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## 数学代写|复分析作业代写Complex function代考|The Cauchy Estimates and Liouville’s Theorem

This section will establish some estimates for the derivatives of holomorphic functions in terms of bounds on the function itself. The possibility of such estimation is a feature of holomorphic function theory that has no analogue in the theory of real variables. For example: The functions $f_{k}(x)=\sin k x, x \in \mathbb{R}, k \in \mathbb{Z}$, are for all $k$ bounded in absolute value by 1 ; but their derivatives at 0 are not bounded since $f_{k}^{\prime}(0)=k$. The reader should think frequently about the ways in which real function theory and complex holomorphic function theory differ. In the subject matter of this section, the differences are particularly pronounced.

Theorem 3.4.1 (The Cauchy estimates). Let $f: U \rightarrow \mathbb{C}$ be a holomorphic function on an open $U, P \in U$, and assume that the closed disc $\bar{D}(P, r), r>0$, is contained in $U$. Set $M=\sup {z \in \bar{D}(P, r)}|f(z)|$. Then for $k=1,2,3 \ldots$ we have $$\left|\frac{\partial^{k} f}{\partial z^{k}}(P)\right| \leq \frac{M k !}{r^{k}} .$$ Proof. By Theorem 3.1.1, $$\frac{\partial^{k} f}{\partial z^{k}}(P)=\frac{k !}{2 \pi i} \oint{|\zeta-P|=r} \frac{f(\zeta)}{(\zeta-P)^{k+1}} d \zeta .$$
Now we use Proposition 2.1.8 to see that
$$\left|\frac{\partial^{k} f}{\partial z^{k}}(P)\right| \leq \frac{k !}{2 \pi} \cdot 2 \pi r \sup _{|\zeta-P|=r} \frac{|f|}{|\zeta-P|^{k+1}} \leq \frac{M k !}{r^{k}} .$$

## 数学代写|复分析作业代写Complex function代考|Uniform Limits of Holomorphic Functions

We have already seen that a convergent power series (in $z$ ) defines a holomorphic function (Lemma 3.2.10). One can think of this fact as the assertion that a certain sequence of holomorphic functions, namely the finite partial sums of the series, has a holomorphic limit. This idea, that a limit of holomorphic functions is holomorphic, holds in almost unrestricted generality.
Theorem 3.5.1. Let $f_{j}: U \rightarrow \mathbb{C}, j=1,2,3 \ldots$, be a sequence of holomorphic functions on an open set $U$ in $\mathbb{C}$. Suppose that there is a function $f: U \rightarrow \mathbb{C}$ such that, for each compact subset $E$ of $U$, the sequence $\left.f_{j}\right|{E}$ converges uniformly to $\left.f\right|{E}$. Then $f$ is holomorphic on $U$. (In particular, $\left.f \in C^{\infty}(U) .\right)$

Before beginning the proof, we again note the contrast with the realvariable situation. Any continuous function from $\mathbb{R}$ to $\mathbb{R}$ is the limit, uniformly on compact subsets of $\mathbb{R}$, of some sequence of polynomials: This is the well-known Weierstrass approximation theorem. But, of course, a continuous function from $\mathbb{R}$ to $\mathbb{R}$ certainly need not be real analytic, or even $C^{\infty}$. The difference between the real-variable situation and that of the theorem is related to the Cauchy estimates. The convergence of a sequence of holomorphic functions implies convergence of their derivatives also. No such estimation holds in the real case, and a sequence of $C^{\infty}$ functions can converge uniformly without their derivatives having any convergence properties at all (see Exercise 1 and [RUD1]).

We shall give one detailed proof of Theorem 3.5.1 and sketch a second. The first method is especially brief.

# 复分析代写

## 数学代写|复分析作业代写Complex function代考|The Cauchy Estimates and Liouville’s Theorem

$$\left|\frac{\partial^{k} f}{\partial z^{k}}(P)\right| \leq \frac{M k !}{r^{k}} .$$

$$\frac{\partial^{k} f}{\partial z^{k}}(P)=\frac{k !}{2 \pi i} \oint|\zeta-P|=r \frac{f(\zeta)}{(\zeta-P)^{k+1}} d \zeta$$

$$\left|\frac{\partial^{k} f}{\partial z^{k}}(P)\right| \leq \frac{k !}{2 \pi} \cdot 2 \pi r \sup _{|\zeta-P|=r} \frac{|f|}{|\zeta-P|^{k+1}} \leq \frac{M k !}{r^{k}}$$

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