76 lines
2.5 KiB
Markdown
76 lines
2.5 KiB
Markdown
# Math4121 Lecture 21
|
|
|
|
## Rolling from last lecture
|
|
|
|
### Convergence of integrals
|
|
|
|
#### Arzela-Osgood Theorem
|
|
|
|
Let $\{f_n\}$ be a sequence of function, $f(x)=\lim_{n\to\infty}f_n(x)$ for every $x\in [0,1]$, if $f\in \mathscr{R}[0,1]$, and $\exists B>0$ such that $|f_n(x)|\leq B \forall x\in [0,1]$. (uniformly bounded and integrable)
|
|
|
|
$$
|
|
\lim_{n\to\infty}\int_0^1 f_n(x) dx = \int_0^1 f(x) dx
|
|
$$
|
|
|
|
If we let $\Gamma_{\alpha}$ be the set of intervals where $f_n$ is not continuous,
|
|
|
|
$$
|
|
\Gamma_{\alpha} = \{x\in [0,1] : \textup{ for any }m\in \mathbb{N}, \delta > 0, \exists n\geq m, y\in (x-\delta, x+\delta) \text{ s.t. } |f_n(y)-f(y)|>\alpha\}
|
|
$$
|
|
|
|
Fact: $\Gamma_{\alpha}$ is closed and nowhere dense.
|
|
|
|
<details>
|
|
<summary>Proof</summary>
|
|
|
|
Without loss of generality, we can assume $f=0$. Given any $\alpha > 0$, $\exists N$ such that
|
|
|
|
$$
|
|
\left|\int_0^1 f_n(x) dx \right| < \alpha
|
|
$$
|
|
|
|
for all $n\geq N$.
|
|
|
|
Consider the set $\Gamma_{\alpha/2} = \bigcup_{n=1}^{\infty} E_n$, for each $g\in \Gamma_{\alpha/2}$, we still have $\lim_{n\to\infty}f_n(g) = 0$.
|
|
|
|
So we define
|
|
|
|
$$
|
|
G_i=\{g\in \Gamma_{\alpha/2} :|f_n(g)|<\frac{\alpha}{2} \text{ for all }n\geq i\}
|
|
$$
|
|
|
|
So $G_1\subset G_2\subset \cdots$ and $\Gamma_{\alpha/2} = \bigcup_{i=1}^{\infty} G_i$.
|
|
|
|
By Osgood Lemma, since $\Gamma_{\alpha/2}$ is closed, $\exists K$ such that $c_e(G_K)>c_e(\Gamma_{\alpha/2})-\frac{\alpha}{4B}$.
|
|
|
|
By definition of $c_e$, we cna find open $I_1,\ldots,I_N$ which cover $\Gamma_{\alpha/2}$ and
|
|
|
|
$$
|
|
\sum_{i=1}^N \ell(I_i) < c_e(\Gamma_{\alpha/2})+\frac{\alpha}{4B}
|
|
$$
|
|
|
|
Let $\mathcal{U}=\bigcup_{i=1}^N I_i$, and $\mathcal{C}=[0,1]\setminus \mathcal{U}$.
|
|
|
|
Part 1: Control the integral on $\mathcal{C}$
|
|
|
|
for each $x\in \mathcal{C}$, $x\notin \Gamma_{\alpha/2}$, so $\exists$ and open interval $I(x)$ and an integer $m(x)$ such that $|f_{m(x)}(x)|<\frac{\alpha}{2}$ and $\forall n\geq m(x), y\in I(x)$
|
|
|
|
So $\mathcal{C}\subset \bigcup_{x\in \mathcal{C}} I(x)$, and $\mathcal{C}$ is closed and bounded, $\exists x_1,\ldots,x_J$ such that $\mathcal{C}\subset \bigcup_{j=1}^J I(x_j)$. So if $n\geq \max_{j=1,\ldots,J} m(x_j)$, and $x\in \mathcal{C}$, then $|f_n(x)|<\frac{\alpha}{2}$.
|
|
|
|
So $\int_\mathcal{C} |f_n(x)| dx < \frac{\alpha}{2} c_e(\mathcal{C})$.
|
|
|
|
Part 2: Control the integral on $\mathcal{U}$
|
|
|
|
If $[x_i,x_{i+1}]\cap G_k\neq \emptyset$, then $\inf_{x\in [x_i,x_{i+1}]} |f_n(x)| < \frac{\alpha}{2}$ for all $n\geq K$. Denote such set as $P_1$.
|
|
|
|
Otherwise, we denote such set as $P_2$.
|
|
|
|
So $\ell(\mathcal{U})=\ell(P_1)+\ell(P_2)\geq c_e(G_K)+\ell(P_2)$.
|
|
|
|
This implies $\ell(P_2)\leq \frac{\alpha}{4B}$.
|
|
|
|
Continue on Friday.
|
|
|
|
</details>
|
|
|