Update Math4121_L12.md

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