Martin Zhai's Review Note
- Content Summary
  - Week 1
  - Week 2
  - Week 3
  - Week 4
  - Week 5
  - Week 6
  - Week 7
  - Week 8
  - Week 9
  - Week 10
  - Week 11
  - Week 12
  - Week 13
  - Week 14
  - Week 15
  - Questions

Martin Zhai's Review Note

Content Summary

Week 1

Lecture 1 (Jan 19) - Covered Ross Section 1.1, 1.2, 1.3

Natural Numbers($\natnums$) $\{ 1, 2, 3, … \}$
Integers($\Z$) $\{…, -2, -1, 0, 1, 2, … \}$ (an example of Ring structure)
Rational Numbers($\mathbb{Q}$) $\{\frac{n}{m}, n,m\isin \Z, m\neq0\}$
- Proposition: if $r\isin\mathbb{Q}$ with gcd(c,d)=1, and is a root of $C\scriptstyle{n} \normalsize {x^n}+ C\scriptstyle{n-1} \normalsize {x^{n-1}}+ … +C\scriptstyle{0}$, $C\scriptstyle{0} \normalsize \, \mathrlap{\,/}{=} \, 0, C\scriptstyle{n} \normalsize \, \mathrlap{\,/}{=}\,0$, then d divides $C\scriptstyle{n}$, c divides $C\scriptstyle{0}$
- Corollary: if $r\,=\,\frac{c}{d}\,\mathrlap{\,/}{=}\,0$ is a root of a “monic polynomial”, i.e. leading term has coefficient 1, then r is an integer.

Lecture 2 (Jan 21) - Covered Ross Section 1.4

Maximum and Minimum: let $S\, \subset\, \Reals$ and $S\, \neq\, \text{\o}$, we say $\alpha\, \isin\, S$ is a maximum if $\forall\beta\, \isin\, S,\, \alpha\, \geq\, \beta$. Similarly, $\alpha\, \isin\, S$ is a minimum if $\forall\beta\, \isin\, S,\, \alpha\, \leq\, \beta$.
- Remark: both maximum and minimum are elements in a set, and they are not guaranteed to exist.
- Examples: $S\, =\, [\sqrt{2},100]\, \subset\, \Reals$, $\max(S)\, =\, 100,\, \min(S)\, =\, \sqrt{2}$
- Non-Examples: $S\, =\, [\sqrt{2},100]\, \subset\, \mathbb{Q}$, $\max(S)\, =\, 100,\, \min(S)$ does not exist (since $\sqrt{2}\, \mathrlap{\,/}{\isin}\, \mathbb{Q}$ and $\mathbb{Q}$ is dense)
Upper and Lower Bounds: let $\text{\o}\, \neq\, S\, \subset\, \Reals$
1. $\alpha\, \isin\, \Reals$, we say $\alpha$ is an upper bound of S, if $\forall\, \beta\, \isin\, S,\, \beta\, \leq\, \alpha$
  - Examples: Define $S={x^2-10x+24,\, x \isin \Reals}$, $10$ would be an upper bound for $S$, since for $x\geq100$, then ${x^2}-10x+24 \geq 0$, hence $x \mathrlap{\,/}{\isin}S$.
  - Non-Examples: With the same set S defined as above, 5 is not an upper bound, since consider $6\, \geq\, 5$ and ${x^2}-10x+24$ evaluated at $x\, =\, 6$ is 0, i.e. $6\, \isin\, S$.
2. $\alpha\, \isin\, \Reals$, we say $\alpha$ is a lower bound of S, if $\forall\, \beta\, \isin\, S,\, \beta\, \geq\, \alpha$
  - Examples: Define $S\, =\, [5,10]\cup\Z$, both 1 and 4 would be a lower bound for S, since for $x\, \isin\, S,\, x\, \geq\, 4$ and $x\, \geq\, 1$.
  - Non-Examples: Consider $S\, =\, \Reals$, there is no such lower bound for $\Reals$, since for any number $s$, we could always find an element $\alpha\, \isin\, S$ and $\alpha\, \leq\, s$.
- Remark: upper bounds and lower bounds are not unique, and may not exist.
Supremum and Infimum: let $\text{\o}\, \neq\, S\, \subset\, \Reals$, i.e. $S$ be a non-empty subset of $\Reals$
1. if $S$ is bounded above (i.e. there exists an upper bound for $S$), and $S$ has a least upper bound, then it is called supremum, denoted as $\sup(S)$
  - Examples: $S\, =\, [0,\sqrt{5}]\cup\Z$, then $\sup(S)\, =\, \sqrt{5}$ regardless of $\sqrt{5}\, \mathrlap{\,/}{\isin}\, S$.
  - Non-Examples: $S\, =\, \Z$, then $S$ does not have a supremum since $S$ is not bounded above, hence no such least upper bound exist.
2. if $S$ is bounded below (i.e. there exists a lower bound for $S$), and $S$ has a greatest lower bound, then it is called infimum, denoted as $\inf(S)$
  - Examples: $S\, =\, \natnums$, then $\inf(S)\, =\, 1$.
  - Non-Examples: $S\, =\, \mathbb{Q}$, then there is no infimum for $S$.
Completeness Axiom: Every non-empty subset $S\, \subset\, \Reals$ that is bounded above has a least upper bound, i.e. a supremum.
- Corollary: if $S$ is bounded from below, then $\inf(S)$ exists.
Archimedean Property: If $a\, >\, 0$, $b\, >\, 0$ are real numbers, then for some $n\, \isin\, \natnums$, we have $n*a\, >\, b$.

Homework 1

if $S\, \subset\, T$, then $\inf(T)\, \leq\, \inf(S)\, \leq\, \sup(S)\, \leq\, \sup(T)$
$\sup({S}\cup{T})\, =\, \max(\sup(S),\, \sup(T))$
$\sup(A+B)\, =\, \sup(A)\, +\, \sup(B)$
$\inf(A+B)\, =\, \inf(A)\, +\, \inf(B)$
if $a\, \leq\, b+\frac{1}{n},\, \forall\, n\, \isin\, \natnums$, then $a\, \leq\, b$

Week 2

Lecture 3 (Jan 26) - Covered Ross Section 2.7, 2.9

Sequence: $a\scriptstyle{1}\normalsize,\, a\scriptstyle{2}\normalsize,\, a\scriptstyle{3}\normalsize,\, …,\, a\scriptstyle{n}\normalsize,\,\isin\, \Reals$, could be denoted as $(a\scriptstyle{n}\normalsize)\scriptstyle{n\isin\natnums}$
- Remark: sequence is not a set where sequence considers the order of all its elements while set only record what element is in it.
Limit of Sequence: we say a sequence $(a\scriptstyle{n}\normalsize)\scriptstyle{n}$ has limit $\alpha\, \isin\, \Reals$, if $\forall\, \epsilon>0$, there exists a real number $N>0$ such that for all integer $n>N$, we have $\lvert\, a\scriptstyle{n}\normalsize\, -\, \alpha\, \rvert\, <\, \epsilon$. We denote this by $\lim_{n\to\infty}a_{n}=\alpha$
- Example: Prove $\lim_{n\to\infty}\frac{1}{n}=0$
  - Fix a $\epsilon>0$, we take $N=\frac{1}{\epsilon}$, then $n>N \iff n>\frac{1}{\epsilon} \iff \frac{1}{n}<\epsilon \iff \lvert \frac{1}{n}-0 \rvert\, <\epsilon$, completing the proof.
- Non-Examples: Determine if $\lim_{n\to\infty}n=5$
  - Consider for all $\epsilon>0$. In order to have $\lvert a_{n}-5 \rvert <\epsilon$, then $a_{n} < 5 + \epsilon$, but since $n\rightarrow\infty$, it is impossible to find a $N$ satisfying $\forall n>N,\, a_{n} <5+\epsilon$, hence the claim is incorrect.
Properties and Tools to find Limit:
- Bounded sequence: $(a_{n})_{n}$ is a bounded sequence if $\exists M>0$ such that $-M\leq a_{n} \leq M,\, \forall n \isin \natnums$
  - Examples: Consider $a_{n} = \frac{1}{n},\, \forall n \isin \natnums$, then it is a bounded sequence where we could take $M=1$ such that $a_{n} \leq M=1\, \forall n\isin \natnums$.
  - Non-Examples: Define $a_{n} = n,\, \forall n \isin \natnums$. For this sequence, there is no such $M$ to bound the elements in the sequence since for arbitrary $M$, it is always possible to find a $a_{n}$ such that $a_{n} > M$ ($a_{n}$ tends to $\infty$ as $n \rightarrow \infty$).
1. Theorem 9.1: All convergent sequences are bounded.
2. Theorem 9.2: If $\lim a_{n} = \alpha$, and if $k\isin \Reals$, then $\lim (k*a_{n})= k* \alpha$.
3. Theorem 9.3, 9.4, 9.6: Let $a_{n},\, b_{n}$ be convergent sequences with $\lim a_{n} =\alpha$ and $\lim b_{n} =\beta$, then:
  1. $\lim (a_{n}+b_{n}) = \alpha + \beta$
  2. $\lim (a_{n}*b_{n}) = \alpha * \beta$
  3. if $a_{n} \neq 0\,\, \forall n$ and $\alpha \neq 0$, then $\lim (\frac{b \scriptscriptstyle n \normalsize}{a \scriptscriptstyle n \normalsize}) = \frac{\beta}{\alpha}$

Lecture 4 (Jan 28) - Covered Ross Section 2.9, 2.10

Properties and Tools to find Limit:
1. Theorem 9.7:
  1. $\lim_{n\to\infty} \frac{1}{n^p} = 0,\, \forall p > 0$
  2. $\lim_{n\to\infty} {a^n} = 0,$ if $\lvert a \rvert < 1$
  3. $\lim_{n\to\infty} {n^\frac{1}{n}} = 1$
  4. $\lim_{n\to\infty} {a^\frac{1}{n}} = 1$ for $a>0$
- Limit of Infinity: we say $\lim s_n = +\infty$ provided for each $M>0$, $\exists N$ such that $n>N \implies s_n >M$; similarly $\lim s_n = -\infty$ provided for each $M>0$, $\exists N$ such that $n>N \implies s_n <M$
  - Example: $\lim (\sqrt{n} + 7)= +\infty$
    - Fix a $M>0$, define $N = (M-7)^2$. Then $n>N \implies n>(M-7)^2 \implies \sqrt{n} > M-7 \implies \sqrt{n} +7>M$, completing the proof.
  - Non-Example: Determine if $\lim \frac{1}{n}= +\infty$
    - No. Consider $M = 1$, in order to have $\frac{1}{n} > 1$, $n$ must be smaller than $1$. Thus it is not possible to define a $N$ such that $\forall n>N \implies \frac{1}{n} > 1$.
  1. Theorem 9.9: Let $(s_n),(t_n)$ be sequences such that $\lim s_n = +\infty,\, \lim t_n > 0$, then $\lim (s_n*t_n) = +\infty$.
  2. Theorem 9.10: Let a sequence $(s_n)$ of positive real numbers, then $\lim s_n = +\infty \iff \lim \frac{1}{s_n} = 0$.
Monotone Sequence and Cauchy Sequence:
- Cauchy Sequence: $(a_n)$ is a Cauchy sequence if $\forall \epsilon >0,\, \exists N>0$ such that $\forall n_1,n_2>N$, we have $\lvert a_{n_1} - a_{n_2} \rvert < \epsilon$ i.e. oscillation amplitude gets smaller and smaller.
  - Example: $a_n = \frac{1}{n^2}\, \forall n \isin \natnums$ is a Cauchy sequence.
    - Fix a $\epsilon >0$. Since we already know from Theorem 9.7a), $\lim \frac{1}{n^2} = 0$, then we could find a $N>0$ such that $n>N \implies \frac{1}{n^2} < \epsilon$. Thus take the same $N$, we get $\forall a,b >N$, $\lvert \frac{1}{a^2} - \frac{1}{b^2} \rvert \leq \lvert \frac{1}{a^2} \rvert < \epsilon$, completing the proof.
  - Non-Example: Consider $a_n = n$, then if we fix $\epsilon = 0.5$, there is no such $N>0$ such that $n_1, n_2>N \implies \lvert a_{n_1} - a_{n_2} \rvert < 0.5$ since the smallest difference between any $a_n$ is 1, thus not a Cauchy sequence.
- Monotone Sequence:
  1. A monotone increasing sequence is such that $\forall n > m,\, a_{n} \geq a_m$.
  2. A monotone decreasing sequence is such that $\forall n > m,\, a_{n} \leq a_m$.
    - Example: $a_n = 1 - \frac{1}{n}$ is a monotone increasing sequence, since $\forall n > m,\, \frac{1}{n} < \frac{1}{m}$, hence $1 - \frac{1}{n} > 1 - \frac{1}{m}$.
    - Non-Example: $a_n = (n-2)^2$ is not a monotone sequence.
      - Consider $n = 2,\, m = 1,\, n>m$, but $a_2 = 0 < a_1 = 1$, hence not monotone increasing.
      - Then consider $n = 2,\, m = 3,\, n<m$, but $a_2 = 0 < a_3 = 1$, hence not monotone decreasing.
1. Theorem 10.2: All bounded monotone sequences are convergent.
2. Theorem 10.11: Let $(a_n)$ be a sequence. Then $(a_n)$ is Cauchy $\iff (a_n)$ converges.

Homework 2

9.9c): If there exists $N_0$ such that $s_n \leq t_n\, \forall n > N_0$,and $\lim s_n$ and $\lim t_n$ exists, then $\lim s_n \leq \lim t_n$.

Week 3

Lecture 5 (Feb 2) - Covered Ross Section 2.10

Monotone and Cauchy Sequences:
- $\limsup$: Let $(a_n)$ be a sequence in $\Reals$, $\limsup a_n = \lim_{N\to\infty}(\sup_{n>N} {a_n})$
- $\liminf$: Let $(a_n)$ be a sequence in $\Reals$, $\liminf a_n = \lim_{N\to\infty}(\inf_{n>N} {a_n})$
- Example:$(a_n) = 0$ if $\frac{n}{2} = 0$, $(a_n) = \frac{1}{n}$ if $\frac{n}{2} = 1$.
  - $\limsup a_n = 0$ since the supremum of the subsequence $(a_n)_{n>N}$ for all $N$ is $\frac{1}{N+1}$ if $N$ is even and $\frac{1}{N}$ if $N$ is odd, which converges to $0$ as $N \rightarrow \infty$.
  - $\liminf a_n = 0$ since the infimum of the subsequence $(a_n)_{n>N}$ for all $N$ is $0$ (there is always an even number $\geq N$)
  - Property of Monotone Sequences: If it is bounded, then its limit exists; if it is unbounded, then $\lim a_n = +/- \infty$.
  - Lemma: If $(a_n)$ is a bounded sequence and $\alpha_+ = \limsup a_n$, then for any $\epsilon >0,\, \exists N$ such that $\forall n>N$, we have $a_n \leq \alpha _+ + \epsilon$
  - Theorem 10.7: Let $(a_n)$ be a bounded sequence, then $\lim a_n$ exists $\iff \limsup a_n = \liminf a_n$.

Lecture 6 (Feb 4) - Covered Ross Section 2.11

Subsequences:
- Subsequence: Let $(s_n)_{n\isin \natnums}$ be a sequence of real numbers. Given a list of indices, $n_1 < n_2 < … <n_k< …$. Take $t_k = s_{n_k}$, then $(t_m)$ is called a subsequence of $(s_n)$, we write $(s_{n_k})_k$ for the subsequence.
  - Example: Let $(s_n)$ denote some sequence with real numbers, define $(t_m)$ as $t_m = s_{2*m}$. Then $(t_m)$ is a subsequence of $(s_n)$ (could be denoted as $(s_{2n})_n$)
- Theorem 11.2: Let $(s_n)$ be a sequence.$\\$ If $t\isin \Reals$, then there is a subsequence of $(s_n)$ converging to $t \iff$ the set $\{n\isin \natnums :\, \lvert s_n - t\rvert < \epsilon\}$ is infinite for all $\epsilon >0$.
- Theorem 11.3: If $(s_n)$ is convergent, then any subsequence converges to the same point.
- Theorem 11.4: Every sequence has a monotonic subsequence.

Homework 3

$(s_n)$ a bounded sequence, then $\liminf s_n \leq \limsup s_n$ and $\limsup s_n = \inf \{\sup_{n\geq N} s_n:\, n\isin \natnums\}$.
If $(a_n), (b_n)$ are two bounded sequences, then $\limsup (a_n+b_n) \leq \limsup (a_n) + \limsup (b_n)$.

Week 4

Lecture 7 (Feb 9) - Covered Ross Section 2.11

Subsequences
- Theorem 11.5 (Bolzano-Weiestrass Theorem): Every bounded sequence has a convergent subsequence.
- Subsequential Limit: Let $(s_n)$ be a sequence in $\Reals$, a subsequential limit is any real number or $+/- \infty$ that is the limit of a subsequence of $(s_n)$.
  - Example: Let $(r_n)$ be the enumeration of $\mathbb{Q}$, then $\forall r \isin \Reals$ is a subsequential limit of $(r_n)$. This is because the denseness of $\mathbb{Q}$ in $\Reals$, which means for any $r\isin \Reals,\, \forall \epsilon >0,\, (r - \epsilon, r+\epsilon)$ is an infinite set. Hence by Theorem 11.2, there is some subsequence of $(s_n)$ that converges to such $r$, i.e. a subsequential limit.
- Theorem 11.7: Let $(s_n)$ be any sequence. Then there exists a monotonic subsequence that converges to $\limsup s_n$ and a monotonic subsequence that converges to $\liminf s_n$.
- Theorem 11.8: Let $(s_n)$ be a sequence, $S$ be the set of subsequential limits of $(s_n)$, then
  - S is non-empty
  - $\sup S = \limsup s_n$, and $\inf S = \liminf s_n$
  - $S=\{ \alpha \} \iff \lim s_n = \alpha$
- Theorem 11.9: Let $S$ be the set of subsequential limits of $(s_n)$. Suppose $(t_n)$ is a sequence in $S\cap \Reals$ and $t = \lim t_n$. Then $t \isin S$.

Lecture 8 (Feb 11) - Covered Ross Section 2.12

lim sup's and lim inf's:
- Recall that in general, $\limsup s_n \geq \liminf s_n$, and if they are equivalent, then $\lim s_n$ exists.
- Theorem 12.1: Let $(s_n)$ be a sequence that converges to a positive real number $s$, and $(t_n)$ be any sequence. Then $\limsup (s_n*t_n) = s*\limsup t_n$. (Here we allow notation of $s*(+\infty) = +\infty$ and $s*(-\infty) = -\infty$ for $s>0$).
- Theorem 12.2: Let $(s_n)$ be a sequence of positive real numbers, then we have$\\$ $\enspace \liminf (\frac{s_{n+1}}{s_n}) \leq \liminf (s_n)^{\frac{1}{n}} \leq \limsup (s_n)^{\frac{1}{n}} \leq \limsup (\frac{s_{n+1}}{s_n})$.

Homework 4

$(s_n)$ bounded $\iff \limsup \lvert s_n \rvert < +\infty$
Let $(s_n)$ be a bounded sequence in $\Reals$. If $A = \{ a \isin \Reals:$ only finitely many $s_n <a\}$ and $B = \{ b \isin \Reals:$ only finitely many $s_n >b\}$, then $\sup A = \liminf s_n$ and $inf B = \limsup s_n$.

Week 5

Lecture 9 (Feb 16) - Covered Ross Section 2.13

Metric Space & Topology:
- Metric Space: A metric space is a set $S$ together with a distance function $d: S \times S \rightarrow \Reals$ such that
  1. $d(x,y) \geq 0,$ and $d(x,y) = 0 \iff x=y$
  2. $d(x,y) = d(y,x)$
  3. $d(x,y) + d(y,z) \geq d(x,z)$ (Triangle Inequality)
  - Example: Let $d(x,y) = \sqrt{\lvert x-y \rvert}$, then with the set $\Reals$, it forms a metric space. The first two criteria is obvious. For the triangle inequality, $(\sqrt{\lvert x-y \rvert} + \sqrt{\lvert y-z \rvert})^2 = \lvert x-y \rvert + \lvert y-z \rvert + 2\sqrt{\lvert x-y \rvert} \sqrt{\lvert y-z \rvert}$. By Euclidean distance metric, we know $\lvert x-y \rvert + \lvert y-z \rvert \geq \lvert x-z \rvert$. Also $2 \sqrt{\lvert x-y \rvert} \sqrt{\lvert y-z \rvert} \geq 0$. Therefore $d(x,y) = \sqrt{\lvert x-y \rvert}$ is a metric, hence with $\Reals$ is a metric space.
  - Non-Example: Let $d(x,y) = (x-y)^2$. It is not a metric since consider $x=2, y = 1, z=0$, then $d(x,y) = 1$ and $d(y,z)=1$ but $d(x,z) = 4$ which means $d(x,y) + d(y,z) < d(x,z)$, thus not a metric.
- Cauchy Sequence (in metric space (S,d)): A sequence $(s_n)$ in $S$ is Cauchy if $\forall \epsilon >0,\, \exists N >0$ such that $\forall n,m >N,\, d(s_n,s_m) < \epsilon$.
- Convergence (in metric space (S,d)): A sequence $(s_n)$ converge to a point $s\isin S$ if $\forall \epsilon >0,\exists N>0$ such that $d(s_n,s)<\epsilon \enspace \forall n>N$.
- Completeness: A metric space $(S,d)$ is complete if every Cauchy sequence is convergent.
  - Example: The Euclidean k-space $\Reals^k$ is complete by Theorem 13.4.
  - Non-Example: $\mathbb{Q}$ with the Euclidean distance function is not complete. Consider a sequence of rational numbers converging to $\sqrt{2}$ (possible by example in Lecture 7). This sequence is clearly Cauchy but not convergent since $\sqrt{2} \mathrlap{\,/}{\isin} \mathbb{Q}$. Thus $\mathbb{Q}$ with Euclidean distance formula is not a complete metric space.
- Induced Distance Function: If $(S,d)$ is a metric space and $A \subset S$ is any subset, then $(A, d|_{A\times A})$ is a metric space.
- Theorem 13.5 (Bolzano-Weiestrass Theorem for $\Reals^n$): Every bounded sequence $(s_m)_m \isin \Reals^n$ has a convergent subsequence.
- Topology: Let $S$ be a set. A topological structure on $S$ is the data of a collection of subsets in S. This collection needs to satisfy:
  1. $S$ and $\text{\o}$ are open
  2. arbitrary union of open subsets is still open
  3. finite intersections of open sets are open

Lecture 10 (Feb 18) - Midterm 1

Homework 5

Every open subset of $\Reals$ is the disjoint union of finite or countably infinite sequence of open intervals.

Week 6

Lecture 11 (Feb 23) - Covered Ross Section 2.13, Rudin Chapter 2

Topology of Metric Space:
- Basic Notions: Let $E \subset S$
  - Interior point: $p\isin E$ is an interior point of E if $\exist \delta>0$ such that $B_{\delta}(p) = \{q \isin S| d(p,q)< \delta\} \subset E$.
    - Example: Consider $E = [0,1] \cap \Reals \subset \Reals$, then $p=0.5$ is an interior point of $E$. If we take $\delta = 0.3$, then $B_{0.3}(0.5) = [0.2, 0.8]\cup \Reals \subset E$.
    - Non-Example: Take the same E as above, consider $p=0$. P is not an interior point since for any $\delta \neq 0$, the points in $\Reals$ on the left side of $0$ is not in $E$, hence not an interior point.
  - Interior: The set of all interior points of E, denote as $E^o$.
- Open: $E\subset S$ is an open subset of $S$ if $E=E^o$, i.e. $\forall p \isin E,\, \exists \delta >0$ such that $B_{\delta}(p) \subset E$.
  - Example: Take $E = (0,1) \cap \Reals \subset \Reals$, $E$ is a subset of $\Reals$. For any point $p \isin E$, let $\delta_o = \max (d(p,0), d(p,1))$. Then we could choose $\delta = \frac{\delta_o}{2}$, then $B_{\delta}(p) \subset E$ for arbitrary point $p \isin E$, hence $E$ is an open subset of $\Reals$.
  - Non-Example: Take $E = [0,1] \cap \Reals \subset \Reals$. As the non-example for interior points shown, $0$ is not an interior point in $E$, hence $E$ is not an open subset of $\Reals$.
- Close: $E\subset S$ is a closed subset of S if the complement $E^c = S \setminus E$ is open.
  - Propositions:
    1. $S$ and $\text{\o}$ are closed
    2. An arbitrary collection of closed sets is closed
    3. A finite intersection of closed sets is closed
- Limit Point: Let $E\subset S$, $p\isin S$ is a limit point of $E \iff \forall \delta >0,\, B_{\delta}(p)$ intersects $E$ non-empty, i.e. $\exists q\isin E,\, q \neq p,\, d(p,q)<\delta$. $E' =$ set of limit points of $E$.
  - Example: Let $S=\Reals,\, E = \{\frac{1}{n}:n\isin \natnums\}$. Then $0$ is a limit point since for any $\delta$, we could always find a $N>0$ such that $\frac{1}{n} < \delta, \forall n>N$, hence $B_{\delta}(0) \cap E \neq \text{\o}$.
  - Non-Example: Consider the same scenario as the previous example. $\frac{1}{2}$ is not a limit point if we take $\delta = \frac{1}{10}$, then $B_{\frac{1}{10}}(\frac{1}{2}) = [\frac{4}{10}, \frac{6}{10}] \cap E = \text{\o}$.
- Closure: $E\subset S$, the closure of $E$ is the intersection of all closed subsets containing $E$. Denote as $E^-$.
  - Proposition: $E^- = E \cup E'$
  - Example: Let $S=\Reals,\, E = \{\frac{1}{n}:n\isin \natnums\}$. Then $E^- = \{\frac{1}{n}:n\isin \natnums\} \cup \{0\}$ by proposition above ($\{0\} = E'$)
- Boundary: The boundary points of $E\subset S$ is the set $E^-\setminus E^o$.
- Proposition 13.9: Let $E$ be a subset of a metric space $(S,d)$
  1. The set $E$ is closed if and only if $E=E^-$
  2. The set $E$ is closed if and only if it contains the limit of every convergent sequence of points in $E$
  3. An element is in $E^-$ if and only if it is the limit of some sequence of points in $E$
  4. A point is in the boundary of $E$ if and only if it belongs to the closure of both $E$ and its complement
- Isolated Point: If $p \isin E$ and $p$ is not a limit point of $E$, then $p$ is called an isolated point.
- Perfect: $E$ is perfect if $E$ is closed and every point of $E$ is a limit point of $E$.
- Dense: $E$ is dense in $S$ if every point of $S$ is a limit point of $E$ or a point of $E$ or both.
- Rudin 2.30: Suppose $Y\subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y\cap G$ for some open subset $G$ of $X$.
Compact Set:
- Open Cover: Let $(S,d)$ be a metric space, $E\subset S$, $\{ G_{\alpha} \}$ is a collection of open sets. We say $\{ G_{\alpha} \}$ is an open cover of $E$ if $E \subset \cup_{\alpha} G_{\alpha}$.
- Compact Set: $K\subset S$ is a compact subset, if for any open cover of $K$, there exists a finite subcover, i.e. if $\{ G_{\alpha} \}$ is an open cover, then $\alpha_1, …, \alpha_n$ indices such that $K\subset G_{\alpha_1} \cup … \cup G_{\alpha_n}$.
  - Example: $K=[0,1]$ is a compact subset of $\Reals$.
  - Non-Example: $K=(0,1]$ is not a compact subset of $\Reals$. Consider the open cover $G_n = B_{\frac{1}{2n}}(\frac{1}{n}), n\isin \natnums,$ then $K \subset \cup_{n\isin \natnums}$. In this case, there is no finite subcover. If we take indices $n_1 < n_2 < … < n_m$, then $x < \frac{1}{2n_m}$ is not in the union.
- Sequentially Compact: $E\subset S$ is sequentially compact if any sequence in $E$ has a convergent subsequence in $E$ (the limit point is also in $E$).
- Theorem: For any metric space $(S,d)$, $E\subset S$, $E$ compact $\iff E$ sequentially compact.
- Theorem 13.12 (Heine-Borel Theorem): Consider $\Reals^n$ with Euclidean metric $d(x,y)= \lvert x-y \rvert$, $E\subset \Reals^n$ is compact $\iff E$ is closed and bounded.
- Rudin 2.33: $K\subset Y\subset X$, then $K$ is compact relative to $Y$ if and only if $K$ is compact relative to $X$.
- Rudin 2.34: Compact subsets of metric spaces are closed.
- Rudin 2.35: Closed subsets of compact sets are compact.

Lecture 12 (Feb 25) - Covered Ross Section 2.14, 2.15

Series:
- Infinite Sum: An infinite sum of sequence $(a_n)$ is defined as $a_1 + a_2 + … = \sum_{n=1}^{\infty} a_n$.
- Partial Sum: Defined as $a_1 + a_2 + … + a_n = \sum_{i=1}^{n} a_i$.
- Convergence: A series converge to $\alpha$ if the corresponding partial sum converges to $\alpha$.
- Cauchy Condition for Series Convergence: $\forall \epsilon>0,\, \exists N>0$ such that $\forall n,m>N,\, \lvert \sum_{i=n+1}^{m} a_i \rvert < \epsilon$.
- Absolute Convergence: If $\sum \lvert a_n \rvert < \infty$, we say $\sum a_n$ converges absolutely.
- Recall Geometric Series: $\sum_{n=0}^{\infty} ar^n$ converges to $a\frac{1}{1-r}$ if $\lvert r \rvert <1$.
Tests for Series Convergence:
- Comparison Test:
  - Suppose $\sum_{n} a_n <\infty,\, a_n >0$, and $b_n \isin \Reals < a_n$, then $\sum_{n} b_n < \infty$.
  - Suppose $\sum_{n} a_n =\infty,\, a_n >0$, and $b_n \geq a_n$, then $\sum_{n} b_n = \infty$.
- Ratio Test (Test for Absolute Convergence):
  - If $\limsup \lvert \frac{a_{n+1}}{a_n} \rvert <1$, then $\sum_{n} \lvert a_n \rvert$ converges.
  - If $\liminf \lvert \frac{a_{n+1}}{a_n} \rvert >1$, then $\sum_{n} \lvert a_n \rvert$ diverges.
  - Otherwise the test does not give any information on its convergence.
- Root Test: Let $\sum_{n} a_n$ be a series, $\alpha = \limsup (\lvert a_n \rvert)^{\frac{1}{n}}$, then $\sum_{n} a_n$:
  - converges absolutely if $\alpha <1$.
  - diverges if $\alpha >1$.
  - gives no information if $\alpha = 1$.
- Alternating Series Test: Let $a_1 \geq a_2 \geq …$ be a monotone decreasing series, $a_n \geq 0$. And assuming $\lim a_n = 0$. Then $\sum_{n=1}^{\infty} (-1)^{n+1}a_n = a_1 - a_2 + a_3 - …$ converges. Moreover, the partial sums $s_n = \sum_{k=1}^{n} (-1)^{k+1}a_k$ satisfy $\lvert s- s_n \rvert \leq a_n$ for all $n$.
- Integral Test: If the terms $a_n$ in $\sum_{n} a_n$ are non-negative and $f(n) = a_n$ is a decreasing function on $[1, \infty)$, then let $\alpha = \lim_{n\to\infty} \int_{1}^{n} f(x)dx$
  - If $\alpha = \infty$, then the series diverge
  - If $\alpha < \infty$, then the series converge

Homework 6

Rudin 2.36: If $\{K_{\alpha}\}$ is a collection of compact subsets of a metric space such that the intersection of every finite subcollection of $\{K_{\alpha}\}$ is non-empty, then $\cap K_{\alpha}$ is non-empty. (This statement is incorrect if compact is replaced by closed or bounded).
If $a_n >0$ and $\sum a_n$ diverges, then $\sum \frac{a_n}{1+a_n}$ also diverges.
If $a_n >0$ and $\sum a_n$ converges, then $\sum \frac{\sqrt{a_n}}{n}$ also converges.

Week 7

Lecture 13 (Mar 2) - Covered Rudin Chapter 4

Continuous Functions:
- Function: A function from set $A$ to set $B$ is an assignment for each element $\alpha \isin A$ an element $f(\alpha) \isin B$.
  1. Injective: A function $f$ is injective/one-to-one if $\forall x,y\isin A,\, x \neq y$, then $f(x) \neq f(y)$
  2. Surjective: A function $f$ is surjective/onto if $\beta \isin B,$ there exists at least one element $\alpha \isin A$ such that $f(\alpha)=\beta$
  3. Bijective: A function $f$ is bijective if $f$ is both injective and surjective
  - Example: $f: \Reals \rightarrow \Reals,\, f(x)=x$ is a bijective function since if $f(x) = f(y)$, then $x = y$. And for any $y \isin \Reals$, we take $x=y \isin \Reals$ as the domain will prove its surjectivity. Hence a bijection.
  - Non-Example:
    1. $f: \Reals \rightarrow \Reals,\, f(x)=x^2$ is not injective since for $x=1$ and $y=-1$, $f(x) = f(y)$ even if $x\neq y$.
    2. $f: \Reals \rightarrow \Reals,\, f(x)=\frac{1}{x}$ is not surjective since for $y = 0 \isin \Reals$, there is no element in domain that get mapped to $0$ under such $f$.
- Pre-image: If $f: A\rightarrow B$. Given a subset $E\subset B$, $f^{-1}(E) = \{\alpha \isin A|\, f(\alpha) \isin E\}$ is called the pre-image of $E$ under $f$, which is a subset of $A$.
  - Example: $f: \Reals \rightarrow \Reals,\, f(x)=\ln x$ and let $E=[0,\infty)$. Then pre-image of $E$ under $f$ is $[1,\infty)$.
- Limit of a Function: Suppose $p\isin E'$(set of limit points of $E$), we write $f(x) \rightarrow q(\isin Y)$ as $x \rightarrow p$ or $\lim_{x\to p} f(x) = q$ if $\forall \epsilon >0,\, \exists \delta >0$ such that $\forall x \isin E,\, 0<d_X(x,p)<\delta \implies d_Y(f(x),q)<\epsilon$.
  - Example: Consider $f: \Reals \rightarrow \Reals$ $f(x)=\frac{x^2-1}{x-1}$. Claim $\lim_{x\to 1} f(x) = 2.
    - Fix a $\epsilon >0$, consider $\delta = \lvert \epsilon - 1\rvert$. Using the Euclidean distance formula, $\lvert x - 1\rvert < \delta = \lvert \epsilon - 1\rvert \implies -\epsilon + 1 < x - 1< \epsilon -1 \implies -\epsilon + 3 < x + 1< \epsilon + 1 \implies -\epsilon + 3 < \frac{(x + 1)(x-1)}{x-1}< \epsilon + 1 \implies -\epsilon + 2 < \frac{(x + 1)(x-1)}{x-1} - 1< \epsilon \implies \lvert \frac{(x + 1)(x-1)}{x-1} - 1 \rvert < \epsilon$, hence proving the claim.
- Rudin 4.2: With the same notation as above, $\lim_{x\to p} f(x) = q$ if and only iff $\lim_{n\to\infty} f(p_n) = q$ for every sequence $(p_n)$ in $E$ such that $p_n \neq p, \lim_{n\to\infty} p_n = p$.
  - Example: $f:\Reals \rightarrow \Reals$ and $f(x)=x^2$. $\lim_{x\to 0} f(x) = 0$ and then consider sequence $p_n = \frac{1}{n}$ which converges to $0$, and $f(p_n) = \frac{1}{n^2}$ which also converges to $0$ when $n \rightarrow \infty$
- Corollary: If $f$ has a limit at point $p$, then it is unique.
- Rudin 4.4: Suppose $f,g: E \rightarrow \Reals$, suppose $p\isin E;$ and $\lim_{x\to p} f(x) = A, \lim_{x\to p} g(x) = B$, then
  - $\lim_{x\to p} f(x) +g(x)= A+B$
  - $\lim_{x\to p} f(x)g(x) = AB$
  - $\lim_{x\to p} \frac{f(x)}{g(x)} = \frac{A}{B}$ if $B\neq 0$ and $g(x)\neq 0 \, \forall x\isin E$
  - $\forall c\isin \Reals$, $\lim_{x\to p} c*f(x)=cA$
Continuity of Functions:
- Continuity at a Point: Let $(X,d_X), (Y,d_Y)$ be metric spaces, $E\subset X$, $f:E\rightarrow Y$, $p \isin E$, $q=f(p)$. We say $f$ is continuous at $p$, if $\forall \epsilon>0, \exists \delta>0$ such taht $\forall x\isin E$ with $d_X(x,p) <\delta \implies d_Y(f(x),q)<\epsilon$.
- Rudin 4.6: If $p\isin E$ is also a limit point of $E$, then $f$ is continuous at $p \iff \lim_{x\to p} f(x) = f(p)$.
- Continuity: We say $f$ is continuous on $E$ if $f$ is continuous at every point in $E$.
- Rudin 4.7: $(X,d_X), (Y,d_Y)$, $f:X\rightarrow Y$ as above. Then $f$ is continuous $\iff$ for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$.
- Lemma: If $f: A\rightarrow B$ is a function and $E\subset A, F\subset B$. The $f(E)=F \iff E\subset f^{-1}(F)$.
- Rudin 4.7: Let $X,Y,Z$ be metric spaces and $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ continuous functions. We define $h:X\rightarrow Z$ by $h(x)=g(f(x)$. Then $h$ is also continuous ($h$ is called the composition of $f$ and $g$).
- Rudin 4.9_: If $f,g:X\rightarrow \Reals$ continuous, then $f+g,\, f-g,\, fg$ are continuous functions, and if $g(x)\neq 0$ for any $x\isin X$, then $\frac{f}{g}$ is also continuous.
  - Result: All polynomials are continuous since $f(x)=x$ is continuous.
- Rudin 4.10: Let $f:X\rightarrow \Reals^n$ with $f(x) = (f_1(x), f_2(x), …, f_n(x))$. Then $f$ is continuous $\iff$ each $f_i$ is continuous.

Lecture 14 (Mar 4) - Covered Rudin Chapter 4

Review of Compact Sets:
- Compactness: $K\subset X$ is compact if $\forall$ open cover of $K$, $\exists$ a finite subcover.
- Propositions:
  - $K$ compact $\implies K$ bounded
  - $K$ compact $\implies K$ closed
  - $E\subset X$ is closed, $K$ is compact, $E\subset K \implies E$ is compact.
- Theorems:
  - Compactness $\iff$ Sequential Compactness
  - Heine-Borel (Rudin 2.41): in $\Reals^n$, $K$ compact $\iff K$ closed and bounded.
- Remark: The notion of “compact” is intrinsic, while open and closed depends on the ambient space.
Continuous Maps and Compactness:
- Three Definitions of Continuous Maps:
  - $f$ is continuous if and only if $\forall p\isin X, \forall \epsilon >0, \exists \delta >0$ such that $f(B_{\delta}(p)) \subset B_{\epsilon}(f(p))$
  - $f$ is continuous if and only if $\forall V\subset Y$ open, $f^{-1}(V)$ is open
  - $f$ is continuous if and only if $\forall x_n \rightarrow x$ in $X$, we have $f(x_n) \rightarrow f(x)$ in $Y$
- Rudin 4.14: Suppose $f$ is a continuous map from a compact metric space $X$ to another compact metric space $Y$, then $f(X) \subset Y$ is compact.
- Rudin 4.16: Suppose $f$ is a continuous real function on a compact metric space $X$, and $M = \sup_{p\isin X} f(p)$, $m=\inf_{p\isin X} f(p)$. Then there exists point $p,q \isin X$ such that $f(p)=M$ and $f(q)=m$.
  - Recall: if $K\subset \Reals$, $K$ is compact, then $\sup(K) \isin K$ and $\inf(K) \isin K$.
  - Remark: If $f:X\rightarrow Y$ is continuous, $f$ sends compact set $X$ to compact set $Y$, but given $E\subset Y$ compact, $f^{-1}(E)$ is not guaranteed to be compact.

Homework 7

If $f:X\rightarrow Y$ is a continuous function from a metric space $X$ to a metric space $Y$, $f(E^-)\subset {f(E)}^-$ for any $E\subset X$.
Let $f,g$ be continuous maps from $X$ to $Y$. Suppose there is a dense set $E\subset X$ such that $f|_E=g|_E$, then $f=g$.

Week 8

Lecture 15 (Mar 9) - Covered Ross Section 3.19 Rudin Chapter 2 and 4

Uniform Continuity:
- Uniform Continuous Function: $f:X\rightarrow Y$. Suppose for all $\epsilon >0$, $\exists \delta >0$ such that $\forall p,q\isin X$ with $d_X(p,q)<\delta$, we have $d_Y(f(p),f(q)) <\epsilon$. Then we say $f$ is a uniform continuous function.
  - Example: $f:[0,1] \rightarrow \Reals$ and $f(x)=x^2$. Then $f$ is uniformly continuous function. $\forall \epsilon >0$, we can take $\delta=\frac{\epsilon}{2}$, then $\forall p,q \isin [0,1]$, $\lvert p-q \rvert < \delta$ we have $\lvert p^2-q^2 \rvert = \lvert (p-q)(p+q) \rvert = \lvert (p-q) \rvert \lvert (p+q) \rvert < \delta * 2 = \epsilon$.
  - Non-Example 1: $f:\Reals \rightarrow \Reals$ and $f(x)=x^2$. Then $f$ is not uniformly continuous. $\delta > 0 $, we could always find $p,q\isin \Reals$ and $\lvert p-q \rvert < \delta$ such that $\lvert f(p) - f(q) \rvert = 1$. If we take $p-q = \frac{\delta}{2}$ and $p+q = \frac{2}{\delta}$, then $\lvert f(p) - f(q)\rvert = \lvert p^2-q^2 \rvert = \lvert (p-q)(p+q) \rvert = \lvert (p-q) \rvert \lvert (p+q) \rvert = \frac{\delta}{2} * \frac{2}{\delta} = 1$.
  - Non-Example 2: $f:(0, \infty) \rightarrow \Reals$, $f(x)=\frac{1}{x}$ is not uniformly continuous. Intuition: when $x\rightarrow 0$, then distance between two points $p,q$ may be close enough but $\lvert \frac{1}{p} - \frac{1}{q} \rvert$ may be large.
- Theorem: Suppose $f:X\rightarrow Y$ is a continuous function between metric spaces. If $X$ is compact, then $f$ is uniformly continuous.
  - Theorem 19.2: If $f$ is continuous on a closed interval $[a,b]$, then $f$ is uniformly continuous on $[a,b]$.
- Proposition: If $f:X\rightarrow Y$ is uniformly continuous and $S\subset X$ subset with induced metric, then the restriction $f|_S:S\rightarrow Y$ is uniformly continuous.
Connectedness:
- Connected: Let $X$ be a set. We say $X$ is connected if $\forall S\subset X$ and $S$ is both open and closed, then $S$ has to be either $X$ or $\text{\o}$.
  - Non-Example (From Midterm 2): $(0,2)\cap \mathbb{Q}$ is not connected. Consider the set $(0,\sqrt{2}) \cap \mathbb{Q}$ and $(\sqrt{2}, 2) \cap \mathbb{Q}$. Both open and closed but none of those two are $\text{\o}$.
- Proposition: $X$ is connected $\iff$ if $X = U\sqcup V$ and $U\&V$ are both open, then one of $U,V$ is empty set.
- Rudin 4.22: If $f:X\rightarrow Y$ is continuous, if $E\subset X$ is connected, then $f(E)$ is connected. (Continuity preserves connectedness)
- Proposition: $[0,1] \subset \Reals$ is a connected subset.

Lecture 16 (Mar 11) - Covered Rudin Chapter 2 and 4

Review:
- Induced Topology: Given a topological space $X$, and $S\subset X$, we endow $S$ with the induced toplogy: $U\subset S$ is open in $S$ if and only if $\exists V \subset X$ open in $X$ such that $U = V\cap S$.
  - Example: $X= \Reals$ and $S= \Z \subset \Reals$, with induced topology on $S$, $\forall n \isin \Z$, $\{ n \}$ is open in $S$ since $\{ n \} = (n-\frac{1}{2}, n+\frac{1}{2}) \cap S$
- Corollary:
  - If $S\subset X$ is open in $X$, then $U\subset S$ is open in $S$ if and only if $U$ is open in $X$.
  - If $S\subset X$ is closed in $X$, then $E\subset S$ is closed in $S$ if and only if $E$ is closed in $X$.
Connectedness:
- Lemma: $E$ is connected if and only if $E$ cannot be written as $A\cup B$ when $A^- \cap B = \text{\o}$ and $A\cap B^- = \text{\o}$ (closure taken with respect to ambient space $X$).
- Rudin 4.27: $E\subset \Reals$ is connected$\iff \forall x,y\isin E, x<y$, we have $[x,y]\subset E$
- Rudin 4.23: Let $f$ be a continuous real function on the interval $[a,b]$. If $f(a)<f(b)$ and if $c$ is a number such that $f(a)<c<f(b)$, then there exists a point $x\isin [a,b]$ such that $f(x)=c$.
Discontinuities：
- Discontinuous: $f:X\rightarrow Y$ is discontinuous at $x\isin X$ if and only if $x$ is a limit point of $X$ and $\lim_{x\to p} f(q)$ either does not exist or $\neq f(x)$.
- Right and Left Limit: Let $f:(a,b)\rightarrow \Reals$ (not necessarily continuous)
  - $\forall x\isin [a,b)$, we say $f(x+) = q$ if for all sequence $(t_n)$ in $(x,b)$ that converge to $x$, we have limit $\lim_n f(t_n) = q$.
    - Example: $f(x) = 1$ if $x\geq 0$, $f(x)=0$ if $x < 0$. Then $f(x+) =1 $.
  - $\forall x\isin (a,b]$, we say $f(x-) = q$ if for all sequence $(t_n)$ in $(a,x)$ that converge to $x$, we have limit $\lim_n f(t_n) = q$.
    - Example: $f(x) = 1$ if $x\geq 0$, $f(x)=0$ if $x < 0$. Then $f(x+) =1$.
- Discontinuity of First and Second Kind: $f:(a,b)\rightarrow \Reals$, $x\isin (a,b)$. Suppose $f$ is discontinuous at $x$
  - We say $f$ has a simple discontinuity or discontinuity of the first kind at $x_o$ if both $f(x_o+)$ and $f(x_o-)$ exists.
    - Example: $f(x) = \frac{1}{n}$ if $x\isin \mathbb{Q}, x= \frac{m}{n}$ with $m,n$ coprime, $f(x) =0$ otherwise. $\forall x\isin \mathbb{Q}$, we have a simple discontinuity.
  - We say $f$ has a discontinuity of second kind, if it is not a simple discontinuity.
    - Example: $f(x) = \sin \frac{1}{x}, x>0$, $f(x) = x, x\leq 0$. Since $f(0+)$ does not exist, thus $f$ has a discontinuity of second kind at $x=0$.

Homework 8

If $K\subset \Reals^n$ is compact and $C\subset \Reals^n$ is closed, then $K+C$ is closed.

Week 9

Lecture 17 (Mar 16) - Covered Rudin Chapter 4 and 7

Monotonic Functions:
- Monotonic Functions: A function $f:(a,b)\rightarrow \Reals$ is monotone increasing if $\forall x>y$, we have $f(x) \geq f(y)$. Similarly one can define monotone decreasing functions.
  - Example: $f(x) = \log(x)$ is a monotone increasing function, $g(x) = 1$ is both a monotone increasing function and a monotone decreasing function.
  - Non-Example: $f(x) = x^2+2x+1$.
- Rudin 4.29: Suppose $f:(a,b)\rightarrow \Reals$ is a monotone increasing function, then $\forall x\isin (a,b)$, the left limit $f(x-)$ and the right limit $f(x+)$ exists, satisfying $\sup\{f(t)|\, t<x\} = f(x-) \leq f(x+) = \inf\{f(t)|\, t>x\}$; and given $x<y$ in $(a,b)$, then $f(x+) \leq f(y-)$.
- Corollary: If $f$ is monotone, then $f(x)$ only has discontinuity of the first kind/simple discontinuity.
- Rudin 4.30: If $f$ is monotone, then there are at most countably many discontinuities.
Sequence and Convergence of Functions:
- Pointwise Convergence of Sequence of Sequences: Let $(x_n)_n$ be a sequence of sequences, $x_n\isin \Reals^{\natnums}$, we say $(x_n)_n$ converges to $x\isin \Reals^{\natnums}$ pointwise if $\forall i\isin \natnums$, we have $\lim_{n\to\infty} x_{ni} = x_i$.
  - Example: $x_{ni} = \frac{i}{n+i}$, then this seq to $0$ pointwise, since for arbitrary fixed $i$, we have $\lim_{n\to\infty} x_{ni} = \lim_{n\to\infty} \frac{i}{n+i} = 0$.
- Uniform Convergence of Sequence of Sequences: Let $(x_n)_n$ be a sequence of sequences, $x_n\isin \Reals^{\natnums}$, we say $x_n \rightarrow x$ uniformly if $\forall \epsilon >0$, $\exists N>0$ such that $\forall n>N$, $\sup\{\lvert x_{ni} - x_i \rvert :\, i\isin\natnums\} <\epsilon$ (also known as $d_{\infty}(x_n, x)$).
  - Non-Example: $x_{ni} = \frac{i}{n+i}$ failed to converge uniformly to $0$, since $d_{\infty}(x_n,0) = \sup \{\lvert x_{ni} \rvert :\, i\isin\natnums\} = \sup \{\frac{i}{n+i}:\, i\isin\natnums\} = 1$ (n is fixed).
- Pointwise Convergence of Sequence of Functions: Given a sequence of functions $f_n \isin$ Map$(\Reals, \Reals)$, we say $f_n$ converge to $f$ pointwise if $\forall x\isin\Reals$, $\lim_{n\to\infty} f_n(x) = f(x) \iff \lim_{n\to\infty} \lvert f_n(x) - f(x) \rvert = 0$.
  - Examples：Running and Shrinking Bumps.

Lecture 18 (Mar 18) - Covered Rudin Chapter 7

Uniform Convergence:
- Uniform Convergence of Sequence of Functions: Given a sequence of functions $(f_n): X\rightarrow Y$, is said to converge uniformly to $f:X \rightarrow Y$, if $\forall \epsilon >0, \exists N>0$ such that $\forall n>N, \forall x\isin X$, we have $\lvert f_n(x) - f(x) \rvert <\epsilon$.
  - Remark: The integer $N$ depends only on $\epsilon$ in uniform convergence while $N$ could depend on both $\epsilon$ and $x$ in pointwise convergence.
  - Example: $f_n(x) = \frac{\sin(x)}{1+nx^2}$ converges uniformly. See Homework 9 Question 4 for the detailed proof.
- Rudin 7.8: Suppose $f_n: X\rightarrow \Reals$ satisfies that $\forall \epsilon >0,\exists N>0$ such that $\forall x\isin X, \lvert f_n(x) - f_m(x) \rvert < \epsilon$, then $f_n$ converges uniformly (Uniform Cauchy $\iff$ Uniform Convergence).
- Rudin 7.9: Suppose $f_n\rightarrow f$ pointwise, then $f_n \rightarrow f$ uniformly $\iff \lim_{n\to\infty} (\sup \lvert f_n(x) - f(x) \rvert) = 0$.
- A sequence of functions $\{ f_n\}$ is uniformly convergent to $f:D\to\Reals\iff \lim_{n\to\infty} \sup \{\lvert f_n(x) - f(x) \rvert : x\isin D\}$.
- Rudin 7.10 (Weiestrass M-Test): Suppose $f(x) = \sum_{n=1}^{\infty} f_n(x)\, \forall x\isin X$. If $\exists M_n>0$ such that $\sup_x \lvert f_n(x) \rvert \leq M_n$ and $\sum_{n} M_n < \infty$, then the partial sum $F_N(x)=\sum_{n=1}^{N} f_n(x)$ converges to $f(x)$ uniformly.
  - Example: See last question on Midterm 2 version A.
Uniform Convergence and Continuity:
- Rudin 7.11: Suppose $f_n \rightarrow f$ uniformly on set $E$ in a metric space. of $E$, and suppose that $\lim_{t\to x} f_n(t) = A_n$. Then $\{ A_n\}$ converges and $\lim_{t\to x} f(t) = \lim_{n\to\infty} A_n$. In conclusion, $\lim_{t\to x} \lim_{n\to\infty} f_n(t) = \lim_{n\to\infty} \lim_{t\to x} f_n(t)$.
- Rudin 7.12: If $\{f_n\}$ is a sequence of continuous functions on $E$, and if $f_n\rightarrow f$ uniformly on $E$, then $f$ is continuous on $E$.
- Rudin 7.13: Suppose $K$ compact and
  1. $\{f_n\}$ is a sequence of continuous functions on $K$
  2. $\{f_n\}$ converges pointwise to a continuous function $f(x)$ on $K$
  3. $f_n(x) \geq f_{n+1}(x) \forall x\isin K,\, \forall n = 1,2,…$
  - Then $f_n\rightarrow f$ uniformly on $K$.

Homework 9

Let $f:X\rightarrow \Reals$ be a function on a metric space. We say $f$ is Lipschitz Continuous if there exists a $K>0$ such that for any $x,y\isin X$, we have $\lvert f(x)-f(y) \rvert \leq K*d(x,y)$. Such $K$ is called a Lipschitz constant for $f$.

Week 10

Spring Break (Mar 23, Mar 25)

Week 11

Lecture 19 (Mar 30) - Review for Midterm 2

Lecture 20 (Apr 1) - Midterm 2

Week 12

Lecture 21 (Apr 6) - Covered Rudin Chapter 5

Derivative:
- Differentiability: Let $f:[a,b]\rightarrow \Reals$ be a real valued function. Define $\forall x\isin [a.b]$, $f'(x) = \lim_{t\to x} \frac{f(t)-f(x)}{t-x}$. If $f'(x)$ exists, we say $f$ is differentiable at this point $x$.
  - Example: $f(x)=x^2$. For $x=3$, define $g(x) = \frac{f(x)-f(3)}{x-3} = \frac{x^2-9}{x-3} = x+3$. Then $f'(3) = \lim_{x\to 3} g(x) = \lim_{x\to 3} x+3 = 6$.
  - Non-Example: $f(x) = x\sin \frac{1}{x}$ if $x>0$ and $f(x)=0$ if $x\leq 0$, then f is not differentiable at $x=0$. For $x>0$, $g(x) = \frac{f(x) - f(0)}{x-0} = \frac{x\sin \frac{1}{x}}{x} = \sin \frac{1}{x}$, and $\lim_{x\to 0^+} g(x)$ does not exist, thus $f'(0)$ does not exist as well.
- Proposition: If $f:[a,b]\rightarrow \Reals$ is differentiable at $x_o\isin [a,b]$, then $f$ is continuous at $x_o$, i.e. $\lim_{x\to x_o} f(x) = f(x_o)$.
- Rudin 5.3: Let $f,g: [a,b]\rightarrow \Reals$. Assume $f,g$ are differentiable at point $x_o\isin [a,b]$, then
  1. $\forall c\isin \Reals$, $(cf)'(x_o)=cf'(x_o)$
  2. $(f+g)'(x_o) = f'(x_o)+g'(x_o)$
  3. $(fg)'(x_o)=f'(x_o)g(x_o)+f(x_o)g'(x_o)$ (Leibniz's Rule)
  4. if $g(x_o)\neq 0$, then $(\frac{f}{g})'(x_o) = \frac{f'(x_o)g(x_o)-f(x_o)g'(x_o)}{(g(x_o))^2}$
- Rudin 5.5 (Chain Rule): Suppose $f:[a,b]\rightarrow I\subset \Reals$ and $g: I\rightarrow \Reals$. Suppose for some $x_o\isin [a,b]$, $f(x_o) = y_o$, $y_o\isin \Reals$, $f'(x_o)$ and $g'(y_o)$ exists. Then, the composition $h=g \circ f:[a,b]\rightarrow \Reals$. $h(x)=g(f(x))$ is differentiable at $x_o$, $h'(x_o) = g'(y_o)f'(x_o)$.
  - Example: $h(x) = \sin(x^2)$, then by Chain rule, $h'(x) = 2x\cos(x^2)$
Mean Value Theorem:
- Local Maximum and Minimum: Let $f:[a,b]\rightarrow \Reals$. We say $f$ has a local maximum at point $p\isin [a,b]$, if $\exists \delta >0$ and $\forall x\isin[a,b]\cap B_{\delta}(p),\, f(x) \leq f(p)$. Local minimum is defined in the similar way.
  - Example: Say $f:[-\pi,\pi] \rightarrow \Reals$ and $f(x)=\sin(x)$. Then we can say $p= \frac{\pi}{2}$ is a local maximum with $\delta = \frac{\pi}{4}$. And similarly, we can say $q=0$ is a local minimum with $\delta = \frac{\pi}{2}$.
- Rudin 5.8: Let $f:[a,b]\rightarrow \Reals$. If $f$ has a local maximum or minimum at $p\isin (a,b)$, and if $f$ is differentiable at $p$, then $f'(p)=0$.
  - Remark: The point $p$ cannot be taken at the endpoints of the domain.
- Rolle's Theorem: Suppose $f:[a,b]\rightarrow \Reals$ is a continuous function and $f$ is differentiable in $(a,b)$. If $f(a)=f(b)$, then there is some $d\isin (a,b)$ such that $f'(d) = 0$.

Lecture 22 (Apr 8) - Covered Rudin Chapter 5

Mean Value Theorem:
- Rudin 5.9 (Mean Value Theorem): Let $f,g: [a,b]\rightarrow \Reals$ be continuous function differentiable on $(a,b)$. Then $\exists d\isin (a,b)$ such that $[f(a)-f(b)]g'(d) = [g(a)-g(b)]f'(d)$.
- Rudin 5.10: Let $f: [a,b]\rightarrow \Reals$ be continuous function differentiable on $(a,b)$. Then $\exists d\isin (a,b)$ such that
  $[f(b)-f(a)] = [b-a]f'(d)$.
  - Remark : Mean Value Theorem relates slope at a point to the difference of values of the functions.
- Corollary: Suppose $f: [a,b]\rightarrow \Reals$ be continuous function, $f'(x)$ exists for all $x\isin (a,b)$, and $\lvert f'(x) \rvert \leq M$ for some constant $M$. Then $f$ is uniformly continuous.
- Rudin 5.11: Suppose $f$ is differentiable in $(a,b)$, then
  1. If $f'(x) \geq 0$ for all $x\isin (a,b)$, then $f$ is monotonically increasing.
  2. If $f'(x) = 0$ for all $x\isin (a,b)$, then $f$ is constant.
  3. If $f'(x) \leq 0$ for all $x\isin (a,b)$, then $f$ is monotonically decreasing.
Intermediate Value Theorem:
- Rudin 5.12: Assume $f$ is differentiable over $[a,b]$ with $f'(a)<f'(b)$. Then from each $\lambda \isin (f'(a),f'(b))$, there exists a $d\isin (a,b)$ such that $f'(d)=\lambda$.
L'Hospital's Rule:
- Rudin 5.13: Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)\neq 0$ for all $x\isin (a,b)$, where $-\infty \leq a<b\leq \infty$. Suppose $\frac{f'(x)}{g'(x)} \to A$ as $x\to a$. Then if $f(x) \to 0$ and $g(x) \to 0$ as $x\to a$, or if $g(x) \to +\infty$ as $x\to a$, then $\frac{f(x)}{g(x)}\to A$ as $x\to a$.
  - Example: $f(x)=\frac{\sin(x)}{x}$. $\lim_{x\to 0} \sin(x) = 0$ and $\lim_{x\to 0} x= 0$, then by L'Hospital's Rule, $\lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{\cos(x)}{1} = 1$.
  - Another Example: $f(x)=\frac{\log(x)}{x}$. $\lim_{x\to\infty} x = +\infty$. Thus by L'Hospital's Rule, $\lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{\frac{1}{x}}{1} = 0$.

Homework 10

Let $f:[a,b]\rightarrow \Reals$ be differentiable, then $f'(x)$ cannot have any simple discontinuities.
Even if a sequence of differentiable functions converges uniformly to $f$, $f$ is not guaranteed to be differentiable.

Week 13

Lecture 23 (Apr 13) - Covered Rudin Chapter 5

Higher Order Derivatives:
- Definition: If $f'(x)$ is differentiable at $x_o$, then we define $f“(x_o)=(f')'(x_o)$. Similarly, if the (n-1)th derivative $f^{(n-1)}$ exists and differentiable at $x_o$, we define $f^{(n)}(x_o) = (f^{(n-1)})'(x_o)$
  - Example: $f(x)=\sin(x)$. Then $f'(x)=\cos(x)$, $f”(x)=-\sin(x)$, $f^{(3)}(x)=-\cos(x)$, $f^{(4)}(x)=\sin(x)$, …
- Smooth Function: $f(x)$ is a smooth function on $(a,b)$ if $\forall x\isin (a,b)$, $\forall k\isin \{1,2,…\}$, $f^{(k)}(x)$ exists.
  - Example: $f(x)=\sin(x)$; $f(x)=e^{\frac{-1}{x^2}}$; smooth step function; smooth bump function.
Taylor Theorem:
- Rudin 5.15: Suppose $f$ is a real function on $[a,b]$, $n$ is a positive integer, $f^{(n-1)}$ is continuous on $[a,b]$, $f^{(n)}(t)$ exists for every $t\isin (a,b)$. Let $\alpha, \beta$ be distinct points of $[a,b]$, and define $P(t) = \sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!} (t-\alpha)^k$. Then there exists a point $x$ between $\alpha$ and $\beta$ such that $f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta - \alpha)^n$.
- Taylor Series for a Smooth Function: If $f$ is a smooth function on $(a,b)$, and $\alpha \isin (a,b)$, we can form the Taylor Series:
  $P_{\alpha}(x)=\sum_{k=0}^{\infty} \frac{f^{(k)}(\alpha)}{k!} (x-\alpha)^k$.
  - Remark: RHS is not guaranteed to converge and even if it converges, it may not equal to $f(x)$.

Lecture 24 (Apr 15) - Covered Rudin Chapter 3 and 6

Taylor Series:
- Remark: Taylor expansion is finite term expansion with remainder while Taylor series is infinite sum with no remainder.
- Nth Order Taylor Expansion: $P_{x_o,N}(x) = \sum_{n=0}^{N} f^{n)}(x_o) * \frac{1}{n!} (x-x_o)^n$.
- Definition: Let $N\to\infty$, we write $P_{x_o}(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_o)}{n!} (x-x_o)^n$ (Taylor Series at $x_o$).
- Rudin 3.39: Consider power series $\sum_{n} c_n z^n$, put $\alpha = \lim_{n\to\infty} \sup \lvert c_n \rvert^{\frac{1}{n}}$. Let $R=\frac{1}{\alpha}$ (if $\alpha = 0$ then $R=+\infty$; if $\alpha =+\infty$ then $R=0$), then the series is convergent if $\lvert z \rvert <R $ and the series is divergent if $\lvert z \rvert >R$. Such $R$ is called the radius of convergence.
  - Remark: If $\lvert z \rvert = R$, it depends.
- Example: $f(x)=\frac{1}{1+x^2}$, find its Taylor series based at $x=0$.
  - Directly manipulate, $\forall \lvert x^2 \rvert < 1$, $\frac{1}{1+x^2} = \frac{1}{1-\alpha}= 1+\alpha +{\alpha}^2 + … = 1 + (-x^2) + (-x^2)^2+…=1-x^2+x^4-… +(-x^2)^n+…$
  - Radius of convergence: $\lvert c_n \rvert = 0$ if odd and $\lvert c_n \rvert = 1$ if even. Hence $\lim_{n\to\infty} \sup \lvert c_n \rvert^{\frac{1}{n}} = 1 =\alpha$, then $R=\frac{1}{\alpha}=1$.
- Remark: Taylor expansion is a way to approximate a smooth function near a given point, but the approximation is not uniform over the entire domain of $f$.
Riemann Integral:
- Partition: Let $[a,b]\subset \Reals$ be a closed interval. A partition $P$ of $[a,b]$ is finite set of number in $[a,b]$: $a=x_0 \leq x_1 \leq … \leq x_n=b$. Define $\Delta x_i=x_i-x_{i-1}$.
  - Example: $[10, 20]\subset \Reals$, then a partition would be $P = \{10, 15, 18, 19, 20\}$.
- U(P,f) and L(P,f): Given $f:[a,b]\to \Reals$ bounded, and partion $p = \{x_0 \leq x_1 \leq … \leq x_n\}$, we define $U(P,f) = \sum_{i=1}^{n} \Delta x_i M_i$ where $M_i= \sup \{f(x), x\isin [x_{i-1},x_i]\}$; $L(P,f) = \sum_{i=1}^{n} \Delta x_i m_i$ where $m_i= \inf \{f(x), x\isin [x_{i-1},x_i]\}$.

U(f) and L(f): Define $U(f) = \inf_{P} U(P,f)$ and $L(f)= \sup_{P} L(P,f)$.
- Since $f$ is bounded, hence $\exists m,M\isin \Reals$ such that $m\leq f(x)\leq M$ for all $x\isin [a,b]$, then $\forall P$ partition of $[a,b]$, $U(P,f) \leq \sum_{i=1}^{n} \Delta x_i M = M(b-a)$, and $L(P,f) \geq m(b-a)$, and $m(b-a)\leq L(P,f) leq U(P,f) \leq M(b-a)$.
Riemann Integrable: We say a function $f$ is Riemann integrable if $U(f)=L(f)$,
- Some sufficient conditions:
  1. If $f$ is continuous, then $f$ is Riemann integrable.
  2. If $f$ is monotone, then $f$ is Riemann integrable.

Homework 11

A function $f$ is convex if for any $x,y\isin\Reals$ and any $t\isin [0,1]$, we have $tf(x)+(1-t)f(y)\geq f(tx + (1-t)y)$.
If $f: \Reals\to\Reals$ is a differentiable and convex function, then $f'(x)$ is monotone increasing.
Real and bounded function $\neq $ Riemann integrable.
- Example: $f(x) =1$ if $x\isin \mathbb{Q}$ and $f(x)=0$ if $x\isin \Reals\setminus\mathbb{Q}$. Then $U(f) = 1$ and $L(f) = 0$, since $U(f) \neq L(f)$, then $f$ is not Riemann integrable even if $f$ is real and bounded.

Week 14

Lecture 25 (Apr 20) - Covered Rudin Chapter 6

Stieltjes Integral:
- Weight Function: Let $\alpha: [a.b]\to\Reals$ be a monotone increasing function, then $\alpha$ could be referred to as a weight function for Stieltjes Integral. We refer to $\Delta \alpha _i = \alpha(x_i) - \alpha(x_{i-1})$.
- Basic Notions: Similar to what we defined in Riemann Integral, we define $U(P,f,\alpha) = \sum_{i=1}^{n} M_i \Delta\alpha_i$ and $L(P,f,\alpha) = \sum_{i=1}^{n} m_i \Delta\alpha_i$.
- Stilejes Integrable: If $U(f,\alpha) = L(f,\alpha)$, we say $f$ is integrable with respect to $\alpha$ and write $f\isin \mathscr{R}(\alpha)$ on $[a,b]$.
  - Remark: If $\forall x\isin [a,b]$, $m\leq f(x)\leq M$, then $m (\alpha(b)-\alpha(a)) \leq L(P,f,\alpha) \leq U(P,f,\alpha) \leq M(\alpha(b) - \alpha(a))$.
- Refinement: Let $P$ and $Q$ be 2 partitions of $[a,b]$, then $P$ and $Q$ can be identifies as a finite subset of $[a,b]$. We say $Q$ is a refinement of $P$ if $P\subset Q$ as subsets of $[a,b]$.
  - Example: $[a,b]=[0,10]$. Let $P = \{ 0, 1, 2,3,4,5,6,7,8,9,10\}$, and $Q=\{0, 0.5, 1, 1.5,2,3,4,5,6,7,8,9,9.9,10\}$. We could claim that $Q$ is a refinement of $P$ on $[0,10]$.
- Common Refinement: Let $P_1$ and $P_2$ be 2 partitions of $[a,b]$, then $P_1 \cup P_2$ is the common refinement of $P_1$ and $P_2$.
- Rudin 6.4: If $P'$ is a refinement of $P$, then $L(P',f,\alpha) \leq L(P,f,\alpha)$ and $U(P',f,\alpha) \leq U(P,f,\alpha)$.
- Rudin 6.5: $L(f,\alpha) \leq U(f,\alpha)$.
- Rudin 6.6(Cauchy Condition): $f\isin \mathscr{R}(\alpha) \iff \forall \epsilon >0, \exists P$ partition such that $U(P,f,\alpha)-L(P,f,\alpha) < \epsilon$.
- Rudin 6.7:
  - If Rudin 6.6 holds for $P$, then for any refinement $Q$ of $P$, $U(Q,f,\alpha)-L(Q,f,\alpha) < \epsilon$.
  - If Rudin 6.6 holds for $P$, and let $s_i, t_i\isin [x_{i-1},x_i] \forall i = 1,2,…,n$, then $\sum_{i=1}^{n} \lvert f(s_i) - f(t_i) \rvert \Delta\alpha_i < \epsilon$.
  - If $f\isin\mathscr{R}(\alpha)$ and the above holds, then $\sum_{i=1}^{n} \lvert f(s_i) \Delta\alpha_i - \int fd\alpha \rvert < \epsilon$.
- Rudin 6.8: If $f$ is continuous on $[a,b]$, then $f\isin\mathscr{R}(\alpha)$ on $[a,b]$.
- Rudin 6.9: If $f$ is monotonic on $[a,b]$ and $\alpha$ is continuous, then $f\isin \mathscr{R}(\alpha)$.

Lecture 26 (Apr 22) - Covered Rudin Chapter 6

More on Integrations:
- Rudin 6.10: If $f$ is discontinuous only at finitely many points, and $\alpha$ is continuous where $f$ is discontinuous, then $f\isin \mathscr{R}(\alpha)$.
- Rudin 6.11: Let $f:[a,b]\to [m,M]$ and $\phi:[m,M]\to \Reals$ is continuous. If $f$ is integrable with respect to $\alpha$, then $h=\phi \circ f$ is integrable with respect to $\alpha$.
  - If $f_1,f_2\isin\mathscr{R}(\alpha)$ and $c\isin\Reals$, then $f_1+f_2,cf_1\isin\mathscr{R}(\alpha$, and $\int f_1+f_2 d\alpha = \int f_1 d\alpha + \int f_2 d\alpha$, $\int cf_1 d\alpha = c \int f_1 d\alpha$.
  - If $f,g\isin\mathscr{R}(\alpha)$ and $f(x)\leq g(x)\forall x\isin [a,b]$, then $\int_{a}^{b} fd\alpha \leq \int_{a}^{b} gd\alpha$.
  - If $f\isin\mathscr{R}(\alpha)$ on $[a,c]$, then $f\isin\mathscr{R}(\alpha)$ on $[a,b]$ and on $[b,c]$ if $a < c< b$, and $\int_{a}^{c} fd\alpha = \int_{a}^{b} fd\alpha + \int_{b}^{c} fd\alpha$.
  - If $f\isin\mathscr{R}(\alpha)$ on $[a,b]$, and $\lvert f(x) \rvert \leq M$ on $[a,b]$, then $\lvert \int_{a}^{b} fd\alpha \rvert \leq M(\alpha(b)-\alpha(a))$.
  - If $f\isin\mathscr{R}(\alpha_1)$ and $f\isin\mathscr{R}(\alpha_2)$ and let $c$ be a positive constant, then $f\isin\mathscr{R}(\alpha_1 +\alpha_2)$ and $f\isin\mathscr{R}(c \alpha_1)$ with $\int fd(\alpha_1 + \alpha_2) = \int fd\alpha_1 + \int fd\alpha_2$ and $\int fd(c \alpha_1) = c \int fd\alpha_1$.
- Rudin 6.13:
  - If $f,g\isin\mathscr{R}(\alpha)$, then $fg\isin\mathscr{R}(\alpha)$.
  - If $f\isin\mathscr{R}(\alpha)$, then $\lvert f\rvert \isin\mathscr{R}(\alpha)$ and $\lvert \int_{a}^{b} fd\alpha \rvert \leq \int_{a}^{b} \lvert f\rvert d\alpha$.
- Unit Step function: The unit step function $I$ is defined by $I(x) = 0$ if $x\leq 0$ and $I(x) = 1$ if $x > 0$.
- Rudin 6.15: If $f:[a,b]\to\Reals$ and is continuous at $s\isin [a,b]$ and $\alpha(x) = I(x-s)$, then $\int fd\alpha = f(s)$.
- Rudin 6.16: Suppose $c_n\geq 0$ for $n=1,2,3,…$, $\sum c_n < \infty$, $\{ s_n \}$ is a sequence of distinct points in $(a,b)$, and $\alpha(x) = \sum_{n=1}^{\infty} c_n I(x-s_n)$. Let $f$ be continuous on $[a,b]$, then $\int fd\alpha = \sum_{n=1}^{\infty} c_n f(s_n)$.

Homework 12

If $f$ is a continuous non-negative function on $[a,b]$ and $\int_{a}^{b} fdx=0$, then $f(x) = 0$ for all $x\isin [a,b]$.
If $f,g\isin\mathscr{R}$ are real and bounded functions and $p,q>0$ such that $\frac{1}{p} + \frac{1}{q} = 1$, then $\int fgdx \leq [\int \lvert f \rvert^p dx]^{\frac{1}{p}} [\int \lvert g \rvert^q dx]^{\frac{1}{q}}$.
- If $u,v>0$, then $uv \leq \frac{u^p}{p} + \frac{v^q}{q}$.
- If $f,g$ are non-negative Riemann integrable functions on $[a,b]$, and $\int f^p dx = \int g^q dx = 1$ then $\int fgdx \leq 1$.

Week 15

Lecture 27 (Apr 27) - Covered Rudin Chapter 6

Integration:
- Rudin 6.17: Assume $\alpha$ is increasing and $\alpha' \isin \mathscr{R}$ on $[a,b]$. $f$ is a bounded real function on $[a,b]$, then $f\isin\mathscr{R}(\alpha) \iff f\alpha' \isin \mathscr{R}$ and if so, $\int_{a}^{b} fd\alpha = \int_{a}^{b} f\alpha' dx$.
- Rudin 6.19 (Change of Variable): Suppose $\alpha$ is increasing on $[a,b]$ and $f\isin\mathscr{R}(\alpha)$. Suppose $\phi :[A,B]\to [a,b]$ is a strictly increasing continuous and surjective function. Define $\beta(y) = \alpha(\phi(y))$ and $g(y) = f(\phi(y))$. Then $g\isin\mathscr{R}(\beta)$ and $\int_{A}^{B} gd\beta = \int_{a}^{b} fd\alpha$.
Relation Between Integration and Differentiation:
- Rudin 6.20: Let $f\isin\mathscr{R}$ on $[a,b]$. For $a\leq x\leq b$, let $F(x) = \int_{a}^{x} f(t)dt$. Then F is continuous on $[a,b]$; furthermore if $f(x)$ is continuous at $x_o\isin [a,b]$, then $F(x)$ is differentiable at $x_o$, $F'(x_o) = f(x_o)$.
- Rudin 6.21 (Fundamental Theorem of Calculus): If $f\isin\mathscr{R}$ on $[a,b]$ and if there is a differentiable function $F$ on $[a,b]$ such that $F'=f$, then $\int_{a}^{b} f(x)dx = F(b) - F(a)$.
- Rudin 6.22 (Integration by Parts): Suppose $F,G$ are differentiable, $F',G'$ are integrable, $f=F'$ and $g=G'$. Then $\int_{a}^{b} F(x)g(x)dx = F(b)G(b) - F(a)G(a) - \int_{a}^{b} f(x)G(x)dx$.

Lecture 28 (Apr 29) - Covered Rudin Chapter 7

Uniform Convergence :
- Rudin 7.16: Let $\alpha$ be monotone increasing on $[a,b]$. Suppose $f_n\isin\mathscr{R}(\alpha)$, and $f_n\to f$ uniformly on $[a,b]$. Then $f$ is integrable and $\int_{a}^{b} fd\alpha = \lim_{n\to\infty} \int_{a}^{b} f_n d\alpha$.
- Corollary: Suppose $f_n\isin\mathscr{R}(\alpha)$ and $F(x) = \sum_{n=1}^{\infty} f_n(x)$, the series converges uniformly, then $F\isin\mathscr{R}(\alpha)$ and $\int_{a}^{b} F(x)d\alpha = \sum_{n=1}^{\infty} \int_{a}^{b} f_n(x)d\alpha$.
- Theorem: Suppose $\{ f_n \}$ is a sequence of differentiable functions on $[a,b]$ such that $f_n'(x)$ converges uniformly to $g(x)$ and $\exists x_o\isin [a,b]$ such that $\{f_n(x_o)\}$ converges. Then $f_n(x)$ converges to some function $f$ uniformly and $f'(x)=g(x)=\lim_{n\to\infty} f_n'(x)$.

Questions

I understand the visualization of this recursive sequence, but to $\sqrt{5}$?
In general, how to prove a set is infinite(in order to use theorem 11.2 in Ross)?
Is there a way/analogy to understand/visualize the closure of a set?
Is there a way to actually test if a set is compact or not instead of merely finding finite subcovers for all open covers?
Rudin 4.6 states that if $p\isin E$, a limit point, and $f$ is continuous at $p$ if and only if $\lim_{x\to p} f(x) = f(p)$. Does this theorem hold for $p\isin E$ but not a limit point of $E$?
How is the claim at the bottom proved?
Could we regard the global maximum as the maximum of all local minimums?
Under what circumstance would Taylor Theorem be essential to our proof, since for the question appeared in HW11 Q2, I really do not think using Taylor Theorem is a better way to prove the claim?
In order for a Taylor series to converge ($\sum_{n} c_n z^n$), $\lvert z \rvert < R$ where $R$ is the radium of convergence. But if $\lvert z \rvert = R$, how can we tell?
If we are claiming $f$ is continuous on $[a,b]$, , i.e. do we just extend our interval to the left side of $a$ and right side of $b$ to do so?
What information can we extract from the line “$f$ has a bounded first derivative (i.e. $\lvert f' \rvert \leq M$ for some $M>0$)”?
How sequentially compact without proving that it is compact? (Starting from ms too complicated to take into account all sequences in the set)
If $a_{n+1} = \cos (a_n)$ and choose $a_1$ such that $0 < a_1 < 1$, is $a_n$ a ?
Does uniform convergence on a sequence of functions $\{f_n\}$ in $F$ to $f$ imply ?
If $\sum f_n$ converges uniformly, does it imply $f_n$ satisfies Weiestrass M-test?
For the alternating series test, if instead of sequence of numbers we have sequence of functions and those functions $\{ f_n \}$ satisfies $f_1 \geq f_2 \geq f_3 …$ and $f_n \geq 0$ for all $x\isin X$, $\lim f_n = 0$, does that mean $\sum_{n} (-1)^n f_n$ converges uniformly?
What is measure zero? (Related to Lebesgue measure and volume of open balls)
Is there any covering for a compact set that satisfies total length of the finite subcovers for the set is less than $\lvert b-a \rvert$? (Answer is no)
Question 16 on Prof Fan's practice exam.
This is my solutions towards the practice exam: practice_solutions.pdf

Table of Contents

Martin Zhai's Review Note

Content Summary

Week 1

Lecture 1 (Jan 19) - Covered Ross Section 1.1, 1.2, 1.3

Lecture 2 (Jan 21) - Covered Ross Section 1.4

Homework 1

Week 2

Lecture 3 (Jan 26) - Covered Ross Section 2.7, 2.9

Lecture 4 (Jan 28) - Covered Ross Section 2.9, 2.10

Homework 2

Week 3

Lecture 5 (Feb 2) - Covered Ross Section 2.10

Lecture 6 (Feb 4) - Covered Ross Section 2.11

Homework 3

Week 4

Lecture 7 (Feb 9) - Covered Ross Section 2.11

Lecture 8 (Feb 11) - Covered Ross Section 2.12

Homework 4

Week 5

Lecture 9 (Feb 16) - Covered Ross Section 2.13

Lecture 10 (Feb 18) - Midterm 1

Homework 5

Week 6

Lecture 11 (Feb 23) - Covered Ross Section 2.13, Rudin Chapter 2

Lecture 12 (Feb 25) - Covered Ross Section 2.14, 2.15

Homework 6

Week 7

Lecture 13 (Mar 2) - Covered Rudin Chapter 4

Lecture 14 (Mar 4) - Covered Rudin Chapter 4

Homework 7

Week 8

Lecture 15 (Mar 9) - Covered Ross Section 3.19 Rudin Chapter 2 and 4

Lecture 16 (Mar 11) - Covered Rudin Chapter 2 and 4

Homework 8

Week 9

Lecture 17 (Mar 16) - Covered Rudin Chapter 4 and 7

Lecture 18 (Mar 18) - Covered Rudin Chapter 7

Homework 9

Week 10

Spring Break (Mar 23, Mar 25)

Week 11

Lecture 19 (Mar 30) - Review for Midterm 2

Lecture 20 (Apr 1) - Midterm 2

Week 12

Lecture 21 (Apr 6) - Covered Rudin Chapter 5

Lecture 22 (Apr 8) - Covered Rudin Chapter 5

Homework 10

Week 13

Lecture 23 (Apr 13) - Covered Rudin Chapter 5

Lecture 24 (Apr 15) - Covered Rudin Chapter 3 and 6

Homework 11

Week 14

Lecture 25 (Apr 20) - Covered Rudin Chapter 6

Lecture 26 (Apr 22) - Covered Rudin Chapter 6

Homework 12

Week 15

Lecture 27 (Apr 27) - Covered Rudin Chapter 6

Lecture 28 (Apr 29) - Covered Rudin Chapter 7

Questions