# Lecture Notes

MATH 334

## Lecture 1: Background and Notation

• Plan for fall – chapters 1 to 4 of the Folland textbook
• Pre-midterm I: chapter 1
• Pre-midterm 2: chapters 2 and 3
• Reading responses, free points, 5%
• HW, posted weekly, assigned on Wednesday and due the following Wednesday at 11:59, 25%
• Midterms: 20%
• Final: 30%
• All exams are inclass.

Historical Background

• Fourier and the crisis of the 19th century
• Fourier’s ideas animate the course
• 1795: joined Napolean’s expedition to Cairo
• Mathematics and imperialism
• Becomes very interested in heat
• How does heat diffuse across an object?
• The total heat distribution should solve $$\frac{\partial^2 u}{\partial x^2} = \partial w^u$$ and $$u(0, x) = f(x)$$
• Fourier found a general method for solving this – partial differential equations
• His method requires assuming a special form: you can write your initial function as a sum of cosines. Then the distribution $$u$$ can be found easily.
• However this assumes that $$f(x) = 0; x = \pm 1$$, and that the function is continuous
• You can’t solve $$f(x) = 1$$, for instance: you can’t solve for heat when the temperature applied to a bar is constant.
• What’s the solution? Bold idea: larger sums of cosines can predictably better approximate functions which cannot themselves be written as a finite sum.
• For instance: write $$1 = \frac{4}{\pi} \left[ \cos \left( \frac{\pi x}{2} \right) - \frac{1}{3} \cos \left( \frac{3\pi x}{2} \right) + \frac{1}{5} \cos \left( \frac{5 \pi x}{2} \right) + ... \right]$$, for $$x \in (-1, 1)$$
• You cannot do this with a Taylor epansion, you can’t write constant functions in terms of nonconstant functions.
• You immediately begin to run into all sorts of funny paradoxes
• Notice that $$\cos \left( \frac{(2n-1) \pi (x+2)}{2} \right) = \cos \left( \frac{(2n-1) \pi x}{2} + (2n-1) \pi \right)$$: this amounts to a phase shift of $$\pi$$ in each instance
• Therefore you could write $$-1$$ simply by shifting, so the formula is the same but the interval is now $$(1, 3)$$
• All you did was add continuous things, but you got something totally noncontinuous…
• Response to Fourier: this is not a function. If you could call that a function and consider it a representation of such a familiar function, maybe Weierstrauss was onto something when he claimed $$W(x) = \cos(\pi x) + \frac{1}{2} \cos (13 \pi x) + \frac{1}{4} \cos (169 \pi x) + ... = \sum_{j=0}^\infty \frac{\cos(13^j \pi x)}{2^j}$$.
• This series converges and is a function
• It is continuous in $$x$$
• It is differentiable nowhere – why can’t I just take the derivatives of each term? – you can’t
• “A monstrosity”
• What does this do to our conception of continuity?
• Several questions follow:
• What is a function in the first place?
• When can a function be represented as a Fourier series, ideally one that converges?
• When can a series be differentiated or integrated, term-by-term?
• What is differentiation? What is integration?
• What is continuity? WHat is convergence?
• This class will be occupied by these questions

Notation

• Basic
• $$\forall$$, for all
• $$\exists$$, there exists
• Summation: $$\sum_{n=1}^k a_n = a_1 + a_2 + ... + a_k$$
• $$\mathbb{N}$$, the naturals / whole numbers. $$\mathbb{N} = \{1, 2, 3, ...\}$$
• $$\mathbb{Z}$$, the integers. $$\mathbb{Z} = \{..., -2, -1, 0, 1, 2, ...\}$$
• $$\mathbb{Q}$$, the rationals. $$\mathbb{Q} = \left\{ \frac{p}{q} \mid p, q \in \mathbb{Z}, q \neq 0 \right\}$$
• $$\mathbb{R}$$, the reals.
• Set theory
• $$\Omega$$, the universal set
• Union: $$A \cup B = \{x \in \Omega \vert x \mid x \in A \vee x \in B\}$$
• Intersection: $$A \cap B = \{x \in \Omega \vert x \mid x \in A \wedge x \in B\}$$
• Complement: $$A^C = \{x \in \Omega \vert x \mid x \notin A\}$$
• Set difference: $$A \setminus B = \{x \in \Omega \vert x \mid x \in A \wedge x \notin B\}$$
• Can also be written as $$A \setminus B = A \cap B^C$$
• Big-Union: $$\bigcup_{j=1}^k A_j = A_1 \cup A_2 \cup ... \cup A_k = \{ x \| \exists j, x \in A_j \}$$
• Big-Intersection: $$\bigcap_{j=1}^k A_j = A_1 \cap A_2 \cap ... \cap A_k = \{ x \| \forall j, x \in A_j \}$$
• Mappings and functions
• Function from a set $$A$$ to a set $$B$$ is an assignment of every element in $$A$$ to exactly one element in $$B$$, denoted by $$f: A \to B$$
• $$A$$: domain of $$f$$
• $$B$$: codomain of $$f$$
• $$f(A) = \{ y \in B \vert \exists x \in A, f(x) = y \} \subset B$$, the image/range of $$f$$
• $$f$$ is injective / 1-to-1 if $$a_1 \neq a_2 \to f(a_1) \neq f(a_2)$$
• $$f$$ is surjective / onto if $$\forall b \in B, \exists A \in A : f(A) = b$$
• $$f$$ is bijective if both injective and surjective. Bijectivity gives you one-on-one pairing. Bijection is pairing but also works for infinite sets.
• If $$f: A \to B$$ and $$g : B \to C$$, we can write their composition as $$g \circ f : A \to C$$, where $$(g \circ f)(x) = g(f(x))$$. Requirement: the codomain of $$f$$ must be a subset of the domain of $$g$$.
• If $$f : A \to B$$ and $$g : B \to A$$, $$f$$ is invertible if $$\exists g: B \to A : g \circ f$$ s.t. $$A \to A$$ satisfies $$\forall x \in A, g \circ f(x) = x$$ an $$f \circ g: B \to B$$ satisfies $$\forall y \in B, f \circ g(y) = y$$. Basically: it sends every element back to itself.
• Bijectivity implies invertability.
• Definition: $$g = f^{-1}$$
• Euclidean spaces and vectors
• The reals $$\mathbb{R}$$
• We can also consider tuples of reals: $$\mathbb{R}^n$$ for $$n \in \mathbb{N}$$, the set of ordered tuples of reals. $$\mathbb{R}^n = \{(x_1, x_2, ..., x_n) \vert \forall j, x_j \in \mathbb{R}\}$$
• Vectors also used to describe $$n$$-tuples
• Vector arrows are menat to capture that an $$n$$-tuple can also describe a direction and magnitude
• The magnitude / norm of a vector / $$n$$-tuple is $$\Vert \vec{x} \Vert = \sqrt{x_1^2 + x_2^2 + ... + x_n^2}$$
• Vector addition: $$\vec{x} + \vec{y} = (x_1 + y_1, x_2 + y_2, ..., x_n + y_n)$$
• Scalar multiplication: $$\lambda \vec{x} = (\lambda x_1, \lambda x_2, ..., \lambda x_n)$$
• Dot product / inner product: $$\vec{x} \cdot \vec{y} = x_1 y_1 + x_2 y_2 + ... + x_n y_n$$
• Note that the norm is defined in terms of a dot product: $$\Vert \vec{x} \Vert = \sqrt{\vec{x} \cdot \vec{x}}$$

The Cauchy-Schwarz Inequality

• If $$\vec{a}, \vec{b} \in \mathbb{R}^n$$, $$\vert \vec{a} \cdot \vec{b} \vert \leq \Vert \vec{a} \Vert \Vert \vec{b} \Vert$$
• Going to be your best friend in this class
• This proof, which seems to work only in Euclidean space, is very much broadly applicable.
• Proof:
1. If $$\vec{b} = \vec{0}$$, then both sides are zero. The inequality holds.
2. Now, assume $$\vec{b} \neq \vec{0}$$. Consider the function $$f : \mathbb{R} \to \mathbb{R}_{\ge 0}$$ where $$f(t) = \vert \vert \vec{a} - t \vec{b} \vert \vert^2 = (\vert{a} \cdot t \vert{b}) \cdot (\vec{a} - t\vec{b}) = \vert \vert \vec{a} \vert \vert^2 - 2t (\vec{a} \cdot \vec{b}) + t^2 \vert \vert \vec{b} \vert \vert^2$$
3. This quadratic function has a minimum at $$t = \frac{\vec{a} \cdot \vec{b}}{\vert \vert b \vert \vert ^2}$$
4. Plugging this into $$f$$ gives you $$\vert \vert \vec{a} \vert \vert^2 - \frac{(\vec{a} \cdot \vec{b})^2}{\vert \vert \vec{b} \vert \vert^2}$$
5. Since $$f(t) \ge 0$$, $$\vert \vert \vec{a} \vert \vert^2 - \frac{(\vec{a} \cdot \vec{b})^2}{\vert \vert \vec{b} \vert \vert^2} \ge 0$$, so $$\vert \vert \vec{a} \vert \vert^2 \vert \vert \vec{b} \vert \vert^2 \ge (\vec{a} \cdot \vec{b})^2$$
6. Take square roots to get our inequality.

## Lecture 2: Euclidean Spaces, Open Sets, Limits and Continuity, Sequences

• Triangle inequality: $$\forall \vec{a}, \vec{b} \in \mathbb{R}^n$$, $$\| \vec{a} + \vec{b} \| \le \| \vec{a} \| + \| \vec{b} \|$$
• Norm of sum is less than or equal to the sum of the norms
• Proof
1. Since $$(\| \vec{a} + \vec{b} \|)^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \|a\|^2 + 2\vec{a}\vec{b} + \|b\|^2$$
2. Appplying CS to the middle term: $$\le \| \vec{a} \|^2 + 2 \| \vec{a} \| \| \vec{b} \| + \| \vec{b} \|^2$$
3. Simplify: $$( \| \vec{a} \| + \| \vec{b} \|)^2$$
• Why is it called the triangle inequality?
• Define distance between two vectors using a norm: it’s faster to go along the direct line (‘hypotenuse’) than the legs of the triangle
• Formally: $$d(\vec{a}, \vec{b}) = \| \vec{a} - \vec{b} \| = \sqrt{(a_1 - y_1)^2 + (x_2 - y_2)^2 + ... }$$
• We use $$\vec{x} \cdot \vec{y}$$ or $$\langle \vec{x}, \vec{y} \rangle$$ to denote the dot product
• The angle between vectors describes some plane. By the Cauchy-Schwartz theorem, $$\frac{\vec{ x} \cdot \vec{y}}{\| \vec{x} \| \|\vec{y} \|}$$ is between $$-1$$ and $$1$$, so it is a cosine. Therefore, $$\cos \theta_{\vec{x}, \vec{y}} = \frac{\vec{x} \cdot \vec{y}}{\| \vec{x} \| \| \vec{y} \|}$$
• $$\vec{x}, \vec{y}$$ are perpendicular iff $$\vec{x} \cdot \vec{y} = 0$$, in which case they are orthogonal.
• As soon as you have a notion of inner product, you can define the angle between vectors.
• Useful inequality: if you have a vector $$\vec{x} \in \mathbb{R}^n$$, $$\vec{x} = (x_1, ..., x_n)$$, then its maximum of the absolute value of the vector is less than the norm of $$\vec{x}$$, and itself less than $$\sqrt{n} max \{ \vert x_1 \vert, ..., \vert x_n \vert \}$$.
• $$\| \vec{x} \|_{\infty} = max \{ \vert x_1 \vert, ..., \vert x_n \vert \}$$, called the $$L^\infty$$ norm.
• This is to be contrasted with the usual L-2 norm, $$\Vert \vec{x} \Vert_2$$, which is the square root of the sum of the squares of the components.
• Norms are generalizable across vector spaces.

Subsets of $$\mathbb{R}^n$$ - introduction to topology

• Topology tries to abstract the notion of “nearness” or “closeness” in a space, without specific notions of angle, length, etc.
• What do you need to describe a space? You need to say when points are near, and you don’t need to do so quantiatively.
• Definition: Let $$\vec{y}$$ be a point in $$\mathbb{R}^n$$ and $$r > 0$$. Define an open ball $$B_r(\vec{y})$$ to be the collection of all points in $$\mathbb{R}^n$$ such that the distance from every point to the center is less than $$r$$.
• Formally: $$B_r(\vec{y}) = \{ \vec{x} \in \mathbb{R}^n \vert \| \vec{x} - \vec{y} \| < r \}$$
• A set $$S \subset \mathbb{R}^n$$ is bounded if it is contained in some ball centered at the origin $$\vec{0}$$. There exists a sufficiently large radius for an enclosing ball.
• Definition: a point $$\vec{x} \in \mathbb{R}^n$$ is an interior point of $$S$$ if there exists a ball centered at some point which is strictly contained in that set, where $$\vec{x}$$ is contained in that ball.
• Definition: the set of points in $$S$$ which are interior points of $$S$$ is called the interior of $$S$$, denoted $$S^\text{int}$$.
$S^\text{int} = \{ \vec{x} \in S \vert \exists r > 0, B_r(\vec{x}) \subset S \}$
• Examples of interiors:
• If $$S = Br(\vec{y})$$, then $$S^\text{int} = S$$
• If $$S = \{ \frac{1}{n} \in \mathbb{R} \vec n \in \mathbb{N} \}$$ (subset of a line), then $$S^\text{int} = \emptyset$$, since there is no ball around any point which is contained in $$S$$. Also note that $$0 \notin S$$.
• Definition: a point $$\vec{x} \in \mathbb{R}^n$$ is a boundary point of $$S$$ if every ball centered at $$\vec{x}$$ contains a point in $$S$$ and a point not in $$S$$.
• Definition: the set of boundary points of $$S$$ is called the boundary of $$S$$, denoted $$\partial S$$.
• Example 1: if $$S = Br(\vec{y})$$, $$\partial S = \{ \vec{x} \in \mathbb{R}^n \vert \| \vec{x} - \vec{y} \| = r \}$$, the sphere of radius $$r$$ centered at $$\vec{y}$$.
• Example 2: if $$S = \{ \frac{1}{n} \in \mathbb{R} \vert n \in \mathbb{N} \}$$, the boundary is $$\{ 0 \}$$. This is because every ball around $$0$$ contains points in $$S$$ and points not in $$S$$. Or maybe not… depends on if you need to look at the center or not. Otherwise the definition is kind of empty… need to clarify.
• Definition: the closure of a set $$\bar{S} = S \cup \partial S$$. The closure is the set of all points in $$S$$ and all boundary points of $$S$$.
• Example 1: if $$S = Br(\vec{y})$$, then $$\bar{S} = \{ \vec{x} \in \mathbb{R}^n \vert \| \vec{x} - \vec{y} \| \le r \}$$, the closed ball of radius $$r$$ centered at $$\vec{y}$$.
• Example 2: if $$S = \{ \frac{1}{n} \in \mathbb{R} \vert n \in \mathbb{N} \}$$, then $$\bar{S} = S \cup \{ 0 \}$$.
• Definition: $$S$$ is open if it contains no boundary points. $$S$$ is closed if it contains all of its boundary points.
• Example 1: $$Br(\vec{y})$$ is open
• Example 2: $$\{ \frac{1}{n} \in \mathbb{R} \vert n \in \mathbb{N} \}$$ is closed
• Example 3: $$\{ \frac{1}{n} \in \mathbb{R} \vert n \in \mathbb{N} \} \cup \{ 0 \}$$ is neither open nor closed
• Proposition: If $$S \subset \mathbb{R}^n$$,
• $$S$$ is open iff it is equal to its interior
• $$S$$ is closed iff its complement is open
• Proof:
1. An element of $$S$$ is iether an interior point or a boundary point.
2. $$S$$ is open iff every point is interior.
3. $$S = \bar{S} = S \cup \partial S \iff \partial S \subset S \implies \partial(S^C) \subset S$$ (boundaries are shared between a set and its complement) $$\iff S^C$$ contains no boundary points
• Example:
• If $$S = \emptyset$$, this is a set which is both open and closed.
• If $$S = \mathbb{Q}$$, the set of fractions, this set is neither open nor closed. The interior is empty, $$S^{\text{int}} = \emptyset$$; yet the closure of $$S$$ is $$\mathbb{R}$$, so $$S$$ is not closed.

Limits

• “You can’t lift your pencil”
• Definition: let $$f: \mathbb{R}^n \to \mathbb{R}$$ be a real-valued function.
$\lim_{\vec{x} \to \vec{a}} = L \text{ if } \forall \epsilon_{> 0}, \exists \delta_{> 0} : \vert f(\vec{x}) - L \vert < \epsilon \text{ whenever } \Vert \vec{x} - \vec{a} \Vert < \delta$
• $$a$$ whenever $$b$$: $$b \to a$$
• We can replace $$\Vert \vec{x} - \vec{a} \Vert < \delta$$ with the $$L^\infty$$ norm:
$\Vert \vec{x} - \vec{a} \Vert_{\infty} = \max \{ \vert x_1 - a_1 \vert, ..., \vert x_n - a_n \vert \} < \frac{\delta}{\sqrt{n}}$
• You can also consider a limit on a subset, $$f : S \sub \mathbb{R}^n \to \mathbb{R}$$.
$\lim_{\vec{x} \to \vec{a}, \vec{x} \in S} f(\vec{x}) = L \text{ if } \forall \epsilon_{> 0}, \exists \delta_{> 0} : \vert f(\vec{x}) - L \vert < \epsilon \text{ whenever } \Vert \vec{x} - \vec{a} \Vert < \delta \text{ and } \vec{x} \in S$

Continuity

• Definition: $$f(\vec{x})$$ is continuous at $$\vec{a}$$ if the limit as $$\vec{x}$$ approaches $$\vec{a}$$ of $$f(\vec{x})$$, and $$f(\vec{a})$$, both exist and are equal.
• Definitions also make sense for vector-valued functions: you can go from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$. All you need is a corresponding notion of norm, which you do in any Euclidean space / dimension. The only thing that changes is to replace $$\Vert f(\vec{x}) - \vec{L} \Vert_n < \epsilon$$, where $$n$$ is the dimension of the value space.
• Alternatively, you can find the limit of each indvididual component, considering the sub-function $$f_{j}: \mathbb{R}^m \to \mathbb{R} \sub \mathbb{R}^n$$.
• The notion of “getting close” is very subtle in high dimensions – it’s harder than just taking left and right limits.
• Example 1: $$f : \mathbb{R}^2 \to \mathbb{R}, f(x, y) = \frac{xy}{x^2 + y^2}$$. This function is everywhere bounded. $$\vec f(x, y) \vec \le 2$$. A consequence of Cauchy-Schwartz. But the function fails to be continuous at the origin. The limit as $$\vec{x} \to \vec{0}$$ does not exist. By taking linear lines of approach, the line is constant, but the line in particular changes this constant value
• Theorem. Let $$f_1(x, y) = x+ y$$, $$f_2(x, y) = xy$$, $$g(x) = \frac{1}{x}$$. $$f_1, f_2 : \mathbb{R}^2 \to \mathbb{R}$$; $$g: \mathbb{R} \setminus \{0\} \to \mathbb{R}$$. Proof:
1. Let $$(a, b) \in \mathbb{R}^2$$
2. Given $$\epsilon > 0$$, we must show that there exists some $$\delta$$ such that if $$\vert x - a \vert < \delta$$ and $$\vert y - b \vert < \delta$$, then $$\vert (x + y) - (a + b) \vert < \epsilon$$.
3. Let us choose $$\delta = \frac{1}{2} \epsilon$$.
4. $$\vert (x + y) - (a + b) \vert = \vert (x - a) + (y - b) \vert \le \vert (x -a ) \vert + \vert (y - b) \vert$$. Via triangle inequality
5. We have that $$\delta + \delta = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$.
6. Revisit proof!
• Theorem. Say $$f : \mathbb{R}^n \to \mathbb{R}^m$$ and $$g = \mathbb{R}^m \to \mathbb{R}^k$$. If $$f$$ and $$g$$ are continuous at $$\vec{a}$$, then $$g \circ f$$ is continuous at $$\vec{a}$$. Proof:
1. Let $$\epsilon > 0$$.
2. Since $$g$$ is continuous at $$f(\vec{a})$$, there exists $$\delta_1 > 0$$ such that $$\Vert g(\vec{y}) - g(f(\vec{a})) \Vert < \epsilon$$ whenever $$\Vert \vec{y} - f(\vec{a}) \Vert < \delta_1$$.
3. Since $$f$$ is continuous at $$\vec{a}$$, there exists $$\delta_2 > 0$$ such that $$\Vert f(\vec{x}) - f(\vec{a}) \Vert < \delta_1$$ whenever $$\Vert \vec{x} - \vec{a} \Vert < \delta_2$$.
4. Let $$\delta = \delta_2$$.
5. Then $$\Vert g(f(\vec{x})) - g(f(\vec{a})) \Vert < \epsilon$$ whenever $$\Vert \vec{x} - \vec{a} \Vert < \delta$$.
• Later, with sequences: $$a_n = 2a_{n-1} - a_n-2 + 2$$. Claim: $$a_n = n^2$$. You can prove this with induction.

## Lecture 3: Sequences, Induction, the Completeness Axiom

Sequences

• A sequence is a collection of mathematical objects which are indexed by the natural numbers
• We denote a sequence by $$\{x_k \}^\infty_{k=1}$$
• A set has no notion of order, whereas a sequence does have order
• e.g. the sequence 1, 4, 9, 16, … can be written as $$\{ n^2 \}_{n=1}^\infty$$
• e.g. a sequence of intervals $$(-1, 1), (-1/2, 1/2), (-1/3, 1/3), ...$$ can be written as $$\{ (-1/n, 1/n) \}_{n=1}^\infty$$
• You can also define sequences inductively: $$F_1 = 1, F_2 = 1, F_n = F_{n+2} = F_{n+1} + F_n$$
• A nice proof technique: an inductive proof. Useful for proving statements of the form $$\forall n \in \mathbb{N}, P(n)$$ is true. First, show $$P(1)$$. Then, show $$P(n) \implies P(n+1)$$ (inductive step)
• Example: consider $$a_1 = 1, a_2 = 4, a_n = 2a_{n-1} - a_{n-2} + 2$$ for $$n \ge 3$$. Claim: $$a_n = n^2$$.
• Sequences of real numbers / vectors, $$\{ x_n \}_{n=1}^\infty \subseteq \mathbb{R}^2$$ converge to $$\vec{a} \in \mathbb{R}^n$$ if $$\forall \epsilon > 0, \exists N > 0$$ such that $$\Vert \vec{x}_n - \vec{a} \vert < \epsilon, \forall n > N$$. If this is so, then $$\vec{x}_n \to \vec{a}$$. if $$\{ \vec{x}_{n=1}^\infty \}$$ does not converge to any vector, the sequence diverges.
• We can describe limits / continuity in the form of sequences
• e.g. the sequence $$\{ 1/k \}_{k=1}^\infty$$ converges to 0.
• e.g. the sequence $$\{ (-1)^k \}_{k=1}^\infty$$ is divergent.
• The following are equivalent.
1. $$f : \mathbb{R}^n \to \mathbb{R}^n$$ is continuous.
2. $$\forall \vec{x}_0 \in \mathbb{R}^n, \forall \{ \vec{x}_n \}_{n=1}^\infty$$ converging to $$\vec{x}_0$$, $$\{ f(\vec{x}_n ) \}_{n=1}^\infty$$ converges to $$f(\vec{x}_0)$$.
3. $$\forall U \subseteq \mathbb{R}^k$$ an open set, the preimage $$f^{-1}(U) = \{ \vec{x} \in \mathbb{R}^n \vert f(\vec{x}) \in U \} \subseteq \mathbb{R}^n$$ is also open.
• Proof that 1 implies 2:
1. Let $$\vec{x}_0 \in \mathbb{R}^n$$ be arbitrary, and assume $$f$$ is continuous at $$\vec{x}_0$$.
2. Let $$\{ \vec{x}_n \}_{n=1}^\infty$$ be a sequence converging to $$\vec{x}_0$$.
3. By continuity, $$\forall \epsilon > 0$$, $$\exists \delta > 0$$ such that $$\Vert f(\vec{x}) - f(\vec{x}_0) \Vert > \epsilon$$ whenever $$\Vert \vec{x} - \vec{x}_0 \Vert < \delta$$.
4. Since $$\vec{x}_n \to \vec{x}_0$$, $$\exists N > 0$$ such that $$\Vert \vec{x}_n - \vec{x}_0 \Vert < \delta, \forall n > N$$.
• Proof that 2 implies 1 (prove the logicallay equivalent contrapositive statement):”not 1 implies not 2”. If $$\exists \vec{x}_0 \in \mathbb{R}^n$$ such that $$f(\vec{x})$$ is not continuous at $$\vec{x}_0$$, we’ll prove $$\exists \vec{x}_n \to \vec{x}_0$$ and $$f(\vec{x}_n)$$ does not converge to $$f(\vec{x}_0)$$.
1. Since $$f$$ is not continuous at $$\vec{x}_0$$, $$\exists \epsilon_0 > 0, \forall \delta > 0, \exists \vec{x}_{\delta} \in \mathbb{R}^n$$ where $$\Vert \vec{x}_\delta - \vec{x}_0 \Vert < \delta$$ and $$\Vert f(\vec{x}_\delta) - f(\vec{x}_0) \Vert > \epsilon_0$$
2. For each of $$\delta = 1, 1/2, 1/3, ..., 1/k, ...$$ choose such a vector $$\vec{x}_k$$ as above.
3. Although the sequence $$\vec{x}_k \to \vec{x}_0$$, $$f(\vec{x}_k)$$ does not converge to $$f(\vec{x}_0)$$.
• Proof that 1 implies 3:
1. Say $$U \subseteq \mathbb{R}^l$$ is open. We will show that every point in $$f^{-1}(U)$$ is contained in a ball, contained in the preimage.
2. If $$\vec{a} \in f^{-1}(u)$$, then $$f(\vec{a}) \in U$$, since $$U$$ is open, $$\exists r > 0$$ such that $$B_r(f(\vec{a})) \subseteq U$$; i.e., $$\forall \vec{y} \in \mathbb{R}^k$$, $$\Vert \vec{y} - f(\vert{a}) \Vert < r, \vec{y} \in U$$.
3. By continuity of $$f$$, for $$r > 0$$, $$\exists \delta > 0$$ such that $$\Vert f(\vec{x}) - f(\vec{a}) \Vert < r$$ whenever $$\Vert \vec{x} - \vec{a} \Vert < \delta$$; i.e., $$B_\delta(\vec{a}) \subseteq f^{-1}(U)$$, so $$f^{-1}(U)$$ is open.
• Preimage definition: $$f^{-1}(U) = \{ \vec{x} \in \mathbb{R}^n \vert f(\vec{x}) \in U \}$$
• Proof that 3 implies 1.
1. Let $$\vec{a} \in \mathbb{R}^n$$ be arbitrary. $$\forall \epsilon > 0, B_\epsilon ( f(\vec{a})) \subseteq \mathbb{R}^k$$ is open.
2. By 3, $$f^{-1}(B_\epsilon (f(\vec{a})))$$ is open. By openness, $$\exists \delta > 0$$ such that $$B_\delta(\vec{a}) \subseteq f^{-1} (B_\epsilon (f(\vec{a})))$$.
3. So, if $$\Vert \vec{x} - \vec{x} \Vert < \delta$$, then $$\Vert f(\vec{x}) - f(\vert{a}) \Vert < \epsilon$$, so $$f$$ is continuous at $$\vec{a}$$.
• Corollary: If $$f : \mathbb{R}^n \to \mathbb{R}^k$$ is continuous and $$V \subset \mathbb{R}^k$$, then $$f^{-1}(V)$$ (the preimage) is also closed.
• Example: if $$C > 0$$, the sequence $$\{ C^n / n! \}_{n=1}^\infty$$ converges to 0.
1. Choose $$N > 2C$$, then $$\forall n > N$$ we have $$0 < C^n / n! = C^N / N! \cdot C / (N+1) \cdot C / (N+2) \cdot ... \cdot C / n$$
2. We can upper-bound the rest of these $$C^n / N! \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot ... \cdot \frac{1}{2}$$
3. These factors are $$n - N$$ in number, and going to zero as $$n \to \infty$$. Therefore the sequence converges to zero.
• Theorem. $$A \subseteq \mathbb{R}^n$$ is a subset. Then $$\vec{x} \in \bar{A} \iff A \text{ intersects every neighborhood of } \vec{x}$$. (A point is in the closure of a set if the set intersects every neighborhood of the point.) Recall that $$\vec{x}$$ has a neighborhood $$U$$ if $$\vec{x}$$ is interior to $$U$$. We will prove the contrapositive: a point is not in the closure of a set iff the point has a neighborhood not intersecting $$A$$.
1. Say $$\vec{x} \notin \bar{A}$$, then $$\mathbb{R}^n \setminus \bar{A} = U$$ is a neighborhood which does not intersect $$A$$.
2. Say $$\exists U$$, a neighborhood of $$\vec{x}$$ not meeting $$A$$. Then $$\mathbb{R}^n \setminus U$$ is a closed set containing $$A$$, implying that the closure of $$A$$ is a subset of $$\mathbb{R}^n \setminus U$$. Since $$\vec{x} \notin \mathbb{R}^n \setminus U, \vec{x} \notin A$$.
• Theorem. If $$S \subseteq \mathbb{R}^n$$, and $$\vec{x}_0 \in \mathbb{R}^n$$, $$\vec{x}_0 \in \bar{S} \iff \exists \{ \vec{x}_n \}_{n=1}^\infty \subseteq S \text{ where } \vec{x}_n \to \vec{x}_0$$.
• “It’s not possible to take a sequence of points and escape a closed ball”
1. Leftwards proof. If $$\vec{x}_n \to \vec{x}_0$$ and $$\forall n, \vec{x}_n \in S$$, then every neighborhood of $$\vec{x}_0$$ contains an element of $$S$$, namely $$\vec{x}_n$$ for $$n$$ sufficiently large.
2. Since $$\vec{x}_n \to \vec{x}_0$$, $$\forall \epsilon > 0$$, $$\exists N > 0$$ such that $$\Vert \vec{x}_n - \vec{x}_0 \Vert < \epsilon$$, $$\forall n > N$$; any neighborhood of $$\vec{x}_0$$ contains such a ball and $$\vec{x}_n \in B_\epsilon ( \vec{x}_0 ) \subseteq U$$, a neighborhood of $$\vec{x}_0$$.
3. So $$\vec{x}_n \in S \cap U$$ for every neighborhood $$U$$, and thus $$\vec{x}_0 \in S$$.
4. Rightwards proof. Say we have a point which is in the closure $$\vec{x}_0 \in \bar{S}$$. If $$\vec{x}_0 \in S$$, then in fact we can take a constant sequence $$\exists \{ \vec{x}_0 \}_{n=1}^\infty$$ which converges to $$\vec{x}_0$$.
5. If $$\vec{x}_0 \in \bar{S}$$ but $$\vec{x}_0 \notin S$$, then $$B_{1/k} (\vec{x}_0)$$ is a neighborhood of $$\vec{x}_0$$, $$\forall k \in \mathbb{N}$$.
6. By theorem, $$\forall k, \exists \vec{x}_k \in B_{1/k}(\vec{x}_0) \cap S$$. Choosing one for each ball produces a sequence. This sequence clearly converges to $$\vec{x}_0$$.
• The rightward implication: you can construct something just outside your set, to the closure, but no further.
• Important property of the reals. Completeness in the reals: every sequence of real numbers which gets close to itself converges to a real number. When we talk about a limit existing, we point towards an existing limit. But even if we don’t know about the limit, the numbers get closer and closer to each other.

If $$S \subseteq \mathbb{R}$$ is a subset, an upper bound $$b$$ for $$S$$ is a real number $$b \in \mathbb{R}$$ such that$$\forall x \in S, x \le b$$. Similarly, a lower bound is a $$y \in \mathbb{R}$$ such that $$\forall x \in S$$, $$y \le x$$. Bounds for a set are not unique. The Completeness Axiom. – one of the defining properties of the reals (proven in Kemp’s notes, theorem 4.13). If $$S \susbsetequals \mathbb{R}$$ and non-empty,

1. If $$S$$ has an upper bound, it has a least upper bound
2. If $$S$$ has a lower boud, it has a greatest lower bound Although this is true for the reals, it is false for the rationals. For example, $$S = \{ x \in \mathbb{R} \vert x \in \mathbb{Q} \text{ and } x^2 < 2 \}$$. There is no least upper bound in the rationals (it is in the reals).

## Lecture 4: Sequences, Extrema

• Monotone Sequence Theorem: every bounded monotone sequence is convergent.
• Since $$l$$ is a least upper bound, $$\forall \epsilon > 0, \exists N > 0$$ s.t. $$l - \epsilon < x_n, \forall n > N$$, as otherwise $$l - \epsilon$$ would be a smaller upper bound than $$l$$. Thus $$\forall \epsilon > 0, \exists N > 0, l - \epsilon \le x_n < l, \forall n > N$$. Therefore, $$x_n \to l$$.

## Lecture 14: 1-Dimensional Integration, Integration in Higher Dimensions

Theorem. If $$f$$ is bounded and monotone on $$[a, b]$$ then $$f$$ is integrable on $$[a, b]$$.

Theorem. If $$f$$ is continuous on $$[a, b]$$, then $$f$$ is integrable on $$[a, b]$$. Proof.

1. Since $$f$$ is bounded on $$[a, b]$$ by the extreme value theorem, then $$U_P f, L_P f$$ exist for any partition $$P$$.
2. Since $$f$$ is continuous on a compact set, $$f$$ is uniformly continuous.
3. For all $$\epsilon > 0$$, $$\exists \delta$$ s.t. $$\forall x, y \in [a, b]$$, $$\vert x - y \vert < \delta \implies \vert f(x) - f(y) \vert < \epsilon$$.
4. Let $$P$$ be a partition of $$[a, b]$$ into equally spaced intervals each with length $$< \delta$$.
5. So we have $$M_j - m_j < \epsilon / (b - a)$$
6. Meaning $$U_p f - L_p f = \sum_{j = 1}^J (M_j - m_j) (x_j - x_{j - 1}) \le \frac{\epsilon}{b - a} \sum_{j = 1}^J (x_j - x_{j - 1}) = \frac{\epsilon}{b - a} (x_J - x_1) = \epsilon / (b -a ) \cdot (b - a) = \epsilon$$

Note that partitions are all finite, but you can bound them by any $$\epsilon$$. So we have $$\forall \epsilon > 0, \exists P : U_P f - L_P f \le \epsilon$$

• Midterm: 1.8 up to end of chapter 2

Function: $$f(x) = \begin{cases} 0 & \text{if } x \notin \mathbb{Q} \\ 1 / q & \text{if } x = p / q \text{ in lowest terms} \end{cases}$$.

• $$f(x)$$ is bounded, but not monotone or continuous
• But it is integrable on $$[0, 1]$$

Theorem. If $$f$$ is bounded on $$[a, b]$$ and is continuous on $$[a, b]$$ except at finitely many points, then $$f$$ is integrable on $$[a, b]$$. Proof.

• Suppose $$y_1, ..., y_L$$ are points of discontinuity.
• Set $$m = \inf \{ f(x) \vert [a, b] \}$$ and $$M = \sup { f(x) \vert [a, b] } • Given $$\delta > 0$$, set $$I_l = [a, b] \cap [y_l - \delta, y_l + \delta]$$ • $$U = \bigcup_{l = 1}^L I_l$$, $$V = [a, b] \setminus U^{\text{int}}$$ •  Notice that the length of $$U$$ is$$ le 2 \delta L$$, so$$f : V \to R$$is continuous • If $$P$$ is any partition containing $$\delta U$$, we can write $$U_P f = U_P^U f + U_P^V f$$ • basically… with time, $$\int_a^b f(x) - g(x) dx = 0$$, at finitely many points, for some surrogate $$g$$ Fundamental Theorem of Calculus. There are two! 1. Let $$f$$ be integrable on $$[a, b]$$. For $$x \in [a, b]$$, define $$F(x) = \int_a^x f(t) dt$$. Then $$F$$ is continuous on $$[a, b]$$ and differentiable on $$(a, b)$$, and $$F'(x) = f(x)$$ on $$[a, b]$$. (It makes sense even if $$f$$ is integrable but not continuous!) Proof: 2. Since $$\forall x, y \in [a, b]$$ it is all true that $$F(y) - F(x) = \int_x^y f(t) dt$$ 3. You can define $$C = \sup \{ \vert f(t) \vert : t \in [a, b] \}$$ 4. We see that $$\vert F(y) - F(x) \vert \le \int_x^y \vert f(t) \vert dt \le C \vert y - x \vert$$. Which means that $$F$$ is Lipschitz continuous. 5. Say that $$f$$ is continuous at $$x_0 \in [a, b]$$. Then $$\forall \epsilon > 0, \exists \delta > 0$$ s.t. $$\vert t - x_0 \vert < \delta \implies \vert f(t) - f(x_0) < \epsilon$$. 6. Further, $$f(x_0) = \frac{f(x_0)}{(y - x_0)} \int_{x_0}^y 1 dt = \int_{x_0}^y f(x_0) dt$$ 7. $$\frac{F(y) - F(x_0)}{y - x_0} - f(x_0) = \frac{1}{y - x_0} \int_{x_0}^y f(t) - f(x_0) dt$$. 8. We get that $$\lim_{y \to x_0} \frac{F(y) - F(x_0)}{y - x_0} = f(x_0)$$ by an epsilon-delta argument. 9. Let $$F$$ be continuous on $$[a, b]$$ and differentiable except at finitely many points of $$[a, b]$$. Let $$f$$ be a function which agrees with $$F'(x)$$ where it is defined. If $$f$$ is integrable on $$[a, b]$$ then $$\int_a^b f(t) dt = F(b) - F(a)$$. 10. Suppose $$P = \{ x_0, ..., x_J \}$$ is a partition of $$[a, b]$$. Then after perhaps refining we can assume each point of $$F$$ being non-differentiable is an end point of a subinterval of $$P$$. 11. Then $$\forall j, F$$ is continuous on $$[x_{j - 1}, x_j]$$ and differentiable on $$(x_{j - 1}, x_j)$$. 12. Thus by the MVT, $$F(x_j) - F(x_{j - 1}) = F'(t_j)(x_j - x_{j - 1}) = f(t_j)(x_j - x_{j-1})$$ for some $$t_j \in (x_{j -1 }, x_j)$$ 13. Then, we get a telescoping sum: $$F(b) - F(a) = F(x_J) - F(x_0) = \sum_{j = 1}^J f(t_j) (x_j - x_{j - 1})$$ 14. But we know this is bounded between the lower and upper Riemann sums: $$L_P f \le F(b) - F(a) \le U_P f$$ 15. We know we can make the difference between the two arbitrarily small. Therefore $$F(b) - F(a) = \int_a^b f(t) dt$$ if $$f$$ is integrable. Integration in higher dimensions. • A product of intervals is a rectangle $$R = [a, b] \times [c, d]$$ • A partition is a grid of rectangles. Definition. If $$f : \mathbb{R}^2 \to P$$ and $$P$$ a partition of $$R$$, as above. Define $$m_{jk} = \inf_{x \in R_{jk}} f(x), M_{jk} = \sup_{x \in R_{jk}} f(x)$$. We can define upper and lower Riemann sums by similar logic. Definition. The characteristic function of $$S$$ is $$\chi_S(x) = \begin{cases} 1 & \text{if } x \in S \\ 0 & \text{if } x \notin S \end{cases}$$ Indicator functions are interesting examples depending on the set. So you can define the integral over a non-rectangle as the integral over a rectangle of the indicator function. Indicator functions can actually fuck up nice behavior of our original sets. So how can you guarantee that the result is still integrable? But the boundary of $$S$$ can have zero content. $$f \times \chi_S$$ will only be discontinuous on the boundary of $$S$$. ## Lecture 15: Higher Dimension Integration • if $$S \in \mathbb{R}^n$$, the characteristic function of $$S$$ is 1 if the point is in $$S$$ and 0 otherwise • If $$f: S \in \mathbb{R}^2 \to \mathbb{R}$$ is bounded, and $$S$$ is bounded and $$R$$ is any rectangle containing $$S$$, then then $$f \cdot \chi_S$$ is integrable on $$R$$ and define $$\int \int_S f dA = \int \int_R f \cdot \chi_S dA$$ Theorem declaring the properties of integrals: • linearity – a linear combination of any to integrals is integrable on $$S$$ • if a function is integrable on bounded disjoint domain sets, it is integrable on their union and the sum of their individual integrals • if $$f, g$$ are integrable and $$f(\vec{x}) \le g(\vec{x})$$ for all $$\vec{x} \in S$$, then you have a bound on the integrals: $$\int \int_S f(\vec{x}) dA \le \int \int_S g(\vec{x}) dA$$ • if $$f$$ is integrable on $$S$$, then $$\vert f(\vec{x}) \vert$$ is integrable on $$S$$ and $$\vert \int \in_S f dx \vert \le \int \int_S \vert f \vert dA$$ Combining a. and d. gives you a form of the triangle inequality $$\vert \int \int f + g dA \vert \le \int \int \vert f \vert dA + \int \int \vert g \vert dA$$ Even if $$f$$ is very nice, $$f \cdot \chi_S$$ won’t be continuous on $$R$$. Lemma. $$\chi_S$$ is dicontinuous at $$\vec{x}$$ means $$\vec{x} \in \partial S$$ 1. If $$x \in S^{\text{int}}$$, there exists a ball of some radius centered at $$\vec{x}$$ which is a subset of $$S$$, therefore the characteristic function is constant and continuous in the neighborhood of that point. 2. If $$x \in (S^C)^{\text{int}}$$, you are also constant. 3. If $$\vec{x} \in \partial S4, then$$\forall \delta > 04, $$B_\delta(\vec{x}) \cap S \neq \emptyset$$ and $$B_\delta(\vec{x}) \cap S^C \neq \emptyset$$. Proposition. 1. If $$Z \subseteq \mathbb{R}^2$$ has zero content and $$U \subseteq Z$$ then $$U$$ has zero content. 2. If $$Z_1, ..., Z_k$$ have zero content, then their union has zero content. 3. If $$\vec{f} : (a_0, b_0) \to \mathbb{R}^2$$ is $$C^1$$, then the image of any subinterval $$\vec{f}((a_0, b_0))$$ has zero content. • Each part of the curve can be contained in some set of rectangles; you can find a set of rectangles whose total area is less than your tolerance. Proof for 3. 1. $$C^1$$ buys us the mean value theorem 2. If $$P_k$$ is some equally spaced partition of $$[a, b]$$, of subintervals of length $$\delta = \frac{b - a}{k}$$, $$C = \sup \{ \vert f'(t) \vert \}$$, writing $$\vec{f}(t) = (x(t), y(t))$$, we apply MVT to each comp to obtain $$\vert x(t) - x(t_j) \vert \le C \delta$$, $$\vert y(t) - y(t_j) \vert \le C \delta$$. 3. Together, these imply that the image of the subinterval is contained in a square of side length $$C \delta$$. 4. Then the area as $$k \to \infty$$ goes to zero. The image of any curve in $$C^1$$ will have zero content. If $$f : S \in \mathbb{R}^k \to \mathbb{R}^n$$, $$k < n$$, is $$C^1$$, and 44S$$is bounded, then$$f(S)$$ has zero content.

Remark. A set $$S \subset \mathbb{R}^2$$ s.t. $$S$$ is bounded and $$\partial S$$ has zero content, then $$S$$ is Jordan-measurable.

Theorem. If $$S \subseteq \mathbb{R}^2$$ is Jordan measurable and $$f : \mathbb{R}^2 \to \mathbb{R}$$ is bounded and continuous on $$S$$ except perhaps on a set of zero content, then $$f$$ is integrable on $$S$$.

Proposition. If $$Z \subseteq \mathbb{R}^2$$ of zero content and $$f : \mathbb{R}^2 \to \mathbb{R}$$ is bounded then $$f$$ is integrable on $$Z$$ and has integral zero.

Proof.

1. If $$\epsilon > 0$$, there exists some collection of rectangles such that $$Z$$ is contained in their union and the sum of their areas is less than $$\epsilon$$
2. After subdividing the rectangles as needed, we can assume they form a partition of a rectangle containing $$Z$$ and interior pairwise disjoint
3. Setting $$C = \sup_Z \vert f(\vec{x})$$ then we have $$-C \epsilon < -C \sum_{j = 1}^M A(R_j) \le L_P (f) \le U_P(f) < C \sum_{j = 1}^M A (R_j) < C\epsilon$$. This tells us both the lower and upper Riemann sums are within $$C \epsilon$$ of each other, and therefore the integral is zero.

Corollary

1. If $$f$$ is integrable on $$S \subseteq \mathbb{R}^2$$ and $$f = g$$ except on a set of zero content, then $$\int \int_S dA = \int \int_S g dA$$
2. If $$f$$ is integrable on $$S, T$$ and $$S \cap T$$ has zero content, then $$f$$ is integrable on $$S \cup T$$ and $$\int \int f dA = \int \int_S f dA + \int \int_T f dA$$

How to generalize to higher dimensions? Every argument, lemma, proposition, and theorem in this section is generalizable in higher-dimensional spaces in a straightforward way. Turn squares into boxes. Riemann sums are defined on partitions of boxes.

Mean Value Theorem for integrals. If $$S \subseteq \mathbb{R}^n$$ is compact, connected, and Jordan-measurable, and $$f, g$$ are continuous on the set. Also $$g$$ is non-negative. Then there exists some $$\vec{a} \in S$$ such that $$\int_S f(\vec{x}) g(\vec{x}) dv = f(\vec{a}) \int_S g(\vec{x}) dV$$.

Proof.

1. Because $$f$$ is continuous on $$S$$, by the extreme value theorem, we have that there exists $$M, m$$ such that $$m \le f(\vec{x}) \le M$$ for all $$\vec{x} \in S$$.
2. That is, we have that $$m \int_S g dV \le \int f \cdot g dV \le M \int_S g dV$$.
3. This implies $$m \le \frac{\int_S f \cdot g dV}{\int_S g d} M$$
4. By the intermediate value theorem, there exists an $$\vec{a} \in S$$ such that $$f(\vec{a}) = \frac{\int_S f \cdot g dV}{\int_S g dV}$$.

Corollary. Average Value. If $$S \subseteq \mathbb{R}^n$$ is compact, connected, Jordan-measurable, and $$f$$ is continuous on $$S$$, then there exists some $$\vec{a} \in S$$ such that $$f(\vec{a}) = \text{Avg}_S(f) = \frac{1}{\text{Vol}(S)} \int_S f dV$$. There is some point in $$S$$ where you are equal to your average across the set.