NoteNextra-origin/content/Math416/Math416_L8.md

Math416 Lecture 8

Review

Sequences of Functions

Let f_n: G \to \mathbb{C} be a sequence of functions.

Convergence Pointwise

Definition:

Let z\in G, \forall \epsilon > 0, \exists N \in \mathbb{N} such that \forall n \geq N, |f_n(z) - f(z)| < \epsilon.

Convergence Uniformly

Definition:

\forall \epsilon > 0, \forall z\in G, \exists N \in \mathbb{N} such that \forall n \geq N, |f_n(z) - f(z)| < \epsilon.

Convergence Locally Uniformly

Definition:

\forall \epsilon > 0, \forall z\in G, \exists N \in \mathbb{N} such that \forall n \geq N, |f_n(z) - f(z)| < \epsilon.

Convergence Uniformly on Compact Sets

Definition: \forall C\subset G that is compact, \forall \epsilon > 0, \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, \forall z\in C, |f_n(z) - f(z)| < \epsilon

Power Series

Definition:

\sum_{n=0}^{\infty} c_n (z - z_0)^n

z_0 is the center of the power series.

Theorem of Power Series

If a power series converges at z_0, then it converges absolutely at every point of \overline{B_r(z_0)} that is strictly inside the disk of convergence.

Continue on Power Series

Review on \limsup

The \limsup(a_n) a_n\in\mathbb{R} is defined as the sup of subsequence of (a_n) as n approaches infinity.

It has the following properties that is useful for proving the remaining parts for this course.

Suppose (a_n)_1^\infty is a sequence of real numbers

1. If \rho\in \mathbb{R} satisfies that \rho<\limsup_{n\to\infty}a_n, then \{a_n : a_n > \rho\} is infinite.
2. If \rho\in \mathbb{R} satisfies that \rho>\limsup_{n\to\infty}a_n, then \{a_n : a_n > \rho\} is finite.

Limits of Power Series

Theorem 5.12

Cauchy-Hadamard Theorem:

The radius of convergence of the power series is given by \sum_{n=0}^{\infty} a_n (z - z_0)^n is given by

\frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}

Proof

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}).

The limit superior of (b_n) is defined as

s_n = \sup_{k\geq n} b_k

s_n is a decreasing sequence, by completeness of \mathbb{R}, every bounded sequence has a limit in \mathbb{R}.

So s_n converges to some limit s\in\mathbb{R}.

Without loss of generality, this also holds for infininum of s_n.

Forward direction:

We want to show that the radius of convergence of \sum_{n=0}^{\infty} a_n (z - z_0)^n is greater than or equal to \frac{1}{\limsup_{n\to\infty} |a_n|^{1/n}}.

Since \sum_{n=0}^{\infty} 1z^n=\frac{1}{1-z} for |z|<1. Assume \limsup_{n\to\infty} |a_n|^{1/n} is finite, then \sum_{n=0}^{\infty} a_n (z - z_0)^n converges absolutely at z_0.

Let \rho>\limsup_{n\to\infty} |a_n|^{1/n}, then \exists N \in \mathbb{N} such that \forall n \geq N, |a_n|^{1/n}\leq \rho. (By property of \limsup)

So \frac{1}{R}=\limsup_{n\to\infty} |a_n|^{1/n}\leq\rho

So R\geq\frac{1}{\rho}

Backward direction:

Suppose |z|>R, then \exists number |z| such that |z|>\frac{1}{\rho}\geq R.

So \rho<\limsup_{n\to\infty} |a_n|^{1/n}

This means that \exists infinitely many $n_j$s such that |a_{n_j}|^{1/n_j}>\rho

So |a_{n_j}z^{n_j}|>\rho^{n_j}|z|^{n_j}

Series \sum_{n=1}^{\infty} a_nz^n diverges, each individual term is not going to 0.

So \sum_{n=0}^{\infty} a_n (z - z_0)^n does not converge at z if |z|> \frac{1}{\rho}\geq R

So R=\frac{1}{\rho}.

What if |z-z_0|=R?

For \sum_{n=0}^{\infty} z^n, the radius of convergence is 1.

It diverges eventually on the circle of convergence.

For \sum_{n=0}^{\infty} \frac{1}{(n+1)^2}z^n, the radius of convergence is 1.

This converges everywhere on the circle of convergence.

For \sum_{n=0}^{\infty} \frac{1}{n+1}z^n, the radius of convergence is 1.

This diverges at z=1 (harmonic series) and converges at z=-1 (alternating harmonic series).

Theorem 5.15

Differentiation of power series

Suppose \sum_{n=0}^{\infty} a_n (z - z_0)^n has a positive radius of convergence R. Define f(z)=\sum_{n=0}^{\infty} a_n (z - z_0)^n, then f is holomorphic on B_R(0) and f'(z)=\sum_{n=1}^{\infty} n a_n (z - z_0)^{n-1}=\sum_{k=0}^{\infty} (k+1)a_{k+1} (z - z_0)^k.

Here below is the proof on book, which will be covered in next lecture.

Proof

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|.

For z,w\in \mathbb{C}, n\in \N, z^n-w^n=(z-w)\sum_{k=0}^{n-1} z^k w^{n-k-1}

Let z_1\in B_R(0), |z_1|<\rho<R for some \rho\in\mathbb{R}.

\begin{aligned}
\frac{f(z)-f(z_1)}{z-z_1}-g(z_1)&=\frac{1}{z-z_1}\left[\sum_{n=0}^\infty a_n z^n -\sum_{n=0}^\infty a_n z_1^n\right]-\sum_{n=1}^{\infty} n a_n z_1 ^{n-1}\\
&=\sum_{n=1}^{\infty} a_n \left[\frac{z^n-z_1^n}{z-z_1}-nz_1^{n-1}\right]\\
&=\sum_{n=1}^{\infty} a_n \left[\left(\sum_{k=0}^{n-1}z^kz_1^{n-k-1}\right)-nz_1^{n-1}\right]\\
&=\sum_{n=2}^{\infty} a_n \left[\sum_{k=1}^{n-1}z_1^{n-k-1}(z^k-z^k_1)\right]
\end{aligned}

Using the lemma again we get

\begin{aligned}
|z^k-z_1^k|&=|z-z_1|\left|\sum_{j=0}^{k-1}z_jz_1^{k-j-1}\right|\\
&\leq |z-z_1| \sum_{j=0}^{k-1}|z_j||z_1^{k-j-1}|\\
&\leq k\rho^{k-1}|z-z_1|
\end{aligned}

Then,

\begin{aligned}
\left|\frac{f(z)-f(z_1)}{z-z_1}-g(z_1)\right|&=\left|\sum_{n=2}^{\infty} a_n \left[\sum_{k=1}^{n-1}z_1^{n-k-1}(z^k-z^k_1)\right]\right|\\
&\leq \sum_{n=2}^{\infty} |a_n| \left[\sum_{k=1}^{n-1}|z_1|^{n-k-1}|z^k-z_1^k|\right]\\
&\leq \sum_{n=2}^{\infty} |a_n|  \left[ \sum_{k=1}^{n-1} \rho^{n-k-1} (k\rho^{k-1}|z-z_1|) \right]\\
&=|z-z_1|\sum_{n=2}^\infty|a_n|\left[\frac{n(n-1)}{2}\rho^{n-2}\right]
\end{aligned}

One can use ratio test to find that \sum_{n=2}^\infty|a_n|\left[\frac{n(n-1)}{2}\rho^{n-2}\right] converges, we denote the sum using M

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).