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Proposition 4.2.3. Suppose \( {U}_{1},{U}_{2},\ldots ,{U}_{n} \) is a finite collection of open sets. Then\n\n\[ \mathop{\bigcap }\limits_{{i = 1}}^{n}{U}_{i} \]\n\n\( \left( {4.2.6}\right) \)\nis open.
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Proof. Let \( x \in \mathop{\bigcap }\limits_{{i = 1}}^{n}{U}_{i} \) . Then \( x \in {U}_{i} \) for every \( i = 1,2,\ldots, n \) . For each \( i \) , choose \( {\epsilon }_{i} > 0 \) such that \( \left( {x - {\epsilon }_{i}, x + {\epsilon }_{i}}\right) \subset {U}_{i} \) . Let \( \epsilon \) be the smallest of \( {\epsilon }_{1},{\epsilon }_{2},\ldots ,{\epsilon }_{n} \) . Then \( \epsilon > 0 \) and\n\n\[ \left( {x - \epsilon, x + \epsilon }\right) \subset \left( {x - {\epsilon }_{i}, x + {\epsilon }_{i}}\right) \subset {U}_{i} \]\n\n\( \left( {4.2.7}\right) \)\n\nfor every \( i = 1,2,\ldots, n \) . Thus\n\n\[ \left( {x - \epsilon, x + \epsilon }\right) \subset \mathop{\bigcap }\limits_{{i = 1}}^{n}{U}_{i} \]\n\n\( \left( {4.2.8}\right) \)\n\nHence \( \mathop{\bigcap }\limits_{{i = 1}}^{n}{U}_{i} \) is an open set.\n\nQ.E.D.
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Yes
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Proposition 4.3.1. If \( A \subset \mathbb{R} \), then \( \bar{A} \) is closed.
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Proof. Suppose \( x \) is a limit point of \( \bar{A} \). We will show that \( x \) is a limit point of \( A \), and hence \( x \in \bar{A} \). Now for any \( \epsilon > 0 \), there exists \( a \in \bar{A}, a \neq x \), such that\n\n\[ a \in \left( {x - \frac{\epsilon }{2}, x + \frac{\epsilon }{2}}\right) . \]\n\nIf \( a \in A \), let \( b = a \). If \( a \notin A \), then \( a \) is a limit point of \( A \), so there exists \( b \in A \), \( b \neq a \) and \( b \neq x \), such that\n\n\[ b \in \left( {a - \frac{\epsilon }{2}, a + \frac{\epsilon }{2}}\right) . \]\n\nIn either case\n\n\[ \left| {x - b}\right| \leq \left| {x - a}\right| + \left| {a - b}\right| < \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon . \]\n\nHence \( x \in {A}^{\prime } \), and so \( \bar{A} \) is closed.\n\nQ.E.D.
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Yes
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Proposition 4.3.2. A set \( C \subset \mathbb{R} \) is closed if and only if for every convergent sequence \( {\left\{ {a}_{k}\right\} }_{k \in K} \) with \( {a}_{k} \in C \) for all \( k \in K \) , \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{a}_{k} \in C \]
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Proof. Suppose \( C \) is closed and \( {\left\{ {a}_{k}\right\} }_{k \in K} \) is a convergent sequence with \( {a}_{k} \in C \) for all \( k \in K \) . Let \( x = \mathop{\lim }\limits_{{k \rightarrow \infty }}{a}_{k} \) . If \( x = {a}_{k} \) for some integer \( k \), then \( x \in C \) . Otherwise, for every \( \epsilon > 0 \), there exists an integer \( N \) such that \( \left| {{a}_{N} - x}\right| < \epsilon \) . Hence \( {a}_{N} \neq x \) and \[ {a}_{N} \in \left( {x - \epsilon, x + \epsilon }\right) . \] Thus \( x \) is a limit point of \( C \), and so \( x \in C \) since \( C \) is closed. Now suppose that for every convergent sequence \( {\left\{ {a}_{k}\right\} }_{k \in K} \) with \( {a}_{k} \in C \) for all \( k \in K,\mathop{\lim }\limits_{{k \rightarrow \infty }}{a}_{k} \in C \) . Let \( x \) be a limit point of \( C \) . For \( k = 1,2,3,\ldots \), choose \( {a}_{k} \in C \) such that \( {a}_{k} \in \left( {x - \frac{1}{k}, x + \frac{1}{k}}\right) \) . Then clearly \[ x = \mathop{\lim }\limits_{{k \rightarrow \infty }}{a}_{k} \] so \( x \in C \) . Thus \( C \) is closed. Q.E.D.
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Yes
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Proposition 4.3.3. Suppose \( A \) is a set and, for each \( \alpha \in A,{C}_{\alpha } \) is a closed set. Then\n\n\[ \mathop{\bigcap }\limits_{{\alpha \in A}}{C}_{\alpha } \]\n\n(4.3.8)\n\nis a closed set.
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Proof. Suppose \( x \) is a limit point of \( \mathop{\bigcap }\limits_{{\alpha \in A}}{C}_{\alpha } \) . Then for any \( \epsilon > 0 \), there exists \( y \in \mathop{\bigcap }\limits_{{\alpha \in A}}{C}_{\alpha } \) such that \( y \neq x \) and \( y \in \left( {x - \epsilon, x + \epsilon }\right) \) . But then for any \( \alpha \in A \) , \( y \in {C}_{\alpha } \), so \( x \) is a limit point of \( {C}_{\alpha } \) . Since \( {C}_{\alpha } \) is closed, it follows that \( x \in {C}_{\alpha } \) for every \( \alpha \in A \) . Thus \( x \in \mathop{\bigcap }\limits_{{\alpha \in A}}{C}_{\alpha } \) and \( \mathop{\bigcap }\limits_{{\alpha \in A}}{C}_{\alpha } \) is closed.\n\nQ.E.D.
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Yes
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Proposition 4.3.4. Suppose \( {C}_{1},{C}_{2},\ldots ,{C}_{n} \) is a finite collection of closed sets. Then\n\n\[ \mathop{\bigcup }\limits_{{i = 1}}^{n}{C}_{i} \]\n\n\( \left( {4.3.9}\right) \)\nis closed.
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Proof. Suppose \( {\left\{ {a}_{k}\right\} }_{k \in K} \) is a convergent sequence with \( {a}_{k} \in \mathop{\bigcup }\limits_{{i = 1}}^{n}{C}_{i} \) for every \( k \in K \) . Let \( L = \mathop{\lim }\limits_{{k \rightarrow \infty }}{a}_{k} \) . Since \( K \) is an infinite set, there must exist an integer \( m \) and a subsequence \( {\left\{ {a}_{{n}_{j}}\right\} }_{j = 1}^{\infty } \) such that \( {a}_{{n}_{j}} \in {C}_{m} \) for \( j = 1,2,\ldots \) Since every subsequence of \( {\left\{ {a}_{k}\right\} }_{k \in K} \) converges to \( L,{\left\{ {a}_{{n}_{j}}\right\} }_{j = 1}^{\infty } \) must converge to \( L \) . Since \( {C}_{m} \) is closed,\n\n\[ L = \mathop{\lim }\limits_{{j \rightarrow \infty }}{a}_{{n}_{j}} \in {C}_{m} \subset \mathop{\bigcup }\limits_{{i = 1}}^{n}{C}_{i} \]\n\n(4.3.10)\n\nThus \( \mathop{\bigcup }\limits_{{i = 1}}^{n}{C}_{i} \) is closed.\n\nQ.E.D.
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Yes
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Proposition 4.3.5. A set \( C \subset \mathbb{R} \) is closed if and only if \( \mathbb{R} \smallsetminus C \) is open.
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Proof. Assume \( C \) is closed and let \( U = \mathbb{R} \smallsetminus C \) . If \( C = \mathbb{R} \), then \( U = \varnothing \), which is open; if \( C = \varnothing \), then \( U = \mathbb{R} \), which is open. So we may assume both \( C \) and \( U \) are nonempty. Let \( x \in U \) . Then \( x \) is not a limit point of \( C \), so there exists an \( \epsilon > 0 \) such that\n\n\[ \left( {x - \epsilon, x + \epsilon }\right) \cap C = \varnothing . \]\n\nThus\n\n\[ \left( {x - \epsilon, x + \epsilon }\right) \subset U \]\n\nso \( U \) is open.\n\nNow suppose \( U = \mathbb{R} \smallsetminus C \) is open. If \( U = \mathbb{R} \), then \( C = \varnothing \), which is closed; if \( U = \varnothing \), then \( C = \mathbb{R} \), which is closed. So we may assume both \( U \) and \( C \) are nonempty. Let \( x \) be a limit point of \( C \) . Then, for every \( \epsilon > 0 \) ,\n\n\[ \left( {x - \epsilon, x + \epsilon }\right) \cap C \neq \varnothing . \]\n\nHence there does not exist \( \epsilon > 0 \) such that\n\n\[ \left( {x - \epsilon, x + \epsilon }\right) \subset U. \]\n\nThus \( x \notin U \), so \( x \in C \) and \( C \) is closed.\n\nQ.E.D.
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Yes
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Proposition 4.4.1. If \( I \) is a closed, bounded interval, then \( I \) is compact.
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Proof. Let \( a \leq b \) be finite real numbers and \( I = \left\lbrack {a, b}\right\rbrack \) . Suppose \( \left\{ {{U}_{\alpha } : \alpha \in A}\right\} \) is an open cover of \( I \) . Let \( \mathcal{O} \) be the set of sets \( \left\{ {{U}_{\beta } : \beta \in B}\right\} \) with the properties that \( B \) is a finite subset of \( A \) and \( a \in \mathop{\bigcup }\limits_{{\beta \in B}}{U}_{\beta } \) . Let\n\n\[ T = \left\{ {x : x \in I,\left\lbrack {a, x}\right\rbrack \subset \mathop{\bigcup }\limits_{{\beta \in B}}{U}_{\beta }\text{ for some }\left\{ {{U}_{\beta } : \beta \in B}\right\} \in \mathcal{O}}\right\} .\ ]\n\nClearly, \( a \in T \) so \( T \neq \varnothing \) . Let \( s = \sup T \) . Suppose \( s < b \) . Since \( \left\{ {{U}_{\alpha } : \alpha \in A}\right\} \) is an open cover of \( I \), there exists an \( \alpha \in A \) for which \( s \in {U}_{\alpha } \) . Hence there exists an \( \epsilon > 0 \) such that\n\n\[ \left( {s - \epsilon, s + \epsilon }\right) \subset {U}_{\alpha } ]\n\nMoreover, there exists a \( \left\{ {{U}_{\beta } : \beta \in B}\right\} \in \mathcal{O} \) for which\n\n\[ \left\lbrack {a, s - \frac{\epsilon }{2}}\right\rbrack \subset \mathop{\bigcup }\limits_{{\beta \in B}}{U}_{\beta } ]\n\nBut then\n\n\[ \left\{ {{U}_{\beta } : \beta \in B}\right\} \cup \left\{ {U}_{\alpha }\right\} \in \mathcal{O} ]\n\nand\n\n\[ \left\lbrack {a, s + \frac{\epsilon }{2}}\right\rbrack \subset \left( {\mathop{\bigcup }\limits_{{\beta \in B}}{U}_{\beta }}\right) \cup {U}_{\alpha } ]\n\ncontradicting the definition of \( s \) . Hence we must have \( s = b \) . Now choose \( {U}_{\alpha } \) such that \( b \in {U}_{\alpha } \) . Then, for some \( \epsilon > 0 \) ,\n\n\[ \left( {b - \epsilon, b + \epsilon }\right) \subset {U}_{\alpha } ]\n\nMoreover, there exists \( \left\{ {{U}_{\beta } : \beta \in B}\right\} \in \mathcal{O} \) such that\n\n\[ \left\lbrack {a, b - \frac{\epsilon }{2}}\right\rbrack \subset \mathop{\bigcup }\limits_{{\beta \in B}}{U}_{\beta } ]\n\nThen\n\n\[ \left\{ {{U}_{\beta } : \beta \in B}\right\} \cup \left\{ {U}_{\alpha }\right\} \in \mathcal{O} ]\n\nis a finite subcover of \( I \) . Thus \( I \) is compact.\n\nQ.E.D.
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Yes
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Proposition 4.4.2. If \( K \) is a closed, bounded subset of \( \mathbb{R} \), then \( K \) is compact.
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Proof. Since \( K \) is bounded, there exist finite real numbers \( a \) and \( b \) such that \( K \subset \left\lbrack {a, b}\right\rbrack \) . Let \( \left\{ {{U}_{\alpha } : \alpha \in A}\right\} \) be an open cover of \( K \) . Let \( V = \mathbb{R} \smallsetminus K \) . Then\n\n\[ \left\{ {{U}_{\alpha } : \alpha \in A}\right\} \cup \{ V\} \]\n\n(4.4.11)\n\nis an open cover of \( \left\lbrack {a, b}\right\rbrack \) . Since \( \left\lbrack {a, b}\right\rbrack \) is compact, there exists a finite subcover of this cover. This subcover is either of the form \( \left\{ {{U}_{\beta } : \beta \in B}\right\} \) or \( \left\{ {{U}_{\beta } : \beta \in }\right. \) \( B\} \cup \{ V\} \) for some \( B \subset A \) . In the former case, we have\n\n\[ K \subset \left\lbrack {a, b}\right\rbrack \subset \mathop{\bigcup }\limits_{{\beta \in B}}{U}_{\beta } \]\n\n(4.4.12)\n\nin the latter case, we have\n\n\[ K \subset \left\lbrack {a, b}\right\rbrack \smallsetminus V \subset \mathop{\bigcup }\limits_{{\beta \in B}}{U}_{\beta } \]\n\n(4.4.13)\n\nIn either case, we have found a finite subcover of \( \left\{ {{U}_{\alpha } : \alpha \in A}\right\} \) .\n\nQ.E.D.
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Yes
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Proposition 4.4.3. If \( K \subset \mathbb{R} \) is compact, then \( K \) is closed.
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Proof. Suppose \( x \) is a limit point of \( K \) and \( x \notin K \) . For \( n = 1,2,3,\ldots \), let\n\n\[ \n{U}_{n} = \left( {-\infty, x - \frac{1}{n}}\right) \cup \left( {x + \frac{1}{n}, + \infty }\right) .\n\]\n\n(4.4.14)\n\nThen\n\n\[ \n\mathop{\bigcup }\limits_{{n = 1}}^{\infty }{U}_{n} = \left( {-\infty, x}\right) \cup \left( {x, + \infty }\right) \supset K.\n\]\n\n\( \left( {4.4.15}\right) \)\n\nHowever, for any \( N \in {\mathbb{Z}}^{ + } \), there exists \( a \in K \) with\n\n\[ \na \in \left( {x - \frac{1}{N}, x + \frac{1}{N}}\right)\n\]\n\n(4.4.16)\n\nand hence\n\n\[ \na \notin \mathop{\bigcup }\limits_{{n = 1}}^{N}{U}_{n} = \left( {-\infty, x - \frac{1}{N}}\right) \cup \left( {x + \frac{1}{N}, + \infty }\right) .\n\]\n\n(4.4.17)\n\nThus the open cover \( \left\{ {{U}_{n} : n \in {\mathbb{Z}}^{ + }}\right\} \) does not have a finite subcover, contradicting the assumption that \( K \) is compact. Q.E.D.
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Yes
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Proposition 4.4.4. If \( K \subset \mathbb{R} \) is compact, then \( K \) is bounded.
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Proof. Suppose \( K \) is not bounded. For \( n = 1,2,3,\ldots \), let \( {U}_{n} = \left( {-n, n}\right) \) . Then\n\n\[ \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{U}_{n} = \left( {-\infty ,\infty }\right) \supset K \]\n\n(4.4.18)\n\nBut, for any integer \( N \), there exists \( a \in K \) such that \( \left| a\right| > N \), from which it follows that\n\n\[ a \notin \mathop{\bigcup }\limits_{{n = 1}}^{N}{U}_{n} = \left( {-N, N}\right) \]\n\n(4.4.19)\n\nThus the open cover \( \left\{ {{U}_{n} : n \in {\mathbb{Z}}^{ + }}\right\} \) does not have a finite subcover, contradicting the assumption that \( K \) is compact. Q.E.D.
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Yes
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Proposition 4.4.6. If \( K \subset \mathbb{R} \) is compact and \( {\left\{ {x}_{n}\right\} }_{n \in I} \) is a sequence with \( {x}_{n} \in K \) for every \( n \in I \), then \( {\left\{ {x}_{n}\right\} }_{n \in I} \) has a convergent subsequence \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) with\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} \in K \]
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Proof. Since \( K \) is bounded, \( {\left\{ {x}_{n}\right\} }_{n \in I} \) has a convergent subsequence \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) . Since \( K \) is closed, we must have \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} \in K \) .\n\nQ.E.D.
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Yes
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Proposition 4.4.7. Suppose \( K \subset \mathbb{R} \) is such that whenever \( {\left\{ {x}_{n}\right\} }_{n \in I} \) is a sequence with \( {x}_{n} \in K \) for every \( n \in I \), then \( {\left\{ {x}_{n}\right\} }_{n \in I} \) has a subsequence \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) with \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} \in K \) . Then \( K \) is compact.
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Proof. Suppose \( K \) is unbounded. Then we may construct a sequence \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) such that \( {x}_{n} \in K \) and \( \left| {x}_{n}\right| > n \) for \( n = 1,2,3,\ldots \) Hence the only possible subsequential limits of \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \) would be \( - \infty \) and \( + \infty \), contradicting our assumptions. Thus \( K \) must be bounded.\n\nNow suppose \( {\left\{ {x}_{n}\right\} }_{n \in I} \) is a convergent sequence with \( {x}_{n} \in K \) for all \( n \in I \) . If \( L = \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} \), then \( L \) is the only subsequential limit of \( {\left\{ {x}_{n}\right\} }_{n \in I} \) . Hence, by the assumptions of the proposition, \( L \in K \) . Hence \( K \) is closed.\n\nSince \( K \) is both closed and bounded, it is compact.\n\nQ.E.D.
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Yes
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Proposition 5.1.1. Suppose \( D \subset \mathbb{R}, f : D \rightarrow \mathbb{R} \), and \( a \) is a limit point of \( D \) . If \( f\left( {a - }\right) = f\left( {a + }\right) = L \), then \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) .
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Proof. Suppose \( {\left\{ {x}_{n}\right\} }_{n = m}^{\infty } \in S\left( {D, a}\right) \) . Let\n\n\[ {J}^{ - } = \left\{ {n : n \in \mathbb{Z},{x}_{n} < a}\right\} \]\n\n\( \left( {5.1.15}\right) \)\n\nand\n\n\[ {J}^{ + } = \left\{ {n : n \in \mathbb{Z},{x}_{n} > a}\right\} . \]\n\n\( \left( {5.1.16}\right) \)\n\nSuppose \( {J}^{ - } \) is empty or finite and let \( k = m - 1 \) if \( {J}^{ - } = \varnothing \) and, otherwise, let \( k \) be the largest integer in \( {J}^{ - } \) . Then \( {\left\{ {x}_{n}\right\} }_{n = k + 1}^{\infty } \in {S}^{ + }\left( {D, a}\right) \), and so\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = f\left( {a + }\right) = L. \]\n\n\( \left( {5.1.17}\right) \)\n\nA similar argument shows that if \( {J}^{ + } \) is empty or finite, then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = f\left( {a - }\right) = L. \]\n\n(5.1.18)\n\nIf neither \( {J}^{ - } \) nor \( {J}^{ + } \) is finite or empty, then \( {\left\{ {x}_{n}\right\} }_{n \in {J}^{ - }} \) and \( {\left\{ {x}_{n}\right\} }_{n \in {J}^{ + }} \) are subsequences of \( {\left\{ {x}_{n}\right\} }_{n = m}^{\infty } \) with \( {\left\{ {x}_{n}\right\} }_{n \in {J}^{ - }} \in {S}^{ - }\left( {D, a}\right) \) and \( {\left\{ {x}_{n}\right\} }_{n \in {J}^{ + }} \in {S}^{ + }\left( {D, a}\right) \) . Hence, given any \( \epsilon > 0 \), we may find integers \( N \) and \( M \) such that\n\n\[ \left| {f\left( {x}_{n}\right) - L}\right| < \epsilon \]\n\n(5.1.19)\n\nwhenever \( n \in \left\{ {j : j \in {J}^{ - }, j > N}\right\} \) and\n\n\[ \left| {f\left( {x}_{n}\right) - L}\right| < \epsilon \]\n\n\( \left( {5.1.20}\right) \)\n\nwhenever \( n \in \left\{ {j : j \in {J}^{ + }, j > M}\right\} \) . Let \( P \) be the larger of \( N \) and \( M \) . Since \( {J}^{ - } \cup {J}^{ + } = \left\{ {j : j \in {\mathbb{Z}}^{ + }, j \geq m}\right\} \), it follows that\n\n\[ \left| {f\left( {x}_{n}\right) - L}\right| < \epsilon \]\n\n\( \left( {5.1.21}\right) \)\n\nwhenever \( n > P \) . Hence \( \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = L \), and so \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) .\n\nQ.E.D.
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Yes
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Proposition 5.1.2. Suppose \( D \subset \mathbb{R}, a \) is a limit point of \( D \), and \( f : D \rightarrow \mathbb{R} \) . If \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) and \( \alpha \in \mathbb{R} \), then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}{\alpha f}\left( x\right) = {\alpha L} \]\n\n\( \left( {5.1.22}\right) \)
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Proof. Suppose \( {\left\{ {x}_{n}\right\} }_{n \in I} \in S\left( {D, a}\right) \) . Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\alpha f}\left( {x}_{n}\right) = \alpha \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = {\alpha L}. \]\n\n\( \left( {5.1.23}\right) \)\n\nHence \( \mathop{\lim }\limits_{{x \rightarrow a}}{\alpha f}\left( x\right) = {\alpha L} \).\n\nQ.E.D.
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Yes
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Proposition 5.1.3. Suppose \( D \subset \mathbb{R}, a \) is a limit point of \( D, f : D \rightarrow \mathbb{R} \), and \( g : D \rightarrow \mathbb{R} \) . If \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) and \( \mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = M \), then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\left( {f\left( x\right) + g\left( x\right) }\right) = L + M. \]
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Proof. Suppose \( {\left\{ {x}_{n}\right\} }_{n \in I} \in S\left( {D, a}\right) \) . Then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {f\left( {x}_{n}\right) + g\left( {x}_{n}\right) }\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) + \mathop{\lim }\limits_{{n \rightarrow \infty }}g\left( {x}_{n}\right) = L + M. \]\n\nHence \( \mathop{\lim }\limits_{{x \rightarrow a}}\left( {f\left( x\right) + g\left( x\right) }\right) = L + M \). \n\nQ.E.D.
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Yes
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Proposition 5.1.5. Suppose \( D \subset \mathbb{R}, a \) is a limit point of \( D, f : D \rightarrow \mathbb{R} \) , \( g : D \rightarrow \mathbb{R} \), and \( g\left( x\right) \neq 0 \) for all \( x \in D \) . If \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L,\mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = M \), and \( M \neq 0 \), then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) }{g\left( x\right) } = \frac{L}{M} \]\n\n\( \left( {5.1.27}\right) \)
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Exercise 5.1.2. Prove the previous proposition.
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No
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Proposition 5.1.7. Suppose \( D \subset \mathbb{R}, E \subset \mathbb{R}, a \) is a limit point of \( D, g : D \rightarrow \mathbb{R} \) , \( f : E \rightarrow \mathbb{R} \), and \( g\left( D\right) \subset E \) . Moreover, suppose \( \lim g\left( x\right) = b \) and, for some \( \epsilon > 0 \) , \( g\left( x\right) \neq b \) for all \( x \in \left( {a - \epsilon, a + \epsilon }\right) \cap D \) . If \( \mathop{\lim }\limits_{{x \rightarrow b}}f\left( x\right) = L \), then \[ \mathop{\lim }\limits_{{x \rightarrow a}}f \circ g\left( x\right) = L. \]
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Proof. Suppose \( {\left\{ {x}_{n}\right\} }_{n \in I} \in S\left( {D, a}\right) \) . Then \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}g\left( {x}_{n}\right) = b \] Let \( N \in {\mathbb{Z}}^{ + } \) such that \( \left| {{x}_{n} - a}\right| < \epsilon \) whenever \( n > N \) . Then \[ {\left\{ g\left( {x}_{n}\right) \right\} }_{n = N + 1}^{\infty } \in S\left( {E, b}\right) , \] so \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {g\left( {x}_{n}\right) }\right) = L \] Thus \( \mathop{\lim }\limits_{{x \rightarrow a}}f \circ g\left( x\right) = L \) . Q.E.D.
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Yes
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Example 5.1.3. Suppose \( f : \mathbb{R} \rightarrow \mathbb{R} \) is defined by \( f\left( x\right) = x \) for all \( x \in \mathbb{R} \) . If, for any \( a \in \mathbb{R},{\left\{ {x}_{n}\right\} }_{n \in I} \in S\left( {\mathbb{R}, a}\right) \), then
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\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = a. \] Hence \( \mathop{\lim }\limits_{{x \rightarrow a}}x = a \) .
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Yes
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Example 5.1.4. Suppose \( n \in {\mathbb{Z}}^{ + } \) and \( f : \mathbb{R} \rightarrow \mathbb{R} \) is defined by \( f\left( x\right) = {x}^{n} \) . Then
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\[ \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow a}}{x}^{n} = \mathop{\prod }\limits_{{i = 1}}^{n}\mathop{\lim }\limits_{{x \rightarrow a}}x = {a}^{n}. \]
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Yes
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Proposition 5.1.8. Suppose \( D \subset \mathbb{R}, a \) is a limit point of \( D \), and \( f : D \rightarrow \mathbb{R} \) . Then \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) if and only if for every \( \epsilon > 0 \) there exists a \( \delta > 0 \) such that\n\n\[ \left| {f\left( x\right) - L}\right| < \epsilon \text{whenever}x \neq a\text{and}x \in \left( {a - \delta, a + \delta }\right) \cap D\text{.} \]\n\n\( \left( {5.1.37}\right) \)
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Proof. Suppose \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) . Suppose there exists an \( \epsilon > 0 \) such that for every \( \delta > 0 \) there exists \( x \in \left( {a - \delta, a + \delta }\right) \cap D, x \neq a \), for which \( \left| {f\left( x\right) - L}\right| \geq \epsilon \) . For \( n = 1,2,3,\ldots \), choose\n\n\[ {x}_{n} \in \left( {a - \frac{1}{n}, a + \frac{1}{n}}\right) \cap D \]\n\n(5.1.38)\n\n\( {x}_{n} \neq a \), such that \( \left| {f\left( {x}_{n}\right) - L}\right| \geq \epsilon \) . Then \( {\left\{ {x}_{n}\right\} }_{n = 1}^{\infty } \in S\left( {D, a}\right) \), but \( {\left\{ f\left( {x}_{n}\right) \right\} }_{n = 1}^{\infty } \) does not converge to \( L \), contradicting the assumption that \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) .\n\nNow suppose that for every \( \epsilon > 0 \) there exists \( \delta > 0 \) such that \( \left| {f\left( x\right) - L}\right| < \epsilon \) whenever \( x \neq a \) and \( x \in \left( {a - \delta, a + \delta }\right) \cap D \) . Let \( {\left\{ {x}_{n}\right\} }_{n \in I} \in S\left( {D, a}\right) \) . Given \( \epsilon > 0 \) , let \( \delta > 0 \) be such that \( \left| {f\left( x\right) - L}\right| < \epsilon \) whenever \( x \neq a \) and \( x \in \left( {a - \delta, a + \delta }\right) \cap D \) . Choose \( N \in \mathbb{Z} \) such that \( \left| {{x}_{n} - a}\right| < \delta \) whenever \( n > N \) . Then \( \left| {f\left( {x}_{n}\right) - L}\right| < \epsilon \) for all \( n > N \) . Hence \( \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = L \), and so \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = L \) . Q.E.D.
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Yes
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Proposition 5.1.9. Suppose \( D \subset \mathbb{R}, a \) is a limit point of \( D, f : D \rightarrow \mathbb{R} \), and \( {S}^{ - }\left( {D, a}\right) \neq \varnothing \) . Then \( \mathop{\lim }\limits_{{x \rightarrow {a}^{ - }}}f\left( x\right) = L \) if and only if for every \( \epsilon > 0 \) there exists a \( \delta > 0 \) such that
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\[ \left| {f\left( x\right) - L}\right| < \epsilon \text{whenever}x \in \left( {a - \delta, a}\right) \cap D\text{.} \]
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Yes
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Proposition 5.1.10. Suppose \( D \subset \mathbb{R}, a \) is a limit point of \( D, f : D \rightarrow \mathbb{R} \), and \( {S}^{ + }\left( {D, a}\right) \neq \varnothing \) . Then \( \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}f\left( x\right) = L \) if and only if for every \( \epsilon > 0 \) there exists a \( \delta > 0 \) such that
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\[ \left| {f\left( x\right) - L}\right| < \epsilon \text{whenever}x \in \left( {a, a + \delta }\right) \cap D\text{.} \]
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Yes
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Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) by\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }x\text{ is rational,} \\ 0, & \text{ if }x\text{ is irrational. } \end{array}\right.\]\n\nLet \( a \in \mathbb{R} \) .
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Since every open interval contains both rational and irrational numbers, for any \( \delta > 0 \) and any choice of \( L \in \mathbb{R} \), there will exist \( x \in \left( {a - \delta, a + \delta }\right) \) , \( x \neq a \), such that\n\n\[ \left| {f\left( x\right) - L}\right| \geq \frac{1}{2} \]\n\nHence \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) \) does not exist for any real number \( a \) .
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Yes
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Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) by\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} x, & \text{ if }x\text{ is rational,} \\ 0, & \text{ if }x\text{ is irrational. } \end{array}\right. \]\n\nThen \( \mathop{\lim }\limits_{{x \rightarrow 0}}f\left( x\right) = 0 \)
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since, given \( \epsilon > 0,\left| {f\left( x\right) }\right| < \epsilon \) provided \( \left| x\right| < \epsilon \)
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Yes
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Define \( \varphi : \left\lbrack {0,1}\right\rbrack \rightarrow \left\lbrack {-1,1}\right\rbrack \) by \[ \varphi \left( x\right) = \left\{ \begin{array}{ll} {4x}, & \text{ if }0 \leq x \leq \frac{1}{4} \\ 2 - {4x}, & \text{ if }\frac{1}{4} < x < \frac{3}{4} \\ {4x} - 4, & \text{ if }\frac{3}{4} \leq x \leq 1 \end{array}\right. \] Next define \( s : \mathbb{R} \rightarrow \mathbb{R} \) by \( s\left( x\right) = \varphi \left( {x-\lfloor x\rfloor }\right) \), where \( \lfloor x\rfloor \) denotes the largest integer less than or equal to \( x \) (that is, \( \lfloor x\rfloor \) is the floor of \( x \) ). The function \( s \) is an example of a sawtooth function. See the graphs of \( \varphi \) and \( s \) in Figure 5.1.1. Note that for any \( n \in \mathbb{Z} \), \[ s\left( \left\lbrack {n, n + 1}\right\rbrack \right) = \left\lbrack {-1,1}\right\rbrack . \]
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Now let \( D = \mathbb{R} \smallsetminus \{ 0\} \) and define \( \sigma : D \rightarrow \mathbb{R} \) by \[ \sigma \left( x\right) = s\left( \frac{1}{x}\right) . \] See the graph of \( \sigma \) in Figure 5.1.2. Note that for any \( n \in {\mathbb{Z}}^{ + } \), \[ \sigma \left( \left\lbrack {\frac{1}{n + 1},\frac{1}{n}}\right\rbrack \right) = s\left( \left\lbrack {n, n + 1}\right\rbrack \right) = \left\lbrack {-1,1}\right\rbrack . \] Hence for any \( \epsilon > 0,\sigma \left( \left( {0,\epsilon }\right) \right) = \left\lbrack {-1,1}\right\rbrack \), and so \( \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\sigma \left( x\right) \) does not exist. Similarly, neither \( \mathop{\lim }\limits_{{x \rightarrow {0}^{ - }}}\sigma \left( x\right) \) nor \( \mathop{\lim }\limits_{{x \rightarrow 0}}\sigma \left( x\right) \) exist.
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Yes
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Example 5.1.8. Let \( s \) be the sawtooth function of the previous example and let \( D = \mathbb{R} \smallsetminus \{ 0\} \) . Define \( \psi : D \rightarrow \mathbb{R} \) by \[ \psi \left( x\right) = {xs}\left( \frac{1}{x}\right) . \] See Figure 5.1.2 for the graph of \( \psi \) . Then for all \( x \in D \) , \[ - \left| x\right| \leq \psi \left( x\right) \leq \left| x\right| \] and so \( \mathop{\lim }\limits_{{x \rightarrow 0}}\psi \left( x\right) = 0 \) by the squeeze theorem.
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and so \( \mathop{\lim }\limits_{{x \rightarrow 0}}\psi \left( x\right) = 0 \) by the squeeze theorem.
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No
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Proposition 5.2.1. If \( f \) is monotonic on \( \left( {a, b}\right) \), then \( f\left( {c + }\right) \) and \( f\left( {c - }\right) \) exist for every \( c \in \left( {a, b}\right) \) .
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Proof. Suppose \( f \) is nondecreasing on \( \left( {a, b}\right) \) . Let \( c \in \left( {a, b}\right) \) and let\n\n\[ \lambda = \sup \{ f\left( x\right) : a < x < c\} . \]\n\nNote that \( \lambda \leq f\left( c\right) < + \infty \) . Given any \( \epsilon > 0 \), there must exist \( \delta > 0 \) such that\n\n\[ \lambda - \epsilon < f\left( {c - \delta }\right) \leq \lambda \]\n\nSince \( f \) is nondecreasing, it follows that\n\n\[ \left| {f\left( x\right) - \lambda }\right| < \epsilon \]\n\nwhenever \( x \in \left( {c - \delta, c}\right) \) . Thus \( f\left( {c - }\right) = \lambda \) . A similar argument shows that \( f\left( {c + }\right) = \kappa \) where\n\n\[ \kappa = \inf \{ f\left( x\right) : c < x < b\} . \]\n\nIf \( f \) is nonincreasing, similar arguments yield\n\n\[ f\left( {c - }\right) = \inf \{ f\left( x\right) : a < x < c\} \]\n\nand\n\n\[ f\left( {c + }\right) = \sup \{ f\left( x\right) : c < x < b\} . \]\n\nQ.E.D.
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Yes
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Proposition 5.2.2. If \( f \) is nondecreasing on \( \\left( {a, b}\\right) \) and \( a < x < y < b \), then\n\n\[ f\\left( {x + }\\right) \\leq f\\left( {y - }\\right) .\n\]
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Proof. By the previous proposition,\n\n\[ f\\left( {x + }\\right) = \\inf \\{ f\\left( t\\right) : x < t < b\\}\n\]\n\nand\n\n\[ f\\left( {y - }\\right) = \\sup \\{ f\\left( t\\right) : a < t < y\\} .\n\]\n\nSince \( f \) is nondecreasing,\n\n\[ \\inf \\{ f\\left( t\\right) : x < t < b\\} = \\inf \\{ f\\left( t\\right) : x < t < y\\}\n\]\n\nand\n\n\[ \\sup \\{ f\\left( t\\right) : a < t < y\\} = \\sup \\{ f\\left( t\\right) : x < t < y\\} .\n\]\n\nThus\n\n\[ f\\left( {x + }\\right) = \\inf \\{ f\\left( t\\right) : x < t < y\\} \\leq \\sup \\{ f\\left( t\\right) : x < t < y\\} = f\\left( {y - }\\right) .\n\]\n\nQ.E.D.
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Yes
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Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) by\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }x\text{ is rational,} \\ 0, & \text{ if }x\text{ is irrational. } \end{array}\right.\n\]
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Then, by Example 5.1.5, \( f \) is discontinuous at every \( x \in \mathbb{R} \) .
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No
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Proposition 5.4.1. Suppose \( D \subset \mathbb{R},\alpha \in \mathbb{R}, f : D \rightarrow \mathbb{R} \), and \( g : D \rightarrow \mathbb{R} \) . If \( f \) and \( g \) are continuous at \( a \), then \( {\alpha f}, f + g \), and \( {fg} \) are all continuous at \( a \) . Moreover, if \( g\left( x\right) \neq 0 \) for all \( x \in D \), then \( \frac{f}{g} \) is continuous at \( a \) .
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Exercise 5.4.1. Prove the previous proposition.
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No
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Proposition 5.4.2. Suppose \( D \subset \mathbb{R}, f : D \rightarrow \mathbb{R}, f\left( x\right) \geq 0 \) for all \( x \in D \), and \( f \) is continuous at \( a \in D \) . If \( g : D \rightarrow \mathbb{R} \) is defined by \( g\left( x\right) = \sqrt{f\left( x\right) } \), then \( g \) is continuous at \( a \) .
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Exercise 5.4.2. Prove the previous proposition.
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No
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Proposition 5.4.3. Suppose \( D \subset \mathbb{R}, f : D \rightarrow \mathbb{R} \), and \( a \in D \) . Then \( f \) is continuous at \( a \) if and only if for every \( \epsilon > 0 \) there exists \( \delta > 0 \) such that\n\n\[ \left| {f\left( x\right) - f\left( a\right) }\right| < \epsilon \text{whenever}x \in \left( {a - \delta, a + \delta }\right) \cap D\text{.} \]
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Proof. Suppose \( f \) is continuous at \( a \) . If \( a \) is an isolated point of \( D \), then there exists a \( \delta > 0 \) such that\n\n\[ \left( {a - \delta, a + \delta }\right) \cap D = \{ a\} . \]\n\nThen for any \( \epsilon > 0 \), if \( x \in \left( {a - \delta, a + \delta }\right) \cap D \), then \( x = a \), and so\n\n\[ \left| {f\left( x\right) - f\left( a\right) }\right| = \left| {f\left( a\right) - f\left( a\right) }\right| = 0 < \epsilon . \]\n\nIf \( a \) is a limit point of \( D \), then \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = f\left( a\right) \) implies that for any \( \epsilon > 0 \) there exists \( \delta > 0 \) such that\n\n\[ \left| {f\left( x\right) - f\left( a\right) }\right| < \epsilon \text{whenever}x \in \left( {a - \delta, a + \delta }\right) \cap D\text{.} \]\n\nNow suppose that for every \( \epsilon > 0 \) there exists \( \delta > 0 \) such that\n\n\[ \left| {f\left( x\right) - f\left( a\right) }\right| < \epsilon \text{whenever}x \in \left( {a - \delta, a + \delta }\right) \cap D\text{.} \]\n\nIf \( a \) is an isolated point, then \( f \) is continuous at \( a \) . If \( a \) is a limit point, then this condition implies \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = f\left( a\right) \), and so \( f \) is continuous at \( a \) . Q.E.D.
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Yes
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Proposition 5.4.4. Suppose \( D \subset \mathbb{R}, E \subset \mathbb{R}, g : D \rightarrow \mathbb{R}, f : E \rightarrow \mathbb{R}, g\left( D\right) \subset E \) , and \( a \in D \) . If \( g \) is continuous at \( a \) and \( f \) is continuous at \( g\left( a\right) \), then \( f \circ g \) is continuous at \( a \) .
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Proof. Let \( {\left\{ {x}_{n}\right\} }_{n \in I} \) be a sequence with \( {x}_{n} \in D \) for every \( n \in I \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = a \) . Then, since \( g \) is continuous at \( a,{\left\{ g\left( {x}_{n}\right) \right\} }_{n \in I} \) is a sequence with \( g\left( {x}_{n}\right) \in E \) for every \( n \in I \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}g\left( {x}_{n}\right) = g\left( a\right) \) . Hence, since \( f \) is continuous at \( g\left( a\right) \) , \( \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {g\left( {x}_{n}\right) }\right) = f\left( {g\left( a\right) }\right) \) . That is,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {f \circ g}\right) \left( {x}_{n}\right) = \left( {f \circ g}\right) \left( a\right) . \]\n\n\( \left( {5.4.10}\right) \)\n\nHence \( f \circ g \) is continuous at \( a \) .\n\nQ.E.D.
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Yes
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Proposition 5.4.5. Suppose \( f \) is monotonic on the interval \( \left( {a, b}\right) \) . Then every discontinuity of \( f \) in \( \left( {a, b}\right) \) is a simple discontinuity. Moreover, if \( E \) is the set of points in \( \left( {a, b}\right) \) at which \( f \) is discontinuous, then either \( E = \varnothing, E \) is finite, or \( E \) is countable.
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Proof. The first statement follows immediately from Proposition 5.2.1. For the second statement, suppose \( f \) is nondecreasing and suppose \( E \) is nonempty. From Exercise 2.1.26 and the the proof of Proposition 5.2.1, it follows that for every \( x \in \left( {a, b}\right) \) ,\n\n\[ f\left( {x - }\right) \leq f\left( x\right) \leq f\left( {x + }\right) . \]\n\n(5.4.11)\n\nHence \( x \in E \) if and only if \( f\left( {x - }\right) < f\left( {x + }\right) \) . Hence for every \( x \in E \), we may choose a rational number \( {r}_{x} \) such that \( f\left( {x - }\right) < {r}_{x} < f\left( {x + }\right) \) . Now if \( x, y \in E \) with \( x < y \), then, by Proposition 5.2.2,\n\n\[ {r}_{x} < f\left( {x + }\right) \leq f\left( {y - }\right) < {r}_{y}, \]\n\n\( \left( {5.4.12}\right) \)\n\nso \( {r}_{x} \neq {r}_{y} \) . Thus we have a one-to-one correspondence between \( E \) and a subset of \( \mathbb{Q} \), and so \( E \) is either finite or countable. A similar argument holds if \( f \) is nonincreasing. Q.E.D.
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Yes
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Proposition 5.4.8. Suppose \( D \subset \mathbb{R} \) and \( f : D \rightarrow \mathbb{R} \). Then \( f \) is continuous on \( D \) if and only if for every open set \( V \subset \mathbb{R},{f}^{-1}\left( V\right) = U \cap D \) for some open set \( U \subset \mathbb{R} \).
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Proof. Suppose \( f \) is continuous on \( D \) and \( V \subset \mathbb{R} \) is an open set. If \( V \cap f\left( D\right) = \varnothing \), then \( {f}^{-1}\left( V\right) = \varnothing \), which is open. So suppose \( V \cap f\left( D\right) \neq \varnothing \) and let \( a \in {f}^{-1}\left( V\right) \). Since \( V \) is open and \( f\left( a\right) \in V \), there exists \( {\epsilon }_{a} > 0 \) such that\n\n\[ \left( {f\left( a\right) - {\epsilon }_{a}, f\left( a\right) + {\epsilon }_{a}}\right) \subset V.\]\n\n(5.4.14)\n\nSince \( f \) is continuous, there exists \( {\delta }_{a} > 0 \) such that\n\n\[ f\left( {\left( {a - {\delta }_{a}, a + {\delta }_{a}}\right) \cap D}\right) \subset \left( {f\left( a\right) - {\epsilon }_{a}, f\left( a\right) + {\epsilon }_{a}}\right) \subset V.\]\n\n\( \left( {5.4.15}\right) \)\n\nThat is, \( \left( {a - {\delta }_{a}, a + {\delta }_{a}}\right) \cap D \subset {f}^{-1}\left( V\right) \). Let\n\n\[ U = \mathop{\bigcup }\limits_{{a \in {f}^{-1}\left( V\right) }}\left( {a - {\delta }_{a}, a + {\delta }_{a}}\right)\]\n\n\( \left( {5.4.16}\right) \)\n\nThen \( U \) is open and \( {f}^{-1}\left( V\right) = U \cap D \).\n\nNow suppose that for every open set \( V \subset \mathbb{R},{f}^{-1}\left( V\right) = U \cap D \) for some open set \( U \subset \mathbb{R} \). Let \( a \in D \) and let \( \epsilon > 0 \) be given. Since \( \left( {f\left( a\right) - \epsilon, f\left( a\right) + \epsilon }\right) \) is open, there exists an open set \( U \) such that\n\n\[ U \cap D = {f}^{-1}\left( \left( {f\left( a\right) - \epsilon, f\left( a\right) + \epsilon }\right) \right) .\]\n\n(5.4.17)\n\nSince \( U \) is open and \( a \in U \), there exists \( \delta > 0 \) such that \( \left( {a - \delta, a + \delta }\right) \subset U \). But then\n\n\[ f\left( {\left( {a - \delta, a + \delta }\right) \cap D}\right) \subset \left( {f\left( a\right) - \epsilon, f\left( a\right) + \epsilon }\right) .\]\n\n(5.4.18)\n\nThat is, if \( x \in \left( {a - \delta, a + \delta }\right) \cap D \), then \( \left| {f\left( x\right) - f\left( a\right) }\right| < \epsilon \). Hence \( f \) is continuous at \( a \). Q.E.D.
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Yes
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Theorem 5.4.9 (Intermediate Value Theorem). Suppose \( a, b \in \mathbb{R}, a < b \), and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) . If \( f \) is continuous and \( s \in \mathbb{R} \) is such that either \( f\left( a\right) \leq s \leq f\left( b\right) \) or \( f\left( b\right) \leq s \leq f\left( a\right) \), then there exists \( c \in \left\lbrack {a, b}\right\rbrack \) such that \( f\left( c\right) = s \) .
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Proof. Suppose \( f\left( a\right) < f\left( b\right) \) and \( f\left( a\right) < s < f\left( b\right) \) . Let\n\n\[ c = \sup \{ x : x \in \left\lbrack {a, b}\right\rbrack, f\left( x\right) \leq s\} . \]\n\n(5.4.19)\n\nSuppose \( f\left( c\right) < s \) . Then \( c < b \) and, since \( f \) is continuous at \( c \), there exists a \( \delta > 0 \) such that \( f\left( x\right) < s \) for all \( x \in \left( {c, c + \delta }\right) \) . But then \( f\left( {c + \frac{\delta }{2}}\right) < s \) , contradicting the definition of \( c \) . Similarly, if \( f\left( c\right) > s \), then \( c > a \) and there exists \( \delta > 0 \) such that \( f\left( x\right) > s \) for all \( x \in \left( {c - \delta, c}\right) \), again contradicting the definition of \( c \) . Hence we must have \( f\left( c\right) = s \) .\n\nQ.E.D.
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Yes
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Example 5.4.3. Suppose \( a \in \mathbb{R}, a > 0 \), and consider \( f\left( x\right) = {x}^{n} - a \) where \( n \in \mathbb{Z}, n > 1 \) . Then \( f\left( 0\right) = - a < 0 \) and
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\[ f\left( {1 + a}\right) = {\left( 1 + a\right) }^{n} - a \] \[ = 1 + {na} + \mathop{\sum }\limits_{{i = 2}}^{n}\left( \begin{matrix} n \\ i \end{matrix}\right) {a}^{i} - a \] \[ = 1 + \left( {n - 1}\right) a + \mathop{\sum }\limits_{{i = 2}}^{n}\left( \begin{matrix} n \\ i \end{matrix}\right) {a}^{i} > 0, \] where \( \left( \begin{matrix} n \\ i \end{matrix}\right) \) is the binomial coefficient \[ \left( \begin{matrix} n \\ i \end{matrix}\right) = \frac{n!}{i!\left( {n - i}\right) !}. \] Hence, by the Intermediate Value Theorem, there exists a real number \( \gamma > 0 \) such that \( {\gamma }^{n} = a \) . Moreover, there is only one such \( \gamma \) since \( f \) is increasing on \( \left( {0, + \infty }\right) \) . We call \( \gamma \) the nth root of \( a \), and write \[ \gamma = \sqrt[n]{a} \] or \[ \gamma = {a}^{\frac{1}{n}}. \]
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Yes
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Theorem 5.4.10. Suppose \( D \subset \mathbb{R} \) is compact and \( f : D \rightarrow \mathbb{R} \) is continuous. Then \( f\left( D\right) \) is compact.
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Proof. Given a sequence \( {\left\{ {y}_{n}\right\} }_{n \in I} \) in \( f\left( D\right) \), choose a sequence \( {\left\{ {x}_{n}\right\} }_{n \in I} \) such that \( f\left( {x}_{n}\right) = {y}_{n} \). Since \( D \) is compact, \( {\left\{ {x}_{n}\right\} }_{n \in I} \) has a convergent subsequence \( {\left\{ {x}_{{n}_{k}}\right\} }_{k = 1}^{\infty } \) with\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{{n}_{k}} = x \in D. \]\n\n\( \left( {5.4.25}\right) \)\n\nLet \( y = f\left( x\right) \). Then \( y \in f\left( D\right) \) and, since \( f \) is continuous,\n\n\[ y = \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{{n}_{k}}\right) = \mathop{\lim }\limits_{{k \rightarrow \infty }}{y}_{{n}_{k}}. \]\n\n\( \left( {5.4.26}\right) \)\n\nHence \( f\left( D\right) \) is compact.\n\nQ.E.D.
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Yes
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Theorem 5.4.11 (Extreme Value Theorem). Suppose \( D \subset \mathbb{R} \) is compact and \( f : D \rightarrow \mathbb{R} \) is continuous. Then there exists \( a \in D \) such that \( f\left( a\right) \geq f\left( x\right) \) for all \( x \in D \) and there exists \( b \in D \) such that \( f\left( b\right) \leq f\left( x\right) \) for all \( x \in D \) .
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Proof. Let \( s = \sup f\left( D\right) \) and \( t = \inf f\left( D\right) \) . Then \( s \in f\left( D\right) \), so there exists \( a \in D \) such that \( f\left( a\right) = s \), and \( t \in f\left( D\right) \), so there exists \( b \in D \) such that \( f\left( b\right) = t \) . Q.E.D.
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No
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Proposition 5.4.12. Suppose \( D \subset \mathbb{R} \) is compact, \( f : D \rightarrow \mathbb{R} \) is continuous and one-to-one, and \( E = f\left( D\right) \) . Then \( {f}^{-1} : E \rightarrow D \) is continuous.
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Proof. Let \( V \subset \mathbb{R} \) be an open set. We need to show that \( f\left( {V \cap D}\right) = U \cap E \) for some open set \( U \subset \mathbb{R} \) . Let \( C = D \cap \left( {\mathbb{R} \smallsetminus V}\right) \) . Then \( C \) is a closed subset of \( D \) , and so is compact. Hence \( f\left( C\right) \) is a compact subset of \( E \) . Thus \( f\left( C\right) \) is closed, and so \( U = \mathbb{R} \smallsetminus f\left( C\right) \) is open. Moreover, \( U \cap E = E \smallsetminus f\left( C\right) = f\left( {V \cap D}\right) \) . Thus \( {f}^{-1} \) is continuous.
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Yes
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Define \( f : \left( {0, + \infty }\right) \) by \( f\left( x\right) = \frac{1}{x} \). Given any \( \delta > 0 \), choose \( n \in {\mathbb{Z}}^{ + } \) such that \( \frac{1}{n\left( {n + 1}\right) } < \delta \). Let \( x = \frac{1}{n} \) and \( y = \frac{1}{n + 1} \). Then
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\[
\left| {x - y}\right| = \frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n\left( {n + 1}\right) } < \delta .
\]
However,
\[
\left| {f\left( x\right) - f\left( y\right) }\right| = \left| {n - \left( {n + 1}\right) }\right| = 1.
\]
Hence, for example, there does not exist a \( \delta > 0 \) such that
\[
\left| {f\left( x\right) - f\left( y\right) }\right| < \frac{1}{2}
\]
whenever \( \left| {x - y}\right| < \delta \) . Thus \( f \) is not uniformly continuous on \( \left( {0, + \infty }\right) \), although \( f \) is continuous on \( \left( {0, + \infty }\right) \).
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Yes
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Example 5.4.5. Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) by \( f\left( x\right) = {2x} \) . Let \( \epsilon > 0 \) be given. If \( \delta = \frac{\epsilon }{2} \) , then
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\[ \left| {f\left( x\right) - f\left( y\right) }\right| = 2\left| {x - y}\right| < \epsilon \] whenever \( \left| {x - y}\right| < \delta \) . Hence \( f \) is uniformly continuous on \( \mathbb{R} \) .
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Yes
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Proposition 5.4.13. Suppose \( D \subset \mathbb{R} \) is compact and \( f : D \rightarrow \mathbb{R} \) is continuous. Then \( f \) is uniformly continuous on \( D \) .
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Proof. Let \( \epsilon > 0 \) be given. For every \( x \in D \), choose \( {\delta }_{x} \) such that\n\n\[ \left| {f\left( x\right) - f\left( y\right) }\right| < \frac{\epsilon }{2} \]\n\n(5.4.28)\n\nwhenever \( y \in D \) and \( \left| {x - y}\right| < {\delta }_{x} \) . Let\n\n\[ {J}_{x} = \left( {x - \frac{{\delta }_{x}}{2}, x + \frac{{\delta }_{x}}{2}}\right) . \]\n\n\( \left( {5.4.29}\right) \)\n\nThen \( \left\{ {{J}_{x} : x \in D}\right\} \) is an open cover of \( D \) . Since \( D \) is compact, there must exist \( {x}_{1},{x}_{2},\ldots ,{x}_{n}, n \in {Z}^{ + } \), such that \( {J}_{{x}_{1}},{J}_{{x}_{2}},\ldots ,{J}_{{x}_{n}} \) is an open cover of \( D \) . Let \( \delta \) be the smallest of\n\n\[ \frac{{\delta }_{{x}_{1}}}{2},\frac{{\delta }_{{x}_{2}}}{2},\ldots ,\frac{{\delta }_{{x}_{n}}}{2} \]\n\n\( \left( {5.4.30}\right) \)\n\nNow let \( x, y \in D \) with \( \left| {x - y}\right| < \delta \) . Then for some integer \( k,1 \leq k \leq n, x \in {J}_{{x}_{k}} \) , that is,\n\n\[ \left| {x - {x}_{k}}\right| < \frac{{\delta }_{{x}_{k}}}{2} \]\n\n\( \left( {5.4.31}\right) \)\n\nMoreover,\n\n\[ \left| {y - {x}_{k}}\right| \leq \left| {y - x}\right| + \left| {x - {x}_{k}}\right| < \delta + \frac{{\delta }_{{x}_{k}}}{2} \leq {\delta }_{{x}_{k}}. \]\n\n\( \left( {5.4.32}\right) \)\n\nHence\n\n\[ \left| {f\left( x\right) - f\left( y\right) }\right| \leq \left| {f\left( x\right) - f\left( {x}_{k}\right) }\right| + \left| {f\left( {x}_{k}\right) - f\left( y\right) }\right| < \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon . \]\n\n\( \left( {5.4.33}\right) \)\n\nQ.E.D.
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Yes
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Proposition 6.1.1. Suppose \( D \subset \mathbb{R}, f : D \rightarrow \mathbb{R} \), and \( a \) is an interior point of \( D \) . Then \( f \) is differentiable at \( a \) if and only if\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) - f\left( a\right) }{x - a} \]\n\n\( \left( {6.1.4}\right) \)\n\nexists, in which case \( d{f}_{a}\left( x\right) = {mx} \) where\n\n\[ m = \mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) - f\left( a\right) }{x - a}. \]\n\n(6.1.5)
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Proof. Let \( m \in \mathbb{R} \) and let \( L : \mathbb{R} \rightarrow \mathbb{R} \) be the linear function \( L\left( x\right) = {mx} \) . Then\n\n\[ \frac{f\left( x\right) - f\left( a\right) - L\left( {x - a}\right) }{x - a} = \frac{f\left( x\right) - f\left( a\right) - m\left( {x - a}\right) }{x - a} \]\n\n\[ = \frac{f\left( x\right) - f\left( a\right) }{x - a} - m. \]\n\n(6.1.6)\n\nHence\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) - f\left( a\right) - L\left( {x - a}\right) }{x - a} = 0 \]\n\n(6.1.7)\n\nif and only if\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) - f\left( a\right) }{x - a} = m. \]\n\n(6.1.8)\n\nQ.E.D.
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Yes
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Let \( n \in {\mathbb{Z}}^{ + } \) and define \( f : \mathbb{R} \rightarrow \mathbb{R} \) by \( f\left( x\right) = {x}^{n} \) . Then
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\[ {f}^{\prime }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{\left( x + h\right) }^{n} - {x}^{n}}{h} \] \[ = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{x}^{n} + n{x}^{n - 1}h + \mathop{\sum }\limits_{{k = 2}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {x}^{n - k}{h}^{k} - {x}^{n}}{h} \] \[ = \mathop{\lim }\limits_{{h \rightarrow 0}}\left( {n{x}^{n - 1} + \mathop{\sum }\limits_{{k = 2}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {x}^{n - k}{h}^{k - 1}}\right) \] \[ = n{x}^{n - 1}\text{.} \]
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Yes
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Define \( f : \mathbb{R} \rightarrow \mathbb{R} \) by \( f\left( x\right) = \left| x\right| \) . Then
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\[ \frac{f\left( {0 + h}\right) - f\left( 0\right) }{h} = \frac{\left| h\right| }{h} = \begin{cases} 1, & \text{ if }h > 0, \\ - 1, & \text{ if }h < 0. \end{cases} \] Hence \[ \mathop{\lim }\limits_{{h \rightarrow {0}^{ - }}}\frac{f\left( {0 + h}\right) - f\left( 0\right) }{h} = - 1 \] and \[ \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\frac{f\left( {0 + h}\right) - f\left( 0\right) }{h} = 1. \] Thus \( f \) is not differentiable at 0 .
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Yes
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Proposition 6.2.1. If \( f \) is differentiable at \( a \), then \( f \) is continuous at \( a \) .
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Proof. If \( f \) is differentiable at \( a \), then\n\n\[\n\mathop{\lim }\limits_{{x \rightarrow a}}\left( {f\left( x\right) - f\left( a\right) }\right) = \mathop{\lim }\limits_{{x \rightarrow a}}\left( \frac{f\left( x\right) - f\left( a\right) }{x - a}\right) \left( {x - a}\right) = {f}^{\prime }\left( a\right) \left( 0\right) = 0.\n\]\n\n\( \left( {6.2.4}\right) \)\n\nHence \( \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = f\left( a\right) \), and so \( f \) is continuous at \( a \) .\n\nQ.E.D.
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Yes
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Proposition 6.2.2. Suppose \( f \) is differentiable at \( a \) and \( \alpha \in \mathbb{R} \) . Then \( {\alpha f} \) is differentiable at \( a \) and \( {\left( \alpha f\right) }^{\prime }\left( a\right) = \alpha {f}^{\prime }\left( a\right) \) .
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Exercise 6.2.5. Prove the previous proposition.
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No
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Proposition 6.2.3. Suppose \( f \) and \( g \) are both differentiable at \( a \) . Then \( f + g \) is differentiable at \( a \) and \( {\left( f + g\right) }^{\prime }\left( a\right) = {f}^{\prime }\left( a\right) + {g}^{\prime }\left( a\right) \) .
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Exercise 6.2.6. Prove the previous proposition.
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No
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Proposition 6.2.4 (Product rule). Suppose \( f \) and \( g \) are both differentiable at \( a \) . Then \( {fg} \) is differentiable at \( a \) and\n\n\[{\left( fg\right) }^{\prime }\left( a\right) = f\left( a\right) {g}^{\prime }\left( a\right) + g\left( a\right) {f}^{\prime }\left( a\right) .
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Proof. We have\n\n\[{\left( fg\right) }^{\prime }\left( a\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {a + h}\right) g\left( {a + h}\right) - f\left( a\right) g\left( a\right) }{h}\]\n\n\[= \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {a + h}\right) g\left( {a + h}\right) - f\left( a\right) g\left( {a + h}\right) + f\left( a\right) g\left( {a + h}\right) - f\left( a\right) g\left( a\right) }{h}\]\n\n\[= \mathop{\lim }\limits_{{h \rightarrow 0}}\left( {g\left( {a + h}\right) \frac{f\left( {a + h}\right) - f\left( a\right) }{h} + f\left( a\right) \frac{g\left( {a + h}\right) - g\left( a\right) }{h}}\right)\]\n\n\[= g\left( a\right) {f}^{\prime }\left( a\right) + f\left( a\right) {g}^{\prime }\left( a\right) ,\]\n\nwhere we know \( \mathop{\lim }\limits_{{h \rightarrow 0}}g\left( {a + h}\right) = g\left( a\right) \) by the continuity of \( g \) at \( a \), which in turn follows from the assumption that \( g \) is differentiable at \( a \) .\n\nQ.E.D.
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Yes
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Proposition 6.2.5 (Quotient rule). Suppose \( D \subset \mathbb{R}, f : D \rightarrow \mathbb{R}, g : D \rightarrow \mathbb{R} \) , \( a \) is in the interior of \( D \), and \( g\left( x\right) \neq 0 \) for all \( x \in D \) . If \( f \) and \( g \) are both differentiable at \( a \), then \( \frac{f}{g} \) is differentiable at \( a \) and\n\n\[{\left( \frac{f}{g}\right) }^{\prime }\left( a\right) = \frac{g\left( a\right) {f}^{\prime }\left( a\right) - f\left( a\right) {g}^{\prime }\left( a\right) }{{\left( g\left( a\right) \right) }^{2}}.\]
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Proof. We have\n\n\[{\left( \frac{f}{g}\right) }^{\prime }\left( a\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\frac{f\left( {a + h}\right) }{g\left( {a + h}\right) } - \frac{f\left( a\right) }{g\left( a\right) }}{h}\]\n\n\[= \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {a + h}\right) g\left( a\right) - f\left( a\right) g\left( {a + h}\right) }{{hg}\left( {a + h}\right) g\left( a\right) }\]\n\n\[= \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {a + h}\right) g\left( a\right) - f\left( a\right) g\left( a\right) + f\left( a\right) g\left( a\right) - f\left( a\right) g\left( {a + h}\right) }{{hg}\left( {a + h}\right) g\left( a\right) }\]\n\n\[= \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{g\left( a\right) \frac{f\left( {a + h}\right) - f\left( a\right) }{h} - f\left( a\right) \frac{g\left( {a + h}\right) - g\left( a\right) }{h}}{g\left( {a + h}\right) g\left( a\right) }\]\n\n\[= \frac{g\left( a\right) {f}^{\prime }\left( a\right) - f\left( a\right) {g}^{\prime }\left( a\right) }{{\left( g\left( a\right) \right) }^{2}}\]\n\nwhere we know \( \mathop{\lim }\limits_{{h \rightarrow 0}}g\left( {a + h}\right) = g\left( a\right) \) by the continuity of \( g \) at \( a \), which in turn follows from the assumption that \( g \) is differentiable at \( a \) .\n\nQ.E.D.
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Yes
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Proposition 6.2.7. Suppose \( D \subset \mathbb{R}, f : D \rightarrow \mathbb{R} \) is one-to-one, \( a \) is in the interior of \( D, f\left( a\right) \) is in the interior of \( f\left( D\right) ,{f}^{-1} \) is continuous at \( f\left( a\right) \), and \( f \) is differentiable at \( a \) with \( {f}^{\prime }\left( a\right) \neq 0 \) . Then \( {f}^{-1} \) is differentiable at \( f\left( a\right) \) and\n\n\[{\left( {f}^{-1}\right) }^{\prime }\left( {f\left( a\right) }\right) = \frac{1}{{f}^{\prime }\left( a\right) }.\]
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Proof. Choose \( \delta > 0 \) so that \( \left( {f\left( a\right) - \delta, f\left( a\right) + \delta }\right) \subset f\left( D\right) \) . For \( h \in \left( {-\delta ,\delta }\right) \), let\n\n\[k = {f}^{-1}\left( {f\left( a\right) + h}\right) - a.\]\n\nThen\n\n\[{f}^{-1}\left( {f\left( a\right) + h}\right) = a + k,\]\n\nso\n\n\[f\left( a\right) + h = f\left( {a + k}\right)\]\n\nand\n\n\[h = f\left( {a + k}\right) - f\left( a\right) .\]\n\nHence\n\n\[\frac{{f}^{-1}\left( {f\left( a\right) + h}\right) - {f}^{-1}\left( {f\left( a\right) }\right) }{h} = \frac{a + k - a}{f\left( {a + k}\right) - f\left( a\right) } = \frac{1}{\frac{f\left( {a + k}\right) - f\left( a\right) }{k}}.\]\n\nNow if \( h \rightarrow 0 \), then \( k \rightarrow 0 \) (since \( {f}^{-1} \) is continuous at \( f\left( a\right) \) ), and so\n\n\[\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{{f}^{-1}\left( {f\left( a\right) + h}\right) - {f}^{-1}\left( {f\left( a\right) }\right) }{h} = \mathop{\lim }\limits_{{k \rightarrow 0}}\frac{1}{\frac{f\left( {a + k}\right) - f\left( a\right) }{k}} = \frac{1}{{f}^{\prime }\left( a\right) }.\]\n\nQ.E.D.
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Yes
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Example 6.2.3. For \( n \in {Z}^{ + } \), define \( f : \lbrack 0, + \infty ) \rightarrow \mathbb{R} \) by \( f\left( x\right) = \sqrt[n]{x} \) . Then \( f \) is the inverse of \( g : \lbrack 0, + \infty ) \rightarrow \mathbb{R} \) defined by \( g\left( x\right) = {x}^{n} \) . Thus, for any \( x \in \left( {0, + \infty }\right) \)
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\[ {f}^{\prime }\left( x\right) = \frac{1}{{g}^{\prime }\left( {f\left( x\right) }\right) } = \frac{1}{n{\left( \sqrt[n]{x}\right) }^{n - 1}} = \frac{1}{n}{x}^{\frac{1}{n} - 1}. \]
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Yes
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Proposition 6.3.1. Suppose \( D \subset \mathbb{R}, f : D \rightarrow \mathbb{R} \), and \( a \) is an interior point of \( D \) at which \( f \) has either a local maximum or a local minimum. If \( f \) is differentiable at \( a \), then \( {f}^{\prime }\left( a\right) = 0 \) .
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Proof. Suppose \( f \) has a local maximum at \( a \) (a similar argument works if \( f \) has a local minimum at \( a \) ). Choose \( \delta > 0 \) so that \( \left( {a - \delta, a + \delta }\right) \subset D \) and \( f\left( a\right) \geq f\left( x\right) \) for all \( x \in \left( {a - \delta, a + \delta }\right) \) . Then\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \geq 0 \]\n\n(6.3.1)\n\nfor all \( x \in \left( {a - \delta, a}\right) \) and\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \leq 0 \]\n\n(6.3.2)\n\nfor all \( x \in \left( {a, a + \delta }\right) \) . Hence\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {a}^{ - }}}\frac{f\left( x\right) - f\left( a\right) }{x - a} \geq 0 \]\n\n(6.3.3)\n\nand\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\frac{f\left( x\right) - f\left( a\right) }{x - a} \leq 0. \]\n\n(6.3.4)\n\nHence\n\n\[ 0 \leq \mathop{\lim }\limits_{{x \rightarrow {a}^{ - }}}\frac{f\left( x\right) - f\left( a\right) }{x - a} = {f}^{\prime }\left( a\right) = \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\frac{f\left( x\right) - f\left( a\right) }{x - a} \leq 0, \]\n\n(6.3.5)\n\nso we must have \( {f}^{\prime }\left( a\right) = 0 \) .\n\nQ.E.D.
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Yes
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Theorem 6.3.2 (Rolle’s Theorem). Let \( a, b \in \mathbb{R} \) and suppose \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . If \( f\left( a\right) = f\left( b\right) \), then there exists a point \( c \in \left( {a, b}\right) \) at which \( {f}^{\prime }\left( c\right) = 0 \) .
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Proof. By the Extreme Value Theorem, we know \( f \) attains a maximum and a minimum value on \( \left\lbrack {a, b}\right\rbrack \) . Let \( m \) be the minimum value and \( M \) the maximum value of \( f \) on \( \left\lbrack {a, b}\right\rbrack \) . If \( m = M = f\left( a\right) = f\left( b\right) \), then \( f\left( x\right) = m \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) , and so \( {f}^{\prime }\left( x\right) = 0 \) for all \( x \in \left( {a, b}\right) \) . Otherwise, one of \( m \) or \( M \) occurs at a point \( c \) in \( \left( {a, b}\right) \) . Hence \( f \) has either a local maximum or a local minimum at \( c \), and so \( {f}^{\prime }\left( c\right) = 0 \) .\n\nQ.E.D.
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Yes
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Theorem 6.3.3 (Generalized Mean Value Theorem). Let \( a, b \in \mathbb{R} \) . If \( f \) and \( g \) are continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \), then there exists a point \( c \in \left( {a, b}\right) \) at which\n\n\[ \left( {f\left( b\right) - f\left( a\right) }\right) {g}^{\prime }\left( c\right) = \left( {g\left( b\right) - g\left( a\right) }\right) {f}^{\prime }\left( c\right) . \]
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Proof. Let\n\n\[ h\left( t\right) = \left( {f\left( b\right) - f\left( a\right) }\right) g\left( t\right) - \left( {g\left( b\right) - g\left( a\right) }\right) f\left( t\right) . \]\n\nThen \( h \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . Moreover,\n\n\[ h\left( a\right) = f\left( b\right) g\left( a\right) - f\left( a\right) g\left( a\right) - f\left( a\right) g\left( b\right) + f\left( a\right) g\left( a\right) \]\n\n\[ = f\left( b\right) g\left( a\right) - f\left( a\right) g\left( b\right) \]\n\nand\n\n\[ h\left( b\right) = f\left( b\right) g\left( b\right) - f\left( a\right) g\left( b\right) - f\left( b\right) g\left( b\right) + f\left( b\right) g\left( a\right) \]\n\n\[ = f\left( b\right) g\left( a\right) - f\left( a\right) g\left( b\right) . \]\n\nHence, by Rolle’s theorem, there exists a point \( c \in \left( {a, b}\right) \) at which \( {h}^{\prime }\left( c\right) = 0 \) . But then\n\n\[ 0 = {h}^{\prime }\left( c\right) = \left( {f\left( b\right) - f\left( a\right) }\right) {g}^{\prime }\left( c\right) - \left( {g\left( b\right) - g\left( a\right) }\right) {f}^{\prime }\left( c\right) ,\]\n\nwhich implies that\n\n\[ \left( {f\left( b\right) - f\left( a\right) }\right) {g}^{\prime }\left( c\right) = \left( {g\left( b\right) - g\left( a\right) }\right) {f}^{\prime }\left( c\right) . \]
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Yes
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Theorem 6.3.4 (Mean Value Theorem). Let \( a, b \in \mathbb{R} \) . If \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \), then there exists a point \( c \in \left( {a, b}\right) \) at which\n\n\[ f\left( b\right) - f\left( a\right) = \left( {b - a}\right) {f}^{\prime }\left( c\right) . \]
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Proof. Apply the previous result with \( g\left( x\right) = x \) .\n\nQ.E.D.
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No
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Proposition 6.3.5. If \( f \) is differentiable on \( \left( {a, b}\right) \) and \( {f}^{\prime }\left( x\right) > 0 \) for all \( x \in \left( {a, b}\right) \) , then \( f \) is increasing on \( \left( {a, b}\right) \) .
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Proof. Let \( x, y \in \left( {a, b}\right) \) with \( x < y \) . By the Mean Value Theorem, there exists a point \( c \in \left( {x, y}\right) \) such that\n\n\[ f\left( y\right) - f\left( x\right) = \left( {y - x}\right) {f}^{\prime }\left( c\right) . \]\n\nSince \( y - x > 0 \) and \( {f}^{\prime }\left( c\right) > 0 \), we have \( f\left( y\right) > f\left( x\right) \), and so \( f \) is increasing on \( \left( {a, b}\right) \) . Q.E.D.
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Yes
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Theorem 6.4.1 (Intermediate Value Theorem for Derivatives). Suppose \( f \) is differentiable on an open interval \( I, a, b \in I \), and \( a < b \) . If \( \lambda \in \mathbb{R} \) and either \( {f}^{\prime }\left( a\right) < \lambda < {f}^{\prime }\left( b\right) \) or \( {f}^{\prime }\left( a\right) > \lambda > {f}^{\prime }\left( b\right) \), then there exists \( c \in \left( {a, b}\right) \) such that \( {f}^{\prime }\left( c\right) = \lambda \) .
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Proof. Suppose \( {f}^{\prime }\left( a\right) < \lambda < {f}^{\prime }\left( b\right) \) and define \( g : I \rightarrow \mathbb{R} \) by \( g\left( x\right) = f\left( x\right) - {\lambda x} \) . Then \( g \) is differentiable on \( I \), and so continuous on \( \left\lbrack {a, b}\right\rbrack \) . Let \( c \) be a point in \( \left\lbrack {a, b}\right\rbrack \) at which \( g \) attains its minimum value. Now\n\n\[ \n{g}^{\prime }\left( a\right) = {f}^{\prime }\left( a\right) - \lambda < 0, \n\] \n\n\( \left( {6.4.1}\right) \) \n\nso there exists \( a < t < b \) such that \n\n\[ \ng\left( t\right) - g\left( a\right) < 0. \n\] \n\n\( \left( {6.4.2}\right) \) \n\nThus \( c \neq a \) . Similarly, \n\n\[ \n{g}^{\prime }\left( b\right) = {f}^{\prime }\left( b\right) - \lambda > 0, \n\] \n\n\( \left( {6.4.3}\right) \) \n\nso there exists \( a < s < b \) such that \n\n\[ \ng\left( s\right) - g\left( b\right) < 0. \n\] \n\n\( \left( {6.4.4}\right) \) \n\nThus \( c \neq b \) . Hence \( c \in \left( {a, b}\right) \), and so \( {g}^{\prime }\left( c\right) = 0 \) . Thus \( 0 = {f}^{\prime }\left( c\right) - \lambda \), and so \( {f}^{\prime }\left( c\right) = \lambda \) . \n\nQ.E.D.
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Yes
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Define \( \varphi : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \) by \( \varphi \left( x\right) = x\left( {{2x} - 1}\right) \left( {x - 1}\right) \) . Define \( \rho : \mathbb{R} \rightarrow \mathbb{R} \) by \( \rho \left( x\right) = 6{x}^{2} - {6x} + 1 \) . Then
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\[ \varphi \left( x\right) = 2{x}^{3} - 3{x}^{2} + x \] so \( {\varphi }^{\prime }\left( x\right) = \rho \left( x\right) \) for all \( x \in \left( {0,1}\right) \) . Next define \( s : \mathbb{R} \rightarrow \mathbb{R} \) by \( s\left( x\right) = \varphi \left( {x-\lfloor x\rfloor }\right) \) . See Figure 6.4.1 for the graphs of \( \varphi \) and \( s \) . Then for any \( n \in \mathbb{Z} \) and \( n < x < n + 1 \) , \[ {s}^{\prime }\left( x\right) = \rho \left( {x - n}\right) = \rho \left( {x-\lfloor x\rfloor }\right) . \] Moreover, if \( x \) is an integer, \[ \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\frac{s\left( {x + h}\right) - s\left( x\right) }{h} = \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\frac{\varphi \left( h\right) }{h} \] \[ = \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\frac{h\left( {{2h} - 1}\right) \left( {h - 1}\right) }{h} \] \[ = \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\left( {{2h} - 1}\right) \left( {h - 1}\right) \] \[ = 1 \] and \[ \mathop{\lim }\limits_{{h \rightarrow {0}^{ - }}}\frac{s\left( {x + h}\right) - s\left( x\right) }{h} = \mathop{\lim }\limits_{{h \rightarrow {0}^{ - }}}\frac{\varphi \left( {h + 1}\right) }{h} \] \[ = \mathop{\lim }\limits_{{h \rightarrow {0}^{ - }}}\frac{\left( {h + 1}\right) \left( {{2h} + 1}\right) h}{h} \] \[ = \mathop{\lim }\limits_{{h \rightarrow {0}^{ - }}}\left( {h + 1}\right) \left( {{2h} + 1}\right) \] \[ = 1\text{.} \] Thus \( {s}^{\prime }\left( x\right) = 1 = \rho \left( {x-\lfloor x\rfloor }\right) \) when \( x \) is an integer, and so \( {s}^{\prime }\left( x\right) = \rho \left( {x-\lfloor x\rfloor }\right) \) for all \( x \in \mathbb{R} \) .
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Yes
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Theorem 6.5.1. Suppose \( a, b \in \mathbb{R}, f \) and \( g \) are differentiable on \( \left( {a, b}\right) ,{g}^{\prime }\left( x\right) \neq 0 \) for all \( x \in \left( {a, b}\right) \), and\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\frac{{f}^{\prime }\left( x\right) }{{g}^{\prime }\left( x\right) } = \lambda \]\n\n(6.5.1)\n\nIf \( \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}f\left( x\right) = 0 \) and \( \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}g\left( x\right) = 0 \), then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\frac{f\left( x\right) }{g\left( x\right) } = \lambda \]\n\n(6.5.2)
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Proof. Given \( \epsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \lambda - \frac{\epsilon }{2} < \frac{{f}^{\prime }\left( x\right) }{{g}^{\prime }\left( x\right) } < \lambda + \frac{\epsilon }{2} \]\n\n(6.5.3)\n\nwhenever \( x \in \left( {a, a + \delta }\right) \) . Now, by the Generalized Mean Value Theorem, for any \( x \) and \( y \) with \( a < x < y < a + \delta \), there exists a point \( c \in \left( {x, y}\right) \) such that\n\n\[ \frac{f\left( y\right) - f\left( x\right) }{g\left( y\right) - g\left( x\right) } = \frac{{f}^{\prime }\left( c\right) }{{g}^{\prime }\left( c\right) }.\]\n\n(6.5.4)\n\nHence\n\n\[ \lambda - \frac{\epsilon }{2} < \frac{f\left( y\right) - f\left( x\right) }{g\left( y\right) - g\left( x\right) } < \lambda + \frac{\epsilon }{2}. \]\n\n(6.5.5)\n\nNow\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\frac{f\left( y\right) - f\left( x\right) }{g\left( y\right) - g\left( x\right) } = \frac{f\left( y\right) }{g\left( y\right) } \]\n\n(6.5.6)\n\nand so we have\n\n\[ \lambda - \epsilon < \lambda - \frac{\epsilon }{2} \leq \frac{f\left( y\right) }{g\left( y\right) } \leq \lambda + \frac{\epsilon }{2} < \lambda + \epsilon \]\n\n\( \left( {6.5.7}\right) \)\n\nfor any \( y \in \left( {a, a + \delta }\right) \) . Hence\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\frac{f\left( x\right) }{g\left( x\right) } = \lambda \]\n\n\( \left( {6.5.8}\right) \)\n\nQ.E.D.
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Yes
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Theorem 6.6.1 (Taylor’s Theorem). Suppose \( f \in {C}^{\left( n\right) }\left( {a, b}\right) \) and \( {f}^{\left( n\right) } \) is differentiable on \( \left( {a, b}\right) \) . Let \( \alpha ,\beta \in \left( {a, b}\right) \) with \( \alpha \neq \beta \), and let\n\n\[ P\left( x\right) = f\left( \alpha \right) + {f}^{\prime }\left( \alpha \right) \left( {x - \alpha }\right) + \frac{{f}^{\prime \prime }\left( \alpha \right) }{2}{\left( x - \alpha \right) }^{2} + \cdots \]\n\n\[ + \frac{{f}^{\left( n\right) }\left( \alpha \right) }{n!}{\left( x - \alpha \right) }^{n} \]\n\n\[ = \mathop{\sum }\limits_{{k = 0}}^{n}\frac{{f}^{\left( k\right) }\left( \alpha \right) }{k!}{\left( x - \alpha \right) }^{k}. \]\n\n(6.6.1)\n\nThen there exists a point \( \gamma \) between \( \alpha \) and \( \beta \) such that\n\n\[ f\left( \beta \right) = P\left( \beta \right) + \frac{{f}^{\left( n + 1\right) }\left( \gamma \right) }{\left( {n + 1}\right) !}{\left( \beta - \alpha \right) }^{n + 1}. \]\n\n(6.6.2)
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Proof. First note that \( {P}^{\left( k\right) }\left( \alpha \right) = {f}^{\left( k\right) }\left( \alpha \right) \) for \( k = 0,1,\ldots, n \) . Let\n\n\[ M = \frac{f\left( \beta \right) - P\left( \beta \right) }{{\left( \beta - \alpha \right) }^{n + 1}}. \]\n\n(6.6.3)\n\nThen\n\n\[ f\left( \beta \right) = P\left( \beta \right) + M{\left( \beta - \alpha \right) }^{n + 1}. \]\n\n(6.6.4)\n\nWe need to show that\n\n\[ M = \frac{{f}^{\left( n + 1\right) }\left( \gamma \right) }{\left( {n + 1}\right) !} \]\n\n(6.6.5)\n\nfor some \( \gamma \) between \( \alpha \) and \( \beta \) . Let\n\n\[ g\left( x\right) = f\left( x\right) - P\left( x\right) - M{\left( x - \alpha \right) }^{n + 1}. \]\n\n(6.6.6)\n\nThen, for \( k = 0,1,\ldots, n \) ,\n\n\[ {g}^{\left( k\right) }\left( \alpha \right) = {f}^{\left( k\right) }\left( \alpha \right) - {P}^{\left( k\right) }\left( \alpha \right) = 0. \]\n\n(6.6.7)\n\nNow \( g\left( \beta \right) = 0 \), so, by Rolle’s theorem, there exists \( {\gamma }_{1} \) between \( \alpha \) and \( \beta \) such that \( {g}^{\prime }\left( {\gamma }_{1}\right) = 0 \) . Using Rolle’s theorem again, we see that there exists \( {\gamma }_{2} \) between \( \alpha \) and \( {\gamma }_{1} \) such that \( {g}^{\prime \prime }\left( {\gamma }_{2}\right) = 0 \) . Continuing for \( n + 1 \) steps, we find \( {\gamma }_{n + 1} \) between \( \alpha \) and \( {\gamma }_{n} \) (and hence between \( \alpha \) and \( \beta \) ) such that \( {g}^{\left( n + 1\right) }\left( {\gamma }_{n + 1}\right) = 0 \) . Hence\n\n\[ 0 = {g}^{\left( n + 1\right) }\left( {\gamma }_{n + 1}\right) = {f}^{\left( n + 1\right) }\left( {\gamma }_{n + 1}\right) - \left( {n + 1}\right) !M. \]\n\n(6.6.8)\n\nLetting \( \gamma = {\gamma }_{n + 1} \), we have\n\n\[ M = \frac{{f}^{\left( n + 1\right) }\left( \gamma \right) }{\left( {n + 1}\right) !} \]\n\n(6.6.9)\n\nas required.
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Yes
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Example 6.6.1. Let \( f\left( x\right) = \sqrt{x} \) . Then the 4th order Taylor polynomial for \( f \) at \( 1 \) is\n\n\[ P\left( x\right) = 1 + \frac{1}{2}\left( {x - 1}\right) - \frac{1}{8}{\left( x - 1\right) }^{2} + \frac{1}{16}{\left( x - 1\right) }^{3} - \frac{5}{128}{\left( x - 1\right) }^{4}. \]
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By Taylor’s theorem, for any \( x > 0 \) there exists \( \gamma \) between 1 and \( x \) such that\n\n\[ \sqrt{x} = P\left( x\right) + \frac{105}{\left( {32}\right) \left( {5!}\right) {\gamma }^{\frac{9}{2}}}{\left( x - 1\right) }^{5} = P\left( x\right) + \frac{7}{{256}{\gamma }^{\frac{9}{2}}}{\left( x - 1\right) }^{5}. \]
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Yes
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Lemma 7.1.1. Suppose \( {P}_{1} = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \) is a partition of \( \left\lbrack {a, b}\right\rbrack, s \in \left( {a, b}\right) \) , \( s \notin {P}_{1} \), and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is bounded. If \( {P}_{2} = {P}_{1} \cup \{ s\} \), then \( L\left( {f,{P}_{1}}\right) \leq L\left( {f,{P}_{2}}\right) \) and \( U\left( {f,{P}_{2}}\right) \leq U\left( {f,{P}_{1}}\right) \) .
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Proof. Suppose \( {x}_{i - 1} < s < {x}_{i} \) and let\n\n\[ \n{w}_{1} = \inf \left\{ {f\left( x\right) : {x}_{i - 1} \leq x \leq s}\right\} \]\n\n(7.1.6)\n\n\[ \n{W}_{1} = \sup \left\{ {f\left( x\right) : {x}_{i - 1} \leq x \leq s}\right\} \]\n\n(7.1.7)\n\n\[ \n{w}_{2} = \inf \left\{ {f\left( x\right) : s \leq x \leq {x}_{i}}\right\} \]\n\n(7.1.8)\n\n\[ \n{W}_{2} = \sup \left\{ {f\left( x\right) : s \leq x \leq {x}_{i}}\right\} \]\n\n\( \left( {7.1.9}\right) \)\n\n\[ \n{m}_{i} = \inf \left\{ {f\left( x\right) : {x}_{i - 1} \leq x \leq {x}_{i}}\right\} \]\n\n\( \left( {7.1.10}\right) \)\n\nand\n\n\[ \n{M}_{i} = \sup \left\{ {f\left( x\right) : {x}_{i - 1} \leq x \leq {x}_{i}}\right\} \]\n\n(7.1.11)\n\nThen \( {w}_{1} \geq {m}_{i},{w}_{2} \geq {m}_{i},{W}_{1} \leq {M}_{i} \), and \( {W}_{2} \leq {M}_{i} \) . Hence\n\n\[ \nL\left( {f,{P}_{2}}\right) - L\left( {f,{P}_{1}}\right) = {w}_{1}\left( {s - {x}_{i - 1}}\right) + {w}_{2}\left( {{x}_{i} - s}\right) - {m}_{i}\left( {{x}_{i} - {x}_{i - 1}}\right) \]\n\n\[ \n= {w}_{1}\left( {s - {x}_{i - 1}}\right) + {w}_{2}\left( {{x}_{i} - s}\right) - {m}_{i}\left( {s - {x}_{i - 1}}\right) \]\n\n\[ \n- {m}_{i}\left( {{x}_{i} - s}\right) \]\n\n\[ \n= \left( {{w}_{1} - {m}_{i}}\right) \left( {s - {x}_{i - 1}}\right) + \left( {{w}_{2} - {m}_{i}}\right) \left( {{x}_{i} - s}\right) \]\n\n\[ \n\geq 0 \]\n\n\( \left( {7.1.12}\right) \)\n\nand\n\n\[ \nU\left( {f,{P}_{1}}\right) - U\left( {f,{P}_{2}}\right) = {M}_{i}\left( {{x}_{i} - {x}_{i - 1}}\right) - {W}_{1}\left( {s - {x}_{i - 1}}\right) - {W}_{2}\left( {{x}_{i} - s}\right) \]\n\n\[ \n= {M}_{i}\left( {s - {x}_{i - 1}}\right) + {M}_{i}\left( {{x}_{i} - s}\right) - {W}_{1}\left( {s - {x}_{i - 1}}\right) \]\n\n\[ \n- {W}_{2}\left( {{x}_{i} - s}\right) \]\n\n\[ \n= \left( {{M}_{i} - {W}_{1}}\right) \left( {s - {x}_{i - 1}}\right) + \left( {{M}_{i} - {W}_{2}}\right) \left( {{x}_{i} - s}\right) \]\n\n\[ \n\geq 0\text{.} \]\n\n\( \left( {7.1.13}\right) \)\n\nThus \( L\left( {f,{P}_{1}}\right) \leq L\left( {f,{P}_{2}}\right) \) and \( U\left( {f,{P}_{2}}\right) \leq U\left( {f,{P}_{1}}\right) \).\n\nQ.E.D.
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Yes
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Proposition 7.1.2. Suppose \( {P}_{1} \) and \( {P}_{2} \) are partitions of \( \left\lbrack {a, b}\right\rbrack \), with \( {P}_{2} \) a refinement of \( {P}_{1} \) . If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is bounded, then \( L\left( {f,{P}_{1}}\right) \leq L\left( {f,{P}_{2}}\right) \) and \( U\left( {f,{P}_{2}}\right) \leq U\left( {f,{P}_{1}}\right) \)
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Proof. The proposition follows immediately from repeated use of the previous lemma.
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No
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Proposition 7.1.3. Suppose \( {P}_{1} \) and \( {P}_{2} \) are partitions of \( \left\lbrack {a, b}\right\rbrack \) . If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is bounded, then \( L\left( {f,{P}_{1}}\right) \leq U\left( {f,{P}_{2}}\right) \) .
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Proof. The result follows immediately from the definitions if \( {P}_{1} = {P}_{2} \) . Otherwise, let \( P \) be the common refinement of \( {P}_{1} \) and \( {P}_{2} \) . Then\n\n\[ L\left( {f,{P}_{1}}\right) \leq L\left( {f, P}\right) \leq U\left( {f, P}\right) \leq U\left( {f,{P}_{2}}\right) . \]\n\n(7.1.14)\n\nQ.E.D.
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Yes
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Proposition 7.1.4. Suppose \( a < b \) and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is bounded. Then\n\n\[{\int }_{a}^{b}f \leq {\bar{\int }}_{a}^{b}f\]
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Proof. Let \( P \) be a partition of \( \left\lbrack {a, b}\right\rbrack \) . Then for any partition \( Q \) of \( \left\lbrack {a, b}\right\rbrack \), we have \( L\left( {f, Q}\right) \leq U\left( {f, P}\right) \) . Hence \( U\left( {f, P}\right) \) is an upper bound for any lower sum, and so\n\n\[{\int }_{a}^{b}f \leq U\left( {f, P}\right)\]\n\nBut this shows that the lower integral is a lower bound for any upper sum. Hence\n\n\[{\int }_{a}^{b}f \leq {\bar{\int }}_{a}^{b}f\]
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Yes
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Define \( f : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \) by\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }x \in \mathbb{Q}, \\ 0, & \text{ if }x \notin \mathbb{Q}. \end{array}\right. \]\n\nIs \( f \) integrable on \( \left\lbrack {0,1}\right\rbrack \)?
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For any partition \( P = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \), we have\n\n\[ L\left( {f, P}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}0\left( {{x}_{i} - {x}_{i - 1}}\right) = 0 \]\n\nand\n\n\[ U\left( {f, P}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{x}_{i} - {x}_{i - 1}}\right) = {x}_{n} - {x}_{0} = 1. \]\n\nThus\n\n\[ {\int }_{0}^{1}f = 0 \]\n\nand\n\n\[ {\int }_{0}^{1}f = 1 \]\n\nHence \( f \) is not integrable on \( \left\lbrack {0,1}\right\rbrack \) .
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Yes
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Example 7.2.2. Let \( \alpha \in \mathbb{R}, a < b \), and define \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) by \( f\left( x\right) = \alpha \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . For any partition \( P = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \), we have
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\[ L\left( {f, P}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\alpha \left( {{x}_{i} - {x}_{i - 1}}\right) = \alpha \left( {{x}_{n} - {x}_{0}}\right) = \alpha \left( {b - a}\right) \] and \[ U\left( {f, P}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\alpha \left( {{x}_{i} - {x}_{i - 1}}\right) = \alpha \left( {{x}_{n} - {x}_{0}}\right) = \alpha \left( {b - a}\right) . \] Thus \[ {\int }_{a}^{b}f = \alpha \left( {b - a}\right) \] and \[ {\int }_{a}^{b}f = \alpha \left( {b - a}\right) \] Hence \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and \[ {\int }_{a}^{b}f = \alpha \left( {b - a}\right) \]
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Yes
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Theorem 7.2.1. Suppose \( a < b \) and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is bounded. Then \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) if and only if for every \( \epsilon > 0 \) there exists a partition \( P \) of \( \left\lbrack {a, b}\right\rbrack \) such that\n\n\[ U\left( {f, P}\right) - L\left( {f, P}\right) < \epsilon . \]\n\n\( \left( {7.2.4}\right) \)
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Proof. If \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and \( \epsilon > 0 \), then we may choose partitions \( {P}_{1} \) and \( {P}_{2} \) such that\n\n\[ {\int }_{a}^{b}f - L\left( {f,{P}_{1}}\right) < \frac{\epsilon }{2} \]\n\n\( \left( {7.2.5}\right) \)\n\nand\n\n\[ U\left( {f,{P}_{2}}\right) - {\int }_{a}^{b}f < \frac{\epsilon }{2} \]\n\n\( \left( {7.2.6}\right) \)\n\nLet \( P \) be the common refinement of \( {P}_{1} \) and \( {P}_{2} \) . Then\n\n\[ U\left( {f, P}\right) - L\left( {f, P}\right) \leq U\left( {f,{P}_{2}}\right) - L\left( {f,{P}_{1}}\right) \]\n\n\[ = \left( {U\left( {f,{P}_{2}}\right) - {\int }_{a}^{b}f}\right) + \left( {{\int }_{a}^{b}f - L\left( {f,{P}_{1}}\right) }\right) \]\n\n\[ < \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon \]\n\n\( \left( {7.2.7}\right) \)\n\nNow suppose for every \( \epsilon > 0 \) there exists a partition \( P \) of \( \left\lbrack {a, b}\right\rbrack \) such that\n\n\[ U\left( {f, P}\right) - L\left( {f, P}\right) < \epsilon . \]\n\n\( \left( {7.2.8}\right) \)\n\nSuppose\n\n\[ {\int }_{a}^{b}f < {\int }_{a}^{b}f \]\n\n\( \left( {7.2.9}\right) \)\n\nIf\n\n\[ \epsilon = {\int }_{a}^{b}f - {\int }_{a}^{b}f \]\n\n(7.2.10)\n\nthen for any partition \( P \) of \( \left\lbrack {a, b}\right\rbrack \) we have\n\n\[ U\left( {f, P}\right) - L\left( {f, P}\right) \geq {\int }_{a}^{b}f - {\int }_{a}^{b}f = \epsilon . \]\n\n(7.2.11)\n\n\nSince this contradicts our assumption, we must have\n\n\[ {\int }_{a}^{b}f = {\bar{\int }}_{a}^{b}f \]\n\n\( \left( {7.2.12}\right) \)\n\nThat is, \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .\n\nQ.E.D.
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Yes
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Example 7.2.3. Suppose \( f : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \) is defined by\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} 0, & \text{ if }x \neq \frac{1}{2} \\ 1, & \text{ if }x = \frac{1}{2} \end{array}\right. \]\n\nIf \( P \) is a partition of \( \left\lbrack {0,1}\right\rbrack \), then clearly \( L\left( {f, P}\right) = 0 \) .
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Given \( \epsilon > 0 \), let\n\n\[ P = \left\{ {0,\frac{1}{2} - \frac{\epsilon }{4},\frac{1}{2} + \frac{\epsilon }{4},1}\right\} \]\n\nThen\n\n\[ U\left( {f, P}\right) = \left( {\frac{1}{2} + \frac{\epsilon }{4}}\right) - \left( {\frac{1}{2} - \frac{\epsilon }{4}}\right) = \frac{\epsilon }{2} < \epsilon .\n\nHence \( U\left( {f, P}\right) - L\left( {f, P}\right) < \epsilon \), so \( f \) is integrable on \( \left\lbrack {0,1}\right\rbrack \) . Moreover,\n\n\[ {\int }_{0}^{1}f = 0 \]
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Yes
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Proposition 7.3.1. If \( a < b \) and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is monotonic, then \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .
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Proof. Suppose \( f \) is nondecreasing. Given \( \epsilon > 0 \), let \( n \in {Z}^{ + } \) be large enough\n\nthat\n\[ \frac{\left( {f\left( b\right) - f\left( a\right) }\right) \left( {b - a}\right) }{n} < \epsilon . \]\n\n(7.3.1)\n\nFor \( i = 0,1,\ldots, n \), let\n\[ {x}_{i} = a + \frac{\left( {b - a}\right) i}{n}. \]\n\n(7.3.2)\n\nLet \( P = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \) . Then\n\[ U\left( {f, P}\right) - L\left( {f, P}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}f\left( {x}_{i}\right) \left( {{x}_{i} - {x}_{i - 1}}\right) - \mathop{\sum }\limits_{{i = 1}}^{n}f\left( {x}_{i - 1}\right) \left( {{x}_{i} - {x}_{i - 1}}\right) \]\n\[ = \mathop{\sum }\limits_{{i = 1}}^{n}\left( {f\left( {x}_{i}\right) - f\left( {x}_{i - 1}\right) }\right) \frac{b - a}{n} \]\n\[ = \frac{b - a}{n}\left( {\left( {f\left( {x}_{1}\right) - f\left( {x}_{0}\right) }\right) + \left( {f\left( {x}_{2}\right) - f\left( {x}_{1}\right) }\right) + \cdots }\right. \]\n\[ \left. {+\left( {f\left( {x}_{n - 1}\right) - f\left( {x}_{n - 2}\right) }\right) + \left( {f\left( {x}_{n}\right) - f\left( {x}_{n - 1}\right) }\right) }\right) \]\n\[ = \frac{b - a}{n}\left( {f\left( b\right) - f\left( a\right) }\right) \]\n\[ < \epsilon \text{.} \]\n\n(7.3.3)\n\nHence \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .\n\nQ.E.D.
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Yes
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Proposition 7.3.2. If \( a < b \) and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is continuous, then \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .
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Proof. Given \( \epsilon > 0 \), let\n\n\[ \gamma = \frac{\epsilon }{b - a} \]\n\n\( \left( {7.3.4}\right) \)\n\nSince \( f \) is uniformly continuous on \( \left\lbrack {a, b}\right\rbrack \), we may choose \( \delta > 0 \) such that\n\n\[ \left| {f\left( x\right) - f\left( y\right) }\right| < \gamma \]\n\n\( \left( {7.3.5}\right) \)\n\nwhenever \( \left| {x - y}\right| < \delta \) . Let \( P = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \) be a partition with\n\n\[ \sup \left\{ {\left| {{x}_{i} - {x}_{i - 1}}\right| : i = 1,2,\ldots, n}\right\} < \delta . \]\n\n(7.3.6)\n\nIf, for \( i = 1,2,\ldots, n \) ,\n\n\[ {m}_{i} = \inf \left\{ {f\left( x\right) : {x}_{i - 1} \leq x \leq {x}_{i}}\right\} \]\n\n(7.3.7)\n\nand\n\n\[ {M}_{i} = \sup \left\{ {f\left( x\right) : {x}_{i - 1} \leq x \leq {x}_{i}}\right\} \]\n\n(7.3.8)\n\nthen \( {M}_{i} - {m}_{i} < \gamma \) . Hence\n\n\[ U\left( {f, P}\right) - L\left( {f, P}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{M}_{i}\left( {{x}_{i} - {x}_{i - 1}}\right) - \mathop{\sum }\limits_{{i = 1}}^{n}{m}_{i}\left( {{x}_{i} - {x}_{i - 1}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{M}_{i} - {m}_{i}}\right) \left( {{x}_{i} - {x}_{i - 1}}\right) \]\n\n\[ < \gamma \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{x}_{i} - {x}_{i - 1}}\right) \]\n\n\[ = \gamma \left( {b - a}\right) \]\n\n\[ = \epsilon \text{.} \]\n\n\( \left( {7.3.9}\right) \)\n\nThus \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .\n\nQ.E.D.
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Yes
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Proposition 7.4.2. Suppose \( f \) and \( g \) are both integrable on \( \left\lbrack {a, b}\right\rbrack \) . Then \( f + g \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[{\int }_{a}^{b}\left( {f + g}\right) = {\int }_{a}^{b}f + {\int }_{a}^{b}g.\]
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Proof. Given \( \epsilon > 0 \), let \( {P}_{1} \) and \( {P}_{2} \) be partitions of \( \left\lbrack {a, b}\right\rbrack \) with\n\n\[U\left( {f,{P}_{1}}\right) - L\left( {f,{P}_{1}}\right) < \frac{\epsilon }{2}\]\n\nand\n\n\[U\left( {g,{P}_{2}}\right) - L\left( {g,{P}_{2}}\right) < \frac{\epsilon }{2}.\]\n\nLet \( P = {P}_{1} \cup {P}_{2} \) . By the previous proposition,\n\n\[U\left( {f + g, P}\right) \leq U\left( {f, P}\right) + U\left( {g, P}\right)\]\n\nand\n\n\[L\left( {f + g, P}\right) \geq L\left( {f, P}\right) + L\left( {g, P}\right) .\]\n\nHence\n\n\[U\left( {f + g, P}\right) - L\left( {f + g, P}\right) \leq \left( {U\left( {f, P}\right) + U\left( {g, P}\right) }\right) - \left( {L\left( {f, P}\right) + L\left( {g, P}\right) }\right)\]\n\n\[= \left( {U\left( {f, P}\right) - L\left( {f, P}\right) }\right) + \left( {U\left( {g, P}\right) - L\left( {g, P}\right) }\right)\]\n\n\[\leq \left( {U\left( {f,{P}_{1}}\right) - L\left( {f,{P}_{1}}\right) }\right) + \left( {U\left( {g,{P}_{2}}\right) - L\left( {g,{}_{2}P}\right) }\right)\]\n\n\[< \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon\]\n\nHence \( f + g \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .\n\nMoreover,\n\n\[{\int }_{a}^{b}\left( {f + g}\right) \leq U\left( {f + g, P}\right)\]\n\n\[\leq U\left( {f, P}\right) + U\left( {g, P}\right)\]\n\n\[\leq \left( {{\int }_{a}^{b}f + \frac{\epsilon }{2}}\right) + \left( {{\int }_{a}^{b}g + \frac{\epsilon }{2}}\right)\]\n\n\[= {\int }_{a}^{b}f + {\int }_{a}^{b}g + \epsilon\]\n\nand\n\n\[{\int }_{a}^{b}\left( {f + g}\right) \geq L\left( {f + g, P}\right)\]\n\n\[\geq L\left( {f, P}\right) + L\left( {g, P}\right)\]\n\n\[\geq \left( {{\int }_{a}^{b}f - \frac{\epsilon }{2}}\right) + \left( {{\int }_{a}^{b}g - \frac{\epsilon }{2}}\right)\]\n\n\[= {\int }_{a}^{b}f + {\int }_{a}^{b}g - \epsilon .\]\n\nSince \( \epsilon > 0 \) was arbitrary, it follows that\n\n\[{\int }_{a}^{b}\left( {f + g}\right) = {\int }_{a}^{b}f + {\int }_{a}^{b}g.\]
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Yes
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Proposition 7.4.3. If \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and \( \alpha \in \mathbb{R} \), then \( {\alpha f} \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[{\int }_{a}^{b}{\alpha f} = \alpha {\int }_{a}^{b}f\]
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Exercise 7.4.5. Prove the previous proposition.
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No
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Proposition 7.4.4. Suppose \( a < b, f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is bounded, and \( c \in \left( {a, b}\right) \) . Then \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) if and only if \( f \) is integrable on both \( \left\lbrack {a, c}\right\rbrack \) and \( \left\lbrack {c, b}\right\rbrack \) .
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Proof. Suppose \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) . Given \( \epsilon > 0 \), let \( Q \) be a partition of \( \left\lbrack {a, b}\right\rbrack \) such that\n\n\[ U\left( {f, Q}\right) - L\left( {f, Q}\right) < \epsilon . \]\n\n\( \left( {7.4.13}\right) \)\n\nLet \( P = Q \cup \{ c\} ,{P}_{1} = P \cap \left\lbrack {a, c}\right\rbrack \), and \( {P}_{2} = P \cap \left\lbrack {c, b}\right\rbrack \) . Then\n\n\[ \left( {U\left( {f,{P}_{1}}\right) - L\left( {f,{P}_{1}}\right) }\right) + \left( {U\left( {f,{P}_{2}}\right) - L\left( {f,{P}_{2}}\right) }\right) = \left( {U\left( {f,{P}_{1}}\right) + U\left( {f,{P}_{2}}\right) }\right) - \left( {L\left( {f,{P}_{1}}\right) + L\left( {f,{P}_{2}}\right) }\right) = U\left( {f, P}\right) - L\left( {f, P}\right) \leq U\left( {f, Q}\right) - L\left( {f, Q}\right) < \epsilon . \]\n\n(7.4.14)\n\nThus we must have both\n\n\[ U\left( {f,{P}_{1}}\right) - L\left( {f,{P}_{1}}\right) < \epsilon \]\n\n\( \left( {7.4.15}\right) \)\n\nand\n\n\[ U\left( {f,{P}_{2}}\right) - L\left( {f,{P}_{2}}\right) < \epsilon . \]\n\n(7.4.16)\n\nHence \( f \) is integrable on both \( \left\lbrack {a, c}\right\rbrack \) and \( \left\lbrack {c, b}\right\rbrack \) .\n\nNow suppose \( f \) is integrable on both \( \left\lbrack {a, c}\right\rbrack \) and \( \left\lbrack {c, b}\right\rbrack \) . Given \( \epsilon > 0 \), let \( {P}_{1} \) and \( {P}_{2} \) be partitions of \( \left\lbrack {a, c}\right\rbrack \) and \( \left\lbrack {c, b}\right\rbrack \), respectively, such that\n\n\[ U\left( {f,{P}_{1}}\right) - L\left( {f,{P}_{1}}\right) < \frac{\epsilon }{2} \]\n\n\( \left( {7.4.17}\right) \)\n\nand\n\n\[ U\left( {f,{P}_{2}}\right) - L\left( {f,{P}_{2}}\right) < \frac{\epsilon }{2}. \]\n\n(7.4.18)\n\nLet \( P = {P}_{1} \cup {P}_{2} \) . Then \( P \) is a partition of \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[ U\left( {f, P}\right) - L\left( {f, P}\right) = \left( {U\left( {f,{P}_{1}}\right) + U\left( {f,{P}_{2}}\right) }\right) - \left( {L\left( {f,{P}_{1}}\right) + L\left( {f,{P}_{2}}\right) }\right) = \left( {U\left( {f,{P}_{1}}\right) - L\left( {f,{P}_{1}}\right) }\right) + \left( {U\left( {f,{P}_{2}}\right) - L\left( {f,{P}_{2}}\right) }\right) < \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon . \]\n\n(7.4.19)\n\nThus \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .\n\nQ.E.D.
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Yes
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Proposition 7.4.5. Suppose \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and \( c \in \left( {a, b}\right) \) . Then\n\n\[{\int }_{a}^{b}f = {\int }_{a}^{c}f + {\int }_{c}^{b}f\]
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Proof. If \( P \) and \( Q \) are partitions of \( \left\lbrack {a, c}\right\rbrack \) and \( \left\lbrack {c, b}\right\rbrack \), respectively, then\n\n\[U\left( {f, P}\right) + U\left( {f, Q}\right) = U\left( {f, P \cup Q}\right) \geq {\int }_{a}^{b}f.\]\n\n(7.4.21)\n\nThus\n\n\[U\left( {f, P}\right) \geq {\int }_{a}^{b}f - U\left( {f, Q}\right)\]\n\n(7.4.22)\n\nso\n\n\[{\int }_{a}^{c}f = {\bar{\int }}_{a}^{c}f \geq {\int }_{a}^{b}f - U\left( {f, Q}\right)\]\n\n(7.4.23)\n\nHence\n\n\[U\left( {f, Q}\right) \geq {\int }_{a}^{b}f - {\int }_{a}^{c}f\]\n\n(7.4.24)\n\nso\n\n\[{\int }_{c}^{b}f = {\bar{\int }}_{c}^{b}f \geq {\int }_{a}^{b}f - {\int }_{a}^{c}f.\]\n\n\( \left( {7.4.25}\right) \)\n\nThus\n\n\[{\int }_{a}^{c}f + {\int }_{c}^{b}f \geq {\int }_{a}^{b}f\]\n\n(7.4.26)\n\nSimilarly, if \( P \) and \( Q \) are partitions of \( \left\lbrack {a, c}\right\rbrack \) and \( \left\lbrack {c, b}\right\rbrack \), respectively, then\n\n\[L\left( {f, P}\right) + L\left( {f, Q}\right) = L\left( {f, P \cup Q}\right) \leq {\int }_{a}^{b}f.\]\n\n(7.4.27)\n\nThus\n\n\[L\left( {f, P}\right) \leq {\int }_{a}^{b}f - L\left( {f, Q}\right)\]\n\n(7.4.28)\n\nso\n\n\[{\int }_{a}^{c}f = {\int }_{a}^{c}f \leq {\int }_{a}^{b}f - L\left( {f, Q}\right)\]\n\n(7.4.29)\n\nHence\n\n\[L\left( {f, Q}\right) \leq {\int }_{a}^{b}f - {\int }_{a}^{c}f\]\n\n\( \left( {7.4.30}\right) \)\n\nso\n\n\[{\int }_{c}^{b}f = {\int }_{c}^{b}f \leq {\int }_{a}^{b}f - {\int }_{a}^{c}f.\]\n\n\( \left( {7.4.31}\right) \)\n\nThus\n\n\[{\int }_{a}^{c}f + {\int }_{c}^{b}f \leq {\int }_{a}^{b}f\]\n\n\( \left( {7.4.32}\right) \)\n\nHence\n\n\[{\int }_{a}^{c}f + {\int }_{c}^{b}f = {\int }_{a}^{b}f\]\n\n\( \left( {7.4.33}\right) \)\n\nQ.E.D.
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Yes
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Proposition 7.4.6. If \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) with \( f\left( x\right) \geq 0 \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) , then \[ {\int }_{a}^{b}f \geq 0 \]
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Proof. The result follows from the fact that \( L\left( {f, P}\right) \geq 0 \) for any partition \( P \) of \( \left\lbrack {a, b}\right\rbrack \) . Q.E.D.
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Yes
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Proposition 7.4.7. Suppose \( f \) and \( g \) are both integrable on \( \left\lbrack {a, b}\right\rbrack \) . If, for all \( x \in \left\lbrack {a, b}\right\rbrack, f\left( x\right) \leq g\left( x\right) \), then\n\n\[{\int }_{a}^{b}f \leq {\int }_{a}^{b}g.\]
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Proof. Since \( g\left( x\right) - f\left( x\right) \geq 0 \) for all \( x \in \left\lbrack {a, b}\right\rbrack \), we have, using Propositions 7.4.2, 7.4.3, and 7.4.6,\n\n\[{\int }_{a}^{b}g - {\int }_{a}^{b}f = {\int }_{a}^{b}\left( {g - f}\right) \geq 0.\]\n\n(7.4.36)
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Yes
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Proposition 7.4.8. Suppose \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack, m \in \mathbb{R}, M \in \mathbb{R} \), and \( m \leq f\left( x\right) \leq M \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ m\left( {b - a}\right) \leq {\int }_{a}^{b}f \leq M\left( {b - a}\right) . \]
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Proof. It follows from the previous proposition that\n\n\[ m\left( {b - a}\right) = {\int }_{a}^{b}{mdx} \leq {\int }_{a}^{b}f\left( x\right) {dx} \leq {\int }_{a}^{b}{Mdx} = M\left( {b - a}\right) . \]\n\n(7.4.38)\n\nQ.E.D.
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Yes
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Proposition 7.4.9. Suppose \( g \) is integrable on \( \left\lbrack {a, b}\right\rbrack, g\left( \left\lbrack {a, b}\right\rbrack \right) \subset \left\lbrack {c, d}\right\rbrack \), and \( f : \left\lbrack {c, d}\right\rbrack \rightarrow \mathbb{R} \) is continuous. If \( h = f \circ g \), then \( h \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .
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Proof. Let \( \epsilon > 0 \) be given. Let\n\n\[ K > \sup \{ f\left( x\right) : x \in \left\lbrack {c, d}\right\rbrack \} - \inf \{ f\left( x\right) : x \in \left\lbrack {c, d}\right\rbrack \}\]\n\n(7.4.39)\n\nand choose \( \delta > 0 \) so that \( \delta < \epsilon \) and\n\n\[ \left| {f\left( x\right) - f\left( y\right) }\right| < \frac{\epsilon }{2\left( {b - a}\right) }\n\n(7.4.40)\n\nwhenever \( \left| {x - y}\right| < \delta \) . Let \( P = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \) be a partition of \( \left\lbrack {a, b}\right\rbrack \) such that\n\n\[ U\left( {g, P}\right) - L\left( {g, P}\right) < \frac{{\delta }^{2}}{2K}\n\n(7.4.41)\n\nFor \( i = 1,2,\ldots, n \), let\n\n\[ {m}_{i} = \inf \left\{ {g\left( x\right) : {x}_{i - 1} \leq x \leq {x}_{i}}\right\}\]\n\n\( \left( {7.4.42}\right) \)\n\n\[ {M}_{i} = \sup \left\{ {g\left( x\right) : {x}_{i - 1} \leq x \leq {x}_{i}}\right\}\]\n\n(7.4.43)\n\n\[ {w}_{i} = \inf \left\{ {h\left( x\right) : {x}_{i - 1} \leq x \leq {x}_{i}}\right\}\]\n\n(7.4.44)\n\nand\n\n\[ {W}_{i} = \sup \left\{ {h\left( x\right) : {x}_{i - 1} \leq x \leq {x}_{i}}\right\}\]\n\n(7.4.45)\n\nFinally, let\n\n\[ I = \left\{ {i : i \in \mathbb{Z},1 \leq i \leq n,{M}_{i} - {m}_{i} < \delta }\right\}\]\n\n\( \left( {7.4.46}\right) \)\n\nand\n\n\[ J = \left\{ {i : i \in \mathbb{Z},1 \leq i \leq n,{M}_{i} - {m}_{i} \geq \delta }\right\} .\n\n\( \left( {7.4.47}\right) \)\n\nNote that\n\n\[ \delta \mathop{\sum }\limits_{{i \in J}}\left( {{x}_{i} - {x}_{i - 1}}\right) \leq \mathop{\sum }\limits_{{i \in J}}\left( {{M}_{i} - {m}_{i}}\right) \left( {{x}_{i} - {x}_{i - 1}}\right)\]\n\n\[ \leq \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{M}_{i} - {m}_{i}}\right) \left( {{x}_{i} - {x}_{i - 1}}\right)\]\n\n\[ < \frac{{\delta }^{2}}{2K}\n\n(7.4.48)\n\nfrom which it follows that\n\n\[ \mathop{\sum }\limits_{{i \in J}}\left( {{x}_{i} - {x}_{i - 1}}\right) < \frac{\delta }{2K}\]\n\n\( \left( {7.4.49}\right) \)\n\nThen\n\n\[ U\left( {h, P}\right) - L\left( {h, P}\right) = \mathop{\sum }\limits_{{i \in I}}\left( {{W}_{i} - {w}_{i}}\right) \left( {{x}_{i} - {x}_{i - 1}}\right) + \mathop{\sum }\limits_{{i \in J}}\left( {{W}_{i} - {w}_{i}}\right) \left( {{x}_{i} - {x}_{i - 1}}\right)\]\n\n\[ < \frac{\epsilon }{2\left( {b - a}\right) }\mathop{\sum }\limits_{{i \in I}}\left( {{x}_{i} - {x}_{i - 1}}\right) + K\mathop{\sum }\limits_{{i \in J}}\left( {{x}_{i} - {x}_{i - 1}}\right)\]\n\n\[ < \frac{\epsilon }{2} + \frac{\delta }{2}\]\n\n\[ < \frac{\epsilon }{2} + \frac{\epsilon }{2}\]\n\n\[ = \epsilon \text{.}\]\n\n\( \left( {7.4.50}\right) \)\n\nThus \( h \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .\n\nQ.E.D.
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Yes
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Proposition 7.4.10. Suppose \( f \) and \( g \) are both integrable on \( \left\lbrack {a, b}\right\rbrack \) . Then \( {fg} \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) .
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Proof. Since \( f \) and \( g \) are both integrable, both \( f + g \) and \( f - g \) are integrable. Hence, by the previous proposition, both \( {\left( f + g\right) }^{2} \) and \( {\left( f - g\right) }^{2} \) are integrable.\n\nThus\n\[ \frac{1}{4}\left( {{\left( f + g\right) }^{2} - {\left( f - g\right) }^{2})}\right) = {fg} \]\n\n\( \left( {7.4.51}\right) \)\n\nis integrable on \( \left\lbrack {a, b}\right\rbrack \) .\n\nQ.E.D.
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Yes
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Proposition 7.4.11. Suppose \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) . Then \( \left| f\right| \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[ \left| {{\int }_{a}^{b}f}\right| \leq {\int }_{a}^{b}\left| f\right| \]
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Proof. The integrability of \( \left| f\right| \) follows from the integrability of \( f \), the continuity of \( g\left( x\right) = \left| x\right| \), and Proposition 7.4.9. For the inequality, note that\n\n\[ - \left| {f\left( x\right) }\right| \leq f\left( x\right) \leq \left| {f\left( x\right) }\right| \]\n\nfor all \( x \in \left\lbrack {a, b}\right\rbrack \) . Hence\n\n\[ - {\int }_{a}^{b}\left| f\right| \leq {\int }_{a}^{b}f \leq {\int }_{a}^{b}\left| f\right| \]\n\nfrom which the result follows.\n\nQ.E.D.
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Yes
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Theorem 7.5.1 (Fundamental Theorem of Calculus). Suppose \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) . If \( F \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) with \( {F}^{\prime }\left( x\right) = f\left( x\right) \) for all \( x \in \left( {a, b}\right) \), then\n\n\[{\int }_{a}^{b}f = F\left( b\right) - F\left( a\right)\]
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Proof. Given \( \epsilon > 0 \), let \( P = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \) be a partition of \( \left\lbrack {a, b}\right\rbrack \) for which\n\n\[U\left( {f, P}\right) - L\left( {f, P}\right) < \epsilon .\](7.5.2)\n\nFor \( i = 1,2,\ldots, n \), let \( {t}_{i} \in \left( {{x}_{i - 1},{x}_{i}}\right) \) be points for which\n\n\[F\left( {x}_{i}\right) - F\left( {x}_{i - 1}\right) = f\left( {t}_{i}\right) \left( {{x}_{i} - {x}_{i - 1}}\right) .\](7.5.3)\n\nThen\n\n\[\mathop{\sum }\limits_{{i = 1}}^{n}f\left( {t}_{i}\right) \left( {{x}_{i} - {x}_{i - 1}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}\left( {F\left( {x}_{i}\right) - F\left( {x}_{i - 1}\right) }\right) = F\left( b\right) - F\left( a\right) .\](7.5.4)\n\nBut\n\n\[L\left( {f, P}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{n}f\left( {t}_{i}\right) \left( {{x}_{i} - {x}_{i - 1}}\right) \leq U\left( {f, P}\right)\](7.5.5)\n\nso\n\n\[\left| {F\left( b\right) - F\left( a\right) - {\int }_{a}^{b}f}\right| < \epsilon .\](7.5.6)\n\nSince \( \epsilon \) was arbitrary, we conclude that\n\n\[{\int }_{a}^{b}f = F\left( b\right) - F\left( a\right)\](7.5.7)\n\nQ.E.D.
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Yes
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Proposition 7.5.2 (Integration by parts). Suppose \( f \) and \( g \) are integrable on \( \left\lbrack {a, b}\right\rbrack \) . If \( F \) and \( G \) are continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) with \( {F}^{\prime }\left( x\right) = f\left( x\right) \) and \( {G}^{\prime }\left( x\right) = g\left( x\right) \) for all \( x \in \left( {a, b}\right) \), then\n\n\[ \n{\int }_{a}^{b}F\left( x\right) g\left( x\right) {dx} = F\left( b\right) G\left( b\right) - F\left( a\right) G\left( a\right) - {\int }_{a}^{b}f\left( x\right) G\left( x\right) {dx}.\n\]
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Proof. By the Fundamental Theorem of Calculus,\n\n\[ \n{\int }_{a}^{b}\left( {F\left( x\right) g\left( x\right) + f\left( x\right) G\left( x\right) }\right) {dx} = F\left( b\right) G\left( b\right) - F\left( a\right) G\left( a\right) .\n\]
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Yes
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Proposition 7.5.3. Suppose \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and \( F : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is defined by\n\n\[ F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \]\n\nThen \( F \) is uniformly continuous on \( \left\lbrack {a, b}\right\rbrack \) .
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Proof. Let \( \epsilon > 0 \) be given and let \( M > 0 \) be such that \( \left| {f\left( x\right) }\right| \leq M \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . Then for any \( x, y \in \left\lbrack {a, b}\right\rbrack \) with \( x < y \) and \( y - x < \frac{\epsilon }{M} \) ,\n\n\[ \left| {F\left( y\right) - F\left( x\right) }\right| = \left| {{\int }_{x}^{y}f\left( t\right) {dt}}\right| \leq M\left( {y - x}\right) < \epsilon . \]\n\nHence \( F \) is uniformly continuous on \( \left\lbrack {a, b}\right\rbrack \) .\n\nQ.E.D.
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Yes
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Theorem 7.5.4. Suppose \( f \) is integrable on \( \left\lbrack {a, b}\right\rbrack \) and continuous at \( u \in \left( {a, b}\right) \) . If \( F : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is defined by\n\n\[ F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \]\n\nthen \( F \) is differentiable at \( u \) and \( {F}^{\prime }\left( u\right) = f\left( u\right) \).
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Proof. Let \( \epsilon > 0 \) be given and choose \( \delta > 0 \) such that \( \left| {f\left( x\right) - f\left( u\right) }\right| < \epsilon \) whenever \( \left| {x - u}\right| < \delta \) . Then if \( 0 < h < \delta \), we have\n\n\[ \left| {\frac{F\left( {u + h}\right) - F\left( u\right) }{h} - f\left( u\right) }\right| = \left| {\frac{1}{h}{\int }_{u}^{u + h}f\left( t\right) {dt} - f\left( u\right) }\right| \]\n\n\[ = \left| {\frac{1}{h}{\int }_{u}^{u + h}\left( {f\left( t\right) - f\left( u\right) }\right) {dt}}\right| \]\n\n\[ < \epsilon \text{.} \]\n\nIf \( - \delta < h < 0 \), then\n\n\[ \left| {\frac{F\left( {u + h}\right) - F\left( u\right) }{h} - f\left( u\right) }\right| = \left| {-\frac{1}{h}{\int }_{u + h}^{u}f\left( t\right) {dt} - f\left( u\right) }\right| \]\n\n\[ = \left| {\frac{1}{h}{\int }_{u + h}^{u}f\left( t\right) {dt} + f\left( u\right) }\right| \]\n\n\[ = \left| {\frac{1}{h}{\int }_{u + h}^{u}f\left( t\right) {dt} - \frac{1}{h}{\int }_{u + h}^{u}f\left( u\right) {dt}}\right| \]\n\n\[ = \left| {\frac{1}{h}{\int }_{u + h}^{u}\left( {f\left( t\right) - f\left( u\right) }\right) {dt}}\right| \]\n\n\[ < \epsilon \text{.} \]\n\nHence\n\n\[ {F}^{\prime }\left( u\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{F\left( {u + h}\right) - F\left( u\right) }{h} = f\left( u\right) . \]
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Yes
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Proposition 7.5.5. If \( a < b \) and \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), then there exists a function \( F : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) which is continuous on \( \left\lbrack {a, b}\right\rbrack \) with \( {F}^{\prime }\left( x\right) = f\left( x\right) \) for all \( x \in \left( {a, b}\right) \) .
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Proof. Let\n\n\[ F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt}. \]\n\n(7.5.16)\n\nQ.E.D.
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Yes
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Proposition 7.5.6 (Integration by substitution). Suppose \( I \) is an open interval, \( \varphi : I \rightarrow \mathbb{R}, a < b,\left\lbrack {a, b}\right\rbrack \subset I \), and \( {\varphi }^{\prime } \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) . If \( f : \varphi \left( \left\lbrack {a, b}\right\rbrack \right) \rightarrow \mathbb{R} \) is continuous, then\n\n\[ \n{\int }_{\varphi \left( a\right) }^{\varphi \left( b\right) }f\left( u\right) {du} = {\int }_{a}^{b}f\left( {\varphi \left( x\right) }\right) {\varphi }^{\prime }\left( x\right) {dx}.\n\]
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Proof. If \( m \) and \( M \) are the minimum and maximum values, respectively, of \( \varphi \) on \( \left\lbrack {a, b}\right\rbrack \), then \( \varphi \left( \left\lbrack {a, b}\right\rbrack \right) = \left\lbrack {m, M}\right\rbrack \) . If \( m = M \), then \( \varphi \left( x\right) = m \) for all \( x \in \left\lbrack {a, b}\right\rbrack \), and both sides of (7.5.17) are 0 . So we may assume \( m < M \) . Let \( F \) be a function which is continuous on \( \left\lbrack {m, M}\right\rbrack \) with \( {F}^{\prime }\left( u\right) = f\left( u\right) \) for every \( u \in \left( {m, M}\right) \) . Let \( g = F \circ \varphi \) . Then\n\n\[ \n{g}^{\prime }\left( x\right) = {F}^{\prime }\left( {\varphi \left( x\right) }\right) {\varphi }^{\prime }\left( x\right) = f\left( {\varphi \left( x\right) }\right) {\varphi }^{\prime }\left( x\right) .\n\]\n\nSo if \( \varphi \left( a\right) \leq \varphi \left( b\right) \),\n\n\[ \n{\int }_{a}^{b}f\left( {\varphi \left( x\right) }\right) {\varphi }^{\prime }\left( x\right) {dx} = g\left( b\right) - g\left( a\right)\n\]\n\n\[ \n= F\left( {\varphi \left( b\right) }\right) - F\left( {\varphi \left( a\right) }\right)\n\]\n\n\[ \n= {\int }_{\varphi \left( a\right) }^{\varphi \left( b\right) }f\left( u\right) {du}.\n\]\n\nIf \( \varphi \left( a\right) > \varphi \left( b\right) \), then\n\n\[ \n{\int }_{a}^{b}f\left( {\varphi \left( x\right) }\right) {\varphi }^{\prime }\left( x\right) {dx} = g\left( b\right) - g\left( a\right)\n\]\n\n\[ \n= F\left( {\varphi \left( b\right) }\right) - F\left( {\varphi \left( a\right) }\right)\n\]\n\n\[ \n= - \left( {F\left( {\varphi \left( a\right) }\right) - F\left( {\varphi \left( b\right) }\right) }\right)\n\]\n\n\[ \n= - {\int }_{\varphi \left( b\right) }^{\varphi \left( a\right) }f\left( u\right) {du}\n\]\n\n\[ \n= {\int }_{\varphi \left( a\right) }^{\varphi \left( b\right) }f\left( u\right) {du}\n\]
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Yes
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Theorem 7.6.1 (Taylor’s Theorem). Suppose \( f \in {C}^{\left( n + 1\right) }\left( {a, b}\right) ,\alpha \in \left( {a, b}\right) \), and\n\n\[ \n{P}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{n}\frac{{f}^{\left( k\right) }\left( \alpha \right) }{k!}{\left( x - \alpha \right) }^{k}.\n\]\n\n(7.6.1)\n\nThen, for any \( x \in \left( {a, b}\right) \) ,\n\n\[ \nf\left( x\right) = {P}_{n}\left( x\right) + {\int }_{\alpha }^{x}\frac{{f}^{\left( n + 1\right) }\left( t\right) }{n!}{\left( x - t\right) }^{n}{dt}.\n\]\n\n(7.6.2)
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Proof. By the Fundamental Theorem of Calculus, we have\n\n\[ \n{\int }_{\alpha }^{x}{f}^{\prime }\left( t\right) {dt} = f\left( x\right) - f\left( \alpha \right)\n\]\n\n(7.6.3)\n\nwhich implies that\n\n\[ \nf\left( x\right) = f\left( \alpha \right) + {\int }_{\alpha }^{x}{f}^{\prime }\left( t\right) {dt}.\n\]\n\n(7.6.4)\n\nHence the theorem holds for \( n = 0 \) . Suppose the result holds for \( n = k - 1 \), that is,\n\n\[ \nf\left( x\right) = {P}_{k - 1}\left( x\right) + {\int }_{\alpha }^{x}\frac{{f}^{\left( k\right) }\left( t\right) }{\left( {k - 1}\right) !}{\left( x - t\right) }^{k - 1}{dt}.\n\]\n\n\( \left( {7.6.5}\right) \)\n\nLet\n\n\[ \nF\left( t\right) = {f}^{\left( k\right) }\left( t\right)\n\]\n\n(7.6.6)\n\n\[ \ng\left( t\right) = \frac{{\left( x - t\right) }^{k - 1}}{\left( {k - 1}\right) !}\n\]\n\n(7.6.7)\n\nand\n\n\[ \nG\left( t\right) = - \frac{{\left( x - t\right) }^{k}}{k!}.\n\]\n\n(7.6.8)\n\nThen\n\n\[ \n{\int }_{\alpha }^{x}\frac{{f}^{\left( k\right) }\left( t\right) }{\left( {k - 1}\right) !}{\left( x - t\right) }^{k - 1}{dt} = {\int }_{\alpha }^{x}F\left( t\right) g\left( t\right) {dt}\n\]\n\n\[ \n= F\left( x\right) G\left( x\right) - F\left( \alpha \right) G\left( \alpha \right) - {\int }_{\alpha }^{x}{F}^{\prime }\left( t\right) G\left( t\right) {dt}\n\]\n\n\[ \n= \frac{{f}^{\left( k\right) }\left( \alpha \right) {\left( x - \alpha \right) }^{k}}{k!} + {\int }_{\alpha }^{x}\frac{{f}^{\left( k + 1\right) }\left( t\right) }{k!}{\left( x - t\right) }^{k}{dt}.\n\]\n\n(7.6.9)\n\nHence\n\n\[ \nf\left( x\right) = {P}_{k}\left( x\right) + {\int }_{\alpha }^{x}\frac{{f}^{\left( k + 1\right) }\left( t\right) }{k!}{\left( x - t\right) }^{k}{dt}\n\]\n\n(7.6.10)\n\nand so the theorem holds for \( n = k \) .\n\nQ.E.D.
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Yes
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Proposition 7.7.1. Suppose \( f \) is continuous on \( \lbrack a,\infty ) \) and \( f\left( x\right) \geq 0 \) for all \( x \geq a \) . If there exists \( g : \lbrack a, + \infty ) \rightarrow \mathbb{R} \) for which\n\n\[ \n{\int }_{a}^{+\infty }g\left( x\right) {dx} \n\]\n\n\( \left( {7.7.5}\right) \)\n\nexists and \( g\left( x\right) \geq f\left( x\right) \) for all \( x \geq a \), then\n\n\[ \n{\int }_{a}^{+\infty }f\left( x\right) {dx} \n\]\n\n\( \left( {7.7.6}\right) \)\n\nexists.
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Exercise 7.7.1. Prove the preceding proposition.
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No
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Suppose\n\n\[ f\left( x\right) = \frac{1}{1 + {x}^{2}} \]\n\nand\n\n\[ g\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }0 \leq x < 1 \\ \frac{1}{{x}^{2}}, & \text{ if }x \geq 1 \end{array}\right. \]\n\nThen, for \( b > 1 \) ,\n\n\[ {\int }_{0}^{b}g\left( x\right) {dx} = {\int }_{0}^{1}{dx} + {\int }_{1}^{b}\frac{1}{{x}^{2}}{dx} = 1 + 1 - \frac{1}{b} = 2 - \frac{1}{b}, \]\n\nso\n\n\[ {\int }_{0}^{+\infty }g\left( x\right) {dx} = \mathop{\lim }\limits_{{b \rightarrow \infty }}\left( {2 - \frac{1}{b}}\right) = 2. \]\n\nSince \( 0 < f\left( x\right) \leq g\left( x\right) \) for all \( x \geq 0 \), it follows that\n\n\[ {\int }_{0}^{+\infty }\frac{1}{1 + {x}^{2}}{dx} \]\n\nexists, and, moreover,\n\n\[ {\int }_{0}^{+\infty }\frac{1}{1 + {x}^{2}}{dx} < 2 \]\n
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Also, the substitution \( u = - x \) shows that\n\n\[ {\int }_{-\infty }^{0}\frac{1}{1 + {x}^{2}}{dx} = - {\int }_{+\infty }^{0}\frac{1}{1 + {u}^{2}}{du} = {\int }_{0}^{+\infty }\frac{1}{1 + {u}^{2}}{du}. \]
|
Yes
|
Proposition 8.1.1. The arctangent function is differentiable at every \( x \in \mathbb{R} \) . Moreover, if \( f\left( x\right) = \arctan \left( x\right) \), then\n\n\[ \n{f}^{\prime }\left( x\right) = \frac{1}{1 + {x}^{2}}. \n\]
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Proof. The result follows immediately from Theorem 7.5.4.\n\nQ.E.D.
|
No
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Proposition 8.1.2. The arctangent is increasing on \( \mathbb{R} \) .
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Proof. The result follows immediately from the previous proposition and the fact that\n\n\[ \frac{1}{1 + {x}^{2}} > 0 \]\n\nfor every \( x \in \mathbb{R} \) .\n\nQ.E.D.
|
Yes
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Proposition 8.1.3. For any \( x \in \mathbb{R} \) , \( \arctan \left( x\right) = - \arctan \left( {-x}\right) \) .
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Proof. Using the substitution \( t = - u \), we have\n\n\[\n\arctan \left( x\right) = {\int }_{0}^{x}\frac{1}{1 + {t}^{2}}{dt} = - {\int }_{0}^{-x}\frac{1}{1 + {u}^{2}}{du} = - \arctan \left( {-x}\right) .\n\]\n\n(8.1.4)\n\nQ.E.D.
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Yes
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Proposition 8.1.4. If \( x > 0 \), then\n\n\[ \arctan \left( x\right) + \arctan \left( \frac{1}{x}\right) = \frac{\pi }{2}. \]\n\n(8.1.6)
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Proof. Using the substitution \( t = \frac{1}{u} \), we have\n\n\[ \arctan \left( \frac{1}{x}\right) = {\int }_{0}^{\frac{1}{x}}\frac{1}{1 + {t}^{2}}{dt} \]\n\n\[ = {\int }_{+\infty }^{x}\frac{1}{1 + \frac{1}{{u}^{2}}}\left( {-\frac{1}{{u}^{2}}}\right) {du} \]\n\n\[ = - {\int }_{+\infty }^{x}\frac{1}{1 + {u}^{2}}{du} \]\n\n\[ = {\int }_{x}^{+\infty }\frac{1}{1 + {u}^{2}}{du} \]\n\n\[ = \frac{\pi }{2} - {\int }_{0}^{x}\frac{1}{1 + {u}^{2}}{du} \]\n\n\[ = \frac{\pi }{2} - \arctan \left( x\right) \]\n\n(8.1.7)\n\nQ.E.D.
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Yes
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Proposition 8.1.5. If \( x < 0 \), then\n\n\[ \arctan \left( x\right) + \arctan \left( \frac{1}{x}\right) = - \frac{\pi }{2}. \]
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Proof. The result follows immediately from the preceding proposition and the fact that arctangent is an odd function. Q.E.D.
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No
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Proposition 8.2.1. The tangent function has domain \( D \) (as defined above), range \( \mathbb{R} \), and is differentiable at every point \( x \in D \) . Moreover, the tangent function is increasing on each interval of the form\n\n\[ \left( {-\frac{\pi }{2} + {n\pi },\frac{\pi }{2} + {n\pi }}\right) \]\n\n\( n \in \mathbb{Z} \), with\n\n\[ \tan \left( {\left( {\frac{\pi }{2} + {n\pi }}\right) + }\right) = - \infty \] \n\n\( \left( {8.2.5}\right) \) \n\nand \n\n\[ \tan \left( {\left( {\frac{\pi }{2} + {n\pi }}\right) - }\right) = + \infty . \]
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Proof. These results follow immediately from our definitions.\n\nQ.E.D.
|
No
|
Proposition 8.2.2. The tangent function has period \( \pi \) .
|
Proof. The result follows immediately from our definitions.\n\nQ.E.D.
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No
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Proposition 8.3.1. The sine and cosine functions are continuous on \( \mathbb{R} \) .
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Proof. From the definitions, it is sufficient to verify continuity at \( \frac{3\pi }{2} \) . Now\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {\frac{3\pi }{2}}^{ - }}}\sin \left( x\right) = \mathop{\lim }\limits_{{x \rightarrow {\frac{3\pi }{2}}^{ - }}}S\left( x\right) = S\left( \frac{3\pi }{2}\right) = - s\left( \frac{\pi }{2}\right) = - 1 \]\n\n(8.3.16)\n\nand\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {\frac{3\pi }{2}}^{ + }}}\sin \left( x\right) = \mathop{\lim }\limits_{{x \rightarrow {\frac{3\pi }{2}}^{ + }}}S\left( {x - {2\pi }}\right) \]\n\n\[ = \mathop{\lim }\limits_{{x \rightarrow - \frac{\pi }{2} + }}s\left( x\right) \]\n\n\[ = \mathop{\lim }\limits_{{y \rightarrow - \infty }}\frac{y}{\sqrt{1 + {y}^{2}}} \]\n\n\[ = \mathop{\lim }\limits_{{y \rightarrow - \infty }}\frac{1}{-\sqrt{1 + \frac{1}{{y}^{2}}}} \]\n\n\[ = - 1 \]\n\n(8.3.17)\n\nand so sine is continuous at \( \frac{3\pi }{2} \) . Similarly,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {\frac{3\pi }{2}}^{ - }}}\cos \left( x\right) = \mathop{\lim }\limits_{{x \rightarrow {\frac{3\pi }{2}}^{ - }}}C\left( x\right) = C\left( \frac{3\pi }{2}\right) = - c\left( \frac{\pi }{2}\right) = 0 \]\n\n(8.3.18)\n\nand\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {\frac{3\pi }{2}}^{ + }}}\cos \left( x\right) = \mathop{\lim }\limits_{{x \rightarrow {\frac{3\pi }{2}}^{ + }}}C\left( {x - {2\pi }}\right) \]\n\n\[ = \mathop{\lim }\limits_{{x \rightarrow - \frac{\pi }{2} + }}c\left( x\right) \]\n\n\[ = \mathop{\lim }\limits_{{y \rightarrow - \infty }}\frac{1}{\sqrt{1 + {y}^{2}}} \]\n\n\[ = \mathop{\lim }\limits_{{y \rightarrow - \infty }}\frac{\frac{1}{y}}{-\sqrt{1 + \frac{1}{{y}^{2}}}} \]\n\n\[ = 0\text{,} \]\n\n(8.3.19)\n\nand so cosine is continuous at \( \frac{3\pi }{2} \) .\n\nQ.E.D.
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No
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