# Friday February 7th

## Sequences of Holomorphic Functions

Theorem (The Uniform Limit of Holomorphic Functions is Holomorphic)
: 	Suppose $\theset{f_n}\to f$ is a sequence of holomorphic functions converging uniformly on any compact subset $K \subset \Omega$.
	Then $f$ is holomorphic.

Proof
: 	Let $D$ be any disc such that $\bar D \subset \Omega$.
	For any rectangle $R \subset D$, we have $$\int_R f_n ~dz = 0.$$
	Since $f_n \to f$ uniformly, $\int_R f ~dz = 0$ and thus $f$ is holomorphic in $D$.

![Image](figures/2020-02-07-13:36.png)\

Theorem (Uniform Convergence of Derivatives)
: 	Under the same hypotheses, $f_n' \to f$ uniformly on any compact subset $K \subset \Omega$.

Proof
: 	See Stein.

Corollary (When Functions Defined by Integrals are Holomorphic)
: 	Suppose $F(z, s): \Omega \cross [a, b] \to \CC$ and

	1. $F(z, s)$ is holomorphic in $z$ for each fixed $s \in [a, b]$.

	2. $F(z, s)$ is continuous in $\Omega \cross [a, b]$.

	Then $f(z) = \int_a^b F(z, s) ~ds$ is holomorphic on $\Omega$.

Proof
: 	Define $f_n(z) = \qty{ \sum_{k=1}^n F(z, s_k) } \frac{b-a}{n}$ where each $s_k = a + \frac{b-a}{n} k \in [a, b]$.
	Need to show $f_n(z)$ converges uniformly on any compact $K \subset \Omega$, i.e. it's uniformly Cauchy.
	Fix $K$ compact, then by a theorem in topology $K \cross [a,b]$ is again compact.
	
	Using the fact that $F$ is continuous on a compact set and thus uniformly continuous, fix $\eps > 0$ and find $\delta>0$ such that $\max_{z\in K} \abs{F(z, s) - F(z, t)} < \eps$ for all $s,t \in [a, b]$ with $\abs{t-s} < \delta$.

	Thus if $\frac{b-a}{n} < \delta$ and $z\in K$, we have an estimate

	\begin{align*}
	\abs{f_n(z) - f(z)} 
	&= \abs{ \sum_{k=1}^n \int_{s_{k-1}}^{s_k} F(z, s_k) - F(z, s) ~ds } \\
	&= \sum_{k=1}^n \int_{s_{k-1}}^{s_k} \abs{ F(z, s_k) - F(z, s) } ~ds \\
	&\leq \eps \qty{b-a} 
	.\end{align*}

	Thus $f_n \converges{u}\to f$.


Remark: this is useful for showing $$\Gamma(z) = \int_0^\infty e^{-s} s^{z-1} ~ds$$ is holomorphic for $\Re z > 0$.

## Uniform Approximation

**Question**: can every function be uniformly approximated by polynomials?

**Answer**: in general, no.
Take $f(z) = \frac 1 z$, which is holomorphic on $\CC \setminus 0$, but $\int_\gamma P_N(z) = 0$ for any polynomial (since )hey are entire) for any loop $\gamma$ around 0, but $\int_\gamma \frac 1 z = 2\pi i$.

Theorem (Uniform Approximation by Polynomials (Stein 5.2))
: 	If $f_n$ is a sequence of holomorphic functions converging uniformly on any compact subset $K$ of $\Omega$ then $f$ is holomorphic in $\Omega$ and if $f(z) = \sum a_n (z- z_0)^n$ then $P_N(z) = \sum^N a_n (z-z_0)^n$.

Theorem (Uniform Approximation by Rational Functions (Stein 5.7))
: 	Any holomorphic function in a neighborhood of a compact set $K$ can be approximated by a *rational* function with singularities only in $K^c$.
		If $K^c$ is connected, it can be approximated by a *polynomial*.

Lemma (5.8, ???)
: 	Suppose $f$ is holomorphic in an open set $\Omega$ with $K\subset \Omega$ compact.
		Then there exist finitely many segments $\theset{\gamma_i}_{i=1}^N$ in $\Omega\setminus K$ such that for all $z\in K$, ???.

Proof (of Lemma, Idea)
: 	Divide region into squares, take $\gamma_i$ to be line segments such that they enclose $K$.
	\begin{align*}
	f(z) 
	&= \frac{1}{2\pi i} \sum_{n=1}^N \int_{\omega_n} \frac{f(\xi)}{z-\xi} ~d\xi \\
	&= \frac{1}{2\pi i} \int_{\Gamma} \frac{f(\xi)}{z-\xi} ~d\xi
	.\end{align*}

	where we can rewrite 
	$$
	\int_{\gamma_n} \cdots = \int_0^1 \frac{f(\gamma_n(t))}{\gamma_n(t) - z_0} \gamma_n'(t) ~dt = \int_0^1 F(z, s) ~ds
	$$

	The idea is that we can then write $\frac 1 {\xi - z} = \frac 1 \xi \frac 1 {1-\frac{z}{\xi}} = \xi\inv \sum_k \qty{\frac{z}{\xi}}^k$, which allows uniform approximation by polynomials.

![Image](figures/2020-02-07-19:51.png)\