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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 \( p{.}^{5} \) | 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 \( {x}_{n} \) i... | No |
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) \( {x}_{m} \pm {y}_{m} \rightarrow q \pm r \) ; | 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 ... | Yes |
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., \( \lim {x}_{m} \leq \lim {y}_{m} \). | 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... | No |
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 \( {x}_{m} \leq c \) (respec... | One can prove this from Corollary 1, with \( {y}_{m} = c \) (or \( {x}_{m} = c \) ) for all \( m \) . | No |
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 \( p{.}^{1} \) | 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( ... | Yes |
(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 \( A \) ). | 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\... | Yes |
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{\l... | 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. | No |
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}\r... | 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}... | No |
Theorem 1. Every convergent sequence \( \left\{ {x}_{m}\right\} \subseteq \left( {S,\rho }\right) \) is Cauchy. | 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\[ \l... | Yes |
Theorem 2. Every Cauchy sequence \( \left\{ {x}_{m}\right\} \subseteq \left( {S,\rho }\right) \) is bounded. | 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... | Yes |
Theorem 3. If a Cauchy sequence \( \left\{ {x}_{m}\right\} \) clusters at a point \( p \), then \( {x}_{m} \rightarrow p \) . | 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}\... | Yes |
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. | 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 ... | Yes |
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., | 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 ... | Yes |
Corollary 2. \( f \) is continuous at \( p\left( {p \in {D}_{f}}\right) \) iff \( f\left( x\right) \rightarrow f\left( p\right) \) as \( x \rightarrow p \) . | The straightforward proof from definitions is left to the reader. | No |
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 \... | 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... | Yes |
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}\\... | 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). | No |
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 \( q = f\left( p\right) \)... | 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(... | Yes |
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\nprov... | 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) > ... | Yes |
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 \( h\left( p\right) \neq 0 \); simi... | 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 assum... | No |
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 \( p\left( {p \in A}\right) \) i... | 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| =... | Yes |
Theorem 3. Any rational function (in particular, every polynomial) in one or several variables is continuous on all of its domain. | 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 \ | No |
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) \l... | 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}^... | Yes |
Theorem 1. If a set \( B \subseteq \left( {S,\rho }\right) \) is compact, so is any closed subset \( A \subseteq B \) . | 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 b... | Yes |
Theorem 2. Every compact set \( A \subseteq \left( {S,\rho }\right) \) is closed. | 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}\ri... | Yes |
Theorem 3. Every compact set \( A \subseteq \left( {S,\rho }\right) \) is bounded. | 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} \i... | Yes |
Theorem 4. In \( {E}^{n}\left( {{}^{ * }\text{and}\left. {C}^{n}\right) }\right. \) a set is compact iff it is closed and bounded. | 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 t... | Yes |
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... | 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 Cau... | No |
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}_{... | 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... | Yes |
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. | 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 \( \varepsi... | Yes |
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 ... | 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 \( {y}_{m} \in f\left\lbrack B\right\rbrack ,{y}_{m} = f\left( {x}_{m}\right) \) f... | Yes |
Lemma 1. Every nonvoid compact set \( F \subseteq {E}^{1} \) has a maximum and a minimum. | 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, ea... | Yes |
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 \( f \)... | 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 \( {E}^{1} \) .... | Yes |
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... | 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 \( \left\{ {y}_{m}\right\} \subseteq f\left\lbrack B\right\rbrack \) , \( {y}_{m} \rightarrow q \in f\left\lbrack B\right\rbrack \),... | Yes |
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 \( B \) . | 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\le... | Yes |
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 ... | Proof. Use Cantor’s theorem (Theorem 5 of \( §6 \) ) and Example (1) in \( §8 \) . | No |
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 \( \left( {a, b}\right) \) . | 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}^{... | Yes |
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). \... | 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} \... | No |
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. | Indeed, this is part of Problem 15 in \( §2 \) with \( \left( {S,\rho }\right) \) replaced by \( \left( {A,\rho }\right) \). | No |
Theorem 2. The only connected sets in \( {E}^{1} \) are exactly all convex sets, i.e., finite and infinite intervals, including \( {E}^{1} \) itself. | 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... | No |
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 \( \left( {T,{\rho }^{\prime }}\right) {}^{4} \) . | 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\nSe... | Yes |
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. | 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 \... | Yes |
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. | Proof. Use Lemma 1 and Example (c) in part I of this section. | No |
Theorem 4. Every open connected set \( A \) in \( {E}^{n} \) (*or in another normed space) is also arcwise connected and even polygon connected. | Proof. If \( A = \varnothing \), this is \ | No |
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 \( B \) . | 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. | Yes |
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 \( S \\times S \) . | 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... | Yes |
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 \(... | 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... | Yes |
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... | 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} \... | Yes |
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}\rig... | 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{\l... | No |
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 +... | 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\} \\) . H... | No |
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... | 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... | Yes |
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 \(... | 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 = ... | Yes |
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). | 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| \righ... | Yes |
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 diverg... | 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... | Yes |
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 ... | 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]{\lef... | No |
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| \... | 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}\... | Yes |
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 \( E = {E}^{ * } \) ). | 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}^{\pri... | Yes |
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) = ... | 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\rig... | Yes |
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}^{\pr... | 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)... | Yes |
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(... | (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) \( {\left( \frac{f}{h}\right) }^{\prime } = \frac{h{f}^{\prime } - {h}^{\prime }f}{{h}^{2}} \) | Yes |
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 | \[ {f}^{\prime }\left( p\right) = \left( {{f}_{1}^{\prime }\left( p\right) ,\ldots ,{f}_{n}^{\prime }\left( p\right) }\right) = \mathop{\sum }\limits_{{k = 1}}^{n}{f}_{k}^{\prime }\left( p\right) {\bar{e}}_{k}, \] with \( {\bar{e}}_{k} \) as in Theorem 2 of Chapter 3, \( §§1 - 3 \) . In particular, a complex function \... | Yes |
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\nSi... | Proof. If \( {f}^{\prime }\left( p\right) > 0 \), the \ | No |
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 \( p \) . | For, by Lemma \( 1,{f}^{\prime }\left( p\right) \neq 0 \) excludes a maximum or minimum at \( p \) . (Why?) | No |
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 th... | 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 ... | Yes |
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 \( p \in \left( {a, b}\right) \) . | 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. | Yes |
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... | 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 relat... | Yes |
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 o... | Proof. Take \( g\left( x\right) = x \) in Theorem 2, so \( {g}^{\prime } = 1 \) on \( {E}^{1} \) . | No |
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 \( \left( {a, b}\right) \) ; | 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... | Yes |
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 \... | 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... | No |
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 \( I \) . | 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. | No |
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... | 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... | No |
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 }... | 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}^{\... | Yes |
Corollary 2. Let \( f \) be as in Corollary 1. Then \( f \) is constant on \( I \) iff \( {f}^{\prime } = 0 \) on \( I - Q \) . | 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 \... | Yes |
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 \( I - Q \) . | Proof. Apply Corollary 2 to the function \( f - g \) . | No |
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 \( I - Q \) . | 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 \( I \), ... | Yes |
Theorem 1. If \( F \) and \( G \) are primitive to \( f \) on \( I \), then \( G - F \) is constant on \( I \). | 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}... | Yes |
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 }_... | 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\nw... | Yes |
Corollary 2. If both \( \int f \) and \( \int \left| f\right| \) exist on \( I = \left\lbrack {a, b}\right\rbrack \), then\n\n\[\n\left| {{\int }_{a}^{b}f}\right| \leq {\int }_{a}^{b}\left| f\right|\n\] | 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| ... | Yes |
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\[ M \leq \mathop{\sup }\limits_{{t \in I - Q}}\left| {f\left( t\right) }\right| \] | This is simply Corollary 1 of \( §4 \), when applied to a primitive, \( F = \int f \) . | No |
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 \( Q \) .) | Proof. Let \( {F}^{\prime } = f \) on \( I - P \) . Then \( {F}^{\prime } = g \) on \( I - \left( {P \cup Q}\right) \) . The rest is clear. | No |
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\[{... | 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 \( \i... | Yes |
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 \( §1 \) ) | \[ {\int }_{a}^{b}f = \left( {{\int }_{a}^{b}{f}_{1},\ldots ,{\int }_{a}^{b}{f}_{n}}\right) = \mathop{\sum }\limits_{{k = 1}}^{n}{\overrightarrow{e}}_{k}{\int }_{a}^{b}{f}_{k}\;\text{ for any }a, b \in I. \] Hence if \( f \) is complex, \[ {\int }_{a}^{b}f = {\int }_{a}^{b}{f}_{\mathrm{{re}}} + i \cdot {\int }_{a}^{b}{... | Yes |
Corollary 8. If \( f = 0 \) on \( I - Q \), then \( \int f \) exists on \( I \), and | \[ \left| {{\int }_{a}^{b}f}\right| = {\int }_{a}^{b}\left| f\right| = 0\;\text{ for }a, b \in I. \] | No |
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}^... | 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 }\... | Yes |
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 \( {\int }_{a}^{b}f \geq 0 \) . | 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 \ | No |
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) . | Proof. Apply Corollary 3 in \( §2 \) to the function \( F = \int f \) . | No |
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) ... | 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 \i... | Yes |
Corollary 1. The sum \( S\left( {f, P}\right) = \ell W \) cannot decrease when \( P \) is refined. | 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 \ | No |
Corollary 2 (monotonicity of \( {V}_{f} \) ). If \( a \leq c \leq d \leq b \), then\n\n\[ \n{V}_{f}\left\lbrack {c, d}\right\rbrack \leq {V}_{f}\left\lbrack {a, b}\right\rbrack \n\] | Proof. By Theorem 1,\n\n\[ \n{V}_{f}\left\lbrack {a, b}\right\rbrack = {V}_{f}\left\lbrack {a, c}\right\rbrack + {V}_{f}\left\lbrack {c, d}\right\rbrack + {V}_{f}\left\lbrack {d, b}\right\rbrack \geq {V}_{f}\left\lbrack {c, d}\right\rbrack . \n\] | Yes |
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 \( \left\lbrack {a, b}\right\rbrack \) . | 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}\righ... | Yes |
Corollary 4. A function \( f \) is finite and constant on \( \left\lbrack {a, b}\right\rbrack \) iff \( {V}_{f}\left\lbrack {a, b}\right\rbrack = 0 \) . | The proof is left to the reader. (Use Corollary 3 and the definitions.) | No |
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}... | 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}\... | No |
(ii) If \( f \) is real and monotone on \( I \), it is of bounded variation there. | 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| = {\De... | Yes |
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. | ## 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} = ... | No |
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( ... | 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( {\\fora... | Yes |
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 \( {v}_{f} \) . | 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... | No |
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 \( §7 \) . | 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). | Yes |
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 \( I \) . | 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... | No |
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\[ \n{V}_{f}\left\lbrack {a, b}\right\rbrack = {\int }_{a}^{b}\left| {f}^{\prime }\right| .\n\] | 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... | Yes |
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... | 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\lef... | No |
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 ... | 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 Theor... | Yes |
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