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# 数学代写|实分析代写Real Analysis代考|MATH350 Uniform Convergence and Continuity

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## 数学代写|实分析代写Real Analysis代考|Uniform Convergence and Continuity

In this section, we will prove that the limit of a uniformly convergent sequence of continuous functions is again continuous. Prior to proving this result, we first prove a stronger result that will have additional applications later.

THEOREM 8.3.1 Suppose $\left{f_n\right}$ is a sequence of real-valued functions that converges uniformly to a function $f$ on a subset $E$ of a metric space $(X, d)$. Let $p$ be a limit point of $E$, and suppose that for each $n \in \mathbb{N}$,
$$\lim {x \rightarrow p} f_n(x)=A_n .$$ Then the sequence $\left{A_n\right}$ converges and $$\lim {x \rightarrow p} f(x)=\lim {n \rightarrow \infty} A_n .$$ Remark. The last statement can be rewritten as $$\lim {x \rightarrow p}\left(\lim {n \rightarrow \infty} f_n(x)\right)=\lim {n \rightarrow \infty}\left(\lim _{x \rightarrow p} f_n(x)\right) .$$
It should be noted that $p$ is not required to be a point of $E$; only a limit point of $E$.

Proof. Let $\epsilon>0$ be given. Since the sequence $\left{f_n\right}$ converges uniformly to $f$ on $E$, there exists a positive integer $n_o$ such that
$$\left|f_n(x)-f_m(x)\right|<\epsilon$$

for all $n, m \geq n_o$ and all $x \in E$. Since (2) holds for all $x \in E$, letting $x \rightarrow p$ gives
$$\left|A_n-A_m\right| \leq \epsilon, \quad \text { for all } n, m \geq n_o$$

## 数学代写|实分析代写Real Analysis代考|Uniform Convergence and Integration

In Example 8.1.2(c) we provided an example of a sequence of Riemann integrable functions that converges pointwise, but for which the limit function is not Riemann integrable. Furthermore, in Example 8.1.2(d) we provided an example of a seuence of continuous function on $[0,1]$ for which $\lim {n \rightarrow \infty} f_n(x)=0$ for all $x \in[0,1]$ but for which $$\int_0^1 f_n(x) d x=\frac{1}{2} \frac{n}{n+1}$$ Thus $\lim {n \rightarrow \infty} \int_0^1 f_n(x) \neq \int_0^1 \lim _{n \rightarrow \infty} f_n(x) d x$. Hence, pointwise convergence, even if the limit function is Riemann integrable, is also not sufficient for the interchange of limits.

In this section, we will prove that uniform convergence of a sequence $\left{f_n\right}$ of Riemann integrable functions is again sufficient for the limit function $f$ to be Riemann integrable, and for convergence of the definite integrals of $f_n$ to the definite integral of $f$. The analogous result for the Riemann-Stieltjes integral is left to the exercises (Exercise 2).

THEOREM 8.4.1 Suppose $f_n \in \mathcal{R}[a, b]$ for all $n \in \mathbb{N}$, and suppose that the sequence $\left{f_n\right}$ converges uniformly to $f$ on $[a, b]$. Then $f \in \mathcal{R}[a, b]$ and
$$\int_a^b f(x) d x=\lim {n \rightarrow \infty} \int_a^b f_n(x) d x .$$ Proof. For each $n \in \mathbb{N}$, set $$\epsilon_n=\max {x \in[a, b]}\left|f_n(x)-f(x)\right| .$$
Since $f_n \rightarrow f$ uniformly on $[a, b]$, by Theorem 8.2.5, $\lim _{n \rightarrow \infty} \epsilon_n=0$. Also, for all $x \in[a, b]$
$$f_n(x)-\epsilon_n \leq f(x) \leq f_n(x)+\epsilon_n$$
Hence
$$\int_a^b\left(f_n-\epsilon_n\right) \leq \underline{\int_a^b f} \leq \overline{\int_a^b} f \leq \int_a^b\left(f_n+\epsilon_n\right)$$
Therefore
$$0 \leq \overline{\int_a^b} f-\underline{\int_a^b} f \leq 2 \epsilon_n[b-a]$$

## 数学代写|实分析代写Real Analysis代考|Uniform Convergence and Continuity

$$\lim x \rightarrow p f_n(x)=A_n .$$

$$\lim x \rightarrow p f(x)=\lim n \rightarrow \infty A_n .$$

$$\lim {x \rightarrow p}\left(\lim n \rightarrow \infty f_n(x)\right)=\lim n \rightarrow \infty\left(\lim {x \rightarrow p} f_n(x)\right)$$

$$\left|f_n(x)-f_m(x)\right|<\epsilon$$

$$\left|A_n-A_m\right| \leq \epsilon, \quad \text { for all } n, m \geq n_o$$

## 数学代写|实分析代写Real Analysis代考|Uniform Convergence and Integration

$$\int_0^1 f_n(x) d x=\frac{1}{2} \frac{n}{n+1}$$

$$f_n(x)-\epsilon_n \leq f(x) \leq f_n(x)+\epsilon_n$$

$$\int_a^b\left(f_n-\epsilon_n\right) \leq \underline{\int_a^b f} \leq \overline{\int_a^b} f \leq \int_a^b\left(f_n+\epsilon_n\right)$$

$$0 \leq \overline{\int_a^b} f-\underline{\int_a^b} f \leq 2 \epsilon_n[b-a]$$

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