# Monday January 27th

> Open question: does a PDE involving analytic functions always have solutions?
> Or does this hold with analytic replaced by smooth?

## Green's Theorem

Fix a connected domain $\Omega$ which is bounded with a piecewise $C^1$ boundary. 

Theorem (Green's)
: 	Given $f, g \in C^1 \bar \Omega$, we can take a vector field $F = \generators{f, g}$ and have

	\begin{align*}
	\int_{\bd \Omega} f~dx + g~dy &= \iint_{\Omega} \qty{ \dd{g}{x} - \dd{f}{y}  } ~dA \\
	\int_{\bd \Omega} - f~dx + g~dy &= \iint_{\Omega} \qty{ \dd{g}{x} + \dd{f}{y}  } ~dA \\
	\int_{\bd \Omega} f~dy - g~dy &= \iint_{\Omega} \qty{ \dd{f}{x} + \dd{g}{y}  } ~dA \\
	\int_{\bd \Omega} F \cdot \vector n ~ds &= \iint_{\Omega} \nabla \cdot F ~dA \\
	\int_{\bd \Omega} \mathrm{curl}(F) ~ds &= \iint_{\Omega} \mathrm{div}(F) ~dA
	,\end{align*}

	where we take $\vector n$ to be orthogonal to $\bd \Omega$.
The quantities appearing on the RHS are referred to as the flux.

For $f(z) \in C^1(\Omega)$ holomorphic, we can then write

\begin{align*}
\int_{\bd \Omega} f ~dz
&= \int_{\bd \Omega} f ~(dx + idy) \\
&= \int_{\bd \Omega} f ~dx + if~dy \\
&= \iint_\Omega \qty{if_x - f_y} ~dA \\
&= 0
,\end{align*}

which follows since $f$ holomorphic, we can write $$f'(z) = f_x = \frac 1 i f_y,$$ so $i f_x = f_y$ and thus $\dd{f}{\bar z} = 0$.

> See Taylor's Introduction to Complex Analysis

Theorem (Cauchy's Integral Formula):
: 	If $f\in C^1(\bar \Omega)$ and $f$ is holomorphic, then for any $z\in \Omega$

	\begin{align*}
	f(z) = \frac{1}{2\pi i} \int_{\bd \Omega} \frac{d(\xi)}{\xi - z} ~d\xi
	.\end{align*}

Proof
: Since $z\in \Omega$ an open set, we can find some $r> 0$ such that $D_r(z) \subset \Omega$.
	Then $\frac{f(\xi)}{\xi - z}$ is holomorphic on $\Omega\setminus D_r(z)$.
	Let $C_r = \bd D_r(z)$.

	**Claim**: 
	$$\int_{\bd \Omega} \frac{f(\xi)}{\xi - z} ~d\xi = \int_{C_r} \frac{f(\xi)}{\xi - z} ~ d\xi.$$

	If we can differentiate through the integral, we can obtain

	\begin{align*}
	\dd{}{z} f(z) = \frac 1 {2\pi i} \int_{\bd \Omega} \frac{f(\xi)}{(\xi - z)^2} ~d\xi
	.\end{align*}

	and thus inductively

	\begin{align*}
	(D_z)^n f(z) = \frac{n!}{2\pi i} \int_{\bd \Omega} \frac{ f(\xi) ~d\xi  }{ (\xi - z)^{n+1}  }
	.\end{align*}

	To prove rigorously, need to write

	\begin{align*}
	\Delta_h f(z) 
	= \frac 1 h \qty{ f(z+h) - f(z) } \\
	= \frac 1 {2\pi i h} \int_{\bd \Omega} f(\xi) \qty{ \frac{1}{\xi - (z+h)} - \frac{1}{\xi - z}  } ~d\xi
	= \frac 1 {2\pi i h} \int_{\bd \Omega} f(\xi) \qty{ \frac{1}{ (\xi - z- h)(\xi - z)  }  } ~d\xi
	,\end{align*}

	and show the integrand converges uniformly, where 
	$$
	\frac{1}{(\xi - z - h)(\xi - z)} \converges{u}\to \frac{1}{(\xi - z)^2}
	.$$

	Continuing inductively yields the integral formula. 

Proof (of claim used in main proof)
: 	Use the parameterization of $C_r$ given by $\xi = z + re^{i\theta}$.
	Then

	\begin{align*}
	\frac{1}{2\pi i} \int_{C_r} \frac{f(\xi)}{\xi - z} ~d\xi 
	&= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(z + re^{i\theta})}{re^{i\theta}} ~ird\theta \\
	&= \frac{1}{2\pi} \int_0^{2\pi} f(z + re^{i\theta}) ~d\theta \\
	&\converges{r \to 0}\to \frac{1}{2\pi} \int_{\bd \Omega} \frac{f(\xi)} {\xi - z}
	.\end{align*}

	where we use the fact that 
	$$f(z + re^{i\theta}) = f(z) + f'(z)re^{i\theta} + o(r) \converges{r\to 0}\to f(z)$$

	Letting $$F(\xi) = \frac{ f(\xi)}{\xi - z},$$ this is holomorphic on $\Omega\setminus D_r(z)$.
	Let $\Omega_r = \bd \Omega \union (-C_r)$.
	Take the following path integral:

	![Image](figures/2020-01-27-13:59.png)\

	Then
	\begin{align*}
	0 = \int_{\bd \Omega_r} F(\xi) ~d\xi = \int_{\bd \Omega} F(\xi) ~d\xi - \int_{C_r} F(\xi) ~d\xi
	,\end{align*}

	which forces these integrals to be equal.

Corollary ($C^1$ implies smooth)
: 	If $f$ is holomorphic, then $f\in C^1(\Omega)$ implies that $f \in C^\infty(\Omega)$.

Theorem (Holomorphic implies analytic)
: 	If $f$ is holomorphic in $\Omega$, then $f$ is equal to its Taylor series (i.e. $f(z_0$ is analytic.)

Proof
: 	Fix $z_0 \in \Omega$ and let $r = \abs{z- z_0}$.

	\begin{align*}
	\frac{1}{\xi - z} 
	&= \frac{1}{ \xi - z_0 - (z-z_0)  } \\
	&= \frac{1}{\xi - z_0} \frac{1}{1 - \qty{ \frac{z-z_0}{\xi - z_0}   }  } \\
	&= \frac{1}{\xi - z_0} \sum_n \qty{ \frac{z - z_0}{\xi - z_0}  }^n \quad\text{for } \abs{z - z_0} < \abs{\xi - z_0}
	.\end{align*}

	Note that $\sum z^n$ converges uniformly for any $\abs{z} < \delta < 1$.

	Thus

	\begin{align*}
	f(z) 
	&= \frac{1}{2\pi i} \int_{\xi \in \bd\Omega} f(\xi) \sum \frac{(z-z_0)^n}{(\xi - z_0)^{n+1}} ~d\xi \\
	&= \sum \qty{ \frac {1}{2\pi i} \int \frac{f(\xi)  }{(\xi - z_0)^{n+1}} ~d\xi} (z-z_0)^n \\
	&= \sum \frac{f^{(n)} (z_0)}{n!} (z-z_0)^n
	.\end{align*}

Corollary
: 	$f$ is holomorphic iff $f$ is analytic.

Counterexample to keep in mind:

\begin{align*}
f(x) = \begin{cases}
x^2 & x > 0 \\
0 & x\leq 0
\end{cases}
.\end{align*}

In the case of $\RR$, smooth and analytic are very different categories of functions.