3.5 KiB
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
\int_{\Gamma}\frac{\phi(\zeta)}{\zeta-z}d\zeta
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.
\int_{\gamma}f(z)dz = 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
f(z) = \frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}d\zeta
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
f_n(z_0) = \frac{1}{2\pi i}\int_{C_r}\frac{f_n(\zeta)}{\zeta-z_0}d\zeta
\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,
f(z_0) = \frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})d\theta
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
g(z) = \begin{cases}
\frac{f(z)}{z} & z\neq 0 \\
f'(0) & z=0
\end{cases}
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,
\frac{|f(z)-f(w)|}{|1-\overline{f(w)}f(z)|}\leq\frac{|z-w|}{|1-\overline{w}z|}=\rho(z,w)
Prove after spring break.