The Real And Complex Number Systems

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3. Here is another proof on (b). Proof: Given any point x S, and thus consider f n x S. Then there is a convergent subsequence f nk x, say its limit p, sinceS is compact. Consider dfp, p d f limf nk x , limf nk x k k and Note that d limff nk x, limf nk x by continuity of f at p k k limdf nk1 x, f nk x 1 k df nk1 x, f nk x ... df 2 f nk1 x, ff nk1 x. 2 limdf 2 f nk1 x, ff nk1 x k d limf 2 f nk1 x, limff nk1 x k k d f 2 limf nk1 x , f limf nk1 x by continuity of f 2 and f at p k k df 2 p, fp. 3 So, by (1)-(3), we know that fp, fp df 2 p, fp p fp by hypothesis dfx, fy dx, y where x y. Hence, f has a unique fixed point p by (a) in Exercise. Note. 1.Ifx n x, andy n y, then dx n , y n dx, y. Thatis, lim n dx n , y n d lim n x n , lim n y n . Proof: Consider dx n , y n dx n , x dx, y dy, y n and dx, y dx, x n dx n , y n dy n , y, then |dx n , y n dx, y| dx, x n dy, y n 0. So,wehaveproveit. 2. The reader should compare the method with Exercise 4.72. 4.72 Assume that f satisfies the condition in Exercise 4.71. If x S, letp 0 x, p n1 fp n ,andc n dp n , p n1 for n 0. (a) Prove that c n is a decreasing sequence, and let c limc n . Proof: Consider c n1 c n dp n1 , p n2 dp n , p n1 dfp n , fp n1 dp n , p n1 dp n , p n1 dp n , p n1 0, so c n is a decreasing sequence. And c n has a lower bound 0, by Completeness of R, we know that c n is a convergent sequence, say c limc n .
(b) Assume there is a subsequence p kn which converges to a point q in S. Prove that c dq, fq dfq, ffq. Deduce that q is a fixed point of f and that p n q. Proof: Since lim n p kn q, and lim n c n c, wehavelim n c kn c. So,we consider c lim n c kn lim n dp kn , p kn1 lim n dp kn , fp kn dq, fq and dp kn , p kn1 dp kn1 , p kn ... dfp kn1 , f 2 p kn1 , we have c dq, fq lim n dfp kn1 , f 2 p kn1 dfq, f 2 q. * So, by (*) and hypoethesis dfx, fy dx, y where x y, we know that q fq c 0, in fact, this q is a unique fixed point. . In order to show that p n p, we consider (let m kn) dp m , q dp m , fq dp m1 , q .. dp kn , q So, given 0, there exists a positive integer N such that as n N, wehave dp kn , q . Hence, as m kN, wehave dp m , q . That is, p n p.
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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
Page 89 and 90: fx, y 1 r cos sinr2 cossin if x
Page 91 and 92: then we have lim fx, y L if x 0an
Page 93 and 94: this is because a can be an isolate
Page 95 and 96: irrational. Prove that: (a) ffx x
Page 97 and 98: In addition, since f1 f1 by f0 0,
Page 99 and 100: lim r n0 fx f lim n x n lim n fx
Page 101 and 102: if x Q, andgx 0ifx Q c . Remark:
Page 103 and 104: number f T supfx fy : x, y T is
Page 105 and 106: Proof: Since the hypothesis says th
Page 107 and 108: (f) S 0, 1 0, 1, T 0, 1 0, 1. S
Page 109 and 110: In Exercises 4.29 through 4.33, we
Page 111 and 112: function. Since f0, 1 0, 1 0, 1,
Page 113 and 114: 4.40 If x is a point in a metric sp
Page 115 and 116: from b to A) are disjoint. So, if e
Page 117 and 118: x F k1 F k A B U V which im
Page 119 and 120: Proof: Assume that B C 1 is discon
Page 121 and 122: Proof: By Mean Value Theorem, wehav
Page 123 and 124: x r y r rz r1 x y ry r1 /2. So,
Page 125 and 126: hx gfx if x S. Iff is uniformly c
Page 127 and 128: dx, y , x, y A, wehave dfx, fy
Page 129 and 130: In addition, part (b) comes from in
Page 131 and 132: continuous at a, a. (2) (x y) Sinc
Page 133 and 134: Proof: Let D denote the set of dico
Page 135 and 136: (a) Provet that the formula df, g
Page 137 and 138: so, by induction, we find dp n1 , p
Page 139: Proof: If p and p are fixed points
Page 143 and 144: 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
Page 147 and 148: h k1 h k k jk f j g kj x j0
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
Page 161 and 162: there is a point p such that fp 0.
Page 163 and 164: Remark: If we can make sure that fx
Page 165 and 166: which implies that Hence, g 1 h g
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
Page 181 and 182: ux, y ey e y cosx 2 and vx, y e
Page 183 and 184: ux, y (1) arctany/x, ifx 0, y R
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
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we know that V f is an increasing
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in Section 6.12. Proof: Sinceft : t
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2n h P 2 hx i hx i1 i1 n 2n
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n k1 n ||fb k | |fa k || |fb k
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Supplement on lim sup and lim inf I
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(1) lim n→ inf a n − a n has
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Remark: The exercise is useful in t
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which implies that c ≤ a b since
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If lim sup n→ a n1 a n , then it
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We first prove lim n→ sup n ≤
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0 ≤ d 1 * by Theorem 8.23 (i). N
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Consider which implies that which i
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So, L 1 5 since L ≥ 1. 2 Remark:
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For case (i), since 1 n log nlog lo
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()Assume that ∑ n1 That is, p!
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n S n ∑−1 k1 1 3k − 2 − 1
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(b) Prove that ∑ n1 −1 n−1 /
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then b k sin kx, a k1 − a k 1
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diverges, then ∑ where By (2) and
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Suppose NOT, it means that lim n→
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then since n lim n→ ∑ k1 n S n
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|sin k|r ∑ k So, ∑ |sink|r dive
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n For the part 1 2 0 that x sint
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lim n→ sin ... sin n 1 ... 1 n
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Also, lim q→ fp, q lim q→ p si
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n c n ∑ a k b n−k k0 n ∑ k0
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So, n s n ∑ cos2k − 1x j1 sin
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n u n ∑−1 k a k k1 n ∑−1
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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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