effect of $f$ near $z_0$ is exactly given by multiplication by $f'(z_0)$. Since multiplication by a nonzero complex number is a similarity transformation, $f$ is conformal at $z_0$.
 
-QED
+</details>
 
 ### Harmonic function
 
@@ -211,7 +214,8 @@ $$
 
 Let $f=u+iv$ be holomorphic function on domain $\Omega\subset \mathbb{C}$. Then $u$ and $v$ are harmonic functions on $\Omega$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 \Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.
@@ -229,7 +233,7 @@ $$
 \Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial y\partial x}=0.
 $$
 
-QED
+</details>
 
 If $v$ is such that $f=u+iv$ is holomorphic on $\Omega$, then $v$ is called harmonic conjugate of $u$ on $\Omega$.
 

content/Math416/Math416_L5.md

@@ -14,7 +14,7 @@ Df(x+iy)=\begin{pmatrix}
 \end{pmatrix}
 $$
 
-So 
+So,
 
 $$
 \begin{aligned}
@@ -53,7 +53,6 @@ $$
 
 ## Chapter 3: Linear fractional Transformations
 
-
 Let $a,b,c,d$ be complex numbers. such that $ad-bc\neq 0$.
 
 The linear fractional transformation is defined as
@@ -185,7 +184,8 @@ So the kernel of $F$ is the set of matrices that represent the identity transfor
 
 If $\phi$ is a non-constant linear fractional transformation, then $\phi$ is conformal.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Know that $\phi_0\circ\phi(z)=z$,
 
@@ -197,13 +197,14 @@ $\phi:\mathbb{C}\cup\{\infty\}\to\mathbb{C}\cup\{\infty\}$ which gives $\phi(\in
 
 So, $\phi$ is conformal.
 
-QED
+</details>
 
 #### Proposition 3.4 of Fixed points
 
 Any non-constant linear fractional transformation except the identity transformation has 1 or 2 fixed points.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\phi(z)=\frac{az+b}{cz+d}$.
 
@@ -221,7 +222,7 @@ Such solutions are $z=\frac{-(d-a)\pm\sqrt{(d-a)^2+4bc}}{2c}$.
 
 So, $\phi$ has 1 or 2 fixed points.
 
-QED
+</details>
 
 #### Proposition 3.5 of triple transitivity
 

content/Math416/Math416_L6.md

@@ -36,7 +36,8 @@ when $\alpha=0$, it is a line.
 
 when $\alpha\neq 0$, it is a circle.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $w=u+iv=\frac{1}{z}$, so $\frac{1}{w}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$.
 
@@ -48,7 +49,7 @@ $$
 
 Which is in the form of circle equation.
 
-QED
+</details>
 
 ## Chapter 4 Elementary functions
 
@@ -83,7 +84,8 @@ $$
 
 $e^z$ is holomorphic on $\mathbb{C}$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 \begin{aligned}
@@ -93,19 +95,20 @@ $$
 \end{aligned}
 $$
 
-QED
+</details>
 
 #### Theorem 4.4 $e^z$ is periodic
 
 $e^z$ is periodic with period $2\pi i$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 e^{z+2\pi i}=e^z e^{2\pi i}=e^z\cdot 1=e^z
 $$
 
-QED
+</details>
 
 #### Theorem 4.5 $e^z$ as a map
 
@@ -185,13 +188,14 @@ A branch of $\log(z)$ in $G$ is a continuous function $\beta$, such that $e^{\be
 
 Note: $G$ has a branch of $\arg(z)$ if and only if it has a branch of $\log(z)$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Suppose there exists $\alpha(z)$ such that $\forall z\in G$, $\alpha(z)\in G$, then $l(z)=\ln|z|+i\alpha(z)$ is a branch of $\log(z)$.
 
 Suppose there exists $l(z)$ such that $\forall z\in G$, $l(z)\in G$, then $\alpha(z)=Im(z)$ is a branch of $\arg(z)$.
 
-QED
+</details>
 
 If $G=\mathbb{C}\setminus\{0\}$, then  not branch of $\arg(z)$ exists.
 
@@ -222,7 +226,8 @@ for some $k\in\mathbb{Z}$.
 
 $\log(z)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
 
-Proof:
+<details>
+<summary>Proof (continue on next lecture)</summary>
 
 Method 1: Use polar coordinates. (See in homework)
 
@@ -238,3 +243,4 @@ $$
 $$
 
 Continue next time.
+</details>

content/Math416/Math416_L7.md

@@ -26,7 +26,8 @@ A branch of logarithm is a continuous function $f$ on a domain $D$ such that $e^
 
 $\log(z)$ is holomorphic on $\mathbb{C}\setminus\{0\}$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 We proved that $\frac{\partial}{\partial\overline{z}}e^{z}=0$ on $\mathbb{C}\setminus\{0\}$.
 
@@ -36,7 +37,7 @@ Since $\frac{d}{dz}e^{z}=e^{z}$, we know that $e^{z}$ is conformal, so any branc
 
 Since $\exp(\log(z))=z$, we know that $\log(z)$ is the inverse of $\exp(z)$, so $\frac{d}{dz}\log(z)=\frac{1}{e^{\log(z)}}=\frac{1}{z}$.
 
-QED
+</details>
 
 We call $\frac{f'}{f}$ the logarithmic derivative of $f$.
 
@@ -78,7 +79,8 @@ If $|c|<1$, then $\lim_{N\to\infty}\sum_{n=0}^{N}c^n=\frac{1}{1-c}$.
 
 otherwise, the series diverges.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 The geometric series converges if $\frac{c^{N+1}}{1-c}$ converges.
 
@@ -90,7 +92,7 @@ If $|c|<1$, then $\lim_{N\to\infty}c^{N+1}=0$, so $\lim_{N\to\infty}(1-c)(1+c+c^
 
 If $|c|\geq 1$, then $c^{N+1}$ does not converge to 0, so the series diverges.
 
-QED
+</details>
 
 #### Theorem 5.4 (Triangle Inequality for Series)
 
@@ -146,7 +148,8 @@ For every power series, there exists a radius of convergence $r$ such that the s
 
 And it diverges pointwise outside $B_r(z_0)$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Without loss of generality, we can assume that $z_0=0$.
 
@@ -166,7 +169,7 @@ So the series converges absolutely and uniformly on $\overline{B_r(0)}$.
 
 If $|z| > r$, then $|c_n z^n|$ does not tend to zero, and the series diverges.
 
-QED
+</details>
 
 We denote this $r$ captialized by te radius of convergence
 

content/Math416/Math416_L8.md

@@ -67,7 +67,8 @@ $$
 \frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Suppose $(b_n)^{\infty}_{n=0}$ is a sequence of real numbers such that $\lim_{n\to\infty} b_n$ may nor may not exists by $(-1)^n(1-\frac{1}{n})$.
 
@@ -111,7 +112,7 @@ So $\sum_{n=0}^{\infty} a_n (z - z_0)^n$ does not converge at $z$ if $|z|> \frac
 
 So $R=\frac{1}{\rho}$.
 
-QED
+</details>
 
 _What if $|z-z_0|=R$?_
 
@@ -135,7 +136,8 @@ Suppose $\sum_{n=0}^{\infty} a_n (z - z_0)^n$ has a positive radius of convergen
 
 > Here below is the proof on book, which will be covered in next lecture.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Without loss of generality, assume $z_0=0$. Let $R$ be the radius of convergence for the two power series: $\sum_{n=0}^{\infty} a_n z^n$ and $\sum_{n=1}^{\infty} n a_n z ^{n-1}$. The two power series have the same radius of convergence $|R|$.
 
@@ -179,4 +181,4 @@ So $\left|\frac{f(z)-f(z_1)}{z-z_1}-g(z_1)\right|\leq M|z-z_1|$ for $|z|<\rho$.
 
 So $\lim_{z\to z_1}\frac{f(z)-f(z_1)}{z-z_1}=g(z_1)$.
 
-QED
+</details>