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本帖最后由 hbghlyj 于 2023-5-17 10:39 编辑 这份讲义的第41页:
Theorem 2.1.15 Consider the power series
$$
\begin{aligned}
f(z) &=\sum_{n=0}^{\infty} a_{n} z^{n} \\
&=a_{0}+a_{1} z+\cdots+a_{n} z^{n}+\cdots .
\end{aligned}
$$
Let $R$ be its convergence radius, and assume that $0<R \leq \infty$. Then
1) The power series obtained by differentiating $f$ term by term
$$
\begin{aligned}
g(z) &=\sum_{n=1}^{\infty} n a_{n} z^{n-1} \\
&=a_{1}+2 a_{2} z \cdots+n a_{n} z^{n-1}+\cdots
\end{aligned}
$$
has the same convergence radius $R$.
2) The [complex] derivative
$$
f^{\prime}(z)=\lim _{w \rightarrow z} \frac{f(w)-f(z)}{w-z}
$$
exists for every $z$ satisfying that $|z|<R$, and $f^{\prime}(z)=g(z)$. That is
$$
\frac{d}{d z} \sum_{n=0}^{\infty} a_{n} z^{n}=\sum_{n=1}^{\infty} n a_{n} z^{n-1} \quad \text { for any }|z|<R
$$
Proof. [This theorem says that we may differentiate a power series term by term. Proof is not examinable in Prelims Paper II - this theorem will be revisited in Paper A2.]
1) Let $|z|<R$. Set $r=\frac{1}{2}(|z|+R)$ (or $r=2|z|+1$ if $\left.R=\infty\right)$. Then $|z|<r<R$ and $q \equiv \frac{|z|}{r} \in[0,1)$. We have the following facts:
(a) $\sum_{n=0}^{\infty}\left|a_{n}\right| r^{n}<\infty$ [Analysis 1: a power series converges absolutely inside its convergence disk],
(b) $\left\{n q^{n-1}\right\}$ is bounded. [Indeed $\sum n q^{n-1}$ converges (by the ratio test), so that $\lim _{n \rightarrow \infty} n q^{n-1}=0$ :
but we don't need these stronger results here].
Let $b_{n}=n q^{n-1}$. Then
$$
\frac{b_{n+1}}{b_{n}}=\frac{n+1}{n} q
$$
which is smaller than 1 for $n$ large enough. Thus $\left(b_{n}\right)$ is decreasing for large $n$, so that $\lim _{n \rightarrow \infty} b_{n}$ exists, and therefore $\left(n q^{n-1}\right)$ is bounded. Let $n q^{n-1} \leq M$ for some $M>0$, for every $n$.
(c) $\sum_{n=1}^{\infty} n a_{n} z^{n-1}$ converges absolutely. Indeed
$$
\begin{aligned}
\left|n a_{n} z^{n-1}\right| & \leq n\left|a_{n}\right||z|^{n-1}=n q^{n-1}\left|a_{n}\right| r^{n-1} \\
& \leq \frac{M}{r}\left|a_{n}\right| r^{n} \quad \forall n \geq 1
\end{aligned}
$$
so that, by the comparison test [Analysis 1]
$$
\sum_{n=1}^{\infty} n\left|a_{n}\right||z|^{n-1} \leq \frac{M}{r} \sum_{n=1}^{\infty}\left|a_{n}\right| r^{n}<\infty .
$$
Similarly we may prove that the convergence radius of $\sum_{n=1}^{\infty} n a_{n} z^{n-1}$ can not be greater than that of $\sum_{n=0}^{\infty} a_{n} z^{n}$.
2) We are going to show that the complex derivative $f^{\prime}(z)$ exists and equals $g(z)$ at every point $z$ such that $|z|<R$. Let $r=\frac{1}{2}(|z|+R)$ (or $r=|z|+1$ if $R=\infty$ ). Then $r<R$, and $|z|<r$. For any point $w \neq z$ such that $|w|<r$, consider
$$
\begin{aligned}
\frac{f(w)-f(z)}{w-z}-g(z) &=\sum_{n=1}^{\infty} a_{n}\left(\frac{w^{n}-z^{n}}{w-z}-n z^{n-1}\right) \\
&=\sum_{n=2}^{\infty} a_{n}\left(\frac{w^{n}-z^{n}}{w-z}-n z^{n-1}\right)
\end{aligned}
$$
where we have added the series $f(w), f(z)$ and $g(z)$ term by term, which is justified as all these series are absolutely convergent [Analysis 1: a power series converges absolutely inside the convergence disk]. Our aim is to show that
$$
\frac{f(w)-f(z)}{w-z}-g(z) \rightarrow 0 \quad \text { as } w \rightarrow z
$$
To this end we use the following identity
$$
\frac{w^{n}-z^{n}}{w-z}=z^{n-1}+z^{n-2} w+\cdots+z w^{n-2}+w^{n-1}
$$
Therefore, for any $w≠z$ and $n≥2$,
$$\begin{aligned} \frac{w^{n}-z^{n}}{w-z}-n z^{n-1}=& z^{n-1}+z^{n-2} w+\cdots+z w^{n-2}+w^{n-1} \\ &-z^{n-1}-z^{n-1}-\cdots-z^{n-1}-z^{n-1} \\=& \sum_{k=1}^{n-1}\left(z^{n-1-k} w^{k}-z^{n-1}\right) \\=& \sum_{k=1}^{n-1} z^{n-1-k}\left(w^{k}-z^{k}\right) \end{aligned}$$
Let
$$
h_{n}(w)=a_{n} \sum_{k=1}^{n-1} z^{n-1-k}\left(w^{k}-z^{k}\right) ; \quad n=2,3, \cdots .
$$
Then
$$
\frac{f(w)-f(z)}{w-z}-g(z)=\sum_{n=2}^{\infty} h_{n}(w)
$$
All $h_{n}$ are continuous in $\mathbb{C}$ (polynomials in $w$ ), and $h_{n}(z)=0$ (for all $n \geq 2$ ). We claim that $\sum_{n=2}^{\infty} h_{n}(w)$ converges uniformly in $|w| \leq r$. In fact
$$
\begin{aligned}
\left|h_{n}(w)\right| & \leq\left|a_{n}\right| \sum_{k=1}^{n-1}|z|^{n-1-k}\left(|w|^{k}+|z|^{k}\right) \\
& \leq 2 n\left|a_{n}\right| r^{n-1}
\end{aligned}
$$
By 1), $\sum n\left|a_{n}\right| r^{n-1}<\infty$, so that $\sum_{n=2}^{\infty} h_{n}(w)$ converges uniformly in closed disk $\{w:|w| \leq r\}$ [Weierstrass M-test, Chapter 2]. Hence $\sum_{n=2}^{\infty} h_{n}(w)$ is continuous in the disk $|w| \leq r$ [Theorem 1.4.11: the uniform limit of continuous functions is continuous]. Therefore
$$
\lim _{w \rightarrow z} \sum_{n=2}^{\infty} h_{n}(w)=\sum_{n=2}^{\infty} h_{n}(z)=0
$$
so that
$$
\begin{aligned}
\lim _{w \rightarrow z} \frac{f(w)-f(z)}{w-z} &=\lim _{w \rightarrow z}\left(\frac{f(w)-f(z)}{w-z}-g(z)\right)+g(z) \\
&=\lim _{w \rightarrow z} \sum_{n=2}^{\infty} h_{n}(w)+g(z) \\
&=g(z) .
\end{aligned}
$$
This completes the proof. |
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