Exercises §17 — Ross, Elementary Analysis, 2nd Edition
Exercise 17.4. Prove that \(\sqrt{x}\) is continuous on its domain \([0, \infty)\).
Proof. Let \(x_0 \in [0, \infty)\), and let \((x_n)\) be a sequence of non-negative real numbers converging to \(x_0\). Note \(x_n \in [0, \infty)\) for all \(n \in \mathbb{N}\). Now consider the sequence \((y_n)\) where \[ y_n = f(x_n) = \sqrt{x_n}. \] We know from a prior exercise that \[ \lim_{n \to \infty} y_n = y_0 = f(x_0) = \sqrt{x_0}, \] thus \(\sqrt{x}\) is continuous on its domain \([0, \infty)\). \(\square\)
Exercise 17.5. (a) Prove that if \(m \in \mathbb{N}\), then \(f(x) = x^m\) is continuous on \(\mathbb{R}\).
(b) Prove that every polynomial function \(p(x) = a_0 + a_1 x + \cdots + a_n x^n\) is continuous on \(\mathbb{R}\).
Proof. (a) We’ll prove this by induction on \(m\).
Base case: \(m = 1\). Let \((x_n)\) be a sequence in \(\mathbb{R}\) converging to \(x_0 \in \mathbb{R}\). Let \(f(x) = x^1 = x\). Clearly \[ \lim_{n \to \infty} x_n = x_0 \implies \lim_{n \to \infty} f(x_n) = f(x_0), \] so \(f(x)\) is continuous on \(\mathbb{R}\) for \(m = 1\).
Inductive hypothesis: Assume this holds for all \(n \in \mathbb{N}\) where \(n < m\).
Inductive step: Note \[ f(x) = x^m = (x) \cdot \bigl( x^{m-1} \bigr). \] Let \(g(x) = x\) and \(h(x) = x^{m-1}\). By our inductive hypothesis both \(g\) and \(h\) are continuous on \(\mathbb{R}\). And from Theorem 17.4 the product \(gh\) of these functions is also continuous on \(\mathbb{R}\). Thus \(f = gh\) is continuous on \(\mathbb{R}\). \(\square\)
(b) Note every polynomial function can be written \[ p(x) = a_0 + a_1 x + \cdots + a_n x^n = p_0(x) + \sum_{m=1}^{n} p_m(x), \] where \(p_m(x) = a_m x^m\) and \(p_0(x) = a_0\). Note each \(p_m(x)\) is continuous on \(\mathbb{R}\), from what we proved in (a) and the fact that \(kf\) is continuous on \(\mathbb{R}\) if \(f\) is continuous on \(\mathbb{R}\) and \(k \in \mathbb{R}\), from Theorem 17.3.
By Theorem 17.4, \[ p_0(x) + \sum_{m=1}^{n} p_m(x) \] is also continuous on \(\mathbb{R}\), hence \(p(x)\) is continuous on \(\mathbb{R}\). \(\square\)
Exercise 17.6. Prove that rational functions \(f = p/q\), where \(p\) and \(q\) are polynomial functions and the domain of \(f\) is \(\{\, x \in \mathbb{R} : q(x) \neq 0 \,\}\), are continuous.
Proof. Note \(p\) and \(q\) are both continuous on the domain of \(f\) from Exercise 17.5. From Theorem 17.4, \(p/q\) is also continuous on the domain of \(f\), thus \(f\) is continuous. \(\square\)
Exercise 17.8. Let \(f\) and \(g\) be real-valued functions.
(a) Show \(\min(f, g) = \tfrac{1}{2}(f + g) - \tfrac{1}{2}|f - g|\).
(b) Show \(\min(f, g) = -\max(-f, -g)\).
(c) Suppose \(f\) and \(g\) are continuous at \(x_0 \in \mathbb{R}\). Prove \(\min(f, g)\) is continuous at \(x_0\).
Proof. (a) Let \(a, b \in \mathbb{R}\). Let’s show \[ \min(a, b) = \tfrac{1}{2}(a + b) - \tfrac{1}{2}|a - b|. \]
Case \(a \ge b\): Then \(\min(a, b) = b\) and \(|a - b| = a - b\). Thus \[ \tfrac{1}{2}(a + b) - \tfrac{1}{2}|a - b| = \tfrac{1}{2}(a + b) - \tfrac{1}{2}(a - b) = b. \]
Case \(a \le b\): Then \(\min(a, b) = a\) and \(|a - b| = -(a - b) = b - a\). Thus \[ \tfrac{1}{2}(a + b) - \tfrac{1}{2}|a - b| = \tfrac{1}{2}(a + b) - \tfrac{1}{2}(b - a) = a. \]
Since \(f\) and \(g\) are real-valued, then \[ \min(f, g) = \tfrac{1}{2}(f + g) - \tfrac{1}{2}|f - g|. \quad \square \]
(b) From Example 5 we know \[ \max(f, g) = \tfrac{1}{2}(f + g) + \tfrac{1}{2}|f - g|. \] So \[ \begin{aligned} -\max(-f, -g) &= -\Bigl[ \tfrac{1}{2}\bigl( (-f) + (-g) \bigr) + \tfrac{1}{2}\bigl| (-f) - (-g) \bigr| \Bigr] \\ &= -\Bigl[ -\tfrac{1}{2}(f + g) + \tfrac{1}{2}|f - g| \Bigr] \\ &= \tfrac{1}{2}(f + g) - \tfrac{1}{2}|f - g| \\ &= \min(f, g). \quad \square \end{aligned} \]
(c) Note \(\min(f, g) = \tfrac{1}{2}(f + g) - \tfrac{1}{2}|f - g|\). From Theorem 17.4, \(\tfrac{1}{2}(f + g)\) is continuous at \(x_0\), and from Theorem 17.3, \(\tfrac{1}{2}|f - g|\) is continuous at \(x_0\). Thus \[ \tfrac{1}{2}(f + g) - \tfrac{1}{2}|f - g| \text{ is continuous at } x_0 \implies \min(f, g) \text{ is continuous at } x_0. \quad \square \]