### 数学代写|实分析作业代写Real analysis代考|MAST20026

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|实分析作业代写Real analysis代考|Continuous Mappings

Here we consider mappings of metric spaces.
1.3.1. Definition. A mapping $f$ from a metric space $\left(X, d_X\right)$ to a metric space $\left(Y, d_Y\right)$ is called continuous at a point $x \in X$ if, for every sequence $\left{x_n\right}$ converging to $x$, the sequence $\left{f\left(x_n\right)\right}$ converges to $f(x)$.
The mapping $f$ is called continuous if it is continuous at every point.
It is clear that the continuity at a point $x$ can be formulated in $(\varepsilon, \delta)$-terms: for every $\varepsilon>0$, there exists $\delta>0$ such that $d_Y(f(z), f(x))<\varepsilon$ for all points $z \in X$ such that $d_X(z, x)<\delta$.

In Exercise 1.9.37 it is suggested to verify that the continuity of the mapping $f$ is equivalent to the property that for every open set $V \subset Y$ the set $f^{-1}(V)$ is open in $X$ (this is also equivalent to the property that for every closed set $Z \subset Y$ the set $f^{-1}(Z)$ is closed in $X$ ).

As in the case of the real line, a stronger mode of continuity can be introduced: the uniform continuity.
1.3.2. Definition. A mapping $f$ from a metric space $\left(X, d_X\right)$ to a metric space $\left(Y, d_Y\right)$ is called uniformly continuous if, for every $\varepsilon>0$, there exists $\delta>0$ such that $d_Y(f(x), f(y)) \leqslant \varepsilon$ whenever $d_X(x, y) \leqslant \delta$.

It is clear that uniformly continuous mappings are continuous. Let $\left(X, d_X\right)$ and $\left(Y, d_Y\right)$ be metric spaces.
1.3.3. Proposition. Let $f_n:\left(X, d_X\right) \rightarrow\left(Y, d_Y\right)$ be continuous mappings uniformly converging to a mapping $f:\left(X, d_X\right) \rightarrow\left(Y, d_Y\right)$ in the following sense: for every $\varepsilon>0$, there exists a number $n_{\varepsilon}$ such that $d_Y\left(f_n(x), f(x)\right) \leqslant \varepsilon$ for all $n \geqslant n_{\varepsilon}$ and $x \in X$. Then the mapping $f$ is continuous.

Proof. Let $x_0 \in X$ and $\varepsilon>0$. Let us take numbers $n_{\varepsilon}$ and $\delta>0$ such that $d_Y\left(f_{n_{\varepsilon}}(x), f_{n_{\varepsilon}}\left(x_0\right)\right)<\varepsilon$ whenever $d_X\left(x, x_0\right)<\delta$. Then for such $x$ we obtain
\begin{aligned} d_Y\left(f(x), f\left(x_0\right)\right) \leqslant d_Y\left(f(x), f_{n_{\varepsilon}}(x)\right) & +d_Y\left(f_{n_{\varepsilon}}(x), f_{n_{\varepsilon}}\left(x_0\right)\right) \ & +d_Y\left(f_{n_{\varepsilon}}\left(x_0\right), f\left(x_0\right)\right) \leqslant 3 \varepsilon, \end{aligned}
which shows the continuity of $f$ at the point $x_0$.
If uniformly convergent mappings $f_n$ are uniformly continuous, then their limit is also uniformly continuous. This is clear from the proof.

## 数学代写|实分析作业代写Real analysis代考|The Contracting Mapping Principle

Lipschitz mappings with constant $L<1$ are called contracting mappings or contractions. The next result is frequently used in applications.
1.4.1. Theorem. (ThE CONTRACting MAPPING PRINCIPLE) Every contraction $f$ of a nonempty complete metric space $X$ has a unique fixed point $\widehat{x}$, i.e., $f(\widehat{x})=\widehat{x}$. In addition, $d\left(\widehat{x}, x_n\right) \leqslant L^n(1-L)^{-1} d\left(x_1, x_0\right)$ for every $x_0 \in X$, where $x_{n+1}:=f\left(x_n\right)$.

Proof. Let $x_0 \in X$. Set $x_n=f\left(x_{n-1}\right), n \in \mathbb{N}$. We show that the sequence $\left{f\left(x_n\right)\right}$ is Cauchy. To this end, we observe that
$$d\left(x_{k+1}, x_k\right) \leqslant \operatorname{Ld}\left(x_k, x_{k-1}\right) \leqslant \cdots \leqslant L^k d\left(x_1, x_0\right) .$$
Hence $d\left(x_{n+m}, x_n\right)$ is estimated by
\begin{aligned} & d\left(x_{n+m}, x_{n+m-1}\right)+d\left(x_{n+m-1}, x_{n+m-2}\right)+\cdots+d\left(x_{n+1}, x_n\right) \leqslant \ & \leqslant L^{n+m-1} d\left(x_1, x_0\right)+L^{n+m-2} d\left(x_1, x_0\right)+\cdots+L^n d\left(x_1, x_0\right), \end{aligned}
which yields $d\left(x_{n+m}, x_n\right) \leqslant L^n(1-L)^{-1} d\left(x_1, x_0\right)$. This estimate and the condition $L<1$ imply that $\left{x_n\right}$ is a Cauchy sequence and that there exists a limit $\widehat{x}=\lim {n \rightarrow \infty} x_n$. Clearly $$f(\widehat{x})=\lim {n \rightarrow \infty} f\left(x_n\right)=\lim {n \rightarrow \infty} x{n+1}=\widehat{x}$$
by the continuity of $f$. The uniqueness of a fixed point is seen from the fact that $d(\widehat{x}, y)=d(f(\widehat{x}), f(y)) \leqslant L d(\widehat{x}, y)$ for any other fixed point $y$. The estimate for the rate of convergence has been also obtained.

# 实分析代写

## 数学代写|实分析作业代写Real analysis代考|Continuous Mappings

1.3.1. 定义。映射 $f$ 从度量空间 $\left(X, d_X\right)$ 到度量空间 $\left(Y, d_Y\right)$ 在一点上称为连续的 $x \in X$ 如 映射 $f$ 如果它在每一点都是连续的，则称为连续的。

1.3.3. 主张。让 $f_n:\left(X, d_X\right) \rightarrow\left(Y, d_Y\right)$ 是一致收敛到一个映射的连续映射 $f:\left(X, d_X\right) \rightarrow\left(Y, d_Y\right)$ 在以下意义上: 对于每个 $\varepsilon>0$, 存在一个数 $n_{\varepsilon}$ 这样
$d_Y\left(f_n(x), f(x)\right) \leqslant \varepsilon$ 对全部 $n \geqslant n_{\varepsilon}$ 和 $x \in X$. 然后映射 $f$ 是连续的。

$$d_Y\left(f(x), f\left(x_0\right)\right) \leqslant d_Y\left(f(x), f_{n_{\varepsilon}}(x)\right)+d_Y\left(f_{n_{\varepsilon}}(x), f_{n_{\varepsilon}}\left(x_0\right)\right) \quad+d_Y\left(f_{n_{\varepsilon}}\left(x_0\right), f\right.$$

## 数学代写|实分析作业代写Real analysis代考|The Contracting Mapping Principle

1.4.1. 定理。（收缩映射原理) 每次收缩 $f$ 非空完备度量空间的 $X$ 有唯一不动点 $\widehat{x}$ ，那是， $f(\widehat{x})=\widehat{x}$. 此外， $d\left(\widehat{x}, x_n\right) \leqslant L^n(1-L)^{-1} d\left(x_1, x_0\right)$ 每一个 $x_0 \in X$ ，在哪里 $x_{n+1}:=f\left(x_n\right)$. 西。为此，我们观察到
$$d\left(x_{k+1}, x_k\right) \leqslant \operatorname{Ld}\left(x_k, x_{k-1}\right) \leqslant \cdots \leqslant L^k d\left(x_1, x_0\right) .$$

$$d\left(x_{n+m}, x_{n+m-1}\right)+d\left(x_{n+m-1}, x_{n+m-2}\right)+\cdots+d\left(x_{n+1}, x_n\right) \leqslant \leqslant L^{n+m-1} d\left(x_1\right.$$

$$f(\widehat{x})=\lim n \rightarrow \infty f\left(x_n\right)=\lim n \rightarrow \infty x n+1=\widehat{x}$$

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## MATLAB代写

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