Limits and Continuity

Definition: A set $U \subseteq \mathbb R$ is open when

$$\begin{align*} \forall a \in U, \exists r \gt 0, (a - r, a + r) \subseteq U \end{align*}$$

Proposition:

1. $\mathbb R$ and $\emptyset$ are open.
2. For any $\lbrace U_\iota \rbrace_{\iota \in I}$ family of open sets, $\bigcup_{\iota \in I} U_\iota$ is open.
3. If $U_1, \dots, U_k$ are open, then $U_1 \cap \dots \cap U_k$ is open.

Proof:

1. $\emptyset$ is open because the definition holds vacuously. $\blacksquare$
2. Let $\mathcal U := \lbrace U_\iota \rbrace_{\iota \in I}$ be a family of open sets and $a \in \bigcup \mathcal U$. Let $\iota_u \in I$ be such that $a \in U_{\iota_u}$. By the openness of $U_{\iota_u}$, we can choose $r \gt 0$ such that $(a - r, a + r) \subseteq \bigcup \mathcal U$. $\blacksquare$.
3. Given $U_1, \dots, U_k$ open sets and $a \in U_1 \cap \dots \cap U_k$ there are $r_1, \dots, r_k \gt 0$ such that $(a - r_j, a + r_j) \subseteq U_j$ for $j = 1\dots k$. Set $r = \min \lbrace r_1, \dots, r_k \rbrace$. Then, $(a - r, a + r) \subseteq (a - r_j, a + r_j) \subseteq U_j, \forall j$. So $(a - r, a + r) \subseteq U_1 \cap \dots \cap U_k$.

Example: The infinite intersection of open sets is not open. Consider $U_n = (-1/n, 1/n)$ for $n \in \mathbb Z_+$. Then $U_n$ is open, $\forall n \geq 1$, but $\bigcap_{n = 1}^\infty U_n = \lbrace 0 \rbrace$, which is not open.

Proposition: Any open interval in $\mathbb R$ is an open set.

Proof: If $I = (-\infty, \infty)$, then $I = \mathbb R$, which is open. So suppose otherwise. Let $a, b \in \mathbb R$ and WLOG, suppose $a \lt b$. Let $c \in (a, b)$. Choose $r = \min \lbrace c - a, b - c \rbrace$. Let $x \in (c - r, c + r)$. Then $a \leq c - r \lt x \lt c + r \leq b$, so $x \in (a, b)$. So $(c - r, c + r) \subseteq (a, b)$, so $(a, b)$ is open. $\blacksquare$

Proposition: Any closed interval is a closed set.

Proof: For any closed interval $[a, b]$, one has $\mathbb R \setminus [a, b] = (-\infty, a) \cup (b, \infty)$, which is an open set. $\blacksquare

If $(a, \infty)$, choose $r = c - a$, and if $(-\infty, b)$, choose $r = b - c$.

Definition: A set $F \subseteq \mathbb R$ is closed, when $\mathbb R \setminus F$ is open.

Proposition:

1. $\mathbb R$ and $\emptyset$ are closed.
2. If $\lbrace F_\iota \rbrace_{\iota \in I}$ are closed sets, then $\bigcap_{\iota \in I} F_\iota$ is closed.
3. If $F_1, \dots, F_k$ are closed, then $F_1 \cup \dots \cup F_k$ is closed.

Proof: TODO.

Definition: A point $a \in \mathbb R$ is a limit point of a set $A \subseteq \mathbb R$ when $\exists (x_n)_{n = 1}^\infty \subseteq A \setminus \lbrace a \rbrace$ such that $\lim_{n \to \infty} x_n = a$.

Theorem 1: A set $X$ is closed if and only if

$$[\forall x_0 \in \mathbb R, \exists (x_n)_{n = 1}^\infty, \lim_{n \to \infty} x_n = x_0] \implies x_0 \in X$$

That is, a set is closed if and only if it contains all its limit points.

Proof: ($\rightarrow$) Assume $X \subseteq \mathbb R$ is closed, and let $x_0 \in \mathbb R$ and $(x_n)_{n = 1}^\infty \subseteq X$ be such that $\lim_{n \to \infty} x_n = x_0$. Suppose $x_0 \not\in X$. Then $x_0 \in \mathbb R \setminus X$, which is open. Choose $\epsilon \gt 0$ such that $(x_0 - \epsilon, x_0 + \epsilon) \subseteq \mathbb R \setminus X$. Then, $\forall n \in \mathbb Z_+$, $\vert x_n - x_0 \vert \geq \epsilon$, contradicting the assumption that $\lim x_n = x_0$. $\blacksquare$

($\leftarrow$) Assume the right side and suppose for the sake of contradiction that $X$ is not closed. Then $\mathbb R \setminus X$ is not open. Choose $x_0 \in \mathbb R \setminus X$ such that a neighborhood $(x_0 - 1/n, x_0 + 1/n) \not\subseteq \mathbb R \setminus X$. Then, $\exists x_n \in X$ such that $\vert x_n - x_0 \vert t 1/n$. So the sequence $(x_n)$ converges to $x_0$, and hence $x_0 \in X$, by assumption; a contradiction. $\blacksquare$

Definition: A set $K \subseteq \mathbb R$ is compact, when $\forall (x_n)_n \subseteq K$, $\exists (x_{n_k})_k \subseteq (x_n)_n$ convergent to an element of $K$.

Proposition: Every closed interval is a compact set.

Proof: Let $I = [a, b]$, with $a \lt b$ since if $a = b$ then $I$ is a singleton and there is nothing to show. Let $(x_n)_n \subseteq [a, b]$. By Bolzano-Weistress, we can choose a subsequence $(x_{n_k})_k \subseteq (x_n)_n$ such that $\lim_{k \to \infty} x_{n_k} = x_0 \in \mathbb R$. By the above theorem and since $I = \bar I$, it follows that $x_0 \in I$. $\blacksquare$

Theorem (Heine-Borel): A set $K \subseteq \mathbb R$ is compact iff $K$ is closed and bounded.

Proof: ($\leftarrow$) Assume $K$ is closed and bounded. Let $S = (x_n)_n \subseteq K$ be an arbitrary sequence. Since $K$ is bounded, $S$ is bounded, and by Bolzano-Weistress, we can choose a convergent subsequence $S' = (x_{n_k})_k \subseteq S$. Let $x_0 = \lim_{k \to \infty} x_{n_k}$. By Theorem 1, $x_0 \in K$.

($\rightarrow$) Assume for the sake of contradiction that $K$ is not bounded above. Then, $\forall n \in \mathbb Z_+$, $\exists x_n \in K$ such that $x_n \geq n$. We claim that $(x_n)_n$ contains no convergent subsequence. Suppose otherwise that there is $(x_{n_k})_k \subseteq (x_n)_n$ convergent to $x_0 \in \mathbb R$. Choose $N_0 \in \mathbb N$ such that $N_0 \geq x_0 + 1$. Then, $\forall n \geq N_0$, $\vert x_n - x_0 \vert \geq 1$. In particular, if $K_0$ is such that $n_{k_0} \geq N_0$, then $\forall k \geq K_0, \vert x_{n_k} - x_0 \vert \geq 1$. So there is no subsequence convergent to an $x_0 \in K$, contradicting the compactness of $K$.

It remains to show that $K$ is closed. Let $x_0 \in \mathbb R, (x_n)_n \subseteq K$ be such that $\lim_{n \to \infty} x_n = x_0$. By assumption of the compcatness of $K$, we can choose a subsequence $(x_{n_k})_k$ convergent to $z_0 \in K$. Since $\lim x_n$ exists, then $x_0 = z_0 \in K$. So $K$ is closed by Theorem 1. $\blacksquare$

Example: The ternary Cantor set is a compact set.

Definition: A set $U \subseteq \mathbb R$ is called an (open) neighborhood of a point $a \in \mathbb R$ when $U$ is open and $a \in U$.

Proposition: $a \in \mathbb R$ is a limit point of $A \subseteq \mathbb R$ when $\forall U$ open neighborhoods of $a$, $\exists x \in U \cap A \setminus \lbrace a \rbrace$.

Limits

Definition: Given a function $f: A \to \mathbb R, A \subseteq \mathbb R$, and a limit point of $A$, $a \in \mathbb R$, we say that the limit of $f$ at $a$ exists and equals $L$, and we write $\lim_{x \to a} f(x) = L$ when

$$\begin{align*} \forall \epsilon \gt 0, \exists \delta \gt 0, \forall x \in A, 0 \lt \vert x - a \vert \lt \delta \implies \vert f(x) - L \vert \lt \epsilon \end{align*}$$

Example: Show that $\lim_{x \to 1} \frac{1}{x} = 1$.

Note: Let $a = 1, f(x) = \frac{1}{x}, A = \mathbb R \setminus \lbrace 0 \rbrace$. We have

$$\begin{align*} \vert 1/x - 1 \vert = \vert \frac{1 - x}{x} \vert = \frac{1}{\vert x \vert} \cdot \vert x - 1 \vert \end{align*}$$

Since we can make $\vert x - 1 \vert$ arbitrarily small, it remains to show that $\frac{1}{\vert x \vert}$ is bounded by some $m$ for all $0 \lt \vert x - 1 \vert \lt \delta$. Assuming that $\delta \leq \frac{1}{2}$, we have $(1 - 1/2 \lt x \lt 1 + 1/2)$. Then, $x \gt \frac{1}{2} \implies \frac{1}{x} \lt 2 \implies \frac{1}{\vert x \vert} \lt 2$, since $x \gt 0$.

Proof: Let $\epsilon \gt 0$ be arbitrary. Choose $\delta = \min\lbrace \frac{1}{2}, \frac{\epsilon}{2} \rbrace$. Then, for all $x \in \mathbb R$ such that $0 \lt \vert x - 1 \vert \lt \delta$, we have $\frac{1}{\vert x \vert} \lt 2$ and so $\left\vert \frac{1}{x} - 1 \right\vert = \frac{1}{\vert x \vert} \cdot \vert x - 1 \vert \lt 2 \delta \leq \epsilon$.

Theorem: Let $f: A \to \mathbb R$ and let $a \in \mathbb R$ be a limit point of $A$. FCAE:

1. $\lim_{x \to a} f(x) = L$
2. For every sequence $(x_n) \subseteq A \setminus \lbrace a \rbrace$ convergent to $a$, $\lim_{n \to \infty} f(x_n) = L$.

Proof: ($1 \to 2$) Let $(x_n)$ be such a subsequence convergent to $a$. Then let $\epsilon \gt 0$ be arbitrary, and let $\delta \gt 0$ such that the limit definition is satisfied. Then choose $N_0 \in \mathbb N$ such that $\forall n \geq N_0$, $\vert x_n - a \vert t \delta$. Since $\delta, \epsilon$ satisfy the limit definition, $\vert f(x_n) - L \vert t \epsilon$ for all $n \geq N_0$, as required. $\blacksquare$

($2 \to 1$) Suppose otherwise. Then we can choose $\epsilon_0 \gt 0$ such that $\forall n \in \mathbb Z_+, \exists x_n \in A$ such that $0 t \vert x_n - a \vert t 1/n$ and $\vert f(x_n) - L \vert \gt \epsilon_0$. But then, $(x_n)$ converges to $a$ so by assumption, $\lim_{n \to \infty} f(x_n) = L$; a contradiction. $\blacksquare$

Theorem (Algebraic Limit Theorem): Let $f, g: A \to \mathbb R$ and let $a \in \mathbb R$ be a limit point of $A$. Suppose $\lim_{x \to a} f(x) = s$, $\lim_{x \to a} g(x) = t$, and $c \in \mathbb R$ is a constant. Then:

1. $\lim_{x \to a}(f \pm g)(x) = s \pm t$
2. $\lim_{x \to a} c \cdot f(x) = c \cdot s$
3. $\lim_{x \to a} f(x) \cdot g(x) = s \cdot t$
4. $\lim_{x \to a} f(x)/g(x) = s/t$, provided $t \neq 0$.

Proof: By the above theorem, it suffices to show that for every sequence $(x_n) \subseteq A \setminus \lbrace a \rbrace$ convergent to $a$, the sequences $(f(x_n) \pm g(x_n)), \dots$ converge to their respective limits. This follows from the Algebriaic Limit THeorem for sequences. $\blacksquare$.

Continuity

Definition: A point $a \in \mathbb R$ is said to be an isolated point of a set $A \subseteq \mathbb R$ when $a \in A$ and $\exists \delta \gt 0$ such that $(a - \delta, a + \delta) \cap A = \lbrace a \rbrace$.

Definition: We say that a function $f: A \to \mathbb R$ is continuous at a point $a \in A$ when $A$ is an isolated point of $A$, or else $\lim_{x \to a} f(x)$ exists and equals $f(a)$.

Definition: We say that $f: A \to \mathbb R$ is continuous when $\forall a \in A$, $f$ is continuous at $a$.

Theorem: Let $f: A \to \mathbb R$ and $a \in A$. The following conditions are equivalent:

1. $f$ is continuous at $a$.
2. $\forall \epsilon \gt 0, \exists \delta \gt 0, \forall x \in A, \vert x - a \vert \lt \delta \implies \vert f(x) - f(a) \vert \lt \epsilon$.
3. For every sequence $(x_n)_n \subseteq A$, $\lim_{n \to \infty} x_n = a \implies \lim_{n \to \infty} f(x) = f(a)$.
4. For every neighborhood $V$ of $f(a)$, there is a neighborhood $U$ of $a$ such that $f(U) \subseteq V$.

Proof:

(1) TODO.

(2) TODO.

(3) TODO.

(4) ($\rightarrow$) Assume $f$ is continuous at $a$. Let $V$ be any neighborhood of $f(a)$. By the openeness of $V$, there is some $\epsilon_0 \gt 0$ such that $(f(a) - \epsilon_0, f(a) + \epsilon_0) \subseteq V$. By continuity of $f$ at $a$, choose $\delta_0 \gt 0$ such that $\forall x \in A, \vert x - a \vert \lt \delta_0 \implies \vert f(x) - f(a) \vert \lt \epsilon_0$. Define $U = (a - \delta_0, a + \delta_0)$. Then, $\forall x \in A \cap U$, so $\vert f(x) - f(a) \vert \lt \epsilon_0$, or $f(x) \in (f(a) - \epsilon_0, f(a) + \epsilon_0) \subseteq V$. So $f(U) \subseteq V$.

($\leftarrow$) Assume next that $\forall V$ open neighborhood of $f(a)$, $\exists U$ open neighborhood of $a$ such that $f(U) \subseteq V$. Let $\epsilon \gt 0$. Consider $V := (f(a) - \epsilon, f(a) + \epsilon)$. By assumption, we can choose $U$ and open neighborhood of $a$ such that $f(x) \in V$ for all $x \in U \cap A$. By openness of $U$, we can choose $\delta \gt 0$ such that $(a - \delta, a + \delta) \subseteq U$. Then, $\forall x \in A, \vert x - a \vert \lt \delta \implies x \in U \cap A \implies f(x) \in V$, so $\vert f(x) - f(a) \vert \lt \epsilon$. $\blacksquare$

Corollary: The following conditions are equivalent:

1. $f$ is continuous
2. $f^{-1}(V)$ is open for any open $V \subseteq \mathbb R$.
3. $f^{-1}(G)$ is closed for any closed $G \subseteq \mathbb R$.

Theorem: If $f, g: A \to \mathbb R$ are continuous at $a \in A, c \in \mathbb R$, then $f + g, f - g, c \cdot f, f \cdot g$ are continuous at $a$ and $f/g$ are continuous at $a$ if $g(a) \neq 0$.

Proof: TODO.

Corollary: Every polynomial function is continuous on $\mathbb R$, and every rational function $P(x)/Q(x)$ is continuous on $\mathbb R \setminus Q^{-1}(0)$.

Proof: Suffices to show that for any $c \in \mathbb R$, $\lbrace x \in \mathbb R \mapsto c \in \mathbb R \rbrace$ is continuous and $\lbrace x \in \mathbb R \mapsto x \in \mathbb R \rbrace$ is continuous. TODO. $\blacksquare$.

Theorem: Given $f: A \to \mathbb R, g: B \to \mathbb R$, $a \in A$, $f(a) \in B$, $f(A) \subseteq B$, if $f$ is continuous at $a$ and $g$ continuous at $f(a)$, then $g \circ f$ is continuous at $a$.

Proof: By assumption, $\forall (a_n)_n \subseteq A$, $\lim_{n \to \infty} a_n = a \implies \lim_{n \to \infty} f(a_n) = f(a)$ and $\forall (b_n)_n \subseteq B$, $\lim_{n \to \infty} b_n = f(a) \implies \lim_{n \to \infty} g(b_n) = g(f(a))$. We want to show that $\forall (a_n)_n \subseteq A$, $\lim_{n \to \infty} a_n = a \implies \lim_{n \to \infty} (g \circ f) (a_n) = (g \circ f)(a)$.

Let $(a_n)_n \subseteq A$ be any such sequence such that $\lim a_n = a$. Then $(f(a_n))_n$ converges to $f(a)$. By assumption, $\lim_{n \to \infty} g(f(a_n)) = g(f(a))$. $\blacksquare$

Theorem: Suppose $f: A \to \mathbb R$ is a continuous function, $K \subseteq A$ is compact. Then $f(K)$ is also compact.

Proof: Let $(y_n)_n$ be any sequence in $f(K)$. Choose, for $n \geq 1$, $x_n \in K$ such that $f(x_n) = y_n$. Then, $(x_n)_n$ is a sequence in $K$. By compactness, we can choose a subsequence $(x_{n_k})_k \in (x_n)_n$ such that $\lim_{k \to \infty} x_{n_k} = x_0 \in K$. Consider $(y_{n_k})_k$ where $y_{n_k} = f(x_{n_k})$. By continuity of $f$ at $x_0$, $\lim_{k \to \infty} f(x_{n_k}) = f(x_0) \in f(K)$. $\blacksquare$

Corollary (Extreme Value Theorem): If $f: K \to \mathbb R$ is continuous and $K \subseteq \mathbb R$ is non-empty and compact, then $f$ admits its maximum and minimum value on $K$.

Proof: By the theorem above, $f(K) \subseteq \mathbb R$ is a compact subset, so by Heine-Borel, it is closed and bounded. Then, $\inf f(K), \sup f(K) \in \mathbb R$. We have $\inf f(K), \sup f(K)$ are limit points of $f(K)$, and $f(K)$ being closed, it follows that $\inf f(K), \sup f(K) \in f(K)$. $\blacksquare$

Definition: We say $f: A \to \mathbb R$ is uniformly continuous when

$$\begin{align*} \forall \epsilon \gt 0, \exists \gt 0, \forall x_1, x_2 \in A, \vert x_1 - x_2 \vert \lt \delta \implies \vert f(x_1) - f(x_2) \vert \lt \epsilon \end{align*}$$

Theorem: If $f: K \to \mathbb R$ is continuous and $K$ is compact, then $f$ is uniformly continuous.

Proof: Suppose for the sake of contradiction that there is $\epsilon_0 \gt 0$ such that $\forall \delta \gt 0$, $\exists x_1, x_2 \in K$ such that $\vert x_1 - x_2 \vert \lt \delta \land \vert f(x_1) - f(x_2) \vert \geq \epsilon_0$. In particular, $\forall n \in \mathbb Z_+$, $\exists x_n^1, x_n^2 \in K$ such that $\vert x_n^1 - x_n^2 \vert \lt 1/n \land \vert f(x_n^1) - f(x_n^2) \vert \geq \epsilon_0$. Hence, we get two sequences $(x_n^1)_n, (x_n^2)_n \subseteq K$. Pick a subsequence $(x^1_{n_k})_k \subseteq (x^1_n)_n$ such that $\lim_{k \to \infty} x^1_{n_k} = x_0 \in K$. Consider $(x^2_{n_k})_k \subseteq (x^2_n)_n$. We have $\forall k, \vert x^1_{n_k} - x^2_{n_k} \vert \lt 1/n_k \to_{n \to \infty} 0$, so $\lim_{k \to \infty} x^2_{n_k} = \lim_{k \to \infty} x^1_{n_k} = x_0$. By continuity of $f$ at $x_0$, $\lim f(x^1_{n_k}) = f(x)_0 = \lim f(x^2_{n_k})$. So $\exists K_0 \in \mathbb N$ such that $\forall k \geq K_0$, $\vert f(x^1_{n_k}) - f(x_0) \vert \lt \epsilon_0/2 \land \vert f(x^2_{n_k}) - f(x_0) \vert \leq \epsilon_0/2$. When adding these two inequalities, we get a contradiction, since we already determined that $\vert f(x^1_n) - f(x^2_n) \vert \geq \epsilon_0$. $\blacksquare$

Examples (Warning): The theorem fails if $K$ is not closed or not bounded.

1. Consider $f(x) = x^2$ on $[0, \infty)$. We show that $f(x)$ is not uniformly continuous.

Proof: Suppose otherwise. Consider $\epsilon = 1$. Let $\delta_0 \gt 0$ such that $\forall 0 \leq x_1 \lt x_2$, $x_2 - x_1 \lt \delta_0$, then $\vert x_2^2 - x_1^2 \vert \lt 1$. Choose $x_1 = 2/\delta_0$ and $x_2 = 2/\delta_0 + \delta_0/2$. Then, $x_2 - x_1 = \delta_0/2 \lt \delta_0$, and $\vert x_2^2 - x_1^2 \vert = (x_2 - x_1)(x_2 + x_1) = \delta_0/2 \cdot (2/\delta_0 + 2/\delta_0 + \delta_0/2) \gt \delta_0/2 \cdot 4/\delta_0 = 2 \gt \epsilon = 1$. $\blacksquare$

1. Consider $f(x) = 1/x$ on $(0, 1]$. Since $(0, 1]$ is not closed, it is not compact. We claim that $f(x)$ is not unifromly continuous.

Proof: Suppose otherwise. Consider $\epsilon = 1$. Choose $0 \lt \delta_0 \lt 1$ such that $\forall 0 \lt x_1 \lt x_2 \leq 1, x_2 - x_1 \lt \delta_0 \implies \vert 1/x_2 - 1/x_1 \vert \lt 1$. Let $x_1 = \delta_0/4$, $x_2 = \delta_0/2$. Then $x_2 - x_1 = \delta_0/2 - \delta_0/4 = \delta_0/4 \lt \delta_0$, and $1/x_1 - 1/x_2 = \frac{x_2 - x_1}{x_1x_2} = \delta_0/4 \cdot 8/\delta_0^2 = 2/\delta_0 \gt 2 \gt 1$; a contradiction. $\blacksquare$

Theorem (Intermediate Value Theorem): Let $f: I \to \mathbb R$ be continuous, where $I$ is any interval. Then, $f(I) \subseteq \mathbb R$ is an interval.

Proof: We want to show that $\forall u, v, w \in \mathbb R; u \lt v \lt w; u, w \in f(I) \implies v \in f(I)$. Let $u, v, w \in \mathbb R$ such that $u \lt v \lt w$ and $u, w \in f(I)$. Choose $x_u, x_w \in I$ be such that $f(x_u) = u$ and $f(x_w) = w$. Either $x_u \lt x_w$ or $x_u \gt x_w$. WLOG, suppose $x_u \lt x_w$. Then, $[x_u, x_w] \subseteq I$. Consider $A = \lbrace x \in [x_u, x_w] : f(x) \lt v \rbrace$, $B = \lbrace x \in [x_u, x_w] : f(x) \gt v \rbrace$. We have $x_u \in A$ and $x_w \in B$. So $A, B$ are non-empty and bounded, so they have an infimum and supremum. Define $\alpha := \sup A \in [x_u, x_w]$. By continuity at $\alpha, f(\alpha) = \lim_{n \to \infty} f(x_n)$ for any $(x_n) \subseteq A$ such that $\lim x_n = \alpha$. It follows that $f(\alpha) \leq v$, since $f(x_n) \lt v$. Now, $\forall n \in \mathbb Z_+$, $\alpha + 1/n \not\in A$. It follows that $\forall n \in \mathbb Z_+, f(\alpha + 1/n) \geq v$. Consider the sequence $(z_n)_n \subseteq I$ where $z_n = \alpha + 1/n$. So $\vert z_n - \alpha \vert = 1/n \to_{n \to \infty} 0$, so $\lim_{n \to \infty} z_n = \alpha$. By continuity, $\lim f(z_n) = f(\alpha)$. Hence, $f(\alpha) \geq v$. So $f(\alpha) = v$. So $v \in f(I)$. $\blacksquare$

Example: Let $g: [0, 1] \to [0, 1]$ be continuous. Show that $\exists a \in [0, 1]$ such that $g(a) + 2a^5 = 3a^7$.

Proof: Consider* $f(x) := g(x) + 2x^5 - 3x^7$. Then $f$ is continuous on $[0, 1]$. We want to show that $\exists a \in [0, 1]$ such that $f(a) = 0$. Note that $f(0) = g(0) \geq 0$. Also, $f(1) = g(1) - 1 \leq 0$. Either $f(0) = 0$ so $a = 0$, or $f(1) = 0$ so $a = 1$, or else $f(0) \gt 0$ and $f(1) \lt 0$. So $0 \in (f(1), f(0))$. By IVT, $\exists a \in (0, 1)$ such that $f(a) = 0$. $\blacksquare$

Corollary: If $f: I \to \mathbb R$ is continuous and $I$ is a closed interval, then $f(I)$ is a closed interval.

Proof: By IVT, $f(I)$ is an interval. By EVT, $f(I)$ is compact, so it is bounded. So $\inf f(I), \sup f(I) \in \mathbb R$. Since $\inf f(I)$ and $\sup f(I)$ are limit points of $f(I)$ and $f(I)$ is closed, $\inf f(I), \sup f(I) \in f(I)$. So $f(I) = [\inf f(I), \sup f(I)]$. $\blacksquare$

Monotone Functions

Definition: Let $f: A \to \mathbb R$ with $A \neq \emptyset$. We say that $f$ is:

1. increasing when $\forall x,y \in A, x \lt y \implies f(x) \leq f(y)$
2. strictly increasing when $\forall x, y \in A, x \lt y \implies f(x) \lt f(y)$
3. decreasing when $\forall x, y \in A, x \lt y \implies f(x) \geq f(y)$
4. strictly decreasing when $\forall x, y \in A, x \lt y \implies f(x) \gt f(y)$

A function is monotone if it is increasing or decreasing, and strictly montoone if it is strictly increasing or strictly decreasing.

Theorem: Let $I \subseteq \mathbb R$ be an interval and let $f: I \to \mathbb R$ be a continuous injection. Then

(1) $f$ is strictly monotone.

(2) $f^{-1}: f(I) \to I$ is a strictly monotone continuous injection.

Proof:

(1) Observe that for any $a, b, c \in I$ such that $a \lt b \lt c$, one cannot have $f(a) \lt f(b) \gt f(c)$x or $f(a) \gt f(b) \lt f(c)$ (by IVT). Therefore, $\forall x \in I, \forall u, v \in I$ such that $u \lt x \lt v$, either (a) $f(u) \lt f(x) \lt f(v)$ or else (b) $f(u) \gt f(x) \gt f(v)$. Let $x_1 \in I$. Suppose that (a) is the case. We claim that $\forall x \in I$, (a) holds at $x$. Suppose otherwise, that there exists $x_2 \in I$ such that (b) holds at $x_2$. If $x_1 \lt x_2$, then $f(x_1) \lt f(x_2)$ (by (a) at $x_1$) and $f(x_1) \gt f(x_2)$ (by (b) at $x_2$), a contradiction. The argument is symmetric for if $x_2 \lt x_1$. So, $f$ is strictly increasing. Case (b) holds symmetrically to show that $f$ is strictly decreasing. $\blacksquare$

(2) We claim that if (and only if) $f$ is strictly increasing, $f^{-1}$ is strictly increasing. If $y_1, y_2 \in f(I), y_1 \lt y_2$, let $x_1, x_2 \in I$ be such that $f(x_1) = y_1, f(x_2) = y_2$. Then $f^{-1}(y_1) = x_1 \lt x_2 = f^{-1}(y_2)$, for otherwise, $x_2 \lt x_1 \implies f(x_2) \lt f(x_1)$; a contradiction.

Let $b \in f(I)$ be arbitrary. We claim that $f^{-1}$ is continuous at $b$. Let $\epsilon \gt 0$. Let $a = f^{-1}(b)$. Consider $y_1 := f(a - \epsilon), y_2 := f(a + \epsilon)$. Choose $\delta = \min \lbrace y_2 - b, b - y_1 \rbrace$. Then $\forall y \in I$,

$$\begin{align*} \vert y - b \vert \lt \delta &\implies y \in (y_1, y_2) \\ &\implies y \in (f(a - \epsilon), f(a + \epsilon)) \\ &\implies \exists x_y \in (a - \epsilon, a + \epsilon), f(x_y) = y \text{ (by IVT)}\\ &\implies f^{-1}(y) \in (a - \epsilon, a + \epsilon) \\ &\implies \vert f^{-1}(y) - f^{-1}(b) \vert \lt \epsilon \end{align*}$$

Definition: A function $f: A \to \mathbb R$ is Lipschitz, when

$$\begin{align*} \exists L \geq 0, \forall x_1, x_2 \in A, \vert f(x_1) - f(x_2) \vert \leq L \cdot \vert x_1 - x_2 \vert \end{align*}$$

Any such $L$ is called a Lipschitz constant for $f$.

Remark: Any Lipschitz function is uniformly continuous (Lipschitz is a stronger condition than uniform continuity).

Proof: Let $L$ be a Lipschitz constant for $f$. Let $\epsilon \gt 0$ be arbitrary. Set $\delta := \epsilon/L$. Ten $\forall x_1, x_2 \in A, \vert x_1 - x_2 \vert \lt \delta \implies \vert f(x_1) - f(x_2) \vert \lt L \cdot \delta = \epsilon$.

Example: Find a function that is uniformly continuous but not Lipschitz.

$$\begin{align*} f(x) = \begin{cases} x\sin(\pi / x^2) & x \in (0, 1] \\ 0 & x = 0 \end{cases} \end{align*}$$

Since $\lim_{x \to 0^+} f(x) = 0 = f(0)$, $f$ is continuous at $0$, so $f$ is continuous on $[0, 1]$, so $f$ is uniformly continuous. Set $x_n = \frac{1}{\sqrt{2n}}$ and $y_n = \frac{1}{\sqrt{2n + \frac{1}{2}}}, n \in \mathbb Z_+$. Then $f(x_n) = 0$ and $f(y_n) = y_n$. Then,

$$\begin{align*} \frac{\vert f(x_n) - f(y_n) \vert}{\vert x_n - y_n \vert} = \frac{1}{\sqrt{2n + \frac{1}{2}}} \cdot \frac{1}{\frac{1}{\sqrt{2n}} - \frac{1}{\sqrt{2n + \frac{1}{2}}}} = 2 \cdot \sqrt{2n} (\sqrt{2n + \frac{1}{2}} + \sqrt{2n}) \to_{n \to \infty} \infty \end{align*}$$

Definition: A function $f: A \to \mathbb R$ is called a contraction when $f$ is Lipschitz with a Lipschitz constant $L \lt 1$.

Theorem (Fixed Point): Let $A \neq \emptyset, A \subseteq \mathbb R$, and let $f: A \to A$ be a contraction. Suppose $A$ is a closed set. Then $\exists!p \in A$ such that $f(p) = p$.

Proof: Pick any $x_1 \in A$. Construct a sequence $(x_n) \subseteq A$ defined as $x_{k + 1} = f(x_k), \forall k \in \mathbb Z_+$. Let $\alpha \in [0, 1)$ be a Lipschitz constant of $f$.

We claim that $\forall k \geq 2$, $\vert x_{k + 1} - x_k \vert \leq \alpha^{k - 1} \cdot \vert x_2 - x_1 \vert$. We proceed by induction. We have $\vert x_3 - x_2 \vert = \vert f(x_2) - f(x_1) \vert \leq \alpha \vert x_2 - x_1 \vert$ since $\alpha$ is a Lipschitz constant of $f$. Now, assuming $\vert x_{k + 1} - x_k \vert \leq \alpha^{k - 1} \vert x_2 - x_1 \vert$, consider $\vert x_{k + 2} - x_{k + 1} \vert = \vert f(x_{k + 1}) - f(x_k) \vert \leq \alpha \cdot \vert x_{k + 1} - x_k \vert \leq \alpha^{(k + 1) - 1} \vert x_2 - x_1 \vert$; proving the claim.

Claim 2: $(x_n)$ is Cauchy. Let $\epsilon \gt 0$. Choose $N_0 \in \mathbb N$ such that $\frac{\alpha^{N_0 - 1}}{1 - \alpha} \cdot \vert x_2 - x_1 \vert \lt \epsilon$. Then, $\forall n \geq m \geq N_0$, and

$$\begin{align*} \vert x_n - x_m \vert &= \vert (x_n - x_{n - 1}) + (x_{n - 1} - x_{n - 2}) + \dots + \vert x_{m + 1} - (x_{m + 1} - x_m) \vert \\ &\leq \vert x_n - x_{n - 1} \vert + \vert x_{n - 1} - x_{n - 2} \vert + \dots + \vert x_{m + 1} - x_m \vert \\ &\leq \alpha^{n - 2} \vert x_2 - x_1 \vert + \alpha^{n - 3} \vert x_2 - x_1 \vert + \dots + \alpha^{m - 1} \vert x_2 - x_1 \vert \\ &= \vert x_2 - x_1 \vert \cdot \alpha^{m - 1} \cdot \alpha^{m - 1} \sum_{k = 0}^{n - m - 1} \alpha^k \\ &\leq \vert x_2 - x_1 \vert \cdot \alpha^{N_0 - 1} \cdot \frac{1}{1 - \alpha}\\ &\lt \epsilon \end{align*}$$

Let $p \in \mathbb R$ be such that $\lim_{n \to \infty} x_n = p$. Since $A$ is closed, $p \in A$. By the continuity of $f$ at $p$, $f(p) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_{n + 1} = \lim_{n \to \infty} x_n = p$ (since $(x_{n + 1}) \subseteq (x_n)$). So suppose there is some other $\exists q \in A$ such that $f(q) = q$. Then $\vert p - q \vert = \vert f(p) - f(q) \vert \leq \alpha \cdot \vert p - q \vert$, so $\vert p - q \vert = 0$, so $p = q$. $\blacksquare$

Definition: Given $f: A \to \mathbb R$, $a \in \mathbb R$ is such that $\forall k \in \mathbb Z_+, \exists x_k \in A, x_k \lt a \land \vert x_k - a \vert \lt 1/k$. Then, we say $L = \lim_{x \to a^-} f(x)$, when

$$\begin{align*} \forall \epsilon \gt 0, \exists \delta \gt 0, \forall x \in A, 0 \lt a - x \lt \delta \implies \vert f(x) - L \vert \lt \epsilon \end{align*}$$