# 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 $\int_a^b (f-f_n)d\alpha\leq \sup_{x\in[a,b]}|f-f_n|(\alpha(b)-\alpha(a))$_