2.8 KiB
Math4121 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.
f_n(x)-\epsilon_n\leq f(x)\leq f_n(x)+\epsilon_n
\int_a^b f_n-\epsilon_nd\alpha\leq\overline{\int_a^b}fd\alpha\leq\int_a^bf_n+\epsilon_nd\alpha
\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
So,
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)]
So f\in\mathscr{R}(\alpha) and
\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
So,
\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
Since \int_a^b \epsilon_n d\alpha\to 0 as n\to\infty, by the squeeze theorem, we have
by the squeeze theorem, we have
\lim_{n\to\infty}\int_a^b f_nd\alpha=\int_a^b fd\alpha
Key is that $\int_a^b (f-f_n)d\alpha\leq \sup{x\in[a,b]}|f-f_n|(\alpha(b)-\alpha(a))$_