Proof. Assume that \(f\) is not uniformly continuous. Then there exists an \(\varepsilon > 0\) such that for every \(n \in {\mathbb{N}}\setminus \{0\}\) and \(\delta_n := \frac{1}{n}\) there exist two points \(x_n,\,\widetilde{x}_n \in [a,b]\) such that \[|x_n - \widetilde{x}_n| < \delta_n, \quad \text{but} \quad |f(x_n) - f(\widetilde{x}_n)| \ge \varepsilon.\] Since \({(x_n)}_{n \ge 1}\) is a bounded sequence, the Bolzano-Weierstraß theorem (Theorem 2.12) implies that there exists a convergent subsequence \(\big( x_{n_k} \big)_{k \in {\mathbb{N}}}\) with limit \(x:= \lim_{k\to\infty}x_{n_k} \in [a,b].\) Since \(|x_n - \widetilde{x}_n| < \frac{1}{n}\) we also have \(\lim_{k \to \infty} \widetilde{x}_{n_k} = x.\) Using the continuity of \(f\) we obtain \[\lim_{k \to \infty} (f(x_{n_k}) - f(\widetilde{x}_{n_k})) = f(x) - f(x) = 0.\] This is a contradiction of \(|f(x_{n_k}) - f(\widetilde{x}_{n_k})| \ge \varepsilon\) for all \(k \in {\mathbb{N}}.\) Thus, our assumption is false and \(f\) is uniformly continuous.

Proof. W. l. o. g. assume that \(f\) is strictly monotonically increasing, otherwise consider \(-f\) instead.

1. By Corollary 4.10, the set \(f(I)\) is an interval. Moreover, \(f\) is injective, since \(x \neq \widetilde{x}\) implies \(f(x) \neq f(\widetilde{x})\) due to the strict montonicity. Hence, \(f:I \to f(I)\) is bijective.

2. Step 1: We show the strict monotonicity of the inverse function. Let \(y,\,\widetilde{y} \in f(I).\) Since \(f\) is (strictly) monotonically increasing, it holds that \[f^{-1}(y) \ge f^{-1}(\widetilde{y}) \quad \Longrightarrow \quad y = f\big(f^{-1}(y)\big) \ge f\big(f^{-1}(\widetilde{y})\big) = \widetilde{y}.\] By contraposition we obtain \[y < \widetilde{y} \quad \Longrightarrow \quad f^{-1}(y) < f^{-1}(\widetilde{y}).\]
   Step 2: We show continuity of \(f^{-1}.\) Let \(y \in f(I)\) be arbitrary and let \({(y_n)}_{n \in {\mathbb{N}}}\) be an arbitrary sequence in \(f(I)\) with \(\lim_{n \to \infty} y_n = y.\) We have to show that \[\lim_{n \to \infty} f^{-1}(y_n) = f^{-1}(y).\] Assume that this is not the case. Then there exists an \(\varepsilon > 0\) such that \(|f^{-1}(y_n) - f^{-1}(y)| \ge \varepsilon\) for infinitely many \(n \in {\mathbb{N}}.\) Let \({\mathcal{N}}\) be the set of such \(n \in {\mathbb{N}}\) and define \[\begin{aligned} {\mathcal{N}}^+ &:= \big\{ n\in{\mathbb{N}}\; \big| \; f^{-1}(y_n) - f^{-1}(y) \ge \varepsilon \big\}, \\ {\mathcal{N}}^- &:= \big\{ n\in{\mathbb{N}}\; \big| \; f^{-1}(y_n) - f^{-1}(y) \le -\varepsilon \big\}. \end{aligned}\] Then \({\mathcal{N}}= {\mathcal{N}}^+ \cup {\mathcal{N}}^-\) and since \({\mathcal{N}}\) has infinitely many elements, this is also the case for at least one of the sets \({\mathcal{N}}^+\) or \({\mathcal{N}}^-.\) Let w. l. o. g. \({\mathcal{N}}^+\) contain infinitely many elements, the other case is analogous. Then for all \(n \in {\mathcal{N}}^+\) we have \[f^{-1}(y_n) \ge f^{-1}(y) + \varepsilon > f^{-1}(y).\] This implies \(f^{-1}(y) + \varepsilon \in I.\) Because of the monotonicity of \(f\) we obtain \[y_n = f\big(f^{-1}(y_n)\big) \ge f\big(f^{-1}(y) + \varepsilon\big) > f\big(f^{-1}(y)\big) = y \quad \text{for all } n\in {\mathcal{N}}^+.\] Hence, \[|y_n - y| \ge \widetilde{\varepsilon} := f\big(f^{-1}(y) + \varepsilon\big) - y > 0\] for infinitely many \(n \in {\mathbb{N}}.\) But this is a contradiction to the convergence of \({(y_n)}_{n \in {\mathbb{N}}}\) towards \(y.\) Thus \(f^{-1}\) is continuous in \(y\) and since \(y \in f(I)\) is arbitrary, the continuity of \(f^{-1}\) follows.

Proof. We know from Theorem 4.3 that \(\exp\) is continuous. Moreover, for \(x > 0\) we have \[\tag{4.2} \exp(x) = 1+x+\sum_{k=2}^\infty \frac{x^k}{k!} > 1+x > 1.\] With this at hand, we can show that \(\exp\) is also strictly monotonically increasing. Let \(x,\,\widetilde{x} \in {\mathbb{R}}\) with \(x > \widetilde{x}.\) Then \(x - \widetilde{x} > 0\) and \[\exp(x) = \exp(x-\widetilde{x}+\widetilde{x}) = \exp(x-\widetilde{x})\exp(\widetilde{x}) > \exp(\widetilde{x}).\] Furthermore, \(\exp({\mathbb{R}}) = (0,\infty)\) since with Corollary 4.10, \(\exp({\mathbb{R}})\) is an interval and because of (4.2) we have \[\lim_{x \to \infty} \exp(x) = \infty \quad \text{and} \quad \lim_{x \to -\infty} \exp(x) = \lim_{y \to \infty} \exp(-y) = \lim_{y \to \infty} \frac{1}{\exp(y)} = 0.\] Since \(\exp(x) > 0\) for all \(x \in {\mathbb{R}},\) the claim follows. The claims about the inverse function then follow directly from the continuous inverse theorem (Theorem 4.14).

Proof. Exercise!

Proof.

1. This follows from the fact that \(\exp_a\) is a composition of two continuous functions.

2. We have \[\begin{aligned} \exp_a(x+y) &= \exp((x+y)\cdot \ln a) = \exp(x\cdot \ln a + y \cdot \ln a) \\ &= \exp(x\cdot\ln a)\cdot\exp(y\cdot\ln a) = \exp_a(x) \cdot\exp_a(y). \end{aligned}\]

3. The proof is as in Corollary 3.14 ii.

4. The proof is as in Corollary 3.14 iv and since \[\exp_a(1) = \exp(1\cdot \ln a) = a.\]

Proof. Exercise!

Proof. Let \(x \in {\mathbb{R}}.\) Then by Example 4.8, \(x\) is an accumulation point of \({\mathbb{Q}},\) i. e., there exists a sequence \({(x_n)}_{n \in {\mathbb{N}}}\) in \({\mathbb{Q}}\setminus\{x\}\) with \(\lim_{n \to \infty} x_n = x.\) By the continuity of \(f\) and \(g\) we obtain \[f(x) = f\left( \lim_{n \to \infty} x_n \right) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} g(x_n) = g\left( \lim_{n \to \infty} x_n \right) = g(x).\]

Proof. “ii \(\Longrightarrow\) i”: This implication is trivial or follows from Theorem 4.17.
“i \(\Longrightarrow\) ii”: Let \(a:=f(1).\) Then because of i we have \[f(1) = f\left(\frac{1}{2}+\frac{1}{2}\right) = f\left(\frac{1}{2}\right)^2 \ge 0,\] i. e., \(a\) is nonnegative. We distinguish two cases:
Case 1: Assume that \(a=0.\) Then for all \(x \in {\mathbb{R}}\) we obtain \[f(x) = f(1+x-1) = f(1)\cdot f(x-1) = 0,\] since \(f(1)=0.\) Hence \(f=0.\)
Case 2: Assume that \(a>0\): As for the exponential functions \(\exp\) and \(\exp_a\) we show that \(f(m) = a^m\) for all \(m \in {\mathbb{Z}}\) by induction. For \(p \in {\mathbb{Z}}\) and \(q \in {\mathbb{N}}\setminus\{0,1\}\) we obtain \[a^p = f(p) = f\left(q\cdot\frac{p}{q}\right) = f\bigg(\underbrace{\frac{p}{q} + \ldots + \frac{p}{q}}_{q \text{ times}} \bigg) = f\left( \frac{p}{q}\right)^q\] by applying i \(q\) times. This implies \(f\left(\frac{p}{q}\right) = \sqrt[q]{a^p}\) since \(f\left(\frac{p}{q}\right) = f\left(\frac{1}{2}\cdot\frac{p}{q}\right)^2 \ge 0.\) We obtain \[f(x) = a^x = \exp_a(x) \quad \text{for all } x \in {\mathbb{Q}}.\] By the continuity of \(f\) and \(\exp_a\) we obtain \(f(x) = \exp_a(x) = a^x\) for all \(x\in {\mathbb{R}}\) by Lemma 4.19.

Proof. Exercise!

Proof.

1. This statements is trivial.

2. Using Remark 5.3 iv it holds that \[|z z'|^2 = (zz')\overline{(zz')} = z\overline{z}\cdot z'\overline{z'} = |z|^2 \cdot |z'|^2.\] From this we obtain \(|zz'| = |z|\cdot|z'|\) since the expressions on both sides are real and positve and an equation \(x^2 = y\) with \(y \ge 0\) and \(x \in {\mathbb{R}}\) has only one nonnegative solution.

3. Similarly as in the previous item we obtain \[\begin{aligned} |z+z'|^2 &= (z+z')\overline{(z+z')} \\ &= z\overline{z} + z\overline{z'} + z'\overline{z} + z'\overline{z'} \\ &= |z|^2 + 2\mathop{\mathrm{Re}}\big(z\overline{z'}\big) + |z'|^2 \\ &\le |z|^2 + 2\big|z\overline{z'}\big| + |z'|^2 \\ &= |z|^2 + 2|z| \cdot \big|\overline{z'}\big| + |z'|^2 = (|z|+|z'|)^2, \end{aligned}\] where we have used Remark 5.3 ii, iv, and vi.

Proof. Let \(a_n := \mathop{\mathrm{Re}}(z_n)\) and \(b_n := \mathop{\mathrm{Im}}(z_n)\) for all \(n \in {\mathbb{N}},\) i. e., \(z_n = a_n + b_n \mathrm{i}\) for all \(n \in {\mathbb{N}}.\)
“\(\Longrightarrow\)”: Let \({(z_n)}_{n \in {\mathbb{N}}}\) and \[\lim_{n \to \infty} z_n =: z = a+b\mathrm{i}\in {\mathbb{C}}.\] Let \(\varepsilon > 0\) be arbitrary. Then there exists an \(N_\varepsilon \in {\mathbb{N}}\) with \[|z_n - z| < \varepsilon \quad \text{for all } n \ge N_\varepsilon.\] Then with Remark 5.3 vi, we obtain \[\begin{aligned} |a_n - a| &= |\mathop{\mathrm{Re}}(z_n - z)| \le |z_n - z| < \varepsilon \quad \text{for all } n \ge N_\varepsilon, \\ |b_n - b| &= |\mathop{\mathrm{Im}}(z_n - z)| \le |z_n - z| < \varepsilon \quad \text{for all } n \ge N_\varepsilon. \end{aligned}\] From this we obtain the convergence of \({(a_n)}_{n \in {\mathbb{N}}}\) and \({(b_n)}_{n \in {\mathbb{N}}}\) with \(\lim_{n \to \infty} a_n = a\) and \(\lim_{n \to \infty} b_n = b,\) hence also (5.1).
“\(\Longleftarrow\)”: Let \({(a_n)}_{n \in {\mathbb{N}}}\) and \({(b_n)}_{n \in {\mathbb{N}}}\) be convergent with limits \(a\) and \(b\) and let \(\varepsilon > 0\) be arbitrary. Then there exists an \(N_\varepsilon \in {\mathbb{N}}\) such that \[|a_n - a| < \frac{\varepsilon}{2}, \quad |b_n - b| < \frac{\varepsilon}{2} \quad \text{for all } n \ge N_\varepsilon.\] With \(z:=a+b\mathrm{i}\) and using the triangle inequality we obtain \[|z_n - z| = |(a_n+b_n\mathrm{i})-(a+b\mathrm{i})| \le |a_n - a| + |\mathrm{i}|\cdot|b_n-b| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} < \varepsilon \quad \text{for all } n \ge N_\varepsilon.\] Since \(\varepsilon > 0\) is chosen arbitrarily, we get \(\lim_{n \to \infty} z_n = z.\)

Proof. Analogously to the proof of Theorem 5.2, we show that the sequences \({(\mathop{\mathrm{Re}}(z_n))}_{n \in {\mathbb{N}}}\) and \({(\mathop{\mathrm{Im}}(z_n))}_{n \in {\mathbb{N}}}\) are Cauchy sequences in \({\mathbb{R}}.\) Then the result follows from the completeness of \({\mathbb{R}}\) and Theorem 5.2.

Proof. The series is absolutely convergent, since \[\sum_{k=0}^\infty \left|\frac{z^k}{k!}\right| = \sum_{k=0}^\infty \frac{|z|^k}{k!}\] is convergent by Example 3.6 ii and by noting that \(|z| \in {\mathbb{R}}.\)

Proof. Statements i, ii, and iv can be shown exactly as in the real case. Now we show iii: By using Remark 5.3 we have \[\overline{\exp(z)} = \overline{\lim_{n \to \infty} \sum_{k=0}^n \frac{z^k}{k!}} = \lim_{n \to \infty} \overline{\sum_{k=0}^n \frac{z^k}{k!}} = \lim_{n \to \infty} \sum_{k=0}^n \frac{\overline{z}^k}{k!} = \exp(\overline{z}).\]