140 lines
2.8 KiB
Markdown
140 lines
2.8 KiB
Markdown
# Math4121 Lecture 12
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## Chapter 7: Uniform Convergence and Integrals
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Our goal is to solve problems like this:
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Let
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$$
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s_{n,m}=\frac{m}{m+n}
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$$
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The different order of computation gives different results:
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$$
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\lim_{n\to\infty}\lim_{m\to\infty}s_{n,m}=1
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$$
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$$
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\lim_{m\to\infty}\lim_{n\to\infty}s_{n,m}=0
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$$
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We cannot always switch the order of limits. We cannot also do this on derivatives.
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### Examples
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#### Example 7.4
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$$
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f_m(x)=\lim_{n\to\infty}cos(m!x\pi)^{2n}
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$$
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If $cos(m!x\pi)^{2n}=\pm 1$, then $f_m(x)=1$.
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If not, then $|cos(m!x\pi)^{2n}|<1$.
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$$
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f_m(x)=\begin{cases}
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1 & \text{if } m!x\text{ is an integer} \\
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0 & \text{if } \text{otherwise}
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\end{cases}
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$$
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This function "raise" the fractions with all denominators less than $m$.
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$$
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\lim_{m\to\infty}f_m(x)=
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\begin{cases}
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1 & \text{if } x\text{ is an rational number} \\
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0 & \text{if } \text{otherwise}
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\end{cases}
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$$
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So this function is not Riemann integrable. (show in homework)
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But
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$$
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g_{n,m}(x)=cos(m!x\pi)^{2n}
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$$
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is continuous, and
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$$
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\lim_{m\to\infty}\lim_{n\to\infty}g_{n,m}(x)=f(x)
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$$
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So the function is not Riemann integrable.
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#### Definition 7.7
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A sequence of functions $\{f_n\}$ **converges uniformly** to $f$ on set $E$ if
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$$
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\forall \epsilon>0, \exists N, \forall n\geq N, \forall x\in E, |f_n(x)-f(x)|<\epsilon
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$$
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If $E$ is just a point, then it is the common definition of convergence.
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_If you have uniform convergence, then you can switch the order of limits._
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### Uniform Convergence and Integrals
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#### Theorem 7.16
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Suppose $\{f_n\}\in\mathscr{R}(\alpha)$ on $[a,b]$ that converges uniformly to $f$ on $[a,b]$. Then $f\in\mathscr{R}(\alpha)$ on $[a,b]$ and
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$$
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\int_a^b f(x)d\alpha=\lim_{n\to\infty}\int_a^b f_n(x)d\alpha
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$$
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#### Proof
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Define $\epsilon_n=\sup_{x\in[a,b]}|f_n(x)-f(x)|$.
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By uniform convergence, $\epsilon_n\to 0$ as $n\to\infty$.
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$$
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f_n(x)-\epsilon_n\leq f(x)\leq f_n(x)+\epsilon_n
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$$
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$$
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\int_a^b f_n-\epsilon_nd\alpha\leq\overline{\int_a^b}fd\alpha\leq\int_a^bf_n+\epsilon_nd\alpha
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$$
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$$
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\int_a^b f_n-\epsilon_n d\alpha\leq\underline{\int_a^b}fd\alpha\leq\int_a^b f_n+\epsilon_n d\alpha
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$$
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So,
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$$
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0\leq\overline{\int_a^b}fd\alpha-\underline{\int_a^b}fd\alpha\leq\int_a^b \epsilon_n d\alpha+\int_a^b \epsilon_n d\alpha=2\epsilon_n[\alpha(b)-\alpha(a)]
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$$
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So $f\in\mathscr{R}(\alpha)$ and
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$$
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\int_a^b fd\alpha\leq \int_a^b f_n d\alpha+\int_a^b \epsilon_n d\alpha\leq \int_a^b fd\alpha+2\epsilon_n d\alpha
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$$
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So,
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$$
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\int_a^b fd\alpha-\int_a^b \epsilon_n d\alpha\leq \int_a^b f_n d\alpha\leq \int_a^b fd\alpha+\int_a^b \epsilon_n d\alpha
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$$
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Since $\int_a^b \epsilon_n d\alpha\to 0$ as $n\to\infty$, by the squeeze theorem, we have
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by the squeeze theorem, we have
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$$
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\lim_{n\to\infty}\int_a^b f_nd\alpha=\int_a^b fd\alpha
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$$
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_Key is that $\int_a^b (f-f_n)d\alpha\leq \sup_{x\in[a,b]}|f-f_n|(\alpha(b)-\alpha(a))$_
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