# Math416 Lecture 17

## Continue on Chapter 7

### Harmonic conjugates

#### Theorem 7.18

Existence of harmonic conjugates.

Let $u$ be a harmonic function on $\Omega$ a convex open subset in $\mathbb{C}$. Then there exists $g\in O(\Omega)$ such that $\text{Re}(g)=u$ on $\Omega$.

Moreover, $g$ is unique up to an imaginary additive constant.

Proof:

Let $f=2\frac{\partial u}{\partial z}=\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}$

$f$ is holomorphic on $\Omega$

Since $\frac{\partial u}{\partial \overline{z}}=0$ on $\Omega$, $f$ is holomorphic on $\Omega$

So $f=g'$, fix $z_0\in \Omega$, we can choose $q(z_0)=u(z_0)$ and $g=u_1+iv_1$, $g'=\frac{\partial u_1}{\partial x}+i\frac{\partial v_1}{\partial x}=\frac{\partial v_1}{\partial y}-i\frac{\partial u_1}{\partial y}=\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}$, given that $\frac{\partial u_1}{\partial x}=\frac{\partial u}{\partial x}$ and $\frac{\partial u_1}{\partial y}=\frac{\partial u}{\partial y}$

So $u_1=u$ on $\Omega$

$\text{Re}(g)=u_1=u$ on $\Omega$

If $u+iv$ is holomorphic, $v$ is harmonic conjugate of $u$

QED

### Corollary For Harmonic functions

#### Theorem 7.19

Harmonic functions are $C^\infty$

$C^\infty$ is a local property.

#### Theorem 7.20

Mean value property for harmonic functions.

Let $u$ be harmonic on an open set of $\Omega$

Then $u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta$

Proof:

$\text{Re}g(z_0)=\frac{1}{2\pi}\int_0^{2\pi}\text{Re}g(z_0+re^{i\theta})d\theta$

QED

#### Theorem 7.21

Identity theorem for harmonic functions.

Let $u$ be harmonic on a domain $\Omega$. If $u=0$ on some open set $G\subset \Omega$, then $u\equiv 0$ on $\Omega$.

_If $u=v$ on $G\subset \Omega$, then $u=v$ on $\Omega$._

Proof:

We proceed by contradiction.

Let $H=\{z\in \Omega:u(z)=0\}$ be the interior of $G$

$H$ is open and nonempty. If $H\neq \Omega$, then $\exists z_0\in \partial H\cap \Omega$. Then $\exists r>0$ such that $B_r(z_0)\subset \Omega$ such that $\exists g\in O(B_r(z_0))$ such that $\text{Re}g=u$ on $B_r(z_0)$

Since $H\cap B_r(z_0)$ is nonempty open set, then $g$ is constant on $H\cap B_r(z_0)$

So $g$ is constant on $B_r(z_0)$

So $u$ is constant on $B_r(z_0)$

So $D(z_0,r)\subset H$. This is a contradiction that $z_0\in \partial H$

QED

#### Theorem 7.22

Maximum principle for harmonic functions.

A non-constant harmonic function on a domain cannot attain a maximum or minimum on the interior of the domain.

Proof:

We proceed by contradiction.

Suppose $u$ attains a maximum at $z_0\in \Omega$.

For all $z$ in the neighborhood of $z_0$, $u(z)<u(z_0)$. We can choose $r>0$ such that $B_r(z_0)\subset \Omega$.

By the mean value property, $u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta$

So $0= \frac{1}{2\pi}\int_0^{2\pi}u[z_0+re^{i\theta}-u(z_0)]d\theta$

We can prove the minimum is similar.

QED

> Maximum/minimum (modulus) principle for holomorphic functions.
>
> If $f$ is holomorphic on a domain $\Omega$ and attains a maximum on the boundary of $\Omega$, then $f$ is constant on $\Omega$.
>
> Except at $z_0\in \Omega$ where $f'(z_0)=0$, if $f$ attains a minimum on the boundary of $\Omega$, then $f$ is constant on $\Omega$.

### Dirichlet problem for domain $D$

Let $h: \partial D\to \mathbb{R}$ be a continuous function. Is there a harmonic function $u$ on $D$ such that $u$ is continuous on $\overline{D}$ and $u|_{\partial D}=h$?

We can always solve the problem for the unit disk.

$$
u(z)=\frac{1}{2\pi}\int_0^{2\pi}h(e^{i t})\text{Re}\left(\frac{e^{it}+z}{e^{it}-z}\right)dt
$$

Let $z=re^{i\theta}$

$$
\text{Re}\left(\frac{e^{it}+re^{i\theta}}{e^{it}-re^{i\theta}}\right)=\frac {1-r^2}{1-2r\cos(\theta-t)+r^2}
$$

_This is called Poisson kernel._

$Pr(\theta, t)>0$ and $\int_0^{2\pi}Pr(\theta, t)dt=1$, $\forall r,t$

## Chapter 8 Laurent series

when $\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$ converges?

Claim $\exists R>0$ such that $\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$ converges if $|z-z_0|<R$ and diverges if $|z-z_0|>R$

Proof:

Let $u=\frac{1}{z-z_0}$

$\sum_{n=0}^{\infty}a_n(z-z_0)^n$ has radius of convergence $\frac{1}{R}$

So the series converges if $|u|<\frac{1}{R}$

So $|z-z_0|=\frac{1}{|u|}>\frac{1}{\frac{1}{R}}=R$

QED

### Laurent series

A Laurent series is a series of the form $\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$

The series converges in some annulus shape $A=\{z:r_1<|z-z_0|<r_2\}$

The annulus is called the region of convergence of the Laurent series.