Math416 Lecture 15

Review on Cauchy Integrals

The cauchy integral of function $\phi$ (may not be holomorphic) on curve $\Gamma$ (may not be closed) is

The Cauchy integral theorem states that if $f$ is holomorphic on a simply connected domain $D$, then the integral of $f$ over any closed curve $\gamma$ in $D$ is 0.

The Cauchy integral formula states that if $f$ is holomorphic on a simply connected domain $D$, then $f$ over any closed curve $\gamma$ in $D$ is

Continue on Cauchy Integrals (Chapter 7)

Convergence of functions

Theorem 7.15 Weierstrass Convergence Theorem

Limit of a sequence of holomorphic functions is holomorphic.

Let $G$ be an open subset of $\mathbb{C}$ and let $\left(f_n\right)_{n\in\mathbb{N}}$ be a sequence of holomorphic functions on $G$ that converges locally uniformly to $f$ on $G$. Then $f$ is holomorphic on $G$.

Proof:

Let $z_0\in G$. There exists a neighborhood $\overline{B_r(z_0)}\subset G$ of $z_0$ such that $\left(f_n\right)_{n\in\mathbb{N}}$ converges uniformly on $\overline{B_r(z_0)}$.

Let $C_r=\partial B_r(z_0)$.

By Cauchy integral formula, we have

$\forall z\in B_r(z_0)$, we have $\frac{f(w)}{w-\zeta}$ converges uniformly on $C_r$.

So $\lim_{n\to\infty}f_n(z_0) = f(z_0) = \frac{1}{2\pi i}\int_{C_r}\frac{f(w)}{w-z_0}dw$

So $f$ is holomorphic on $G$.

QED

Theorem 7.16 Maximum Modulus Principle

If $f$ is a non-constant holomorphic function on a domain $D$ (open and connected subset of $\mathbb{C}$), then $|f|$ does not attain a local maximum value on $D$.

Proof:

Assume at some point $z_0\in D$, $|f(z_0)|$ is a local maximum. $\exists r>0$ such that $\forall z\in \overline{B_r(z_0)}$, $|f(z)|\leq |f(z_0)|$.

If $f(z_0)=0$, then $f(z)$ is identically 0 on $B_r(z_0)$. (by identity theorem)

Else, we can assume that without loss of generality that $f(z_0)>0$. By mean value theorem,

So $f(z_0) =$

/* TRACK LOST */

Corollary 7.16.1 Minimum Modulus Principle

If $f$ is a non-constant holomorphic function on a domain $D$ (open and connected subset of $\mathbb{C}$), and $f$ is non zero on $D$, then $\frac{1}{f}$ does not attain a local minimum value on $D$.

Proof:

Let $g(z) = \frac{1}{f(z)}$. $g$ is holomorphic on $D$.

QED

Theorem 7.17 Schwarz Lemma

Let $f$ be a holomorphic map of the open unit disk $D$ into itself, and $f(0)=0$. Then $\forall z\in D$, $|f(z)|\leq |z|$ and $|f'(0)|\leq 1$.

And the equality holds if and only if $f$ is a rotation, that is, $f(z)=e^{i\theta}z$ for some $\theta\in\mathbb{R}$.

Proof:

Let

We claim that $g$ is holomorphic on $D$.

For $z\neq 0$, $g(z)$ is holomorphic since $f$ is holomorphic on $D$.

For $z=0$, $g(z)$ is holomorphic since $f$‘s power series expansion has $c_0=f(0)=0$. $g'(0)=f'(0)=c_1+c_2z+c_3z^2+\cdots$.

So $g$ is (analytic) thus holomorphic on $D$.

On the boundary of $D$, $|g(z)|\leq\frac{1}{r} \cdot 1$. By maximum modulus principle, $|g(z)|\leq 1$ on $D$.

So $|f(z)|\leq |z|$ on $D$.

And $|f'(0)|\leq 1$.

QED

Schwarz-Pick Lemma

Let $f$ be a holomorphic map of the open unit disk $D$ into itself, then for any $z,w\in D$,

Prove after spring break.