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本帖最后由 hbghlyj 于 2022-8-10 07:06 编辑 这份讲义的28-29页:
Theorem 1.4.7 (Weierstrass $M$-Test [for Uniform Convergence of Series]) Let $\left(f_{n}\right)$ be a sequence of (real or complex) functions defined on $E$. If there is a sequence of real numbers $\left(M_{n}\right)$ such that
$$
\left|f_{n}(x)\right| \leq M_{n}\quad \text { for all } x \in E
$$
[i.e. $M_{n}$ is an upper bound of $\left|f_{n}\right|$ on $E$] for $n=1,2, \cdots$, and $\sum_{n=1}^{\infty} M_{n}$ converges, then $\sum_{n=1}^{\infty} f_{n}$ converges uniformly on $E$. Moreover
$$
\left|\sum_{n=1}^{\infty} f_{n}(x)\right| \leq \sum_{n=1}^{\infty}\left|f_{n}(x)\right| \leq \sum_{n=1}^{\infty} M_{n}
$$
Proof. The proof of the last inequality, though obvious, is left as an exercise. By Cauchy's criterion for series of numbers, for every $\varepsilon>0$, there exists an integer $N$ such that
$$
\sum_{k=m+1}^{n} M_{k}<\varepsilon \quad \text { for all } n>m \geq N
$$
Let $s_{n}=\sum_{k=1}^{n} f_{k}$ be the partial sum sequence of $\sum_{n=1}^{\infty} f_{n}$. Then for any $n>m \geq N$ and for every $x \in E$
$$
\begin{aligned}
\left|s_{n}(x)-s_{m}(x)\right| &=\left|\sum_{k=m+1}^{n} f_{k}(x)\right| \\
& \leq \sum_{k=m+1}^{n}\left|f_{k}(x)\right| \quad \text { [Triangle Inequality] } \\
& \leq \sum_{k=m+1}^{n} M_{k}
\end{aligned}
$$
That is, $\left|s_{n}-s_{m}\right|$ is bounded above by $\sum_{k=m+1}^{n} M_{k}$ and therefore
$$
\sup _{x \in E}\left|s_{n}(x)-s_{m}(x)\right| \leq \sum_{k=m+1}^{n} M_{k}<\varepsilon
$$
Hence, according to Cauchy’s criterion for uniform convergence, $(s_n)$ converges uniformly in $E$.$□$
Example 1.4.8 Let $E=[0,1]$ and
$$
f_{n}(x)=\frac{x}{1+n^{2} x^{2}} .
$$
Then $\lim _{n \rightarrow \infty} f_{n}(x)=0$ for every $x \in E$. Since
$$
0 \leq f_{n}(x)=\frac{1}{2 n} \frac{2 n x}{1+n^{2} x^{2}} \leq \frac{1}{2 n} \rightarrow 0
$$
so that $f_{n} \rightarrow f$ uniformly on $[0,1]$.
Example 1.4.9 Let
$$
f_{n}(x)=\frac{n x}{1+n^{2} x^{2}} \quad \text { for } x \in[0,1] .
$$
Then $\lim _{n \rightarrow \infty} f_{n}(x)=0$ for every $x \in[0,1]$. While $f_{n}(1 / n)=1 / 2$, so that $\sup _{x \in[0,1]}\left|f_{n}(x)-f(x)\right| \geq \frac{1}{2} \nrightarrow 0$ as $n \rightarrow \infty$
and therefore $f_{n}$ converges point-wise but not uniformly in $[0,1]$.
Example 1.4.10 $\sum_{n=0}^{\infty} x^{n}$ converges to $\frac{1}{1-x}$ for $x \in(-1,1)$, but not uniformly. [$\sum_{n=0}^{\infty} x^{n}$ converges uniformly on $[-r, r]$ for any $0<r<1$, see also Theorem 2.1.15 below].
Indeed, $s_{n}(x)=\sum_{k=0}^{n} x^{k}=\frac{1-x^{n+1}}{1-x}$ tends to $\frac{1}{1-x}$ for any $|x|<1$. On the other hand
$$
\left|s_{n}(x)-\frac{1}{1-x}\right|=\frac{|x|^{n+1}}{|1-x|}
$$
so that
$$
\begin{aligned}
\sup _{x \in(-1,1)}\left|s_{n}(x)-\frac{1}{1-x}\right| & \geq \frac{\left(\frac{n+1}{n+2}\right)^{n+1}}{\left|1-\frac{n+1}{n+2}\right|} \\
&=\frac{n+2}{\left(1+\frac{1}{n+1}\right)^{n+1}} \rightarrow \infty
\end{aligned}
$$
Hence $\sum_{n=0}^{\infty} x^{n}$ does not converge uniformly in $(-1,1)$.
Theorem 1.4.11 Let $f_{n}, f: E \rightarrow \mathbb{R}$ (or $\left.\mathbb{C}\right)$, and $f_{n} \rightarrow f$ uniformly in $E$. Suppose all $f_{n}$ are continuous at $x_{0} \in E$, then the limit function $f$ is also continuous at $x_{0}$. Therefore
$$
\lim _{x \rightarrow x_{0}} \lim _{n \rightarrow \infty} f_{n}(x)=\lim _{n \rightarrow \infty} f_{n}\left(x_{0}\right)=\lim _{n \rightarrow \infty} \lim _{x \rightarrow x_{0}} f_{n}(x) .
$$
[The uniform limit of continuous functions is continuous.]
Proof. For $\varepsilon>0$, since $f_{n} \rightarrow f$ uniformly on $E$, there is $N$, such that for every $n>N$ and $x \in E$
$$
\left|f_{n}(x)-f(x)\right|<\frac{\varepsilon}{3} .
$$
Since $f_{N+1}$ is continuous at $x_{0}$, there is $\delta>0$ (depending on $x_{0}$ and $\varepsilon$ ) such that for $x \in E$ satisfying $\left|x-x_{0}\right|<\delta$, we have
$$
\left|f_{N+1}(x)-f_{N+1}\left(x_{0}\right)\right|<\frac{\varepsilon}{3} .
$$
Hence, for every $x \in E$ such that $\left|x-x_{0}\right|<\delta$, by using the Triangle Inequality,
$$
\begin{aligned}
\left|f(x)-f\left(x_{0}\right)\right| \leq &\left|f(x)-f_{N+1}(x)\right|+\left|f\left(x_{0}\right)-f_{N+1}\left(x_{0}\right)\right| \\
&+\left|f_{N+1}(x)-f_{N+1}\left(x_{0}\right)\right| \\
<& \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon
\end{aligned}
$$
According to definition, $f$ is continuous at $x_{0}$.
Remark 1.4.12 [Version for series] If $\sum_{n=1}^{\infty} f_{n}$ converges uniformly on $E$ and every $f_{n}$ is continuous at $x_{0} \in E$, then
$$
\lim _{x \rightarrow x_{0}} \sum_{n=1}^{\infty} f_{n}(x)=\sum_{n=1}^{\infty} f_{n}\left(x_{0}\right)
$$
In particular, if $f_{n}$ is continuous on $E$ for all $n$ and $\sum_{n=1}^{\infty} f_{n}$ converges uniformly on $E$, then $\sum_{n=1}^{\infty} f_{n}$ is continuous on $E$.
Corollary 1.4.13 Suppose the convergence radius of the power series $\sum_{n=1}^{\infty} a_{n} x^{n}$ is $0<R \leq \infty$, then for every $0 \leq r<R, \sum_{n=1}^{\infty} a_{n} x^{n}$ converges uniformly on the closed disk $\{x:|x| \leq r\}$. Therefore, $\sum_{n=1}^{\infty} a_{n} x^{n}$ is continuous on the open ball $\{x:|x|<R\}$.
Proof. According to the definition of convergence radius, $\sum_{n=1}^{\infty} a_{n} x^{n}$ is absolutely convergent for $|x|<R$. In particular, $\sum_{n=1}^{\infty}\left|a_{n}\right| r^{n}$ is convergent. Since for any $x$ such that $|x| \leq r$
$$
\left|a_{n} x^{n}\right| \leq\left|a_{n}\right| r^{n}
$$
therefore, by Weierstrass M-test, $\sum_{n=1}^{\infty} a_{n} x^{n}$ converges uniformly on $\{x:|x| \leq r\}$. It follows that, according to Theorem 1.4.11, as the uniform limit of continuous functions, $f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$ is continuous on $\{x:|x|<r\}$ for any $0 \leq r<R$. Suppose $\left|x_{0}\right|<R$, then we may choose $r$ such that $\left|x_{0}\right|<r<R$, so that $f(x)$ is continuous at $x_{0}$. Since $x_{0} \in\{x:|x|<R\}$ is arbitrary, $f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$ is continuous on $\{x:|x|<R\}$.
In general a power series $\sum_{n=0}^{\infty} a_{n} x^{n}$ is not uniformly convergent on the disk $\{x:|x|<R\}$, where $R$ is its convergence radius, but the previous corollary implies that it is continuous on $\{x:|x|<R\}$. The end points $R$ and $-R$ need to be handled differently.
Theorem 1.4.14 (Abel's theorem) If the series $\sum_{n=0}^{\infty} a_{n}$ converges, then $\sum_{n=0}^{\infty} a_{n} x^{n}$ converges uniformly on $[0,1]$. Therefore, $\sum_{n=0}^{\infty} a_{n} x^{n}$ is continuous on $[0,1]$, and
$$
\lim _{x \uparrow 1} \sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} a_{n} .
$$
Proof. Let $s_{n}(x)=\sum_{l=0}^{n} a_{l} x^{l}$ be the partial sum sequence associated with the power series $\sum a_{n} x^{n}$. We want to show that $\left(s_{n}\right)$ satisfies the uniform Cauchy principle on $[0,1]$. We have already seen that for $n>m$ we have
$$
\left|s_{n}(x)-s_{m}(x)\right|=\left|\sum_{k=m+1}^{n} a_{k} x^{k}\right|
$$
and we want to control the right-hand side uniformly in $x \in[0,1]$.
Since $\sum a_{n}$ is convergent, its partial sum sequence $\sum_{k=0}^{n} a_{k}$ is a Cauchy sequence, according to the General Principle of Convergence Sequences, from Analysis I. Thus, for every $\varepsilon>0$, there is $N$ such that, for every $n>m>N$ we have
$$
\left|\sum_{k=m+1}^{n} a_{k}\right|<\varepsilon .
$$
Fix $m>N$, set
$$
c_{k}=\sum_{j=m+1}^{k} a_{j} \text { for } k \geq m+1, \quad c_{m}=0
$$
[We may use the following observation - at this stage, from now on, we will only deal with the series with the terms $a_{k} x^{k}$ for $k \geq m+1$, while these terms for $k \leq m$ will not play any role in our argument afterwards. Thus we can employ a trick that we can simply assume that all $a_{k}=0$ for $k \leq m !$].
Then (1.4.2) implies that $\left|c_{k}\right|<\varepsilon$ whenever $k \geq m$, and $a_{k}=c_{k}-c_{k-1}$. We have
$$
\begin{aligned}
\sum_{k=m+1}^{n} a_{k} x^{k} &=\sum_{k=m+1}^{n}\left(c_{k}-c_{k-1}\right) x^{k} \\
&=\sum_{k=m+1}^{n} c_{k} x^{k}-\sum_{k=m+1}^{n} c_{k-1} x^{k} \\
&=\sum_{k=m+1}^{n-1} c_{k}\left(x^{k}-x^{k+1}\right)+c_{n} x^{n}
\end{aligned}
$$
[The last equality is called the Abel's summation formula - which is a discrete version of integration by parts]. Hence, for every $x \in[0,1]$,
$$
\begin{aligned}
\left|\sum_{k=m+1}^{n} a_{k} x^{k}\right| & \leq \sum_{k=m+1}^{n-1}\left|c_{k}\right|\left(x^{k}-x^{k+1}\right)+\left|c_{n}\right| x^{n} \\
&<\varepsilon \sum_{k=m+1}^{n-1}\left(x^{k}-x^{k+1}\right)+\varepsilon x^{n} \\
&=\varepsilon x^{m+1} \leq \varepsilon .
\end{aligned}
$$
According to definition, $\sum_{n=0}^{\infty} a_{n} x^{n}$ converges uniformly on $[0,1]$. Therefore $\sum_{n=0}^{\infty} a_{n} x^{n}$ continuous on $[0,1]$. In particular
$$
\lim _{x \uparrow 1} \sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} a_{n} .
$$
The following Dini's theorem is interesting, but not examinable in paper M2.
Theorem 1.4.15 (Dini's Theorem). Let $f_{n}$ be a sequence of real continuous functions on $[a, b]$. Suppose $\lim _{n \rightarrow \infty} f_{n}(x)=f(x)$ for any $x \in[a, b]$, where $f$ is a continuous function on $[a, b]$, and suppose that
$$
f_{n}(x) \geq f_{n+1}(x) \quad \forall n \text { and } \forall x \in[a, b],
$$
then $f_{n} \rightarrow f$ uniformly in $[a, b]$.
Proof. Let $g_{n}(x)=f_{n}(x)-f(x)$. Then $g_{n}$ is continuous for every $n, g_{n} \geq 0$ and $\lim _{n \rightarrow \infty} g_{n}(x)=0$ for any $x \in[a, b]$. Suppose $\left(g_{n}\right)$ were not uniformly convergent on $[a, b]$. Then there is an $\varepsilon>0$, such that for each $k$ there are a natural number $n_{k}>k$ and a point $x_{k} \in[a, b]$ such that
$$
\left|g_{n_{k}}\left(x_{k}\right)\right|=g_{n_{k}}\left(x_{k}\right) \geq \varepsilon .
$$
[which is the contra-positive to that $\left(g_{n}\right)$ converges to 0 uniformly on $[a, b]$.]
We may choose $n_{k}$ so that $k \rightarrow n_{k}$ is increasing, and may assume that $x_{k} \rightarrow p$. [Otherwise we may argue with a convergent subsequence of $\left(x_{k}\right)$, according to Bolzano-Weierstrass' Theorem]. Then $p \in[a, b]$. Since $g_{n}(x)$ is decreasing in $n$ for every $x \in[a, b]$, thus for every $k$ fixed, for all $l>k$, we have
$$
\varepsilon \leq g_{n_{l}}\left(x_{l}\right) \leq g_{n_{k}}\left(x_{l}\right)
$$
Letting $l \rightarrow \infty$ in the above inequality, we obtain
$\varepsilon \leq \lim _{l \rightarrow \infty} g_{n_{k}}\left(x_{l}\right)=g_{n_{k}}(p) \quad$ [since $g_{n_{k}}$ is continuous at $p$ ]
which is a contradicts with the assumption that $\lim _{k \rightarrow \infty} g_{n_{k}}(p)=0$.
Corollary 1.4.16 Suppose the series of functions $\sum_{n=1}^{\infty} g_{n}(x)$ converges to its sum $S(x)$ for $x \in[a, b]$, suppose $g_{n}(x) \geq 0$ for every $n$ and every $x \in[a, b]$, and suppose all $g_{n}$ and its limit function $S$ are continuous on $[a, b]$, then $\sum_{n=1}^{\infty} g_{n}$ converges to $S$ uniformly on $[a, b]$.
Proof. Apply Dini's Theorem to $f_{n}=S-\sum_{k=1}^{n} g_{k}$ to conclude that $f_{n} \downarrow 0$ uniformly on $[a, b]$.
Example 1.4.17 Let $f_{n}(x)=\frac{1}{1+n x}$ for $x \in(0,1)$. Then $\lim _{n \rightarrow \infty} f_{n}(x)=0$ for every $x \in(0,1), f_{n}$ is decreasing in $n$, but $f_{n}$ does not converge uniformly. Dini's theorem does not apply for this case, since $(0,1)$ is not compact.
The proofs of the following two theorems related to the concept of uniform convergence will be given in the Trinity term.
Theorem 1.4.18 If $f_{n} \rightarrow f$ uniformly in $[a, b]$ and if every $f_{n}$ is continuous in $[a, b]$, then
$$
\int_{a}^{b} f=\int_{a}^{b} \lim _{n \rightarrow \infty} f_{n}=\lim _{n \rightarrow \infty} \int_{a}^{b} f_{n}
$$
Similarly, if the series $\sum_{n=1}^{\infty} f_{n}$ converges uniformly in $[a, b]$ and if all $f_{n}$ are continuous, then we may integrate the series term by term
$$
\int_{a}^{b} \sum_{n=1}^{\infty} f_{n}=\sum_{n=1}^{\infty} \int_{a}^{b} f_{n}
$$
Let us however immediately point out that the notion of uniform convergence is not the right condition for integrating a series term by term: we may exchange the order of integration $\int_{a}^{b}$ (which involves a limiting procedure) and $\lim _{n \rightarrow \infty}$ under much weaker conditions. The search for correct conditions for term-by-term integration led to the discovery of Lebesgue's integration [Second year A4 paper: Integration]. For details, see W. Rudin's Principles, Chapter 11 (page 300).
Theorem 1.4.19 Let $f_{n} \rightarrow f$ in $(a, b)$ (convergence point-wisely). Suppose $f_{n}^{\prime}$ exists and is continuous on $(a, b)$ for every $n$, and if $f_{n}^{\prime} \rightarrow g$ uniformly in $(a, b)$. Then $f^{\prime}$ exists and is continuous in $(a, b)$, and
$$
\frac{d}{d x} \lim _{n \rightarrow \infty} f_{n}(x)=\lim _{n \rightarrow \infty} \frac{d}{d x} f_{n}(x) .
$$
Similarly, if $\sum f_{n}$ converges in $(a, b)$, if every $f_{n}^{\prime}$ exists and is continuous in $(a, b)$, and if $\sum f_{n}^{\prime}$ converges uniformly in $(a, b)$, then
$$
\frac{d}{d x} \sum_{n=1}^{\infty} f_{n}=\sum_{n=1}^{\infty} f_{n}^{\prime}
$$ |
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