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Corollary 6. A set \( A \subseteq \left( {S,\rho }\right) \) clusters at \( p \) iff each globe \( {G}_{p} \) (about \( p \) ) contains at least one point of \( A \) other than \( p{.}^{5} \)
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Indeed, assume the latter. Then, in particular, each globe\n\n\[ \n{G}_{p}\left( \frac{1}{n}\right) ,\;n = 1,2,\ldots ,\n\]\n\ncontains some point of \( A \) other than \( p \) ; call it \( {x}_{n} \) . We can make the \( {x}_{n} \) distinct by choosing each time \( {x}_{n + 1} \) closer to \( p \) than \( {x}_{n} \) is. It easily follows that each \( {G}_{p}\left( \varepsilon \right) \) contains infinitely many points of \( A \) (the details are left to the reader), as required. The converse is obvious.
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No
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Theorem 1. Let \( {x}_{m} \rightarrow q,{y}_{m} \rightarrow r \), and \( {a}_{m} \rightarrow a \) in \( {E}^{1} \) or \( C \) (the complex field). Then\n\n(i) \( {x}_{m} \pm {y}_{m} \rightarrow q \pm r \) ;
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Proof. (i) By formula (2) of \( §{14} \), we must show that\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists k}\right) \left( {\forall m > k}\right) \;\left| {{x}_{m} \pm {y}_{m} - \left( {q \pm r}\right) }\right| < \varepsilon . \]\n\nThus we fix an arbitrary \( \varepsilon > 0 \) and look for a suitable \( k \) . Since \( {x}_{m} \rightarrow q \) and \( {y}_{m} \rightarrow r \), there are \( {k}^{\prime } \) and \( {k}^{\prime \prime } \) such that\n\n\[ \left( {\forall m > {k}^{\prime }}\right) \;\left| {{x}_{m} - q}\right| < \frac{\varepsilon }{2} \]\n\nand\n\n\[ \left( {\forall m > {k}^{\prime \prime }}\right) \;\left| {{y}_{m} - r}\right| < \frac{\varepsilon }{2} \]\n\n(as \( \varepsilon \) is arbitrary, we may as well replace it by \( \frac{1}{2}\varepsilon \) ). Then both inequalities hold for \( m > k, k = \max \left( {{k}^{\prime },{k}^{\prime \prime }}\right) \) . Adding them, we obtain\n\n\[ \left( {\forall m > k}\right) \;\left| {{x}_{m} - q}\right| + \left| {{y}_{m} - r}\right| < \varepsilon . \]\n\nHence by the triangle law,\n\n\[ \left| {{x}_{m} - q \pm \left( {{y}_{m} - r}\right) }\right| < \varepsilon \text{, i.e.,}\left| {{x}_{m} \pm {y}_{m} - \left( {q \pm r}\right) }\right| < \varepsilon \text{for}m > k\text{,} \]\n\nas required.\n\nThis proof of (i) applies to sequences of vectors as well, without any change.
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Yes
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Corollary 1. Suppose \( \lim {x}_{m} = p \) and \( \lim {y}_{m} = q \) exist in \( {E}^{ * } \). (a) If \( p > q \), then \( {x}_{m} > {y}_{m} \) for all but finitely many \( m \). (b) If \( {x}_{m} \leq {y}_{m} \) for infinitely many \( m \), then \( p \leq q \) ; i.e., \( \lim {x}_{m} \leq \lim {y}_{m} \).
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This is known as passage to the limit in inequalities. Caution: The strict inequalities \( {x}_{m} < {y}_{m} \) do not imply \( p < q \) but only \( p \leq q \). For example, let \[ {x}_{m} = \frac{1}{m}\text{ and }{y}_{m} = 0 \] Then \[ \left( {\forall m}\right) \;{x}_{m} > {y}_{m} \] yet \( \lim {x}_{m} = \lim {y}_{m} = 0 \) .
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No
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Corollary 2. Let \( {x}_{m} \rightarrow p \) in \( {E}^{ * } \), and let \( c \in {E}^{ * } \) (finite or not). Then the following are true:\n\n(a) If \( p > c \) (respectively, \( p < c \) ), we have \( {x}_{m} > c\left( {{x}_{m} < c}\right) \) for all but finitely many \( m \) .\n\n(b) If \( {x}_{m} \leq c \) (respectively, \( {x}_{m} \geq c \) ) for infinitely many \( m \), then \( p \leq c\left( {p \geq c}\right) \) .
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One can prove this from Corollary 1, with \( {y}_{m} = c \) (or \( {x}_{m} = c \) ) for all \( m \) .
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No
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A sequence \( \left\{ {x}_{m}\right\} \subseteq \left( {S,\rho }\right) \) clusters at a point \( p \in S \) iff it has a subsequence \( \left\{ {x}_{{m}_{n}}\right\} \) converging to \( p{.}^{1} \)
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If \( p = \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{{m}_{n}} \), then by definition each globe about \( p \) contains all but finitely many \( {x}_{{m}_{n}} \), hence infinitely many \( {x}_{m} \) . Thus \( p \) is a cluster point.\n\nConversely, if so, consider in particular the globes\n\n\[ \n{G}_{p}\left( \frac{1}{n}\right) ,\;n = 1,2,\ldots \n\]\n\nBy assumption, \( {G}_{p}\left( 1\right) \) contains some \( {x}_{m} \) . Thus fix\n\n\[ \n{x}_{{m}_{1}} \in {G}_{p}\left( 1\right) \n\]\n\nNext, choose a term\n\n\[ \n{x}_{{m}_{2}} \in {G}_{p}\left( \frac{1}{2}\right) \text{ with }{m}_{2} > {m}_{1}. \n\]\n\n(Such terms exist since \( {G}_{p}\left( \frac{1}{2}\right) \) contains infinitely many \( {x}_{m} \) .) Next, fix\n\n\[ \n{x}_{{m}_{3}} \in {G}_{p}\left( \frac{1}{3}\right) \text{, with }{m}_{3} > {m}_{2} > {m}_{1}, \n\]\n\nand so on.\n\nThus, step by step (inductively), select a sequence of subscripts\n\n\[ \n{m}_{1} < {m}_{2} < \cdots < {m}_{n} < \cdots \n\]\n\nthat determines a subsequence (see Chapter 1, §8) such that\n\n\[ \n\left( {\forall n}\right) \;{x}_{{m}_{n}} \in {G}_{p}\left( \frac{1}{n}\right) \text{, i.e.,}\rho \left( {{x}_{{m}_{n}}, p}\right) < \frac{1}{n} \rightarrow 0\text{,} \n\]\n\n--- \n\n\( {}^{1} \) Therefore, cluster points of \( \left\{ {x}_{m}\right\} \) are also called subsequential limits.\n\n--- \n\nwhence \( \rho \left( {{x}_{{m}_{n}}, p}\right) \rightarrow 0 \), or \( {x}_{{m}_{n}} \rightarrow p \) . (Why?) Thus we have found a subsequence \( {x}_{{m}_{n}} \rightarrow p \), and assertion (i) is proved.
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Yes
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(i) Each bounded infinite set or sequence \( A \) in \( {E}^{n}\left( {{}^{ * }\text{or}\left. {C}^{n}\right) }\right. \) has at least one cluster point \( \bar{p} \) there (possibly outside \( A \) ).
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Proof. Take first a bounded sequence \( \left\{ {z}_{m}\right\} \subseteq \left\lbrack {a, b}\right\rbrack \) in \( {E}^{1} \) . Let\n\n\[ p = \overline{\lim }{z}_{m} \]\n\nBy Theorem 2(i) of Chapter 2, \( §{13},\left\{ {z}_{m}\right\} \) clusters at \( p \) . Moreover, as\n\n\[ a \leq {z}_{m} \leq b \]\n\nwe have\n\n\[ a \leq \inf {z}_{m} \leq p \leq \sup {z}_{m} \leq b \]\n\nby Corollary 1 of Chapter 2, §13. Thus\n\n\[ p \in \left\lbrack {a, b}\right\rbrack \subseteq {E}^{1} \]\n\nand so \( \left\{ {z}_{m}\right\} \) clusters in \( {E}^{1} \) .\n\nAssertion (ii) now follows—for \( {E}^{1} \) —by Theorem 1(i) above.
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Yes
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Theorem 3. We have \( p \in \bar{A} \) in \( \left( {S,\rho }\right) \) iff each globe \( {G}_{p}\left( \delta \right) \) about \( p \) meets \( A \) , i.e.,\n\n\[ \left( {\forall \delta > 0}\right) \;A \cap {G}_{p}\left( \delta \right) \neq \varnothing . \]\n\nEquivalently, \( p \in \bar{A} \) iff\n\n\[ p = \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}\text{ for some }\left\{ {x}_{n}\right\} \subseteq A. \]
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The proof is as in Corollary 6 of \( §{14} \) and Theorem 1. (Here, however, the \( {x}_{n} \) need not be distinct or different from \( p \) .) The details are left to the reader.
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No
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Theorem 4. A set \( A \subseteq \left( {S,\rho }\right) \) is closed iff one of the following conditions holds.\n\n(i) \( A \) contains all its cluster points (or has none); i.e., \( A \supseteq {A}^{\prime } \) .\n\n(ii) \( A = \bar{A} \) .\n\n(iii) A contains the limit of each convergent sequence \( \left\{ {x}_{n}\right\} \subseteq A \) (if any). \( {}^{2} \)
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Proof. Parts (i) and (ii) are equivalent since\n\n\[ A \supseteq {A}^{\prime } \Leftrightarrow A = A \cup {A}^{\prime } = \bar{A}.\;\text{ (Explain!) } \]\n\nNow let \( A \) be closed. If \( p \notin A \), then \( p \in - A \) ; therefore, by Definition 3 in \( §{12} \), some \( {G}_{p} \) fails to meet \( A\left( {{G}_{p} \cap A = \varnothing }\right) \) . Hence no \( p \in - A \) is a cluster point, or the limit of a sequence \( \left\{ {x}_{n}\right\} \subseteq A \) . (This would contradict Definitions 1 and 2 of \( §{14} \) .) Consequently, all such cluster points and limits must be in \( A \), as claimed.\n\nConversely, suppose \( A \) is not closed, so \( - A \) is not open. Then \( - A \) has a noninterior point \( p \) ; i.e., \( p \in - A \) but no \( {G}_{p} \) is entirely in \( - A \) . This means that each \( {G}_{p} \) meets \( A \) . Thus\n\n\[ p \in \bar{A}\text{(by Theorem 3),} \]\n\nand\n\n\[ p = \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n}\text{ for some }\left\{ {x}_{n}\right\} \subseteq A\text{ (by the same theorem),} \]\n\neven though \( p \notin A \) (for \( p \in - A \) ).\n\nWe see that (iii) and (ii), hence also (i), fail if \( A \) is not closed and hold if \( A \) is closed. (See the first part of the proof.) Thus the theorem is proved.
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No
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Theorem 1. Every convergent sequence \( \left\{ {x}_{m}\right\} \subseteq \left( {S,\rho }\right) \) is Cauchy.
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Proof. Let \( {x}_{m} \rightarrow p \) . Then given \( \varepsilon > 0 \), there is a \( k \) such that\n\n\[ \left( {\forall m > k}\right) \;\rho \left( {{x}_{m}, p}\right) < \frac{\varepsilon }{2}. \]\n\nAs this holds for any \( m > k \), it also holds for any other term \( {x}_{n} \) with \( n > k \) . Thus\n\n\[ \left( {\forall m, n > k}\right) \;\rho \left( {{x}_{m}, p}\right) < \frac{\varepsilon }{2}\text{ and }\rho \left( {p,{x}_{n}}\right) < \frac{\varepsilon }{2}. \]\n\nAdding and using the triangle inequality, we get\n\n\[ \rho \left( {{x}_{m},{x}_{n}}\right) \leq \rho \left( {{x}_{m}, p}\right) + \rho \left( {p,{x}_{n}}\right) < \varepsilon , \]\n\nand (1) is proved.
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Yes
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Theorem 2. Every Cauchy sequence \( \left\{ {x}_{m}\right\} \subseteq \left( {S,\rho }\right) \) is bounded.
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Proof. We must show that all \( {x}_{m} \) are in some globe. First we try an arbitrary radius \( \varepsilon \) . Then by (1), there is \( k \) such that \( \rho \left( {{x}_{m},{x}_{n}}\right) < \varepsilon \) for \( m, n > k \) . Fix some \( n > k \) . Then\n\n\[ \left( {\forall m > k}\right) \rho \left( {{x}_{m},{x}_{n}}\right) < \varepsilon \text{, i.e.,}{x}_{m} \in {G}_{{x}_{n}}\left( \varepsilon \right) \text{.} \]\n\nThus the globe \( {G}_{{x}_{n}}\left( \varepsilon \right) \) contains all \( {x}_{m} \) except possibly the \( k \) terms \( {x}_{1},\ldots ,{x}_{k} \) . To include them as well, we only have to take a larger radius \( r \), greater than \( \rho \left( {{x}_{m},{x}_{n}}\right), m = 1,\ldots, k \) . Then all \( {x}_{m} \) are in the enlarged globe \( {G}_{{x}_{n}}\left( r\right) \) .
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Yes
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Theorem 3. If a Cauchy sequence \( \left\{ {x}_{m}\right\} \) clusters at a point \( p \), then \( {x}_{m} \rightarrow p \) .
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Proof. We want to show that \( {x}_{m} \rightarrow p \), i.e., that\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists k}\right) \left( {\forall m > k}\right) \;\rho \left( {{x}_{m}, p}\right) < \varepsilon . \]\n\nThus we fix \( \varepsilon > 0 \) and look for a suitable \( k \) . Now as \( \left\{ {x}_{m}\right\} \) is Cauchy, there is\na \( k \) such that\n\n\[ \left( {\forall m, n > k}\right) \;\rho \left( {{x}_{m},{x}_{n}}\right) < \frac{\varepsilon }{2}. \]\n\nAlso, as \( p \) is a cluster point, the globe \( {G}_{p}\left( \frac{\varepsilon }{2}\right) \) contains infinitely many \( {x}_{n} \), so we can fix one with \( n > k \) ( \( k \) as above). Then \( \rho \left( {{x}_{n}, p}\right) < \frac{\varepsilon }{2} \) and, as noted above, also \( \rho \left( {{x}_{m},{x}_{n}}\right) < \frac{\varepsilon }{2} \) for \( m > k \) . Hence\n\n\[ \left( {\forall m > k}\right) \;\rho \left( {{x}_{m},{x}_{n}}\right) + \rho \left( {{x}_{n}, p}\right) < \varepsilon , \]\n\nimplying \( \rho \left( {{x}_{m}, p}\right) \leq \rho \left( {{x}_{m},{x}_{n}}\right) + \rho \left( {{x}_{n}, p}\right) < \varepsilon \), as required.
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Yes
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Theorem 4 (Cauchy’s convergence criterion). A sequence \( \left\{ {\bar{x}}_{m}\right\} \) in \( {E}^{n} \) (*or \( \left. {C}^{n}\right) \) converges if and only if it is a Cauchy sequence.
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Proof. If \( \left\{ {x}_{m}\right\} \) converges, it is Cauchy by Theorem 1.\n\nConversely, let \( \left\{ {x}_{m}\right\} \) be a Cauchy sequence. Then by Theorem 2, it is bounded. Hence by the Bolzano-Weierstrass theorem (Theorem 2 of \( §{16} \) ), it has a cluster point \( \bar{p} \) . Thus by Theorem 3 above, it converges to \( \bar{p} \), and all is proved.
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Yes
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Corollary 1. If \( A \) clusters at \( p \) in \( \left( {S,\rho }\right) \), then a function \( f : A \rightarrow \left( {T,{p}^{\prime }}\right) \) can have at most one limit at \( p \) ; i.e.,
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Proof. Suppose \( f \) has two limits, \( q \) and \( r \), at \( p \) . By the Hausdorff property, \[ {G}_{q}\left( \varepsilon \right) \cap {G}_{r}\left( \varepsilon \right) = \varnothing \;\text{ for some }\varepsilon > 0. \] Also, by (2), there are \( {\delta }^{\prime },{\delta }^{\prime \prime } > 0 \) such that \[ \left( {\forall x \in A \cap {G}_{\neg p}\left( {\delta }^{\prime }\right) }\right) \;f\left( x\right) \in {G}_{q}\left( \varepsilon \right) \text{ and } \] \[ \left( {\forall x \in A \cap {G}_{\neg p}\left( {\delta }^{\prime \prime }\right) }\right) \;f\left( x\right) \in {G}_{r}\left( \varepsilon \right) . \] Let \( \delta = \min \left( {{\delta }^{\prime },{\delta }^{\prime \prime }}\right) \) . Then for \( x \in A \cap {G}_{\neg p}\left( \delta \right), f\left( x\right) \) is in both \( {G}_{q}\left( \varepsilon \right) \) and \( {G}_{r}\left( \varepsilon \right) \) , and such an \( x \) exists since \( A \cap {G}_{\neg p}\left( \delta \right) \neq \varnothing \) by assumption. But this is impossible since \( {G}_{q}\left( \varepsilon \right) \cap {G}_{r}\left( \varepsilon \right) = \varnothing \) (a contradiction!).
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Yes
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Corollary 2. \( f \) is continuous at \( p\left( {p \in {D}_{f}}\right) \) iff \( f\left( x\right) \rightarrow f\left( p\right) \) as \( x \rightarrow p \) .
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The straightforward proof from definitions is left to the reader.
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No
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Theorem 1 (sequential criterion of continuity). (i) A function \[ f : A \rightarrow \left( {T,{\rho }^{\prime }}\right) ,\text{ with }A \subseteq \left( {S,\rho }\right) ,\] is continuous at a point \( p \in A \) iff for every sequence \( \left\{ {x}_{m}\right\} \subseteq A \) such that \( {x}_{m} \rightarrow p \) in \( \left( {S,\rho }\right) \), we have \( f\left( {x}_{m}\right) \rightarrow f\left( p\right) \) in \( \left( {T,{\rho }^{\prime }}\right) \). In symbols, \[ \left( {\forall \left\{ {x}_{m}\right\} \subseteq A \mid {x}_{m} \rightarrow p}\right) \;f\left( {x}_{m}\right) \rightarrow f\left( p\right) . \] \( \left( {1}^{\prime }\right) \)
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Proof. We first prove (ii). Suppose \( q \) is a limit of \( f \) at \( p \), i.e. (see \( §1 \) ), \[ \left( {\forall \varepsilon > 0}\right) \left( {\exists \delta > 0}\right) \left( {\forall x \in A \cap {G}_{\neg p}\left( \delta \right) }\right) \;f\left( x\right) \in {G}_{q}\left( \varepsilon \right) . \] (2) Thus, given \( \varepsilon > 0 \), there is \( \delta > 0 \) (henceforth fixed) such that \[ f\left( x\right) \in {G}_{q}\left( \varepsilon \right) \text{ whenever }x \in A, x \neq p,\text{ and }x \in {G}_{p}\left( \delta \right) . \] (3) We want to deduce \( \left( {2}^{\prime }\right) \). Thus we fix any sequence \[ \left\{ {x}_{m}\right\} \subseteq A - \{ p\} ,{x}_{m} \rightarrow p.{}^{1} \] Then \[ \left( {\forall m}\right) \;{x}_{m} \in A\text{ and }{x}_{m} \neq p, \] and \( {G}_{p}\left( \delta \right) \) contains all but finitely many \( {x}_{m} \). Then these \( {x}_{m} \) satisfy the con- ditions stated in (3). Hence \( f\left( {x}_{m}\right) \in {G}_{q}\left( \varepsilon \right) \) for all but finitely many \( m \). As \( \varepsilon \) is arbitrary, this implies \( f\left( {x}_{m}\right) \rightarrow q \) (by the definition of \( \left. {\mathop{\lim }\limits_{{m \rightarrow \infty }}f\left( {x}_{m}\right) }\right) \), as is required in \( \left( {2}^{\prime }\right) \). Thus \( \left( 2\right) \Rightarrow \left( {2}^{\prime }\right) \). Conversely, suppose (2) fails, i.e., its negation holds. (See the rules for forming negations of such formulas in Chapter 1, §§1-3.) Thus \[ \left( {\exists \varepsilon > 0}\right) \left( {\forall \delta > 0}\right) \left( {\exists x \in A \cap {G}_{\neg p}\left( \delta \right) }\right) \;f\left( x\right) \notin {G}_{q}\left( \varepsilon \right) \] (4) --- \( {}^{1} \) If no such sequence exists, then \( \left( {2}^{\prime }\right) \) is vacuously true and there is nothing to prove. --- by the rules for quantifiers. We fix an \( \varepsilon \) satisfying (4), and let \[ {\delta }_{m} = \frac{1}{m},\;m = 1,2,\ldots \] By (4), for each \( {\delta }_{m} \) there is \( {x}_{m} \) (depending on \( {\delta }_{m} \) ) such that \[ {x}_{m} \in A \cap {G}_{\neg p}\left( \frac{1}{m}\right) \] (5) and \[ f\left( {x}_{m}\right) \notin {G}_{q}\left( \varepsilon \right) ,\;m = 1,2,3,\ldots \] (6) We fix these \( {x}_{m} \). As \( {x}_{m} \in A \) and \( {x}_{m} \neq p \), we obtain a sequence \[ \left\{ {x}_{m}\right\} \subseteq A - \{ p\} \] Also, as \( {x}_{m} \in {G}_{p}\left( \frac{1}{m}\right) \), we have \( \rho \left( {{x}_{m}, p}\right) < 1/m \rightarrow 0 \), and hence \( {x}_{m} \rightarrow p \). On the other hand, by (6), the image sequence \( \left\{ {f\left( {x}_{m}\right) }\right\} \) cannot converge to \( q \) (why?), i.e., \( \left( {2}^{\prime }\right) \) fails. Thus we see that \( \left( {2}^{\prime }\right) \) fails or holds accordingly as (2) does. This proves assertion (ii). Now, by setting \( q = f\left
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Yes
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Corollary 1. Let \( \\left( {T,{\\rho }^{\\prime }}\\right) \) be complete, such as \( {E}^{n} \) . Let a map \( f : A \\rightarrow T \) with \( A \\subseteq \\left( {S,\\rho }\\right) \) and a point \( p \\in S \) be given. Then for \( f \) to have a limit at \( p \) , it suffices that \( \\left\\{ {f\\left( {x}_{m}\\right) }\\right\\} \) be Cauchy in \( \\left( {T,{\\rho }^{\\prime }}\\right) \) whenever \( \\left\\{ {x}_{m}\\right\\} \\subseteq A - \\{ p\\} \) and \( {x}_{m} \\rightarrow p \) in \( \\left( {S,\\rho }\\right) \) .
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Indeed, as noted above, all such \( \\left\\{ {f\\left( {x}_{m}\\right) }\\right\\} \) converge. Thus it only remains to show that they tend to one and the same limit \( q \), as is required in part (ii) of Theorem 1. We leave this as an exercise (Problem 1 below).
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No
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Theorem 3. Let \( \left( {S,\rho }\right) ,\left( {T,{\rho }^{\prime }}\right) \), and \( \left( {U,{\rho }^{\prime \prime }}\right) \) be metric spaces. If a function \( f : S \rightarrow T \) is continuous at a point \( p \in S \), and if \( g : T \rightarrow U \) is continuous at the point \( q = f\left( p\right) \), then the composite function \( g \circ f \) is continuous at \( p \) .
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Proof. The domain of \( g \circ f \) is \( S \) . So take any sequence\n\n\[ \left\{ {x}_{m}\right\} \subseteq S\text{with}{x}_{m} \rightarrow p\text{.} \]\n\nAs \( f \) is continuous at \( p \), formula \( \left( {1}^{\prime }\right) \) yields \( f\left( {x}_{m}\right) \rightarrow f\left( p\right) \), where \( f\left( {x}_{m}\right) \) is in \( T = {D}_{g} \) . Hence, as \( g \) is continuous at \( f\left( p\right) \), we have\n\n\[ g\left( {f\left( {x}_{m}\right) }\right) \rightarrow g\left( {f\left( p\right) }\right) \text{, i.e.,}\left( {g \circ f}\right) \left( {x}_{m}\right) \rightarrow \left( {g \circ f}\right) \left( p\right) \text{,} \]\n\nand this holds for any \( \left\{ {x}_{m}\right\} \subseteq S \) with \( {x}_{m} \rightarrow p \) . Thus \( g \circ f \) satisfies condition \( \left( {1}^{\prime }\right) \) and is continuous at \( p \) .
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Yes
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Corollary 2. With the notation of Theorem 3, suppose\n\n\\[ \nf\\left( x\\right) \\rightarrow q\\text{as}x \\rightarrow p\\text{, and}g\\left( y\\right) \\rightarrow r\\text{as}y \\rightarrow q\\text{.}\n\\]\n\nThen\n\n\\[ \ng\\left( {f\\left( x\\right) }\\right) \\rightarrow r\\text{ as }x \\rightarrow p,\n\\]\n\nprovided, however, that\n\n(i) \\( g \\) is continuous at \\( q \\), or\n\n(ii) \\( f\\left( x\\right) \\neq q \\) for \\( x \\) in some deleted globe about \\( p \\), or\n\n(iii) \\( f \\) is one to one, at least when restricted to some \\( {G}_{\\neg p}\\left( \\delta \\right) \\) .
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Indeed, (i) and (ii) suffice, as was explained above. Thus assume (iii). Then \\( f \\) can take the value \\( q \\) at most once, say, at some point\n\n\\[ \n{x}_{0} \\in {G}_{\\neg p}\\left( \\delta \\right)\n\\]\n\nAs \\( {x}_{0} \\neq p \\), let\n\n\\[ \n{\\delta }^{\\prime } = \\rho \\left( {{x}_{0}, p}\\right) > 0.\n\\]\n\nThen \\( {x}_{0} \\notin {G}_{\\neg p}\\left( {\\delta }^{\\prime }\\right) \\), so \\( f\\left( x\\right) \\neq q \\) on \\( {G}_{\\neg p}\\left( {\\delta }^{\\prime }\\right) \\), and case (iii) reduces to (ii).
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Yes
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Theorem 1. For any functions \( f, g, h : A \rightarrow {E}^{1}\left( C\right), A \subseteq \left( {S,\rho }\right) \), we have the following:\n\n(i) If \( f, g, h \) are continuous at \( p\left( {p \in A}\right) \), so are \( f \pm g \) and \( {fh} \). So also is \( f/h \), provided \( h\left( p\right) \neq 0 \); similarly for relative continuity over \( B \subseteq A \).\n\n(ii) If \( f\left( x\right) \rightarrow q, g\left( x\right) \rightarrow r \), and \( h\left( x\right) \rightarrow a \) (all, as \( x \rightarrow p \) over \( B \subseteq A \) ), then\n\n(a) \( f\left( x\right) \pm g\left( x\right) \rightarrow q \pm r \);\n\n(b) \( f\left( x\right) h\left( x\right) \rightarrow {qa} \);\n\n(c) \( \frac{f\left( x\right) }{h\left( x\right) } \rightarrow \frac{q}{a} \), provided \( a \neq 0 \).\n\nAll this holds also if \( f \) and \( g \) are vector valued and \( h \) is scalar valued.
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For a simple proof, one can use Theorem 1 of Chapter 3, §15. (An independent proof is sketched in Problems 1-7 below.)\n\nWe can also use the sequential criterion (Theorem 1 in \( §2 \) ). To prove (ii), take any sequence\n\n\[ \left\{ {x}_{m}\right\} \subseteq B - \{ p\} ,{x}_{m} \rightarrow p. \]\n\nThen by the assumptions made,\n\n\[ f\left( {x}_{m}\right) \rightarrow q, g\left( {x}_{m}\right) \rightarrow r\text{, and}h\left( {x}_{m}\right) \rightarrow a\text{.} \]\n\nThus by Theorem 1 of Chapter 3, §15,\n\n\[ f\left( {x}_{m}\right) \pm g\left( {x}_{m}\right) \rightarrow q \pm r, f\left( {x}_{m}\right) g\left( {x}_{m}\right) \rightarrow {qa},\text{ and }\frac{f\left( {x}_{m}\right) }{g\left( {x}_{m}\right) } \rightarrow \frac{q}{a}. \]\n\nAs this holds for any sequence \( \left\{ {x}_{m}\right\} \subseteq B - \{ p\} \) with \( {x}_{m} \rightarrow p \), our assertion (ii) follows by the sequential criterion; similarly for (i).
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No
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Theorem 2 (componentwise continuity and limits). For any function \( f : A \rightarrow \) \( {E}^{n}\left( {{}^{ * }{C}^{n}}\right) \), with \( A \subseteq \left( {S,\rho }\right) \) and with \( f = \left( {{f}_{1},\ldots ,{f}_{n}}\right) \), we have that\n\n(i) \( f \) is continuous at \( p\left( {p \in A}\right) \) iff all its components \( {f}_{k} \) are, and\n\n(ii) \( f\left( x\right) \rightarrow \bar{q} \) as \( x \rightarrow p\left( {p \in S}\right) \) iff\n\n\[{f}_{k}\left( x\right) \rightarrow {q}_{k}\text{ as }x \rightarrow p\;\left( {k = 1,2,\ldots, n}\right) ,\]\ni.e., iff each \( {f}_{k} \) has, as its limit at \( p \), the corresponding component of \( \bar{q} \) .
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We prove (ii). If \( f\left( x\right) \rightarrow \bar{q} \) as \( x \rightarrow p \) then, by definition,\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists \delta > 0}\right) \left( {\forall x \in A \cap {G}_{\neg p}\left( \delta \right) }\right) \;\varepsilon > \left| {f\left( x\right) - \bar{q}}\right| = \sqrt{\mathop{\sum }\limits_{{k = 1}}^{n}{\left| {f}_{k}\left( x\right) - {q}_{k}\right| }^{2}}; \]\n\nin turn, the right-hand side of the inequality given above is no less than each\n\n\[ \left| {{f}_{k}\left( x\right) - {q}_{k}}\right| ,\;k = 1,2,\ldots, n. \]\n\nThus\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists \delta > 0}\right) \left( {\forall x \in A \cap {G}_{\neg p}\left( \delta \right) }\right) \;\left| {{f}_{k}\left( x\right) - {q}_{k}}\right| < \varepsilon \]\n\ni.e., \( {f}_{k}\left( x\right) \rightarrow {q}_{k}, k = 1,\ldots, n \) .\n\nConversely, if each \( {f}_{k}\left( x\right) \rightarrow {q}_{k} \), then Theorem 1(ii) yields\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{n}{\bar{e}}_{k}{f}_{k}\left( x\right) \rightarrow \mathop{\sum }\limits_{{k = 1}}^{n}{\bar{e}}_{k}{q}_{k}.{}^{1} \]\n\nBy formula (1), then, \( f\left( x\right) \rightarrow \bar{q} \) (for \( \mathop{\sum }\limits_{{k = 1}}^{n}{\bar{e}}_{k}{q}_{k} = \bar{q} \) ). Thus (ii) is proved;\n\nsimilarly for (i) and for relative limits and continuity.
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Yes
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Theorem 3. Any rational function (in particular, every polynomial) in one or several variables is continuous on all of its domain.
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Proof. Consider first a monomial of the form\n\n\[ f\left( \bar{x}\right) = {x}_{k}\;\left( {k\text{ fixed }}\right) \]\n\nit is called the kth projection map because it \
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No
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Theorem 2. If a function \( f : A \rightarrow {E}^{ * }\left( {A \subseteq {E}^{ * }}\right) \) is nondecreasing on a finite or infinite interval \( B = \left( {a, b}\right) \subseteq A \) and if \( p \in \left( {a, b}\right) \), then\n\n\[ f\left( {a}^{ + }\right) \leq f\left( {p}^{ - }\right) \leq f\left( p\right) \leq f\left( {p}^{ + }\right) \leq f\left( {b}^{ - }\right) ,\]\n\nand for no \( x \in \left( {a, b}\right) \) do we have\n\n\[ f\left( {p}^{ - }\right) < f\left( x\right) < f\left( p\right) \text{ or }f\left( p\right) < f\left( x\right) < f\left( {p}^{ + }\right) {;}^{1} \]\n\nsimilarly in case \( f \downarrow \) (with all inequalities reversed).
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Proof. By Theorem \( 1, f \uparrow \) on \( \left( {a, p}\right) \) implies\n\n\[ f\left( {a}^{ + }\right) = \mathop{\inf }\limits_{{a < x < p}}f\left( x\right) \text{ and }f\left( {p}^{ - }\right) = \mathop{\sup }\limits_{{a < x < p}}f\left( x\right) ; \]\n\nthus certainly \( f\left( {a}^{ + }\right) \leq f\left( {p}^{ - }\right) \) . As \( f \uparrow \), we also have \( f\left( p\right) \geq f\left( x\right) \) for all \( x \in \) \( \left( {a, p}\right) \) ; hence\n\n\[ f\left( p\right) \geq \mathop{\sup }\limits_{{a < x < p}}f\left( x\right) = f\left( {p}^{ - }\right) . \]\n\nThus\n\n\[ f\left( {a}^{ + }\right) \leq f\left( {p}^{ - }\right) \leq f\left( p\right) \]\n\nsimilarly for the rest of (1).\n\nMoreover, if \( a < x < p \), then \( f\left( x\right) \leq f\left( {p}^{ - }\right) \) since\n\n\[ f\left( {p}^{ - }\right) = \mathop{\sup }\limits_{{a < x < p}}f\left( x\right) \]\n\nIf, however, \( p \leq x < b \), then \( f\left( p\right) \leq f\left( x\right) \) since \( f \uparrow \) . Thus we never have \( f\left( {p}^{ - }\right) < f\left( x\right) < f\left( p\right) \) . Similarly, one excludes \( f\left( p\right) < f\left( x\right) < f\left( {p}^{ + }\right) \) . This completes the proof.
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Yes
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Theorem 1. If a set \( B \subseteq \left( {S,\rho }\right) \) is compact, so is any closed subset \( A \subseteq B \) .
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Proof. We must show that each sequence \( \left\{ {x}_{m}\right\} \subseteq A \) clusters at some \( p \in A \) . However, as \( A \subseteq B,\left\{ {x}_{m}\right\} \) is also in \( B \), so by the compactness of \( B \), it clusters at some \( p \in B \) . Thus it remains to show that \( p \in A \) as well.\n\nNow by Theorem 1 of Chapter \( 3,§{16},\left\{ {x}_{m}\right\} \) has a subsequence \( {x}_{{m}_{k}} \rightarrow p \) . As \( \left\{ {x}_{{m}_{k}}\right\} \subseteq A \) and \( A \) is closed, this implies \( p \in A \) (Theorem 4 in Chapter 3, §16). \( ▱ \)
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Yes
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Theorem 2. Every compact set \( A \subseteq \left( {S,\rho }\right) \) is closed.
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Proof. Given that \( A \) is compact, we must show (by Theorem 4 in Chapter 3, §16) that \( A \) contains the limit of each convergent sequence \( \left\{ {x}_{m}\right\} \subseteq A \) . Thus let \( {x}_{m} \rightarrow p,\left\{ {x}_{m}\right\} \subseteq A \) . As \( A \) is compact, the sequence \( \left\{ {x}_{m}\right\} \) clusters at some \( q \in A \), i.e., has a subsequence \( {x}_{{m}_{k}} \rightarrow q \in A \) . However, the limit of the subsequence must be the same as that of the entire sequence. Thus \( p = q \in A \) ; i.e., \( p \) is in \( A \), as required.
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Yes
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Theorem 3. Every compact set \( A \subseteq \left( {S,\rho }\right) \) is bounded.
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Proof. By Problem 3 in Chapter 3, \( §{13} \), it suffices to show that \( A \) is contained in some finite union of globes. Thus we fix some arbitrary radius \( \varepsilon > 0 \) and, seeking a contradiction, assume that \( A \) cannot be covered by any finite number of globes of that radius.\n\nThen if \( {x}_{1} \in A \), the globe \( {G}_{{x}_{1}}\left( \varepsilon \right) \) does not cover \( A \), so there is a point \( {x}_{2} \in A \) such that\n\n\[{x}_{2} \notin {G}_{{x}_{1}}\left( \varepsilon \right) \text{, i.e.,}\rho \left( {{x}_{1},{x}_{2}}\right) \geq \varepsilon \text{.}\]\n\nBy our assumption, \( A \) is not even covered by \( {G}_{{x}_{1}}\left( \varepsilon \right) \cup {G}_{{x}_{2}}\left( \varepsilon \right) \) . Thus there is a point \( {x}_{3} \in A \) with\n\n\[{x}_{3} \notin {G}_{{x}_{1}}\left( \varepsilon \right) \text{and}{x}_{3} \notin {G}_{{x}_{2}}\left( \varepsilon \right) \text{, i.e.,}\rho \left( {{x}_{3},{x}_{1}}\right) \geq \varepsilon \text{and}\rho \left( {{x}_{3},{x}_{2}}\right) \geq \varepsilon \text{.}\]\n\nAgain, \( A \) is not covered by \( \mathop{\bigcup }\limits_{{i = 1}}^{3}{G}_{{x}_{i}}\left( \varepsilon \right) \), so there is a point \( {x}_{4} \in A \) not in that union; its distances from \( {x}_{1},{x}_{2} \), and \( {x}_{3} \) must therefore be \( \geq \varepsilon \) .\n\nSince \( A \) is never covered by any finite number of \( \varepsilon \) -globes, we can continue this process indefinitely (by induction) and thus select an infinite sequence \( \left\{ {x}_{m}\right\} \subseteq A \), with all its terms at least \( \varepsilon \) -apart from each other.\n\nNow as \( A \) is compact, this sequence must have a convergent subsequence \( \left\{ {x}_{{m}_{k}}\right\} \), which is then certainly Cauchy (by Theorem 1 of Chapter 3,§17). This is impossible, however, since its terms are at distances \( \geq \varepsilon \) from each other, contrary to Definition 1 in Chapter 3, §17. This contradiction completes the proof.
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Yes
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Theorem 4. In \( {E}^{n}\left( {{}^{ * }\text{and}\left. {C}^{n}\right) }\right. \) a set is compact iff it is closed and bounded.
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Proof. In fact, if a set \( A \subseteq {E}^{n}\left( {{}^{ * }{C}^{n}}\right) \) is bounded, then by the Bolzano-Weierstrass theorem, each sequence \( \left\{ {x}_{m}\right\} \subseteq A \) has a convergent subsequence \( {x}_{{m}_{k}} \rightarrow p \) . If \( A \) is also closed, the limit point \( p \) must belong to \( A \) itself.\n\nThus each sequence \( \left\{ {x}_{m}\right\} \subseteq A \) clusters at some \( p \) in \( A \), so \( A \) is compact.\n\nThe converse is obvious.
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Yes
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Theorem 5 (Cantor's principle of nested closed sets). Every contracting sequence of nonvoid compact sets,\n\n\[ \n{F}_{1} \supseteq {F}_{2} \supseteq \cdots \supseteq {F}_{m} \supseteq \cdots ,\n\]\n\nin a metric space \( \left( {S,\rho }\right) \) has a nonvoid intersection; i.e., some \( p \) belongs to all \( {F}_{m} \) .
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Proof. We prove the theorem for complete sets first.\n\nAs \( {F}_{m} \neq \varnothing \), we can pick a point \( {x}_{m} \) from each \( {F}_{m} \) to obtain a sequence \( \left\{ {x}_{m}\right\} ,{x}_{m} \in {F}_{m} \) . Since \( d{F}_{m} \rightarrow 0 \), it is easy to see that \( \left\{ {x}_{m}\right\} \) is a Cauchy sequence. (The details are left to the reader.) Moreover,\n\n\[ \n\left( {\forall m}\right) \;{x}_{m} \in {F}_{m} \subseteq {F}_{1}\n\]\n\nThus \( \left\{ {x}_{m}\right\} \) is a Cauchy sequence in \( {F}_{1} \), a complete set (by assumption).\n\nTherefore, by the definition of completeness (Chapter \( 3,§{17} \) ), \( \left\{ {x}_{m}\right\} \) has a limit \( p \in {F}_{1} \) . This limit remains the same if we drop a finite number of terms, say, the first \( m - 1 \) of them. Then we are left with the sequence \( {x}_{m},{x}_{m + 1},\ldots \) , which, by construction, is entirely contained in \( {F}_{m} \) (why?), with the same limit \( p \) . Then, however, the completeness of \( {F}_{m} \) implies that \( p \in {F}_{m} \) as well. As \( m \) is arbitrary here, it follows that \( \left( {\forall m}\right) p \in {F}_{m} \), i.e.,\n\n\[ \np \in \mathop{\bigcap }\limits_{{m = 1}}^{\infty }{F}_{m}\text{, as claimed.}\n\]\n\nThe proof for compact sets is analogous and even simpler. Here \( \left\{ {x}_{m}\right\} \) need not be a Cauchy sequence. Instead, using the compactness of \( {F}_{1} \), we select from \( \left\{ {x}_{m}\right\} \) a subsequence \( {x}_{{m}_{k}} \rightarrow p \in {F}_{1} \) and then proceed as above.
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No
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Theorem 1 (Lebesgue). Every open covering \( \left\{ {G}_{j}\right\} \) of a sequentially compact set \( F \subseteq \left( {S,\rho }\right) \) has at least one Lebesgue number \( \varepsilon \) . In symbols,\n\n\[ \left( {\exists \varepsilon > 0}\right) \left( {\forall x \in F}\right) \left( {\exists i}\right) \;{G}_{x}\left( \varepsilon \right) \subseteq {G}_{i}. \]
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Proof. Seeking a contradiction, assume that (1) fails, i.e., its negation holds. As was explained in Chapter 1, §§1-3, this negation is\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists {x}_{\varepsilon } \in F}\right) \left( {\forall i}\right) \;{G}_{{x}_{\varepsilon }}\left( \varepsilon \right) \nsubseteq {G}_{i} \]\n\n(where we write \( {x}_{\varepsilon } \) for \( x \) since here \( x \) may depend on \( \varepsilon \) ). As this is supposed to hold for all \( \varepsilon > 0 \), we take successively\n\n\[ \varepsilon = 1,\frac{1}{2},\ldots ,\frac{1}{n},\ldots \]\n\nThen, replacing \
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Yes
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Theorem 2 (generalized Heine-Borel theorem). A set \( F \subseteq \left( {S,\rho }\right) \) is compact iff every open covering of \( F \) has a finite subcovering.
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Proof. Let \( F \) be sequentially compact, and let \( F \subseteq \bigcup {G}_{i} \), all \( {G}_{i} \) open. We have to show that \( \left\{ {G}_{i}\right\} \) reduces to a finite subcovering.\n\nBy Theorem \( 1,\left\{ {G}_{i}\right\} \) has a Lebesgue number \( \varepsilon \) satisfying (1). We fix this \( \varepsilon > 0 \) . Now by Note 1 in \( §6 \), we can cover \( F \) by a finite number of \( \varepsilon \) -globes,\n\n\[ F \subseteq \mathop{\bigcup }\limits_{{k = 1}}^{n}{G}_{{x}_{k}}\left( \varepsilon \right) ,\;{x}_{k} \in F.\]\n\nAlso by (1), each \( {G}_{{x}_{k}}\left( \varepsilon \right) \) is contained in some \( {G}_{i} \) ; call it \( {G}_{{i}_{k}} \) . With the \( {G}_{{i}_{k}} \) so fixed, we have\n\n\[ F \subseteq \mathop{\bigcup }\limits_{{k = 1}}^{n}{G}_{{x}_{k}}\left( \varepsilon \right) \subseteq \mathop{\bigcup }\limits_{{k = 1}}^{n}{G}_{{i}_{k}}\]\n\nThus the sets \( {G}_{{i}_{k}} \) constitute the desired finite subcovering, and the \
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Yes
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Theorem 1. If a function \( f : A \rightarrow \left( {T,{\rho }^{\prime }}\right), A \subseteq \left( {S,\rho }\right) \), is relatively continuous on a compact set \( B \subseteq A \), then \( f\left\lbrack B\right\rbrack \) is a compact set in \( \left( {T,{\rho }^{\prime }}\right) \) . Briefly, the continuous image of a compact set is compact.
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Proof. To show that \( f\left\lbrack B\right\rbrack \) is compact, we take any sequence \( \left\{ {y}_{m}\right\} \subseteq f\left\lbrack B\right\rbrack \) and prove that it clusters at some \( q \in f\left\lbrack B\right\rbrack \) . As \( {y}_{m} \in f\left\lbrack B\right\rbrack ,{y}_{m} = f\left( {x}_{m}\right) \) for some \( {x}_{m} \) in \( B \) . We pick such an \( {x}_{m} \in B \) for each \( {y}_{m} \), thus obtaining a sequence \( \left\{ {x}_{m}\right\} \subseteq B \) with \[ f\left( {x}_{m}\right) = {y}_{m},\;m = 1,2,\ldots \] Now by the assumed compactness of \( B \), the sequence \( \left\{ {x}_{m}\right\} \) must cluster at some \( p \in B \) . Thus it has a subsequence \( {x}_{{m}_{k}} \rightarrow p \) . As \( p \in B \), the function \( f \) is relatively continuous at \( p \) over \( B \) (by assumption). Hence by the sequential criterion (§2), \( {x}_{{m}_{k}} \rightarrow p \) implies \( f\left( {x}_{{m}_{k}}\right) \rightarrow f\left( p\right) \) ; i.e., \[ {y}_{{m}_{k}} \rightarrow f\left( p\right) \in f\left\lbrack B\right\rbrack \] Thus \( q = f\left( p\right) \) is the desired cluster point of \( \left\{ {y}_{m}\right\} \) .
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Yes
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Lemma 1. Every nonvoid compact set \( F \subseteq {E}^{1} \) has a maximum and a minimum.
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Proof. By Theorems 2 and 3 of \( §6, F \) is closed and bounded. Thus \( F \) has an infimum and a supremum in \( {E}^{1} \) (by the completeness axiom), say, \( p = \inf F \) and \( q = \sup F \) . It remains to show that \( p, q \in F \) . Assume the opposite, say, \( q \notin F \) . Then by properties of suprema, each globe \( {G}_{q}\left( \delta \right) = \left( {q - \delta, q + \delta }\right) \) contains some \( x \in B \) (specifically, \( q - \delta < x < q \) ) other than \( q \) (for \( q \notin B \), while \( x \in B \) ). Thus \[ \left( {\forall \delta > 0}\right) \;F \cap {G}_{\neg q}\left( \delta \right) \neq \varnothing \] i.e., \( F \) clusters at \( q \) and hence must contain \( q \) (being closed). However, since \( q \notin F \), this is the desired contradiction, and the lemma is proved.
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Yes
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Theorem 2 (Weierstrass).\n\n(i) If a function \( f : A \rightarrow \left( {T,{\rho }^{\prime }}\right) \) is relatively continuous on a compact set \( B \subseteq A \), then \( f \) is bounded on \( B \) ; i.e., \( f\left\lbrack B\right\rbrack \) is bounded.\n\n(ii) If, in addition, \( B \neq \varnothing \) and \( f \) is real \( \left( {f : A \rightarrow {E}^{1}}\right) \), then \( f\left\lbrack B\right\rbrack \) has a maximum and a minimum; i.e., \( f \) attains a largest and a least value at some points of \( B \) .
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Proof. Indeed, by Theorem \( 1, f\left\lbrack B\right\rbrack \) is compact, so it is bounded, as claimed in (i).\n\nIf further \( B \neq \varnothing \) and \( f \) is real, then \( f\left\lbrack B\right\rbrack \) is a nonvoid compact set in \( {E}^{1} \), so by Lemma 1, it has a maximum and a minimum in \( {E}^{1} \) . Thus all is proved.
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Yes
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Theorem 3. If a function \( f : A \rightarrow \left( {T,{\rho }^{\prime }}\right), A \subseteq \left( {S,\rho }\right) \), is relatively continuous on a compact set \( B \subseteq A \) and is one to one on \( B \) (i.e., when restricted to \( B \) ), then its inverse, \( {f}^{-1} \), is continuous on \( f\left\lbrack B\right\rbrack {\text{.}}^{1} \)
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Proof. To show that \( {f}^{-1} \) is continuous at each point \( q \in f\left\lbrack B\right\rbrack \), we apply the sequential criterion (Theorem 1 in \( §2 \) ). Thus we fix a sequence \( \left\{ {y}_{m}\right\} \subseteq f\left\lbrack B\right\rbrack \) , \( {y}_{m} \rightarrow q \in f\left\lbrack B\right\rbrack \), and prove that \( {f}^{-1}\left( {y}_{m}\right) \rightarrow {f}^{-1}\left( q\right) \) .\n\n\( {}^{1} \) Note that \( f \) need not be one to one on \( {all} \) of its domain \( A \), only on \( B \) . Thus \( {f}^{-1} \) need not be a mapping on \( f\left\lbrack A\right\rbrack \), but it is one on \( f\left\lbrack B\right\rbrack \) . (We use \
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Yes
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Theorem 4. If a function \( f : A \rightarrow \left( {T,{\rho }^{\prime }}\right), A \subseteq \left( {S,\rho }\right) \), is relatively continuous on a compact set \( B \subset A \), then \( f \) is also uniformly continuous on \( B \) .
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Proof (by contradiction). Suppose \( f \) is relatively continuous on \( B \), but (4) fails. Then there is an \( \varepsilon > 0 \) such that\n\n\[ \left( {\forall \delta > 0}\right) \left( {\exists p, x \in B}\right) \;\rho \left( {x, p}\right) < \delta ,\text{ and yet }{\rho }^{\prime }\left( {f\left( x\right), f\left( p\right) }\right) \geq \varepsilon ; \]\n\nhere \( p \) and \( x \) depend on \( \delta \) . We fix such an \( \varepsilon \) and let\n\n\[ \delta = 1,\frac{1}{2},\ldots ,\frac{1}{m},\ldots \]\n\nThen for each \( \delta \) (i.e., each \( m \) ), we get two points \( {x}_{m},{p}_{m} \in B \) with\n\n\[ \rho \left( {{x}_{m},{p}_{m}}\right) < \frac{1}{m} \]\n\n(5)\n\nand\n\n\[ {\rho }^{\prime }\left( {f\left( {x}_{m}\right), f\left( {p}_{m}\right) }\right) \geq \varepsilon ,\;m = 1,2,\ldots \]\n\n(6)\n\nThus we obtain two sequences, \( \left\{ {x}_{m}\right\} \) and \( \left\{ {p}_{m}\right\} \), in \( B \) . As \( B \) is compact, \( \left\{ {x}_{m}\right\} \) has a subsequence \( {x}_{{m}_{k}} \rightarrow q\left( {q \in B}\right) \) . For simplicity, let it be \( \left\{ {x}_{m}\right\} \) itself; thus\n\n\[ {x}_{m} \rightarrow q,\;q \in B. \]\n\nHence by (5), it easily follows that also \( {p}_{m} \rightarrow q \) (because \( \rho \left( {{x}_{m},{p}_{m}}\right) \rightarrow 0 \) ; see Problem 4 in Chapter 3,§17). By the assumed relative continuity of \( f \) on \( B \) , it follows that\n\n\[ f\left( {x}_{m}\right) \rightarrow f\left( q\right) \text{ and }f\left( {p}_{m}\right) \rightarrow f\left( q\right) \text{ in }\left( {T,{\rho }^{\prime }}\right) . \]\n\nThis, in turn, implies that \( {\rho }^{\prime }\left( {f\left( {x}_{m}\right), f\left( {p}_{m}\right) }\right) \rightarrow 0 \), which is impossible, in view of (6). This contradiction completes the proof.
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Yes
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Lemma 1 (principle of nested line segments). Every contracting sequence of closed line segments \( L\left\lbrack {{\bar{p}}_{m},{\bar{q}}_{m}}\right\rbrack \) in \( {E}^{n} \) (*or in any other normed space) has a nonvoid intersection; i.e., there is a point\n\n\[ \bar{p} \in \mathop{\bigcap }\limits_{{m = 1}}^{\infty }L\left\lbrack {{\bar{p}}_{m},{\bar{q}}_{m}}\right\rbrack \]
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Proof. Use Cantor’s theorem (Theorem 5 of \( §6 \) ) and Example (1) in \( §8 \) .
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No
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Theorem 2. If a function \( f : A \rightarrow {E}^{1} \) is monotone and has the Darboux property on a finite or infinite interval \( \left( {a, b}\right) \subseteq A \subseteq {E}^{1} \), then it is continuous on \( \left( {a, b}\right) \) .
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Proof. Seeking a contradiction, suppose \( f \) is discontinuous at some \( p \in \left( {a, b}\right) \) . For definiteness, let \( f \uparrow \) on \( \left( {a, b}\right) \) . Then by Theorems 2 and 3 in \( §5 \), we have either \( f\left( {p}^{ - }\right) < f\left( p\right) \) or \( f\left( p\right) < f\left( {p}^{ + }\right) \) or both, with no function values in between. On the other hand, since \( f \) has the Darboux property, the function values \( f\left( x\right) \) for \( x \) in \( \left( {a, b}\right) \) fill an entire interval (see Note 1). Thus it is impossible for \( f\left( p\right) \) to be the only function value between \( f\left( {p}^{ - }\right) \) and \( f\left( {p}^{ + }\right) \) unless \( f \) is constant near \( p \), but then it is also continuous at \( p \), which we excluded. This contradiction completes the proof. \( {}^{3} \)
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Yes
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Theorem 3. If \( f : A \rightarrow {E}^{1} \) is strictly monotone and continuous when restricted to a finite or infinite interval \( B \subseteq A \subseteq {E}^{1} \), then its inverse \( {f}^{-1} \) has the same properties on the set \( f\left\lbrack B\right\rbrack \) (itself an interval, by Note 1 and Theorem 1). \( {}^{4} \)
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Proof. It is easy to see that \( {f}^{-1} \) is increasing (decreasing) if \( f \) is; the proof is left as an exercise. Thus \( {f}^{-1} \) is monotone on \( f\left\lbrack B\right\rbrack \) if \( f \) is so on \( B \) . To prove the relative continuity of \( {f}^{-1} \), we use Theorem 2, i.e., show that \( {f}^{-1} \) has the Darboux property on \( f\left\lbrack B\right\rbrack \) . Thus let \( {f}^{-1}\left( p\right) < c < {f}^{-1}\left( q\right) \) for some \( p, q \in f\left\lbrack B\right\rbrack \) . We look for an \( r \in f\left\lbrack B\right\rbrack \) such that \( {f}^{-1}\left( r\right) = c \), i.e., \( r = f\left( c\right) \) . Now since \( p, q \in f\left\lbrack B\right\rbrack \), the numbers \( {f}^{-1}\left( p\right) \) and \( {f}^{-1}\left( q\right) \) are in \( B \), an interval. Hence also the intermediate value \( c \) is in \( B \) ; thus it belongs to the domain of \( f \), and so the function value \( f\left( c\right) \) exists. It thus suffices to put \( r = f\left( c\right) \) to get the result.
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No
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Theorem 1. A function \( f : \left( {A,\rho }\right) \rightarrow \left( {T,{\rho }^{\prime }}\right) \) is continuous on \( A \) iff \( {f}^{-1}\left\lbrack B\right\rbrack \) is closed in \( \left( {A,\rho }\right) \) for each closed set \( B \subseteq \left( {T,{\rho }^{\prime }}\right) \) ; similarly for open sets.
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Indeed, this is part of Problem 15 in \( §2 \) with \( \left( {S,\rho }\right) \) replaced by \( \left( {A,\rho }\right) \).
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No
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Theorem 2. The only connected sets in \( {E}^{1} \) are exactly all convex sets, i.e., finite and infinite intervals, including \( {E}^{1} \) itself.
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Proof. The proof that such intervals are exactly all convex sets in \( {E}^{1} \) is left as an exercise.\n\nWe now show that each connected set \( A \subseteq {E}^{1} \) is convex, i.e., that \( a, b \in A \) implies \( \left( {a, b}\right) \subseteq A \) .\n\nSeeking a contradiction, suppose \( p \notin A \) for some \( p \in \left( {a, b}\right), a, b \in A \) . Let\n\n\[ P = A \cap \left( {-\infty, p}\right) \text{ and }Q = A \cap \left( {p, + \infty }\right) . \]\n\nThen \( A = P \cup Q, a \in P, b \in Q \), and \( P \cap Q = \varnothing \) . Moreover, \( \left( {-\infty, p}\right) \) and \( \left( {p, + \infty }\right) \) are open sets in \( {E}^{1} \) . (Why?) Hence \( P \) and \( Q \) are open in \( A \), each being the intersection of \( A \) with a set open in \( {E}^{1} \) (see Note 1 above). As \( A = P \cup Q \), with \( P \cap Q = \varnothing \), it follows that \( A \) is disconnected. This shows that if \( A \) is connected in \( {E}^{1} \), it must be convex.
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No
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Theorem 3. If a function \( f : A \rightarrow \left( {T,{\rho }^{\prime }}\right) \) with \( A \subseteq \left( {S,\rho }\right) \) is relatively continuous on a connected set \( B \subseteq A \), then \( f\left\lbrack B\right\rbrack \) is a connected set in \( \left( {T,{\rho }^{\prime }}\right) {}^{4} \) .
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Proof. By definition (§1), relative continuity on \( B \) becomes ordinary continuity when \( f \) is restricted to \( B \) . Thus we may treat \( f \) as a mapping of \( B \) into \( f\left\lbrack B\right\rbrack \), replacing \( S \) and \( T \) by their subspaces \( B \) and \( f\left\lbrack B\right\rbrack \) .\n\nSeeking a contradiction, suppose \( f\left\lbrack B\right\rbrack \) is disconnected, i.e.,\n\n\[ f\left\lbrack B\right\rbrack = P \cup Q \]\n\nfor some disjoint sets \( P, Q \neq \varnothing \) closed in \( \left( {f\left\lbrack B\right\rbrack ,{\rho }^{\prime }}\right) \) . Then by Theorem 1, with \( T \) replaced by \( f\left\lbrack B\right\rbrack \), the sets \( {f}^{-1}\left\lbrack P\right\rbrack \) and \( {f}^{-1}\left\lbrack Q\right\rbrack \) are closed in \( \left( {B,\rho }\right) \) . They also are nonvoid and disjoint (as are \( P \) and \( Q \) ) and satisfy\n\n\[ B = {f}^{-1}\left\lbrack {P \cup Q}\right\rbrack = {f}^{-1}\left\lbrack P\right\rbrack \cup {f}^{-1}\left\lbrack Q\right\rbrack \]\n\n(see Chapter 1, \( §§4 - 7 \), Problem 6). Thus \( B \) is disconnected, contrary to assumption.
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Yes
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Lemma 1. A set \( A \subseteq \left( {S,\rho }\right) \) is connected iff any two points \( p, q \in A \) are in some connected subset \( B \subseteq A \) . Hence any arcwise connected set is connected.
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Proof. Seeking a contradiction, suppose the condition stated in Lemma 1 holds but \( A \) is disconnected, so \( A = P \cup Q \) for some disjoint sets \( P \neq \varnothing, Q \neq \varnothing \) , both closed in \( \left( {A,\rho }\right) \) .\n\nPick any \( p \in P \) and \( q \in Q \) . By assumption, \( p \) and \( q \) are in some connected set \( B \subseteq A \) . Treat \( \left( {B,\rho }\right) \) as a subspace of \( \left( {A,\rho }\right) \), and let\n\n\[ \n{P}^{\prime } = B \cap P\text{ and }{Q}^{\prime } = B \cap Q.\n\]\n\nThen by Theorem 4 of Chapter \( 3,§{12},{P}^{\prime } \) and \( {Q}^{\prime } \) are closed in \( B \) . Also, they are disjoint (for \( P \) and \( Q \) are) and nonvoid (for \( p \in {P}^{\prime }, q \in {Q}^{\prime } \) ), and\n\n\[ \nB = B \cap A = B \cap \left( {P \cup Q}\right) = \left( {B \cap P}\right) \cup \left( {B \cap Q}\right) = {P}^{\prime } \cup {Q}^{\prime }.\n\]\n\nThus \( B \) is disconnected, contrary to assumption. This contradiction proves the lemma (the converse proof is trivial).\n\nIn particular, if \( A \) is arcwise connected, then any points \( p, q \) in \( A \) are in some arc \( B \subseteq A \), a connected set by Corollary 2 . Thus all is proved.
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Yes
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Corollary 3. Any convex or polygon-connected set (e.g., a globe) in \( {E}^{n} \) (or in any other normed space) is arcwise connected, hence connected.
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Proof. Use Lemma 1 and Example (c) in part I of this section.
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No
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Theorem 4. Every open connected set \( A \) in \( {E}^{n} \) (*or in another normed space) is also arcwise connected and even polygon connected.
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Proof. If \( A = \varnothing \), this is \
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No
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Theorem 5. If a function \( f : A \rightarrow {E}^{1} \) is relatively continuous on a connected set \( B \subseteq A \subseteq \left( {S,\rho }\right) \), then \( f \) has the Darboux property on \( B \) .
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In fact, by Theorems 3 and \( 2, f\left\lbrack B\right\rbrack \) is a connected set in \( {E}^{1} \), i.e., an interval. This, however, implies the Darboux property.
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Yes
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Theorem 3. In every metric space \( \\left( {S,\\rho }\\right) \), the metric \( \\rho : \\left( {S \\times S}\\right) \\rightarrow {E}^{1} \) is a continuous function on the product space \( S \\times S \) .
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Proof. Fix any \( \\left( {p, q}\\right) \\in S \\times S \) . By Theorem 1 of \( §2,\\rho \) is continuous at \( \\left( {p, q}\\right) \) iff\n\n\[ \n\\rho \\left( {{x}_{m},{y}_{m}}\\right) \\rightarrow \\rho \\left( {p, q}\\right) \\text{ whenever }\\left( {{x}_{m},{y}_{m}}\\right) \\rightarrow \\left( {p, q}\\right) ,\n\]\n\ni.e., whenever \( {x}_{m} \\rightarrow p \) and \( {y}_{m} \\rightarrow q \) . However, this follows by Theorem 4 in Chapter 3, §15. Thus continuity is proved.
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Yes
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Theorem 1. Given a sequence of functions \( {f}_{m} : A \rightarrow \left( {T,{\rho }^{\prime }}\right) \), let \( B \subseteq A \) and\n\n\[ \n{Q}_{m} = \mathop{\sup }\limits_{{x \in B}}{\rho }^{\prime }\left( {{f}_{m}\left( x\right), f\left( x\right) }\right) .\n\]\n\nThen \( {f}_{m} \rightarrow f \) (uniformly on \( B \) ) iff \( {Q}_{m} \rightarrow 0 \) .
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Proof. If \( {Q}_{m} \rightarrow 0 \), then by definition\n\n\[ \n\left( {\forall \varepsilon > 0}\right) \left( {\exists k}\right) \left( {\forall m > k}\right) \;{Q}_{m} < \varepsilon .\n\]\n\nHowever, \( {Q}_{m} \) is an upper bound of all distances \( {\rho }^{\prime }\left( {{f}_{m}\left( x\right), f\left( x\right) }\right), x \in B \) . Hence (2) follows.\n\nConversely, if\n\n\[ \n\left( {\forall x \in B}\right) \;{\rho }^{\prime }\left( {{f}_{m}\left( x\right), f\left( x\right) }\right) < \varepsilon ,\n\]\n\nthen\n\n\[ \n\varepsilon \geq \mathop{\sup }\limits_{{x \in B}}{\rho }^{\prime }\left( {{f}_{m}\left( x\right), f\left( x\right) }\right)\n\]\n\ni.e., \( {Q}_{m} \leq \varepsilon \) . Thus (2) implies\n\n\[ \n\left( {\forall \varepsilon > 0}\right) \left( {\exists k}\right) \left( {\forall m \geq k}\right) \;{Q}_{m} \leq \varepsilon \n\]\n\nand \( {Q}_{m} \rightarrow 0 \) .
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Yes
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Theorem 2. Let \( {f}_{m} : A \rightarrow \left( {T,{\rho }^{\prime }}\right) \) be a sequence of functions on \( A \subseteq \left( {S,\rho }\right) \) . If \( {f}_{m} \rightarrow f \) (uniformly) on a set \( B \subseteq A \), and if the \( {f}_{m} \) are relatively (or uniformly) continuous on \( B \), then the limit function \( f \) has the same property.
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Proof. Fix \( \varepsilon > 0 \) . As \( {f}_{m} \rightarrow f \) (uniformly) on \( B \), there is a \( k \) such that\n\n\[ \left( {\forall x \in B}\right) \left( {\forall m \geq k}\right) \;{\rho }^{\prime }\left( {{f}_{m}\left( x\right), f\left( x\right) }\right) < \frac{\varepsilon }{4}. \]\n\nTake any \( {f}_{m} \) with \( m > k \), and take any \( p \in B \) . By continuity, there is \( \delta > 0 \), with\n\n\[ \left( {\forall x \in B \cap {G}_{p}\left( \delta \right) }\right) \;{\rho }^{\prime }\left( {{f}_{m}\left( x\right) ,{f}_{m}\left( p\right) }\right) < \frac{\varepsilon }{4}. \]\n\nSetting \( x = p \) in (3) gives \( {\rho }^{\prime }\left( {{f}_{m}\left( p\right), f\left( p\right) }\right) < \frac{\varepsilon }{4} \) . Combining this with (4) and (3), we obtain \( \left( {\forall x \in B \cap {G}_{p}\left( \delta \right) }\right) \)\n\n\[ {\rho }^{\prime }\left( {f\left( x\right), f\left( p\right) }\right) \leq {\rho }^{\prime }\left( {f\left( x\right) ,{f}_{m}\left( x\right) }\right) + {\rho }^{\prime }\left( {{f}_{m}\left( x\right) ,{f}_{m}\left( p\right) }\right) + {\rho }^{\prime }\left( {{f}_{m}\left( p\right), f\left( p\right) }\right) \]\n\n\[ < \frac{\varepsilon }{4} + \frac{\varepsilon }{4} + \frac{\varepsilon }{4} < \varepsilon \]\n\nWe thus see that for \( p \in B \), \n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists \delta > 0}\right) \left( {\forall x \in B \cap {G}_{p}\left( \delta \right) }\right) \;{\rho }^{\prime }\left( {f\left( x\right), f\left( p\right) }\right) < \varepsilon ,\]\n\ni.e., \( f \) is relatively continuous at \( p \) (over \( B \) ), as claimed.\n\nQuite similarly, the reader will show that \( f \) is uniformly continuous if the \( {f}_{n} \) are.
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Yes
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Theorem 3 (Cauchy criterion for uniform convergence). Let \( \left( {T,{\rho }^{\prime }}\right) \) be complete. Then a sequence \( {f}_{m} : A \rightarrow T, A \subseteq \left( {S,\rho }\right) \), converges uniformly on a set \( B \subseteq A \) iff\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists k}\right) \left( {\forall x \in B}\right) \left( {\forall m, n > k}\right) \;{\rho }^{\prime }\left( {{f}_{m}\left( x\right) ,{f}_{n}\left( x\right) }\right) < \varepsilon . \]
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Proof. If (5) holds then, for any (fixed) \( x \in B,\left\{ {{f}_{m}\left( x\right) }\right\} \) is a Cauchy sequence of points in \( T \), so by the assumed completeness of \( T \), it has a limit \( f\left( x\right) \) . Thus we can define a function \( f : B \rightarrow T \) with\n\n\[ f\left( x\right) = \mathop{\lim }\limits_{{m \rightarrow \infty }}{f}_{m}\left( x\right) \text{ on }B. \]\n\nTo show that \( {f}_{m} \rightarrow f \) (uniformly) on \( B \), we use (5) again. Keeping \( \varepsilon, k \) , \( x \), and \( m \) temporarily fixed, we let \( n \rightarrow \infty \) so that \( {f}_{n}\left( x\right) \rightarrow f\left( x\right) \) . Then by Theorem 4 of Chapter \( 3,§{15},{\rho }^{\prime }\left( {{f}_{m}\left( x\right) ,{f}_{n}\left( x\right) }\right) \rightarrow {p}^{\prime }\left( {f\left( x\right) ,{f}_{m}\left( x\right) }\right) \) . Passing to the limit in (5), we thus obtain (2).\n\nThe easy proof of the converse is left to the reader (cf. Chapter 3, §17, Theorem 1).
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No
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Theorem 4. Let\n\n\\[ \nf = \\mathop{\\sum }\\limits_{{m = 1}}^{\\infty }{f}_{m}\\text{ (pointwise) on }B{.}^{3} \n\\]\n\nLet \\( {m}_{1} < {m}_{2} < \\cdots < {m}_{n} < \\cdots \\) in \\( N \\), and define\n\n\\[ \n{g}_{1} = {s}_{{m}_{1}},{g}_{n} = {s}_{{m}_{n}} - {s}_{{m}_{n - 1}},\\;n > 1.\n\\]\n\n(Thus \\( {g}_{n + 1} = {f}_{{m}_{n} + 1} + \\cdots + {f}_{{m}_{n + 1}} \\) .) Then\n\n\\[ \nf = \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }{g}_{n}\\text{ (pointwise) on }B\\text{ as well; }\n\\]\n\nsimilarly for uniform convergence.
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Proof. Let\n\n\\[ \n{s}_{n}^{\\prime } = \\mathop{\\sum }\\limits_{{k = 1}}^{n}{g}_{k},\\;n = 1,2,\\ldots\n\\]\n\nThen \\( {s}_{n}^{\\prime } = {s}_{{m}_{n}} \\) (verify!), so \\( \\left\{ {s}_{n}^{\\prime }\\right\} \\) is a subsequence, \\( \\left\{ {s}_{{m}_{n}}\\right\} \\), of \\( \\left\{ {s}_{m}\\right\} \\) . Hence \\( {s}_{m} \\rightarrow f \\) (pointwise) implies \\( {s}_{n}^{\\prime } \\rightarrow f \\) (pointwise); i.e.,\n\n\\[ \nf = \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }{g}_{n}\\text{ (pointwise). }\n\\]\n\nFor uniform convergence, see Problem 13 (cf. also Problem 19).
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No
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Theorem 1. Let the range space of the functions \( {f}_{m} \) (all defined on \( A \) ) be \( {E}^{1} \) , \( C \), or \( {E}^{n} \) (*or another complete normed space). Then for \( B \subseteq A \), we have the following:\n\n(i) If \( \sum \left| {f}_{m}\right| \) converges on \( B \) (pointwise or uniformly), so does \( \sum {f}_{m} \) itself. Moreover,\n\n\[ \left| {\mathop{\sum }\limits_{{m = 1}}^{\infty }{f}_{m}}\right| \leq \mathop{\sum }\limits_{{m = 1}}^{\infty }\left| {f}_{m}\right| \;\text{ on }B \]
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Proof.\n\n(i) If \( \sum \left| {f}_{m}\right| \) converges uniformly on \( B \), then by Theorem \( {3}^{\prime } \) of \( §{12} \),\n\n\[ \left( {\forall \varepsilon > 0}\right) \left( {\exists k}\right) \left( {\forall n > m > k}\right) \left( {\forall x \in B}\right) \]\n\n\[ \varepsilon > \mathop{\sum }\limits_{{i = m}}^{n}\left| {{f}_{i}\left( x\right) }\right| \geq \left| {\mathop{\sum }\limits_{{i = m}}^{n}{f}_{i}\left( x\right) }\right| \;\text{ (triangle law). } \]\n\n(1)\n\nHowever, this shows that \( \sum {f}_{n} \) satisfies Cauchy’s criterion (6) of \( §{12} \), so it converges uniformly on \( \bar{B} \).\n\nMoreover, letting \( n \rightarrow \infty \) in the inequality\n\n\[ \left| {\mathop{\sum }\limits_{{m = 1}}^{n}{f}_{m}}\right| \leq \mathop{\sum }\limits_{{m = 1}}^{n}\left| {f}_{m}\right| \]\n\nwe get\n\n\[ \left| {\mathop{\sum }\limits_{{m = 1}}^{\infty }{f}_{m}}\right| \leq \mathop{\sum }\limits_{{m = 1}}^{\infty }\left| {f}_{m}\right| < + \infty \text{ on }B\text{, as claimed. } \]\n\nBy Note 1, this also proves the theorem for pointwise convergence.
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Yes
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Theorem 2 (comparison test). Suppose\n\n\[ \n\\left( {\\forall m}\\right) \\;\\left| {f}_{m}\\right| \\leq \\left| {g}_{m}\\right| \\text{ on }B.\n\]\n\nThen\n\n(i) \( \\mathop{\\sum }\\limits_{{m = 1}}^{\\infty }\\left| {f}_{m}\\right| \\leq \\mathop{\\sum }\\limits_{{m = 1}}^{\\infty }\\left| {g}_{m}\\right| \) on \( B \) ;\n\n(ii) \( \\mathop{\\sum }\\limits_{{m = 1}}^{\\infty }\\left| {f}_{m}\\right| = + \\infty \) implies \( \\mathop{\\sum }\\limits_{{m = 1}}^{\\infty }\\left| {g}_{m}\\right| = + \\infty \) on \( B \) ; and\n\n(iii) If \( \\sum \\left| {g}_{m}\\right| \) converges (pointwise or uniformly) on \( B \), so does \( \\sum \\left| {f}_{m}\\right| \) .
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Proof. Conclusion (i) follows by letting \( n \\rightarrow \\infty \) in\n\n\[ \n\\mathop{\\sum }\\limits_{{m = 1}}^{n}\\left| {f}_{m}\\right| \\leq \\mathop{\\sum }\\limits_{{m = 1}}^{n}\\left| {g}_{m}\\right|\n\]\n\nIn turn, (ii) is a direct consequence of (i).\n\nAlso, by (i),\n\n\[ \n\\mathop{\\sum }\\limits_{{m = 1}}^{\\infty }\\left| {g}_{m}\\right| < + \\infty \\text{ implies }\\mathop{\\sum }\\limits_{{m = 1}}^{\\infty }\\left| {f}_{m}\\right| < + \\infty .\n\]\n\nThis proves (iii) for the pointwise case (see Note 1). The uniform case follows exactly as in Theorem 1(i) on noting that\n\n\[ \n\\mathop{\\sum }\\limits_{{k = m}}^{n}\\left| {f}_{k}\\right| \\leq \\mathop{\\sum }\\limits_{{k = m}}^{n}\\left| {g}_{k}\\right|\n\]\n\nand that the functions \( \\left| {f}_{k}\\right| \) and \( \\left| {g}_{k}\\right| \) are real (so Theorem \( {3}^{\\prime } \) in \( §{12} \) does apply).
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Yes
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Theorem 4 (necessary condition of convergence). If \( \sum {f}_{m} \) or \( \sum \left| {f}_{m}\right| \) converges on \( B \) (pointwise or uniformly), then \( \left| {f}_{m}\right| \rightarrow 0 \) on \( \bar{B} \) (in the same sense).
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Proof. If \( \sum {f}_{m} = f \), say, then \( {s}_{m} \rightarrow f \) and also \( {s}_{m - 1} \rightarrow f \) . Hence\n\n\[ \n{s}_{m} - {s}_{m - 1} \rightarrow f - f = \overline{0}.\n\]\n\nHowever, \( {s}_{m} - {s}_{m - 1} = {f}_{m} \) . Thus \( {f}_{m} \rightarrow \overline{0} \), and \( \left| {f}_{m}\right| \rightarrow 0 \), as claimed.\n\nThis holds for pointwise and uniform convergence alike (see Problem 14 in §12).
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Yes
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Theorem 5 (root and ratio tests). A series of constants \( \sum {a}_{n}\left( {\left| {a}_{n}\right| \neq 0}\right) \) converges absolutely if\n\n\[ \overline{\lim }\sqrt[n]{\left| {a}_{n}\right| } < 1\text{ or }\overline{\lim }\left( \frac{\left| {a}_{n + 1}\right| }{\left| {a}_{n}\right| }\right) < 1. \]\n\nIt diverges if\n\n\[ \overline{\lim }\sqrt[n]{\left| {a}_{n}\right| } > 1\text{ or }\underline{\lim }\left( \frac{\left| {a}_{n + 1}\right| }{\left| {a}_{n}\right| }\right) > 1. \]\n\nIt may converge or diverge if\n\n\[ \overline{\lim }\sqrt[n]{\left| {a}_{n}\right| } = 1 \]\n\nor if\n\n\[ \underline{\lim }\left( \frac{\left| {a}_{n + 1}\right| }{\left| {a}_{n}\right| }\right) \leq 1 \leq \overline{\lim }\left( \frac{\left| {a}_{n + 1}\right| }{\left| {a}_{n}\right| }\right) . \]\n\n(The \( {a}_{n} \) may be scalars or vectors.)
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Proof. If \( \overline{\lim }\sqrt[n]{\left| {a}_{n}\right| } < 1 \), choose \( r > 0 \) such that\n\n\[ \overline{\lim }\sqrt[n]{\left| {a}_{n}\right| } < r < 1 \]\n\nThen by Corollary 2 of Chapter 2, \( §{13},\sqrt[n]{\left| {a}_{n}\right| } < r \) for all but finitely many \( n \) . Thus, dropping a finite number of terms (§12, Problem 17), we may assume that\n\n\[ \left| {a}_{n}\right| < {r}^{n}\text{for all}n\text{.} \]\n\nAs \( 0 < r < 1 \), the geometric series \( \sum {r}^{n} \) converges. Hence so does \( \sum \left| {a}_{n}\right| \) by Theorem 2.\n\nIn the case\n\n\[ \overline{\lim }\left( \frac{\left| {a}_{n + 1}\right| }{\left| {a}_{n}\right| }\right) < 1 \]\n\nwe similarly obtain \( \left( {\exists m}\right) \left( {\forall n \geq m}\right) \left| {a}_{n + 1}\right| < \left| {a}_{n}\right| r \) ; hence by induction,\n\n\[ \left( {\forall n \geq m}\right) \;\left| {a}_{n}\right| \leq \left| {a}_{m}\right| {r}^{n - m}.\;\text{ (Verify!) } \]\n\nThus \( \sum \left| {a}_{n}\right| \) converges, as before.\n\nIf \( \overline{\lim }\sqrt[n]{\left| {a}_{n}\right| } > 1 \), then by Corollary 2 of Chapter 2, \( §{13},\left| {a}_{n}\right| > 1 \) for infinitely many \( n \) . Hence \( \left| {a}_{n}\right| \) cannot tend to 0, and so \( \sum {a}_{n} \) diverges by Theorem 4 .\n\nSimilarly, if\n\n\[ \underline{\lim }\left( \frac{\left| {a}_{n + 1}\right| }{\left| {a}_{n}\right| }\right) > 1 \]\n\nthen \( \left| {a}_{n + 1}\right| > \left| {a}_{n}\right| \) for all but finitely many \( n \), so \( \left| {a}_{n}\right| \) cannot tend to 0 again.
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Yes
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Theorem 6. For any power series \( \sum {a}_{n}{\left( x - p\right) }^{n} \), there is a unique \( r \in {E}^{ * } \) \( \left( {0 \leq r \leq + \infty }\right) \), called its convergence radius, such that the series converges absolutely for each \( x \) with \( \left| {x - p}\right| < r \) and does not converge (even conditionally) if \( \left| {x - p}\right| > r{.}^{8} \)
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Proof. Fix any \( x = {x}_{0} \) . By Theorem 5, the series \( \sum {a}_{n}{\left( {x}_{0} - p\right) }^{n} \) converges absolutely if \( \overline{\lim }\sqrt[n]{\left| {a}_{n}\right| }\left| {{x}_{0} - p}\right| < 1 \), i.e., if\n\n\[ \left| {{x}_{0} - p}\right| < r\;\left( {r = \frac{1}{\overline{\lim }\sqrt[n]{\left| {a}_{n}\right| }} = \frac{1}{d}}\right) ,\]\n\nand diverges if \( \left| {{x}_{0} - p}\right| > r \) . (Here we assumed \( d \neq 0 \) ; but if \( d = 0 \), the condition \( d\left| {{x}_{0} - p}\right| < 1 \) is trivial for any \( {x}_{0} \), so \( r = + \infty \) in this case.) Thus \( r \) is the required radius, and clearly there can be only one such \( r \) . (Why?)
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No
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Theorem 7. If a power series \( \sum {a}_{n}{\left( x - p\right) }^{n} \) converges absolutely for some \( x = {x}_{0} \neq p \), then \( \sum \left| {{a}_{n}{\left( x - p\right) }^{n}}\right| \) converges uniformly on the closed globe \( {\bar{G}}_{p}\left( \delta \right) \) , \( \delta = \left| {{x}_{0} - p}\right| \) . So does \( \sum {a}_{n}{\left( x - p\right) }^{n} \) if the range space is complete (Theorem 1).
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Proof. Suppose \( \sum \left| {{a}_{n}{\left( {x}_{0} - p\right) }^{n}}\right| \) converges. Let\n\n\[ \delta = \left| {{x}_{0} - p}\right| \text{ and }{M}_{n} = \left| {a}_{n}\right| {\delta }^{n} \]\n\nthus \( \sum {M}_{n} \) converges.\n\nNow if \( x \in {\bar{G}}_{p}\left( \delta \right) \), then \( \left| {x - p}\right| \leq \delta \), so\n\n\[ \left| {{a}_{n}{\left( x - p\right) }^{n}}\right| \leq \left| {a}_{n}\right| {\delta }^{n} = {M}_{n} \]\n\nHence by Theorem 3, \( \sum \left| {{a}_{n}{\left( x - p\right) }^{n}}\right| \) converges uniformly on \( {\bar{G}}_{p}\left( \delta \right) \) .
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Yes
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Theorem 1. If a function \( f : {E}^{1} \rightarrow E \) is differentiable at a point \( p \in {E}^{1} \), it is continuous at \( p \), and \( f\left( p\right) \) is finite (even if \( E = {E}^{ * } \) ).
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Proof. Setting \( {\Delta x} = x - p \) and \( {\Delta f} = f\left( x\right) - f\left( p\right) \), we have the identity\n\n\[ \left| {f\left( x\right) - f\left( p\right) }\right| = \left| {\frac{\Delta f}{\Delta x} \cdot \left( {x - p}\right) }\right| \;\text{ for }x \neq p. \]\n\n(2)\n\nBy assumption,\n\n\[ {f}^{\prime }\left( p\right) = \mathop{\lim }\limits_{{x \rightarrow p}}\frac{\Delta f}{\Delta x} \]\n\nexists and is finite. Thus as \( x \rightarrow p \), the right side of (2) (hence the left side as well) tends to 0 , so\n\n\[ \mathop{\lim }\limits_{{x \rightarrow p}}\left| {f\left( x\right) - f\left( p\right) }\right| = 0\text{, or }\mathop{\lim }\limits_{{x \rightarrow p}}f\left( x\right) = f\left( p\right) ,\]\n\nproving continuity at \( p \) .\n\nAlso, \( f\left( p\right) \neq \pm \infty \), for otherwise \( \left| {f\left( x\right) - f\left( p\right) }\right| = + \infty \) for all \( x \), and so \( \left| {f\left( x\right) - f\left( p\right) }\right| \) cannot tend to 0 .\n\nNote 1. Similarly, the existence of a finite left (right) derivative at \( p \) implies left (right) continuity at \( p \) . The proof is the same.\n\nNote 2. The existence of an infinite derivative does not imply continuity, nor does it exclude it. For example, consider the two cases\n\n(i) \( f\left( x\right) = \frac{1}{x} \), with \( f\left( 0\right) = 0 \), and\n\n(ii) \( f\left( x\right) = \sqrt[3]{x} \).\n\nGive your comments for \( p = 0 \) .
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Yes
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Theorem 2. A function \( f : {E}^{1} \rightarrow E \) is differentiable at \( p \), and \( {f}^{\prime }\left( p\right) = c \), iff there is a finite \( c \in E \) and a function \( \delta : {E}^{1} \rightarrow E \) such that \( \mathop{\lim }\limits_{{x \rightarrow p}}\delta \left( x\right) = \delta \left( p\right) = 0 \) , and such that\n\n\[{\Delta f} = \left\lbrack {c + \delta \left( x\right) }\right\rbrack {\Delta x}\;\text{ for all }x \in {E}^{1}.\]
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Proof. If \( f \) is differentiable at \( p \), put \( c = {f}^{\prime }\left( p\right) \) . Define \( \delta \left( p\right) = 0 \) and\n\n\[ \delta \left( x\right) = \frac{\Delta f}{\Delta x} - {f}^{\prime }\left( p\right) \text{ for }x \neq p.\]\n\nThen \( \mathop{\lim }\limits_{{x \rightarrow p}}\delta \left( x\right) = {f}^{\prime }\left( p\right) - {f}^{\prime }\left( p\right) = 0 = \delta \left( p\right) \) . Also,(3) follows.\n\nConversely, if (3) holds, then\n\n\[ \frac{\Delta f}{\Delta x} = c + \delta \left( x\right) \rightarrow c\text{ as }x \rightarrow p\text{ (since }\delta \left( x\right) \rightarrow 0\text{ ). }\]\n\nThus by definition,\n\n\[ c = \mathop{\lim }\limits_{{x \rightarrow p}}\frac{\Delta f}{\Delta x} = {f}^{\prime }\left( p\right) \text{ and }{f}^{\prime }\left( p\right) = c\text{ is finite. }\]
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Yes
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Theorem 3 (chain rule). Let the functions \( g : {E}^{1} \rightarrow {E}^{1} \) (real) and \( f : {E}^{1} \rightarrow E \) (real or not) be differentiable at \( p \) and \( q \), respectively, where \( q = g\left( p\right) \). Then the composite function \( h = f \circ g \) is differentiable at \( p \), and \[ {h}^{\prime }\left( p\right) = {f}^{\prime }\left( q\right) {g}^{\prime }\left( p\right) \]
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Proof. Setting \[ {\Delta h} = h\left( x\right) - h\left( p\right) = f\left( {g\left( x\right) }\right) - f\left( {g\left( p\right) }\right) = f\left( {g\left( x\right) }\right) - f\left( q\right) , \] we must show that \[ \mathop{\lim }\limits_{{x \rightarrow p}}\frac{\Delta h}{\Delta x} = {f}^{\prime }\left( q\right) {g}^{\prime }\left( p\right) \neq \pm \infty . \] Now as \( f \) is differentiable at \( q \), Theorem 2 yields a function \( \delta : {E}^{1} \rightarrow E \) such that \( \mathop{\lim }\limits_{{x \rightarrow q}}\delta \left( x\right) = \delta \left( q\right) = 0 \) and such that \[ \left( {\forall y \in {E}^{1}}\right) \;f\left( y\right) - f\left( q\right) = \left\lbrack {{f}^{\prime }\left( q\right) + \delta \left( y\right) }\right\rbrack {\Delta y},{\Delta y} = y - q. \] Taking \( y = g\left( x\right) \), we get \[ \left( {\forall x \in {E}^{1}}\right) \;f\left( {g\left( x\right) }\right) - f\left( q\right) = \left\lbrack {{f}^{\prime }\left( q\right) + \delta \left( {g\left( x\right) }\right) }\right\rbrack \left\lbrack {g\left( x\right) - g\left( p\right) }\right\rbrack , \] where \[ g\left( x\right) - g\left( p\right) = y - q = {\Delta y}\text{ and }f\left( {g\left( x\right) }\right) - f\left( q\right) = {\Delta h}, \] as noted above. Hence \[ \frac{\Delta h}{\Delta x} = \left\lbrack {{f}^{\prime }\left( q\right) + \delta \left( {g\left( x\right) }\right) }\right\rbrack \cdot \frac{g\left( x\right) - g\left( p\right) }{x - p}\;\text{ for all }x \neq p. \] Let \( x \rightarrow p \) . Then we obtain \( {h}^{\prime }\left( p\right) = {f}^{\prime }\left( q\right) {g}^{\prime }\left( p\right) \), for, by the continuity of \( \delta \circ g \) at \( p \) (Chapter 4, \( §2 \), Theorem 3), \[ \mathop{\lim }\limits_{{x \rightarrow p}}\delta \left( {g\left( x\right) }\right) = \delta \left( {g\left( p\right) }\right) = \delta \left( q\right) = 0. \]
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Yes
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Theorem 4. If \( f, g \), and \( h \) are real or complex and are differentiable at \( p \), so are \[ f \pm g,{hf},\text{ and }\frac{f}{h} \] (the latter if \( h\left( p\right) \neq 0 \) ), and at the point \( p \) we have (i) \( {\left( f \pm g\right) }^{\prime } = {f}^{\prime } \pm {g}^{\prime } \) ; (ii) \( {\left( hf\right) }^{\prime } = h{f}^{\prime } + {h}^{\prime }f \) ; and (iii) \( {\left( \frac{f}{h}\right) }^{\prime } = \frac{h{f}^{\prime } - {h}^{\prime }f}{{h}^{2}} \) .
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(i) \( {\left( f \pm g\right) }^{\prime } = {f}^{\prime } \pm {g}^{\prime } \) ; (ii) \( {\left( hf\right) }^{\prime } = h{f}^{\prime } + {h}^{\prime }f \) ; and (iii) \( {\left( \frac{f}{h}\right) }^{\prime } = \frac{h{f}^{\prime } - {h}^{\prime }f}{{h}^{2}} \)
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Yes
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Theorem 5 (componentwise differentiation). A function \( f : {E}^{1} \rightarrow {E}^{n}\left( {{}^{ * }{C}^{n}}\right) \) is differentiable at \( p \) iff each of its \( n \) components \( \left( {{f}_{1},\ldots ,{f}_{n}}\right) \) is, and then
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\[ {f}^{\prime }\left( p\right) = \left( {{f}_{1}^{\prime }\left( p\right) ,\ldots ,{f}_{n}^{\prime }\left( p\right) }\right) = \mathop{\sum }\limits_{{k = 1}}^{n}{f}_{k}^{\prime }\left( p\right) {\bar{e}}_{k}, \] with \( {\bar{e}}_{k} \) as in Theorem 2 of Chapter 3, \( §§1 - 3 \) . In particular, a complex function \( f : {E}^{1} \rightarrow C \) is differentiable iff its real and imaginary parts are, and \( {f}^{\prime } = {f}_{\mathrm{{re}}}^{\prime } + i \cdot {f}_{\mathrm{{im}}}^{\prime } \) (Chapter \( 4,§3 \), Note 5).
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Yes
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Lemma 1. If \( {f}^{\prime }\left( p\right) > 0 \) at some \( p \in {E}^{1} \), then\n\n\[ x < p < y \]\n\nimplies\n\n\[ f\left( x\right) < f\left( p\right) < f\left( y\right) \]\n\nfor all \( x, y \) in a sufficiently small globe \( {G}_{p}\left( \delta \right) = \left( {p - \delta, p + \delta }\right) {}^{1} \)\n\nSimilarly, if \( {f}^{\prime }\left( p\right) < 0 \), then \( x < p < y \) implies \( f\left( x\right) > f\left( p\right) > f\left( y\right) \) for \( x, y \) in some \( {G}_{p}\left( \delta \right) \) .
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Proof. If \( {f}^{\prime }\left( p\right) > 0 \), the \
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No
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Corollary 1. If \( f\left( p\right) \) is the maximum or minimum value of \( f\left( x\right) \) for \( x \) in some \( {G}_{p}\left( \delta \right) \), then \( {f}^{\prime }\left( p\right) = 0 \) ; i.e., \( f \) has a zero derivative, or none at all, at \( p \) .
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For, by Lemma \( 1,{f}^{\prime }\left( p\right) \neq 0 \) excludes a maximum or minimum at \( p \) . (Why?)
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No
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Theorem 1. Let \( f : {E}^{1} \rightarrow {E}^{ * } \) be relatively continuous on an interval \( \left\lbrack {a, b}\right\rbrack \) , with \( {f}^{\prime } \neq 0 \) on \( \left( {a, b}\right) \) . Then \( f \) is strictly monotone on \( \left\lbrack {a, b}\right\rbrack \), and \( {f}^{\prime } \) is sign-constant there (possibly 0 at a and b), with \( {f}^{\prime } \geq 0 \) if \( f \uparrow \), and \( {f}^{\prime } \leq 0 \) if \( f \downarrow \) .
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Proof. By Theorem 2 of Chapter \( 4,§8, f \) attains a least value \( m \), and a largest value \( M \), at some points of \( \left\lbrack {a, b}\right\rbrack \) . However, neither can occur at an interior point \( p \in \left( {a, b}\right) \), for, by Corollary 1, this would imply \( {f}^{\prime }\left( p\right) = 0 \), contrary to our assumption.\n\n\( {}^{1} \) This does not mean that \( f \) is monotone on any \( {G}_{p} \) (see Problem 6). We shall only say in such cases that \( f \) increases at the point \( p \) .\n\n\n\nFIGURE 22\n\nThus \( M = f\left( a\right) \) or \( M = f\left( b\right) \) ; for the moment we assume \( M = f\left( b\right) \) and \( m = f\left( a\right) \) . We must have \( m < M \), for \( m = M \) would make \( f \) constant on \( \left\lbrack {a, b}\right\rbrack \) , implying \( {f}^{\prime } = 0 \) . Thus \( m = f\left( a\right) < f\left( b\right) = M \).\n\nNow let \( a \leq x < y \leq b \) . Applying the previous argument to each of the intervals \( \left\lbrack {a, x}\right\rbrack ,\left\lbrack {a, y}\right\rbrack ,\left\lbrack {x, y}\right\rbrack \), and \( \left\lbrack {x, b}\right\rbrack \) (now using that \( m = f\left( a\right) < f\left( b\right) = M \) ), we find that\n\n\[ f\left( a\right) \leq f\left( x\right) < f\left( y\right) \leq f\left( b\right) .\;\text{ (Why?) } \]\n\nThus \( a \leq x < y \leq b \) implies \( f\left( x\right) < f\left( y\right) \) ; i.e., \( f \) increases on \( \left\lbrack {a, b}\right\rbrack \) . Hence \( {f}^{\prime } \) cannot be negative at any \( p \in \left\lbrack {a, b}\right\rbrack \), for, otherwise, by Lemma \( 1, f \) would decrease at \( p \) . Thus \( {f}^{\prime } \geq 0 \) on \( \left\lbrack {a, b}\right\rbrack \).\n\nIn the case \( M = f\left( a\right) > f\left( b\right) = m \), we would obtain \( {f}^{\prime } \leq 0 \) .\n\nCaution: The function \( f \) may increase or decrease at \( p \) even if \( {f}^{\prime }\left( p\right) = 0 \) . See Note 1.
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Yes
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Corollary 2 (Rolle’s theorem). If \( f : {E}^{1} \rightarrow {E}^{ * } \) is relatively continuous on \( \left\lbrack {a, b}\right\rbrack \) and if \( f\left( a\right) = f\left( b\right) \), then \( {f}^{\prime }\left( p\right) = 0 \) for at least one interior point \( p \in \left( {a, b}\right) \) .
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For, if \( {f}^{\prime } \neq 0 \) on all of \( \left( {a, b}\right) \), then by Theorem \( 1, f \) would be strictly monotone on \( \left\lbrack {a, b}\right\rbrack \), so the equality \( f\left( a\right) = f\left( b\right) \) would be impossible.
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Yes
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Theorem 2 (Cauchy’s law of the mean). Let the functions \( f, g : {E}^{1} \rightarrow {E}^{ * } \) be relatively continuous and finite on \( \left\lbrack {a, b}\right\rbrack \) and have derivatives on \( \left( {a, b}\right) \), with \( {f}^{\prime } \) and \( {g}^{\prime } \) never both infinite at the same point \( p \in \left( {a, b}\right) \) . Then\n\n\[ \n{g}^{\prime }\left( q\right) \left\lbrack {f\left( b\right) - f\left( a\right) }\right\rbrack = {f}^{\prime }\left( q\right) \left\lbrack {g\left( b\right) - g\left( a\right) }\right\rbrack \text{for at least one}q \in \left( {a, b}\right) \text{.} \n\]
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Proof. Let \( A = f\left( b\right) - f\left( a\right) \) and \( B = g\left( b\right) - g\left( a\right) \) . We must show that \( A{g}^{\prime }\left( q\right) = \) \( B{f}^{\prime }\left( q\right) \) for some \( q \in \left( {a, b}\right) \) . For this purpose, consider the function \( h = {Ag} - {Bf} \) . It is relatively continuous and finite on \( \left\lbrack {a, b}\right\rbrack \), as are \( g \) and \( f \) . Also,\n\n\[ \nh\left( a\right) = f\left( b\right) g\left( a\right) - g\left( b\right) f\left( a\right) = h\left( b\right) .\;\text{ (Verify!) } \n\]\n\nThus by Corollary \( 2,{h}^{\prime }\left( q\right) = 0 \) for some \( q \in \left( {a, b}\right) \) . Here, by Theorem 4 of \( §1,{h}^{\prime } = {\left( Ag - Bf\right) }^{\prime } = A{g}^{\prime } - B{f}^{\prime } \) . (This is legitimate, for, by assumption, \( {f}^{\prime } \) and \( {g}^{\prime } \) never both become infinite, so no indeterminate limits occur.) Thus \( {h}^{\prime }\left( q\right) = A{g}^{\prime }\left( q\right) - B{f}^{\prime }\left( q\right) = 0 \), and (1) follows.
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Yes
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Corollary 3 (Lagrange’s law of the mean). If \( f : {E}^{1} \rightarrow {E}^{1} \) is relatively continuous on \( \left\lbrack {a, b}\right\rbrack \) with a derivative on \( \left( {a, b}\right) \), then\n\n\[ f\left( b\right) - f\left( a\right) = {f}^{\prime }\left( q\right) \left( {b - a}\right) \text{ for at least one }q \in \left( {a, b}\right) . \]
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Proof. Take \( g\left( x\right) = x \) in Theorem 2, so \( {g}^{\prime } = 1 \) on \( {E}^{1} \) .
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No
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Corollary 4. Let \( f \) be as in Corollary 3. Then\n\n(i) \( f \) is constant on \( \left\lbrack {a, b}\right\rbrack \) iff \( {f}^{\prime } = 0 \) on \( \left( {a, b}\right) \) ;
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Proof. Let \( {f}^{\prime } = 0 \) on \( \left( {a, b}\right) \) . If \( a \leq x \leq y \leq b \), apply Corollary 3 to the interval \( \left\lbrack {x, y}\right\rbrack \) to obtain\n\n\[ f\left( y\right) - f\left( x\right) = {f}^{\prime }\left( q\right) \left( {y - x}\right) \text{ for some }q \in \left( {a, b}\right) \text{ and }{f}^{\prime }\left( q\right) = 0. \]\n\nThus \( f\left( y\right) - f\left( x\right) = 0 \) for \( x, y \in \left\lbrack {a, b}\right\rbrack \), so \( f \) is constant.
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Yes
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Theorem 3 (inverse functions). Let \( f : {E}^{1} \rightarrow {E}^{1} \) be relatively continuous and strictly monotone on an interval \( I \subseteq {E}^{1} \). Let \( {f}^{\prime }\left( p\right) \neq 0 \) at some interior point \( p \in I \). Then the inverse function \( g = {f}^{-1} \) (with \( f \) restricted to \( I \)) has a derivative at \( q = f\left( p\right) \), and\n\n\[ \n{g}^{\prime }\left( q\right) = \frac{1}{{f}^{\prime }\left( p\right) }.\n\]\n\n(If \( {f}^{\prime }\left( p\right) = \pm \infty \), then \( {g}^{\prime }\left( q\right) = 0 \).)
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Proof. By Theorem 3 of Chapter \( 4,§9, g = {f}^{-1} \) is strictly monotone and relatively continuous on \( f\left\lbrack I\right\rbrack \), itself an interval. If \( p \) is interior to \( I \), then \( q = f\left( p\right) \) is interior to \( f\left\lbrack I\right\rbrack \) . (Why?)\n\nNow if \( y \in f\left\lbrack I\right\rbrack \), we set\n\n\[ \n{\Delta g} = g\left( y\right) - g\left( q\right) ,{\Delta y} = y - q, x = {f}^{-1}\left( y\right) = g\left( y\right) \text{, and}f\left( x\right) = y\n\]\n\nand obtain\n\n\[ \n\frac{\Delta g}{\Delta y} = \frac{g\left( y\right) - g\left( q\right) }{y - q} = \frac{x - p}{f\left( x\right) - f\left( p\right) } = \frac{\Delta x}{\Delta f}\text{ for }x \neq p.\n\]\n\nNow if \( y \rightarrow q \), the continuity of \( g \) at \( q \) yields \( g\left( y\right) \rightarrow g\left( q\right) \) ; i.e., \( x \rightarrow p \). Also, \( x \neq p \) iff \( y \neq q \), for \( f \) and \( g \) are one-to-one functions. Thus we may substitute \( y = f\left( x\right) \) or \( x = g\left( y\right) \) to get\n\n\[ \n{g}^{\prime }\left( q\right) = \mathop{\lim }\limits_{{y \rightarrow q}}\frac{\Delta g}{\Delta y} = \mathop{\lim }\limits_{{x \rightarrow p}}\frac{\Delta x}{\Delta f} = \frac{1}{\mathop{\lim }\limits_{{x \rightarrow p}}\left( {{\Delta f}/{\Delta x}}\right) } = \frac{1}{{f}^{\prime }\left( p\right) }\n\]\n\n\( \left( {2}^{\prime }\right) \)\n\nwhere we use the convention \( \frac{1}{\infty } = 0 \) if \( {f}^{\prime }\left( p\right) = \infty \) .
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No
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Theorem 4 (Darboux). If \( f : {E}^{1} \rightarrow {E}^{ * } \) is relatively continuous and has a derivative on an interval \( I \), then \( {f}^{\prime } \) has the Darboux property (Chapter 4,§9) on \( I \) .
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Proof. Let \( p, q \in I \) and \( {f}^{\prime }\left( p\right) < c < {f}^{\prime }\left( q\right) \) . Put \( g\left( x\right) = f\left( x\right) - {cx} \) . Assume \( {g}^{\prime } \neq 0 \) on \( \left( {p, q}\right) \) and find a contradiction to Theorem 1. Details are left to the reader.
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No
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Theorem 1 (finite increments law). Let \( f : {E}^{1} \rightarrow E \) and \( g : {E}^{1} \rightarrow {E}^{ * } \) be relatively continuous and finite on a closed interval \( I = \left\lbrack {a, b}\right\rbrack \subseteq {E}^{1} \), and have derivatives \( {}^{2} \) with \( \left| {f}^{\prime }\right| \leq {g}^{\prime } \), on \( I - Q \) where \( Q \subseteq \left\{ {{p}_{1},{p}_{2},\ldots ,{p}_{m},\ldots }\right\} \) . Then \[ \left| {f\left( b\right) - f\left( a\right) }\right| \leq g\left( b\right) - g\left( a\right) . \]
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The proof is somewhat laborious, but worthwhile. (At a first reading, one may omit it, however.) We outline some preliminary ideas. Given any \( x \in I \), suppose first that \( x > {p}_{m} \) for at least one \( {p}_{m} \in Q \) . In this case, we put \[ Q\left( x\right) = \mathop{\sum }\limits_{{{p}_{m} < x}}{2}^{-m} \] here the summation is only over those \( m \) for which \( {p}_{m} < x \) . If, however, there are no \( {p}_{m} \in Q \) with \( {p}_{m} < x \), we put \( Q\left( x\right) = 0 \) . Thus \( Q\left( x\right) \) is defined for all \( x \in I \) . It gives an idea as to \
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No
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If \( f : {E}^{1} \rightarrow E \) is relatively continuous and finite on \( I = \left\lbrack {a, b}\right\rbrack \subseteq \) \( {E}^{1} \), and has a derivative on \( I - Q \), then there is a real \( M \) such that \n\n\[ \n\left| {f\left( b\right) - f\left( a\right) }\right| \leq M\left( {b - a}\right) \text{ and }M \leq \mathop{\sup }\limits_{{t \in I - Q}}\left| {{f}^{\prime }\left( t\right) }\right| .\n\]
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Proof. Let \n\n\[ \n{M}_{0} = \mathop{\sup }\limits_{{t \in I - Q}}\left| {{f}^{\prime }\left( t\right) }\right| \n\] \n\nIf \( {M}_{0} < + \infty \), put \( M = {M}_{0} \geq \left| {f}^{\prime }\right| \) on \( I - Q \), and take \( g\left( x\right) = {Mx} \) in Theorem 1 . Then \( {g}^{\prime } = M \geq \left| {f}^{\prime }\right| \) on \( I - Q \), so formula (1) yields (5) since \n\n\[ \ng\left( b\right) - g\left( a\right) = {Mb} - {Ma} = M\left( {b - a}\right) .\n\] \n\nIf, however, \( {M}_{0} = + \infty \), let \n\n\[ \nM = \left| \frac{f\left( b\right) - f\left( a\right) }{b - a}\right| < {M}_{0}.\n\] \n\nThen (5) clearly is true. Thus the required \( M \) exists always. \( {}^{3} \)
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Yes
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Corollary 2. Let \( f \) be as in Corollary 1. Then \( f \) is constant on \( I \) iff \( {f}^{\prime } = 0 \) on \( I - Q \) .
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Proof. If \( {f}^{\prime } = 0 \) on \( I - Q \), then \( M = 0 \) in Corollary 1, so Corollary 1 yields, for any subinterval \( \left\lbrack {a, x}\right\rbrack \left( {x \in I}\right) ,\left| {f\left( x\right) - f\left( a\right) }\right| \leq 0 \) ; i.e., \( f\left( x\right) = f\left( a\right) \) for all \( x \in I \) . Thus \( f \) is constant on \( I \) .\n\nConversely, if so, then \( {f}^{\prime } = 0 \), even on all of \( I \) .
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Yes
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Corollary 3. Let \( f, g : {E}^{1} \rightarrow E \) be relatively continuous and finite on \( I = \) \( \left\lbrack {a, b}\right\rbrack \), and differentiable on \( I - Q \) . Then \( f - g \) is constant on \( I \) iff \( {f}^{\prime } = {g}^{\prime } \) on \( I - Q \) .
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Proof. Apply Corollary 2 to the function \( f - g \) .
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No
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Theorem 2. Let \( f \) be real and have the properties stated in Corollary 1. Then\n\n(i) \( f \uparrow \) on \( I = \left\lbrack {a, b}\right\rbrack \) iff \( {f}^{\prime } \geq 0 \) on \( I - Q \) ; and\n\n(ii) \( f \downarrow \) on \( I \) iff \( {f}^{\prime } \leq 0 \) on \( I - Q \) .
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Proof. Let \( {f}^{\prime } \geq 0 \) on \( I - Q \) . Fix any \( x, y \in I\left( {x < y}\right) \) and define \( g\left( t\right) = 0 \) on \( {E}^{1} \) . Then \( \left| {g}^{\prime }\right| = 0 \leq {f}^{\prime } \) on \( I - Q \) . Thus \( g \) and \( f \) satisfy Theorem 1 (with their roles reversed) on \( I \), and certainly on the subinterval \( \left\lbrack {x, y}\right\rbrack \) . Thus we have\n\n\[ f\left( y\right) - f\left( x\right) \geq \left| {g\left( y\right) - g\left( x\right) }\right| = 0\text{, i.e.,}f\left( y\right) \geq f\left( x\right) \text{whenever}y > x\text{in}I\text{,}\]\n\nso \( f \uparrow \) on \( I \) .\n\nConversely, if \( f \uparrow \) on \( I \), then for every \( p \in I \), we must have \( {f}^{\prime }\left( p\right) \geq 0 \), for otherwise, by Lemma 1 of \( §2, f \) would decrease at \( p \) . Thus \( {f}^{\prime } \geq 0 \), even on all of \( I \), and (i) is proved. Assertion (ii) is proved similarly.
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Yes
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Theorem 1. If \( F \) and \( G \) are primitive to \( f \) on \( I \), then \( G - F \) is constant on \( I \).
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Proof. By assumption, \( F \) and \( G \) are relatively continuous and finite on \( I \) ; hence so is \( G - F \) . Also, \( {F}^{\prime } = f \) on \( I - Q \) and \( {G}^{\prime } = f \) on \( I - P \) . ( \( Q \) and \( P \) are countable, but possibly \( Q \neq P \) .)\n\nHence both \( {F}^{\prime } \) and \( {G}^{\prime } \) equal \( f \) on \( I - S \), where \( S = P \cup Q \), and \( S \) is countable itself by Theorem 2 of Chapter 1, §9.\n\nThus by Corollary 3 in \( §4,{F}^{\prime } = {G}^{\prime } \) on \( I - S \) implies \( G - F = c \) (constant) on each \( \left\lbrack {x, y}\right\rbrack \subseteq I \) ; hence \( G - F = c \) (or \( G = F + c \) ) on \( I \).
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Yes
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Corollary 1 (linearity). If \( \int f \) and \( \int g \) exist on \( I \), so does \( \int \left( {{pf} + {qg}}\right) \) for any scalars \( p, q \) (in the scalar field of \( E{)}^{.3} \) Moreover, for any \( a, b \in I \), we obtain\n\n(i) \( {\int }_{a}^{b}\left( {{pf} + {qg}}\right) = p{\int }_{a}^{b}f + q{\int }_{a}^{b}g \) ;\n\n(ii) \( {\int }_{a}^{b}\left( {f \pm g}\right) = {\int }_{a}^{b}f \pm {\int }_{a}^{b}g \) ; and\n\n(iii) \( {\int }_{a}^{b}{pf} = p{\int }_{a}^{b}f \) .
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Proof. By assumption, there are \( F \) and \( G \) such that\n\n\[ \n{F}^{\prime } = f\text{ on }I - Q\text{ and }{G}^{\prime } = g\text{ on }I - P.\n\]\n\nThus, setting \( S = P \cup Q \) and \( H = {pF} + {qG} \), we have\n\n\[ \n{H}^{\prime } = p{F}^{\prime } + q{G}^{\prime } = {pf} + {qg}\text{ on }I - S,\n\]\n\nwith \( P, Q \), and \( S \) countable. Also, \( H = {pF} + {qG} \) is relatively continuous and finite on \( I \), as are \( F \) and \( G \) .\n\nThus by definition, \( H = \int \left( {{pf} + {qg}}\right) \) exists on \( I \), and by (1),\n\n\[ \n{\int }_{a}^{b}\left( {{pf} + {qg}}\right) = H\left( b\right) - H\left( a\right) = {pF}\left( b\right) + {qG}\left( b\right) - {pF}\left( a\right) - {qG}\left( a\right) = p{\int }_{a}^{b}f + q{\int }_{a}^{b}g,\n\]\n\nproving \( \left( {\mathrm{i}}^{ * }\right) \) .\n\nWith \( p = 1 \) and \( q = \pm 1 \), we obtain \( \left( {\mathrm{{ii}}}^{ * }\right) \) .\n\nTaking \( q = 0 \), we get \( \left( {\mathrm{{iii}}}^{ * }\right) \) .
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Yes
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Corollary 2. If both \( \int f \) and \( \int \left| f\right| \) exist on \( I = \left\lbrack {a, b}\right\rbrack \), then\n\n\[\n\left| {{\int }_{a}^{b}f}\right| \leq {\int }_{a}^{b}\left| f\right|\n\]
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Proof. As before, let\n\n\[{F}^{\prime } = f\text{and}{G}^{\prime } = \left| f\right| \text{on}I - S\left( {S = Q \cup P\text{, all countable}}\right) \text{,}\]\n\nwhere \( F \) and \( G \) are relatively continuous and finite on \( I \) and \( G = \int \left| f\right| \) is real. Also, \( \left| {F}^{\prime }\right| = \left| f\right| = {G}^{\prime } \) on \( I - S \) . Thus by Theorem 1 of \( §4 \) ,\n\n\[ \left| {F\left( b\right) - F\left( a\right) }\right| \leq G\left( b\right) - G\left( a\right) = {\int }_{a}^{b}\left| f\right| .\]
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Yes
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Corollary 3. If \( \int f \) exists on \( I = \left\lbrack {a, b}\right\rbrack \), exact on \( I - Q \), then\n\n\[ \left| {{\int }_{a}^{b}f}\right| \leq M\left( {b - a}\right) \]\n\nfor some real\n\n\[ M \leq \mathop{\sup }\limits_{{t \in I - Q}}\left| {f\left( t\right) }\right| \]
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This is simply Corollary 1 of \( §4 \), when applied to a primitive, \( F = \int f \) .
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No
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Corollary 4. If \( F = \int f \) on \( I \) and \( f = g \) on \( I - Q \), then \( F \) is also a primitive of \( g \), and\n\n\[{\int }_{a}^{b}f = {\int }_{a}^{b}g\;\text{ for }a, b \in I\]\n\n(Thus we may arbitrarily redefine \( f \) on a countable \( Q \) .)
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Proof. Let \( {F}^{\prime } = f \) on \( I - P \) . Then \( {F}^{\prime } = g \) on \( I - \left( {P \cup Q}\right) \) . The rest is clear.
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No
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Corollary 5 (integration by parts). Let \( f \) and \( g \) be real or complex (or let \( f \) be scalar valued and \( g \) vector valued), both relatively continuous on \( I \) and differentiable on \( I - Q \) . Then if \( \int {f}^{\prime }g \) exists on \( I \), so does \( \int f{g}^{\prime } \), and we have\n\n\[{\int }_{a}^{b}f{g}^{\prime } = f\left( b\right) g\left( b\right) - f\left( a\right) g\left( a\right) - {\int }_{a}^{b}{f}^{\prime }g\;\text{ for any }a, b \in I.\]
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Proof. By assumption, \( {fg} \) is relatively continuous and finite on \( I \), and\n\n\[{\left( fg\right) }^{\prime } = f{g}^{\prime } + {f}^{\prime }g\text{ on }I - Q.\]\n\nThus, setting \( H = {fg} \), we have \( H = \int \left( {f{g}^{\prime } + {f}^{\prime }g}\right) \) on \( I \) . Hence by Corollary 1, if \( \int {f}^{\prime }g \) exists on \( I \), so does \( \int \left( {\left( {f{g}^{\prime } + {f}^{\prime }g}\right) - {f}^{\prime }g}\right) = \int f{g}^{\prime } \), and\n\n\[{\int }_{a}^{b}f{g}^{\prime } + {\int }_{a}^{b}{f}^{\prime }g = {\int }_{a}^{b}\left( {f{g}^{\prime } + {f}^{\prime }g}\right) = H\left( b\right) - H\left( a\right) = f\left( b\right) g\left( b\right) - f\left( a\right) g\left( a\right) .\]\n\nThus (2) follows.
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Yes
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A function \( f : {E}^{1} \rightarrow {E}^{n}\left( {{}^{ * }{C}^{n}}\right) \) is integrable on \( I \) iff all its components \( \left( {{f}_{1},{f}_{2},\ldots ,{f}_{n}}\right) \) are, and then (by Theorem 5 in \( §1 \) )
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\[ {\int }_{a}^{b}f = \left( {{\int }_{a}^{b}{f}_{1},\ldots ,{\int }_{a}^{b}{f}_{n}}\right) = \mathop{\sum }\limits_{{k = 1}}^{n}{\overrightarrow{e}}_{k}{\int }_{a}^{b}{f}_{k}\;\text{ for any }a, b \in I. \] Hence if \( f \) is complex, \[ {\int }_{a}^{b}f = {\int }_{a}^{b}{f}_{\mathrm{{re}}} + i \cdot {\int }_{a}^{b}{f}_{\mathrm{{im}}} \] (see Chapter 4, §3, Note 5).
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Yes
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Corollary 8. If \( f = 0 \) on \( I - Q \), then \( \int f \) exists on \( I \), and
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\[ \left| {{\int }_{a}^{b}f}\right| = {\int }_{a}^{b}\left| f\right| = 0\;\text{ for }a, b \in I. \]
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No
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Theorem 2 (change of variables). Suppose \( g : {E}^{1} \rightarrow {E}^{1} \) (real) is differentiable on \( I \), while \( f : {E}^{1} \rightarrow E \) has a primitive on \( g\left\lbrack I\right\rbrack ,{}^{4} \) exact on \( g\left\lbrack {I - Q}\right\rbrack \) . Then \[ \int f\left( {g\left( x\right) }\right) {g}^{\prime }\left( x\right) {dx}\;\left( {\text{ i.e.,}\int \left( {f \circ g}\right) {g}^{\prime }}\right) \] exists on \( I \), and for any \( a, b \in I \), we have \[ {\int }_{a}^{b}f\left( {g\left( x\right) }\right) {g}^{\prime }\left( x\right) {dx} = {\int }_{p}^{q}f\left( y\right) {dy},\text{ where }p = g\left( a\right) \text{ and }q = g\left( b\right) . \]
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Proof. Let \( F = \int f \) on \( g\left\lbrack I\right\rbrack \), and \( {F}^{\prime } = f \) on \( g\left\lbrack {I - Q}\right\rbrack \) . Then the composite function \( H = F \circ g \) is relatively continuous and finite on \( I \) . (Why?) By Theorem 3 of \( §1 \) , \[ {H}^{\prime }\left( x\right) = {F}^{\prime }\left( {g\left( x\right) }\right) {g}^{\prime }\left( x\right) \text{ for }x \in I - Q; \] i.e., \[ {H}^{\prime } = \left( {{F}^{\prime } \circ g}\right) {g}^{\prime }\text{ on }I - Q. \] Thus \( H = \int \left( {f \circ g}\right) {g}^{\prime } \) exists on \( I \), and \[ {\int }_{a}^{b}\left( {f \circ g}\right) {g}^{\prime } = H\left( b\right) - H\left( a\right) = F\left( {g\left( b\right) }\right) - F\left( {g\left( a\right) }\right) = F\left( q\right) - F\left( p\right) = {\int }_{p}^{q}f. \]
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Yes
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Theorem 3. If \( f, g : {E}^{1} \rightarrow {E}^{1} \) are integrable on \( I = \left\lbrack {a, b}\right\rbrack \), then we have the following:\n\n(i) \( f \geq 0 \) on \( I - Q \) implies \( {\int }_{a}^{b}f \geq 0 \) .
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Proof. By Corollary 4, we may redefine \( f \) on \( Q \) so that our assumptions in (i)-(iv) hold on all of \( I \) . Thus we write \
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No
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Corollary 9 (first law of the mean). If \( f \) is real and \( \int f \) exists on \( \left\lbrack {a, b}\right\rbrack \), exact on \( \left( {a, b}\right) \), then\n\n\[{\int }_{a}^{b}f = f\left( q\right) \left( {b - a}\right) \text{ for some }q \in \left( {a, b}\right) .
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Proof. Apply Corollary 3 in \( §2 \) to the function \( F = \int f \) .
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No
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Theorem 2. Let \( f : {E}^{1} \rightarrow {E}^{ * } \) be of class \( {\mathrm{{CD}}}^{n} \) on \( {G}_{p}\left( \delta \right) \) for an even number \( n \geq 2 \), and let\n\n\[ \n{f}^{\left( k\right) }\left( p\right) = 0\text{ for }k = 1,2,\ldots, n - 1, \n\] \n\nwhile \n\n\[ \n{f}^{\left( n\right) }\left( p\right) < 0\text{ (respectively,}{f}^{\left( n\right) }\left( p\right) > 0\text{). } \n\] \n\nThen \( f\left( p\right) \) is the maximum (respectively, minimum) value of \( f \) on some \( {G}_{p}\left( \varepsilon \right) \) , \( \varepsilon \leq \delta \) .\n\nIf, however, these conditions hold for some odd \( n \geq 1 \) (i.e., the first nonvanishing derivative at \( p \) is of odd order), \( f \) has no maximum or minimum at \( p \) .
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Proof. As \n\n\[ \n{f}^{\left( k\right) }\left( p\right) = 0,\;k = 1,2,\ldots, n - 1, \n\] \n\nTheorem \( {1}^{\prime } \) (with \( n \) replaced by \( n - 1 \) ) yields \n\n\[ \nf\left( x\right) = f\left( p\right) + {f}^{\left( n\right) }\left( {q}_{n}\right) \frac{{\left( x - p\right) }^{n}}{n!}\;\text{ for all }x \in {G}_{p}\left( \delta \right) , \n\] \n\nwith \( {q}_{n} \) between \( x \) and \( p \) .\n\nAlso, as \( f \in {\mathrm{{CD}}}^{n},{f}^{\left( n\right) } \) is continuous at \( p \) . Thus if \( {f}^{\left( n\right) }\left( p\right) < 0 \), then \( {f}^{\left( n\right) } < 0 \) on some \( {G}_{p}\left( \varepsilon \right) ,0 < \varepsilon \leq \delta \) . However, \( x \in {G}_{p}\left( \varepsilon \right) \) implies \( {q}_{n} \in {G}_{p}\left( \varepsilon \right) \), so \n\n\[ \n{f}^{\left( n\right) }\left( {q}_{n}\right) < 0 \n\] \n\nwhile \n\n\[ \n{\left( x - p\right) }^{n} \geq 0\text{ if }n\text{ is even. \n\] \n\nIt follows that \n\n\[ \n{f}^{\left( n\right) }\left( {q}_{n}\right) \frac{{\left( x - p\right) }^{n}}{n!} \leq 0 \n\] \n\nand so \n\n\[ \nf\left( x\right) = f\left( p\right) + {f}^{\left( n\right) }\left( {q}_{n}\right) \frac{{\left( x - p\right) }^{n}}{n!} \leq f\left( p\right) \;\text{ for }x \in {G}_{p}\left( \varepsilon \right) , \n\] \n\ni.e., \( f\left( p\right) \) is the maximum value of \( f \) on \( {G}_{p}\left( \varepsilon \right) \), as claimed.\n\nSimilarly, in the case \( {f}^{\left( n\right) }\left( p\right) > 0 \), a minimum would result.\n\nIf, however, \( n \) is odd, then \( {\left( x - p\right) }^{n} \) is negative for \( x < p \) but positive for \( x > p \) . The same argument then shows that \( f\left( x\right) < f\left( p\right) \) on one side of \( p \) and \( f\left( x\right) > f\left( p\right) \) on the other side; thus no local maximum or minimum can exist at \( p \) . This completes the proof.
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Yes
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Corollary 1. The sum \( S\left( {f, P}\right) = \ell W \) cannot decrease when \( P \) is refined.
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Thus when new partition points are added, \( S\left( {f, P}\right) \) grows in general; i.e., it approaches some supremum value (finite or not). Roughly speaking, the inscribed polygon \( W \) gets \
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No
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Corollary 2 (monotonicity of \( {V}_{f} \) ). If \( a \leq c \leq d \leq b \), then\n\n\[ \n{V}_{f}\left\lbrack {c, d}\right\rbrack \leq {V}_{f}\left\lbrack {a, b}\right\rbrack \n\]
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Proof. By Theorem 1,\n\n\[ \n{V}_{f}\left\lbrack {a, b}\right\rbrack = {V}_{f}\left\lbrack {a, c}\right\rbrack + {V}_{f}\left\lbrack {c, d}\right\rbrack + {V}_{f}\left\lbrack {d, b}\right\rbrack \geq {V}_{f}\left\lbrack {c, d}\right\rbrack . \n\]
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Yes
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For each \( t \in \left\lbrack {a, b}\right\rbrack \) , \n\n\[ \left| {f\left( t\right) - f\left( a\right) }\right| \leq {V}_{f}\left\lbrack {a, b}\right\rbrack \] \n\nHence if \( f \) is of bounded variation on \( \left\lbrack {a, b}\right\rbrack \), it is bounded on \( \left\lbrack {a, b}\right\rbrack \) .
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Proof. If \( t \in \left\lbrack {a, b}\right\rbrack \), let \( P = \{ a, t, b\} \), so \n\n\[ \left| {f\left( t\right) - f\left( a\right) }\right| \leq \left| {f\left( t\right) - f\left( a\right) }\right| + \left| {f\left( b\right) - f\left( t\right) }\right| = S\left( {f, P}\right) \leq {V}_{f}\left\lbrack {a, b}\right\rbrack , \] \n\nproving our first assertion. \( {}^{3} \) Hence \n\n\[ \left( {\forall t \in \left\lbrack {a, b}\right\rbrack }\right) \;\left| {f\left( t\right) }\right| \leq \left| {f\left( t\right) - f\left( a\right) }\right| + \left| {f\left( a\right) }\right| \leq {V}_{f}\left\lbrack {a, b}\right\rbrack + \left| {f\left( a\right) }\right| . \] \n\nThis proves the second assertion.
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Yes
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Corollary 4. A function \( f \) is finite and constant on \( \left\lbrack {a, b}\right\rbrack \) iff \( {V}_{f}\left\lbrack {a, b}\right\rbrack = 0 \) .
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The proof is left to the reader. (Use Corollary 3 and the definitions.)
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No
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Theorem 2. Let \( f, g, h \) be real or complex (or let \( f \) and \( g \) be vector valued and \( h \) scalar valued). Then on any interval \( I = \left\lbrack {a, b}\right\rbrack \), we have\n\n(i) \( {V}_{\left| f\right| } \leq {V}_{f} \)\n\n(ii) \( {V}_{f \pm g} \leq {V}_{f} + {V}_{g} \) ; and\n\n(iii) \( {V}_{hf} \leq s{V}_{f} + r{V}_{h} \), with \( r = \mathop{\sup }\limits_{{t \in I}}\left| {f\left( t\right) }\right| \) and \( s = \mathop{\sup }\limits_{{t \in I}}\left| {h\left( t\right) }\right| \) .\n\nHence if \( f, g \), and \( h \) are of bounded variation on \( I \), so are \( f \pm g,{hf} \), and \( \left| f\right| \) .
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Proof. We first prove (iii).\n\nTake any partition \( P = \left\{ {{t}_{0},\ldots ,{t}_{m}}\right\} \) of \( I \) . Then\n\n\[ \left| {{\Delta }_{i}{hf}}\right| = \left| {h\left( {t}_{i}\right) f\left( {t}_{i}\right) - h\left( {t}_{i - 1}\right) f\left( {t}_{i - 1}\right) }\right| \]\n\n\[ \leq \left| {h\left( {t}_{i}\right) f\left( {t}_{i}\right) - h\left( {t}_{i - 1}\right) f\left( {t}_{i}\right) }\right| + \left| {h\left( {t}_{i - 1}\right) f\left( {t}_{i}\right) - h\left( {t}_{i - 1}\right) f\left( {t}_{i - 1}\right) }\right| \]\n\n\[ = \left| {f\left( {t}_{i}\right) }\right| \left| {{\Delta }_{i}h}\right| + \left| {h\left( {t}_{i - 1}\right) }\right| \left| {{\Delta }_{i}f}\right| \]\n\n\[ \leq r\left| {{\Delta }_{i}h}\right| + s\left| {{\Delta }_{i}f}\right| \]\n\nAdding these inequalities, we obtain\n\n\[ S\left( {{hf}, P}\right) \leq r \cdot S\left( {h, P}\right) + s \cdot S\left( {f, P}\right) \leq r{V}_{h} + s{V}_{f}. \]\n\n\( {}^{3} \) By our conventions, it also follows that \( \left| {f\left( a\right) }\right| \) is a finite constant, and so is \( {V}_{f}\left\lbrack {a, b}\right\rbrack + \left| {f\left( a\right) }\right| \) if \( {V}_{f}\left\lbrack {a, b}\right\rbrack < + \infty \) . As this holds for all sums \( S\left( {{hf}, P}\right) \), it holds for their supremum, so\n\n\[ {V}_{hf} = \sup S\left( {{hf}, P}\right) \leq r{V}_{h} + s{V}_{f} \]\n\n as claimed.\n\nSimilarly, (i) follows from\n\n\[ \left| \right| f\left( {t}_{i}\right) \left| -\right| f\left( {t}_{i - 1}\right) \left| \right| \leq \left| {f\left( {t}_{i}\right) - f\left( {t}_{i - 1}\right) }\right| . \]\n\nThe analogous proof of (ii) is left to the reader.\n\nFinally,(i)-(iii) imply that \( {V}_{f},{V}_{f \pm g} \), and \( {V}_{hf} \) are finite if \( {V}_{f},{V}_{g} \), and \( {V}_{h} \) are. This proves our last assertion.
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No
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(ii) If \( f \) is real and monotone on \( I \), it is of bounded variation there.
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Proof. We prove (ii) first.\n\nLet \( f \uparrow \) on \( I \) . If \( P = \left\{ {{t}_{0},\ldots ,{t}_{m}}\right\} \), then\n\n\[ \n{t}_{i} \geq {t}_{i - 1}\text{ implies }f\left( {t}_{i}\right) \geq f\left( {t}_{i - 1}\right) .\n\]\n\nHence \( {\Delta }_{i}f \geq 0 \) ; i.e., \( \left| {{\Delta }_{i}f}\right| = {\Delta }_{i}f \) . Thus\n\n\[ \nS\left( {f, P}\right) = \mathop{\sum }\limits_{{i = 1}}^{m}\left| {{\Delta }_{i}f}\right| = \mathop{\sum }\limits_{{i = 1}}^{m}{\Delta }_{i}f = \mathop{\sum }\limits_{{i = 1}}^{m}\left\lbrack {f\left( {t}_{i}\right) - f\left( {t}_{i - 1}\right) }\right\rbrack \n\]\n\n\[ \n= f\left( {t}_{m}\right) - f\left( {t}_{0}\right) = f\left( b\right) - f\left( a\right) \n\]\n\nfor any \( P \) . (Verify!) This implies that also\n\n\[ \n{V}_{f}\left\lbrack I\right\rbrack = \sup S\left( {f, P}\right) = f\left( b\right) - f\left( a\right) < + \infty .\n\]\n\nThus (ii) is proved.
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Yes
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Theorem 4. (i) A function \( f : {E}^{1} \rightarrow {E}^{n}\left( {{}^{ * }{C}^{n}}\right) \) is of bounded variation on \( I = \left\lbrack {a, b}\right\rbrack \) iff all of its components \( \left( {{f}_{1},{f}_{2},\ldots ,{f}_{n}}\right) \) are.
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## Proof. (i) Take any partition \( P = \left\{ {{t}_{0},\ldots ,{t}_{m}}\right\} \) of \( I \) . Then \[ {\left| {f}_{k}\left( {t}_{i}\right) - {f}_{k}\left( {t}_{i - 1}\right) \right| }^{2} \leq \mathop{\sum }\limits_{{j = 1}}^{n}{\left| {f}_{j}\left( {t}_{i}\right) - {f}_{j}\left( {t}_{i - 1}\right) \right| }^{2} = {\left| f\left( {t}_{i}\right) - f\left( {t}_{i - 1}\right) \right| }^{2}; \] i.e., \( \left| {{\Delta }_{i}{f}_{k}}\right| \leq \left| {{\Delta }_{i}f}\right|, i = 1,2,\ldots, m \) . Thus \[ \left( {\forall P}\right) \;S\left( {{f}_{k}, P}\right) \leq S\left( {f, P}\right) \leq {V}_{f}, \] and \( {V}_{{f}_{k}} \leq {V}_{f} \) follows. Thus \[ {V}_{f} < + \infty \text{ implies }{V}_{{f}_{k}} < + \infty ,\;k = 1,2,\ldots, n. \] The converse follows by Theorem 2 since \( f = \mathop{\sum }\limits_{{k = 1}}^{n}{f}_{k}{\overrightarrow{e}}_{k} \) . (Explain!)
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No
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Theorem 1. The following are equivalent:\n\n(i) \( f \) is (weakly) absolutely continuous on \( I = \\left\\lbrack {a, b}\\right\\rbrack \) ;\n\n(ii) \( {v}_{f} \) is finite and relatively continuous on \( I \) ; and\n\n(iii) \( \\left( {\\forall \\varepsilon > 0}\\right) \\left( {\\exists \\delta > 0}\\right) \\left( {\\forall x, y \\in I \\mid 0 \\leq y - x < \\delta }\\right) {V}_{f}\\left\\lbrack {x, y}\\right\\rbrack < \\varepsilon \) .
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Proof. We shall show that (ii) \( \\Rightarrow \) (iii) \( \\Rightarrow \) (i) \( \\Rightarrow \) (ii).\n\n(ii) \( \\Rightarrow \) (iii). As \( I = \\left\\lbrack {a, b}\\right\\rbrack \) is compact,(ii) implies that \( {v}_{f} \) is uniformly continuous on \( I \) (Theorem 4 of Chapter 4,§8). Thus\n\n\[\\left( {\\forall \\varepsilon > 0}\\right) \\left( {\\exists \\delta > 0}\\right) \\left( {\\forall x, y \\in I \\mid 0 \\leq y - x < \\delta }\\right) \\;{v}_{f}\\left( y\\right) - {v}_{f}\\left( x\\right) < \\varepsilon .\]\n\nHowever,\n\n\[{v}_{f}\\left( y\\right) - {v}_{f}\\left( x\\right) = {V}_{f}\\left\\lbrack {a, y}\\right\\rbrack - {V}_{f}\\left\\lbrack {a, x}\\right\\rbrack = {V}_{f}\\left\\lbrack {x, y}\\right\\rbrack\]\n\nby additivity (Theorem 1 in \( §7 \) ). Thus (iii) follows.\n\n(iii) \( \\Rightarrow \) (i). By Corollary 3 of \( §7,\\left| {f\\left( x\\right) - f\\left( y\\right) }\\right| \\leq {V}_{f}\\left\\lbrack {x, y}\\right\\rbrack \) . Therefore,(iii) implies that\n\n\[\\left( {\\forall \\varepsilon > 0}\\right) \\left( {\\exists \\delta > 0}\\right) \\left( {\\forall x, y \\in I\\left| \\right| x - y \\mid < \\delta }\\right) \\;\\left| {f\\left( x\\right) - f\\left( y\\right) }\\right| < \\varepsilon ,\]\n\nand so \( f \) is relatively (even uniformly) continuous on \( I \) .\n\nNow with \( \\varepsilon \) and \( \\delta \) as in (iii), take a partition \( P = \\left\{ {{t}_{0},\\ldots ,{t}_{m}}\\right\} \) of \( I \) so fine that\n\n\[{t}_{i} - {t}_{i - 1} < \\delta ,\\;i = 1,2,\\ldots, m.\]\n\nThen \( \\left( {\\forall i}\\right) {V}_{f}\\left\\lbrack {{t}_{i - 1},{t}_{i}}\\right\\rbrack < \\varepsilon \) . Adding up these \( m \) inequalities and using the additivity of \( {V}_{f} \), we obtain\n\n\[{V}_{f}\\left\\lbrack I\\right\\rbrack = \\mathop{\\sum }\\limits_{{i = 1}}^{m}{V}_{f}\\left\\lbrack {{t}_{i - 1},{t}_{i}}\\right\\rbrack < {m\\varepsilon } < + \\infty .\]\n\nThus (i) follows, by definition.\n\nThat (i) \( \\Rightarrow \) (ii) is given as the next theorem.
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Yes
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Theorem 2. If \( {V}_{f}\left\lbrack I\right\rbrack < + \infty \) and if \( f \) is relatively continuous at some \( p \in I \) (over \( I = \left\lbrack {a, b}\right\rbrack \) ), then the same applies to the length function \( {v}_{f} \) .
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Proof. We consider left continuity first, with \( a < p \leq b \) .\n\nLet \( \varepsilon > 0 \) . By assumption, there is \( \delta > 0 \) such that\n\n\[ \left| {f\left( x\right) - f\left( p\right) }\right| < \frac{\varepsilon }{2}\text{ when }\left| {x - p}\right| < \delta \text{ and }x \in \left\lbrack {a, p}\right\rbrack . \]\n\nFix any such \( x \) . Also, \( {V}_{f}\left\lbrack {a, p}\right\rbrack = \mathop{\sup }\limits_{P}S\left( {f, P}\right) \) over \( \left\lbrack {a, p}\right\rbrack \) . Thus\n\n\[ {V}_{f}\left\lbrack {a, p}\right\rbrack - \frac{\varepsilon }{2} < \mathop{\sum }\limits_{{i = 1}}^{k}\left| {{\Delta }_{i}f}\right| \]\n\nfor some partition\n\n\[ P = \left\{ {{t}_{0} = a,\ldots ,{t}_{k - 1},{t}_{k} = p}\right\} \text{of}\left\lbrack {a, p}\right\rbrack \text{. (Why?)} \]\n\nWe may assume \( {t}_{k - 1} = x, x \) as above. (If \( {t}_{k - 1} \neq x \), add \( x \) to \( P \) .) Then\n\n\[ \left| {{\Delta }_{k}f}\right| = \left| {f\left( p\right) - f\left( x\right) }\right| < \frac{\varepsilon }{2} \]\n\nand hence\n\n\[ {V}_{f}\left\lbrack {a, p}\right\rbrack - \frac{\varepsilon }{2} < \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}\left| {{\Delta }_{i}f}\right| + \left| {{\Delta }_{k}f}\right| < \mathop{\sum }\limits_{{i = 1}}^{{k - 1}}\left| {{\Delta }_{i}f}\right| + \frac{\varepsilon }{2} \leq {V}_{f}\left\lbrack {a,{t}_{k - 1}}\right\rbrack + \frac{\varepsilon }{2}. \]\n\n(1)\n\nHowever,\n\n\[ {V}_{f}\left\lbrack {a, p}\right\rbrack = {v}_{f}\left( p\right) \]\n\nand\n\n\[ {V}_{f}\left\lbrack {a,{t}_{k - 1}}\right\rbrack = {V}_{f}\left\lbrack {a, x}\right\rbrack = {v}_{f}\left( x\right) . \]\n\nThus (1) yields\n\n\[ \left| {{v}_{f}\left( p\right) - {v}_{f}\left( x\right) }\right| = {V}_{f}\left\lbrack {a, p}\right\rbrack - {V}_{f}\left\lbrack {a, x}\right\rbrack < \varepsilon \text{for}x \in \left\lbrack {a, p}\right\rbrack \text{with}\left| {x - p}\right| < \delta \text{.} \]\n\nThis shows that \( {v}_{f} \) is left continuous at \( p \) .\n\nRight continuity is proved similarly on noting that\n\n\[ {v}_{f}\left( x\right) - {v}_{f}\left( p\right) = {V}_{f}\left\lbrack {p, b}\right\rbrack - {V}_{f}\left\lbrack {x, b}\right\rbrack \text{ for }p \leq x < b.\text{ (Why?) } \]\n\nThus \( {v}_{f} \) is, indeed, relatively continuous at \( p \) . Observe that \( {v}_{f} \) is also of bounded variation on \( I \), being monotone and finite (see Theorem 3(ii) of \( §7 \) ).\n\nThis completes the proof of both Theorem 2 and Theorem 1.
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No
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Corollary 1. If \( f \) is real and absolutely continuous on \( I = \left\lbrack {a, b}\right\rbrack \) (weakly), so are the nondecreasing functions \( g \) and \( h\left( {f = g - h}\right) \) defined in Theorem 3 of \( §7 \) .
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Indeed, the function \( g \) as defined there is simply \( {v}_{f} \) . Thus it is relatively continuous and finite on \( I \) by Theorem 1. Hence so also is \( h = f - g \) . Both are of bounded variation (monotone!) and hence absolutely continuous (weakly).
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Yes
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Corollary 2. If \( F = \int f \) on \( I = \left\lbrack {a, b}\right\rbrack \) and if \( f \) is bounded \( \left( {\left| f\right| \leq K \in {E}^{1}}\right) \) on \( I - Q \) ( \( Q \) countable), then \( F \) is weakly absolutely continuous on \( I \) .
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Proof. By definition, \( F = \int f \) is finite and relatively continuous on \( I \), so we only have to show that \( {V}_{F}\left\lbrack I\right\rbrack < + \infty \) . This, however, easily follows by Problem 3 of \( §7 \) on noting that \( {F}^{\prime } = f \) on \( I - S \) ( \( S \) countable). Details are left to the reader.
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No
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Theorem 3. If \( f : {E}^{1} \rightarrow E \) is continuously differentiable on \( I = \left\lbrack {a, b}\right\rbrack \left( {§6}\right) \) , then \( {v}_{f} = \int \left| {f}^{\prime }\right| \) on \( I \) and\n\n\[ \n{V}_{f}\left\lbrack {a, b}\right\rbrack = {\int }_{a}^{b}\left| {f}^{\prime }\right| .\n\]
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Proof. Let \( a < p < x \leq b,{\Delta x} = x - p \), and\n\n\[ \n\Delta {v}_{f} = {v}_{f}\left( x\right) - {v}_{f}\left( p\right) = {V}_{f}\left\lbrack {p, x}\right\rbrack .\;\text{ (Why?) }\n\]\n\nAs a first step, we shall show that\n\n\[ \n\frac{\Delta {v}_{f}}{\Delta x} \leq \mathop{\sup }\limits_{\left\lbrack p, x\right\rbrack }\left| {f}^{\prime }\right| \n\]\n\n(2)\n\nFor any partition \( P = \left\{ {p = {t}_{0},\ldots ,{t}_{m} = x}\right\} \) of \( \left\lbrack {p, x}\right\rbrack \), we have\n\n\[ \nS\left( {f, P}\right) = \mathop{\sum }\limits_{{i = 1}}^{m}\left| {{\Delta }_{i}f}\right| \leq \mathop{\sum }\limits_{{i = 1}}^{m}\mathop{\sup }\limits_{\left\lbrack {t}_{i - 1},{t}_{i}\right\rbrack }\left| {f}^{\prime }\right| \left( {{t}_{i} - {t}_{i - 1}}\right) \leq \mathop{\sup }\limits_{\left\lbrack p, x\right\rbrack }\left| {f}^{\prime }\right| {\Delta x}.\n\]\n\nSince this holds for any partition \( P \), we have\n\n\[ \n{V}_{f}\left\lbrack {p, x}\right\rbrack \leq \mathop{\sup }\limits_{\left\lbrack p, x\right\rbrack }\left| {f}^{\prime }\right| {\Delta x} \n\]\n\nwhich implies (2).\n\nOn the other hand,\n\n\[ \n\Delta {v}_{f} = {V}_{f}\left\lbrack {p, x}\right\rbrack \geq \left| {f\left( x\right) - f\left( p\right) }\right| = \left| {\Delta f}\right| .\n\]\n\nCombining, we get\n\n\[ \n\left| \frac{\Delta f}{\Delta x}\right| \leq \frac{\Delta {v}_{f}}{\Delta x} \leq \mathop{\sup }\limits_{\left\lbrack p, x\right\rbrack }\left| {f}^{\prime }\right| < + \infty \n\]\n\n(3)\n\nsince \( {f}^{\prime } \) is relatively continuous on \( \left\lbrack {a, b}\right\rbrack \), hence also uniformly continuous and bounded. (Here we assumed \( a < p < x \leq b \) . However,(3) holds also if \( a \leq x < p < b \), with \( \Delta {v}_{f} = - V\left\lbrack {x, p}\right\rbrack \) and \( {\Delta x} < 0 \) . Verify!)\n\nNow\n\n\[ \n\left| \right| {f}^{\prime }\left( p\right) \left| -\right| {f}^{\prime }\left( x\right) \left| \right| \leq \left| {{f}^{\prime }\left( p\right) - {f}^{\prime }\left( x\right) }\right| \rightarrow 0\;\text{ as }x \rightarrow p,\n\]\n\nso, taking limits as \( x \rightarrow p \), we obtain\n\n\[ \n\mathop{\lim }\limits_{{x \rightarrow p}}\frac{\Delta {v}_{f}}{\Delta x} = \left| {{f}^{\prime }\left( p\right) }\right| \n\]\n\nThus \( {v}_{f} \) is differentiable at each \( p \) in \( \left( {a, b}\right) \), with \( {v}_{f}^{\prime }\left( p\right) = \left| {{f}^{\prime }\left( p\right) }\right| \) . Also, \( {v}_{f} \) is relatively continuous and finite on \( \left\lbrack {a, b}\right\rbrack \) (by Theorem 1). \( {}^{2} \) Hence \( {v}_{f} = \int \left| {f}^{\prime }\right| \) on \( \left\lbrack {a, b}\right\rbrack \), and we obtain\n\n\[ \n{\int }_{a}^{b}\left| {f}^{\prime }\right| = {v}_{f}\left( b\right) - {v}_{f}\left( a\right) = {V}_{f}\left\lbrack {a, b}\right\rbrack ,\text{ as asserted. }\n\]\n\n(4)
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Yes
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Theorem 1. Let \( {F}_{n} : {E}^{1} \rightarrow E\left( {n = 1,2,\ldots }\right) \) be finite and relatively continuous on \( I \) and differentiable on \( I - Q \) . Suppose that\n\n(a) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{F}_{n}\left( p\right) \) exists for some \( p \in I \) ;\n\n(b) \( {F}_{n}^{\prime } \rightarrow f \neq \pm \infty \) (uniformly) on \( J - Q \) for each finite subinterval \( J \subseteq I \) ;\n\n(c) \( E \) is complete.\n\nThen\n\n(i) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{F}_{n} = F \) exists uniformly on each finite subinterval \( J \subseteq I \) ;\n\n(ii) \( F = \int f \) on \( I \) ; and\n\n(iii) \( {\left( \lim {F}_{n}\right) }^{\prime } = {F}^{\prime } = f = \mathop{\lim }\limits_{{n \rightarrow \infty }}{F}_{n}^{\prime } \) on \( I - Q \) .
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Proof. Fix \( \varepsilon > 0 \) and any subinterval \( J \subseteq I \) of length \( \delta < \infty \), with \( p \in J(p \) as in (a)). By (b), \( {F}_{n}^{\prime } \rightarrow f \) (uniformly) on \( J - Q \), so there is a \( k \) such that for \( m, n > k \) ,\n\n\[ \left| {{F}_{n}^{\prime }\left( t\right) - f\left( t\right) }\right| < \frac{\varepsilon }{2},\;t \in J - Q; \]\n\n(1)\n\nhence\n\n\[ \mathop{\sup }\limits_{{t \in J - Q}}\left| {{F}_{m}^{\prime }\left( t\right) - {F}_{n}^{\prime }\left( t\right) }\right| \leq \varepsilon .\;\text{ (Why?) } \]\n\n(2)\n\nNow apply Corollary 1 in \( §4 \) to the function \( h = {F}_{m} - {F}_{n} \) on \( J \) . Then for each \( x \in J,\left| {h\left( x\right) - h\left( p\right) }\right| \leq M\left| {x - p}\right| \), where\n\n\[ M \leq \mathop{\sup }\limits_{{t \in J - Q}}\left| {{h}^{\prime }\left( t\right) }\right| \leq \varepsilon \]\n\nby (2). Hence for \( m, n > k, x \in J \) and\n\n\[ \left| {{F}_{m}\left( x\right) - {F}_{n}\left( x\right) - {F}_{m}\left( p\right) + {F}_{n}\left( p\right) }\right| \leq \varepsilon \left| {x - p}\right| \leq {\varepsilon \delta }. \]\n\n(3)\n\nAs \( \varepsilon \) is arbitrary, this shows that the sequence\n\n\[ \left\{ {{F}_{n} - {F}_{n}\left( p\right) }\right\} \]\n\nsatisfies the uniform Cauchy criterion (Chapter 4, \( §{12} \), Theorem 3). Thus as \( E \) is complete, \( \left\{ {{F}_{n} - {F}_{n}\left( p\right) }\right\} \) converges uniformly on \( J \) . So does \( \left\{ {F}_{n}\right\} \), for \( \left\{ {{F}_{n}\left( p\right) }\right\} \) converges, by (a). Thus we may set\n\n\[ F = \lim {F}_{n}\text{ (uniformly) on }J, \]\nproving assertion (i). \( {}^{1} \)
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No
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Theorem 2. Let the functions \( {f}_{n} : {E}^{1} \rightarrow E, n = 1,2,\ldots \), have antiderivatives, \( {F}_{n} = \int {f}_{n} \), on \( I \). Suppose \( E \) is complete and \( {f}_{n} \rightarrow f \) (uniformly) on each finite subinterval \( J \subseteq I \), with \( f \) finite there. Then \( \int f \) exists on \( I \), and \[ {\int }_{p}^{x}f = {\int }_{p}^{x}\mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{p}^{x}{f}_{n}\text{ for any }p, x \in I. \]
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Proof. Fix any \( p \in I \). By Note 2 in \( §5 \), we may choose \[ {F}_{n}\left( x\right) = {\int }_{p}^{x}{f}_{n}\text{ for }x \in I. \] Then \( {F}_{n}\left( p\right) = {\int }_{p}^{p}{f}_{n} = 0 \), and so \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{F}_{n}\left( p\right) = 0 \) exists, as required in Theorem 1(a). Also, by definition, each \( {F}_{n} \) is relatively continuous and finite on \( I \) and differentiable, with \( {F}_{n}^{\prime } = {f}_{n} \), on \( I - {Q}_{n} \). The countable sets \( {Q}_{n} \) need not be the same, so we replace them by \[ Q = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{Q}_{n} \] (including in \( Q \) also the endpoints of \( I \), if any). Then \( Q \) is countable (see Chapter 1,§9, Theorem 2), and \( I - Q \subseteq I - {Q}_{n} \), so all \( {F}_{n} \) are differentiable on \( I - Q \), with \( {F}_{n}^{\prime } = {f}_{n} \) there. Additionally, by assumption, \[ {f}_{n} \rightarrow f\text{ (uniformly) } \] on finite subintervals \( J \subseteq I \). Hence \[ {F}_{n}^{\prime } \rightarrow f\text{ (uniformly) on }J - Q \] for all such \( J \), and so the conditions of Theorem 1 are satisfied. By that theorem, then, \[ F = \int f = \lim {F}_{n}\text{ exists on }I \] and, recalling that \[ {F}_{n}\left( x\right) = {\int }_{p}^{x}{f}_{n} \] we obtain for \( x \in I \) \[ {\int }_{p}^{x}f = F\left( x\right) - F\left( p\right) = \lim {F}_{n}\left( x\right) - \lim {F}_{n}\left( p\right) = \lim {F}_{n}\left( x\right) - 0 = \lim {\int }_{p}^{x}{f}_{n}. \] As \( p \in I \) was arbitrary, and \( f = \lim {f}_{n} \) (by assumption), all is proved.
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Yes
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