### 数学代写|数学分析代写Mathematical Analysis代考|MATH2060

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
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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|数学分析代写Mathematical Analysis代考|Uniform Convergence and Monotonicity

In this section we shall discuss two classical results regarding uniform convergence under a monotonicity hypothesis. The first theorem (by Dini) assumes monotonicity in the parameter $k$, the second one supposes monotonicity in the variable $x$.

Theorem 1 (Dini) Let $I=[a, b]$ be a closed and bounded interval and consider a sequence $f_k: I \rightarrow \mathbb{R}$ of continuous functions, monotone in $k$ (for instance, increasing: $f_k(x) \leq f_{k+1}(x)$ for any $\left.k \in \mathbb{N}, x \in I\right)$, and pointwise convergent on $[a, b]$ to some continuous function $f$. Then $f_k$ converges uniformly to $f$ on $[a, b]$.
Proof Consider, for example, an increasing sequence $f_k$, that is to say $f_k(x) \leq$ $f_{k+1}(x) \leq f(x)$ for any $k \in \mathbb{N}$ and any $x \in I=[a, b]$.

Suppose, by contradiction, that $f_k$ does not converge uniformly to $f$ on $[a, b]$. This means there exists $\varepsilon_0>0$ such that for any $v \in \mathbb{N}$ we can find $k>v$ and $x \in[a, b]$ for which
$$\left|f_k(x)-f(x)\right|=f(x)-f_k(x) \geq \varepsilon_0 .$$
Hence for any $v=h \in \mathbb{N}$, there exist $k_h \rightarrow+\infty$ and $x_h \in[a, b]$ such that
$$f\left(x_h\right)-f_{k_h}\left(x_h\right) \geq \varepsilon_0 .$$
But the monotonicity of $f_k$ in $k$ forces $f_{k_h} \geq f_i$ when $k_h \geq i$. So we obtain
$$f\left(x_h\right)-f_i\left(x_h\right) \geq \varepsilon_0, \quad \forall h \in \mathbb{N}, \quad \forall i \leq k_h .$$
The sequence $x_h$, being bounded, admits a subsequence $x_{h_j}$ converging to a point $x_0$ of the interval $[a, b]$. Taking the limit as $j \rightarrow+\infty$ in
$$f\left(x_{h_j}\right)-f_i\left(x_{h_j}\right) \geq \varepsilon_0, \quad \forall h_j \in \mathbb{N}, \quad \forall i \leq k_{h_j},$$
due to the continuity of $f$ and $f_i$ we have
$$f\left(x_0\right)-f_i\left(x_0\right) \geq \varepsilon_0 \quad \forall i \in \mathbb{N} .$$
Taking the limit when $i \rightarrow+\infty$ we reach the contradiction $0 \geq \varepsilon_0$.

## 数学代写|数学分析代写Mathematical Analysis代考|Series of Functions

If $f_k$ is a sequence of real functions defined on the subset $I$ of $\mathbb{R}$, we indicate by $s_k$ the sequence of partial sums
\begin{aligned} & s_1=f_1 \ & s_2=f_1+f_2 \ & \ldots \ldots \ldots \ & s_k=f_1+f_2+\ldots+f_k \ & \ldots \ldots \ldots \ldots \end{aligned}
The sequence of functions $s_k$ is called series (of functions) with general term $f_k$, and we shall also use for it the expression
$$f_1+f_2+\ldots+f_k+\ldots$$
If, for any $x \in I$, the numerical series with general term $f_k(x)$
$$f_1(x)+f_2(x)+\ldots+f_k(x)+\ldots$$
is convergent, i.e. if the sequence $s_k(x)$ converges (it has finite limit) for every $x \in I$, one says that the series of functions (1.18) converges pointwise on $I$.

When the sequence of functions $s_k$ converges uniformly on $I$, we say the series of functions (1.18) converges uniformly on $I$. In either case, the limit of $s_k$ as $k \rightarrow+\infty$ is called sum of the series of general term $f_k$, and we denote it by
$$\sum_{k=1}^{\infty} f_k$$
At times, (1.19) also indicates the series of general term $f_k$, apart from its sum. Often, as in the case of numerical series, one uses distinct summation indices for a series’ general term and (for example) the sequence of partial sums of a convergent series:
\begin{aligned} s_k & =\sum_{i=1}^k f_i, \quad \forall k \in \mathbb{N} \ f & =\lim {k \rightarrow+\infty} s_k=\lim {k \rightarrow+\infty} \sum_{i=1}^k f_i=\sum_{i=1}^{\infty} f_i \end{aligned}

# 数学分析代考

## 数学代写|数学分析代写Mathematical Analysis代考|Uniform Convergence and Monotonicity

$$\left|f_k(x)-f(x)\right|=f(x)-f_k(x) \geq \varepsilon_0 .$$

$$f\left(x_h\right)-f_{k_h}\left(x_h\right) \geq \varepsilon_0 .$$

$$f\left(x_h\right)-f_i\left(x_h\right) \geq \varepsilon_0, \quad \forall h \in \mathbb{N}, \quad \forall i \leq k_h .$$

$$f\left(x_{h_j}\right)-f_i\left(x_{h_j}\right) \geq \varepsilon_0, \quad \forall h_j \in \mathbb{N}, \quad \forall i \leq k_{h_j},$$

$$f\left(x_0\right)-f_i\left(x_0\right) \geq \varepsilon_0 \quad \forall i \in \mathbb{N} .$$

## 数学代写|数学分析代写Mathematical Analysis代考|Series of Functions

$$s_1=f_1 \quad s_2=f_1+f_2 \ldots \ldots . \quad s_k=f_1+f_2+\ldots+f_k \ldots \ldots \ldots \ldots$$

$$f_1+f_2+\ldots+f_k+\ldots$$

$$f_1(x)+f_2(x)+\ldots+f_k(x)+\ldots$$

$$\sum_{k=1}^{\infty} f_k$$

$$s_k=\sum_{i=1}^k f_i, \quad \forall k \in \mathbb{N} f \quad=\lim k \rightarrow+\infty s_k=\lim k \rightarrow+\infty \sum_{i=1}^k f_i=\sum_{i=1}^{\infty} f_i$$

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## MATLAB代写

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