The Real And Complex Number Systems

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n |fb k fa k | 1. k1 For this , and let K be the smallest positive integer such that K/2 b a. So,we partition a, b into K closed subintervals, i.e., P y 0 a, y 1 a /2,....,y K1 a K 1/2, y K b. So, it is clear that f is of bounded variation y i , y i1 ,wherei 0, 1, . . . , K. It implies that f is of bounded variation on a, b. Note: There exists functions which are continuous and of bounded variation but not absolutely continuous. Remark: 1. The standard example is called Cantor-Lebesgue function. The reader can see this in the book, Measure and Integral, An Introduction to Real Analysis by Richard L. Wheeden and Antoni Zygmund, pp 35 and pp 115. 2. If we wrtie ”absolutely continuous” by ABC, ”continuous” by C, and ”bounded variation” by B, then it is clear that by preceding result, ABC implies B and C, andB and C do NOT imply ABC. 6.12 Prove that f is absolutely continuous if it satisfies a uniform Lipschitz condition of order 1 on a, b. (See Exercise 6.2) Proof: Letf satisfy a uniform Lipschitz condition of order 1 on a, b, i.e., |fx fy| M|x y| where x, y a, b. Thengiven 0, there is a /M such that n as k1 b k a k , wherea k , b k s are disjoint open subintervals on a, b, k 1, . . , n, we have n k1 |fb k fa k | M|b k a k | n k1 n k1 Mb k a k M . Hence, f is absolutely continuous on a, b. 6.13 If f and g are absolutely continunous on a, b, prove that each of the following is also: |f|, cf (c constant), f g, f g; alsof/g if g is bounded away from zero. Proof: (1) (|f| is absolutely continuous on a, b): Given 0, we want to find a n 0, such that as k1 b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave n ||fb k | |fa k || . 1* k1 Since f is absolutely continunous on a, b, for this , there is a 0 such that as n k1 b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave n |fb k fa k | k1 which implies that (1*) holds by the following
n k1 n ||fb k | |fa k || |fb k fa k | . So, we know that |f| is absolutely continuous on a, b. (2) (cf is absolutely continuous on a, b): If c 0, it is clear. So, we may assume that n c 0. Given 0, we want to find a 0, such that as k1 b k a k , where a k , b k s are disjoint open intervals on a, b, wehave n k1 k1 |cfb k cfa k | . 2* Since f is absolutely continunous on a, b, for this , there is a 0 such that as n k1 b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave n |fb k fa k | /|c| k1 which implies that (2*) holds by the following n k1 |cfb k cfa k | |c| |fb k fa k | . So, we know that cf is absolutely continuous on a, b. (3) (f g is absolutely continuous on a, b): Given 0, we want to find a 0, n such that as k1 b k a k , wherea k , b k s are disjoint open intervals on a, b, we have n |f gb k f ga k | . 3* k1 Since f and g are absolutely continunous on a, b, for this , there is a 0 such that as n b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave k1 n k1 |fb k fa k | /2 and |gb k ga k | /2 which implies that (3*) holds by the following n k1 n n k1 n k1 |f gb k f ga k | |fb k fa k gb k ga k | k1 n k1 n |fb k fa k | k1 |gb k ga k | . So, we know that f g is absolutely continuous on a, b. (4) (f g is absolutely continuous on a, b.): Let M f sup xa,b |fx| and M g sup xa,b |gx|. Given 0, we want to find a 0, such that as n b k a k , wherea k , b k s are disjoint open intervals on a, b, wehave k1
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The Real And Complex Number Systems
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Proof: Consider 2n−2 ∑ (x + 1)
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Then S = N. Proof: (A ⇒ B): If S
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Note: There are many and many metho
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In addition, (√ a ) 2 − − b (
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Proof: Say {ar + b : a ∈ Z, b ∈
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(⇐)It is clear since every a n is
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Proof: Choose a 0 = [x], and thus c
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Proof: Use Cauchy-Schwarz inequalit
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In addition, ∑ a k b k + a j b j
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So, z 1 z 2 = ¯z 1¯z 2 (c) z¯z =
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so, |a − b| 2 ≤ ( 1 + |a| 2) (
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Proof: Note that in this text book,
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1.39 State and prove a theorem anal
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1.43 (a) Prove that Log (z w ) = wL
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For complex θ, we show that it hol
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So, ix−y = i (x + iy) = iz = log
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and the sum of their squares is giv
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Some Basic Notations Of Set Theory
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(i) If x ∈ R, it is clear that (x
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(a) A ∪ (B ∪ C) = (A ∪ B) ∪
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(e) A ∩ (B − C) = (A ∩ B) −
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Proof: Given x ∈ X, then f (x)
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By hyppothesis, we get T ⊆ f (S)
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Proof: Since S is an infinite set,
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Proof: Assume S˜R and let f be a o
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2.22 Let S denote the collection of
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Charpter 3 Elements of Point set To
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And thus by the remark in (e), it i
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3.7 Prove that a nonempty, bounded
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Bx, r x S Bx, r x T . So, at
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Claim that Bp, r S as follows. Let
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Covering theorems in R n 3.17 If S
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uncountable. Hence, by exercise 3.2
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Then we have (a) (b) (c) (d) x y
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(b) Give an example of a metric spa
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Hence from (1)-(4), we know that i
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3.44 intM A M A . Proof: Let B
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That is, intB which is absurb. He
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Limits And Continuity Limits of seq
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x n1 x n 1 1 x n x n 1 2 . Rema
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|a m a n| |a m a m1 a m1 a m2
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and use the fact if a n converges
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lim y0 fx, y 0ifx 0, 1ifx 0. and
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fx, y 1 r cos sinr2 cossin if x
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then we have lim fx, y L if x 0an
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this is because a can be an isolate
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irrational. Prove that: (a) ffx x
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In addition, since f1 f1 by f0 0,
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lim r n0 fx f lim n x n lim n fx
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if x Q, andgx 0ifx Q c . Remark:
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number f T supfx fy : x, y T is
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Proof: Since the hypothesis says th
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(f) S 0, 1 0, 1, T 0, 1 0, 1. S
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In Exercises 4.29 through 4.33, we
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function. Since f0, 1 0, 1 0, 1,
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4.40 If x is a point in a metric sp
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from b to A) are disjoint. So, if e
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x F k1 F k A B U V which im
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Proof: Assume that B C 1 is discon
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Proof: By Mean Value Theorem, wehav
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x r y r rz r1 x y ry r1 /2. So,
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hx gfx if x S. Iff is uniformly c
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dx, y , x, y A, wehave dfx, fy
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In addition, part (b) comes from in
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continuous at a, a. (2) (x y) Sinc
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Proof: Let D denote the set of dico
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(a) Provet that the formula df, g
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so, by induction, we find dp n1 , p
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Proof: If p and p are fixed points
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(b) Assume there is a subsequence p
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Hence, f is increasing on , a/3 a
Page 145 and 146: f k x f k 0 x 0 fk x x by induct
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Page 149 and 150: g x g x 3 2 a d f x a d f x 3 2
Page 151 and 152: which implies that F 0, where
Page 153 and 154: g fx gfx. Assume that there is a
Page 155 and 156: Remark: For x 0, we can show that
Page 157 and 158: Proof: Look at the Generalized Mean
Page 159 and 160: Proof: By Generalized Mean Value Th
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Page 163 and 164: Remark: If we can make sure that fx
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Page 167 and 168: fx fy x y f x /2 *’ Combi
Page 169 and 170: lim xa fx gx L. Remark: 1. The pr
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Page 173 and 174: f x 1 ...x n n f x 1 ...x n1 n x
Page 175 and 176: f x 1 ...x n n fx 1 ...fx n n . C
Page 177 and 178: lx fc f cx c fx. (Exercise 2)
Page 179 and 180: So, we have Partial derivatives fb
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Page 185 and 186: Functions of Bounded Variation and
Page 187 and 188: For the part fx x fy 1 |fx fy| 1
Page 189 and 190: Claim that 1 sup A. Suppose NOT, t
Page 191 and 192: we know that V f is an increasing
Page 193 and 194: in Section 6.12. Proof: Sinceft : t
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Page 199 and 200: Supplement on lim sup and lim inf I
Page 201 and 202: (1) lim n→ inf a n − a n has
Page 203 and 204: Remark: The exercise is useful in t
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Page 207 and 208: If lim sup n→ a n1 a n , then it
Page 209 and 210: We first prove lim n→ sup n ≤
Page 211 and 212: 0 ≤ d 1 * by Theorem 8.23 (i). N
Page 213 and 214: Consider which implies that which i
Page 215 and 216: So, L 1 5 since L ≥ 1. 2 Remark:
Page 217 and 218: For case (i), since 1 n log nlog lo
Page 219 and 220: ()Assume that ∑ n1 That is, p!
Page 221 and 222: n S n ∑−1 k1 1 3k − 2 − 1
Page 223 and 224: (b) Prove that ∑ n1 −1 n−1 /
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Page 229 and 230: Suppose NOT, it means that lim n→
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Page 233 and 234: |sin k|r ∑ k So, ∑ |sink|r dive
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Page 237 and 238: lim n→ sin ... sin n 1 ... 1 n
Page 239 and 240: Also, lim q→ fp, q lim q→ p si
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Page 243 and 244: So, n s n ∑ cos2k − 1x j1 sin
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which implies that which implies th
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So, by above sayings, we have prove
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(a) If fn is multiplicative and if
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Sequences of Functions Uniform conv
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(b) Prove that h n (x) does not con
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Proof: Since g is continuous on a c
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Hence, {g (x) x n } converges unifo
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y continuity of f k(x0 ) (x) − f
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In addition, as c ≥ 1/2, if f n
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(Lemma) If {a n } and {b n } are tw
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9.16 Let {f n } be a sequence of re
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which implies that ∣ lim sup 1 k
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have ∫ x 0 ∞∑ t 2n − 1 2 n=
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[ ] π And as x ∈ , π , then n+p
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0.1 Supplement on some results on W
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which implies that h (x + t) − h
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we have ∫ d c |f n − g| 2 dx
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we complete it. 9.31 Given that two
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if x = λ (≠ 0) is a root of 1 +
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So, the series diverges. (ii) As
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9.38 For each real t, define f t (x
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So, B 0 = P 0 (0) = C 0 = 1, B 1 =
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Remark: (1) The reader can see the
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That is, lim n inf a n sup c n . n
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and lim n infa n b n lim n a n
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q! kq1 1 k! kq1 q! k! 1 q 1
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g x log 1 1 x x a x 2 x log
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The Real And Complex Number Systems

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