### 数学代写|复变函数作业代写Complex function代考|The Cauchy Estimates and Liouville’s Theorem

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## 数学代写|复变函数作业代写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 set $U, P \in U$, 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
$$\left|\frac{\partial^{k} f}{\partial z^{k}}(P)\right| \leq \frac{k !}{2 \pi} \cdot 2 \pi r \sup {|\zeta-P|=r} \frac{|f|}{|\zeta-P|^{k+1}} \leq \frac{M k !}{r^{k}}$$ Notice that the Cauchy estimates enable one to estimate directly the radius of convergence of the power series $$\sum \frac{f^{(k)}(P)}{k !}(z-P)^{k}$$ Namely $$\limsup {k \rightarrow+\infty}\left|\frac{f^{(k)}(P)}{k !}\right|^{1 / k} \leq \limsup {k \rightarrow+\infty}\left|\frac{M \cdot k !}{r^{k}} \cdot \frac{1}{k !}\right|^{1 / k}=\frac{1}{r}$$ for any $r$ such that $\bar{D}(P, r) \subseteq U$. In particular, the radius of convergence, which equals $$\left[\limsup {k \rightarrow+\infty}\left|\frac{f^{(k)}(P)}{k !}\right|^{1 / k}\right]^{-1}$$
is at least $1 /(1 / r)=r$ for all $r$ such that $\bar{D}(P, r) \subseteq U$. Hence the radius of convergence is at least as large as the distance from $P$ to $\mathbb{C} \backslash U$ (this result was considered from a different perspective in Theorem 3.3.1).

The derivative bounds of Theorem 3.4.1 have some remarkable consequences. The first one is the fact that a holomorphic function on $\mathbb{C}$ that is bounded in absolute value is in fact constant. We shall need the following lemma to prove this.

## 数学代写|复变函数作业代写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.

Proof of Theorem 3.5.1. Let $P \in U$ be arbitrary. Then choose $r>0$ such that $\bar{D}(P, r) \subseteq U$. Since $\left{f_{j}\right}$ converges to $f$ uniformly on $\bar{D}(P, r)$ and since each $f_{j}$ is continuous, $f$ is also continuous on $\bar{D}(P, r)$. For any $z \in D(P, r)$

## 数学代写|复变函数作业代写Complex function代考|The Zeros of a Holomorphic Function

Let $f$ be a holomorphic function. If $f$ is not identically zero, then it turns out that $f$ cannot vanish at too many points. This once again bears out the dictum that holomorphic functions are a lot like polynomials. The idea has a precise formulation as follows:

Theorem 3.6.1. Let $U \subseteq \mathbb{C}$ be a connected open set and let $f: U \rightarrow \mathbb{C}$ be holomorphic. Let $\mathbf{Z}={z \in U: f(z)=0}$. If there are a $z_{0} \in \mathbf{Z}$ and $\left{z_{j}\right}_{j=1}^{\infty} \subseteq \mathbf{Z} \backslash\left{z_{0}\right}$ such that $z_{j} \rightarrow z_{0}$, then $f \equiv 0$.

Let us formulate Theorem $3.6 .1$ in topological terms. We recall that a point $z_{0}$ is said to be an accumulation point of a set $Z$ if there is a sequence $\left{z_{j}\right} \subseteq Z \backslash\left{z_{0}\right}$ with $\lim {j \rightarrow \infty} z{j}=z_{0}$. Then Theorem $3.6 .1$ is equivalent to the statement: If $f: U \rightarrow \mathbb{C}$ is a holomorphic function on a connected open set $U$ and if $Z={z \in U: f(z)=0}$ has an accumulation point in $U$, then $f \equiv 0$.

There is still more terminology attached to the situation in Theorem 3.6.1. A set $S$ is said to be discrete if for each $s \in S$ there is an $\epsilon>0$ such that $D(s, \epsilon) \cap S={s}$. People also say, in an abuse of language, that a discrete set has points which are “isolated” or that $S$ contains only “isolated points.” Theorem 3.6.1 thus asserts that if $f$ is a nonconstant holomorphic function on a connected open set, then its zero set is discrete or, less formally, the zeros of $f$ are isolated. It is important to realize that Theorem 3.6.1 does not rule out the possibility that the zero set of $f$ can have accumulation points in $\mathbb{C} \backslash U$; in particular, a nonconstant holomorphic function on an open set $U$ can indeed have zeros accumulating at a point of $\partial U$. For example, the function $f(z)=\sin (1 /(1-z))$ is holomorphic on $U=D(0,1)$ and vanishes on the set
$$\mathbf{Z}=\left{1-\frac{1}{\pi n}: n=1,2,3, \ldots\right} .$$
Plainly $\mathbf{Z}$ has no accumulation points in $U$; however the point $1 \in \partial U$ is an accumulation point of $\mathbf{Z}$.

Proof of Theorem 3.6.1. We first claim that, under the hypotheses of the theorem, $(\partial / \partial z)^{n} f\left(z_{0}\right)=0$ for every nonnegative integer $n$. If this is not the case, let $n_{0}$ be the least nonnegative integer $n$ such that
$$\left(\frac{\partial}{\partial z}\right)^{n_{0}} f\left(z_{0}\right) \neq 0$$
Then we have, on some small disc $D\left(z_{0}, r\right)$, the power series expansion
$$f(z)=\sum_{j=n_{0}}^{\infty}\left(\frac{\partial^{j}}{\partial z^{j}} f\left(z_{0}\right)\right) \frac{\left(z-z_{0}\right)^{j}}{j !} .$$
Therefore the function $g$ defined by
$$g(z) \equiv \sum_{j=n_{0}}^{\infty}\left(\frac{\partial}{\partial z}\right)^{j} f\left(z_{0}\right) \frac{\left(z-z_{0}\right)^{j-n_{0}}}{j !}$$
is holomorphic on $D\left(z_{0}, r\right)$ and $g\left(z_{0}\right) \neq 0$ since $\left(\frac{\partial}{\partial z}\right)^{n_{0}} f\left(z_{0}\right) \neq 0$. Notice that the indicated power series has the same radius of convergence as that for $f$ itself. Furthermore, $g\left(z_{l}\right)=0$ for $l=1,2,3, \ldots$. But then, by the continuity of $g, g\left(z_{0}\right)=0$. This contradiction proves our claim.

## 数学代写|复变函数作业代写Complex function代考|The Cauchy Estimates and Liouville’s Theorem

|∂ķF∂和ķ(磷)|≤ķ!2圆周率⋅2圆周率r支持|G−磷|=r|F||G−磷|ķ+1≤米ķ!rķ请注意，柯西估计使人们能够直接估计幂级数的收敛半径∑F(ķ)(磷)ķ!(和−磷)ķ即林汤ķ→+∞|F(ķ)(磷)ķ!|1/ķ≤林汤ķ→+∞|米⋅ķ!rķ⋅1ķ!|1/ķ=1r对于任何r这样D¯(磷,r)⊆在. 特别是，收敛半径等于[林汤ķ→+∞|F(ķ)(磷)ķ!|1/ķ]−1

## 数学代写|复变函数作业代写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 fj:U→C,j=1,2,3…, be a sequence of holomorphic functions on an open set U in C. Suppose that there is a function f:U→C such that, for each compact subset E of U, the sequence fj|E converges uniformly to f|E. Then f is holomorphic on U. (In particular, f∈C∞(U).)

Before beginning the proof, we again note the contrast with the realvariable situation. Any continuous function from R to R是限制，一致地在紧凑子集上R, 一些多项式序列：这是著名的 Weierstrass 逼近定理。但是，当然，一个连续的函数R到R当然不必是真正的分析，甚至C∞. 实变量情况与定理的区别与柯西估计有关。一系列全纯函数的收敛也意味着它们的导数的收敛。在实际情况中没有这样的估计，并且一系列C∞函数可以均匀收敛，而它们的导数根本不具有任何收敛特性（参见练习 1 和 [RUD1]）。

## 数学代写|复变函数作业代写Complex function代考|The Zeros of a Holomorphic Function

\mathbf{Z}=\left{1-\frac{1}{\pi n}: n=1,2,3, \ldots\right} 。\mathbf{Z}=\left{1-\frac{1}{\pi n}: n=1,2,3, \ldots\right} 。

(∂∂和)n0F(和0)≠0

F(和)=∑j=n0∞(∂j∂和jF(和0))(和−和0)jj!.

G(和)≡∑j=n0∞(∂∂和)jF(和0)(和−和0)j−n0j!

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