Math4121 Exam 1 Review

Range: Chapter 5 and 6 of Rudin. We skipped (and so you will not be tested on)

• Differentiation of Vector Valued Functions (pp. 111-113)
• Integration of Vector-Valued Function and Rectifiable Curves (pp.135-137)

You will also not be tested on Uniform Convergence and Integration, which we cover in class on Monday 2/10.

Chapter 5: Differentiation

Definition of the Derivative

Let $f$ be a real function defined on an closed interval $[a,b]$. We say that $f$ is differentiable at a point $x \in [a,b]$ if the following limit exists:

If the limit exists, we call it the derivative of $f$ at $x$ and denote it by $f'(x)$.

Theorem 5.2

Every differentiable function is continuous .

The converse is not true, consider $f(x) = |x|$.

Theorem 5.3

If $f,g$ are differentiable at $x$, then

1. $(f+g)'(x) = f'(x) + g'(x)$
2. $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$
3. If $g(x) \neq 0$, then $(f/g)'(x) = (f'(x)g(x) - f(x)g'(x))/g(x)^2$

Theorem 5.4

Constant function is differentiable and its derivative is $0$.

Theorem 5.5

Chain rule: If $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$, then the composite function $g\circ f$ is differentiable at $x$ and

Theorem 5.8

The derivative of local extremum ($\exists \delta > 0$ s.t. $f(x)\geq f(y)$ or $f(x)\leq f(y)$ for all $y\in (x-\delta,x+\delta)$) is $0$.

Theorem 5.9

Generalized mean value theorem: If $f,g$ are differentiable on $(a,b)$, then there exists a point $x\in (a,b)$ such that

If we put $g(x) = x$, we get the mean value theorem.

for some $x\in (a,b)$.

Theorem 5.12

Intermediate value theorem:

If $f$ is differentiable on $[a,b]$, for all $\lambda$ between $f'(a)$ and $f'(b)$, there exists a $c\in (a,b)$ such that $f'(x) = \lambda$.

Theorem 5.13

L’Hôpital’s rule: If $f,g$ are differentiable in $(a,b)$ and $g'(x) \neq 0$ for all $x\in (a,b)$, where $-\infty \leq a < b \leq \infty$,

Suppose

If

or if

then

Theorem 5.15

Taylor’s theorem: If $f$ is $n$ times differentiable on $[a,b]$, $f^{(n-1)}$ is continuous on $[a,b]$, and $f^{(n)}$ exists on $(a,b)$, for any distinct points $\alpha, \beta \in [a,b]$, there exists a point $x\in (\alpha, \beta)$ such that

Chapter 6: Riemann-Stieltjes Integration

Definition of the Integral

Let $\alpha$ be a monotonically increasing function on $[a,b]$.

A partition of $[a,b]$ is a set of points $P = \{x_0, x_1, \cdots, x_n\}$ such that

Let $\Delta \alpha_i = \alpha(x_{i}) - \alpha(x_{i-1})$ for $i = 1, \cdots, n$.

Let $m_i = \inf \{f(x) : x_{i-1} \leq x \leq x_{i}\}$ and $M_i = \sup \{f(x) : x_{i-1} \leq x \leq x_{i}\}$ for $i = 1, \cdots, n$.

The lower sum of $f$ with respect to $\alpha$ is

$L(f,P,\alpha) = \sum_{i=1}^{n} m_i \Delta \alpha_i$

The upper sum of $f$ with respect to $\alpha$ is

$U(f,P,\alpha) = \sum_{i=1}^{n} M_i \Delta \alpha_i$

Let $\overline{\int_a^b} f(x) d\alpha(x)=\sup_P L(f,P,\alpha)$ and $\underline{\int_a^b} f(x) d\alpha(x)=\inf_P U(f,P,\alpha)$.

If $\overline{\int_a^b} f(x) d\alpha(x) = \underline{\int_a^b} f(x) d\alpha(x)$, we say that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ and we write

Theorem 6.4

Refinement of partition will never make the lower sum smaller or the upper sum larger.

Theorem 6.5

$\underline{\int_a^b} f(x) d\alpha(x) \leq \overline{\int_a^b} f(x) d\alpha(x)$

Theorem 6.6

$f\in \mathscr{R}(\alpha)$ on $[a,b]$ if and only if for every $\epsilon > 0$, there exists a partition $P$ of $[a,b]$ such that

Theorem 6.8

Every continuous function on a closed interval is Riemann-Stieltjes integrable with respect to any monotonically increasing function.

Theorem 6.9

If $f$ is monotonically increasing on $[a,b]$ and $\alpha$ is continuous on $[a,b]$, then $f\in \mathscr{R}(\alpha)$ on $[a,b]$.

Key: We can repartition the interval $[a,b]$ using $f$.

Theorem 6.10

If $f$ is bounded on $[a,b]$ and has only finitely many discontinuities on $[a,b]$, then $f\in \mathscr{R}(\alpha)$ on $[a,b]$.

Key: We can use the bound and partition around the points of discontinuity to make the error arbitrary small.

Theorem 6.11

If $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and $m\leq f(x) \leq M$ for all $x\in [a,b]$, and $\phi$ is a continuous function on $[m,M]$, then $\phi\circ f\in \mathscr{R}(\alpha)$ on $[a,b]$.

Composition of bounded integrable functions and continuous functions is integrable.

Theorem 6.12

Properties of the integral:

Let $f,g\in \mathscr{R}(\alpha)$ on $[a,b]$, and $c$ be a constant. Then

1. $f+g\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\int_a^b (f(x) + g(x)) d\alpha(x) = \int_a^b f(x) d\alpha(x) + \int_a^b g(x) d\alpha(x)$
2. $cf\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\int_a^b cf(x) d\alpha(x) = c\int_a^b f(x) d\alpha(x)$
3. $f\in \mathscr{R}(\alpha)$ on $[a,b]$ and $c\in [a,b]$, then $\int_a^b f(x) d\alpha(x) = \int_a^c f(x) d\alpha(x) + \int_c^b f(x) d\alpha(x)$.
4. Favorite Estimate: If $|f(x)| \leq M$ for all $x\in [a,b]$, then $\left|\int_a^b f(x) d\alpha(x)\right| \leq M(\alpha(b)-\alpha(a))$.
5. If $f\in \mathscr{R}(\beta)$ on $[a,b]$, then $\int_a^b f(x) d(\alpha+\beta) = \int_a^b f(x) d\alpha + \int_a^b f(x) d\beta$.

Theorem 6.13

If $f,g\in \mathscr{R}(\alpha)$ on $[a,b]$, then

1. $fg\in \mathscr{R}(\alpha)$ on $[a,b]$
2. $|f|\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\left|\int_a^b f(x) d\alpha(x)\right| \leq \int_a^b |f(x)| d\alpha(x)$

Key: (1), use Theorem 6.12, 6.11 to build up $fg$ from $(f+g)^2-f^2-g^2$. (2), take $\phi(x) = |x|$ in Theorem 6.11.

Theorem 6.14

Integration over indicator functions:

If $a<s<b$, $f$ is bounded on $[a,b]$, and $f$ is continuous at $s$, and $\alpha(x)=I(x-s)$, then

Key: Note the max difference can be made only occurs at $s$.

Theorem 6.15

Integration over step functions:

If $\alpha(x) = \sum_{i=1}^{n} c_i I(x-x_i)$ for $x\in [a,b]$, then

Theorem 6.21

Fundamental theorem of calculus:

Let $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and $F(x) = \int_a^x f(t) d\alpha(t)$. Then

1. $F$ is continuous on $[a,b]$
2. If $f$ is continuous at $x\in [a,b]$, then $F$ is differentiable at $x$ and $F'(x) = f(x)$

Chapter 7: Sequence and Series of Functions

Example of non-Riemann integrable function

$\lim_{m\to \infty} \lim_{n\to \infty} (\cos(m!\pi x))^{2n}=\begin{cases} 1 & x\in \mathbb{Q} \\ 0 & x\notin \mathbb{Q} \end{cases}$

This function is everywhere discontinuous and not Riemann integrable.

Uniform Convergence

Definition 7.7

A sequence of functions $\{f_n\}$ converges uniformly to $f$ on $E$ if for every $\epsilon > 0$, there exists a positive integer $N$ such that

If $E$ is a point, then that’s the common definition of convergence.

If we have uniform convergence, then we can swap the order of limits.

Theorem 7.16

If $\{f_n\}\in \mathscr{R}(\alpha)$ on $[a,b]$, and $\{f_n\}$ converges uniformly to $f$ on $[a,b]$, then

Key: Use the definition of uniform convergence to bound the difference between the integral of the limit and the limit of the integral. $\int_a^b (f-f_n)d\alpha \leq |f-f_n| \int_a^b d\alpha = |f-f_n| (\alpha(b)-\alpha(a))$.