# Wednesday February 5th

## Cauchy/Morera Theorems 
Recall last time:
We have Cauchy's theorem, which says that if $f: \Omega \to \CC$ is holomorphic then $$\int_\gamma f~dz = 0.$$

We have a partial converse:

Theorem (Morera)
: 	If $g: \Omega \to \CC$ is continuous and $\int_R g~dz = 0$ for every rectangle $R\subset \Omega$ with sides parallel to the axes, then $g$ is holomorphic.

Proof (Morera)
: 	Fix a point $a\in\Omega$, then for any $z\in\Omega$ define $f(z) = \int_{\gamma_{a, z}} g(\xi) d\xi = \int_{\sigma_{a, z}} g(\xi) d\xi$.

	![Image](figures/2020-02-05-13:42.png)\

	Then $\dd{f}{z} = \dd{f}{x} = \frac 1 i \dd{f}{y} = g(z)$, making $g$ holomorphic.

## Schwarz Reflection

Theorem (Schwarz Reflection, Extending Holomorphic Functions Across Reflected Regions)
: 	Let $\Omega = \Omega^+ \union L \union \Omega^-$ be a region of the following form:

	![Image](figures/2020-02-05-13:45.png)\

	I.e., $L = \theset{z\in \Omega \suchthat \im z = 0}$, $\Omega^{\pm} = \theset{\pm \im z > 0}$ where $\Omega$ is symmetric about the real axis, i.e. $z\in \Omega \implies \bar z \in \Omega$.

	Assume that $f: \Omega^+ \union L \to \CC$ is continuous and holomorphic in $\Omega^+$ and real-valued on $L$.
	Define

	\begin{align*}
	g(z) = 
	\begin{cases}
	f(z) & z\in \Omega^+ \union L \\
	\bar{f(z)} & z\in \Omega^-
	\end{cases}
	.\end{align*}

	Then $g(z)$ is defined and holomorphic on $\Omega$.

Proof (Schwarz Reflection)
: 	Since $g$ is $C^1$ in $\Omega^-$, check that $g$ satisfies the Cauchy-Riemann equations on $\Omega^-$ and thus holomorphic there.
	To see that $g$ is holomorphic on all of $\Omega$, we'll show the integral over every rectangle is zero.

	It's clear that if $R\subset \Omega^{\pm}$, $\int_R g = 0$ since $g$ is holomorphic there, so it suffices to check rectangles intersecting the real axis.
	Write $R = R^+ \union R^-$:

	![Image](figures/2020-02-05-13:59.png)\

	We then have $R^+ = \lim_{\eps\to 0} R_\eps$  and $R^- = \lim_{\eps \to 0} R_{-\eps}$, and $\int_{R_{\pm \eps}} g = 0$ for all $\eps > 0$.
	By continuity of $f$ on $L$, we have $\lim \int_{R_{\eps}} g(z) ~dz = 0$.

## Goursat's Theorem

Theorem (Goursat, $C^1$ implies smooth)
: 	If $f: \Omega \to \CC$ is complex differentiable at each point of $\Omega$, then $f$ is holomorphic. 
	I.e., $$f\in C^1(\Omega) \implies f\in C^\infty(\Omega).$$

Proof (Goursat)
: 	We have $\int_R f ~dz = 0$ for all rectangles $R$.
	Write $I = \int_R f ~dz$.
	Break $R$ into 4 sub-rectangles:

	![Image](figures/2020-02-05-14:08.png)\

	Then rewriting the integral and applying the triangle inequality yields

	\begin{align*}
	I = \int_R f = \sum_{j=1}^4 \int_{R_j} f  = \sum_{j=1}^4 I_j \implies \abs{I} \leq \sum_j \abs{I_j}
	.\end{align*}

	So for at least one $j$, we have $\abs{I_j} \geq \frac 1 4 \abs{I}$; wlog call it $R_1$.
	By continuing to subdivide, we can write 

	\begin{align*}
	\abs I \leq 4 \abs {I_k} = 4 \abs{\int_{R_1} f} \leq 4\qty{4 \abs{ \int_{R_2} f }  } \cdots \leq 4^k \abs{\int_{R_k} f}
	.\end{align*}

	This is a sequence of nested compact intervals, so there is some $z_0 \in \intersect R_k$.

	Write $f(z) = f(z_0) + f'(z_0)(z-z_0) + \delta(z, z_0)$, and since

	\begin{align*}
	\lim_{z\to z_0} \frac{\abs{\delta(z, z_0)}}{z-z_0} = 0
	,\end{align*}

	we have $\delta(z, z_0) = o(z-z_0)$.
	Then $\abs I \leq 4^k \frac{1}{2^k} \abs{R}$.
	We then try to estimate the integral using the fact that $\abs{\delta(z, z_0)} \leq \delta_k \abs{z-z_0}$ for some constant $\delta_k \to 0$ as $k\to \infty$.

	\begin{align*}
	\int_{R_k} f i
	&= \int f(z_0) + f'(z_0)(z-z_0) + \delta(z, z_0) \\
	&= \int_{R_k} \delta(z, z_0) \quad\text{since the first two terms are holomorphic} \\
	&\leq \frac{1}{2^k} \abs{R} \delta_k \frac{C}{2^k} \abs{R} \\
	& = c/4^k \abs{R}^2 \delta_k  \\
	&\converges{k\to\infty}\to 0
	,\end{align*}

	where we use the fact that in $R_k$ we have

	![Image](figures/2020-02-05-14:23.png)\

	\begin{align*}
	R_k = 2(x+y) \implies R^2/4 = x^2 + y^2 + x + y \leq_{CS} x^2 + y^2 + x^2 + y^2 = 2(x^2 + y^2) \\
	\implies x^2 + y^2 \leq R^2 / 8 \implies L = \sqrt{x^2 + y^2} \leq R^8 /2\sqrt{2} \\
	\implies \abs{z-z_0} \leq \sqrt{x^2+y^2} \leq R_k/2\sqrt{2} \text{ and } R_k = \frac 1 {2^k} \abs{R}
	.\end{align*}

	> Note that triangles implies rectangles, but think about how to use triangles to prove it for rectangles (note that sides should be parallel to axes!)