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Corollary 3.4.4 If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is continuous, then it has an absolute minimum and an absolute maximum on \( \left\lbrack {a, b}\right\rbrack \) .
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Corollary 3.4.4 is sometimes referred to as the Extreme Value Theorem. It follows immediately from Theorem 3.4.2, and the fact that the interval \( \left\lbrack {a, b}\right\rbrack \) is compact (see Example 2.6.4).
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Yes
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Lemma 3.4.5 Let \( f : D \rightarrow \mathbb{R} \) be continuous at \( c \in D \) . Suppose \( f\left( c\right) > 0 \) . Then there exists \( \delta > 0 \) such that\n\n\[ f\left( x\right) > 0\text{ for every }x \in B\left( {c;\delta }\right) \cap D. \]
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Proof: Let \( \varepsilon = f\left( c\right) > 0 \) . By the continuity of \( f \) at \( c \), there exists \( \delta > 0 \) such that if \( x \in D \) and \( \left| {x - c}\right| < \delta \), then\n\n\[ \left| {f\left( x\right) - f\left( c\right) }\right| < \varepsilon \]\n\nThis implies, in particular, that \( f\left( x\right) > f\left( c\right) - \varepsilon = 0 \) for every \( x \in B\left( {c;\delta }\right) \cap D \) . The proof is now complete. \( ▱ \)
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Yes
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Theorem 3.4.7 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Suppose \( f\left( a\right) \cdot f\left( b\right) < 0 \) (this means either \( f\left( a\right) < 0 < f\left( b\right) \) or \( f\left( a\right) > 0 > f\left( b\right) ) \) . Then there exists \( c \in \left( {a, b}\right) \) such that \( f\left( c\right) = 0 \) .
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Proof: We prove only the case \( f\left( a\right) < 0 < f\left( b\right) \) (the case \( f\left( a\right) > 0 > f\left( b\right) \) is completely analogous). Define\n\n\[ A = \{ x \in \left\lbrack {a, b}\right\rbrack : f\left( x\right) \leq 0\} . \]\n\nThis set is nonempty since \( a \in A \) . This set is also bounded since \( A \subset \left\lbrack {a, b}\right\rbrack \) . Therefore, \( c = \sup A \) exists and \( a \leq c \leq b \) . We are going to prove that \( f\left( c\right) = 0 \) by showing that \( f\left( c\right) < 0 \) and \( f\left( c\right) > 0 \) lead to contradictions.\n\nSuppose \( f\left( c\right) < 0 \) . Then there exists \( \delta > 0 \) such that\n\n\[ f\left( x\right) < 0\text{for all}x \in B\left( {c;\delta }\right) \cap \left\lbrack {a, b}\right\rbrack \text{.} \]\n\nBecause \( c < b \) (since \( f\left( b\right) > 0 \) ), we can find \( s \in \left( {c, b}\right) \) such that \( f\left( s\right) < 0 \) (indeed \( s = \min \{ c + \) \( \delta /2,\left( {c + b}\right) /2\} \) will do). This is a contradiction because \( s \in A \) and \( s > c \) .\n\nSuppose \( f\left( c\right) > 0 \) . Then there exists \( \delta > 0 \) such that\n\n\[ f\left( x\right) > 0\text{for all}x \in B\left( {c;\delta }\right) \cap \left\lbrack {a, b}\right\rbrack \text{.} \]\n\nSince \( a < c \) (because \( f\left( a\right) < 0 \) ), there exists \( t \in \left( {a, c}\right) \) such that \( f\left( x\right) > 0 \) for all \( x \in \left( {t, c}\right) \) (in fact, \( t = \max \{ c - \delta /2,\left( {a + c}\right) /2\} \) will do). On the other hand, since \( t < c = \sup A \), there exists \( {t}^{\prime } \in A \) with \( t < {t}^{\prime } \leq c \) . But then \( t < {t}^{\prime } \) and \( f\left( {t}^{\prime }\right) \leq 0 \) . This is a contradiction. We conclude that \( f\left( c\right) = 0 \) . \( ▱ \)
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Yes
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Theorem 3.4.8 — Intermediate Value Theorem. Let \( f \) : \( \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function. Suppose \( f\left( a\right) < \gamma < f\left( b\right) \) . Then there exists a number \( c \in \left( {a, b}\right) \) such that \( f\left( c\right) = \gamma \) .
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Proof: Define\n\n\[ \varphi \left( x\right) = f\left( x\right) - \gamma, x \in \left\lbrack {a, b}\right\rbrack . \]\n\nThen \( \varphi \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) . Moreover,\n\n\[ \varphi \left( a\right) \varphi \left( b\right) = \left\lbrack {f\left( a\right) - \gamma }\right\rbrack \left\lbrack {f\left( b\right) - \gamma }\right\rbrack < 0. \]\n\nBy Theorem 3.4.7, there exists \( c \in \left( {a, b}\right) \) such that \( \varphi \left( c\right) = 0 \) . This is equivalent to \( f\left( c\right) = \gamma \) . The proof is now complete.
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Yes
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Theorem 3.4.10 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be strictly increasing and continuous on \( \left\lbrack {a, b}\right\rbrack \) . Let \( c = f\left( a\right) \) and \( d = f\left( b\right) \) . Then \( f \) is one-to-one, \( f\left( \left\lbrack {a, b}\right\rbrack \right) = \left\lbrack {c, d}\right\rbrack \), and the inverse function \( {f}^{-1} \) defined on \( \left\lbrack {c, d}\right\rbrack \) by\n\n\[ \n{f}^{-1}\left( {f\left( x\right) }\right) = x\text{ where }x \in \left\lbrack {a, b}\right\rbrack ,\n\]\n\nis a continuous function from \( \left\lbrack {c, d}\right\rbrack \) onto \( \left\lbrack {a, b}\right\rbrack \) .
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Proof: The first two assertions follow from the monotonicity of \( f \) and the Intermediate Value Theorem (see also Corollary 3.4.9). We will prove that \( {f}^{-1} \) is continuous on \( \left\lbrack {c, d}\right\rbrack \) . Fix any \( \bar{y} \in \left\lbrack {c, d}\right\rbrack \) and fix any sequence \( \left\{ {y}_{k}\right\} \) in \( \left\lbrack {c, d}\right\rbrack \) that converges to \( \bar{y} \) . Let \( \bar{x} \in \left\lbrack {a, b}\right\rbrack \) and \( {x}_{k} \in \left\lbrack {a, b}\right\rbrack \) be such that\n\n\[ \nf\left( \bar{x}\right) = \bar{y}\text{and}f\left( {x}_{k}\right) = {y}_{k}\text{for every}k\text{.}\n\]\n\nThen \( {f}^{-1}\left( \bar{y}\right) = \bar{x} \) and \( {f}^{-1}\left( {y}_{k}\right) = {x}_{k} \) for every \( k \) . Suppose by contradiction that \( \left\{ {x}_{k}\right\} \) does not converge to \( \bar{x} \) . Then there exist \( {\varepsilon }_{0} > 0 \) and a subsequence \( \left\{ {x}_{{k}_{\ell }}\right\} \) of \( \left\{ {x}_{k}\right\} \) such that\n\n\[ \n\left| {{x}_{{k}_{\ell }} - \bar{x}}\right| \geq {\varepsilon }_{0}\text{ for every }\ell .\n\]\n\n(3.7)\n\nSince the sequence \( \left\{ {x}_{{k}_{\ell }}\right\} \) is bounded, it has a further subsequence that converges to \( {x}_{0} \in \left\lbrack {a, b}\right\rbrack \) . To simplify the notation, we will again call the new subsequence \( \left\{ {x}_{{k}_{\ell }}\right\} \) . Taking limits in (3.7), we get\n\n\[ \n\left| {{x}_{0} - \bar{x}}\right| \geq {\varepsilon }_{0} > 0\n\]\n\n(3.8)\n\nOn the other hand, by the continuity of \( f,\left\{ {f\left( {x}_{{k}_{\ell }}\right) }\right\} \) converges to \( f\left( {x}_{0}\right) \) . Since \( f\left( {x}_{{k}_{\ell }}\right) = {y}_{{k}_{\ell }} \rightarrow \bar{y} \) as \( \ell \rightarrow \infty \), it follows that \( f\left( {x}_{0}\right) = \bar{y} = f\left( \bar{x}\right) \) . This implies \( {x}_{0} = \bar{x} \), which contradicts (3.8). \( ▱ \)
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Yes
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Theorem 3.5.2 If a function \( f : D \rightarrow \mathbb{R} \) is Hölder continuous, then it is uniformly continuous.
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Proof: Since \( f \) is Hölder continuous, there are constants \( \ell \geq 0 \) and \( \alpha > 0 \) such that\n\n\[ \left| {f\left( u\right) - f\left( v\right) }\right| \leq \ell {\left| u - v\right| }^{\alpha }\text{ for every }u, v \in D.\]\n\nIf \( \ell = 0 \), then \( f \) is constant and, thus, uniformly continuous. Suppose next that \( \ell > 0 \) . For any \( \varepsilon > 0 \), let \( \delta = {\left( \frac{\varepsilon }{\ell }\right) }^{1/\alpha } \) . Then, whenever \( u, v \in D \), with \( \left| {u - v}\right| < \delta \) we have\n\n\[ \left| {f\left( u\right) - f\left( v\right) }\right| \leq \ell {\left| u - v\right| }^{\alpha } < \ell {\delta }^{\alpha } = \varepsilon .\]\n\nThe proof is now complete. \( ▱ \)
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Yes
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Theorem 3.5.3 Let \( D \) be a nonempty subset of \( \mathbb{R} \) and \( f : D \rightarrow \mathbb{R} \) . Then \( f \) is uniformly continuous on \( D \) if and only if the following condition holds\n\n(C) for every two sequences \( \left\{ {u}_{n}\right\} ,\left\{ {v}_{n}\right\} \) in \( D \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{u}_{n} - {v}_{n}}\right) = 0 \), it follows that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {f\left( {u}_{n}\right) - f\left( {v}_{n}\right) }\right) = 0.
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Proof: Suppose first that \( f \) is uniformly continuous and let \( \left\{ {u}_{n}\right\} ,\left\{ {v}_{n}\right\} \) be sequences in \( D \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{u}_{n} - {v}_{n}}\right) = 0 \) . Let \( \varepsilon > 0 \) . Choose \( \delta > 0 \) such that \( \left| {f\left( u\right) - f\left( v\right) }\right| < \varepsilon \) whenever \( u, v \in D \) and \( \left| {u - v}\right| < \delta \) . Let \( N \in \mathbb{N} \) be such that \( \left| {{u}_{n} - {v}_{n}}\right| < \delta \) for \( n \geq N \) . For such \( n \), we have \( \left| {f\left( {u}_{n}\right) - f\left( {v}_{n}\right) }\right| < \varepsilon \) . This shows \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {f\left( {u}_{n}\right) - f\left( {v}_{n}\right) }\right) = 0 \).\n\nTo prove the converse, assume condition (C) holds and suppose, by way of contradiction, that \( f \) is not uniformly continuous. Then there exists \( {\varepsilon }_{0} > 0 \) such that for any \( \delta > 0 \), there exist \( u, v \in D \) with\n\n\[ \left| {u - v}\right| < \delta \text{ and }\left| {f\left( u\right) - f\left( v\right) }\right| \geq {\varepsilon }_{0} \]\n\nThus, for every \( n \in \mathbb{N} \), there exist \( {u}_{n},{v}_{n} \in D \) with\n\n\[ \left| {{u}_{n} - {v}_{n}}\right| \leq 1/n\text{ and }\left| {f\left( {u}_{n}\right) - f\left( {v}_{n}\right) }\right| \geq {\varepsilon }_{0}. \]\n\nIt follows that for such sequences, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{u}_{n} - {v}_{n}}\right) = 0 \), but \( \left\{ {f\left( {u}_{n}\right) - f\left( {v}_{n}\right) }\right\} \) does not converge to zero, which contradicts the assumption. \( ▱ \)
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Yes
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Theorem 3.5.4 Let \( f : D \rightarrow \mathbb{R} \) be a continuous function. Suppose \( D \) is compact. Then \( f \) is uniformly continuous on \( D \) .
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Proof: Suppose by contradiction that \( f \) is not uniformly continuous on \( D \) . Then there exists \( {\varepsilon }_{0} > 0 \) such that for any \( \delta > 0 \), there exist \( u, v \in D \) with\n\n\[ \left| {u - v}\right| < \delta \text{ and }\left| {f\left( u\right) - f\left( v\right) }\right| \geq {\varepsilon }_{0}. \]\n\nThus, for every \( n \in \mathbb{N} \), there exist \( {u}_{n},{v}_{n} \in D \) with\n\n\[ \left| {{u}_{n} - {v}_{n}}\right| \leq 1/n\text{ and }\left| {f\left( {u}_{n}\right) - f\left( {v}_{n}\right) }\right| \geq {\varepsilon }_{0}. \]\n\nSince \( D \) is compact, there exist \( {u}_{0} \in D \) and a subsequence \( \left\{ {u}_{{n}_{k}}\right\} \) of \( \left\{ {u}_{n}\right\} \) such that\n\n\[ {u}_{{n}_{k}} \rightarrow {u}_{0}\text{ as }k \rightarrow \infty . \]\n\nThen\n\n\[ \left| {{u}_{{n}_{k}} - {v}_{{n}_{k}}}\right| \leq \frac{1}{{n}_{k}} \]\n\nfor all \( k \) and, hence, we also have\n\n\[ {v}_{{n}_{k}} \rightarrow {u}_{0}\text{as}k \rightarrow \infty \text{.} \]\n\nBy the continuity of \( f \) ,\n\n\[ f\left( {u}_{{n}_{k}}\right) \rightarrow f\left( {u}_{0}\right) \text{ and }f\left( {v}_{{n}_{k}}\right) \rightarrow f\left( {u}_{0}\right) . \]\n\nTherefore, \( \left\{ {f\left( {u}_{{n}_{k}}\right) - f\left( {v}_{{n}_{k}}\right) }\right\} \) converges to zero, which is a contradiction. The proof is now complete. \( ▱ \)
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Yes
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Theorem 3.5.5 Let \( a, b \in \mathbb{R} \) and \( a < b \) . A function \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is uniformly continuous if and only if \( f \) can be extended to a continuous function \( \widetilde{f} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) (that is, there is a continuous function \( \widetilde{f} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) such that \( f = {\widetilde{f}}_{\mid \left( {a, b}\right) } \) ).
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Proof: Suppose first that there exists a continuous function \( \widetilde{f} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) such that \( f = {\widetilde{f}}_{\mid \left( {a, b}\right) } \) . By Theorem 3.5.4, the function \( \widetilde{f} \) is uniformly continuous on \( \left\lbrack {a, b}\right\rbrack \) . Therefore, it follows from our early observation that \( f \) is uniformly continuous on \( \left( {a, b}\right) \) .\n\nFor the converse, suppose \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is uniformly continuous. We will show first that \( \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}f\left( x\right) \) exists. Note that the one sided limit corresponds to the limit in Theorem 3.2.2. We will check that the \( \varepsilon - \delta \) condition of Theorem 3.2.2 holds.\n\nLet \( \varepsilon > 0 \) . Choose \( {\delta }_{0} > 0 \) so that \( \left| {f\left( u\right) - f\left( v\right) }\right| < \varepsilon \) whenever \( u, v \in \left( {a, b}\right) \) and \( \left| {u - v}\right| < {\delta }_{0} \) . Set \( \delta = {\delta }_{0}/2 \) . Then, if \( u, v \in \left( {a, b}\right) ,\left| {u - a}\right| < \delta \), and \( \left| {v - a}\right| < \delta \) we have\n\n\[ \left| {u - v}\right| \leq \left| {u - a}\right| + \left| {a - v}\right| < \delta + \delta = {\delta }_{0} \]\n\nand, hence, \( \left| {f\left( u\right) - f\left( v\right) }\right| < \varepsilon \) . We can now invoke Theorem 3.2.2 to conclude \( \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}f\left( x\right) \) exists. In a similar way we can show that \( \mathop{\lim }\limits_{{x \rightarrow {b}^{ - }}}f\left( x\right) \) exists. Now define, \( \widetilde{f} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) by\n\n\[ \widetilde{f}\left( x\right) = \left\{ \begin{array}{ll} f\left( x\right) , & \text{ if }x \in \left( {a, b}\right) \\ \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}f\left( x\right) , & \text{ if }x = a \\ \mathop{\lim }\limits_{{x \rightarrow {b}^{ - }}}f\left( x\right) , & \text{ if }x = b \end{array}\right. \]\n\nBy its definition \( {\widetilde{f}}_{\mid \left( {a, b}\right) } = f \) and, so, \( \widetilde{f} \) is continuous at every \( x \in \left( {a, b}\right) \) . Moreover, \( \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\widetilde{f}\left( x\right) = \) \( \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}f\left( x\right) = \widetilde{f}\left( a\right) \) and \( \mathop{\lim }\limits_{{x \rightarrow {b}^{ - }}}\widetilde{f}\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow {b}^{ - }}}f\left( x\right) = \widetilde{f}\left( b\right) \), so \( \widetilde{f} \) is also continuous at \( a \) and \( b \) by Theorem 3.3.2. Thus \( \widetilde{f} \) is the desired continuous extension of \( f \) . \( ▱ \)
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Yes
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Theorem 3.6.1 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then \( \ell = \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) if and only if the following two conditions hold:\n\n(1) For every \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ f\left( x\right) < \ell + \varepsilon \text{ for all }x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D; \]\n\n(2) For every \( \varepsilon > 0 \) and for every \( \delta > 0 \), there exists \( {x}_{\delta } \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D \) such that\n\n\[ \ell - \varepsilon < f\left( {x}_{\delta }\right) \]
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Proof: Suppose \( \ell = \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) . Then\n\n\[ \ell = \mathop{\inf }\limits_{{\delta > 0}}g\left( \delta \right) \]\n\nwhere \( g \) is defined in (3.10). For any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \ell \leq g\left( \delta \right) = \mathop{\sup }\limits_{{x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D}}f\left( x\right) < \ell + \varepsilon . \]\n\nThus,\n\n\[ f\left( x\right) < \ell + \varepsilon \text{ for all }x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D, \]\n\nwhich proves condition (1). Next note that for any \( \varepsilon > 0 \) and \( \delta > 0 \), we have\n\n\[ \ell - \varepsilon < \ell \leq g\left( \delta \right) = \mathop{\sup }\limits_{{x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D}}f\left( x\right) \]\n\nThus, there exists \( {x}_{\delta } \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D \) with\n\n\[ \ell - \varepsilon < f\left( {x}_{\delta }\right) \]\n\nThis proves (2).\n\nLet us now prove the converse. Suppose (1) and (2) are satisfied. Fix any \( \varepsilon > 0 \) and let \( \delta > 0 \) satisfy (1). Then\n\n\[ g\left( \delta \right) = \mathop{\sup }\limits_{{x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D}}f\left( x\right) \leq \ell + \varepsilon . \]\n\nThis implies\n\n\[ \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \mathop{\inf }\limits_{{\delta > 0}}g\left( \delta \right) \leq \ell + \varepsilon . \]\n\nSince \( \varepsilon \) is arbitrary, we get\n\n\[ \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \leq \ell \]\n\nAgain, let \( \varepsilon > 0 \) . Given \( \delta > 0 \), let \( {x}_{\delta } \) be as in (2). Therefore,\n\n\[ \ell - \varepsilon < f\left( {x}_{\delta }\right) \leq \mathop{\sup }\limits_{{x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D}}f\left( x\right) = g\left( \delta \right) . \]\n\nThis implies\n\n\[ \ell - \varepsilon \leq \mathop{\inf }\limits_{{\delta > 0}}g\left( \delta \right) = \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \]\n\nIt follows that \( \ell \leq \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) . Therefore, \( \ell = \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) . \( ▱ \)
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Yes
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Corollary 3.6.2 Suppose \( \ell = \lim \mathop{\sup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) . Then there exists a sequence \( \left\{ {x}_{k}\right\} \) in \( D \) such that \( \left\{ {x}_{k}\right\} \) converges to \( \bar{x},{x}_{k} \neq \bar{x} \) for every \( k \), and\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) = \ell \]\n\nMoreover, if \( \left\{ {y}_{k}\right\} \) is a sequence in \( D \) that converges to \( \bar{x},{y}_{k} \neq \bar{x} \) for every \( k \), and \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {y}_{k}\right) = {\ell }^{\prime } \) , then \( {\ell }^{\prime } \leq \ell \) .
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Proof: For each \( k \in \mathbb{N} \), take \( {\varepsilon }_{k} = \frac{1}{k} \) . By (1) of Theorem 3.6.1, there exists \( {\delta }_{k} > 0 \) such that\n\n\[ f\left( x\right) < \ell + {\varepsilon }_{k}\text{ for all }x \in {B}_{0}\left( {\bar{x};{\delta }_{k}}\right) \cap D. \]\n\n(3.11)\n\nLet \( {\delta }_{k}^{\prime } = \min \left\{ {{\delta }_{k},\frac{1}{k}}\right\} \) . Then \( {\delta }_{k}^{\prime } \leq {\delta }_{k} \) and \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{\delta }_{k}^{\prime } = 0 \) . From (2) of Theorem 3.6.1, there exists \( {x}_{k} \in {B}_{0}\left( {\bar{x};{\delta }_{k}^{\prime }}\right) \cap D \) such that\n\n\[ \ell - {\varepsilon }_{k} < f\left( {x}_{k}\right) \]\n\nMoreover, \( f\left( {x}_{k}\right) < \ell + {\varepsilon }_{k} \) by (3.11). Therefore, \( \left\{ {x}_{k}\right\} \) is a sequence that satisfies the conclusion of the corollary.\n\nNow let \( \left\{ {y}_{k}\right\} \) be a sequence in \( D \) that converges to \( \bar{x},{y}_{k} \neq \bar{x} \) for every \( k \), and \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {y}_{k}\right) = {\ell }^{\prime } \) . For any \( \varepsilon > 0 \), let \( \delta > 0 \) be as in (1) of Theorem 3.6.1. Since \( {y}_{k} \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D \) when \( k \) is sufficiently large, we have\n\n\[ f\left( {y}_{k}\right) < \ell + \varepsilon \]\n\nfor such \( k \) . This implies \( {\ell }^{\prime } \leq \ell + \varepsilon \) . It follows that \( {\ell }^{\prime } \leq \ell \) . \( ▱ \)
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Yes
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Theorem 3.6.4 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then\n\n\[ \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \infty \]\n\nif and only if there exists a sequence \( \left\{ {x}_{k}\right\} \) in \( D \) such that \( \left\{ {x}_{k}\right\} \) converges to \( \bar{x},{x}_{k} \neq \bar{x} \) for every \( k \), and \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) = \infty \) .
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Proof: Suppose \( \lim \mathop{\sup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \infty \) . Then\n\n\[ \mathop{\inf }\limits_{{\delta > 0}}g\left( \delta \right) = \infty \]\n\nwhere \( g \) is the extended real-valued function defined in (3.10). Thus, \( g\left( \delta \right) = \infty \) for every \( \delta > 0 \) . Given \( k \in \mathbb{N} \), for \( {\delta }_{k} = \frac{1}{k} \), since\n\n\[ g\left( {\delta }_{k}\right) = \mathop{\sup }\limits_{{x \in {B}_{0}\left( {\bar{x};{\delta }_{k}}\right) \cap D}}f\left( x\right) = \infty ,\]\n\nthere exists \( {x}_{k} \in {B}_{0}\left( {\bar{x};{\delta }_{k}}\right) \cap D \) such that \( f\left( {x}_{k}\right) > k \) . Therefore, \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) = \infty \) .\n\nLet us prove the converse. Since \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) = \infty \), for every \( M \in \mathbb{R} \), there exists \( K \in \mathbb{N} \) such that\n\n\[ f\left( {x}_{k}\right) \geq M\text{for every}k \geq K\text{.} \]\n\nFor any \( \delta > 0 \), we have\n\n\[ {x}_{k} \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D \]\n\nwhenever \( k \) is sufficiently large. Thus,\n\n\[ g\left( \delta \right) = \mathop{\sup }\limits_{{x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D}}f\left( x\right) \geq M. \]\n\nThis implies \( g\left( \delta \right) = \infty \), and hence \( \lim \mathop{\sup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \infty \) .
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Yes
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Corollary 3.6.7 Suppose \( \ell = \mathop{\liminf }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) . Then there exists a sequence \( \left\{ {x}_{k}\right\} \) in \( D \) such that \( {x}_{k} \) converges to \( \bar{x},{x}_{k} \neq \bar{x} \) for every \( k \), and
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\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) = \ell \] Moreover, if \( \left\{ {y}_{k}\right\} \) is a sequence in \( D \) that converges to \( \bar{x},{y}_{k} \neq \bar{x} \) for every \( k \), and \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {y}_{k}\right) = {\ell }^{\prime } \) , then \( {\ell }^{\prime } \geq \ell \) .
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Yes
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Theorem 3.6.11 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \]\n\nif and only if\n\n\[ \mathop{\limsup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \mathop{\liminf }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell . \]\n
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Proof: Suppose\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \]\n\nThen for every \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \ell - \varepsilon < f\left( x\right) < \ell + \varepsilon \text{ for all }x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D. \]\n\nSince this also holds for every \( 0 < {\delta }^{\prime } < \delta \), we get\n\n\[ \ell - \varepsilon < g\left( {\delta }^{\prime }\right) \leq \ell + \varepsilon \]\n\nIt follows that\n\n\[ \ell - \varepsilon \leq \mathop{\inf }\limits_{{{\delta }^{\prime } > 0}}g\left( {\delta }^{\prime }\right) \leq \ell + \varepsilon \]\n\nTherefore, \( \lim \mathop{\sup }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \) since \( \varepsilon \) is arbitrary. The proof for the limit inferior is similar. The converse follows directly from (1) of Theorem 3.6.1 and Theorem 3.6.6. \( ▱ \)
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Yes
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Theorem 3.7.1 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \in D \) be a limit point of \( D \) . Then \( f \) is lower semicontinuous at \( \bar{x} \) if and only if\n\n\[ \mathop{\liminf }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \geq f\left( \bar{x}\right) \]
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Proof: Suppose \( f \) is lower semicontinuous at \( \bar{x} \) . Let \( \varepsilon > 0 \) . Then there exists \( {\delta }_{0} > 0 \) such that\n\n\[ f\left( \bar{x}\right) - \varepsilon < f\left( x\right) \text{ for all }x \in B\left( {\bar{x};{\delta }_{0}}\right) \cap D. \]\n\nThis implies\n\n\[ f\left( \bar{x}\right) - \varepsilon \leq h\left( {\delta }_{0}\right) \]\n\nwhere\n\n\[ h\left( \delta \right) = \mathop{\inf }\limits_{{x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D}}f\left( x\right) . \]\n\nThus,\n\n\[ \mathop{\liminf }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \mathop{\sup }\limits_{{\delta > 0}}h\left( \delta \right) \geq h\left( {\delta }_{0}\right) \geq f\left( \bar{x}\right) - \varepsilon . \]\n\nSince \( \varepsilon \) is arbitrary, we obtain \( \mathop{\liminf }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \geq f\left( \bar{x}\right) \) .\n\nWe now prove the converse. Suppose\n\n\[ \mathop{\liminf }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \mathop{\sup }\limits_{{\delta > 0}}h\left( \delta \right) \geq f\left( \bar{x}\right) \]\n\nand let \( \varepsilon > 0 \) . Since\n\n\[ \mathop{\sup }\limits_{{\delta > 0}}h\left( \delta \right) > f\left( \bar{x}\right) - \varepsilon \]\n\nthere exists \( \delta > 0 \) such that \( h\left( \delta \right) > f\left( \bar{x}\right) - \varepsilon \) . This implies\n\n\[ f\left( x\right) > f\left( \bar{x}\right) - \varepsilon \text{ for all }x \in {B}_{0}\left( {\bar{x};\delta }\right) \cap D. \]\n\nSince this is also true for \( x = \bar{x} \), the function \( f \) is lower semicontinuous at \( \bar{x} \) .\n\nThe proof for the upper semicontinuous case is similar. \( ▱ \)
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Yes
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Theorem 3.7.2 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \in D \) . Then \( f \) is l.s.c. at \( \bar{x} \) if and only if for every sequence \( \left\{ {x}_{k}\right\} \) in \( D \) that converges to \( \bar{x} \) , \[ \mathop{\liminf }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) \geq f\left( \bar{x}\right) \]
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Proof: Suppose \( f \) is l.s.c. at \( \bar{x} \) . Then for any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that (3.12) holds. Since \( \left\{ {x}_{k}\right\} \) converges to \( \bar{x} \), we have \( {x}_{k} \in B\left( {\bar{x};\delta }\right) \) when \( k \) is sufficiently large. Thus, \[ f\left( \bar{x}\right) - \varepsilon < f\left( {x}_{k}\right) \] for such \( k \) . It follows that \( f\left( \bar{x}\right) - \varepsilon \leq \mathop{\liminf }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) \) . Since \( \varepsilon \) is arbitrary, it follows that \( f\left( \bar{x}\right) \leq \) \( \mathop{\liminf }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) \) . We now prove the converse. Suppose \( \mathop{\liminf }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) \geq f\left( \bar{x}\right) \) and assume, by way of contradiction, that \( f \) is not l.s.c. at \( \bar{x} \) . Then there exists \( \bar{\varepsilon } > 0 \) such that for every \( \delta > 0 \), there exists \( {x}_{\delta } \in B\left( {\bar{x};\delta }\right) \cap D \) with \[ f\left( \bar{x}\right) - \bar{\varepsilon } \geq f\left( {x}_{\delta }\right) \] Applying this for \( {\delta }_{k} = \frac{1}{k} \), we obtain a sequence \( \left\{ {x}_{k}\right\} \) in \( D \) that converges to \( \bar{x} \) with \[ f\left( \bar{x}\right) - \bar{\varepsilon } \geq f\left( {x}_{k}\right) \text{ for every }\mathrm{k}. \] This implies \[ f\left( \bar{x}\right) - \bar{\varepsilon } \geq \mathop{\liminf }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) \] This is a contradiction.
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Yes
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Theorem 3.7.3 Suppose \( D \) is a compact set of \( \mathbb{R} \) and \( f : D \rightarrow \mathbb{R} \) is lower semicontinuous. Then \( f \) has an absolute minimum on \( D \) . That means there exists \( \bar{x} \in D \) such that\n\n\[ f\left( x\right) \geq f\left( \bar{x}\right) \text{ for all }x \in D. \]
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Proof: We first prove that \( f \) is bounded below. Suppose by contradiction that for every \( k \in \mathbb{N} \), there exists \( {x}_{k} \in D \) such that\n\n\[ f\left( {x}_{k}\right) < - k.\]\n\nSince \( D \) is compact, there exists a subsequence \( \left\{ {x}_{{k}_{\ell }}\right\} \) of \( \left\{ {x}_{k}\right\} \) that converges to \( {x}_{0} \in D \) . Since \( f \) is l.s.c., by Theorem 3.7.2\n\n\[ \mathop{\liminf }\limits_{{\ell \rightarrow \infty }}f\left( {x}_{{k}_{\ell }}\right) \geq f\left( {x}_{0}\right) \]\n\nThis is a contraction because \( \mathop{\liminf }\limits_{{\ell \rightarrow \infty }}f\left( {x}_{{k}_{\ell }}\right) = - \infty \) . This shows \( f \) is bounded below. Define\n\n\[ \gamma = \inf \{ f\left( x\right) : x \in D\} . \]\n\nSince the set \( \{ f\left( x\right) : x \in D\} \) is nonempty and bounded below, \( \gamma \in \mathbb{R} \) .\n\nLet \( \left\{ {u}_{k}\right\} \) be a sequence in \( D \) such that \( \left\{ {f\left( {u}_{k}\right) }\right\} \) converges to \( \gamma \) . By the compactness of \( D \), the sequence \( \left\{ {u}_{k}\right\} \) has a convergent subsequence \( \left\{ {u}_{{k}_{\ell }}\right\} \) that converges to some \( \bar{x} \in D \) . Then\n\n\[ \gamma = \mathop{\lim }\limits_{{\ell \rightarrow \infty }}f\left( {u}_{{k}_{\ell }}\right) = \mathop{\liminf }\limits_{{\ell \rightarrow \infty }}f\left( {u}_{{k}_{\ell }}\right) \geq f\left( \bar{x}\right) \geq \gamma . \]\n\nThis implies \( \gamma = f\left( \bar{x}\right) \) and, hence,\n\n\[ f\left( x\right) \geq f\left( \bar{x}\right) \text{ for all }x \in D. \]\n\nThe proof is now complete. \( ▱ \)
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Yes
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Theorem 3.7.5 Let \( f : D \rightarrow \mathbb{R} \) . Then \( f \) is lower semicontinuous if and only if \( {\mathcal{L}}_{a}\left( f\right) \) is closed in \( D \) for every \( a \in \mathbb{R} \) . Similarly, \( f \) is upper semicontinuous if and only if \( {\mathcal{U}}_{a}\left( f\right) \) is closed in \( D \) for every \( a \in \mathbb{R} \) .
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Proof: Suppose \( f \) is lower semicontinuous. Using Corollary 2.6.10, we will prove that for every sequence \( \left\{ {x}_{k}\right\} \) in \( {\mathcal{L}}_{a}\left( f\right) \) that converges to a point \( \bar{x} \in D \), we get \( \bar{x} \in {\mathcal{L}}_{a}\left( f\right) \) . For every \( k \), since \( {x}_{k} \in {\mathcal{L}}_{a}\left( f\right), f\left( {x}_{k}\right) \leq a. \n\nSince \( f \) is lower semicontinuous at \( \bar{x} \) ,\n\n\[ \nf\left( \bar{x}\right) \leq \mathop{\liminf }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) \leq a.\n\]\n\nThus, \( \bar{x} \in {\mathcal{L}}_{a}\left( f\right) \) . It follows that \( {\mathcal{L}}_{a}\left( f\right) \) is closed.\n\nWe now prove the converse. Fix any \( \bar{x} \in D \) and \( \varepsilon > 0 \) . Then the set\n\n\[ \nG = \{ x \in D : f\left( x\right) > f\left( \bar{x}\right) - \varepsilon \} = D \smallsetminus {\mathcal{L}}_{f\left( \bar{x}\right) - \varepsilon \left( f\right) }\n\]\n\n is open in \( D \) and \( \bar{x} \in G \) . Thus, there exists \( \delta > 0 \) such that\n\n\[ \nB\left( {\bar{x};\delta }\right) \cap D \subset G.\n\]\n\nIt follows that\n\n\[ \nf\left( \bar{x}\right) - \varepsilon < f\left( x\right) \text{ for all }x \in B\left( {\bar{x};\delta }\right) \cap D.\n\]\n\nTherefore, \( f \) is lower semicontinuous. The proof for the upper semicontinuous case is similar. \( ▱ \)
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Yes
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Theorem 3.7.7 Let \( f : D \rightarrow \mathbb{R} \) . Then \( f \) is continuous if and only if for every \( a, b \in \mathbb{R} \) with \( a < b \) , the set\n\n\[ \n{O}_{a, b} = \{ x \in D : a < f\left( x\right) < b\} = {f}^{-1}\left( \left( {a, b}\right) \right)\n\]\n\nis an open set in \( D \) .
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Proof: Suppose \( f \) is continuous. Then \( f \) is lower semicontinuous and upper semicontinuos. Fix \( a, b \in \mathbb{R} \) with \( a < b \) . Then\n\n\[ \n{O}_{a, b} = {L}_{b} \cap {U}_{a}\n\]\n\nBy Theorem 3.7.6, the set \( {O}_{a, b} \) is open since it is the intersection of two open sets \( {L}_{a} \) and \( {U}_{b} \) .\n\nLet us prove the converse. We will only show that \( f \) is lower semicontinuous since the proof of upper semicontinuity is similar. For every \( a \in \mathbb{R} \), we have\n\n\[ \n{U}_{a}\left( f\right) = \{ x \in D : f\left( x\right) > a\} = { \cup }_{n \in \mathbb{N}}{f}^{-1}\left( \left( {a, a + n}\right) \right)\n\]\n\nThus, \( {U}_{a}\left( f\right) \) is open in \( D \) as it is a union of open sets in \( D \) . Therefore, \( f \) is lower semicontinuous by Corollary 3.7.6. \( ▱ \)
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Yes
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Theorem 4.1.1 Let \( G \) be an open subset of \( \mathbb{R} \) and let \( f \) be defined on \( G \) . If \( f \) is differentiable at \( a \in G \), then \( f \) is continuous at this point.
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Proof: We have the following identity for \( x \in G \smallsetminus \{ a\} \) :\n\n\[ f\left( x\right) = f\left( x\right) - f\left( a\right) + f\left( a\right) \]\n\n\[ = \frac{f\left( x\right) - f\left( a\right) }{x - a}\left( {x - a}\right) + f\left( a\right) . \]\n\nThus,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow a}}\left\lbrack {\frac{f\left( x\right) - f\left( a\right) }{x - a}\left( {x - a}\right) + f\left( a\right) }\right\rbrack = {f}^{\prime }\left( a\right) \cdot 0 + f\left( a\right) = f\left( a\right) . \]\n\nTherefore, \( f \) is continuous at \( a \) by Theorem 3.3.2. \( ▱ \)
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Yes
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Theorem 4.1.3 Let \( G \) be an open subset of \( \mathbb{R} \) and let \( f, g : G \rightarrow \mathbb{R} \) . Suppose both \( f \) and \( g \) are differentiable at \( a \in G \) . Then the following hold.\n\n(a) The function \( f + g \) is differentiable at \( a \) and\n\n\[{\left( f + g\right) }^{\prime }\left( a\right) = {f}^{\prime }\left( a\right) + {g}^{\prime }\left( a\right) .
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Proof: The proofs of (a) and (b) are straightforward and we leave them as exercises. Let us prove (c). For every \( x \in G \smallsetminus \{ a\} \), we can write\n\n\[ \frac{\left( {fg}\right) \left( x\right) - \left( {fg}\right) \left( a\right) }{x - a} = \frac{f\left( x\right) g\left( x\right) - f\left( a\right) g\left( x\right) + f\left( a\right) g\left( x\right) - f\left( a\right) g\left( a\right) }{x - a}\]\n\n\[= \frac{\left( {f\left( x\right) - f\left( a\right) }\right) g\left( x\right) }{x - a} + \frac{f\left( a\right) \left( {g\left( x\right) - g\left( a\right) }\right) }{x - a}. \]\n\nBy Theorem 4.1.1, the function \( g \) is continuous at \( a \) and, hence,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) = g\left( a\right) \]\n\n(4.1)\n\nThus,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{\left( {fg}\right) \left( x\right) - \left( {fg}\right) \left( a\right) }{x - a} = {f}^{\prime }\left( a\right) g\left( a\right) + f\left( a\right) {g}^{\prime }\left( a\right) . \]\n\nThis implies (c).
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No
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Lemma 4.1.4 Let \( G \) be an open subset of \( \mathbb{R} \) and let \( f : G \rightarrow \mathbb{R} \) . Suppose \( f \) is differentiable at \( a \) . Then there exists a function \( u : G \rightarrow \mathbb{R} \) satisfying\n\n\[ f\left( x\right) - f\left( a\right) = \left\lbrack {{f}^{\prime }\left( a\right) + u\left( x\right) }\right\rbrack \left( {x - a}\right) \text{ for all }x \in G \]\n\nand \( \mathop{\lim }\limits_{{x \rightarrow a}}u\left( x\right) = 0 \) .
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Proof: Define\n\n\[ u\left( x\right) = \left\{ \begin{array}{ll} \frac{f\left( x\right) - f\left( a\right) }{x - a} - {f}^{\prime }\left( a\right) , & x \in G \smallsetminus \{ a\} \\ 0, & x = a. \end{array}\right. \]\n\nSince \( f \) is differentiable at \( a \), we have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}u\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) - f\left( a\right) }{x - a} - {f}^{\prime }\left( a\right) = {f}^{\prime }\left( a\right) - {f}^{\prime }\left( a\right) = 0. \]\n\nTherefore, the function \( u \) satisfies the conditions of the lemma. \( ▱ \)
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Yes
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Theorem 4.1.5 - Chain rule. Let \( f : {G}_{1} \rightarrow \mathbb{R} \) and let \( g : {G}_{2} \rightarrow \mathbb{R} \), where \( {G}_{1} \) and \( {G}_{2} \) are two open subsets of \( \mathbb{R} \) with \( f\left( {G}_{1}\right) \subset {G}_{2} \) . Suppose \( f \) is differentiable at \( a \) and \( g \) is differentiable at \( f\left( a\right) \) . Then the function \( g \circ f \) is differentiable at \( a \) and \[ {\left( g \circ f\right) }^{\prime }\left( a\right) = {g}^{\prime }\left( {f\left( a\right) }\right) {f}^{\prime }\left( a\right) . \]
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Proof: Since \( f \) is differentiable at \( a \), by Lemma 4.1.4, there exists a function \( u \) defined on \( {G}_{1} \) with \[ f\left( x\right) - f\left( a\right) = \left\lbrack {{f}^{\prime }\left( a\right) + u\left( x\right) }\right\rbrack \left( {x - a}\right) \text{ for all }x \in {G}_{1}, \] and \( \mathop{\lim }\limits_{{x \rightarrow a}}u\left( x\right) = 0 \) . Similarly, since \( g \) is differentiable at \( f\left( a\right) \), there exists a function \( v \) defined on \( {G}_{2} \) with \[ g\left( t\right) - g\left( {f\left( a\right) }\right) = \left\lbrack {{g}^{\prime }\left( {f\left( a\right) }\right) + v\left( t\right) }\right\rbrack \left\lbrack {t - f\left( a\right) }\right\rbrack \text{ for all }t \in {G}_{2}, \] and \( \mathop{\lim }\limits_{{t \rightarrow f\left( a\right) }}v\left( t\right) = 0 \) . Applying (4.2) for \( t = f\left( x\right) \), we have \[ g\left( {f\left( x\right) }\right) - g\left( {f\left( a\right) }\right) = \left\lbrack {{g}^{\prime }\left( {f\left( a\right) }\right) + v\left( {f\left( x\right) }\right) }\right\rbrack \left\lbrack {f\left( x\right) - f\left( a\right) }\right\rbrack . \] Thus, \[ g\left( {f\left( x\right) }\right) - g\left( {f\left( a\right) }\right) = \left\lbrack {{g}^{\prime }\left( {f\left( a\right) }\right) + v\left( {f\left( x\right) }\right) }\right\rbrack \left\lbrack {{f}^{\prime }\left( a\right) + u\left( x\right) }\right\rbrack \left( {x - a}\right) \text{ for all }x \in {G}_{1}. \] This implies \[ \frac{g\left( {f\left( x\right) }\right) - g\left( {f\left( a\right) }\right) }{x - a} = \left\lbrack {{g}^{\prime }\left( {f\left( a\right) }\right) + v\left( {f\left( x\right) }\right) }\right\rbrack \left\lbrack {{f}^{\prime }\left( a\right) + u\left( x\right) }\right\rbrack \text{ for all }x \in {G}_{1} \smallsetminus \{ a\} . \] By the continuity of \( f \) at \( a \) and the property of \( v \), we have \( \mathop{\lim }\limits_{{x \rightarrow a}}v\left( {f\left( x\right) }\right) = 0 \) and, hence, \[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{g\left( {f\left( x\right) }\right) - g\left( {f\left( a\right) }\right) }{x - a} = {g}^{\prime }\left( {f\left( a\right) }\right) {f}^{\prime }\left( a\right) . \] The proof is now complete.
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Yes
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Theorem 4.2.1 — Fermat’s Rule. Let \( I \) be an open interval and \( f : I \rightarrow \mathbb{R} \) . If \( f \) has a local minimum or maximum at \( a \in I \) and \( f \) is differentiable at \( a \), then \( {f}^{\prime }\left( a\right) = 0 \) .
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Proof: Suppose \( f \) has a local minimum at \( a \) . Then there exists \( \delta > 0 \) sufficiently small such that\n\n\[ f\left( x\right) \geq f\left( a\right) \text{ for all }x \in B\left( {a;\delta }\right) . \]\n\nSince \( {B}_{ + }\left( {a;\delta }\right) \) is a subset of \( B\left( {a;\delta }\right) \), we have\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \geq 0\text{ for all }x \in {B}_{ + }\left( {a;\delta }\right) . \]\n\nTaking into account the differentiability of \( f \) at \( a \) yields\n\n\[ {f}^{\prime }\left( a\right) = \mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) - f\left( a\right) }{x - a} = \mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\frac{f\left( x\right) - f\left( a\right) }{x - a} \geq 0. \]\n\nSimilarly,\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \leq 0\text{ for all }x \in {B}_{ - }\left( {a;\delta }\right) . \]\n\nIt follows that\n\n\[ {f}^{\prime }\left( a\right) = \mathop{\lim }\limits_{{x \rightarrow a}}\frac{f\left( x\right) - f\left( a\right) }{x - a} = \mathop{\lim }\limits_{{x \rightarrow {a}^{ - }}}\frac{f\left( x\right) - f\left( a\right) }{x - a} \leq 0. \]\n\nTherefore, \( {f}^{\prime }\left( a\right) = 0 \) . The proof is similar for the case where \( f \) has a local maximum at \( a \) .
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Yes
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Theorem 4.2.2 - Rolle’s Theorem. Let \( a, b \in \mathbb{R} \) with \( a < b \) and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) . Suppose \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) with \( f\left( a\right) = f\left( b\right) \) . Then there exists \( c \in \left( {a, b}\right) \) such that\n\n\[ \n{f}^{\prime }\left( c\right) = 0 \n\]
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Proof: Since \( f \) is continuous on the compact set \( \left\lbrack {a, b}\right\rbrack \), by the extreme value theorem (Theorem 3.4.2) there exist \( {\bar{x}}_{1} \in \left\lbrack {a, b}\right\rbrack \) and \( {\bar{x}}_{2} \in \left\lbrack {a, b}\right\rbrack \) such that\n\n\[ \nf\left( {\bar{x}}_{1}\right) = \min \{ f\left( x\right) : x \in \left\lbrack {a, b}\right\rbrack \} \text{ and }f\left( {\bar{x}}_{2}\right) = \max \{ f\left( x\right) : x \in \left\lbrack {a, b}\right\rbrack \} .\n\]\n\nThen\n\n\[ \nf\left( {\bar{x}}_{1}\right) \leq f\left( x\right) \leq f\left( {\bar{x}}_{2}\right) \text{ for all }x \in \left\lbrack {a, b}\right\rbrack .\n\]\n\nIf \( {\bar{x}}_{1} \in \left( {a, b}\right) \) or \( {\bar{x}}_{2} \in \left( {a, b}\right) \), then \( f \) has a local minimum at \( {\bar{x}}_{1} \) or \( f \) has a local maximum at \( {\bar{x}}_{2} \) . By Theorem 4.2.1, \( {f}^{\prime }\left( {\bar{x}}_{1}\right) = 0 \) or \( {f}^{\prime }\left( {\bar{x}}_{2}\right) = 0 \), and (4.3) holds with \( c = {\bar{x}}_{1} \) or \( c = {\bar{x}}_{2} \) .\n\nIf both \( {\bar{x}}_{1} \) and \( {\bar{x}}_{2} \) are the endpoints of \( \left\lbrack {a, b}\right\rbrack \), then \( f\left( {\bar{x}}_{1}\right) = f\left( {\bar{x}}_{2}\right) \) because \( f\left( a\right) = f\left( b\right) \) . By (4.4), \( f \) is a constant function, so \( {f}^{\prime }\left( c\right) = 0 \) for any \( c \in \left( {a, b}\right) \) . \( ▱ \)
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Yes
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Theorem 4.2.3 — Mean Value Theorem. Let \( a, b \in \mathbb{R} \) with \( a < b \) and \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R}. \) Suppose \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . Then there exists \( c \in \left( {a, b}\right) \) such that\n\n\[ \n{f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\left( a\right) }{b - a}.\n\]
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Proof: The linear function whose graph goes through \( \left( {a, f\left( a\right) }\right) \) and \( \left( {b, f\left( b\right) }\right) \) is\n\n\[ \ng\left( x\right) = \frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - a}\right) + f\left( a\right) .\n\]\nDefine\n\n\[ \nh\left( x\right) = f\left( x\right) - g\left( x\right) = f\left( x\right) - \left\lbrack {\frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - a}\right) + f\left( a\right) }\right\rbrack \text{ for }x \in \left\lbrack {a, b}\right\rbrack .\n\]\n\nThen \( h\left( a\right) = h\left( b\right) \), and \( h \) satisfies the assumptions of Theorem 4.2.2. Thus, there exists \( c \in \left( {a, b}\right) \) such that \( {h}^{\prime }\left( c\right) = 0 \) . Since\n\n\[ \n{h}^{\prime }\left( x\right) = {f}^{\prime }\left( x\right) - \frac{f\left( b\right) - f\left( a\right) }{b - a},\n\]\n\nit follows that\n\n\[ \n{f}^{\prime }\left( c\right) - \frac{f\left( b\right) - f\left( a\right) }{b - a} = 0.\n\]\n\nThus, (4.5) holds. \( ▱ \)
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Yes
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Theorem 4.2.4 — Cauchy’s Theorem. Let \( a, b \in \mathbb{R} \) with \( a < b. \) Suppose \( f \) and \( g \) are continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . Then there exists \( c \in \left( {a, b}\right) \) such that\n\n\[ \left\lbrack {f\left( b\right) - f\left( a\right) }\right\rbrack {g}^{\prime }\left( c\right) = \left\lbrack {g\left( b\right) - g\left( a\right) }\right\rbrack {f}^{\prime }\left( c\right) . \]\n\n(4.6)
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Proof: Define\n\n\[ h\left( x\right) = \left\lbrack {f\left( b\right) - f\left( a\right) }\right\rbrack g\left( x\right) - \left\lbrack {g\left( b\right) - g\left( a\right) }\right\rbrack f\left( x\right) \text{ for }x \in \left\lbrack {a, b}\right\rbrack . \]\n\nThen \( h\left( a\right) = f\left( b\right) g\left( a\right) - f\left( a\right) g\left( b\right) = h\left( b\right) \), and \( h \) satisfies the assumptions of Theorem 4.2.2. Thus, there exists \( c \in \left( {a, b}\right) \) such that \( {h}^{\prime }\left( c\right) = 0 \) . Since\n\n\[ {h}^{\prime }\left( x\right) = \left\lbrack {f\left( b\right) - f\left( a\right) }\right\rbrack {g}^{\prime }\left( x\right) - \left\lbrack {g\left( b\right) - g\left( a\right) }\right\rbrack {f}^{\prime }\left( x\right) , \]\n\nthis implies (4.6). \( ▱ \)
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Yes
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Theorem 4.2.5 — Intermediate Value Theorem for Derivatives. Let \( a, b \in \mathbb{R} \) with \( a < b. \) Suppose \( f \) is differentiable on \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[ \n{f}_{ + }^{\prime }\left( a\right) < \lambda < {f}_{ - }^{\prime }\left( b\right) \n\]\n\nThen there exists \( c \in \left( {a, b}\right) \) such that\n\n\[ \n{f}^{\prime }\left( c\right) = \lambda \n\]
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Proof: Define the function \( g : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) by\n\n\[ \ng\left( x\right) = f\left( x\right) - {\lambda x} \n\]\n\nThen \( g \) is differentiable on \( \left\lbrack {a, b}\right\rbrack \) and\n\n\[ \n{g}_{ + }^{\prime }\left( a\right) < 0 < {g}_{ - }^{\prime }\left( b\right) \n\]\n\nThus,\n\n\[ \n\mathop{\lim }\limits_{{x \rightarrow {a}^{ + }}}\frac{g\left( x\right) - g\left( a\right) }{x - a} < 0. \n\]\n\nIt follows that there exists \( {\delta }_{1} > 0 \) such that\n\n\[ \ng\left( x\right) < g\left( a\right) \text{ for all }x \in \left( {a, a + {\delta }_{1}}\right) \cap \left\lbrack {a, b}\right\rbrack . \n\]\n\nSimilarly, there exists \( {\delta }_{2} > 0 \) such that\n\n\[ \ng\left( x\right) < g\left( b\right) \text{ for all }x \in \left( {b - {\delta }_{2}, b}\right) \cap \left\lbrack {a, b}\right\rbrack . \n\]\n\nSince \( g \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), it attains its minimum at a point \( c \in \left\lbrack {a, b}\right\rbrack \) . From the observations above, it follows that \( c \in \left( {a, b}\right) \) . This implies \( {g}^{\prime }\left( c\right) = 0 \) or, equivalently, that \( {f}^{\prime }\left( c\right) = \lambda \) . \( ▱ \)
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Yes
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Proposition 4.3.1 Let \( f \) be continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . If \( {f}^{\prime }\left( x\right) = 0 \) for all \( x \in \left( {a, b}\right) \), then \( f \) is constant on \( \left\lbrack {a, b}\right\rbrack \) .
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Proof: Suppose by contradiction that \( f \) is not constant on \( \left\lbrack {a, b}\right\rbrack \) . Then there exist \( {a}_{1} \) and \( {b}_{1} \) such that \( a \leq {a}_{1} < {b}_{1} \leq b \) and \( f\left( {a}_{1}\right) \neq f\left( {b}_{1}\right) \) . By Theorem 4.2.3, there exists \( c \in \left( {{a}_{1},{b}_{1}}\right) \) such that\n\n\[ \n{f}^{\prime }\left( c\right) = \frac{f\left( {b}_{1}\right) - f\left( {a}_{1}\right) }{{b}_{1} - {a}_{1}} \neq 0,\n\]\n\nwhich is a contradiction. \( ▱ \)
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Yes
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Proposition 4.3.2 Let \( f \) be differentiable on \( \\left( {a, b}\\right) \) .\n\n(i) If \( {f}^{\\prime }\\left( x\\right) > 0 \) for all \( x \\in \\left( {a, b}\\right) \), then \( f \) is strictly increasing on \( \\left( {a, b}\\right) \) .
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Proof: Let us prove (i). Fix any \( {x}_{1},{x}_{2} \\in \\left( {a, b}\\right) \) with \( {x}_{1} < {x}_{2} \) . By Theorem 4.2.3, there exists \( c \\in \\left( {{x}_{1},{x}_{2}}\\right) \) such that\n\n\[ \n\\frac{f\\left( {x}_{2}\\right) - f\\left( {x}_{1}\\right) }{{x}_{2} - {x}_{1}} = {f}^{\\prime }\\left( c\\right) > 0.\n\]\n\nThis implies \( f\\left( {x}_{1}\\right) < f\\left( {x}_{2}\\right) \) . Therefore, \( f \) is strictly increasing on \( \\left( {a, b}\\right) \) . The proof of (ii) is similar. \( ▱ \)
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Yes
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Theorem 4.3.3 — Inverse Function Theorem. Suppose \( f \) is differentiable on \( I = \\left( {a, b}\\right) \) and \( {f}^{\\prime }\\left( x\\right) \\neq 0 \) for all \( x \\in \\left( {a, b}\\right) \). Then \( f \) is one-to-one, \( f\\left( I\\right) \) is an open interval, and the inverse function \( {f}^{-1} : f\\left( I\\right) \\rightarrow I \) is differentiable. Moreover,\n\n\[{\\left( {f}^{-1}\\right) }^{\\prime }\\left( y\\right) = \\frac{1}{{f}^{\\prime }\\left( x\\right) }\](4.7)\n\nwhere \( f\\left( x\\right) = y \).
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Proof: It follows from Theorem 4.2.5 that\n\n\[{f}^{\\prime }\\left( x\\right) > 0\\text{for all}x \\in \\left( {a, b}\\right) \\text{, or}{f}^{\\prime }\\left( x\\right) < 0\\text{for all}x \\in \\left( {a, b}\\right) \\text{.}\]\n\nSuppose \( {f}^{\\prime }\\left( x\\right) > 0 \) for all \( x \\in \\left( {a, b}\\right) \). Then \( f \) is strictly increasing on this interval and, hence, it is one-to-one. It follows from Theorem 3.4.10 and Remark 3.4.11 that \( f\\left( I\\right) \) is an open interval and \( {f}^{-1} \) is continuous on \( f\\left( I\\right) \).\n\nIt remains to prove the differentiability of the inverse function \( {f}^{-1} \) and the representation of its derivative (4.7). Fix any \( \\bar{y} \\in f\\left( I\\right) \) with \( \\bar{y} = f\\left( \\bar{x}\\right) \). Let \( g = {f}^{-1} \). We will show that\n\n\[\\mathop{\\lim }\\limits_{{y \\rightarrow \\bar{y}}}\\frac{g\\left( y\\right) - g\\left( \\bar{y}\\right) }{y - \\bar{y}} = \\frac{1}{{f}^{\\prime }\\left( \\bar{x}\\right) }.\]\n\nFix any sequence \( \\left\{ {y}_{k}\\right\} \) in \( f\\left( I\\right) \) that converges to \( \\bar{y} \) and \( {y}_{k} \\neq \\bar{y} \) for every \( k \). For each \( {y}_{k} \), there exists \( {x}_{k} \\in I \) such that \( f\\left( {x}_{k}\\right) = {y}_{k} \). That is, \( g\\left( {y}_{k}\\right) = {x}_{k} \) for all \( k \). It follows from the continuity of \( g \) that \( \\left\{ {x}_{k}\\right\} \) converges to \( \\bar{x} \). Then\n\n\[\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}\\frac{g\\left( {y}_{k}\\right) - g\\left( \\bar{y}\\right) }{{y}_{k} - \\bar{y}} = \\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}\\frac{{x}_{k} - \\bar{x}}{f\\left( {x}_{k}\\right) - f\\left( \\bar{x}\\right) }\](4.8)\n\n\[= \\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}\\frac{1}{\\frac{f\\left( {x}_{k}\\right) - f\\left( \\bar{x}\\right) }{{x}_{k} - \\bar{x}}} = \\frac{1}{{f}^{\\prime }\\left( \\bar{x}\\right) }.\]\n\nThe proof is now complete. \( ▱ \)
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Yes
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Theorem 4.4.1 Suppose \( f \) and \( g \) are continuous on \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \) . Suppose \( f\left( \bar{x}\right) = g\left( \bar{x}\right) = 0 \), where \( \bar{x} \in \left\lbrack {a, b}\right\rbrack \) . Suppose further that there exists \( \delta > 0 \) such that \( {g}^{\prime }\left( x\right) \neq 0 \) for all \( x \in B\left( {\bar{x};\delta }\right) \cap \left\lbrack {a, b}\right\rbrack, x \neq \bar{x}. \) If \[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}\frac{{f}^{\prime }\left( x\right) }{{g}^{\prime }\left( x\right) } = \ell \] then \[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}\frac{f\left( x\right) }{g\left( x\right) } = \ell \]
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Proof: Let \( \left\{ {x}_{k}\right\} \) be a sequence in \( \left\lbrack {a, b}\right\rbrack \) that converges to \( \bar{x} \) and such that \( {x}_{k} \neq \bar{x} \) for every \( k \) . By Theorem 4.2.4, for each \( k \), there exists a sequence \( \left\{ {c}_{k}\right\} \), with \( {c}_{k} \) between \( {x}_{k} \) and \( \bar{x} \), such that \[ \left\lbrack {f\left( {x}_{k}\right) - f\left( \bar{x}\right) }\right\rbrack {g}^{\prime }\left( {c}_{k}\right) = \left\lbrack {g\left( {x}_{k}\right) - g\left( \bar{x}\right) }\right\rbrack {f}^{\prime }\left( {c}_{k}\right) . \] Since \( f\left( \bar{x}\right) = g\left( \bar{x}\right) = 0 \), and \( {g}^{\prime }\left( {c}_{k}\right) \neq 0 \) for sufficiently large \( k \), we have \[ \frac{f\left( {x}_{k}\right) }{g\left( {x}_{k}\right) } = \frac{{f}^{\prime }\left( {c}_{k}\right) }{{g}^{\prime }\left( {c}_{k}\right) }. \] Under the assumptions that \( {g}^{\prime }\left( x\right) \neq 0 \) for \( x \) near \( \bar{x} \) and \( g\left( \bar{x}\right) = 0 \), we also have \( g\left( {x}_{k}\right) \neq 0 \) for sufficiently large \( k \) . By the squeeze theorem (Theorem 2.1.6), \( \left\{ {c}_{k}\right\} \) converges to \( \bar{x} \) . Thus, \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}\frac{f\left( {x}_{k}\right) }{g\left( {x}_{k}\right) } = \mathop{\lim }\limits_{{k \rightarrow \infty }}\frac{{f}^{\prime }\left( {c}_{k}\right) }{{g}^{\prime }\left( {c}_{k}\right) } = \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}\frac{{f}^{\prime }\left( x\right) }{{g}^{\prime }\left( x\right) } = \ell . \] Therefore, (4.9) follows from Theorem 3.1.2. \( ▱ \)
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Yes
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Theorem 4.4.5 Let \( f \) and \( g \) be differentiable on \( \left( {a,\infty }\right) \) . Suppose \( {g}^{\prime }\left( x\right) \neq 0 \) for all \( x \in \left( {a,\infty }\right) \) and\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow \infty }}g\left( x\right) = \infty . \]\n\nIf \( \ell \in \mathbb{R} \) and\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{{f}^{\prime }\left( x\right) }{{g}^{\prime }\left( x\right) } = \ell \]\n\nthen\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{f\left( x\right) }{g\left( x\right) } = \ell \]
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Example 4.4.7 Consider the limit\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{\ln x}{x} \]\n\nClearly the functions \( f\left( x\right) = \ln x \) and \( g\left( x\right) = x \) satisfy the conditions of Theorem 4.4.5. We have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{{f}^{\prime }\left( x\right) }{{g}^{\prime }\left( x\right) } = \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{1/x}{1} = 0 \]\n\nIt follows from Theorem 4.4.5 that \( \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{\ln x}{x} = 0 \) .
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No
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Theorem 4.5.1 - Taylor’s Theorem. Let \( n \) be a positive integer. Suppose \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is a function such that \( {f}^{\left( n\right) } \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), and \( {f}^{\left( n + 1\right) }\left( x\right) \) exists for all \( x \in \left( {a, b}\right) \) . Let \( \bar{x} \in \left\lbrack {a, b}\right\rbrack \) . Then for any \( x \in \left\lbrack {a, b}\right\rbrack \) with \( x \neq \bar{x} \), there exists a number \( c \) in between \( \bar{x} \) and \( x \) such that\n\n\[ f\left( x\right) = {P}_{n}\left( x\right) + \frac{{f}^{\left( n + 1\right) }\left( c\right) }{\left( {n + 1}\right) !}{\left( x - \bar{x}\right) }^{n + 1}, \]\n\nwhere\n\n\[ {P}_{n}\left( x\right) = \mathop{\sum }\limits_{{k = 0}}^{n}\frac{{f}^{\left( k\right) }\left( \bar{x}\right) }{k!}{\left( x - \bar{x}\right) }^{k}. \]
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Proof: Let \( \bar{x} \) be as in the statement and let us fix \( x \neq \bar{x} \) . Since \( x - \bar{x} \neq 0 \), there exists a number \( \lambda \in \mathbb{R} \) such that\n\n\[ f\left( x\right) = {P}_{n}\left( x\right) + \frac{\lambda }{\left( {n + 1}\right) !}{\left( x - \bar{x}\right) }^{n + 1}. \]\n\nWe will now show that\n\n\[ \lambda = {f}^{\left( n + 1\right) }\left( c\right) \]\nfor some \( c \) in between \( \bar{x} \) and \( x \) .\n\nConsider the function\n\n\[ g\left( t\right) = f\left( x\right) - \mathop{\sum }\limits_{{k = 0}}^{n}\frac{{f}^{\left( k\right) }\left( t\right) }{k!}{\left( x - t\right) }^{k} - \frac{\lambda }{\left( {n + 1}\right) !}{\left( x - t\right) }^{n + 1}. \]\n\nThen\n\n\[ g\left( \bar{x}\right) = f\left( x\right) - \mathop{\sum }\limits_{{k = 0}}^{n}\frac{{f}^{\left( k\right) }\left( \bar{x}\right) }{k!}{\left( x - \bar{x}\right) }^{k} - \frac{\lambda }{\left( {n + 1}\right) !}{\left( x - \bar{x}\right) }^{n + 1} = f\left( x\right) - {P}_{n}\left( x\right) - \frac{\lambda }{\left( {n + 1}\right) !}{\left( x - \bar{x}\right) }^{n + 1} = 0. \]\n\nand\n\n\[ g\left( x\right) = f\left( x\right) - \mathop{\sum }\limits_{{k = 0}}^{n}\frac{{f}^{\left( k\right) }\left( x\right) }{k!}{\left( x - x\right) }^{k} - \frac{\lambda }{\left( {n + 1}\right) !}{\left( x - x\right) }^{n + 1} = f\left( x\right) - f\left( x\right) = 0. \]\n\nBy Rolle’s theorem, there exists \( c \) in between \( \bar{x} \) and \( x \) such that \( {g}^{\prime }\left( c\right) = 0 \) . Taking the derivative of \( g \) (keeping in mind that \( x \) is fixed and the independent variable is \( t \) ) and using the product rule for derivatives, we have\n\n\[ {g}^{\prime }\left( c\right) = - {f}^{\prime }\left( c\right) + \mathop{\sum }\limits_{{k = 1}}^{n}\left( {-\frac{{f}^{\left( k + 1\right) }\left( c\right) }{k!}{\left( x - c\right) }^{k} + \frac{{f}^{\left( k\right) }\left( c\right) }{\left( {k - 1}\right) !}{\left( x - c\right) }^{k - 1}}\right) + \frac{\lambda }{n!}{\left( x - c\right) }^{n} \]\n\n\[ = \frac{\lambda }{n!}{\left( x - c\right) }^{n} - \frac{1}{n!}{f}^{\left( n + 1\right) }\left( c\right) {\left( x - c\right) }^{n} \]\n\n\[ = 0\text{.} \]\n\nThis implies \( \lambda = {f}^{\left( n + 1\right) }\left( c\right) \) . The proof is now complete. \( ▱ \)
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Yes
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Theorem 4.5.3 Let \( n \) be an even positive integer. Suppose \( {f}^{\left( n\right) } \) exists and continuous on \( \left( {a, b}\right) \) . Let \( \bar{x} \in \left( {a, b}\right) \) satisfy\n\n\[ \n{f}^{\prime }\left( \bar{x}\right) = \ldots = {f}^{\left( n - 1\right) }\left( \bar{x}\right) = 0\text{ and }{f}^{\left( n\right) }\left( \bar{x}\right) \neq 0.\n\]\n\nThe following hold:\n\n(a) \( {f}^{\left( n\right) }\left( \bar{x}\right) > 0 \) if and only if \( f \) has a local minimum at \( \bar{x} \) .\n\n(b) \( {f}^{\left( n\right) }\left( \bar{x}\right) < 0 \) if and only if \( f \) has a local maximum at \( \bar{x} \) .
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Proof: We will prove (a). Suppose \( {f}^{\left( n\right) }\left( \bar{x}\right) > 0 \) . Since \( {f}^{\left( n\right) }\left( \bar{x}\right) > 0 \) and \( {f}^{\left( n\right) } \) is continuous at \( \bar{x} \), there exists \( \delta > 0 \) such that\n\n\[ \n{f}^{\left( n\right) }\left( t\right) > 0\text{ for all }t \in B\left( {\bar{x};\delta }\right) \subset \left( {a, b}\right) .\n\]\n\nFix any \( x \in B\left( {\bar{x};\delta }\right) \) . By Taylor’s theorem and the given assumption, there exists \( c \) in between \( \bar{x} \) and \( x \) such that\n\n\[ \nf\left( x\right) = f\left( \bar{x}\right) + \frac{{f}^{\left( n\right) }\left( c\right) }{n!}{\left( x - \bar{x}\right) }^{n}.\n\]\n\nSince \( n \) is even and \( c \in B\left( {\bar{x};\delta }\right) \), we have \( f\left( x\right) \geq f\left( \bar{x}\right) \) . Thus, \( f \) has a local minimum at \( \bar{x} \) .\n\nNow, for the converse, suppose that \( f \) has a local minimum at \( \bar{x} \) . Then there exists \( \delta > 0 \) such that\n\n\[ \nf\left( x\right) \geq f\left( \bar{x}\right) \text{ for all }x \in B\left( {\bar{x};\delta }\right) \subset \left( {a, b}\right) .\n\]\n\nFix a sequence \( \left\{ {x}_{k}\right\} \) in \( \left( {a, b}\right) \) that converges to \( \bar{x} \) with \( {x}_{k} \neq \bar{x} \) for every \( k \) . By Taylor’s theorem, there exists a sequence \( \left\{ {c}_{k}\right\} \), with \( {c}_{k} \) between \( {x}_{k} \) and \( \bar{x} \) for each \( k \), such that\n\n\[ \nf\left( {x}_{k}\right) = f\left( \bar{x}\right) + \frac{{f}^{\left( n\right) }\left( {c}_{k}\right) }{n!}{\left( {x}_{k} - \bar{x}\right) }^{n}.\n\]\n\nSince \( {x}_{k} \in B\left( {\bar{x};\delta }\right) \) for sufficiently large \( k \), we have\n\n\[ \nf\left( {x}_{k}\right) \geq f\left( \bar{x}\right) \n\]\n\nfor such \( k \) . It follows that\n\n\[ \nf\left( {x}_{k}\right) - f\left( \bar{x}\right) = \frac{{f}^{\left( n\right) }\left( {c}_{k}\right) }{n!}{\left( {x}_{k} - \bar{x}\right) }^{n} \geq 0.\n\]\n\nThis implies \( {f}^{\left( n\right) }\left( {c}_{k}\right) \geq 0 \) for such \( k \) . Since \( \left\{ {c}_{k}\right\} \) converges to \( \bar{x},{f}^{\left( n\right) }\left( \bar{x}\right) = \mathop{\lim }\limits_{{k \rightarrow \infty }}{f}^{\left( n\right) }\left( {c}_{k}\right) \geq 0 \) .\n\nThe proof of (b) is similar.
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Yes
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Theorem 4.6.1 Let \( I \) be an interval of \( \mathbb{R} \) . A function \( f : I \rightarrow \mathbb{R} \) is convex if and only if for every \( {\lambda }_{i} \geq 0, i = 1,\ldots, n \), with \( \mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{i} = 1\left( {n \geq 2}\right) \) and for every \( {x}_{i} \in I, i = 1,\ldots, n \) , \[ f\left( {\mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{i}{x}_{i}}\right) \leq \mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{i}f\left( {x}_{i}\right) \]
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Proof: Since the converse holds trivially, we only need to prove the implication by induction. The conclusion holds for \( n = 2 \) by the definition of convexity. Let \( k \) be such that the conclusion holds for any \( n \) with \( 2 \leq n \leq k \) . We will show that it also holds for \( n = k + 1 \) . Fix \( {\lambda }_{i} \geq 0, i = 1,\ldots, k + 1 \), with \( \mathop{\sum }\limits_{{i = 1}}^{{k + 1}}{\lambda }_{i} = 1 \) and fix every \( {x}_{i} \in I, i = 1,\ldots, k + 1 \) . Then \[ \mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i} = 1 - {\lambda }_{k + 1} \] If \( {\lambda }_{k + 1} = 1 \), then \( {\lambda }_{i} = 0 \) for all \( i = 1,\ldots, k \), and (4.13) holds. Suppose \( 0 \leq {\lambda }_{k + 1} < 1 \) . Then, for each \( i = 1,\ldots, k,{\lambda }_{i}/\left( {1 - {\lambda }_{k + 1}}\right) \geq 0 \) and \[ \mathop{\sum }\limits_{{i = 1}}^{k}\frac{{\lambda }_{i}}{1 - {\lambda }_{k + 1}} = 1 \] It follows that \[ f\left( {\mathop{\sum }\limits_{{i = 1}}^{{k + 1}}{\lambda }_{i}{x}_{i}}\right) = f\left\lbrack {\left( {1 - {\lambda }_{k + 1}}\right) \frac{\mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{x}_{i}}{1 - {\lambda }_{k + 1}} + {\lambda }_{k + 1}{x}_{k + 1}}\right\rbrack \] \[ \leq \left( {1 - {\lambda }_{k + 1}}\right) f\left( \frac{\mathop{\sum }\limits_{{i = 1}}^{k}{\lambda }_{i}{x}_{i}}{1 - {\lambda }_{k + 1}}\right) + {\lambda }_{k + 1}f\left( {x}_{k + 1}\right) \] \[ = \left( {1 - {\lambda }_{k + 1}}\right) f\left( {\mathop{\sum }\limits_{{i = 1}}^{k}\frac{{\lambda }_{i}}{1 - {\lambda }_{k + 1}}{x}_{i}}\right) + {\lambda }_{k + 1}f\left( {x}_{k + 1}\right) \] \[ \leq \left( {1 - {\lambda }_{k + 1}}\right) \mathop{\sum }\limits_{{i = 1}}^{k}\frac{{\lambda }_{i}}{1 - {\lambda }_{k + 1}}f\left( {x}_{i}\right) + {\lambda }_{k + 1}f\left( {x}_{k + 1}\right) \] \[ = \mathop{\sum }\limits_{{i = 1}}^{{k + 1}}{\lambda }_{i}f\left( {x}_{i}\right) \] where the first inequality follows from the definition of convexity (or is trivial if \( {\lambda }_{k + 1} = 0 \) ) and the last inequality follows from the inductive assumption. The proof is now complete. \( ▱ \)
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Yes
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Theorem 4.6.2 Let \( I \) be an interval and let \( f : I \rightarrow \mathbb{R} \) be a convex function. Then \( f \) has a local minimum at \( \bar{x} \) if and only if \( f \) has an absolute minimum at \( \bar{x} \) .
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Proof: Clearly if \( f \) has a global minimum at \( \bar{x} \), then it also has a local minimum at \( \bar{x} \). Conversely, suppose that \( f \) has a local minimum at \( \bar{x} \). Then there exists \( \delta > 0 \) such that \[ f\left( u\right) \geq f\left( \bar{x}\right) \text{ for all }u \in B\left( {\bar{x};\delta }\right) \cap I. \] For any \( x \in I \), we have \( {x}_{n} = \left( {1 - \frac{1}{n}}\right) \bar{x} + \frac{1}{n}x \rightarrow \bar{x} \). Thus, \( {x}_{n} \in B\left( {\bar{x};\delta }\right) \cap I \) when \( n \) is sufficiently large. Thus, for such \( n \), \[ f\left( \bar{x}\right) \leq f\left( {x}_{n}\right) \leq \left( {1 - \frac{1}{n}}\right) f\left( \bar{x}\right) + \frac{1}{n}f\left( x\right) . \] This implies that for a sufficient large \( n \), we have \[ \frac{1}{n}f\left( \bar{x}\right) \leq \frac{1}{n}f\left( x\right) \] and, hence, \( f\left( \bar{x}\right) \leq f\left( x\right) \). Since \( x \) was arbitrary, this shows \( f \) has an absolute minimum at \( \bar{x} \). \( ▱ \)
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Yes
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Theorem 4.6.3 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a convex function. Suppose \( f \) is differentiable at \( \bar{x} \) . Then\n\n\[ \n{f}^{\prime }\left( \bar{x}\right) \left( {x - \bar{x}}\right) \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x \in I.\n\]
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Proof: For any \( x \in I \) and \( t \in \left( {0,1}\right) \), we have\n\n\[ \n\frac{f\left( {\bar{x} + t\left( {x - \bar{x}}\right) }\right) - f\left( \bar{x}\right) }{t} = \frac{f\left( {{tx} + \left( {1 - t}\right) \bar{x}}\right) - f\left( \bar{x}\right) }{t}\n\]\n\n\[ \n\leq \frac{{tf}\left( x\right) + \left( {1 - t}\right) f\left( \bar{x}\right) - f\left( \bar{x}\right) }{t}\n\]\n\n\[ \n= f\left( x\right) - f\left( \bar{x}\right) .\n\]\n\nSince \( f \) is differentiable at \( \bar{x} \) ,\n\n\[ \n{f}^{\prime }\left( \bar{x}\right) \left( {x - \bar{x}}\right) = \mathop{\lim }\limits_{{t \rightarrow {0}^{ + }}}\frac{f\left( {\bar{x} + t\left( {x - \bar{x}}\right) }\right) - f\left( \bar{x}\right) }{t} \leq f\left( x\right) - f\left( \bar{x}\right) ,\n\]\n\nwhich completes the proof. \( ▱ \)
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Yes
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Corollary 4.6.4 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a convex function. Suppose \( f \) is differentiable at \( \bar{x} \) . Then \( f \) has an absolute minimum at \( \bar{x} \) if and only if \( {f}^{\prime }\left( \bar{x}\right) = 0 \) .
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Proof: Suppose \( f \) has an absolute minimum at \( \bar{x} \) . By Theorem 4.2.1, \( {f}^{\prime }\left( \bar{x}\right) = 0 \) . Let us prove the converse. Suppose \( {f}^{\prime }\left( \bar{x}\right) = 0 \) . It follows from Theorem 4.6.3 that\n\n\[ 0 = {f}^{\prime }\left( \bar{x}\right) \left( {x - \bar{x}}\right) \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x \in I. \]\n\nThis implies\n\n\[ f\left( x\right) \geq f\left( \bar{x}\right) \text{ for all }x \in I. \]\n\nThus, \( f \) has an absolute minimum at \( \bar{x} \) . \( ▱ \)
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Yes
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Lemma 4.6.5 Let \( I \) be an open interval and suppose \( f : I \rightarrow \mathbb{R} \) is a convex function. Fix \( a, b, x \in I \) with \( a < x < b \) . Then\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \leq \frac{f\left( b\right) - f\left( a\right) }{b - a} \leq \frac{f\left( b\right) - f\left( x\right) }{b - x}. \]\n
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Proof: Let\n\n\[ t = \frac{x - a}{b - a} \]\n\nThen \( t \in \left( {0,1}\right) \) and\n\n\[ f\left( x\right) = f\left( {a + \left( {x - a}\right) }\right) = f\left( {a + \frac{x - a}{b - a}\left( {b - a}\right) }\right) = f\left( {a + t\left( {b - a}\right) }\right) = f\left( {{tb} + \left( {1 - t}\right) a}\right) . \]\n\nBy convexity of \( f \), we obtain\n\n\[ f\left( x\right) \leq {tf}\left( b\right) + \left( {1 - t}\right) f\left( a\right) . \]\n\nThus,\n\n\[ f\left( x\right) - f\left( a\right) \leq {tf}\left( b\right) + \left( {1 - t}\right) f\left( a\right) - f\left( a\right) = t\left\lbrack {f\left( b\right) - f\left( a\right) }\right\rbrack = \frac{x - a}{b - a}\left( {f\left( b\right) - f\left( a\right) }\right) . \]\nEquivalently,\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \leq \frac{f\left( b\right) - f\left( a\right) }{b - a}. \]\n\nSimilarly,\n\n\[ f\left( x\right) - f\left( b\right) \leq {tf}\left( b\right) + \left( {1 - t}\right) f\left( a\right) - f\left( b\right) = \left( {1 - t}\right) \left\lbrack {f\left( a\right) - f\left( b\right) }\right\rbrack = \frac{x - b}{b - a}\left\lbrack {f\left( b\right) - f\left( a\right) }\right\rbrack . \]\n\nIt follows that\n\n\[ \frac{f\left( b\right) - f\left( a\right) }{b - a} \leq \frac{f\left( b\right) - f\left( x\right) }{b - x}. \]\n\nThe proof is now complete.
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Yes
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Theorem 4.6.6 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a differentiable function. Then \( f \) is convex if and only if \( {f}^{\prime } \) is increasing on \( I \) .
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Proof: Suppose \( f \) is convex. Fix \( a < b \) with \( a, b \in I \) . By Lemma 4.6.5, for any \( x \in \left( {a, b}\right) \), we have\n\n\[ \frac{f\left( x\right) - f\left( a\right) }{x - a} \leq \frac{f\left( b\right) - f\left( a\right) }{b - a}. \]\n\nThis implies, taking limits, that\n\n\[ {f}^{\prime }\left( a\right) \leq \frac{f\left( b\right) - f\left( a\right) }{b - a}. \]\n\nSimilarly,\n\n\[ \frac{f\left( b\right) - f\left( a\right) }{b - a} \leq {f}^{\prime }\left( b\right) \]\n\nTherefore, \( {f}^{\prime }\left( a\right) \leq {f}^{\prime }\left( b\right) \), and \( {f}^{\prime } \) is an increasing function.\n\nLet us prove the converse. Suppose \( {f}^{\prime } \) is increasing. Fix \( {x}_{1} < {x}_{2} \) and \( t \in \left( {0,1}\right) \) . Then\n\n\[ {x}_{1} < {x}_{t} < {x}_{2} \]\n\nwhere \( {x}_{t} = t{x}_{1} + \left( {1 - t}\right) {x}_{2} \) . By the Mean Value Theorem (Theorem 4.2.3), there exist \( {c}_{1} \) and \( {c}_{2} \) such that\n\n\[ {x}_{1} < {c}_{1} < {x}_{t} < {c}_{2} < {x}_{2} \]\n\nwith\n\n\[ f\left( {x}_{t}\right) - f\left( {x}_{1}\right) = {f}^{\prime }\left( {c}_{1}\right) \left( {{x}_{t} - {x}_{1}}\right) = {f}^{\prime }\left( {c}_{1}\right) \left( {1 - t}\right) \left( {{x}_{2} - {x}_{1}}\right) ; \]\n\n\[ f\left( {x}_{t}\right) - f\left( {x}_{2}\right) = {f}^{\prime }\left( {c}_{2}\right) \left( {{x}_{t} - {x}_{2}}\right) = {f}^{\prime }\left( {c}_{2}\right) t\left( {{x}_{1} - {x}_{2}}\right) . \]\n\nThis implies\n\n\[ {tf}\left( {x}_{t}\right) - {tf}\left( {x}_{1}\right) = {f}^{\prime }\left( {c}_{1}\right) t\left( {1 - t}\right) \left( {{x}_{2} - {x}_{1}}\right) \]\n\n\[ \left( {1 - t}\right) f\left( {x}_{t}\right) - \left( {1 - t}\right) f\left( {x}_{2}\right) = {f}^{\prime }\left( {c}_{2}\right) t\left( {1 - t}\right) \left( {{x}_{1} - {x}_{2}}\right) . \]\n\nSince \( {f}^{\prime }\left( {c}_{1}\right) \leq {f}^{\prime }\left( {c}_{2}\right) \), we have\n\n\[ {tf}\left( {x}_{t}\right) - {tf}\left( {x}_{1}\right) = {f}^{\prime }\left( {c}_{1}\right) t\left( {1 - t}\right) \left( {{x}_{2} - {x}_{1}}\right) \leq {f}^{\prime }\left( {c}_{2}\right) t\left( {1 - t}\right) \left( {{x}_{2} - {x}_{1}}\right) = \left( {1 - t}\right) f\left( {x}_{2}\right) - \left( {1 - t}\right) f\left( {x}_{t}\right) . \]\n\nRearranging terms, we get\n\n\[ f\left( {x}_{t}\right) \leq {tf}\left( {x}_{1}\right) + \left( {1 - t}\right) f\left( {x}_{2}\right) \]\n\nTherefore, \( f \) is convex. The proof is now complete.
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Yes
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Corollary 4.6.7 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a function. Suppose \( f \) is twice differentiable on \( I \) . Then \( f \) is convex if and only if \( {f}^{\prime \prime }\left( x\right) \geq 0 \) for all \( x \in I \) .
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Proof: It follows from Proposition 4.3.2 that \( {f}^{\prime \prime }\left( x\right) \geq 0 \) for all \( x \in I \) if and only if the derivative function \( {f}^{\prime } \) is increasing on \( I \) . The conclusion then follows directly from Theorem 4.6.6. \( ▱ \)
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Yes
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Theorem 4.6.8 Let \( I \) be an open interval and let \( f : I \rightarrow \mathbb{R} \) be a convex function. Then it is locally Lipschitz continuous in the sense that for any \( \bar{x} \in I \), there exist \( \ell \geq 0 \) and \( \delta > 0 \) such that\n\n\[ \left| {f\left( u\right) - f\left( v\right) }\right| \leq \ell \left| {u - v}\right| \text{ for all }u, v \in B\left( {\bar{x};\delta }\right) . \]\n\n(4.15)\n\nIn particular, \( f \) is continuous.
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Proof: Fix any \( \bar{x} \in I \) . Choose four numbers \( a, b, c, d \) satisfying\n\n\[ a < b < \bar{x} < c < d\text{with}a, d \in I\text{.} \]\n\nChoose \( \delta > 0 \) such that \( B\left( {\bar{x};\delta }\right) \subset \left( {b, c}\right) \) . Let \( u, v \in B\left( {\bar{x};\delta }\right) \) with \( v < u \) . Then by Lemma 4.6.5, we see that\n\n\[ \frac{f\left( b\right) - f\left( a\right) }{b - a} \leq \frac{f\left( u\right) - f\left( a\right) }{u - a} \leq \frac{f\left( u\right) - f\left( v\right) }{u - v} \leq \frac{f\left( d\right) - f\left( v\right) }{d - v} \leq \frac{f\left( d\right) - f\left( c\right) }{d - c}. \]\n\nUsing a similar approach for the case \( u < v \), we get\n\n\[ \frac{f\left( b\right) - f\left( a\right) }{b - a} \leq \frac{f\left( u\right) - f\left( v\right) }{u - v} \leq \frac{f\left( d\right) - f\left( c\right) }{d - c}\text{ for all }u, v \in B\left( {\bar{x};\delta }\right) . \]\n\nChoose \( \ell \geq 0 \) sufficiently large so that\n\n\[ - \ell \leq \frac{f\left( b\right) - f\left( a\right) }{b - a} \leq \frac{f\left( u\right) - f\left( v\right) }{u - v} \leq \frac{f\left( d\right) - f\left( c\right) }{d - c} \leq \ell \text{ for all }u, v \in B\left( {\bar{x};\delta }\right) . \]\n\nThen (4.15) holds. The proof is now complete.
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Yes
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Lemma 4.7.1 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function. Fix \( a \in \mathbb{R} \) . Define the slope function \( {\phi }_{a} \) by\n\n\[ \n{\phi }_{a}\left( x\right) = \frac{f\left( x\right) - f\left( a\right) }{x - a} \]\n\n(4.17)\n\nfor \( x \in \left( {-\infty, a}\right) \cup \left( {a,\infty }\right) \) . Then, for \( {x}_{1},{x}_{2} \in \left( {-\infty, a}\right) \cup \left( {a,\infty }\right) \) with \( {x}_{1} < {x}_{2} \), we have\n\n\[ \n{\phi }_{a}\left( {x}_{1}\right) \leq {\phi }_{a}\left( {x}_{2}\right) \n\]
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Proof: This lemma follows directly from Lemma 4.6.5. \( ▱ \)
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No
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Theorem 4.7.2 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function and let \( \bar{x} \in \mathbb{R} \) . Then \( f \) has left derivative and right derivative at \( \bar{x} \) . Moreover, \[ \mathop{\sup }\limits_{{x < \bar{x}}}{\phi }_{\bar{x}}\left( x\right) = {f}_{ - }^{\prime }\left( \bar{x}\right) \leq {f}_{ + }^{\prime }\left( \bar{x}\right) = \mathop{\inf }\limits_{{x > \bar{x}}}{\phi }_{\bar{x}}\left( x\right) , \] where \( {\phi }_{\bar{x}} \) is defined in (4.17).
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Proof: By Lemma 4.7.1, the slope function \( {\phi }_{\bar{x}} \) defined by (4.17) is increasing on the interval \( \left( {\bar{x},\infty }\right) \) and bounded below by \( {\phi }_{\bar{x}}\left( {\bar{x} - 1}\right) \) . By Theorem 3.2.4, the limit \[ \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ + }}}{\phi }_{\bar{x}}\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ + }}}\frac{f\left( x\right) - f\left( \bar{x}\right) }{x - \bar{x}} \] exists and is finite. Moreover, \[ \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ + }}}{\phi }_{\bar{x}}\left( x\right) = \mathop{\inf }\limits_{{x > \bar{x}}}{\phi }_{\bar{x}}\left( x\right) \] Thus, \( {f}_{ + }^{\prime }\left( \bar{x}\right) \) exists and \[ {f}_{ + }^{\prime }\left( \bar{x}\right) = \mathop{\inf }\limits_{{x > \bar{x}}}{\phi }_{\bar{x}}\left( x\right) \] Similarly, \( {f}_{ - }^{\prime }\left( \bar{x}\right) \) exists and \[ {f}_{ - }^{\prime }\left( \bar{x}\right) = \mathop{\sup }\limits_{{x < \bar{x}}}{\phi }_{\bar{x}}\left( x\right) \] Applying Lemma 4.7.1 again, we see that \[ {\phi }_{\bar{x}}\left( x\right) \leq {\phi }_{\bar{x}}\left( y\right) \text{whenever}x < \bar{x} < y\text{.} \] This implies \( {f}_{ - }^{\prime }\left( \bar{x}\right) \leq {f}_{ + }^{\prime }\left( \bar{x}\right) \) . The proof is complete. \( ▱ \)
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Yes
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Theorem 4.7.3 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function and let \( \bar{x} \in \mathbb{R} \) . Then\n\n\[ \partial f\left( \bar{x}\right) = \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack . \]
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Proof: Suppose \( u \in \partial f\left( \bar{x}\right) \) . By the definition (4.16), we have\n\n\[ u \cdot \left( {x - \bar{x}}\right) \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x > \bar{x}. \]\n\nThis implies\n\n\[ u \leq \frac{f\left( x\right) - f\left( \bar{x}\right) }{x - \bar{x}}\text{ for all }x > \bar{x} \]\n\nThus,\n\n\[ u \leq \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ + }}}\frac{f\left( x\right) - f\left( \bar{x}\right) }{x - \bar{x}} = {f}_{ + }^{\prime }\left( \bar{x}\right) . \]\n\nSimilarly, we have\n\n\[ u \cdot \left( {x - \bar{x}}\right) \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x < \bar{x}. \]\n\nThus,\n\n\[ u \geq \frac{f\left( x\right) - f\left( \bar{x}\right) }{x - \bar{x}}\text{ for all }x < \bar{x} \]\n\nThis implies \( u \geq {f}_{ - }^{\prime }\left( \bar{x}\right) \) . So\n\n\[ \partial f\left( \bar{x}\right) \subset \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack . \]\n\nTo prove the opposite inclusion, take \( u \in \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack \) . By Theorem 4.7.2\n\n\[ \mathop{\sup }\limits_{{x < \bar{x}}}{\phi }_{\bar{x}}\left( x\right) = {f}_{ - }^{\prime }\left( \bar{x}\right) \leq u \leq {f}_{ + }^{\prime }\left( \bar{x}\right) = \mathop{\inf }\limits_{{x > \bar{x}}}{\phi }_{\bar{x}}\left( x\right) . \]\n\nUsing the upper estimate by \( {f}_{ + }^{\prime }\left( \bar{x}\right) \) for \( u \), one has\n\n\[ u \leq {\phi }_{\bar{x}}\left( x\right) = \frac{f\left( x\right) - f\left( \bar{x}\right) }{x - \bar{x}}\text{ for all }x > \bar{x}. \]\n\nIt follows that\n\n\[ u \cdot \left( {x - \bar{x}}\right) \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x \geq \bar{x}. \]\n\nSimilarly, one also has\n\n\[ u \cdot \left( {x - \bar{x}}\right) \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x < \bar{x}. \]\n\nThus,(4.16) holds and, hence, \( u \in \partial f\left( \bar{x}\right) \) . Therefore,(4.18) holds.
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Yes
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Corollary 4.7.4 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function and \( \bar{x} \in \mathbb{R} \) . Then \( f \) is differentiable at \( \bar{x} \) if and only if \( \partial f\left( \bar{x}\right) \) is a singleton. In this case, \[ \partial f\left( \bar{x}\right) = \left\{ {{f}^{\prime }\left( \bar{x}\right) }\right\} \]
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Proof: Suppose \( f \) is differentiable at \( \bar{x} \) . Then \[ {f}_{ - }^{\prime }\left( \bar{x}\right) = {f}_{ + }^{\prime }\left( \bar{x}\right) = {f}^{\prime }\left( \bar{x}\right) . \] By Theorem 4.7.3, \[ \partial f\left( \bar{x}\right) = \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack = \left\{ {{f}^{\prime }\left( \bar{x}\right) }\right\} . \] Thus, \( \partial f\left( \bar{x}\right) \) is a singleton. Conversely, if \( \partial f\left( \bar{x}\right) \) is a singleton, we must have \( {f}_{ - }^{\prime }\left( \bar{x}\right) = {f}_{ + }^{\prime }\left( \bar{x}\right) \) . Thus, \( f \) is differentiable at \( \bar{x} \) .
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Yes
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Theorem 4.7.5 Let \( f, g : \mathbb{R} \rightarrow \mathbb{R} \) be convex functions and let \( \alpha > 0 \) . Then \( f + g \) and \( {\alpha f} \) are convex functions and\n\n\[ \partial \left( {f + g}\right) \left( \bar{x}\right) = \partial f\left( \bar{x}\right) + \partial g\left( \bar{x}\right) \]\n\n\[ \partial \left( {\alpha f}\right) \left( \bar{x}\right) = \alpha \partial f\left( \bar{x}\right) . \]
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Proof: It is not hard to see that \( f + g \) is a convex function and\n\n\[ {\left( f + g\right) }_{ + }^{\prime }\left( \bar{x}\right) = {f}_{ + }^{\prime }\left( \bar{x}\right) + {g}_{ + }^{\prime }\left( \bar{x}\right) \]\n\n\[ {\left( f + g\right) }_{ - }^{\prime }\left( \bar{x}\right) = {f}_{ - }^{\prime }\left( \bar{x}\right) + {g}_{ - }^{\prime }\left( \bar{x}\right) . \]\n\nBy Theorem 4.7.3,\n\n\[ \partial \left( {f + g}\right) \left( \bar{x}\right) = \left\lbrack {{\left( f + g\right) }_{ - }^{\prime }\left( \bar{x}\right) ,{\left( f + g\right) }_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack \]\n\n\[ = \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) + {g}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) + {g}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack \]\n\n\[ = \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack + \left\lbrack {{g}_{ - }^{\prime }\left( \bar{x}\right) ,{g}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack \]\n\n\[ = \partial f\left( \bar{x}\right) + \partial g\left( \bar{x}\right) \]\n\nThe proof for the second formula is similar. \( ▱ \)
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Yes
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Theorem 4.7.6 Let \( {f}_{i} : \mathbb{R} \rightarrow \mathbb{R}, i = 1,\ldots, n \), be convex functions. Define\n\n\[ f\left( x\right) = \max \left\{ {{f}_{i}\left( x\right) : i = 1,\ldots, n}\right\} \text{and}I\left( u\right) = \left\{ {i = 1,\ldots, n : {f}_{i}\left( u\right) = f\left( u\right) }\right\} \text{.} \]\n\nThen \( f \) is a convex function. Moreover,\n\n\[ \partial f\left( \bar{x}\right) = \left\lbrack {m, M}\right\rbrack \]\n\nwhere\n\n\[ m = \mathop{\min }\limits_{{i \in I\left( \bar{x}\right) }}{f}_{i - }^{\prime }\left( \bar{x}\right) \text{ and }M = \mathop{\max }\limits_{{i \in I\left( \bar{x}\right) }}{f}_{i + }^{\prime }\left( \bar{x}\right) . \]
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Proof: Fix \( u, v \in \mathbb{R} \) and \( \lambda \in \left( {0,1}\right) \) . For any \( i = 1,\ldots, n \), we have\n\n\[ {f}_{i}\left( {{\lambda u} + \left( {1 - \lambda }\right) v}\right) \leq \lambda {f}_{i}\left( u\right) + \left( {1 - \lambda }\right) {f}_{i}\left( v\right) \leq {\lambda f}\left( u\right) + \left( {1 - \lambda }\right) f\left( v\right) .\n\nThis implies\n\n\[ f\left( {{\lambda u} + \left( {1 - \lambda }\right) v}\right) = \mathop{\max }\limits_{{1 \leq i \leq n}}{f}_{i}\left( {{\lambda u} + \left( {1 - \lambda }\right) v}\right) \leq {\lambda f}\left( u\right) + \left( {1 - \lambda }\right) f\left( v\right) .\n\nThus, \( f \) is a convex function. Similarly we verify that \( {f}_{ + }^{\prime }\left( \bar{x}\right) = M \) and \( {f}_{ - }^{\prime }\left( \bar{x}\right) = m \) . By Theorem 4.7.3,\n\n\[ \partial f\left( \bar{x}\right) = \left\lbrack {m, M}\right\rbrack \]\n\nThe proof is now complete. \( ▱ \)
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Yes
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Theorem 4.7.8 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function. Then \( f \) has an absolute minimum at \( \bar{x} \) if and only if\n\n\[ 0 \in \partial f\left( \bar{x}\right) = \left\lbrack {{f}_{ - }^{\prime }\left( \bar{x}\right) ,{f}_{ + }^{\prime }\left( \bar{x}\right) }\right\rbrack . \]\n
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Proof: Suppose \( f \) has an absolute minimum at \( \bar{x} \) . Then\n\n\[ f\left( \bar{x}\right) \leq f\left( x\right) \text{ for all }x \in \mathbb{R}. \]\n\nThis implies\n\n\[ 0 \cdot \left( {x - \bar{x}}\right) = 0 \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x \in \mathbb{R}. \]\n\nIt follows from (4.16) that \( 0 \in \partial f\left( \bar{x}\right) \) .\n\nConversely, if \( 0 \in \partial f\left( \bar{x}\right) \), again, by (4.16),\n\n\[ 0 \cdot \left( {x - \bar{x}}\right) = 0 \leq f\left( x\right) - f\left( \bar{x}\right) \text{ for all }x \in \mathbb{R}. \]\n\nThus, \( f \) has an absolute minimum at \( \bar{x} \) . \( ▱ \)
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Yes
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Theorem 4.7.9 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function and let \( a < b \) . Then there exists \( c \in \left( {a, b}\right) \) such that\n\n\[ \frac{f\left( b\right) - f\left( a\right) }{b - a} \in \partial f\left( c\right) . \]
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Proof: Define\n\n\[ g\left( x\right) = f\left( x\right) - \left\lbrack {\frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - a}\right) + f\left( a\right) }\right\rbrack . \]\n\nThen \( g \) is a convex function and \( g\left( a\right) = g\left( b\right) \) . Thus, \( g \) has a local minimum at some \( c \in \left( {a, b}\right) \) and, hence, \( g \) also has an absolute minimum at \( c \) . Observe that the function\n\n\[ h\left( x\right) = - \left\lbrack {\frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - a}\right) + f\left( a\right) }\right\rbrack \]\n\nis differentiable at \( c \) and, hence,\n\n\[ \partial h\left( c\right) = \left\{ {{h}^{\prime }\left( c\right) }\right\} = \left\{ {-\frac{f\left( b\right) - f\left( a\right) }{b - a}}\right\} . \]\n\nBy Theorem 4.7.8 and the subdifferential sum rule,\n\n\[ 0 \in \partial g\left( c\right) = \partial f\left( c\right) - \left\{ \frac{f\left( b\right) - f\left( a\right) }{b - a}\right\} . \]\n\nThis implies (4.19). The proof is now complete. \( ▱ \)
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Yes
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Corollary 4.7.10 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function. Then \( f \) is Lipschitz continuous if and only if there exists \( \ell \geq 0 \) such that\n\n\[ \partial f\left( x\right) \subset \left\lbrack {-\ell ,\ell }\right\rbrack \text{for all}x \in \mathbb{R}\text{.} \]\n
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Proof: Suppose \( f \) is Lipschitz continuous on \( \mathbb{R} \) . Then there exists \( \ell \geq 0 \) such that\n\n\[ \left| {f\left( u\right) - f\left( v\right) }\right| \leq \ell \left| {u - v}\right| \text{ for all }u, v \in \mathbb{R}. \]\n\nThen for any \( x \in \mathbb{R} \) ,\n\n\[ {f}_{ + }^{\prime }\left( x\right) = \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\frac{f\left( {x + h}\right) - f\left( x\right) }{h} \leq \mathop{\lim }\limits_{{h \rightarrow {0}^{ + }}}\frac{\ell \left| h\right| }{h} = \ell . \]\n\nSimilarly, \( {f}_{ - }^{\prime }\left( x\right) \geq - \ell \) . Thus,\n\n\[ \partial f\left( x\right) = \left\lbrack {{f}_{ - }^{\prime }\left( x\right) ,{f}_{ + }^{\prime }\left( x\right) }\right\rbrack \subset \left\lbrack {-\ell ,\ell }\right\rbrack . \]\n\nConversely, fix any \( u, v \in \mathbb{R} \) with \( u \neq v \) . Applying Theorem 4.7.9, we get\n\n\[ \frac{f\left( v\right) - f\left( u\right) }{v - u} \in \partial f\left( c\right) \subset \left\lbrack {-\ell ,\ell }\right\rbrack \]\n\nfor some \( c \) in between \( u \) and \( v \) . This implies\n\n\[ \left| {f\left( u\right) - f\left( v\right) }\right| \leq \ell \left| {u - v}\right| \]\n\nThis inequality obviously holds for \( u = v \) . Therefore, \( f \) is Lipschitz continuous. \( ▱ \)
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Yes
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Example 1.1 Here are some further illustrations of set-builder notation.
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1. \( \{ n : n \) is a prime number \( \} = \{ 2,3,5,7,{11},{13},{17},\ldots \} \)\n\n2. \( \{ n \in \mathbb{N} : n \) is prime \( \} = \{ 2,3,5,7,{11},{13},{17},\ldots \} \)\n\n3. \( \left\{ {{n}^{2} : n \in \mathbb{Z}}\right\} = \{ 0,1,4,9,{16},{25},\ldots \} \)\n\n4. \( \left\{ {x \in \mathbb{R} : {x}^{2} - 2 = 0}\right\} = \{ \sqrt{2}, - \sqrt{2}\} \)\n\n5. \( \left\{ {x \in \mathbb{Z} : {x}^{2} - 2 = 0}\right\} = \varnothing \)\n\n6. \( \{ x \in \mathbb{Z} : \left| x\right| < 4\} = \{ - 3, - 2, - 1,0,1,2,3\} \)\n\n7. \( \{ {2x} : x \in \mathbb{Z},\left| x\right| < 4\} = \{ - 6, - 4, - 2,0,2,4,6\} \)\n\n8. \( \{ x \in \mathbb{Z} : \left| {2x}\right| < 4\} = \{ - 1,0,1\} \)
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Yes
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Example 1.2 Describe the set \( A = \{ {7a} + {3b} : a, b \in \mathbb{Z}\} \) .
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Solution: This set contains all numbers of form \( {7a} + {3b} \), where \( a \) and \( b \) are integers. Each such number \( {7a} + {3b} \) is an integer, so \( A \) contains only integers. But which integers? If \( n \) is any integer, then \( n = {7n} + 3\left( {-{2n}}\right) \), so \( n = {7a} + {3b} \) where \( a = n \) and \( b = - {2n} \) . Therefore \( n \in A \) . We’ve now shown that \( A \) contains only integers, and also that every integer is an element of \( A \) . Consequently \( A = \mathbb{Z} \) .
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Yes
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This brings us to a significant fact: If \( B \) is any set whatsoever, then \( \varnothing \subseteq B \) . To see why this is true, look at the last sentence of Definition 1.3. It says that \( \varnothing \notin B \) would mean that there is at least one element of \( \varnothing \) that is not an element of \( B \) . But this cannot be so because \( \varnothing \) contains no elements! Thus it is not the case that \( \varnothing \nsubseteq B \), so it must be that \( \varnothing \subseteq B \) .
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Fact 1.2 The empty set is a subset of all sets, that is, \( \varphi \subseteq B \) for any set \( B \) . Here is another way to look at it. Imagine a subset of \( B \) as a thing you make by starting with braces \( \{ \} \), then filling them with selections from \( B \) . For example, to make one particular subset of \( B = \{ a, b, c\} \), start with \( \{ \} \) , select \( b \) and \( c \) from \( B \) and insert them into \( \{ \} \) to form the subset \( \{ b, c\} \) . Alternatively, you could have chosen just \( a \) to make \( \{ a\} \), and so on. But one option is to simply select nothing from \( B \) . This leaves you with the subset \( \{ \} \) . Thus \( \{ \} \subseteq B \) . More often we write it as \( \varnothing \subseteq B \) .
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Yes
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1. \( 1 \in \{ 1,\{ 1\} \} \) .
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1 is the first element listed in \( \{ 1,\{ 1\} \} \)
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Yes
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Example 1.7 You should examine the following statements and make sure you understand how the answers were obtained. In particular, notice that in each instance the equation \( \left| {\mathcal{P}\left( A\right) }\right| = {2}^{\left| A\right| } \) is true.
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1. \( \mathcal{P}\left( {\{ 0,1,3\} }\right) = \{ \varnothing ,\{ 0\} ,\{ 1\} ,\{ 3\} ,\{ 0,1\} ,\{ 0,3\} ,\{ 1,3\} ,\{ 0,1,3\} \} \)\n\n2. \( \mathcal{P}\left( {\{ 1,2\} }\right) = \{ \varnothing ,\{ 1\} ,\{ 2\} ,\{ 1,2\} \} \)\n\n3. \( \mathcal{P}\left( {\{ 1\} }\right) = \{ \varnothing ,\{ 1\} \} \)\n\n4. \( \mathcal{P}\left( \varnothing \right) = \{ \varnothing \} \)\n\n5. \( \mathcal{P}\left( {\{ a\} }\right) = \{ \varnothing ,\{ a\} \} \)\n\n6. \( \mathcal{P}\left( {\{ \varnothing \} }\right) = \{ \varnothing ,\{ \varnothing \} \} \)\n\n7. \( \mathcal{P}\left( {\{ a\} }\right) \times \mathcal{P}\left( {\{ \varnothing \} }\right) = \{ \left( {\varnothing ,\varnothing }\right) ,\left( {\varnothing ,\{ \varnothing \} }\right) ,\left( {\{ a\} ,\varnothing }\right) ,\left( {\{ a\} ,\{ \varnothing \} }\right) \} \)\n\n8. \( \mathcal{P}\left( {\mathcal{P}\left( {\{ \varnothing \} }\right) }\right) = \{ \varnothing ,\{ \varnothing \} ,\{ \{ \varnothing \} \} ,\{ \varnothing ,\{ \varnothing \} \} \} \)\n\n9. \( \mathcal{P}\left( {\{ 1,\{ 1,2\} \} }\right) = \{ \varnothing ,\{ 1\} ,\{ \{ 1,2\} \} ,\{ 1,\{ 1,2\} \} \} \)\n\n10. \( \mathcal{P}\left( {\{ \mathbb{Z},\mathbb{N}\} }\right) = \{ \varnothing ,\{ \mathbb{Z}\} ,\{ \mathbb{N}\} ,\{ \mathbb{Z},\mathbb{N}\} \} \)
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Yes
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Example 1.8 Suppose \( A = \{ a, b, c, d, e\}, B = \{ d, e, f\} \) and \( C = \{ 1,2,3\} \).
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1. \( A \cup B = \{ a, b, c, d, e, f\} \)\n2. \( A \cap B = \{ d, e\} \)\n3. \( A - B = \{ a, b, c\} \)\n4. \( B - A = \{ f\} \)\n5. \( \;\left( {A - B}\right) \cup \left( {B - A}\right) = \{ a, b, c, f\} \)\n6. \( A \cup C = \{ a, b, c, d, e,1,2,3\} \)\n7. \( A \cap C = \varnothing \)\n8. \( A - C = \{ a, b, c, d, e\} \)\n9. \( \;\left( {A \cap C}\right) \cup \left( {A - C}\right) = \{ a, b, c, d, e\} \)\n10. \( \left( {A \cap B}\right) \times B = \{ \left( {d, d}\right) ,\left( {d, e}\right) ,\left( {d, f}\right) ,\left( {e, d}\right) ,\left( {e, e}\right) ,\left( {e, f}\right) \} \)\n11. \( \left( {A \times C}\right) \cap \left( {B \times C}\right) = \{ \left( {d,1}\right) ,\left( {d,2}\right) ,\left( {d,3}\right) ,\left( {e,1}\right) ,\left( {e,2}\right) ,\left( {e,3}\right) \} \)
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Yes
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Example 1.12 Suppose \( {A}_{1} = \{ 0,2,5\} ,{A}_{2} = \{ 1,2,5\} \) and \( {A}_{3} = \{ 2,5,7\} \) . Then
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\[ \mathop{\bigcup }\limits_{{i = 1}}^{3}{A}_{i} = {A}_{1} \cup {A}_{2} \cup {A}_{3} = \{ 0,1,2,5,7\} \;\text{ and }\;\mathop{\bigcap }\limits_{{i = 1}}^{3}{A}_{i} = {A}_{1} \cap {A}_{2} \cap {A}_{3} = \{ 2,5\} . \]
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Yes
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This example involves the following infinite list of sets. \( \begin{matrix} {A}_{1} = \{ - 1,0,1\} , & {A}_{2} = \{ - 2,0,2\} , & {A}_{3} = \{ - 3,0,3\} , & \cdots , & {A}_{i} = \{ - i,0, i\} , & \cdots \end{matrix} \)
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Observe that \( \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{A}_{i} = \mathbb{Z} \), and \( \mathop{\bigcap }\limits_{{i = 1}}^{\infty }{A}_{i} = \{ 0\} \)
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Yes
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In this example, all sets \( {A}_{\alpha } \) are all subsets of the plane \( {\mathbb{R}}^{2} \) . Each \( \alpha \) belongs to the index set \( I = \left\lbrack {0,2}\right\rbrack = \{ x \in \mathbb{R} : 0 \leq x \leq 2\} \), which is the set of all real numbers between 0 and 2 . For each number \( \alpha \in I \), define \( {A}_{\alpha } \) to be the set \( {A}_{\alpha } = \left\lbrack {\alpha ,2}\right\rbrack \times \left\lbrack {0,\alpha }\right\rbrack \), which is the rectangle on the \( {xy} \) -plane whose base runs from \( \alpha \) to 2 on the \( x \) -axis, and whose height is \( \alpha \) . Some of these are shown shaded below. (The dotted diagonal line \( y = x \) is not a part of any of the sets, but is shown for clarity, as the upper left corner of each \( {A}_{\alpha } \) touches it.) Note that these sets are not indexed with just integers. For example, as \( \sqrt{2} \in I \), there is a set \( {A}_{\sqrt{2}} \), which shown below on the right.
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Now consider the infinite union \( \mathop{\bigcup }\limits_{{\alpha \in I}}{A}_{\alpha } \) . It is the shaded triangle shown below, because any point \( \left( {x, y}\right) \) on this triangle belongs to the set \( {A}_{x} \), and is therefore in the union. (And any point not on the triangle is not in any \( {A}_{x} \) .)
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Yes
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Example 1.15 Here our sets are indexed by \( {\mathbb{R}}^{2} \) . For any \( \left( {a, b}\right) \in {\mathbb{R}}^{2} \), let \( {P}_{\left( a, b\right) } \) be the following subset of \( {\mathbb{R}}^{3} \) :\n\n\[ \n{P}_{\left( a, b\right) } = \left\{ {\left( {x, y, z}\right) \in {\mathbb{R}}^{3} : {ax} + {by} = 0}\right\} .\n\]\n\nIn words, given a point \( \left( {a, b}\right) \in {\mathbb{R}}^{2} \), the corresponding set \( {P}_{\left( a, b\right) } \) consists of all points \( \left( {x, y, z}\right) \) in \( {\mathbb{R}}^{3} \) that satisfy the equation \( {ax} + {by} = 0 \) . From previous math courses you will recognize this as a plane in \( {\mathbb{R}}^{3} \), that is, \( {P}_{\left( a, b\right) } \) is a plane in \( {\mathbb{R}}^{3} \) . Moreover, since any point \( \left( {0,0, z}\right) \) on the \( z \) -axis automatically satisfies \( {ax} + {by} = 0 \), each \( {P}_{\left( a, b\right) } \) contains the \( z \) -axis.
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For any point \( \left( {a, b}\right) \in {\mathbb{R}}^{2} \) with \( \left( {a, b}\right) \neq \left( {0,0}\right) \), we can visualize \( {P}_{\left( a, b\right) } \) as the vertical plane that cuts the \( {xy} \) -plane at the line \( {ax} + {by} = 0 \) . Figure 1.11 (right) shows a few of the \( {P}_{\left( a, b\right) } \) . Since any two such planes intersect along the \( z \) -axis, and because the \( z \) -axis is a subset of every \( {P}_{\left( a, b\right) } \), it is immediately clear that\n\n\[ \n\mathop{\bigcap }\limits_{{\left( {a, b}\right) \in {\mathbb{R}}^{2}}}{P}_{\left( a, b\right) } = \{ \left( {0,0, z}\right) : z \in \mathbb{R}\} = \text{\
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Yes
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Every integer that is not odd is even.
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\( \forall n \in \mathbb{Z}, \sim \left( {n\text{ is odd }}\right) \Rightarrow \left( {n\text{ is even }}\right) ,\; \) or \( \;\forall n \in \mathbb{Z}, \sim O\left( n\right) \Rightarrow E\left( n\right) . \)
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Yes
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Every integer is even.
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\( \forall n \in \mathbb{Z}, E\left( n\right) \)
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No
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\[ \forall x \in \mathbb{R},\exists y \in \mathbb{R},{y}^{3} = x. \]
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This statement is true, for no matter what number \( x \) is there exists a number \( y = \sqrt[3]{x} \) for which \( {y}^{3} = x \) .
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Yes
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Example 2.8 Consider the Mean Value Theorem from Calculus:\n\nIf \( f \) is continuous on the interval \( \left\lbrack {a, b}\right\rbrack \) and differentiable on \( \left( {a, b}\right) \), then there is a number \( c \in \left( {a, b}\right) \) for which \( {f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\left( a\right) }{b - a} \) .
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Here is a translation to symbolic form:\n\n\( \left( {\left( {f\text{ cont. on }\left\lbrack {a, b}\right\rbrack }\right) \land \left( {f\text{ is diff. on }\left( {a, b}\right) }\right) }\right) \Rightarrow \left( {\exists \;c \in \left( {a, b}\right) ,{f}^{\prime }\left( c\right) = \frac{f\left( b\right) - f\left( a\right) }{b - a}}\right) . \)
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Yes
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Example 2.10 Consider negating the following statement.\n\n\( R \) : You can solve it by factoring or with the quadratic formula.
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Now, \( R \) means (You can solve it by factoring) \( \vee \) (You can solve it with Q.F.), which we will denote as \( P \vee Q \) . The negation of this is\n\n\[ \sim \left( {P \vee Q}\right) = \left( { \sim P}\right) \land \left( { \sim Q}\right) . \]\n\nTherefore, in words, the negation of \( R \) is\n\n\( \sim R \) : You can’t solve it by factoring and you can’t solve it with the quadratic formula.
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Yes
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Example 2.11 We will negate the following sentence.\n\n\( R \) : The numbers \( x \) and \( y \) are both odd.
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This statement means \( \left( {x\text{is odd}}\right) \land \left( {y\text{is odd}}\right) \), so its negation is\n\n\[ \sim \left( {\left( {x\text{ is odd}}\right) \land \left( {y\text{ is odd}}\right) }\right) \; = \; \sim \left( {x\text{ is odd}}\right) \vee \sim \left( {y\text{ is odd}}\right) \]\n\n\[ = \;\left( {x\text{is even}}\right) \vee \left( {y\text{is even}}\right) \text{.} \]\n\nTherefore the negation of \( R \) can be expressed in the following ways:\n\n\( \sim R \) : The number \( x \) is even or the number \( y \) is even.\n\n\( \sim R \) : At least one of \( x \) and \( y \) is even.
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Yes
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Example 2.13 If a statement has multiple quantifiers, negating it involves several iterations of Equations (2.8) and (2.9). Consider the following:\n\n\( S \) : For every real number \( x \) there is a real number \( y \) for which \( {y}^{3} = x \) .
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This statement asserts any real number \( x \) has a cube root \( y \), so it’s true. Symbolically \( S \) can be expressed as\n\n\[ \forall x \in \mathbb{R},\exists y \in \mathbb{R},{y}^{3} = x. \]\n\nLet's work out the negation of this statement.\n\n\[ \sim \left( {\forall x \in \mathbb{R},\exists y \in \mathbb{R},{y}^{3} = x}\right) \; = \;\exists x \in \mathbb{R}, \sim \left( {\exists y \in \mathbb{R},{y}^{3} = x}\right) \]\n\n\[ = \exists x \in \mathbb{R},\forall y \in \mathbb{R}, \sim \left( {{y}^{3} = x}\right) \]\n\n\[ = \;\exists x \in \mathbb{R},\forall y \in \mathbb{R},{y}^{3} \neq x. \]\n\nThus the negation is a (false) statement that can be written in either of the following ways.\n\n\( \sim S \) : There is a real number \( x \) such that for all real numbers \( y,{y}^{3} \neq x \) .\n\n\( \sim S \) : There is a real number \( x \) for which \( {y}^{3} \neq x \) for all real numbers \( y \) .
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Yes
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Example 2.14 Negate the following statement about a particular (i.e., constant) number \( a \) .\n\n\( R \) : If \( a \) is odd then \( {a}^{2} \) is odd.
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Using Equation (2.10), we get the following negation.\n\n\( \sim R : a \) is odd and \( {a}^{2} \) is not odd.
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Yes
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Example 2.15 This example is like the previous one, but the constant \( a \) is replaced by a variable \( x \) . We will negate the following statement.\n\n\( R \) : If \( x \) is odd then \( {x}^{2} \) is odd.
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As in Section 2.8, we interpret this as the universally quantified statement\n\n\[ R : \forall x \in \mathbb{Z},\left( {x\text{ odd }}\right) \Rightarrow \left( {{x}^{2}\text{ odd }}\right) .\n\]\n\nBy Equations (2.8) and (2.10), we get the following negation for \( R \) .\n\n\[ \sim \left( {\forall x \in \mathbb{Z},\left( {x\text{ odd}}\right) \Rightarrow \left( {{x}^{2}\text{ odd}}\right) }\right) \; = \;\exists x \in \mathbb{Z}, \sim \left( {\left( {x\text{ odd}}\right) \Rightarrow \left( {{x}^{2}\text{ odd}}\right) }\right)\n\]\n\n\[ = \exists x \in \mathbb{Z},\left( {x\text{ odd }}\right) \land \sim \left( {{x}^{2}\text{ odd }}\right) .\n\]\n\nTranslating back into words, we have\n\n\( \sim R \) : There is an odd integer \( x \) whose square is not odd.\n\nNotice that \( R \) is true and \( \sim R \) is false.
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Yes
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In ordering a café latte, you have a choice of whole, skim or soy milk; small, medium or large; and either one or two shots of espresso. How many choices do you have in ordering one drink?
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Solution: Your choice is modeled by a list of form (milk, size, shots). There are 3 choices for the first entry, 3 for the second and 2 for the third. By the multiplication principle, the number of choices is \( 3 \cdot 3 \cdot 2 = {18} \).
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Yes
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Example 3.3 Consider lists of length 4 made with symbols \( A, B, C, D, E, F, G \) .\n\n(a) How many such lists are possible if repetition is allowed?
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(a) Imagine the list as containing four boxes that we fill with selections from the letters \( A, B, C, D, E, F \) and \( G \), as illustrated below.\n\n\n\nWe have 7 choices in filling each box. The multiplication principle says the total number of lists that can be made this way is \( 7 \cdot 7 \cdot 7 \cdot 7 = \mathbf{{2401}} \) .
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Yes
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Example 3.4 A non-repetitive list of length 5 is to be made from the symbols \( A, B, C, D, E, F, G \) . The first entry must be either a \( B, C \) or \( D \), and the last entry must be a vowel. How many such lists are possible?
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Solution: Start by making a list of five boxes. The first box must contain either \( B, C \) or \( D \), so there are three choices for it.\n\n\n\nNow there are 6 letters left for the remaining 4 boxes. The knee-jerk action is to fill them in, one at a time, using up an additional letter each time.\n\n\n\nBut when we get to the last box, there is a problem. It is supposed to contain a vowel, but for all we know we have already used up one or both vowels in the previous boxes. The multiplication principle breaks down because there is no way to tell how many choices there are for the last box.\n\nThe correct way to solve this problem is to fill in the first and last boxes (the ones that have restrictions) first.\n\n\n\nThen fill the remaining middle boxes with the 5 remaining letters.\n\n\n\nBy the multiplication principle, there are \( 3 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = \textbf{360 lists}. \)
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Yes
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Example 3.5 How many length-4 non-repetitive lists can be made from the symbols \( A, B, C, D, E, F, G \), if the list must contain an \( E \) ?
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In Example 3.3 (c) our approach was to divide these lists into four types, depending on whether the \( E \) is in the first, second, third or fourth position. Then we used the multiplication principle to count the lists of type 1. There are 6 choices for the second entry, 5 for the third, and 4 for the fourth. This is indicated above, where the number below a box is the number of choices we have for that position. The multiplication principle implies that there are \( 6 \cdot 5 \cdot 4 = {120} \) lists of type 1. Similarly there are \( 6 \cdot 5 \cdot 4 = {120} \) lists of types 2, 3, and 4. We then used the addition principle intuitively, conceiving of the lists to be counted as the elements of a set \( X \), broken up into parts \( {X}_{1},{X}_{2},{X}_{3} \) and \( {X}_{4} \) , which are the lists of types 1,2,3 and 4, respectively. The addition principle says that the number of lists that contain an \( E \) is \( \left| X\right| = \left| {X}_{1}\right| + \left| {X}_{2}\right| + \left| {X}_{3}\right| + \left| {X}_{4}\right| = {120} + {120} + {120} + {120} = \mathbf{{480}}. \)
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Yes
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Example 3.6 How many even 5-digit numbers are there for which no digit is 0, and the digit 6 appears exactly once? For instance, 55634 and 16118 are such numbers, but not 63304 (has a 0), nor 63364 (too many 6’s), nor 55637 (not even).
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Solution: Let \( X \) be the set of all such numbers. The answer will be \( \left| X\right| \), so our task is to find \( \left| X\right| \) . Put \( X = {X}_{1} \cup {X}_{2} \cup {X}_{3} \cup {X}_{4} \cup {X}_{5} \), where \( {X}_{i} \) is the set of those numbers in \( X \) whose \( i \) th digit is 6, as diagramed below. Note \( {X}_{i} \cap {X}_{j} = \varnothing \) whenever \( i \neq j \) because the numbers in \( {X}_{i} \) have their 6 in a different position than the numbers in \( {X}_{j} \) . Our plan is to use the multiplication principle to compute each \( \left| {X}_{i}\right| \), and follow this with the addition principle.\n\n\n\nThe first digit of any number in \( {X}_{1} \) is 6, and the three digits following it can be any of the ten digits except 0 (not allowed) or 6 (already appears). Thus there are eight choices for each of three digits following the first 6. But because any number in \( {X}_{1} \) is even, its final digit must be one of 2,4 or 8 , so there are just three choices for this final digit. By the multiplication principle, \( \left| {X}_{1}\right| = 8 \cdot 8 \cdot 8 \cdot 3 = {1536} \) . Likewise \( \left| {X}_{2}\right| = \left| {X}_{3}\right| = \left| {X}_{4}\right| = 8 \cdot 8 \cdot 8 \cdot 3 = {1536} \) .\n\nBut \( {X}_{5} \) is slightly different because we do not choose the final digit, which is already 6. The multiplication principle gives \( \left| {X}_{5}\right| = 8 \cdot 8 \cdot 8 \cdot 8 = {4096} \) .\n\nThe addition principle gives our final answer. The number of even 5- digit numbers with no 0’s and one 6 is \( \left| X\right| = \left| {X}_{1}\right| + \left| {X}_{2}\right| + \left| {X}_{3}\right| + \left| {X}_{4}\right| + \left| {X}_{5}\right| = \) \( {1536} + {1536} + {1536} + {1536} + {4096} = \mathbf{{10},{240}}. \)
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Yes
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Example 3.7 How many length-4 lists can be made from the symbols \( A, B, C, D, E, F, G \) if the list has at least one \( E \), and repetition is allowed?
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Solution: Such a list might contain one, two, three or four \( E \) ’s, which could occur in various positions. This is a fairly complex situation.\n\nBut it is very easy to count the set \( U \) of all lists of length 4 made from \( A, B, C, D, E, F, G \) if we don’t care whether or not the lists have any \( E \) ’s. The multiplication principle says \( \left| U\right| = 7 \cdot 7 \cdot 7 \cdot 7 = {2401} \) .\n\nIt is equally easy to count the set \( X \) of those lists that contain no \( E \) ’s. The multiplication principle says \( \left| X\right| = 6 \cdot 6 \cdot 6 \cdot 6 = {1296}. \)\n\nWe are interested in those lists that have at least one \( E \), and this is the set \( U - X \) . By the subtraction principle, the answer to our question is \( \left| {U - X}\right| = \left| U\right| - \left| X\right| = {2401} - {1296} = \mathbf{{1105}}. \)
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Yes
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How many such lists are there if repetition is not allowed?
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To answer the first question, note that there are seven letters, so the number of lists is \( 7! = \mathbf{5040} \) .
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Yes
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Example 3.9 Ten contestants run a marathon. All finish, and there are no ties. How many different possible rankings are there for first-, second-and third-place?
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Solution: Call the contestants \( A, B, C, D, E, F, G, H, I \) and \( J \) . A ranking of winners can be regarded as a 3-permutation of the set of 10 contestants. For example, \( E{CH} \) means \( E \) in first-place, \( C \) in second-place and \( H \) in third. Thus there are \( P\left( {{10},3}\right) = {10} \cdot 9 \cdot 8 = {720} \) possible rankings.
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Yes
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Example 3.10 You deal five cards off of a standard 52-card deck, and line them up in a row. How many such lineups are there that either consist of all red cards, or all clubs?
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Solution: There are 26 red cards. The number of ways to line up five of them is \( P\left( {{26},5}\right) = {26} \cdot {25} \cdot {24} \cdot {23} \cdot {22} = 7,{893},{600} \) . \n\nThere are 13 club cards (which are black). The number of ways to line up five of them is \( P\left( {{13},5}\right) = {13} \cdot {12} \cdot {11} \cdot {10} \cdot 9 = {154},{440} \) . \n\nBy the addition principle, the answer to our question is that there are \( P\left( {{26},5}\right) + P\left( {{13},5}\right) = 8,{048},{040} \) lineups that are either all red cards, or all club cards.
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Yes
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Example 3.11 How many size-4 subsets does \( \{ 1,2,3,4,5,6,7,8,9\} \) have?
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The answer is \( \begin{aligned} \left( \begin{aligned} 9 \\ 4 \end{aligned}\right) = \frac{9!}{4!\left( {9 - 4}\right) !} & = \frac{9!}{4!5!} = \frac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{4!5!} \\ & = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4!} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{24} = \mathbf{{126}}. \end{aligned} \)
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Yes
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Example 3.13 A single 5-card hand is dealt off of a standard 52-card deck. How many different 5-card hands are possible?
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Solution: Think of the deck as a set \( D \) of 52 cards. Then a 5-card hand is just a 5-element subset of \( D \) . There are many such subsets, such as \n\nThus the number of 5-card hands is the number of 5-element subsets of \( D \) , which is\n\n\[ \left( \begin{matrix} {52} \\ 5 \end{matrix}\right) = \frac{{52}!}{5! \cdot {47}!} = \frac{{52} \cdot {51} \cdot {50} \cdot {49} \cdot {48} \cdot {47}!}{5! \cdot {47}!} = \frac{{52} \cdot {51} \cdot {50} \cdot {49} \cdot {48}}{5!} = 2,{598},{960}. \]\n\nAnswer: There are 2,598,960 different five-card hands that can be dealt from a deck of 52 cards.
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Yes
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Example 3.14 This problem concerns 5-card hands that can be dealt off of a 52-card deck. How many such hands are there in which two of the cards are clubs and three are hearts?
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Solution: Such a hand is described by a list of length two of the form\n\n\[ \left( {\left\{ {\left\lbrack \begin{matrix} * \\ * \\ * \end{matrix}\right\rbrack ,\left\lbrack \begin{matrix} * \\ * \\ * \end{matrix}\right\rbrack }\right\} ,\left\{ {\left\lbrack \begin{matrix} * \\ \heartsuit \end{matrix}\right\rbrack ,\left\lbrack \begin{matrix} * \\ \heartsuit \end{matrix}\right\rbrack ,\left\lbrack \begin{matrix} * \\ \heartsuit \end{matrix}\right\rbrack }\right\} }\right) ,\]\n\nwhere the first entry is a 2-element subset of the set of 13 club cards, and the second entry is a 3-element subset of the set of 13 heart cards. There are \( \left( \begin{matrix} {13} \\ 2 \end{matrix}\right) \) choices for the first entry and \( \left( \begin{matrix} {13} \\ 3 \end{matrix}\right) \) choices for the second, so by the multiplication principle there are \( \left( \begin{aligned} {13} \\ 2 \end{aligned}\right) \left( \begin{aligned} {13} \\ 3 \end{aligned}\right) = \frac{{13}!}{2!{11}!}\frac{{13}!}{3!{10}!} = {22},{308} \) such lists. Thus there are 22,308 such 5-card hands.
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Yes
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Example 3.15 A lottery features a bucket of 36 balls numbered 1 through 36. Six balls will be drawn randomly. For \$1 you buy a ticket with six blanks: \( ▱▱▱▱▱▱▱ \) . You fill in the blanks with six different numbers between 1 and 36. You win \( \$ 1,{000},{000} \) if you chose the same numbers that are drawn, regardless of order. What are your chances of winning?
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Solution: In filling out the ticket you are choosing six numbers from a set of 36 numbers. Thus there are \( \left( \begin{matrix} {36} \\ 6 \end{matrix}\right) = \frac{{36}!}{6!\left( {{36} - 6}\right) !} = 1,{947},{792} \) different combinations of numbers you might write. Only one of these will be a winner. Your chances of winning are one in 1,947,792.
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Yes
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Example 3.16 How many 7-digit binary strings (0010100,1101011, etc.) have an odd number of 1’s?
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Solution: Let \( A \) be the set of all 7-digit binary strings with an odd number of 1’s, so the answer will be \( \left| A\right| \) . To find \( \left| A\right| \), we break \( A \) into smaller parts. Notice any string in \( A \) will have either one, three, five or seven 1’s. Let \( {A}_{1} \) be the set of 7-digit binary strings with only one 1 . Let \( {A}_{3} \) be the set of 7-digit binary strings with three 1’s. Let \( {A}_{5} \) be the set of 7-digit binary strings with five 1’s, and let \( {A}_{7} \) be the set of 7-digit binary strings with seven 1’s. Then \( A = {A}_{1} \cup {A}_{3} \cup {A}_{5} \cup {A}_{7} \) . Any two of the sets \( {A}_{i} \) have empty intersection, so the addition principle gives \( \left| A\right| = \left| {A}_{1}\right| + \left| {A}_{3}\right| + \left| {A}_{5}\right| + \left| {A}_{7}\right| \) .\n\nNow we must compute the individual terms of this sum. Take \( {A}_{3} \), the set of 7-digit binary strings with three 1's. Such a string can be formed by selecting three out of seven positions for the 1's and putting 0's in the other spaces. Thus \( \left| {A}_{3}\right| = \left( \begin{array}{l} 7 \\ 3 \end{array}\right) \) . Similarly \( \left| {A}_{1}\right| = \left( \begin{array}{l} 7 \\ 1 \end{array}\right) ,\left| {A}_{5}\right| = \left( \begin{array}{l} 7 \\ 5 \end{array}\right) \), and \( \left| {A}_{7}\right| = \left( \begin{array}{l} 7 \\ 7 \end{array}\right) \) .\n\n\( \textbf{Answer: }\left| A\right| = \left| {A}_{1}\right| + \left| {A}_{3}\right| + \left| {A}_{5}\right| + \left| {A}_{7}\right| = \left( \begin{aligned} 7 \\ 1 \end{aligned}\right) + \left( \begin{aligned} 7 \\ 3 \end{aligned}\right) + \left( \begin{aligned} 7 \\ 5 \end{aligned}\right) + \left( \begin{aligned} 7 \\ 7 \end{aligned}\right) = 7 + {35} + {21} + 1 = {64}. \n\nThere are 64 7-digit binary strings with an odd number of 1's.
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Yes
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Theorem 3.1 (Binomial Theorem) If \( n \) is a non-negative integer, then \( {\left( x + y\right) }^{n} = \left( \begin{aligned} n \\ 0 \end{aligned}\right) {x}^{n} + \left( \begin{aligned} n \\ 1 \end{aligned}\right) {x}^{n - 1}y + \left( \begin{aligned} n \\ 2 \end{aligned}\right) {x}^{n - 2}{y}^{2} + \left( \begin{aligned} n \\ 3 \end{aligned}\right) {x}^{n - 3}{y}^{3} + \cdots + \left( \begin{aligned} n \\ {n - 1} \end{aligned}\right) x{y}^{n - 1} + \left( \begin{aligned} n \\ n \end{aligned}\right) {y}^{n}. \)
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For now we will be content to accept the binomial theorem without proof. (You will be asked to prove it in an exercise in Chapter 10.)
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No
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Example 3.17 A 3-card hand is dealt off of a standard 52-card deck. How many different such hands are there for which all three cards are red or all three cards are face cards?
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Solution: Let \( A \) be the set of 3-card hands where all three cards are red (i.e., either \( \circ \) or \( \diamondsuit \) ). Let \( B \) be the set of 3-card hands in which all three cards are face cards (i.e., \( J, K \) or \( Q \) of any suit). These sets are illustrated below.\n\n\[ A = \left\{ {\left\{ {\left\lbrack \begin{array}{l} 5 \\ 9 \end{array}\right\rbrack ,\left\lbrack \begin{array}{l} K \\ \phi \end{array}\right\rbrack ,\left\lbrack \begin{array}{l} 2 \\ 0 \end{array}\right\rbrack }\right\} ,\left\{ {\left\lbrack \begin{array}{l} K \\ 0 \end{array}\right\rbrack ,\left\lbrack \begin{array}{l} J \\ 0 \end{array}\right\rbrack ,\left\lbrack \begin{array}{l} Q \\ 0 \end{array}\right\rbrack }\right\} ,\left\{ {\left\lbrack \begin{array}{l} A \\ \phi \end{array}\right\rbrack ,\left\lbrack \begin{array}{l} 6 \\ \phi \end{array}\right\rbrack ,\left\lbrack \begin{array}{l} 6 \\ 0 \end{array}\right\rbrack }\right\} ,\ldots }\right\} \;\text{ (Red cards) }\n\]\n\nWe seek the number of 3-card hands that are all red or all face cards, and this number is \( \left| {A \cup B}\right| \) . By Fact 3.6, \( \left| {A \cup B}\right| = \left| A\right| + \left| B\right| - \left| {A \cap B}\right| \) . Let’s examine \( \left| A\right| ,\left| B\right| \) and \( \left| {A \cap B}\right| \) separately. Any hand in \( A \) is formed by selecting three cards from the 26 red cards in the deck, so \( \left| A\right| = \left( \begin{matrix} {26} \\ 3 \end{matrix}\right) \) . Similarly, any hand in \( B \) is formed by selecting three cards from the 12 face cards in the deck, so \( \left| B\right| = \left( \begin{matrix} {12} \\ 3 \end{matrix}\right) \) . Now think about \( A \cap B \) . It contains all the 3-card hands made up of cards that are red face cards. \n\nThe deck has only 6 red face cards, so \( \left| {A \cap B}\right| = \left( \begin{array}{l} 6 \\ 3 \end{array}\right) \) .\n\nNow we can answer our question. The number of 3-card hands that are all red or all face cards is \( \left| {A \cup B}\right| = \left| A\right| + \left| B\right| - \left| {A \cap B}\right| = \left( \begin{matrix} {26} \\ 3 \end{matrix}\right) + \left( \begin{matrix} {12} \\ 3 \end{matrix}\right) - \left( \begin{matrix} 6 \\ 3 \end{matrix}\right) = {2600} + {220} - {20} = \mathbf{{2800}} .
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Yes
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Example 3.18 A 3-card hand is dealt off of a standard 52-card deck. How many different such hands are there for which it is not the case that all 3 cards are red or all three cards are face cards?
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Solution: We will use the subtraction principle combined with our answer to Example 3.17, above. The total number of 3-card hands is \( \left( \begin{matrix} {52} \\ 3 \end{matrix}\right) = \frac{{52}!}{3!\left( {{52} - 3}\right) !} = \) \( \frac{{52}!}{3!{49}!} = \frac{{52} \cdot {51} \cdot {50}}{3!} = {26} \cdot {17} \cdot {50} = {22},{100} \) . To get our answer, we must subtract from this the number of 3-card hands that are all red or all face cards, that is, we must subtract the answer from Example 3.17. Thus the answer to our question is \( {22},{100} - {2800} = \mathbf{{19},{300}} \) .
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Yes
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A bag contains 20 identical red marbles, 20 identical green marbles, and 20 identical blue marbles. You reach in and grab 20 marbles. There are many possible outcomes. You could have 11 reds, 4 greens and 5 blues. Or you could have 20 reds, 0 greens and 0 blues, etc. All together, how many outcomes are possible?
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Each outcome can be thought of as a 20-element multiset made from the elements of the 3-element set \( X = \{ \mathrm{R},\mathrm{G},\mathrm{B}\} \) . For example,11 reds,4 greens and 5 blues would correspond to the multiset\n\n\[ \left\lbrack {R, R, R, R, R, R, R, R, R, R, R, G, G, G, G, B, B, B, B, B}\right\rbrack . \]\n\nThe outcome consisting of 10 reds and 10 blues corresponds to the multiset\n\n\[ \left\lbrack {R, R, R, R, R, R, R, R, R, R, B, B, B, B, B, B, B, B, B, B, B}\right\rbrack . \]\n\nThus the total number of outcomes is the number of 20-element multisets made from the elements of the 3-element set \( X = \{ \mathrm{R},\mathrm{G},\mathrm{B}\} \) . By Fact 3.7, the answer is \( \left( \begin{aligned} {{20} + 3 - 1} \\ {20} \end{aligned}\right) = \left( \begin{aligned} {22} \\ {20} \end{aligned}\right) = \mathbf{{231}} \) possible outcomes.
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Yes
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Example 3.20 How many non-negative integer solutions does the equation \( w + x + y + z = {20} \) have?
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Solution: We can model a solution with stars and bars. For example, encode the solution \( \left( {w, x, y, z}\right) = \left( {3,4,5,8}\right) \) as\n\n\n\nIn general, any solution \( \left( {w, x, y, z}\right) = \left( {a, b, c, d}\right) \) gets encoded as\n\n\n\nwhere all together there are 20 stars and 3 bars. So, for instance the solution \( \left( {w, x, y, z}\right) = \left( {0,0,{10},{10}}\right) \) gets encoded as \( \left| \right| * * * * * * * * * * * * * * * * * * , \) and the solution \( \left( {w, x, y, z}\right) = \left( {7,3,5,5}\right) \) is encoded as \( * * * * * * * * \left| {* * * }\right| * * * * * \) | \( * * * * * \) . Thus we can describe any non-negative integer solution to the equation as a list of length \( {20} + 3 = {23} \) that has 20 stars and 3 bars. We can make any such list by choosing 3 out of 23 spots for the bars, and filling the remaining 20 spots with stars. The number of ways to do this is \( \left( \begin{matrix} {23} \\ 3 \end{matrix}\right) = \frac{{23}!}{3!{20}!} = \frac{{23} \cdot {22} \cdot {21}}{3 \cdot 2} = \) \( {23} \cdot {11} \cdot 7 = {1771} \) . Thus there are \( {1771} \) non-negative integer solutions of \( w + x + y + z = {20} \) .
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Yes
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Example 3.21 This problem concerns the lists \( \left( {w, x, y, z}\right) \) of integers with the property that \( 0 \leq w \leq x \leq y \leq z \leq {10} \) . That is, each entry is an integer between 0 and 10, and the entries are ordered from smallest to largest. For example, \( \left( {0,3,3,7}\right) ,\left( {1,1,1,1}\right) \) and \( \left( {2,3,6,9}\right) \) have this property, but \( \left( {2,3,6,4}\right) \) does not. How many such lists are there?
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Solution: We can encode such a list with 10 stars and 4 bars, where \( w \) is the number of stars to the left of the first bar, \( x \) is the number of stars to the left of the second bar, \( y \) is the number of stars to the left of the third bar, and \( z \) is the number of stars to the left of the fourth bar.\n\nFor example, \( \left( {2,3,6,9}\right) \) is encoded as \( * * \left| *\right| * * * \left| {* * * }\right| * \), and \( \left( {1,2,3,4}\right) \) is encoded as \( * \left| *\right| * \left| *\right| * * * * * * \) .\n\nHere are some other examples of lists paired with their encodings.\n\n\[ \text{| * * * | | * * * * | * * *} \]\n\n\( \left( {1,1,1,1}\right) \)\n\n\[ \text{* | | | | | * * * * * * * * *} \]\n\n\[ \text{10)}\; * * * * * * * * * * \mid \mid \mid * \mid \]\n\nSuch encodings are lists of length 14, with 10 stars and 4 bars. We can make such a list by choosing 4 of the 14 slots for the bars and filling the remaining slots with stars. The number of ways to do this is \( \left( \begin{matrix} {14} \\ 4 \end{matrix}\right) = {1001} \) . Answer: There are 1001 such lists.
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Yes
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Example 3.22 Count the permutations of the letters in MISSISSIPPI.
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Solution: Think of this word as an 11-element multiset with one M, four I’s, four S’s and two P’s. By Fact 3.8, it has \( \frac{{11}!}{1!4!4!2!} = {34},{650} \) permutations.
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Yes
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Example 3.23 Determine the number of permutations of the multiset \( \left\lbrack {1,1,1,1,5,5,{10},{25},{25}}\right\rbrack \) .
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Solution: By Fact 3.8 the answer is \( \frac{9!}{4!2!1!2!} = {3780} \) .
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Yes
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Example 3.24 Pick six integers between 0 and 9 (inclusive). Show that two of them must add up to 9.
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Solution: Pick six numbers between 0 and 9. Here's why two of them sum to 9: Imagine five boxes, each marked with two numbers, as shown below. Each box is labeled so that the two numbers written on it sum to 9.\n\n\n\nFor each number that was picked, put it in the box having that number written on it. For example, if we picked 7, it goes in the box labeled \
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No
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A store has a gumball machine containing a large number of red, green, blue and white gumballs. You get one gumball for each nickel you put into the machine. The store offers the following deal: You agree to buy some number of gumballs, and if 13 or more of them have the same color you get $5. What is the fewest number of gumballs you need to buy to be 100% certain that you will make money on the deal?
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Let \( n \) be the number of gumballs that you buy. Imagine sorting your \( n \) gumballs into four boxes labeled RED, GREEN, BLUE, and WHITE. (That is, red balls go in the red box, green balls go in the green box, etc.)\n\nThe division principle says that one box contains \( \lceil \frac{n}{4}\rceil \) or more gumballs. Provided \( \lceil \frac{n}{4}\rceil \geq 13 \), you will know you have 13 gumballs of the same color. This happens if \( \frac{n}{4} > 12 \) (so the ceiling of \( \frac{n}{4} \) rounds to a value larger than 12). Therefore you need \( n > 4 \cdot 12 = 48 \), so if \( n = 49 \) you know you have at least \( \lceil \frac{49}{4}\rceil = \lceil 12.25\rceil = 13 \) gumballs of the same color.\n\nAnswer: Buy 49 gumballs for 49 nickels, which is \$2.45. You get \$5, and therefore have made \$2.55.
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Yes
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Example 3.26 Nine points are randomly placed on the right triangle shown below. Show that three of these points form a triangle whose area is \( \frac{1}{8} \) square unit or less. (We allow triangles with zero area, in which case the three points lie on a line.)
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Solution: Divide the triangle into four smaller triangles, as indicated by the dashed lines below. Each of these four triangles has an area of \( \frac{1}{2}{bh} = \frac{1}{2}\frac{1}{2}\frac{1}{2} = \frac{1}{8} \) square units. Think of these smaller triangles as \
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No
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Example 3.27 Use combinatorial proof to show \( \left( \begin{array}{l} n \\ k \end{array}\right) = \left( \begin{matrix} n \\ n - k \end{matrix}\right) \) .
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Solution: First, by definition, if \( k < 0 \) or \( k > n \), then both sides are 0, and thus equal. Therefore for the rest of the proof we can assume \( 0 \leq k \leq n \). The left-hand side \( \left( \begin{array}{l} n \\ k \end{array}\right) \) is the number of \( k \)-element subsets of \( S = \{ 1,2,\ldots, n\} \). Every \( k \)-element subset \( X \subseteq S \) pairs with a unique \( \left( {n - k}\right) \)-element subset \( \bar{X} = S - X \subseteq S \). Thus the number of \( k \)-element subsets of \( S \) equals the number of \( \left( {n - k}\right) \)-element subsets of \( S \), which is to say \( \left( \begin{array}{l} n \\ k \end{array}\right) = \left( \begin{matrix} n \\ n - k \end{matrix}\right) \).
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Yes
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Example 3.28 Use a combinatorial proof to show that \( \mathop{\sum }\limits_{{k = 0}}^{n}{\left( \begin{array}{l} n \\ k \end{array}\right) }^{2} = \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \) .
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Solution: First, the right-hand side \( \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \) is the number of ways to select \( n \) things from a set \( S \) that has \( {2n} \) elements.\n\nNow let’s count this a different way. Divide \( S \) into two equal-sized parts, \( S = A \cup B \), where \( \left| A\right| = n \) and \( \left| B\right| = n \), and \( A \cap B = \varnothing \).\n\nFor any fixed \( k \) with \( 0 \leq k \leq n \), we can select \( n \) things from \( S \) by taking \( k \) things from \( A \) and \( n - k \) things from \( B \) for a total of \( k + \left( {n - k}\right) = n \) things. By the multiplication principle, we get \( \left( \begin{array}{l} n \\ k \end{array}\right) \left( \begin{matrix} n \\ n - k \end{matrix}\right) n \) -element subsets of \( S \) this way.\n\nAs \( k \) could be any number from 0 to \( n \), the number of ways to select \( n \) things from \( S \) is thus\n\n\n\nBut because \( \left( \begin{aligned} n \\ {n - k} \end{aligned}\right) = \left( \begin{aligned} n \\ k \end{aligned}\right) , \) this expression equals \( \left( \begin{aligned} n \\ 0 \end{aligned}\right) \left( \begin{aligned} n \\ 0 \end{aligned}\right) + \left( \begin{aligned} n \\ 1 \end{aligned}\right) \left( \begin{aligned} n \\ 1 \end{aligned}\right) + \left( \begin{aligned} n \\ 2 \end{aligned}\right) \left( \begin{aligned} n \\ 2 \end{aligned}\right) + \cdots + \left( \begin{aligned} n \\ n \end{aligned}\right) \left( \begin{aligned} n \\ n \end{aligned}\right) , \) which is \( {\left( \begin{aligned} n \\ 0 \end{aligned}\right) }^{2} + {\left( \begin{aligned} n \\ 1 \end{aligned}\right) }^{2} + {\left( \begin{aligned} n \\ 2 \end{aligned}\right) }^{2} + \cdots + {\left( \begin{aligned} n \\ n \end{aligned}\right) }^{2} = \mathop{\sum }\limits_{{k = 0}}^{n}{\left( \begin{aligned} n \\ k \end{aligned}\right) }^{2} .\n\nIn summary, we’ve counted the ways to choose \( n \) elements from the set \( S \) with two methods. One method gives \( \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \), and the other gives \( \mathop{\sum }\limits_{{k = 0}}^{n}{\left( \begin{array}{l} n \\ k \end{array}\right) }^{2} \).\n\nTherefore \( \mathop{\sum }\limits_{{k = 0}}^{n}{\left( \begin{array}{l} n \\ k \end{array}\right) }^{2} = \left( \begin{matrix} {2n} \\ n \end{matrix}\right) \).
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Yes
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Proposition Let \( a, b \in \mathbb{Z} \) and \( n \in \mathbb{N} \) . If \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \), then \( {a}^{2} \equiv {b}^{2}\left( {\;\operatorname{mod}\;n}\right) \) .
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Proof. We will use direct proof. Suppose \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \).\n\nBy definition of congruence of integers, this means \( n \mid \left( {a - b}\right) \).\n\nThen by definition of divisibility, there is an integer \( c \) for which \( a - b = {nc} \).\n\nNow multiply both sides of this equation by \( a + b \).\n\n\[ a - b = {nc} \]\n\n\[ \left( {a - b}\right) \left( {a + b}\right) = {nc}\left( {a + b}\right) \]\n\n\[ {a}^{2} - {b}^{2} = {nc}\left( {a + b}\right) \]\n\nSince \( c\left( {a + b}\right) \in \mathbb{Z} \), the above equation tells us \( n \mid \left( {{a}^{2} - {b}^{2}}\right) \).\n\nAccording to Definition 5.1, this gives \( {a}^{2} \equiv {b}^{2}\left( {\;\operatorname{mod}\;n}\right) \).
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Yes
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Proposition 7.1 If \( a, b \in \mathbb{N} \), then there exist integers \( k \) and \( \ell \) for which \( \gcd \left( {a, b}\right) = {ak} + b\ell \) .
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Proof. (Direct) Suppose \( a, b \in \mathbb{N} \) . Consider the set \( A = \{ {ax} + {by} : x, y \in \mathbb{Z}\} \) . This set contains both positive and negative integers, as well as 0 . (Reason: Let \( y = 0 \) and let \( x \) range over all integers. Then \( {ax} + {by} = {ax} \) ranges over all multiples of \( a \), both positive, negative and zero.) Let \( d \) be the smallest positive element of \( A \) . Then, because \( d \) is in \( A \), it must have the form \( d = {ak} + b\ell \) for some specific \( k,\ell \in \mathbb{Z} \) .\n\nTo finish, we will show \( d = \gcd \left( {a, b}\right) \) . We will first argue that \( d \) is a common divisor of \( a \) and \( b \), and then that it is the greatest common divisor.\n\nTo see that \( d \mid a \), use the division algorithm (page 30) to write \( a = {qd} + r \) for integers \( q \) and \( r \) with \( 0 \leq r < d \) . The equation \( a = {qd} + r \) yields\n\n\[ r = a - {qd} \]\n\n\[ = a - q\left( {{ak} + b\ell }\right) \]\n\n\[ = a\left( {1 - {qk}}\right) + b\left( {-q\ell }\right) \text{.} \]\n\nTherefore \( r \) has form \( r = {ax} + {by} \), so it belongs to \( A \) . But \( 0 \leq r < d \) and \( d \) is the smallest positive number in \( A \), so \( r \) can’t be positive; hence \( r = 0 \) . Updating our equation \( a = {qd} + r \), we get \( a = {qd} \), so \( d \mid a \) . Repeating this argument with \( b = {qd} + r \) shows \( d \mid b \) . Thus \( d \) is indeed a common divisor of \( a \) and \( b \) . It remains to show that it is the greatest common divisor.\n\nAs \( \gcd \left( {a, b}\right) \) divides \( a \) and \( b \), we have \( a = \gcd \left( {a, b}\right) \cdot m \) and \( b = \gcd \left( {a, b}\right) \cdot n \) for some \( m, n \in \mathbb{Z} \) . So \( d = {ak} + b\ell = \gcd \left( {a, b}\right) \cdot {mk} + \gcd \left( {a, b}\right) \cdot n\ell = \gcd \left( {a, b}\right) \left( {{mk} + n\ell }\right) \) , and thus \( d \) is a multiple of \( \gcd \left( {a, b}\right) \) . Therefore \( d \geq \gcd \left( {a, b}\right) \) . But \( d \) can’t be a larger common divisor of \( a \) and \( b \) than \( \gcd \left( {a, b}\right) \), so \( d = \gcd \left( {a, b}\right) \) .
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Yes
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