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# 实分析代写| 连续函数代写|continue function代写|Math115A代写

##### 函数在一点连续的精确定义

1. 大家更加适应和习惯自然语言, 而第一次认真的接触逻辑语言是会不适应的, 就像你第一次学匀 编程一样, 肯定既兴奋又不太适应;
2 . 练习或是接触的频率不够, 接触多了你就会觉得相对直观和舒服多了。其实我们学习中文或是英文这种自然语言也是如此, 母语从小到大一直学习和使用你就觉得亲切, 而第二第三语言要是接触少一些你 自然就会有不少心理障碍, 逻辑语言就更是了。

$$\left|\sum_{n=1}^{\infty} x_{n}\right| \leq \sum_{n=1}^{\infty}\left|x_{n}\right|$$

$$\left|\sum_{n=1}^{N} x_{n}\right| \leq \sum_{n=1}^{N}\left|x_{n}\right|$$
Taking $N \rightarrow \infty$ and noticing that
$$\lim _{N \rightarrow \infty}\left|\sum_{n=1}^{N} x_{n}\right|=\lim _{N \rightarrow \infty} \sum_{n=1}^{N} x_{n} \mid$$
(since the limit of the right hand side exists) gives the inequality.

Cesàro summable if $C=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} s_{k}$ exists. The value $C$ is called the Cesàro sum of the sequence.
a) Show that if $\left\{a_{n}\right\}_{n=1}^{\infty}$ is a convergent sequence with $\lim _{n \rightarrow \infty} a_{n}=A,$ then the sequence $\left\{b_{n}\right\}_{n=1}^{\infty}$ defined by $b_{n}=\frac{1}{n} \sum_{k=1}^{n} a_{k}$ is also convergent and $\lim _{n \rightarrow \infty} b_{n}=A$.
b) Show that if $S=\sum_{n=1}^{\infty} x_{n}$ converges, then the sequence is Cesàro summable and $S=C .$ Hint: use a).
c) Show that $x_{n}=(-1)^{n}$ is Cesàro summable, even though $\sum_{n=1}^{\infty}(-1)^{n}$ is not convergent.

$$\left|a_{n}-A\right|<\varepsilon / 2$$
Since $a_{n}$ is convergent there is $M>0$ such that $\left|a_{n}\right| \leq M$ for all $n .$ Now given $n>N$ we have by triangle inequality
$$\left|b_{n}-A\right| \leq \sum_{k=1}^{n} \frac{1}{n}\left|a_{n}-A\right| \leq \sum_{k=1}^{N} \frac{1}{n}\left|a_{n}-A\right|+\sum_{k=N}^{n} \frac{1}{n}\left|a_{n}-A\right| \leq \sum_{k=1}^{N} \frac{2 M}{n}+\sum_{k=N}^{n} \frac{\varepsilon}{2 n}<\frac{2 M N}{n}+\frac{\varepsilon}{2}$$
Now let $N_{1}$ be such that
$$\frac{2 M N}{N_{1}}<\varepsilon / 2$$
It follows that for $n>\max \left\{N, N_{1}\right\}$ we have
$$\left|b_{n}-A\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$
which means $b_{n} \rightarrow A$ as well.
(b) This follows immediately by taking $a_{n}=s_{n}$ in part (a).
(c) The partial sums are $s_{k}=0$ if $k$ is even and -1 if $k$ is odd. So we have
$$\sum_{k=1}^{n} s_{k}=\left\{\begin{array}{ll} -\frac{n}{2} & n \text { even } \\ -\frac{n+1}{2} & n \text { odd } \end{array}\right.$$
and it follows that
$$\frac{1}{n} \sum_{k=1}^{n} s_{k}=\left\{\begin{array}{ll} -\frac{1}{2} & n \text { even } \\ -\frac{1}{2}-\frac{1}{2 n} & n \text { odd } \end{array}\right.$$
In either case the limit is $-\frac{1}{2}$ as $n \rightarrow \infty,$ so the series is Cesàro summable with $S=-\frac{1}{2}$

$(c-\varepsilon, c+\varepsilon) \backslash\{c\} \cap S,$ so the latter set is also not empty for any $\varepsilon>0$

Problem #4. Let $c_{1}$ be a cluster point of $A$ and $c_{2} \in B$ be a cluster point of $B$. Suppose $f: A \rightarrow \mathbb{R}$ and $g: B \rightarrow \mathbb{R}$ are functions with $\lim _{x \rightarrow c_{1}} f(x)=c_{2}$ and $\lim _{x \rightarrow c_{2}} g(x)=L=g\left(c_{2}\right) .$ Show $\lim _{x \rightarrow c_{1}} g(f(x))=L$

have
$$\begin{array}{l} \qquad|g(f(c))-L|=\left|g(f(c))-g\left(c_{2}\right)\right|<\varepsilon \\ \text { since }\left|f(c)-c_{2}\right|<\delta_{1} \end{array}$$

Problem $\# 5 .$ Suppose $S \subset \mathbb{R}$ and $c$ is a cluster point of $S .$ Suppose $f: S \rightarrow \mathbb{R}$ is bounded. Show that there exists a sequence $\left\{x_{n}\right\}_{n=1}^{\infty}$ with $x_{n} \in S \backslash\{c\}$ and $\lim _{n \rightarrow \infty} x_{n}=c$ such that $\left\{f\left(x_{n}\right)\right\}_{n=1}^{\infty}$ converges.

Since $c$ is a cluster point of $S,$ there is $x_{n} \in(c-1 / n, c+1 / n) \backslash\{c\}$ for every $n \in \mathbb{N}$. Clearly $x_{n} \rightarrow c$. Since $f$ is bounded, by Bolzano-Weierstrass we may extract a subsequence $\left\{x_{n_{i}}\right\} \subset\left\{x_{n}\right\}$ such that $\left\{f\left(x_{n_{i}}\right)\right\}$ converges. Since $x_{n} \rightarrow c$ it follows $x_{n_{i}} \rightarrow c$ as well, so the sequnce $\left\{x_{n_{j}}\right\}$ is what we wanted.

Using the $\epsilon-\delta$ -definition of continuity directly prove that $f(x)=\frac{1}{x}$ is continuous on $(0, \infty)$.

$$\left|\frac{1}{x}-\frac{1}{x_{0}}\right|=\frac{\left|x-x_{0}\right|}{x x_{0}}$$
Let us first agree that $\left|x-x_{0}\right|<\frac{1}{2}\left|x_{0}\right|,$ so that $x>\frac{x_{0}}{2}$ and
$$\frac{\left|x-x_{0}\right|}{x x_{0}}<\frac{2\left|x-x_{0}\right|}{x_{0}^{2}}$$
If further $\left|x-x_{0}\right|<\frac{\varepsilon x_{0}^{2}}{2}$ we will have
$$\frac{2\left|x-x_{0}\right|}{x_{0}^{2}}<\varepsilon$$
Thus it suffices to pick $\delta=\min \left\{\frac{1}{2}\left|x_{0}\right|, \frac{\varepsilon\left|x_{0}\right|^{2}}{2}\right\}$ (so that both of the above inequalities hold).

Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous and $f(c)>0$, then there is a $\delta>0$ so that $f(x)>0$ for all $x \in(c-\delta, c+\delta)$.

$$|f(x)-f(c)|<\frac{|f(c)|}{2}$$
which in turn (by triangle inequality) implies $f(x)>f(c)-\frac{f(c)}{2}=\frac{f(c)}{2}>0$ for $|x-c|<\delta$.

A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is said to be Lipschitz if there is a constant $L>0$ so $|f(x)-f(y)| \leq$ $L|x-y|$ for all $x, y \in \mathbb{R} .$ Show that $f$ is continuous.

$$|f(x)-f(y)| \leq L|x-y|<L \frac{\varepsilon}{L}=\varepsilon$$
which shows the continuity at $x$. Since $x$ is arbitrary, $f$ is continuous.

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