MATH 447: Real Variables — Lecture Notes
Compiled by Jerich Lee
December 08, 2024
Peano Axioms
[Peano Axioms] The natural numbers $ \mathbb{N} $ are defined by the following postulates:- $ \mathbb{N} $ contains a distinguished element $ 1 $.
- Every $ n \in \mathbb{N} $ has its successor in $ \mathbb{N} $, denoted $ S(n) $.
- $ 1 $ is not the successor of any element in $ \mathbb{N} $.
- If $ m $ and $ n $ have the same successor, then $ m = n $.
- If $ A \subseteq \mathbb{N} $ such that $ 1 \in A $ and $ S(n) \in A $ whenever $ n \in A $, then $ A = \mathbb{N} $.
Mathematical Induction
[Principle of Mathematical Induction] Suppose $ (P_n)_{n \in{} \mathbb{N}} $ is a sequence of statements such that:- $ P_1 $ is true.
- For any $ n \in \mathbb{N} $, $ P_n $ implies $ P_{n+1} $.
Properties of Integers
[Addition Properties of $ \mathbb{Z} $] The integers $ \mathbb{Z} $ satisfy:- {Associativity:} $ a + (b + c) = (a + b) + c $ for all $ a, b, c \in \mathbb{Z} $.
- {Commutativity:} $ a + b = b + a $ for all $ a, b \in \mathbb{Z} $.
- {Neutral Element:} $ \exists 0 \in \mathbb{Z} $ such that $ a + 0 = a $.
- {Existence of Opposites:} For every $ a \in \mathbb{Z} $, $ \exists -a \in \mathbb{Z} $ such that $ a + (-a) = 0 $.
- The neutral element $ 0 $ is unique.
- For any $ a \in \mathbb{Z} $, the opposite $ -a $ is unique.
Proof.
1. Suppose $ 0 $ and $ 0' $ are both neutral elements. Then:
$$\begin{align}
0 = 0 + 0' = 0'.
\end{align}$$
2. Suppose $ a + x = 0 $ and $ a + y = 0 $. Then:
$$\begin{align}
x = x + 0 = x + (a + y) = (x + a) + y = 0 + y = y.
\end{align}$$
Thus, $ -a $ is unique.
∎
[Addition Properties of $ \mathbb{Z} $]
- {Associativity:} $ a + (b + c) = (a + b) + c $ for all $ a, b, c \in \mathbb{Z} $.
- {Commutativity:} $ a + b = b + a $ for all $ a, b \in \mathbb{Z} $.
- {Neutral Element:} $ \exists 0 \in \mathbb{Z} $ such that $ a + 0 = a $ for all $ a \in \mathbb{Z} $.
- {Existence of Opposites:} $ \forall a \in \mathbb{Z}, \exists -a \in \mathbb{Z} $ such that $ a + (-a) = 0 $.
- The neutral element $ 0 $ is unique.
- For each $ a \in \mathbb{Z} $, the opposite $ -a $ is unique.
- {Associativity:} $ a \cdot (b \cdot c) = (a \cdot b) \cdot c $ for all $ a, b, c \in \mathbb{Z} $.
- {Commutativity:} $ a \cdot b = b \cdot a $ for all $ a, b \in \mathbb{Z} $.
- {Neutral Element:} $ \exists 1 \in \mathbb{Z} $ such that $ 1 \cdot a = a $ for all $ a \in \mathbb{Z} $.
- {Distributive Law:} $ (a + b) \cdot c = a \cdot c + b \cdot c $ for all $ a, b, c \in \mathbb{Z} $.
- {Inverse:} $ \forall a \in \mathbb{Q} \setminus \{0\}, \exists a^{-1} \in \mathbb{Q} $ such that $ a \cdot a^{-1} = 1 $.
Ordered Fields
[Ordered Fields] A field $ F $ is ordered if equipped with a linear order $ \leq{} $ such that:- If $ a \leq b $, then $ a + c \leq b + c $ for all $ a, b, c \in F $.
- If $ a \leq b $ and $ c \geq 0 $, then $ ac \leq bc $.
- If $ a \leq b $, then $ -b \leq -a $.
- If $ a \leq b $ and $ c \leq 0 $, then $ bc \leq ac $.
- If $ 0 \leq a $ and $ 0 \leq b $, then $ 0 \leq ab $.
- $ 0 \leq a^2 $ for all $ a \in F $.
Rational Zeros Theorem
[Rational Zeros Theorem] Suppose $ p(x) = c_nx^n + \ldots{} + c_1x + c_0 $, with $ c_0, \ldots{}, c_n \in{} \mathbb{Z} $, $ c_0 \neq{} 0 $, $ c_n \neq{} 0 $. If $ p(r) = 0 $ for $ r = \frac{c}{d} $ (where $ c, d \in{} \mathbb{Z} $, $ d \neq{} 0 $, $ \gcd{}(c, d) = 1 $), then $ c \mid{} c_0 $ and $ d \mid{} c_n $. [Irrationality of $ \sqrt{2} $] No rational number $ r $ satisfies $ r^2 = 2 $.Ordered Fields and Completeness
Fields and Order
If $ F $ is a field with more than one element, then $ 0 \neq{} 1 $.Proof.
Let $ x \in{} F $ be distinct from $ 0 $. Then $ 0 = x \cdot{} 0 \neq{} x \cdot{} 1 = x $, hence $ 0 \neq{} 1 $.
∎
[Ordered Fields] A field $ F $ is called ordered if it is equipped with a linear order $ \leq{} $ satisfying:
- If $ a \leq b $, then $ a + c \leq b + c $ for all $ a, b, c \in F $.
- If $ a \leq b $ and $ c \geq 0 $, then $ ac \leq bc $.
Properties of Ordered Fields
Let $ F $ be an ordered field. Then for all $ a, b, c \in{} F $:- If $ a \leq b $, then $ -b \leq -a $.
- If $ a \leq b $ and $ c \leq 0 $, then $ bc \leq ac $.
- If $ 0 \leq a $ and $ 0 \leq b $, then $ 0 \leq ab $.
- $ 0 \leq a^2 $ for all $ a \in F $.
Absolute Value and Distance
[Absolute Value] For $ a \in{} F $, the absolute value $ |a| $ is defined as: $$\begin{align} |a| = \begin{cases} a & \text{if } a \geq 0, \\ -a & \text{if } a < 0. \end{cases} \end{align}$$ [Distance] The distance between $ a, b \in{} F $ is defined as: $$\begin{align} \text{dist}(a, b) = |a - b|. \end{align}$$Completeness Axiom
[Completeness Axiom] If $ S \subset{} \mathbb{R} $ is non-empty and bounded above, then it has a unique least upper bound (supremum), denoted $ \sup{} S $. [Archimedean Property] If $ a, b > 0 $ in $ \mathbb{R} $, then there exists $ n \in{} \mathbb{N} $ such that $ n \cdot{} a > b $. [Denseness of $ \mathbb{Q} $] The rational numbers $ \mathbb{Q} $ are dense in $ \mathbb{R} $, meaning that for any $ a, b \in{} \mathbb{R} $ with $ a < b $, there exists $ r \in{} \mathbb{Q} $ such that $ a < r < b $.Sequences and Limits (Sections 7-9)
Definitions and Examples
[Sequence] A sequence is a function $ s : \{m, m + 1, \dots{} \} \to{} \mathbb{R} $ (for $ m \in{} \mathbb{Z} $). We denote the sequence as $ (s_n)_{n \geq{} m} $, where $ s_n = s(n) $. [Convergence and Limit] A sequence $ (s_n) $ converges to $ L \in{} \mathbb{R} $ if: $$\begin{align} \forall \varepsilon > 0, \exists N \in \mathbb{R} \text{ such that } |s_n - L| < \varepsilon \text{ for } n > N. \end{align}$$ We write $ \lim_{n \to \infty} s_n = L $ or $ s_n \to{} L $. [Uniqueness of Limits] A sequence cannot have more than one limit.Examples of Limits
- $ \lim_{n \to \infty} \frac{1}{n} = 0 $.
- The sequence $ ((-1)^n)_{n \in \mathbb{N}} $ does not converge.
- $ \lim_{n \to \infty} \frac{3n + 1}{2n + 1} = \frac{3}{2} $.
Facts about Limits
If $ (s_n) $ converges, and $ s_n \geq{} a $ for all but finitely many $ n $, then $ \lim{} s_n \geq{} a $. If $ s_n \geq{} 0 $ for all $ n $ and $ \lim{} s_n = s $, then $ \lim{} \sqrt{s_n} = \sqrt{s} $.Convergent Sequences are Bounded
[Bounded Sequence] A sequence $ (s_n) $ is called bounded if $ \exists{} A \in{} \mathbb{R} $ such that $ |s_n| \leq{} A $ for all $ n $. Convergent sequences are bounded.Arithmetic of Limits
[Limits of Sums, Products, and Ratios] Suppose $ \lim{} s_n = s $ and $ \lim{} t_n = t $. Then:- $ \lim (s_n + t_n) = s + t $,
- $ \lim (a \cdot s_n) = a \cdot s $ for any $ a \in \mathbb{R} $,
- $ \lim (s_n \cdot t_n) = s \cdot t $,
- If $ t \neq 0 $, then $ \lim \frac{s_n}{t_n} = \frac{s}{t} $.
Sums, Products, Ratios of Limits (Section 9)
Arithmetic of Limits
[Arithmetic of Limits] Suppose $ \lim{} s_n = s $ and $ \lim{} t_n = t $. Then:- $ \lim (s_n + t_n) = s + t $,
- $ \lim (a \cdot s_n) = a \cdot s $ for any $ a \in \mathbb{R} $,
- $ \lim (s_n \cdot t_n) = s \cdot t $,
- If $ t \neq 0 $, then $ \lim \frac{s_n}{t_n} = \frac{s}{t} $.
Basic Examples of Limits
[Basic Examples]- $ \lim \frac{1}{n^p} = 0 $ for $ p > 0 $,
- $ \lim a^n = 0 $ if $ |a| < 1 $,
- $ \lim n^{1/n} = 1 $,
- $ \lim a^{1/n} = 1 $ for $ a > 0 $.
Diverging Sequences
[Divergence to Infinity] We say $ \lim{} s_n = +\infty{} $ if for all $ A > 0 $, there exists $ N \in{} \mathbb{R} $ such that $ s_n > A $ for $ n > N $. Similarly, $ \lim{} s_n = -\infty{} $ is defined. [Product Rule for Divergence] If $ \lim{} s_n = +\infty{} $ and $ \lim{} t_n > 0 $, then $ \lim{} (s_n \cdot{} t_n) = +\infty{} $. If $ s_n > 0 $ for all $ n $, then $ \lim{} s_n = +\infty{} $ if and only if $ \lim{} \frac{1}{s_n} = 0 $.Monotone Sequences (Section 10)
[Monotone Sequences] A sequence $ (s_n) $ is:- \emph{Increasing} if $ s_n \leq s_{n+1} $ for all $ n $,
- \emph{Decreasing} if $ s_n \geq s_{n+1} $ for all $ n $,
- \emph{Monotone} if it is either increasing or decreasing.
- $ x_n = \sum_{k=1}^n \frac{1}{k^2} $ is increasing because $ x_{n+1} = x_n + \frac{1}{(n+1)^2} > x_n $.
- $ y_n = \frac{(-1)^n}{n^2} $ is not monotone because $ y_{n+1} > y_n $ if $ n $ is odd, and $ y_{n+1} < y_n $ if $ n $ is even.
Monotone Sequences and Convergence (Section 10)
Monotone Sequences
[Monotone Sequences] A sequence $ (s_n) $ is:- \emph{Increasing} if $ s_n \leq s_{n+1} $ for all $ n $,
- \emph{Decreasing} if $ s_n \geq s_{n+1} $ for all $ n $,
- \emph{Monotone} if it is either increasing or decreasing.
- Increasing, since $ s_{n+1} = s_n + \frac{1}{n!} > s_n $.
- Bounded, since $ s_n \leq 3 $ (using an induction-based proof that $ k! \geq 2^{k-1} $ for $ k \geq 1 $).
Unbounded Monotone Sequences
[Theorem 10.4] If $ (s_n)_{n \geq{} m} $ is an unbounded increasing (decreasing) sequence, then $ \lim{} s_n = +\infty{} $ (resp. $ \lim{} s_n = -\infty{} $). [Harmonic Sequence] Let $ s_n = \sum_{k=1}^n \frac{1}{k} $. This sequence is:- Increasing, since $ s_{n+1} = s_n + \frac{1}{n+1} > s_n $.
- Unbounded, as shown using a lower bound argument: $$\begin{align} s_{2^m} \geq \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{2^{m-1}} \geq m. \end{align}$$
Lim Sup and Lim Inf
[Lim Sup and Lim Inf] Let $ u_N = \sup{}\{s_n : n > N\} $ and $ v_N = \inf{}\{s_n : n > N\} $. Then: $$\begin{align} \limsup s_n = \lim_{N \to \infty} u_N, \quad \liminf s_n = \lim_{N \to \infty} v_N. \end{align}$$ [Properties of Lim Sup and Lim Inf]- $ \limsup s_n \geq \liminf s_n $.
- If $ \lim s_n $ exists, then $ \limsup s_n = \lim s_n = \liminf s_n $.
- If $ \limsup s_n = \liminf s_n = s $, then $ \lim s_n = s $.
- $ \limsup s_n = 0 $,
- $ \liminf s_n = -\infty $.
Decimal Expansions
Any real number can be expressed as a decimal expansion $ K.d_1d_2d_3\ldots{} $, where $ K \in{} \{0, 1, 2, \dots{} \} $ and $ d_k \in{} \{0, \ldots{}, 9\} $. For instance: $$\begin{align} 1 = 1.000\ldots = 0.999\ldots \end{align}$$Lim Inf, Lim Sup, and Cauchy Sequences (Sections 10-11)
Lim Inf and Lim Sup
[Lim Sup and Lim Inf] Let $ u_N = \sup{}\{s_n : n > N\} $ and $ v_N = \inf{}\{s_n : n > N\} $. Then: $$\begin{align} \limsup s_n = \lim_{N \to \infty} u_N, \quad \liminf s_n = \lim_{N \to \infty} v_N. \end{align}$$ [Theorem 10.7]- If $ \lim s_n $ is defined, then $ \liminf s_n = \lim s_n = \limsup s_n $.
- If $ \liminf s_n = s = \limsup s_n $, then $ \lim s_n = s $.
- $ \limsup s_n = 0 $,
- $ \liminf s_n = -\infty $.
Cauchy Sequences
[Cauchy Sequence] A sequence $ (s_n) $ is called Cauchy if: $$\begin{align} \forall \varepsilon > 0, \exists N \in \mathbb{N} \text{ such that } |s_n - s_m| < \varepsilon \text{ for } n, m > N. \end{align}$$ [Theorem 10.11] A sequence $ (s_n) $ converges if and only if it is Cauchy. [Lemma 10.9] Any convergent sequence is Cauchy. [Lemma 10.10] Any Cauchy sequence is bounded.Subsequences
[Subsequence] A sequence $ (t_k) $ is a subsequence of $ (s_n) $ if there exists a strictly increasing sequence $ n_1 < n_2 < \ldots{} $ such that $ t_k = s_{n_k} $ for any $ k $. [Subsequence Examples]- $ s_n = \frac{1}{n} $: A subsequence $ t_k = \frac{1}{k^2} $, where $ n_k = k^2 $.
- $ s_n = (-1)^n + \frac{1}{n} $: The sequence diverges, but $ s_{2k} = 1 + \frac{1}{2k} $ converges.
Subsequences and Subsequential Limits (Section 11)
Subsequences
[Subsequence] A sequence $ (t_k) $ is a subsequence of $ (s_n) $ if there exists a strictly increasing sequence $ n_1 < n_2 < \ldots{} $ such that $ t_k = s_{n_k} $ for any $ k $. [Subsequences of Subsequences] Any subsequence of a subsequence of $ (s_n) $ is a subsequence of $ (s_n) $.Convergence of Subsequences
[Theorem 11.3] If $ \lim{} s_n = s $ (finite or $ \pm{}\infty{} $), then any subsequence $ (t_k) $ has the same limit.Proof.
Let $ \lim{} s_n = s $, and let $ t_k = s_{n_k} $. Then for $ \varepsilon{} > 0 $, there exists $ N $ such that $ |s_n - s| < \varepsilon{} $ for $ n > N $. Since $ n_k \to{} \infty{} $, we can find $ K $ such that $ n_k > N $ for $ k > K $. Thus, $ |t_k - s| < \varepsilon{} $ for $ k > K $, implying $ \lim{} t_k = s $.
∎
Monotone Subsequences
[Theorem 11.4] Every sequence has a monotone subsequence. [Bolzano-Weierstrass Theorem] Every bounded sequence has a convergent subsequence. [Divergent Sequence with Convergent Subsequence] Let $ s_n = (-1)^n (1 + \frac{1}{n}) $. This sequence is bounded but divergent. The subsequence $ s_{2k} = 1 + \frac{1}{2k} $ converges to $ 1 $.Subsequential Limits
[Subsequential Limit] A subsequential limit of $ (s_n) $ is any limit of a subsequence, possibly $ \pm{}\infty{} $. [Theorem 11.2] Suppose $ (s_n) $ is a sequence.- $ t \in \mathbb{R} $ is a subsequential limit if and only if $ \forall \varepsilon > 0, \{n : |s_n - t| < \varepsilon\} $ is infinite.
- $ t = +\infty $ (or $ t = -\infty $) is a subsequential limit if $ (s_n) $ is not bounded above (or below).
Lim Inf and Lim Sup as Subsequential Limits
[Theorem 11.7] For any sequence $ (s_n) $, $ \limsup{} s_n $ and $ \liminf{} s_n $ are limits of monotone subsequences. [Theorem 11.8] Let $ S $ be the set of subsequential limits of $ (s_n) $. Then:- $ S $ is non-empty.
- $ \inf S = \liminf s_n, \quad \sup S = \limsup s_n $.
- $ \lim s_n $ exists if and only if $ S $ consists of a single point, $ S = \{\lim s_n\} $.
Lim Sup, Lim Inf, and Metric Spaces (Sections 11-13)
Subsequential Limits
[Subsequential Limit (Definition 11.6)] For a sequence $ (s_n) $, a subsequential limit is any limit of a subsequence (in $ \mathbb{R} \cup{} \{\pm{}\infty{}\} $). [Properties of Subsequential Limits (Theorem 11.2)] Suppose $ (s_n) $ is a sequence.- $ t \in \mathbb{R} $ is a subsequential limit if and only if $ \forall \varepsilon > 0, \{n : |s_n - t| < \varepsilon\} $ is infinite.
- $ t = +\infty $ ($ t = -\infty $) is a subsequential limit if $ (s_n) $ is not bounded above (resp. below).
The Set of Subsequential Limits
[Theorem 11.8] Suppose $ (s_n) $ is a sequence, and $ S $ is the set of subsequential limits. Then:- $ S $ is non-empty.
- $ \inf S = \liminf s_n $ and $ \sup S = \limsup s_n $.
- $ \lim s_n $ exists if and only if $ S $ consists of a single point. Then $ \{\lim s_n\} = S $.
Lim Sup and Lim Inf Revisited
[Theorem 12.1] If $ \lim{} s_n = s \in{} (0, \infty{}) $, then for any sequence $ (t_n) $: $$\begin{align} \limsup (s_n t_n) = s \cdot \limsup t_n. \end{align}$$ [Corollary 12.3] If $ (s_n) $ is a sequence of positive numbers and $ \lim{} \frac{s_{n+1}}{s_n} $ exists, then: $$\begin{align} \lim s_n^{1/n} \text{ also exists, and } \lim s_n^{1/n} = \lim \frac{s_{n+1}}{s_n}. \end{align}$$- $ \lim (n!)^{1/n} = +\infty $.
- $ \lim \frac{1}{n} (n!)^{1/n} = \frac{1}{e} $.
Metric Spaces
[Metric (Definition 13.1)] A metric $ d : S \times{} S \to{} [0, \infty{}) $ satisfies:- {Non-degeneracy:} $ d(x, y) = 0 \iff x = y $.
- {Symmetry:} $ d(x, y) = d(y, x) $ for all $ x, y \in S $.
- {Triangle Inequality:} $ d(x, y) + d(y, z) \geq d(x, z) $ for all $ x, y, z \in S $.
- On $ \mathbb{R} $: $ d(x, y) = |x - y| $.
- On $ \mathbb{R}^n $: $ d(\vec{x}, \vec{y}) = \sqrt{\sum_{i=1}^n (x_i - y_i)^2} $.
Convergence in Metric Spaces
[Convergence (Definition 13.2)] A sequence $ (s_n) \subset{} S $ converges to $ s \in{} S $ if: $$\begin{align} \lim_{n \to \infty} d(s_n, s) = 0. \end{align}$$ [Cauchy Sequence (Definition 13.2)] A sequence $ (s_n) \subset{} S $ is Cauchy if: $$\begin{align} \forall \varepsilon > 0, \exists N \text{ such that } d(s_n, s_m) < \varepsilon \text{ for all } n, m > N. \end{align}$$ [Cauchy and Convergence] If $ (s_n) $ converges in $ (S, d) $, then $ (s_n) $ is Cauchy. [Complete Metric Spaces] A metric space $ (S, d) $ is complete if every Cauchy sequence in $ S $ converges to a point in $ S $. The space $ \mathbb{R}^n $ with the Euclidean metric is complete.Metric Spaces (Section 13)
Definition and Examples
[Metric (Definition 13.1)] Suppose $ S $ is a set. A function $ d : S \times{} S \to{} [0, \infty{}) $ is called a metric if the following hold:- {Non-degeneracy:} $ d(x, y) = 0 \iff x = y $ (hence $ d(x, y) > 0 $ when $ x \neq y $).
- {Symmetry:} $ d(x, y) = d(y, x) $ for all $ x, y \in S $.
- {Triangle Inequality:} $ d(x, y) + d(y, z) \geq d(x, z) $ for all $ x, y, z \in S $.
- On $ \mathbb{R} $: $ d(x, y) = |x - y| $.
- On $ \mathbb{R}^n $: $ d(\vec{x}, \vec{y}) = \sqrt{\sum_{i=1}^n (x_i - y_i)^2} $ (Euclidean metric).
- Discrete Metric: For $ x, y \in S $, define: $$\begin{align} d(x, y) = \begin{cases} 0 & \text{if } x = y, \\ 1 & \text{if } x \neq y. \end{cases} \end{align}$$
Convergence and Completeness
[Convergence (Definition 13.2)] Suppose $ (S, d) $ is a metric space. A sequence $ (s_n) \subset{} S $ converges to $ s \in{} S $ if $ \lim_{n \to \infty} d(s_n, s) = 0 $; that is, $ \forall{} \varepsilon{} > 0, \exists{} N \text{such that} d(s_n, s) < \varepsilon{} \text{for} n > N $. [Cauchy Sequence (Definition 13.2 continued)] A sequence $ (s_n) \subset{} S $ is called Cauchy if: $$\begin{align} \forall \varepsilon > 0, \exists N \text{ such that } d(s_n, s_m) < \varepsilon \text{ for } n, m > N. \end{align}$$ [Cauchy Sequences are Convergent in Complete Spaces] If $ (S, d) $ is complete, then every Cauchy sequence in $ S $ converges to a point in $ S $. [Completeness of $ \mathbb{R} $] $ \mathbb{R} $ with the standard metric $ d(x, y) = |x - y| $ is complete. Any Cauchy sequence in $ \mathbb{R} $ converges to a real number. [Non-Completeness of $ \mathbb{Q} $] Consider $ \mathbb{Q} $ with $ d(x, y) = |x - y| $. The sequence $ r_n \in{} (\sqrt{2} - \frac{1}{n}, \sqrt{2}) $ is Cauchy in $ \mathbb{Q} $ but does not converge in $ \mathbb{Q} $ because $ \sqrt{2} \notin{} \mathbb{Q} $.Special Metrics
[Manhattan (Taxicab) Metric] For $ \vec{x} = (x_1, \ldots{}, x_n) $ and $ \vec{y} = (y_1, \ldots{}, y_n) $ in $ \mathbb{R}^n $, the taxicab metric is defined as: $$\begin{align} d_1(\vec{x}, \vec{y}) = \sum_{i=1}^n |x_i - y_i|. \end{align}$$ [Completeness of $ (\mathbb{R}^n, d_1) $] The metric space $ (\mathbb{R}^n, d_1) $ is complete.Inner Product and Triangle Inequality
[Inner Product] For $ \vec{x}, \vec{y} \in{} \mathbb{R}^n $, define the inner product: $$\begin{align} \langle \vec{x}, \vec{y} \rangle = \sum_{i=1}^n x_i y_i. \end{align}$$ The magnitude of $ \vec{x} $ is: $$\begin{align} \|\vec{x}\| = \sqrt{\langle \vec{x}, \vec{x} \rangle}. \end{align}$$ [Bunyakovsky-Cauchy-Schwarz Inequality] For all $ \vec{x}, \vec{y} \in{} \mathbb{R}^n $: $$\begin{align} |\langle \vec{x}, \vec{y} \rangle| \leq \|\vec{x}\| \|\vec{y}\|. \end{align}$$ [Triangle Inequality Lite] For $ \vec{x}, \vec{y} \in{} \mathbb{R}^n $: $$\begin{align} \|\vec{x} + \vec{y}\| \leq \|\vec{x}\| + \|\vec{y}\|. \end{align}$$Metric Spaces: Bounded Sets, Open and Closed Sets, and Closure (Section 13)
Bounded Sets
[Bounded Sets] A set $ E $ in a metric space $ (S, d) $ is bounded if there exists $ y \in{} S $ such that: $$\begin{align} \sup_{x \in E} d(y, x) < \infty. \end{align}$$ If such a $ y $ exists, then for any $ z \in{} S $, $ \sup_{x \in E} d(z, x) < \infty{} $. This follows from the triangle inequality: $$\begin{align} d(z, x) \leq d(y, x) + d(z, y). \end{align}$$ A sequence $ (x_k) $ is bounded if the set $ \{x_1, x_2, \ldots{}\} $ is bounded. That is, for some (or any) $ y \in{} S $: $$\begin{align} \sup_k d(y, x_k) < \infty. \end{align}$$Bolzano-Weierstrass Theorem
[Bolzano-Weierstrass for $ \mathbb{R}^n $] Any bounded sequence in $ \mathbb{R}^n $ has a convergent subsequence. [Failure of Bolzano-Weierstrass in Discrete Metrics] Consider $ \mathbb{N} $ equipped with the discrete metric $ d(x, y) = 1 $ for $ x \neq{} y $ and $ d(x, y) = 0 $ for $ x = y $. The sequence $ x_n = n $ is bounded but has no convergent subsequences because convergent sequences are eventually constant in discrete metrics.Interior Points and Open Sets
[Open Ball] An open ball with center $ s_0 $ and radius $ r > 0 $ is: $$\begin{align} B_r^o(s_0) = \{s \in S : d(s, s_0) < r\}. \end{align}$$ [Interior Points] A point $ s_0 \in{} S $ is interior to $ E \subset{} S $ if there exists $ r > 0 $ such that $ B_r^o(s_0) \subset{} E $. The set of all interior points is denoted $ E^o $, called the interior of $ E $. [Open Sets] A set $ E \subset{} S $ is open if $ E = E^o $.- In $ S = \mathbb{R} $ with the usual metric, $ [0, \infty) $ is not open, but $ (0, \infty) $ is.
- In $ S = \mathbb{R}^2 $, the set $ E = \{(x, 0) : x \geq 0\} $ has $ E^o = \emptyset $.
Properties of Open and Closed Sets
[Facts about Open Sets]- $ S $ and $ \emptyset $ are open.
- A union of any collection of open sets is open.
- A finite intersection of open sets is open.
- $ S $ and $ \emptyset $ are closed.
- An intersection of any collection of closed sets is closed.
- A finite union of closed sets is closed.
Closure and Boundary
[Closure] The closure of $ E \subset{} S $, denoted $ \overline{E} $, is the intersection of all closed sets containing $ E $. [Boundary] The boundary of $ E \subset{} S $ is: $$\begin{align} \partial E = \overline{E} \setminus E^o. \end{align}$$ [Closure in $ \mathbb{R} $] Let $ E = \{\frac{1}{n} : n \in{} \mathbb{N}\} \subset{} \mathbb{R} $. Then: $$\begin{align} \overline{E} = E \cup \{0\}. \end{align}$$Closure, Boundary, and Open/Closed Sets (Section 13)
Open Balls and Open Sets
[Open Ball] For $ s_0 \in{} S $ and $ r > 0 $, the open ball with center $ s_0 $ and radius $ r $ is: $$\begin{align} B_r^o(s_0) = \{s \in S : d(s, s_0) < r\}. \end{align}$$ A set $ E \subset{} S $ is open if and only if it is a union of open balls.Closed Sets and De Morgan's Laws
[Closed Sets] A set $ E \subset{} S $ is closed if $ S \setminus{} E $ is open. [Properties of Closed Sets]- $ S $ and $ \emptyset $ are closed.
- Any intersection of closed sets is closed.
- A finite union of closed sets is closed.
Examples of Open and Closed Sets
[Intervals in $ \mathbb{R} $]- $ (a, b) $ is open, but not closed.
- $ [a, b] $ is closed, but not open.
- $ (a, b], [a, b) $ are neither open nor closed.
- Every set is both open and closed.
Closure and Boundary
[Closure] The closure of $ E \subset{} S $, denoted $ \overline{E} $, is the intersection of all closed sets containing $ E $. [Boundary] The boundary of $ E \subset{} S $ is: $$\begin{align} \partial E = \overline{E} \setminus E^o. \end{align}$$Properties of Closure and Boundary
- $ E = \overline{E} $ if and only if $ E $ is closed.
- $ s \in \overline{E} $ if and only if $ s $ is a limit of a sequence in $ E $.
- $ \partial E = \overline{E} \cap (S \setminus E)^- $.
Compactness (Section 13)
Definition of Compactness
[Compactness (Definition 13.11)] Suppose $ E \subset{} S $. A family $ \mathcal{U} $ of open sets is an open cover for $ E $ if: $$\begin{align} E \subset \bigcup_{U \in \mathcal{U}} U. \end{align}$$ A subcover is a subfamily of $ \mathcal{U} $ which is also an open cover. $ E $ is called compact if every open cover has a finite subcover. A cover $ \mathcal{U} $ is a collection of sets, not their union. Thus, a cover is a subset of $ \mathcal{P}(S) $ (the power set of $ S $), not $ S $.Examples in $ \mathbb{R
$ with Usual Metric}- $ E = [0, \infty) $ is not compact. For example:
- $ U_k = (-1, k) $ ($ k \in \mathbb{N} $) is an open cover with no finite subcover.
- $ E = (0, 1) $ is not compact. For example:
- $ U_k = (1/k, 1) $ ($ k \in \mathbb{N} $) is an open cover with no finite subcover.
- $ E = [a, b] $ ($ a, b \in \mathbb{R} $) is compact (proof to follow).
Proof.
Let $ E = \{e_1, \ldots{}, e_N\} $. For any open cover $ \mathcal{U} $ of $ E $, select $ U_i \in{} \mathcal{U} $ containing $ e_i $. Then $ \{U_1, \ldots{}, U_N\} $ is a finite subcover.
∎
Compactness and Boundedness
Any compact set is bounded.Proof.
If $ E $ is not bounded, then for $ s \in{} S $, the sets $ B_k^o(s) $ ($ k \in{} \mathbb{N} $) form an open cover of $ E $ with no finite subcover.
∎
[Bounded but Not Compact] Equip $ \mathbb{N} $ with the discrete metric $$\begin{align} d(x, y) = \begin{cases} 0 & x = y, \\ 1 & x \neq y. \end{cases} \end{align}$$ Then $ \mathbb{N} $ is bounded, but it is not compact because the open cover $ U_n = \{n\} $ ($ n \in{} \mathbb{N} $) has no finite subcover.
Properties of Compact Sets
- A closed subset of a compact set is compact.
- A finite union of compact sets is compact.
Nested Sequences of Closed Sets
Suppose $ F_1 \supset{} F_2 \supset{} \cdots{} $ are closed non-empty subsets of a compact set $ E $. Then: $$\begin{align} \bigcap_{n} F_n \neq \emptyset, \quad \text{and it is compact.} \end{align}$$Heine-Borel Theorem and Cantor Set
[Heine-Borel Theorem] A subset of $ \mathbb{R}^n $ is compact if and only if it is closed and bounded. [Cantor Set] Define: $$\begin{align} F_0 = [0, 1], \quad F_1 = [0, 1/3] \cup [2/3, 1], \quad F_2 = [0, 1/9] \cup [2/9, 1/3] \cup [2/3, 7/9] \cup [8/9, 1], \dots \end{align}$$ The Cantor set $ C = \bigcap_n{} F_n $ is non-empty, closed, and compact. It contains no intervals, and its interior is empty.Compactness and Total Boundedness (Section 13)
Definition of Compactness
[Compactness (Definition 13.11)] Suppose $ E \subset{} S $. A family $ \mathcal{U} $ of open sets is an open cover for $ E $ if: $$\begin{align} E \subset \bigcup_{U \in \mathcal{U}} U. \end{align}$$ A subcover is a subfamily of $ \mathcal{U} $ which is also an open cover. $ E $ is called compact if every open cover has a finite subcover. A cover $ \mathcal{U} $ is a collection of sets, not their union. Thus, a cover is a subset of $ \mathcal{P}(S) $ (the power set of $ S $), not $ S $.Compactness and Completeness
Suppose $ E \subset{} S $. If $ E $ is compact, then $ E $ is complete.Proof.
Suppose $ E $ is not complete. Then there exists a Cauchy sequence $ (s_n) \subset{} E $ which does not converge in $ E $. For $ k \in{} \mathbb{N} $, find $ n_k $ such that $ d(s_m, s_\ell{}) < 2^{-k} $ for $ m, \ell{} \geq{} n_k $. Construct open sets $ U_k = \{s \in{} S : d(s, s_{n_k}) > 2^{-k}\} $, which form an open cover for $ E $ with no finite subcover. Contradiction.
∎
Any compact set is closed. [Compactness Criterion] $ E \subset{} S $ is compact if and only if it is complete and totally bounded.
Total Boundedness
[Total Boundedness] A set $ E \subset{} S $ is called totally bounded if $ \forall{} \varepsilon{} > 0 $, there exist $ s_1, \ldots{}, s_n \in{} S $ such that: $$\begin{align} E \subset \bigcup_{i=1}^n B_\varepsilon^o(s_i). \end{align}$$ A set is totally bounded if and only if any sequence in the set has a Cauchy subsequence.Characterization of Compactness
For a subset $ E $ of a metric space, the following are equivalent:- $ E $ is compact.
- $ E $ is complete and totally bounded.
- Any sequence in $ E $ has a subsequence with a limit in $ E $.
Heine-Borel Theorem
[Heine-Borel Theorem] A subset of $ \mathbb{R}^n $ is compact if and only if it is closed and bounded. [Cantor Set] The Cantor set $ C $, constructed as: $$\begin{align} F_0 = [0, 1], \quad F_1 = [0, 1/3] \cup [2/3, 1], \quad F_2 = \cdots, \end{align}$$ is compact, closed, and totally bounded but has no interior.A Note on Compactness
Total Boundedness
[Total Boundedness (Definition 1.1)] A set $ S \subset{} E $ is called totally bounded if for every $ \varepsilon{} > 0 $, there exist $ p_1, \ldots{}, p_n \in{} E $ such that: $$\begin{align} S \subset \bigcup_{i=1}^n B_\varepsilon^o(p_i). \end{align}$$ [Intrinsic Nature of Total Boundedness (Proposition 1.2)] A set $ S \subset{} E $ is totally bounded if and only if for every $ \varepsilon{} > 0 $, there exist $ q_1, \ldots{}, q_m \in{} S $ such that: $$\begin{align} S \subset \bigcup_{j=1}^m B_\varepsilon^o(q_j). \end{align}$$ [Total Boundedness and Cauchy Subsequences (Proposition 1.3)] A set $ S $ is totally bounded if and only if any sequence in $ S $ has a Cauchy subsequence. [Characterization of Compactness (Corollary 1.4)] A set $ S $ is totally bounded and complete if and only if any sequence in $ S $ has a subsequence converging to a limit in $ S $.Compactness
[Characterization of Compactness (Theorem 2.1)] For a subset $ S \subset{} E $, the following are equivalent:- $ S $ is compact.
- Any sequence in $ S $ has a convergent subsequence.
- $ S $ is complete and totally bounded.
Compactness and Series (Sections 13-14)
Total Boundedness
[Total Boundedness] A set $ E \subset{} S $ is totally bounded if: $$\begin{align} \forall \varepsilon > 0, \exists s_1, \ldots, s_n \in S \text{ such that } E \subset \bigcup_{i=1}^n B_\varepsilon^o(s_i). \end{align}$$ A set $ E $ is totally bounded if and only if any sequence in $ E $ has a Cauchy subsequence.Proof.
For $ (s_i) \subset{} E $, construct $ \{s_{i_k}\} $ with $ s_{i_k} \in{} B_{2^{-k}}^o(x_{kj_k}) $. Using the triangle inequality, show $ (s_{i_k}) $ is Cauchy.
∎
Characterization of Compactness
For a subset $ E $ of a metric space, the following are equivalent:- $ E $ is compact.
- $ E $ is complete and totally bounded.
- Any sequence in $ E $ has a subsequence with a limit in $ E $.
Proof.
- $ (1) \implies (2) $: Shown in the last lecture.
- $ (2) \implies (3) $: Total boundedness guarantees a Cauchy subsequence, and completeness ensures convergence.
- $ (3) \implies (1) $: Contraposition: if $ E $ is not compact, construct an open cover with no finite subcover.
∎
[Heine-Borel] A subset of $ \mathbb{R}^n $ is compact if and only if it is closed and bounded. [Non-Compact Set] In $ \mathbb{N} $ with the discrete metric, $ \mathbb{N} $ is closed and bounded but not compact.
Series
[Series and Convergence] The $ n $-th partial sum of a series $ \sum_{j=k_0}^\infty{} a_j $ is: $$\begin{align} s_n = \sum_{j=k_0}^n a_j. \end{align}$$ The series converges if $ \lim_{n \to \infty} s_n $ exists, diverges otherwise. [Geometric Series] $$\begin{align} \sum_{j=0}^\infty r^j = \begin{cases} \frac{1}{1-r}, & |r| < 1, \\ \infty, & r \geq 1. \end{cases} \end{align}$$Cauchy Criterion for Convergence
[Cauchy Criterion] A series $ \sum_j{} a_j $ satisfies the Cauchy criterion if: $$\begin{align} \forall \varepsilon > 0, \exists N \text{ such that } \left| \sum_{j=m}^n a_j \right| < \varepsilon \text{ for } n \geq m > N. \end{align}$$ A series converges if and only if it satisfies the Cauchy criterion.Comparison Test for Convergence
[Comparison Test]- If $ |b_n| \leq a_n $ and $ \sum a_n $ converges, then $ \sum b_n $ converges.
- If $ 0 \leq a_n \leq b_n $ and $ \sum b_n = \infty $, then $ \sum a_n = \infty $.
Series and Decimal Expansions (Sections 14 and 16)
Series
[Partial Sums and Convergence of Series] The $ n $-th partial sum of a series $ \sum_{j=k_0}^\infty{} a_j $ is: $$\begin{align} s_n = \sum_{j=k_0}^n a_j. \end{align}$$ The series $ \sum_{j=k_0}^\infty{} a_j $ converges if $ \lim_{n \to \infty} s_n $ exists. Otherwise, it diverges. [Geometric Series] $$\begin{align} \sum_{j=0}^\infty r^j = \begin{cases} \frac{1}{1-r}, & |r| < 1, \\ \text{diverges}, & r \geq 1 \text{ or } r \leq -1. \end{cases} \end{align}$$Cauchy Criterion for Convergence
[Cauchy Criterion (Definition 14.3)] A series $ \sum_j{} a_j $ satisfies the Cauchy Criterion if: $$\begin{align} \forall \varepsilon > 0, \exists N \text{ such that } \left|\sum_{j=m}^n a_j\right| < \varepsilon \text{ for } n \geq m > N. \end{align}$$ [Cauchy Criterion (Theorem 14.4)] A series converges if and only if it satisfies the Cauchy Criterion.Properties of Convergence
[Necessary Condition for Convergence (Corollary 14.5)] If $ \sum_j{} a_j $ converges, then $ \lim_{n \to \infty} a_n = 0 $. If $ a_n = \frac{1}{n} $, then $ \lim_{n \to \infty} a_n = 0 $, but $ \sum_{n=1}^\infty{} \frac{1}{n} $ diverges.Tests for Convergence
[Comparison Test (Theorem 14.6)]- If $ 0 \leq |b_n| \leq a_n $ and $ \sum a_n $ converges, then $ \sum b_n $ converges.
- If $ 0 \leq a_n \leq b_n $ and $ \sum b_n = \infty $, then $ \sum a_n = \infty $.
- The series converges absolutely if $ \alpha < 1 $.
- The series diverges if $ \alpha > 1 $.
- If $ \alpha = 1 $, the test gives no information.
- The series converges absolutely if $ \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1 $.
- The series diverges if $ \liminf_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| > 1 $.
- If $ \liminf_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \leq 1 \leq \limsup_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| $, the test gives no information.
Decimal Expansions
[Decimal Expansions (Theorem 16.2)] Any real number $ x \geq{} 0 $ has at least one decimal expansion: $$\begin{align} x = K.d_1d_2d_3\ldots = K + \sum_{j=1}^\infty \frac{d_j}{10^j}, \end{align}$$ where $ K \in{} \mathbb{Z} $ and $ d_j \in{} \{0, 1, \ldots{}, 9\} $. [Uniqueness of Decimal Expansions (Theorem 16.3)] Any $ x \geq{} 0 $ has either exactly one decimal expansion or exactly two, one ending in $ \ldots{} d000\ldots{} $ and the other in $ \ldots{} [d-1]999\ldots{} $. [Repeating Decimals (Theorem 16.5)] A real number $ x $ is rational if and only if its decimal expansion is repeating.Series and Decimal Expansions (Sections 17 and 21)
Root and Ratio Tests for Series Convergence
[Root Test] For a series $ \sum_{n} a_n $, let $ \alpha{} = \limsup_{n \to \infty} |a_n|^{1/n} $. Then:- The series converges absolutely if $ \alpha < 1 $.
- The series diverges if $ \alpha > 1 $.
- If $ \alpha = 1 $, the test gives no information.
- The series converges absolutely if $ \limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 $.
- The series diverges if $ \liminf_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1 $.
- If $ \liminf_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \leq 1 \leq \limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $, the test gives no information.
- Consider $ \sum_{k=1}^\infty \frac{k^4}{2^k} $. Using the Root Test: $$\begin{align} a_k^{1/k} = \left( k^{1/k} \right)^4 \frac{1}{2}, \quad \lim_{k \to \infty} a_k^{1/k} = \frac{1}{2} < 1, \end{align}$$ so the series converges.
- The $ p $-series $ \sum_{k=1}^\infty \frac{1}{k^p} $ converges if and only if $ p > 1 $. The Root and Ratio Tests are inconclusive for this series.
Decimal Expansions
[Decimal Expansion] For $ x \in{} [0, \infty{}) $, the decimal expansion of $ x $ is: $$\begin{align} x = K .d_1 d_2 d_3 \ldots = K + \sum_{j=1}^\infty \frac{d_j}{10^j}, \end{align}$$ where $ K \in{} \{0, 1, 2, \ldots{}\} $ and $ d_1, d_2, \ldots{} \in{} \{0, 1, \ldots{}, 9\} $. [Existence of Decimal Expansions (Theorem 16.2)] Any real number $ x \geq{} 0 $ has at least one decimal expansion. [Uniqueness of Decimal Expansions (Theorem 16.3)] Any $ x \geq{} 0 $ has either exactly one decimal expansion or exactly two:- One ending in $ \ldots d000 \ldots $, where $ d \in \{1, \ldots, 9\} $,
- Another ending in $ \ldots (d-1)999 \ldots $.
Repeating Decimal Expansions
[Repeating Decimals (Definition 16.4)] A repeating decimal expansion is one of the form: $$\begin{align} K .d_1 \ldots d_\ell d_{\ell+1} \ldots d_{\ell+r} = K .d_1 \ldots d_\ell \overline{d_{\ell+1} \ldots d_{\ell+r}}, \end{align}$$ where the sequence $ d_{\ell{}+1} \ldots{} d_{\ell{}+r} $ repeats. [Repeating Decimals and Rational Numbers (Theorem 16.5)] A real number $ x $ is rational if and only if its decimal expansion is repeating.Proof.
- ($ x $ is rational $ \implies $ repeating): Follows from performing long division.
- (Repeating $ \implies x $ is rational): Suppose $ x = K .d_1 \ldots d_\ell \overline{d_{\ell+1} \ldots d_{\ell+r}} $. Then: $$\begin{align} x = K + \sum_{j=1}^\ell \frac{d_j}{10^j} + 10^{-\ell} \left( \frac{z}{1 - 10^{-r}} \right), \end{align}$$ where $ z = \sum_{j=1}^r d_{\ell+j} 10^{-j} \in \mathbb{Q} $, so $ x \in \mathbb{Q} $.
∎
Continuity in Metric Spaces (Sections 17 and 21)
Definition of Continuity
[Continuity (Definition 21.1)] Suppose $ (S, d) $ and $ (S^*, d^*) $ are metric spaces. The function $ f : \text{dom}(f) \to{} S^* $ (with $ \text{dom}(f) \subset{} S $) is continuous at $ x \in{} \text{dom}(f) $ if: $$\begin{align} \forall \varepsilon > 0, \exists \delta > 0 \text{ such that } d^*(f(x), f(y)) < \varepsilon \text{ whenever } d(x, y) < \delta. \end{align}$$ $ f $ is called continuous on $ E \subset{} S $ if it is continuous at every $ x \in{} E $. [Sequential Criterion for Continuity (Theorem 17.1 + 17.2)] $ f : S \to{} S^* $ is continuous at $ x \in{} S $ if and only if $ f(x_n) \to{} f(x) $ whenever $ x_n \to{} x $.Examples of Continuity and Discontinuity
[Discontinuous Everywhere] The Dirichlet function $ f : \mathbb{R} \to{} \mathbb{R} $, defined as: $$\begin{align} f(x) = \begin{cases} 1 & x \in \mathbb{Q}, \\ 0 & x \notin \mathbb{Q}, \end{cases} \end{align}$$ is discontinuous at every $ x \in{} \mathbb{R} $. This is because for any $ x \in{} \mathbb{R} $, one can find sequences $ (x_n) \subset{} \mathbb{Q} $ and $ (y_n) \not{}\subset{} \mathbb{Q} $ such that $ x_n, y_n \to{} x $, but $ f(x_n) \to{} 1 $ and $ f(y_n) \to{} 0 $, which do not match $ f(x) $. [Continuous at a Single Point] The modified Dirichlet function $ g : \mathbb{R} \to{} \mathbb{R} $, defined as: $$\begin{align} g(x) = \begin{cases} x & x \in \mathbb{Q}, \\ 0 & x \notin \mathbb{Q}, \end{cases} \end{align}$$ is continuous only at $ x = 0 $. At other points, similar reasoning as the Dirichlet function applies. [Continuous on $ \mathbb{R} \setminus{} \mathbb{Q} $] The Thomae function $ h : \mathbb{R} \to{} \mathbb{R} $, defined as: $$\begin{align} h(x) = \begin{cases} \frac{1}{b} & x = \frac{a}{b}, \, \gcd(a, b) = 1, \, b > 0, \, x \neq 0, \\ 0 & x \notin \mathbb{Q}, \end{cases} \end{align}$$ is continuous at $ x \notin{} \mathbb{Q} $ and discontinuous at $ x \in{} \mathbb{Q} $.Operations on Continuous Functions
Suppose $ f, g $ are continuous at $ x_0 $ in a metric space $ (S, d) $. Then the following functions are also continuous at $ x_0 $:- $ |f| $,
- $ kf $ ($ k \in \mathbb{R} $),
- $ f + g $,
- $ f \cdot g $,
- $ f / g $ (if $ g(x_0) \neq 0 $).
Composition of Continuous Functions
Suppose $ (S_1, d_1), (S_2, d_2), (S_3, d_3) $ are metric spaces, and $ f : \text{dom}(f) \to{} S_2 $, $ g : \text{dom}(g) \to{} S_3 $ are functions such that $ f $ is continuous at $ x_0 $, $ g $ is continuous at $ f(x_0) $, and $ x_0 \in{} \text{dom}(f) $. Then $ g \circ{} f $ is continuous at $ x_0 $.Characterization of Continuity
[Characterization of Continuity (Theorem 21.3)] Suppose $ (S, d) $ and $ (S^*, d^*) $ are metric spaces. $ f : S \to{} S^* $ is continuous if and only if $ f^{-1}(U) $ is open for every open $ U \subset{} S^* $, where: $$\begin{align} f^{-1}(U) = \{s \in S : f(s) \in U\}. \end{align}$$ $ f $ is continuous at $ s_0 \in{} S $ if for any open set $ U $ containing $ f(s_0) $, there exists an open set $ V $ containing $ s_0 $ such that $ f(V) \subset{} U $.Continuity and the Intermediate Value Theorem (Sections 18 and 21)
Another Characterization of Continuity
[Theorem 21.3] Suppose $ (S, d) $ and $ (S^*, d^*) $ are metric spaces. A function $ f : S \to{} S^* $ is continuous if and only if $ f^{-1}(U) $ is open for every open $ U \subset{} S^* $. Here: $$\begin{align} f^{-1}(U) = \{s \in S : f(s) \in U\}. \end{align}$$ [Exercise 21.2] $ f $ is continuous at $ s_0 \in{} S $ if and only if for any open set $ U \ni{} f(s_0) $, there exists an open set $ V \ni{} s_0 $ such that $ f(V) \subset{} U $. [Exercise 21.4] Suppose $ (S, d) $ is a metric space. A function $ f : S \to{} \mathbb{R} $ is continuous if and only if $ f^{-1}((a, b)) $ is open whenever $ a < b $.Continuous Image of a Compact Set
If $ f : S \to{} S^* $ is continuous, and $ E \subset{} S $ is compact, then $ f(E) \subset{} S^* $ is compact. If $ f : S \to{} \mathbb{R} $ is continuous, and $ E \subset{} S $ is compact, then $ f(E) $ is bounded. Moreover, $ f $ attains its maximum and minimum values, i.e., there exist $ x, y \in{} E $ such that: $$\begin{align} f(x) = \sup_{e \in E} f(e), \quad f(y) = \inf_{e \in E} f(e). \end{align}$$Intermediate Value Theorem (IVT)
[Theorem 18.2] Suppose $ I \subset{} \mathbb{R} $ is an interval, and $ f : I \to{} \mathbb{R} $ is continuous. Then $ f $ has the Intermediate Value Property (IVP) on $ I $: if $ a, b \in{} I $ with $ a < b $, and $ y $ lies between $ f(a) $ and $ f(b) $, then there exists $ x \in{} (a, b) $ such that $ f(x) = y $. If $ I $ is an interval, and $ f : I \to{} \mathbb{R} $ has the IVP, then $ f(I) $ is either an interval or a single point.Applications of IVT
[Roots of Polynomials] Any polynomial of odd degree has at least one real root. [Existence of Fixed Points] Any continuous function $ f : [0, 1] \to{} [0, 1] $ has a fixed point, i.e., $ x \in{} [0, 1] $ such that $ f(x) = x $. [Existence of $ m $-th Roots] For any $ m \in{} \mathbb{N} $ and $ y > 0 $, there exists $ x > 0 $ such that $ x^m = y $.Continuity of Inverse Functions
[Theorem 18.4] Suppose $ I \subset{} \mathbb{R} $ is an interval, and $ f : I \to{} \mathbb{R} $ is strictly increasing and continuous. Then $ J = f(I) $ is an interval, and $ f^{-1} : J \to{} I $ is strictly increasing and continuous. The function $ x \mapsto{} x^{1/m} $, taking $ [0, \infty{}) $ to itself, is continuous.Continuity and Compactness (Section 18)
Continuous Image of a Compact Set
[Theorem 21.4(i)] Suppose $ f : S \to{} S^* $ is continuous, where $ (S, d) $ and $ (S^*, d^*) $ are metric spaces, and $ E \subset{} S $ is compact. Then $ f(E) \subset{} S^* $ is compact.Proof.
Let $ (U_i)_{i \in{} I} $ be an open cover for $ f(E) $. Define $ V_i = f^{-1}(U_i) $, which are open sets forming a cover for $ E $. By compactness of $ E $, there exist $ i_1, \ldots{}, i_n \in{} I $ such that $ E \subset{} \bigcup_{k=1}^n V_{i_k} $. It follows that $ f(E) \subset{} \bigcup_{k=1}^n U_{i_k} $, proving compactness of $ f(E) $.
∎
Maximum and Minimum of Continuous Functions
[Similar to 18.1] If $ f : S \to{} \mathbb{R} $ is continuous and $ E \subset{} S $ is compact, then $ f(E) $ is bounded. Moreover, $ f $ attains its maximum and minimum values, i.e., there exist $ x, y \in{} E $ such that: $$\begin{align} f(x) = \sup_{e \in E} f(e), \quad f(y) = \inf_{e \in E} f(e). \end{align}$$Intermediate Value Property
[IVP] Suppose $ I \subset{} \mathbb{R} $ is an interval, and $ f : I \to{} \mathbb{R} $ is a function. $ f $ has the Intermediate Value Property (IVP) on $ I $ if for any $ a, b \in{} I $ with $ a < b $, and any $ y $ between $ f(a) $ and $ f(b) $, there exists $ x \in{} (a, b) $ such that $ f(x) = y $. [Theorem 18.2] Any continuous function has the IVP.Applications of IVP
[18.3] If $ I $ is an interval, and $ f : I \to{} \mathbb{R} $ has the IVP, then $ f(I) $ is either an interval or a single point. [Roots of Polynomials] Any polynomial of odd degree has at least one real root. [Existence of Fixed Points] Any continuous function $ f : [0, 1] \to{} [0, 1] $ has a fixed point, i.e., a point $ x \in{} [0, 1] $ such that $ f(x) = x $. [Existence of $ m $-th Root] For any $ m \in{} \mathbb{N} $ and $ y > 0 $, there exists $ x > 0 $ such that $ x^m = y $.Continuity of Inverse Functions
[Theorem 18.4] Suppose $ I \subset{} \mathbb{R} $ is an interval, and $ f : I \to{} \mathbb{R} $ is strictly increasing and continuous. Then $ f(I) $ is an interval, and $ f^{-1} : f(I) \to{} I $ is strictly increasing and continuous. The function $ x \mapsto{} x^{1/m} $, taking $ [0, \infty{}) $ to itself, is continuous.Monotonicity of Injective Functions
[Theorem 18.6] Suppose $ f : I \to{} \mathbb{R} $ is a continuous one-to-one function on an interval $ I $. Then $ f $ is strictly monotone.Proof. [Sketch]
For $ a, b \in{} I $ with $ a < b $, if $ f(a) < f(b) $, then $ f $ is strictly increasing. Otherwise, by the IVP, there would exist $ x \in{} (a, b) $ such that $ f(x) = f(a) $, contradicting injectivity.
∎
Uniform Continuity and Lipschitz Functions (Sections 18-19)
Uniform Continuity
[Uniform Continuity (Definition 21.1)] Suppose $ (S, d) $ and $ (S^*, d^*) $ are metric spaces. A function $ f : S \to{} S^* $ is uniformly continuous on $ E \subset{} S $ if: $$\begin{align} \forall \varepsilon > 0, \exists \delta > 0 \text{ such that } d^*(f(x), f(y)) < \varepsilon \text{ whenever } d(x, y) < \delta. \end{align}$$ Here, $ \delta{} $ depends only on $ \varepsilon{} $ and not on the specific point $ x $. [Sequential Criterion for Uniform Continuity (Theorem 19.4)] If $ f : S \to{} S^* $ is uniformly continuous, then $ f $ maps Cauchy sequences in $ S $ to Cauchy sequences in $ S^* $. [Non-Uniformly Continuous Function] The function $ f(x) = \frac{1}{x} $ is not uniformly continuous on $ (0, \infty{}) $, as the Cauchy sequence $ x_n = \frac{1}{n} $ is mapped to $ f(x_n) = n $, which is not Cauchy.Lipschitz Functions
[Lipschitz Continuity] A function $ f : S \to{} S^* $ is Lipschitz if there exists $ K > 0 $ (called the Lipschitz constant) such that: $$\begin{align} d^*(f(s), f(t)) \leq K \cdot d(s, t), \quad \forall s, t \in S. \end{align}$$ Any Lipschitz function is uniformly continuous.Proof.
Let $ \varepsilon{} > 0 $ and set $ \delta{} = \frac{\varepsilon}{K} $. Then if $ d(s, t) < \delta{} $, it follows that:
$$\begin{align}
d^*(f(s), f(t)) \leq K \cdot d(s, t) < K \cdot \frac{\varepsilon}{K} = \varepsilon.
\end{align}$$
∎
For $ a > 0 $, $ f(x) = \frac{1}{x} $ is Lipschitz (hence uniformly continuous) on $ [a, \infty{}) $. [Uniformly Continuous but Not Lipschitz] The function $ f(x) = \sqrt{x} $ is uniformly continuous on $ [0, \infty{}) $ but not Lipschitz.
Uniform Continuity on Compact Sets
[Uniform Continuity on Compact Sets (Theorem 21.4(ii))] If $ f : S \to{} S^* $ is continuous, and $ E \subset{} S $ is compact, then $ f $ is uniformly continuous on $ E $.Proof. [Sketch of Proof]
Assume $ f $ is not uniformly continuous. Then $ \exists{} \varepsilon{} > 0 $ and sequences $ (x_n), (y_n) \subset{} E $ such that $ d(x_n, y_n) \to{} 0 $ but $ d^*(f(x_n), f(y_n)) \geq{} \varepsilon{} $. By compactness, $ (x_n) $ has a subsequence $ (x_{n_k}) $ converging to some $ x \in{} E $, and $ (y_{n_k}) $ also converges to $ x $. Continuity of $ f $ implies $ d^*(f(x_{n_k}), f(y_{n_k})) \to{} 0 $, contradicting $ d^*(f(x_{n_k}), f(y_{n_k})) \geq{} \varepsilon{} $.
∎
Uniform Continuity and Connectedness (Section 22)
Uniform Continuity on Compact Sets
[Theorem 21.4(ii)] Suppose $ (S, d) $ and $ (S^*, d^*) $ are metric spaces, and $ f : S \to{} S^* $ is continuous. If $ E \subset{} S $ is compact, then $ f|_E $ is uniformly continuous.Proof.
For $ \varepsilon{} > 0 $, find $ \delta{} > 0 $ such that $ d^*(f(s), f(t)) < \varepsilon{} $ whenever $ d(s, t) < \delta{} $. For $ s \in{} S $, find $ \delta_s{} > 0 $ such that $ d^*(f(s), f(t)) < \varepsilon{}/2 $ whenever $ d(s, t) < \delta_s{} $. Since $ E $ is compact:
$$\begin{align}
E \subset \bigcup_{s \in E} B_{\delta_s/2}^o(s),
\end{align}$$
there exist $ s_1, \ldots{}, s_n $ such that $ E \subset{} \bigcup_{k=1}^n B_{\delta_{s_k}/2}^o(s_k) $. Define $ \delta{} = \frac{1}{2} \min_{1 \leq k \leq n} \delta_{s_k} $. For $ s, t \in{} E $ with $ d(s, t) < \delta{} $, choose $ s_k $ such that $ s \in{} B_{\delta_{s_k}/2}^o(s_k) $. Then:
$$\begin{align}
d(t, s_k) \leq d(t, s) + d(s, s_k) < \delta + \frac{\delta_{s_k}}{2} \leq \delta_{s_k},
\end{align}$$
implying $ d^*(f(s), f(t)) \leq{} d^*(f(s), f(s_k)) + d^*(f(t), f(s_k)) < \varepsilon{} $.
∎
Extension of Uniformly Continuous Functions
Suppose $ E \subset{} S $ is compact, $ f : E \to{} S^* $, and $ S^* $ is complete. Then $ f $ is uniformly continuous if and only if it extends to a continuous $ \tilde{f} : \overline{E} \to{} S^* $.Proof. [Sketch]
If $ f $ is uniformly continuous, define $ \tilde{f}(x) = \lim_{n \to \infty} f(x_n) $ for $ x \in{} \overline{E} $, where $ x_n \subset{} E $ and $ x_n \to{} x $. The limit exists by completeness and does not depend on the sequence. Continuity of $ \tilde{f} $ follows from the uniform continuity of $ f $.
∎
Connectedness
[Connected and Disconnected Sets] Suppose $ (S, d) $ is a metric space. A set $ E \subset{} S $ is disconnected if there exist open sets $ U_1, U_2 \subset{} S $ such that:- $ E \subset U_1 \cup U_2 $,
- $ (E \cap U_1) \cap (E \cap U_2) = \emptyset $,
- $ E \cap U_1 \neq \emptyset $ and $ E \cap U_2 \neq \emptyset $.
Connectedness of Intervals
Any interval $ I \subset{} \mathbb{R} $ is connected.Proof.
Suppose, for contradiction, that $ I = A \cup{} B $, where $ A, B \neq{} \emptyset{} $, $ \overline{A} \cap{} B = \emptyset{} $, and $ A \cap{} \overline{B} = \emptyset{} $. Choose $ a \in{} A $, $ b \in{} B $, with $ a < b $. Define:
$$\begin{align}
c = \sup \{x \in A : x < b\}.
\end{align}$$
Then $ c \in{} I $ and $ c < b $. If $ c \in{} A $, there exists $ \sigma{} > 0 $ such that $ (c - \sigma{}, c + \sigma{}) \subset{} A $, contradicting the definition of $ c $. If $ c \in{} B $, there exists $ \sigma{} > 0 $ such that $ (c - \sigma{}, c + \sigma{}) \subset{} B $, contradicting $ c = \sup{} \{x \in{} A : x < b\} $.
∎
Connectedness and Path Connectedness (Section 22)
Connectedness
[Connected Set (Definition 22.1)] Suppose $ (S, d) $ is a metric space. A set $ E \subset{} S $ is called disconnected if there exist open sets $ U_1, U_2 \subset{} S $ such that:- $ E \subset U_1 \cup U_2 $,
- $ (E \cap U_1) \cap (E \cap U_2) = \emptyset $,
- $ E \cap U_1 \neq \emptyset $ and $ E \cap U_2 \neq \emptyset $.
Continuous Images of Connected Sets
[Theorem 22.2] Suppose $ (S, d) $ and $ (S^*, d^*) $ are metric spaces. If $ E \subset{} S $ is connected and $ f : S \to{} S^* $ is continuous, then $ f(E) $ is connected.Proof. [Sketch of Contrapositive Proof]
If $ f(E) \subset{} S^* $ is disconnected, write $ f(E) = C \cup{} D $, where $ C, D $ are disjoint, non-empty, closed subsets. Define $ A = f^{-1}(C) \cap{} E $, $ B = f^{-1}(D) \cap{} E $. Then $ E = A \cup{} B $, $ A \cap{} \overline{B} = \emptyset{} $, and $ \overline{A} \cap{} B = \emptyset{} $, so $ E $ is disconnected.
∎
Path Connectedness
[Path Connectedness (Definition 22.4)] A set $ E \subset{} S $ is path connected if for all $ a, b \in{} E $, there exists a continuous function $ \gamma{} : [0, 1] \to{} E $ such that $ \gamma{}(0) = a $ and $ \gamma{}(1) = b $. [Path Connected Sets Are Connected (Theorem 22.5)] Every path connected set is connected.Proof.
If $ E $ is disconnected, then there exist open sets $ U_1, U_2 $ such that $ E \subset{} U_1 \cup{} U_2 $, $ E \cap{} U_1 \neq{} \emptyset{} $, $ E \cap{} U_2 \neq{} \emptyset{} $, and $ (E \cap{} U_1) \cap{} (E \cap{} U_2) = \emptyset{} $. Let $ a \in{} E \cap{} U_1 $, $ b \in{} E \cap{} U_2 $. A path $ \gamma{} : [0, 1] \to{} E $ with $ \gamma{}(0) = a $, $ \gamma{}(1) = b $ would imply $ \gamma{}([0, 1]) $ is connected, contradicting the disconnectedness of $ E $.
∎
Connected but Not Path Connected Sets
Consider $ E \subset{} \mathbb{R}^2 $, where: $$\begin{align} E_1 = \{(0, y) : y \in (0, 1]\}, \quad E_2 = \{(x, 0) : x \in (0, 1]\} \cup \bigcup_{n \in \mathbb{N}} \{(1/n, y) : y \in (0, 1]\}. \end{align}$$ Then $ E = E_1 \cup{} E_2 $ is connected but not path connected.Convex Sets
[Convex Sets] A set $ E \subset{} \mathbb{R}^n $ is convex if for all $ \vec{x}, \vec{y} \in{} E $ and $ t \in{} [0, 1] $, the point: $$\begin{align} \vec{z} = (1-t)\vec{x} + t\vec{y} \in E. \end{align}$$ Any convex set is path connected.Path Connectedness of Graphs
The graph of a function $ f : I \to{} \mathbb{R} $, where $ I \subset{} \mathbb{R} $ is an interval, is path connected if and only if $ f $ is continuous.Graphs of Functions and Path Connectedness (Sections 22-23-24)
Graphs and Path Connectedness
[Graph of a Function] The graph of a function $ f : I \to{} \mathbb{R} $ (where $ I \subset{} \mathbb{R} $ is an interval) is: $$\begin{align} G(f) = \{(x, f(x)) : x \in I\}. \end{align}$$ [Example 4 from Section 22] $ G(f) $ is path connected if and only if $ f $ is continuous on $ I $. [Discontinuous $ f $ with Connected $ G(f) $] Exercise 22.4 describes a function $ f $ such that $ G(f) $ is connected but $ f $ is discontinuous. [Multivariate Continuity] The function $ f : S \to{} \mathbb{R}^n $, $ x \mapsto{} (f_1(x), \ldots{}, f_n(x)) $, is continuous if and only if each $ f_i : S \to{} \mathbb{R} $ is continuous for $ 1 \leq{} i \leq{} n $.Proof. [Sketch]
If $ f $ is continuous, then $ G(f) $ is path connected. For $ \vec{x} = (a, f(a)) $, $ \vec{y} = (b, f(b)) \in{} G(f) $, define a path:
$$\begin{align}
\gamma(t) = ((1-t)a + tb, f((1-t)a + tb)), \quad t \in [0, 1].
\end{align}$$
If $ G(f) $ is path connected, continuity of $ f $ follows from the textbook proof.
∎
Power Series
[Power Series] A power series is a series of the form: $$\begin{align} \sum_{n=0}^\infty a_n x^n, \end{align}$$ where $ x $ is a variable. [Radius of Convergence] Let $ \beta{} = \limsup{} |a_n|^{1/n} $ and $ R = 1/\beta{} $. The series: $$\begin{align} \sum_{n=0}^\infty a_n x^n \end{align}$$ converges for $ |x| < R $, diverges for $ |x| > R $. $ R $ is called the radius of convergence. If $ \lim{} |a_{n+1}/a_n| $ exists, it equals $ \beta{} $. The series may converge or diverge at $ \pm{} R $. The interval of convergence is one of: $$\begin{align} (-R, R), \, [-R, R), \, (-R, R], \, \text{or} \, [-R, R]. \end{align}$$Examples of Power Series
- $ \sum_{n=0}^\infty \frac{x^n}{n!} $: $ a_n = \frac{1}{n!} $, $ \beta = 0 $, $ R = \infty $. Interval: $ (-\infty, \infty) $. $$\begin{align} \sum_{n=0}^\infty \frac{x^n}{n!} = e^x. \end{align}$$
- $ \sum_{n=0}^\infty x^n $: $ a_n = 1 $, $ \beta = 1 $, $ R = 1 $. Diverges for $ x = \pm 1 $. Interval: $ (-1, 1) $. $$\begin{align} \sum_{n=0}^\infty x^n = \frac{1}{1-x}. \end{align}$$
- $ \sum_{n=0}^\infty \frac{x^n}{n+1} $: $ a_n = \frac{1}{n+1} $, $ \beta = 1 $, $ R = 1 $. Diverges at $ x = 1 $, converges for $ x \in [-1, 1) $. $$\begin{align} \sum_{n=0}^\infty \frac{x^n}{n+1} = \ln(1-x). \end{align}$$
- $ \sum_{n=0}^\infty \frac{x^n}{(n+1)^2} $: $ a_n = \frac{1}{(n+1)^2} $, $ \beta = 1 $, $ R = 1 $. Converges for $ x \in [-1, 1] $.
- $ \sum_{n=0}^\infty n! x^n $: $ a_n = n! $, $ \beta = \infty $, $ R = 0 $. Diverges for all $ x \neq 0 $.
Uniform Convergence
[Uniform Convergence (Definition 24.1-2)] A sequence $ f_n \to{} f $ pointwise on $ S $ if: $$\begin{align} \forall x \in S, \, \forall \varepsilon > 0, \exists N \text{ such that } |f_n(x) - f(x)| < \varepsilon \text{ for } n \geq N. \end{align}$$ It converges uniformly if: $$\begin{align} \forall \varepsilon > 0, \exists N \text{ such that } \sup_{x \in S} |f_n(x) - f(x)| < \varepsilon \text{ for } n \geq N. \end{align}$$ [Preservation of Continuity (Theorem 24.3)] If $ f_n \to{} f $ uniformly on $ S $ and each $ f_n $ is continuous, then $ f $ is continuous. Consider $ f_n(x) = n^2 x^n (1-x) $ on $ [0, 1] $. It converges pointwise to $ f(x) = 0 $, but not uniformly.Uniform Convergence and Series of Functions (Sections 24-25)
Uniform Convergence
[Pointwise and Uniform Convergence (24.1-2)] Suppose $ f, f_1, f_2, \ldots{} $ are functions $ S \to{} \mathbb{R} $.- $ f_n \to f $ pointwise on $ S $ if: $$\begin{align} \forall x \in S, \, \forall \varepsilon > 0, \, \exists N \in \mathbb{N} \text{ such that } |f_n(x) - f(x)| < \varepsilon \text{ for } n \geq N. \end{align}$$
- $ f_n \to f $ uniformly on $ S $ if: $$\begin{align} \forall \varepsilon > 0, \, \exists N \in \mathbb{N} \text{ such that } |f_n(x) - f(x)| < \varepsilon \text{ for } n \geq N, \, \forall x \in S. \end{align}$$ Equivalently, $ \lim_{n \to \infty} \sup_{x \in S} |f_n(x) - f(x)| = 0 $.
Proof. [Sketch of Proof]
Using an $ \varepsilon{}/3 $ argument, fix $ \varepsilon{} > 0 $. For $ f_n \to{} f $ uniformly, find $ n $ such that $ |f_n(x) - f(x)| < \varepsilon{}/3 $. By continuity of $ f_n $, there exists $ \delta{} > 0 $ such that $ |f_n(x_0) - f_n(x)| < \varepsilon{}/3 $ for $ |x - x_0| < \delta{} $. Combine inequalities to conclude $ |f(x_0) - f(x)| < \varepsilon{} $.
∎
Uniformly Cauchy Sequences
[Uniformly Cauchy (25.3)] A sequence $ (f_n) $ of functions $ S \to{} \mathbb{R} $ is uniformly Cauchy if: $$\begin{align} \forall \varepsilon > 0, \, \exists N \in \mathbb{N} \text{ such that } |f_i(x) - f_j(x)| < \varepsilon \, \forall x \in S \text{ for } i, j \geq N. \end{align}$$ Equivalently, $ \sup_{x \in S} |f_i(x) - f_j(x)| < \varepsilon{} $ for $ i, j \geq{} N $. [Uniformly Cauchy $ \iff{} $ Uniform Convergence (25.4)] A sequence $ (f_n) $ is uniformly Cauchy if and only if it converges uniformly to some $ f $.Series of Functions
[Convergence of Series] A series $ \sum_{n=1}^\infty{} g_n(x) $ converges (uniformly) if the sequence of partial sums $ s_k(x) = \sum_{n=1}^k g_n(x) $ converges (uniformly). [Uniform Convergence Preserves Continuity (25.5)] If $ g_n : S \to{} \mathbb{R} $ are continuous and $ \sum_{n=1}^\infty{} g_n(x) $ converges uniformly on $ S $, then $ \sum_{n=1}^\infty{} g_n(x) $ is continuous.Weierstrass $ M $-Test
[Weierstrass $ M $-Test (25.7)] Suppose $ M_1, M_2, \ldots{} \geq{} 0 $ and $ \sum_{k=1}^\infty{} M_k < \infty{} $. If $ |g_k(x)| \leq{} M_k $ for all $ x \in{} S $ and $ k $, then $ \sum_{k=1}^\infty{} g_k(x) $ converges uniformly on $ S $. A power series $ \sum_{k=0}^\infty{} a_k x^k $ converges uniformly (to a continuous function) on $ [-b, b] $ if $ b < R $, where $ R = (\limsup{} |a_k|^{1/k})^{-1} $. Convergence need not be uniform on $ (-R, R) $. For example, $ \sum_{k=0}^\infty{} x^k = \frac{1}{1-x} $ converges on $ (-1, 1) $, but not uniformly because the partial sums are bounded while $ \frac{1}{1-x} $ is not.Limits and Differentiation (Sections 20, 28-29)
Limits
[Limit (20.1, slightly modified)] Suppose $ S \subset{} \mathbb{R} $, $ a \in{} S^- $, $ f : S \to{} \mathbb{R} $, and $ L \in{} \mathbb{R} \cup{} \{\pm{} \infty{}\} $. Then: $$\begin{align} \lim_{x \to a, S} f = L \end{align}$$ if $ \lim{} f(x_n) = L $ for any sequence $ (x_n) \subset{} S $ with $ \lim{} x_n = a $. Such sequences $ (x_n) $ exist because $ a \in{} S^- $. [Connection Between Limits and Continuity] If $ a \in{} S $, then $ f : S \to{} \mathbb{R} $ is continuous at $ a $ if and only if $ \lim_{x \to a, S} f = f(a) $.Common Set-Ups for Limits
- **Usual Limit**: Let $ I $ be an interval, $ a $ be interior to $ I $, and $ S = I \setminus \{a\} $. Write $ \lim_{x \to a} f $ instead of $ \lim_{x \to a, S} f $.
- **One-Sided Limit**: For $ S = (a, b) $, write $ \lim_{x \to a^+} f $ (right-hand limit). Define $ \lim_{x \to a^-} f $ similarly.
Useful Theorems About Limits
[Equivalent Definition of Limits (20.6, Simplified)] Suppose $ a \in{} S^- $. For $ f : S \to{} \mathbb{R} $ and $ L \in{} \mathbb{R} $, the following are equivalent:- $ \lim_{x \to a, S} f = L $.
- For all $ \varepsilon > 0 $, there exists $ \delta > 0 $ such that $ |f(x) - L| < \varepsilon $ whenever $ x \in (a - \delta, a + \delta) \cap S \setminus \{a\} $.
- $ \lim_{x \to a, S} (f_1 + f_2) = L_1 + L_2 $,
- $ \lim_{x \to a, S} (f_1 \cdot f_2) = L_1 \cdot L_2 $,
- If $ L_2 \neq 0 $, $ \lim_{x \to a, S} \frac{f_1}{f_2} = \frac{L_1}{L_2} $.
Differentiation
[Derivative (28.1)] Suppose $ I $ is an open interval, $ a \in{} I $, and $ f : I \to{} \mathbb{R} $. The derivative of $ f $ at $ a $ is: $$\begin{align} f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}, \end{align}$$ if the limit exists and is finite.Rules of Differentiation
[Product Rule] If $ f $ and $ g $ are differentiable at $ a $, then $ (fg)'(a) = f'(a)g(a) + f(a)g'(a) $. [Chain Rule (28.4)] If $ f $ is differentiable at $ a $, and $ g $ is differentiable at $ f(a) $, then $ g \circ{} f $ is differentiable at $ a $, with: $$\begin{align} (g \circ f)'(a) = g'(f(a)) \cdot f'(a). \end{align}$$Carathéodory’s Theorem
[Carathéodory (Exercise 28.16)] Suppose $ I $ is an interval, $ f : I \to{} \mathbb{R} $. $ f $ is differentiable at $ a \in{} I $ if and only if there exists a function $ \phi{} : I \to{} \mathbb{R} $, continuous at $ a $, such that: $$\begin{align} f(x) - f(a) = \phi(x) \cdot (x - a), \quad \forall x \in I, \end{align}$$ and $ \phi{}(a) = f'(a) $.Rules of Differentiation and Mean Value Theorem (Sections 28-29)
Definition of Derivative
[Derivative (28.1)] Suppose $ I $ is an open interval and $ a \in{} I $. A function $ f : I \to{} \mathbb{R} $ is differentiable at $ a $ if the derivative: $$\begin{align} f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \end{align}$$ exists and is finite.Rules of Differentiation
[Product Rule (28.3)] Suppose $ f $ and $ g $ are differentiable at $ a $. Then $ fg $ is differentiable at $ a $, with: $$\begin{align} (fg)'(a) = f'(a)g(a) + f(a)g'(a). \end{align}$$ If $ f(x) = x^m $ for $ m \in{} \mathbb{N} $, then: $$\begin{align} (x^m)' = mx^{m-1}. \end{align}$$ [Chain Rule (28.4)] Suppose $ f $ is differentiable at $ a $ and $ g $ is differentiable at $ f(a) $. Then $ g \circ{} f $ is differentiable at $ a $, with: $$\begin{align} (g \circ f)'(a) = g'(f(a))f'(a). \end{align}$$Examples of Differentiation
- If $ f(x) = x^n $ for $ n \in \mathbb{N} $, then $ f'(x) = nx^{n-1} $.
- If $ f(x) = x^{-n} $ for $ n \in \mathbb{N} $, then $ f'(x) = -nx^{-n-1} $.
- For $ f(x) = 1/g(x) $, if $ g(a) \neq 0 $, then: $$\begin{align} \left( \frac{1}{g} \right)'(a) = -\frac{g'(a)}{g(a)^2}. \end{align}$$
Criterion for Extrema
[Extrema Criterion (29.1)] Suppose $ f $ is defined on an open interval $ I $ and has a maximum or minimum at $ x_0 \in{} I $. If $ f $ is differentiable at $ x_0 $, then: $$\begin{align} f'(x_0) = 0. \end{align}$$ Suppose $ f : [a, b] \to{} \mathbb{R} $ is continuous and attains its maximum or minimum at $ x_0 $. Then one of the following holds:- $ x_0 \in \{a, b\} $,
- $ f $ is not differentiable at $ x_0 $,
- $ f'(x_0) = 0 $.
Rolle’s Theorem
[Rolle’s Theorem (29.2)] Suppose $ f : [a, b] \to{} \mathbb{R} $ is continuous, differentiable on $ (a, b) $, and $ f(a) = f(b) $. Then there exists $ c \in{} (a, b) $ such that: $$\begin{align} f'(c) = 0. \end{align}$$Mean Value Theorem
[Mean Value Theorem (29.3)] Suppose $ f : [a, b] \to{} \mathbb{R} $ is continuous, differentiable on $ (a, b) $. Then there exists $ c \in{} (a, b) $ such that: $$\begin{align} f'(c) = \frac{f(b) - f(a)}{b - a}. \end{align}$$Examples and Consequences of MVT
[Constant Function (29.4)] If $ f $ is differentiable on $ (a, b) $ and $ f'(x) = 0 $ for all $ x \in{} (a, b) $, then $ f $ is a constant function. [Equality of Derivatives (29.5)] If $ f $ and $ g $ are differentiable on $ (a, b) $ and $ f'(x) = g'(x) $ for all $ x \in{} (a, b) $, then $ f(x) - g(x) = c $ for some constant $ c $.Rolle’s Theorem, Mean Value Theorem, and Applications (Section 29)
Rolle’s Theorem
[Rolle’s Theorem (29.2)] Suppose $ f : [a, b] \to{} \mathbb{R} $ is continuous, differentiable on $ (a, b) $, and $ f(a) = f(b) $. Then there exists $ x \in{} (a, b) $ such that: $$\begin{align} f'(x) = 0. \end{align}$$Proof.
The function $ f $ attains its maximum and minimum on $ [a, b] $. Let $ x_0, y_0 \in{} [a, b] $ such that $ f(y_0) \leq{} f(x) \leq{} f(x_0) $ for all $ x \in{} [a, b] $. If $ f(y_0) = f(a) = f(b) = f(x_0) $, then $ f $ is constant, so $ f' = 0 $ on $ (a, b) $. Otherwise:
- If $ f(x_0) > f(a) = f(b) $, then $ x_0 \in (a, b) $, and $ f'(x_0) = 0 $.
- If $ f(y_0) < f(a) = f(b) $, then $ y_0 \in (a, b) $, and $ f'(y_0) = 0 $.
∎
Mean Value Theorem
[Mean Value Theorem (29.3)] Suppose $ f : [a, b] \to{} \mathbb{R} $ is continuous, differentiable on $ (a, b) $. Then there exists $ x \in{} (a, b) $ such that: $$\begin{align} f'(x) = \frac{f(b) - f(a)}{b - a}. \end{align}$$Proof.
Define $ L(x) = f(a) + \frac{f(b) - f(a)}{b - a}(x - a) $ and $ g(x) = f(x) - L(x) $. The function $ g $ is continuous on $ [a, b] $, differentiable on $ (a, b) $, and $ g(a) = g(b) = 0 $. By Rolle’s Theorem, there exists $ x \in{} (a, b) $ such that $ g'(x) = 0 $. Thus:
$$\begin{align}
f'(x) = g'(x) + L'(x) = 0 + \frac{f(b) - f(a)}{b - a}.
\end{align}$$
∎
[MVT Application] For $ x, y \in{} \mathbb{R} $, $ |\sin{} x - \sin{} y| \leq{} |x - y| $. Apply MVT to $ f(t) = \sin{} t $ on $ [x, y] $: $ \exists{} z \in{} (x, y) $ such that: $$\begin{align} \frac{f(x) - f(y)}{x - y} = f'(z) = \cos z. \end{align}$$ Since $ |\cos{} z| \leq{} 1 $, $ | \frac{f(x) - f(y)}{x - y} | = |\cos{} z| \leq{} 1 $, hence $ |\sin{} x - \sin{} y| \leq{} |x - y| $.
Corollaries of MVT
[Constant Functions (29.4)] If $ f $ is differentiable on $ (a, b) $ and $ f' = 0 $ on $ (a, b) $, then $ f $ is constant.Proof.
If $ f $ is not constant, then there exist $ x < y $ such that $ f(x) \neq{} f(y) $. By MVT, $ \exists{} z \in{} (x, y) $ such that:
$$\begin{align}
f'(z) = \frac{f(y) - f(x)}{y - x} \neq 0,
\end{align}$$
contradicting $ f'(z) = 0 $.
∎
[Equality of Derivatives (29.5)] If $ f, g $ are differentiable on $ (a, b) $ and $ f' = g' $ on $ (a, b) $, then $ \exists{} c \in{} \mathbb{R} $ such that $ f(x) - g(x) = c $ for all $ x \in{} (a, b) $.
Proof.
Define $ h(x) = f(x) - g(x) $. Then $ h' = f' - g' = 0 $. By Corollary 29.4, $ h $ is constant.
∎
Increasing and Decreasing Functions
[Monotonicity (29.6)] A function $ f $ on an interval $ I $ is:- {Increasing} if $ f(x_1) \leq f(x_2) $ for $ x_1 < x_2 $,
- {Strictly increasing} if $ f(x_1) < f(x_2) $ for $ x_1 < x_2 $.
- $ f $ is increasing if and only if $ f' \geq 0 $ on $ (a, b) $,
- If $ f' > 0 $ on $ (a, b) $, then $ f $ is strictly increasing.
Differentiating Inverse Functions and Integration (Sections 29, 32)
Differentiating Inverse Functions
[Derivative of an Inverse Function (29.9)] Suppose $ I $ is an interval, $ f : I \to{} \mathbb{R} $ is a continuous, strictly monotone function. Let $ J = f(I) $, and $ g = f^{-1} : J \to{} I $. If $ f $ is differentiable at $ c \in{} I $, and $ f'(c) \neq{} 0 $, then $ g $ is differentiable at $ d = f(c) $, and: $$\begin{align} g'(d) = \frac{1}{f'(c)} = \frac{1}{f'(g(d))}. \end{align}$$Proof. [Proof Sketch]
Using Carathéodory’s Theorem:
$$\begin{align}
f(x) - f(c) = \phi(x)(x - c), \quad \phi(c) = f'(c),
\end{align}$$
where $ \phi{} $ is continuous at $ c $. For $ y = f(g(y)) $, differentiate both sides to find $ g'(d) = 1 / \phi{}(g(d)) $.
∎
Derivatives of Rational Powers
[Derivative of Rational Powers] Let $ f(x) = x^n $ (strictly increasing on $ (0, \infty{}) $) with $ f'(x) = nx^{n-1} $. The inverse function is $ g(y) = y^{1/n} $. For $ y > 0 $: $$\begin{align} g'(y) = \frac{1}{f'(g(y))} = \frac{1}{n(y^{1/n})^{n-1}} = \frac{1}{n} y^{1/n - 1}. \end{align}$$ If $ n \in{} \mathbb{Z} $ is odd, extend $ f, g $ to $ \mathbb{R} $. Then $ g'(y) = \frac{1}{n} y^{1/n - 1} $ for $ y < 0 $. [Derivative of $ h(x) = x^r $, $ r \in{} \mathbb{Q} $] Write $ r = m/n $, $ h(x) = x^{m/n} $. Use the chain rule: $$\begin{align} h'(x) = \frac{m}{n} x^{r-1}. \end{align}$$Inverse Trigonometric Functions
[Arcsine] For $ f(x) = \sin{} x $ on $ [-\pi{}/2, \pi{}/2] $, $ g = \arcsin{} : [-1, 1] \to{} [-\pi{}/2, \pi{}/2] $. Since $ f'(x) = \cos{} x $, for $ y \in{} (-1, 1) $: $$\begin{align} (\arcsin y)' = \frac{1}{\sqrt{1-y^2}}. \end{align}$$ [Arctangent] For $ f(x) = \tan{} x $ on $ (-\pi{}/2, \pi{}/2) $, $ g = \arctan{} : \mathbb{R} \to{} (-\pi{}/2, \pi{}/2) $. Since $ f'(x) = 1 + x^2 $, for $ y \in{} \mathbb{R} $: $$\begin{align} (\arctan y)' = \frac{1}{1 + y^2}. \end{align}$$Integration: Concepts and Definitions
[Darboux Sums] Let $ f : [a, b] \to{} \mathbb{R} $ be bounded.- Partition $ P = \{t_0, t_1, \ldots, t_n\} $ of $ [a, b] $ gives subintervals $ [t_{k-1}, t_k] $.
- Lower Darboux sum: $$\begin{align} L(f, P) = \sum_{k=1}^n m(f, [t_{k-1}, t_k])(t_k - t_{k-1}), \end{align}$$ where $ m(f, [t_{k-1}, t_k]) = \inf_{x \in [t_{k-1}, t_k]} f(x) $.
- Upper Darboux sum: $$\begin{align} U(f, P) = \sum_{k=1}^n M(f, [t_{k-1}, t_k])(t_k - t_{k-1}), \end{align}$$ where $ M(f, [t_{k-1}, t_k]) = \sup_{x \in [t_{k-1}, t_k]} f(x) $.
Examples of Integrability
[Constant Function] If $ f(x) = c $, then: $$\begin{align} \int_a^b f(x) dx = c(b-a). \end{align}$$ [Discontinuous Function] Let $ g(x) = 1 $ if $ x \in{} \mathbb{Q} $, $ g(x) = 0 $ otherwise. Then: $$\begin{align} \sup_P L(g, P) = 0, \quad \inf_P U(g, P) = 1. \end{align}$$ Since $ \sup_P{} L \neq{} \inf_P{} U $, $ g $ is not integrable. [Linear Function] If $ h(x) = x $, then: $$\begin{align} \int_0^b h(x) dx = \frac{b^2}{2}. \end{align}$$Darboux Sums, Integrability, and Riemann Integration (Sections 32-33)
Darboux Sums and Integrals
[Darboux Sums] Let $ f : [a, b] \to{} \mathbb{R} $ be bounded. For a partition $ P = \{a = t_0 < t_1 < \cdots{} < t_n = b\} $:- Lower Darboux sum: $$\begin{align} L(f, P) = \sum_{k=1}^n m(f, [t_{k-1}, t_k])(t_k - t_{k-1}), \end{align}$$ where $ m(f, [t_{k-1}, t_k]) = \inf_{x \in [t_{k-1}, t_k]} f(x) $.
- Upper Darboux sum: $$\begin{align} U(f, P) = \sum_{k=1}^n M(f, [t_{k-1}, t_k])(t_k - t_{k-1}), \end{align}$$ where $ M(f, [t_{k-1}, t_k]) = \sup_{x \in [t_{k-1}, t_k]} f(x) $.
Integrals: Example
[Linear Function] Is $ h(x) = x $ integrable on $ [0, b] $? Compute $ \int_0{}^b h(x) dx $. For $ P = \{0, \frac{b}{n}, \frac{2b}{n}, \ldots{}, b\} $: $$\begin{align} L(h, P_n) = \frac{b^2}{2}\left(1 - \frac{1}{n}\right), \quad U(h, P_n) = \frac{b^2}{2}. \end{align}$$ Thus: $$\begin{align} L(h) \geq \sup_n L(h, P_n) = \frac{b^2}{2}, \quad U(h) \leq \lim_n U(h, P_n) = \frac{b^2}{2}. \end{align}$$ Since $ L(h) = U(h) $, $ h(x) $ is integrable with: $$\begin{align} \int_0^b h(x) dx = \frac{b^2}{2}. \end{align}$$Criterion for Integrability
[32.5] A bounded function $ f : [a, b] \to{} \mathbb{R} $ is integrable if and only if for all $ \varepsilon{} > 0 $, there exists a partition $ P $ such that: $$\begin{align} U(f, P) - L(f, P) < \varepsilon. \end{align}$$Monotone and Continuous Functions
[33.1] Any monotone function on $ [a, b] $ is integrable. [33.2] Any continuous function on $ [a, b] $ is integrable.Mesh of a Partition
[Mesh (32.6)] The mesh of a partition $ P = \{t_0, t_1, \ldots{}, t_n\} $ is: $$\begin{align} \text{mesh}(P) = \max_{1 \leq k \leq n} (t_k - t_{k-1}). \end{align}$$ [32.7] A bounded $ f : [a, b] \to{} \mathbb{R} $ is integrable if and only if for all $ \varepsilon{} > 0 $, there exists $ \delta{} > 0 $ such that: $$\begin{align} U(f, P) - L(f, P) < \varepsilon \quad \text{whenever mesh}(P) < \delta. \end{align}$$Riemann Integration
[Riemann Integral (32.8)] Let $ f : [a, b] \to{} \mathbb{R} $ be bounded. For a partition $ P = \{t_0, t_1, \ldots{}, t_n\} $ and $ x_k \in{} [t_{k-1}, t_k] $, define the Riemann sum: $$\begin{align} S = \sum_{k=1}^n f(x_k)(t_k - t_{k-1}). \end{align}$$ $ f $ is Riemann integrable if there exists $ r \in{} \mathbb{R} $ such that for all $ \varepsilon{} > 0 $, there exists $ \delta{} > 0 $ such that: $$\begin{align} |S - r| < \varepsilon \quad \text{whenever mesh}(P) < \delta. \end{align}$$ [32.9] A bounded function $ f : [a, b] \to{} \mathbb{R} $ is Riemann integrable if and only if it is Darboux integrable. In this case: $$\begin{align} \int_a^b f = R\int_a^b f. \end{align}$$Integrability and Riemann Integration (Sections 32-33)
Monotone and Continuous Functions
[Integrability Criterion (32.5)] A bounded function $ f : [a, b] \to{} \mathbb{R} $ is integrable if and only if: $$\begin{align} \forall \varepsilon > 0, \exists \text{ a partition } P \text{ such that } U(f, P) - L(f, P) < \varepsilon. \end{align}$$ [Monotone Functions Are Integrable (33.1)] Any monotone function on $ [a, b] $ is integrable.Proof.
Assume $ f $ is increasing. Fix $ \varepsilon{} > 0 $. Choose $ n \in{} \mathbb{N} $ such that:
$$\begin{align}
\frac{(f(b) - f(a))(b - a)}{n} < \varepsilon.
\end{align}$$
Consider the partition $ P $ with $ t_k = a + kh $ for $ 0 \leq{} k \leq{} n $, where $ h = \frac{b-a}{n} $. Then:
$$\begin{align}
U(f, P) - L(f, P) = h \sum_{k=1}^n (f(t_k) - f(t_{k-1})) = \frac{(f(b) - f(a))(b - a)}{n} < \varepsilon.
\end{align}$$
∎
[Continuous Functions Are Integrable (33.2)] Any continuous function on $ [a, b] $ is integrable.
Proof.
Fix $ \varepsilon{} > 0 $. By uniform continuity, $ \exists{} \delta{} > 0 $ such that $ |f(x) - f(y)| < \frac{\varepsilon}{b-a} $ whenever $ |x - y| < \delta{} $. Choose $ n \in{} \mathbb{N} $ such that $ h = \frac{b-a}{n} < \delta{} $, and partition $ P $ with $ t_k = a + kh $. Then:
$$\begin{align}
M(f, [t_{k-1}, t_k]) - m(f, [t_{k-1}, t_k]) < \frac{\varepsilon}{b-a}.
\end{align}$$
Thus:
$$\begin{align}
U(f, P) - L(f, P) < hn \cdot \frac{\varepsilon}{b-a} = \varepsilon.
\end{align}$$
∎
Mesh of a Partition
[Mesh (32.6)] The mesh of a partition $ P = \{t_0, t_1, \ldots{}, t_n\} $ is: $$\begin{align} \text{mesh}(P) = \max_{1 \leq k \leq n} (t_k - t_{k-1}). \end{align}$$ [Integrability and Mesh (32.7)] A bounded function $ f : [a, b] \to{} \mathbb{R} $ is integrable if and only if: $$\begin{align} \forall \varepsilon > 0, \exists \delta > 0 \text{ such that } U(f, P) - L(f, P) < \varepsilon \text{ whenever mesh}(P) < \delta. \end{align}$$Riemann Integration
[Riemann Integral (32.8)] Let $ f : [a, b] \to{} \mathbb{R} $ be bounded. For a partition $ P = \{t_0, t_1, \ldots{}, t_n\} $ and $ x_k \in{} [t_{k-1}, t_k] $, define the Riemann sum: $$\begin{align} S = \sum_{k=1}^n f(x_k)(t_k - t_{k-1}). \end{align}$$ $ f $ is Riemann integrable if: $$\begin{align} \exists r \in \mathbb{R} \text{ such that } \forall \varepsilon > 0, \exists \delta > 0 \text{ such that } |S - r| < \varepsilon \text{ whenever mesh}(P) < \delta. \end{align}$$ [Equivalence of Riemann and Darboux Integrability (32.9)] A bounded $ f : [a, b] \to{} \mathbb{R} $ is Riemann integrable if and only if it is Darboux integrable. In this case: $$\begin{align} R\int_a^b f = \int_a^b f. \end{align}$$Properties of Integrable Functions
[Exercise 32.7] If $ f $ is integrable on $ [a, b] $, and $ f = g $ except at finitely many points, then $ g $ is integrable on $ [a, b] $ and: $$\begin{align} \int_a^b f = \int_a^b g. \end{align}$$ The statement fails if the set of exceptions is countably infinite. For example: $$\begin{align} f(x) = 0, \quad g(x) = \begin{cases} 1 & x \in \mathbb{Q}, \\ 0 & x \notin \mathbb{Q}. \end{cases} \end{align}$$ $ f $ is integrable, but $ g $ is not.Properties of Integrals and Convergence Theorems (Section 33)
Properties of Integrals
[Linearity and Comparison of Integrals (33.3, 33.4(i))] Suppose $ f, g $ are integrable on $ [a, b] $, and $ c \in{} \mathbb{R} $. Then:- $ cf $ is integrable, and: $$\begin{align} \int_a^b cf = c \int_a^b f. \end{align}$$
- $ f + g $ is integrable, and: $$\begin{align} \int_a^b (f + g) = \int_a^b f + \int_a^b g. \end{align}$$
- If $ f \geq g $, then: $$\begin{align} \int_a^b f \geq \int_a^b g. \end{align}$$
Integrability of Products and Piecewise Functions
If $ f $ is integrable on $ [a, b] $, then $ f^2 $ is integrable. If $ f, g $ are integrable on $ [a, b] $, then $ fg $ is integrable.Proof.
Express $ fg $ as:
$$\begin{align}
fg = \frac{1}{4} \left( (f + g)^2 - (f - g)^2 \right).
\end{align}$$
Since $ f + g $ and $ f - g $ are integrable, their squares are integrable, and hence $ fg $ is integrable.
∎
[Piecewise Monotone and Continuous Functions (33.8)] Suppose $ f : [a, b] \to{} \mathbb{R} $ is either:
- Piecewise monotone and bounded, or
- Piecewise continuous.
Proof.
Partition $ [a, b] $ such that $ f $ is monotone or uniformly continuous on each subinterval. On each subinterval, $ f $ is integrable. By additivity of the integral, $ f $ is integrable on $ [a, b] $.
∎
Convergence and Interchange of Limits and Integrals
Suppose $ (f_n) $ is a sequence of integrable functions on $ [a, b] $ that converges uniformly to $ f $. Then $ f $ is integrable, and: $$\begin{align} \int_a^b f = \lim_{n \to \infty} \int_a^b f_n. \end{align}$$Convergence Theorems
[Bounded Convergence (33.11)] Suppose $ (f_n) $ are integrable on $ [a, b] $, $ |f_n| \leq{} M $ for all $ n $, $ f_n \to{} f $ pointwise on $ [a, b] $, and $ f $ is integrable. Then: $$\begin{align} \lim_{n \to \infty} \int_a^b f_n = \int_a^b f. \end{align}$$ [Monotone Convergence (33.12)] Suppose $ (f_n) $ are integrable on $ [a, b] $, $ f_1 \leq{} f_2 \leq{} \cdots{} $, $ f_n \to{} f $ pointwise on $ [a, b] $, and $ f $ is integrable. Then: $$\begin{align} \lim_{n \to \infty} \int_a^b f_n = \int_a^b f. \end{align}$$ [Application of Monotone Convergence] Let $ f_n(x) = \frac{1}{1 + nx^3} $ on $ [0, 1] $. Then: $$\begin{align} \int_0^1 f_n(x) dx \to \int_0^1 f(x) dx = 0, \end{align}$$ where $$\begin{align} f(x) = \begin{cases} 1, & x = 0, \\ 0, & x \in (0, 1]. \end{cases} \end{align}$$Fundamental Theorems of Calculus and Change of Variable (Section 34)
Fundamental Theorem of Calculus I
[Fundamental Theorem of Calculus I (34.1)] Suppose $ g : [a, b] \to{} \mathbb{R} $ is continuous, differentiable on $ (a, b) $, and $ g' $ is integrable on $ [a, b] $. Then: $$\begin{align} \int_a^b g'(x) \, dx = g(b) - g(a). \end{align}$$ Compute $ \int_a{}^b x^n \, dx $. Use $ g(x) = \frac{x^{n+1}}{n+1} $, so: $$\begin{align} \int_a^b x^n \, dx = \frac{b^{n+1} - a^{n+1}}{n+1}. \end{align}$$Proof.
Partition $ P = \{a = t_0 < t_1 < \ldots{} < t_n = b\} $. By the Mean Value Theorem:
$$\begin{align}
g'(x_k) = \frac{g(t_k) - g(t_{k-1})}{t_k - t_{k-1}},
\end{align}$$
for some $ x_k \in{} (t_{k-1}, t_k) $. Then:
$$\begin{align}
L(g', P) \leq \sum_{k=1}^n g'(x_k)(t_k - t_{k-1}) = g(b) - g(a) \leq U(g', P).
\end{align}$$
Thus $ \int_a{}^b g'(x) \, dx = g(b) - g(a) $.
∎
Integration by Parts
[Integration by Parts (34.2)] Suppose $ u, v : [a, b] \to{} \mathbb{R} $ are continuous, differentiable on $ (a, b) $, and $ u', v' $ are integrable on $ [a, b] $. Then: $$\begin{align} \int_a^b u(x)v'(x) \, dx + \int_a^b u'(x)v(x) \, dx = u(b)v(b) - u(a)v(a). \end{align}$$ Compute $ \int_0{}^\pi{} x \cos{} x \, dx $. Let $ u(x) = x $, $ v'(x) = \cos{} x $: $$\begin{align} \int_0^\pi x \cos x \, dx = x \sin x \big|_0^\pi - \int_0^\pi \sin x \, dx = 0 - (-2) = -2. \end{align}$$Fundamental Theorem of Calculus II
[Fundamental Theorem of Calculus II (34.3)] Suppose $ f : [a, b] \to{} \mathbb{R} $ is integrable. Define: $$\begin{align} F(x) = \int_a^x f(t) \, dt. \end{align}$$ If $ f $ is continuous at $ c $, then $ F $ is differentiable at $ c $, with $ F'(c) = f(c) $. Let $ G(x) = \int_{x^2}^2 \sin{}(t^2) \, dt $. Then $ G'(x) = -2x \sin{}(x^4) $ by the Chain Rule.Proof.
Let $ F(x) = \int_a{}^x f(t) \, dt $. Then:
$$\begin{align}
F'(c) = \lim_{x \to c} \frac{F(x) - F(c)}{x - c} = \lim_{x \to c} \frac{\int_c^x f(t) \, dt}{x - c}.
\end{align}$$
Since $ f $ is continuous at $ c $, $ |f(t) - f(c)| \leq{} \varepsilon{} $ for $ |t - c| < \delta{} $, and thus:
$$\begin{align}
\lim_{x \to c} \frac{\int_c^x f(t) \, dt}{x - c} = f(c).
\end{align}$$
∎
Change of Variable in Integrals
[Change of Variable (34.4)] Suppose $ u : J \to{} I $, $ u' $ is continuous, and $ f : I \to{} \mathbb{R} $ is continuous. Then for $ a, b \in{} J $: $$\begin{align} \int_a^b f(u(x)) u'(x) \, dx = \int_{u(a)}^{u(b)} f(t) \, dt. \end{align}$$ Compute $ \int_1{}^4 \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx $. Let $ u(x) = \sqrt{x} $, then $ u'(x) = \frac{1}{2\sqrt{x}} $: $$\begin{align} \int_1^4 \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx = \int_1^2 2 \sin t \, dt = 2(-\cos t) \big|_1^2 = 2(\cos 1 - \cos 2). \end{align}$$Interchanging Integration, Differentiation, and Power Series (Section 26)
Interchanging Integration with Limits and Sums
[Exercise 33.9, Lecture 32] Suppose $ (f_n) $ is a sequence of integrable functions on $ [a, b] $ converging uniformly to $ f $. Then: $$\begin{align} \lim_{n \to \infty} \int_a^b f_n = \int_a^b f. \end{align}$$ If $ g_n $ are integrable on $ [a, b] $, and $ f = \sum_{n=0}^\infty{} g_n $ converges uniformly, then $ f $ is integrable, and: $$\begin{align} \int_a^b f = \sum_{n=0}^\infty \int_a^b g_n. \end{align}$$Interchanging Differentiation with Limits and Sums
Let $ f_n(x) = \frac{1}{n} \sin{}(n^2 x) $. Then $ f_n \to{} 0 $ uniformly on $ \mathbb{R} $. However: $$\begin{align} f_n'(x) = n \cos(n^2 x). \end{align}$$ If $ x = \frac{p}{q} \pi{} $, then $ f_n'(x) $ does not converge, even pointwise.Power Series and Radius of Convergence
[Radius of Convergence] For a power series $ \sum_{n=0}^\infty{} a_n x^n $, let: $$\begin{align} \beta = \limsup_{n \to \infty} |a_n|^{1/n}. \end{align}$$ The radius of convergence is $ R = \frac{1}{\beta{}} $. [Uniform Convergence (26.1)] The series $ \sum_{n=0}^\infty{} a_n x^n $ converges uniformly on $ [-R_1, R_1] $ for $ R_1 < R $. The series $ \sum_{n=0}^\infty{} a_n x^n $ converges to a continuous function on $ (-R, R) $.Differentiation and Integration of Power Series
[Differentiation and Integration (26.3)] If $ \sum_{n=0}^\infty{} a_n x^n $ has radius of convergence $ R $, then:- $ \sum_{n=1}^\infty n a_n x^{n-1} $ has radius of convergence $ R $,
- $ \sum_{n=0}^\infty \frac{a_n}{n+1} x^{n+1} $ has radius of convergence $ R $.
Abel’s Theorem
[Abel’s Theorem (26.6)] Suppose $ f(x) = \sum_{n=0}^\infty{} a_n x^n $ has radius of convergence $ R > 0 $. If the series converges at $ R $ (or $ -R $), then $ f $ is continuous at $ R $ (or $ -R $).Abel’s Theorem, Convexity, and Inequalities (Section 26)
Abel Summation Theorem
[Abel’s Theorem (26.6)] Let $ f(x) = \sum_{n=0}^\infty{} a_n x^n $ have radius of convergence $ R > 0 $. If the series converges at $ R $ (or $ -R $), then $ f $ is continuous at $ R $ (or $ -R $. $$\begin{align} 1 - \frac{1}{2} + \frac{1}{3} - \cdots = \ln 2. \end{align}$$ Let $ g(t) = \frac{1}{1+t} = \sum_{n=0}^\infty{} (-1)^n t^n $ ($ R = 1 $), and $ f(x) = \sum_{k=1}^\infty{} \frac{(-1)^{k-1}}{k} x^k $. The series diverges at $ -1 $ but converges at $ 1 $: $$\begin{align} \ln(1+x) = \int_0^x g(t) \, dt = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} x^k. \end{align}$$ Thus: $$\begin{align} f(1) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} = \ln 2. \end{align}$$Alternating Series
[Alternating Series Test] Suppose $ a_1 \geq{} a_2 \geq{} \cdots{} \geq{} 0 $. Then: $$\begin{align} \sum_{k=1}^\infty (-1)^{k-1} a_k = a_1 - a_2 + a_3 - \cdots \end{align}$$ converges if and only if $ \lim_{k \to \infty} a_k = 0 $. $$\begin{align} 1 - \frac{1}{3} + \frac{1}{5} - \cdots = \frac{\pi}{4}. \end{align}$$ Let $ f(x) = \arctan{} x $, so $ f'(x) = \frac{1}{1+x^2} $. For $ |x| < 1 $: $$\begin{align} f'(x) = \sum_{n=0}^\infty (-1)^n x^{2n}, \quad f(x) = \int_0^x f'(t) \, dt = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1}. \end{align}$$ At $ x = 1 $: $$\begin{align} \frac{\pi}{4} = \arctan 1 = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}. \end{align}$$Convex Functions
[Convexity] A function $ f $ on $ I $ is convex if: $$\begin{align} f\left(\frac{x+y}{2}\right) \leq \frac{f(x) + f(y)}{2}, \quad \forall x, y \in I. \end{align}$$ If $ f $ is convex, then: $$\begin{align} f((1-t)x + ty) \leq (1-t)f(x) + tf(y), \quad \forall t \in (0, 1). \end{align}$$Criteria for Convexity
If $ f $ is differentiable on $ I $, and $ f' $ is increasing, then $ f $ is convex. If $ f $ is twice differentiable on $ I $, and $ f'' \geq{} 0 $, then $ f $ is convex.- $ f(x) = e^x $ is convex on $ \mathbb{R} $ because $ f''(x) = e^x > 0 $.
- $ g(x) = \ln x $ is concave on $ (0, \infty) $ because $ g''(x) = -\frac{1}{x^2} < 0 $.
Inequalities
[Jensen’s Inequality] If $ f $ is convex on $ I $, $ x_1, \ldots{}, x_n \in{} I $, $ t_1, \ldots{}, t_n \geq{} 0 $, and $ \sum_{i=1}^n t_i = 1 $, then: $$\begin{align} f\left(\sum_{i=1}^n t_i x_i\right) \leq \sum_{i=1}^n t_i f(x_i). \end{align}$$ [Arithmetic-Geometric Means Inequality] If $ x_1, \ldots{}, x_n > 0 $, $ t_1, \ldots{}, t_n > 0 $, and $ \sum_{i=1}^n t_i = 1 $, then: $$\begin{align} \sum_{i=1}^n t_i x_i \geq \prod_{i=1}^n x_i^{t_i}. \end{align}$$ [Special Case] If $ x_1, \ldots{}, x_n > 0 $, then: $$\begin{align} \frac{x_1 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}. \end{align}$$Convexity, Inequalities, and Nowhere Differentiable Functions (Section 36)
Jensen’s Inequality for Convex Functions
[Jensen’s Inequality] Let $ f $ be a convex function on an interval $ I $, and let $ x_1, \ldots{}, x_n \in{} I $ with $ t_1, \ldots{}, t_n \geq{} 0 $ and $ \sum_{i=1}^n t_i = 1 $. Then: $$\begin{align} f\left(\sum_{i=1}^n t_i x_i\right) \leq \sum_{i=1}^n t_i f(x_i). \end{align}$$ If $ f $ is concave, the inequality is reversed.Inequalities Between Means
[Power Mean Inequality] Suppose $ r > 1 $ and $ x_1, \ldots{}, x_n \geq{} 0 $. Then: $$\begin{align} \frac{x_1 + \cdots + x_n}{n} \leq \left(\frac{x_1^r + \cdots + x_n^r}{n}\right)^{1/r}. \end{align}$$ If $ r = 2 $, this gives the inequality between arithmetic and quadratic means: $$\begin{align} \frac{x_1 + \cdots + x_n}{n} \leq \sqrt{\frac{x_1^2 + \cdots + x_n^2}{n}}. \end{align}$$Proof.
On $ [0, \infty{}) $, $ f(x) = x^r $ is convex because $ f'(x) = rx^{r-1} $ is increasing. Apply Jensen’s Inequality with $ t_i = \frac{1}{n} $:
$$\begin{align}
\left(\frac{x_1 + \cdots + x_n}{n}\right)^r \leq \frac{x_1^r + \cdots + x_n^r}{n}.
\end{align}$$
Taking the $ r $-th root gives the result.
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Arithmetic and Harmonic Means
[Arithmetic-Harmonic Mean Inequality] If $ x_1, \ldots{}, x_n > 0 $, then: $$\begin{align} \frac{x_1 + \cdots + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}. \end{align}$$Proof.
Let $ g(x) = \frac{1}{x} $, which is convex on $ (0, \infty{}) $. Let $ y_i = \frac{1}{x_i} $. By Jensen’s Inequality with $ t_i = \frac{1}{n} $:
$$\begin{align}
\frac{1}{n} \sum_{i=1}^n g(y_i) = \frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n} \geq g\left(\frac{1}{n} \sum_{i=1}^n y_i\right) = \frac{n}{x_1 + \cdots + x_n}.
\end{align}$$
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Nowhere Differentiable Functions
There exists a bounded, uniformly continuous function $ f : \mathbb{R} \to{} \mathbb{R} $ that is differentiable nowhere.Proof. [Sketch]
Construct a $ 1 $-periodic function $ f(x) = \sum_{k=0}^\infty{} 8^{-k}s(64^k x) $, where $ s(x) $ is the sawtooth function:
$$\begin{align}
s(x) = \phi(x - \lfloor x \rfloor), \quad \phi(t) = \min\{t, 1-t\}.
\end{align}$$
- $ f $ is bounded and uniformly continuous by the Weierstrass $ M $-test.
- For any $ x, \delta > 0, A > 0 $, there exists $ y $ with $ |x - y| \leq \delta $ and $ |f(x) - f(y)| \geq A|x - y| $, showing $ f $ is nowhere differentiable.
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Infinite Primes and Divergence of Series
Let $ p_1 < p_2 < \cdots{} $ be the increasing sequence of prime numbers. Then: $$\begin{align} \sum_{n=1}^\infty \frac{1}{p_n} \quad \text{diverges}. \end{align}$$Proof.
Assume $ \sum_{n=1}^\infty{} \frac{1}{p_n} $ converges. Let $ \alpha{} = \sum_{n=1}^\infty{} \frac{1}{p_{2n}^2} < 1 $, and $ \beta{} = \sum_{n=K+1}^\infty{} \frac{1}{p_n} < 1 - \alpha{} $ for some $ K $. Consider $ N $ such that $ N > 2K/(1 - \alpha{} - \beta{}) $. Counting arguments on $ \{1, 2, \ldots{}, N\} $ lead to a contradiction.
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