Math4121 Lecture 12

Chapter 7: Uniform Convergence and Integrals

Our goal is to solve problems like this:

Let

The different order of computation gives different results:

We cannot always switch the order of limits. We cannot also do this on derivatives.

Examples

Example 7.4

If $cos(m!x\pi)^{2n}=\pm 1$, then $f_m(x)=1$.

If not, then $|cos(m!x\pi)^{2n}|<1$.

This function “raise” the fractions with all denominators less than $m$.

So this function is not Riemann integrable. (show in homework)

But

is continuous, and

So the function is not Riemann integrable.

Definition 7.7

A sequence of functions $\{f_n\}$ converges uniformly to $f$ on set $E$ if

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

Proof

Define $\epsilon_n=\sup_{x\in[a,b]}|f_n(x)-f(x)|$.

By uniform convergence, $\epsilon_n\to 0$ as $n\to\infty$.

So,

So $f\in\mathscr{R}(\alpha)$ and

So,

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

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