Math 331, Nov. 11

Due Monday:

Reading assignment: Section 4.6, Sequences of Functions

For next week: Section 4.7, Series of Functions

Last time

Pointwise convergence of $\{f_n\}_{n=1}^\infty$ on $I$: $$\forall x\in I, \forall \varepsilon>0, \exists N\in\N, \forall n\in\N\ \ (n\ge N \Rightarrow |f_n(x)-f(x)|<\varepsilon)$$

Uniform convergence of $\{f_n\}_{n=1}^\infty$ on $I$: $$\forall \varepsilon>0, \exists N\in\N, \forall x\in i, \forall n\in\N\ \ (n\ge N \Rightarrow |f_n(x)-f(x)|<\varepsilon)$$

Example: Find the pointwise limit of $\{\root n \of x\}_{n=1}^\infty$ on [0,2]. (Is the convergence uniform?)

Do we have any examples of uniform convergence? How about $f_n(x) = \frac{1}{n+x^2}$ on $\R$?

(Look at examples on Prof. Mitchell's web page.)

Integrability

Theorem: Suppose that $\{f_n\}_{n=1}^\infty$ is a sequence of integrable functions on an interval $[a,b]$, and that the sequence converges uniformly to an integrable function $f$ on $[a,b]$. Then $$\int_a^b f = \lim_{n\to\infty}\int_a^b f_n$$

Proof: (Note: The proof is pretty easy, assuming the integrability of $f$. However, as the book notes, that hypothesis is not actually necessary. See Problem 4.5.9.)

Note that we are trying to prove that the sequence of numbers $\{\int_a^b f_n\}_{n=1}^\infty$ convereges to the number $\int_a^b f$. We apply the definition of a sequence of numbers. Let $\eps>0$. We want to find $N\in\N$ such that for all $n\ge N$, $\big|\int_a^b f_n - \int_a^b\big|<\eps.$

By uniform convergence of the sequence of functions, the is an $N\in\N$ such that for all $n\ge N$ and all $x\in[a,b]$, $$|f_n(x)-f(x)|<\frac{\eps}{b-a}$$ Using this same $N$, $$\begin{align*} \left|\int_a^b f_n - \int_a^b f\;\right| &= \left|\int_a^b \big(f_n - f\big)\right|\\ &\le \int_a^b \big|f_n-f|\\ &\le \int_a^b \frac{\eps}{b-a}\\ &=\frac{\eps}{b-a}\cdot(b-a)\\ &=\eps \end{align*}$$ which finishes the proof.

Why care?

Uniform convergence will be important for series of functions. As an example, consider the series $$\sum_{n=0}^\infty\frac{x^n}{n!}$$ which we know converges to $e^x$ for every $x\in\R$. Let $s_n(x)$ be the $n$-th partial sum, $$s_n(x) = \sum_{k=0}^n \frac{x^k}{k!}$$

Why is the convergence uniform? $s_n(x)$ is the n-th Taylor polynomial for $f(x)=e^x$ at 0. So at least for $x>0$, we can consider the Legrange form of Taylor's remainder for $s_n(x)$, which is, for some $t\le x,$ $$\frac{f^{(n+1)}(t)}{(n+1)!}x^{n+1} = \frac{e^tx^{n+1}}{(n+1)!}\le \frac{e^xx^{n+1}}{(n+1)!}$$ As $n\to\infty$, the remainder goes to 0. [$e_x$ is constant as $n\to\infty$, and as soon as $n+1>x$, each term is multiplied by a fraction that becomes smaller as $n$ increases.]

This means that we don't only have convergence, we have uniform convergence on the interval $[0,x]$, because $\frac{e^xx^{n+1}}{(n+1)!}$ is a uniform limit for the remainder term on this interval. So, we should find that $$\lim_{n\to\infty}\int_0^x s_n = \int_0^x f = \int_0^x e^t\,dt = e^x-e^0=e^x-1$$ Let's try it: $$\begin{align*} \lim_{n\to\infty}\left(\int_0^x s_n\right) &= \lim_{n\to\infty}\left(\int_0^x 1 + t + t^2/2 + t^3/3! + t^n/n!\,dt\right)\\ &= \lim_{n\to\infty}\left(t + (1/2)t^2 + (1/3)t^3/2 + (1/4)t^4/3! + (1/(n+1))t^{n+1}/n!\right)\\ &= \lim_{n\to\infty}\left.\left(t + t^2/2 + t^3/3! + t^4/4! + t^{n+1}/(n+1)!\right)\right|_0^x\\ &= \lim_{n\to\infty}\left(x + x^2/2 + x^3/3! + x^4/4! + x^{n+1}/(n+1)!\right)\\ &= \sum_{k=1}^\infty \frac{x^k}{k!}\\ &= \left(\sum_{k=0}^\infty \frac{x^k}{k!}\right) - 1\\ &= e^x-1 \end{align*}$$

If only we could show uniform convergence of series of functions in general!!

Differentiability

Theorem: Suppose that $\{f_n\}_{n=1}^\infty$ is a sequence of differentiable functions on an interval $[a,b],$ and that the sequence converges pointwise to a function $f$ on $[a,b]$. Suppose also that $f_n'$ is continuous on $[a,b]$ and that the sequence $\{f_n'\}_{n=1}^\infty$ converges uniformly on $[a,b]$. Then $f$ is differentiable on $[a,b]$, and the limit of the sequence $\{f_n'\}_{n=1}^\infty$ is $f'$. That is, for all $x\in[a,b]$, and $$\lim_{n\to\infty} f_n'(x)=f'(x)$$

Proof: Let $g(x) = \lim_{n\to\infty}f_n'x)$. We want to show $f'(x)=g(x)$. Now, $f_n'$ is continuous by assumption, and therefor integrable. Since $g$ is the uniform limit of continuous functions, it is continuous and hence integrable. We then have for any $x\in[a,b]$, $$\begin{align*} \int_a^x f(t)\,dt &= \lim_{n\to\infty}\int_a^x f_n'(t)\,dt\text{, by the preceding Theorem}\\ &= \lim_{n\to\infty}\big(f_n(x)-f_n(a)\big)\text{, by the first FToC}\\ &= f(x)-f(a)\text{, by the convergence of $f_n$ to $f$} \end{align*}$$ Solving for $f(x)$, $$f(x)=f(a)+\int_a^x g(t)\,dt$$ Then, by the second FToC, $f$ is differentiable and $f'(x)=g(x)$.

Problem 4.5.13

Let $\displaystyle f_n(x)=\begin{cases} {\textstyle\frac{1}{n}}&\text{ if }0\le x\le n\\ 0&\text{ if }x>n. \end{cases}$

(a) Show that $f_n(x)$ converges uniformly to $f(x)=0$ on $[0,\infty)$.

(b) Show that ${\displaystyle\lim_{n\to\infty}} \int_0^\infty f_n(t)\,dt \ne \int_0^\infty f(t)\,dt$.

Why doesn't this contradict the first theorem on this page.

On to Infinite Series of Functions, Section 4.6

What should the definitions be for series of functions and convergence of a series of functions??

Not on the Takehome Exam

Let $\{x_n\}_{n=1}^\infty$ be an infinite sequence. We define a rearrangement of the sequence as follows: Let $s\colon \N\to\N$ be a bijective function. Then $\{x_{s(i)}\}_{i=1}^\infty$ is a rearrangement of the sequence $\{x_n\}_{n=1}^\infty$. The rearranged sequence has exactly the same terms as the original sequence, just in a different order.

Suppose that $\{x_n\}_{n=1}^\infty$ is a convergent sequence and that $\displaystyle\lim_{n\to\infty}x_n=L$, and let $\{x_{s(i)}\}_{i=1}^\infty$ be a rearrangement of the sequence. Show that the rearranged sequence $\{x_{s(i)}\}_{i=1}^\infty$ is convergent and converges to the same limit, $\displaystyle\lim_{i\to\infty}x_{s(i)}=L$. [Hint: This is easier than it looks. Note that for any $N\in\N$, the set $\{x_1,x_2,\dots,x_N\}$ is finite. Use the definition of convergence.]