The Real And Complex Number Systems

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Note: In remark 2, we CANNOT use the relation fx gx since the difference "" are not necessarily defined on the metric space T, d 2 . 4.23 Given a function f : R R, define two sets A and B in R 2 as follows: A x, y : y fx, B x, y : y fx. and Prove that f is continuous on R if, and only if, both A and B are open subsets of R 2 . Proof: () Suppose that f is continuous on R. Leta, b A, then fa b. Sincef is continuous at a, thengiven fab 0, there exists a 0 such that as 2 |x a| , wehave fa b fa fx fa . * 2 Consider x, y Ba, b; , then |x a| 2 |y b| 2 2 which implies that 1. |x a| fx fa b by (*) and 2 2. |y b| y b b fa b . 2 Hence, we have fx y. Thatis,Ba, b; A. So,A is open since every point of A is interior. Similarly for B. () Suppose that A and B are open in R 2 . Trivially, a, fa /2 : p 1 A, and a, fa /2 : p 2 B. SinceA and B are open in R 2 , there exists a /2 0 such that Bp 1 , A and Bp 2 , B. Hence, if x, y Bp 1 , , then x a 2 y fa /2 2 2 and y fx. So, it implies that |x a| , |y fa /2| , andy fx. Hence, as |x a| , wehave y fa /2 fa /2 y fx fa y fx fa fx. ** And if x, y Bp 2 , , then x a 2 y fa /2 2 2 and y fx. So, it implies that |x a| , |y fa /2| , andy fx. Hence, as |x a| , wehave fx y fa /2 fa . *** So, given 0, there exists a 0 such that as |x a| , we have by (**) and (***) fa fx fa . That is, f is continuous at a. Sincea is arbitrary, we know that f is continuous on R. 4.24 Let f be defined and bounded on a compact interval S in R. IfT S, the
number f T supfx fy : x, y T is called the oscillation (or span) of f on T. Ifx S, the oscillation of f at x is defined to be the number f x lim h0 fBx; h S. Prove that this limit always exists and that f x 0if,andonlyif,f is continuous at x. Proof: 1. Note that since f is bounded, say |fx| M for all x, wehave |fx fy| 2M for all x, y S. So, f T, the oscillation of f on any subset T of S, exists. In addition, we define gh f Bx; h S. Note that if T 1 T 2 S, wehave f T 1 f T 2 . Hence, the oscillation of f at x, f x lim h0 gh g0 since g is an increasing function. That is, the limit of f Bx; h S always exists as h 0 . 2. Suppose that f x 0, then given 0, there exists a 0 such that as h 0, , wehave |gh| gh f Bx; h S /2. That is, as h 0, , wehave supft fs : t, s Bx; h S /2 which implies that /2 ft fx /2 as t x , x S. So, as t x , x S. wehave |ft fx| . That is, f is continuous at x. Suppose that f is continous at x, thengiven 0, there exists a 0 such that as t x , x S, wehave |ft fx| /3. So, as t, s x , x S, wehave |ft fs| |ft fx| |fx fs| /3 /3 2/3 which implies that supt fs : t, s x , x S 2/3 . So, as h 0, , wehave f Bx; h S supt fs : t, s x , x S . Hence, the oscillation of f at x, f x 0. Remark: 1. The compactness of S is not used here, we will see the advantage of the oscillation of f in text book, Theorem 7.48, in page 171. (On Lebesgue’s Criterion for Riemann-Integrability.) 2. One of advantage of the oscillation of f is to show the statement: Let f be defined on a, b Prove that a bounded f does NOT have the properties: f is continuous on Q a, b, and discontinuous on R Q a, b. Proof: Since f x 0if,andonlyif,f is continuous at x, we know that f r 0for r R Q a, b. DefineJ 1/n r : f r 1/n, then by hypothesis, we know that n1 J 1/n R Q a, b. It is easy to show that J 1/n is closed in a, b. Hence, intclJ 1/n intJ 1/n for all n N. It means that J 1/n is nowhere dense for all n N. Hence,
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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,
Page 51 and 52: Proof: Assume S˜R and let f be a o
Page 53 and 54: 2.22 Let S denote the collection of
Page 55 and 56: Charpter 3 Elements of Point set To
Page 57 and 58: And thus by the remark in (e), it i
Page 59 and 60: 3.7 Prove that a nonempty, bounded
Page 61 and 62: Bx, r x S Bx, r x T . So, at
Page 63 and 64: Claim that Bp, r S as follows. Let
Page 65 and 66: Covering theorems in R n 3.17 If S
Page 67 and 68: uncountable. Hence, by exercise 3.2
Page 69 and 70: Then we have (a) (b) (c) (d) x y
Page 71 and 72: (b) Give an example of a metric spa
Page 73 and 74: Hence from (1)-(4), we know that i
Page 75 and 76: 3.44 intM A M A . Proof: Let B
Page 77 and 78: That is, intB which is absurb. He
Page 79 and 80: Limits And Continuity Limits of seq
Page 81 and 82: x n1 x n 1 1 x n x n 1 2 . Rema
Page 83 and 84: |a m a n| |a m a m1 a m1 a m2
Page 85 and 86: and use the fact if a n converges
Page 87 and 88: 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: if x Q, andgx 0ifx Q c . Remark:
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 and 140: Proof: If p and p are fixed points
Page 141 and 142: (b) Assume there is a subsequence p
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
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g fx gfx. Assume that there is a
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Remark: For x 0, we can show that
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Proof: Look at the Generalized Mean
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Proof: By Generalized Mean Value Th
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there is a point p such that fp 0.
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Remark: If we can make sure that fx
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which implies that Hence, g 1 h g
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fx fy x y f x /2 *’ Combi
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lim xa fx gx L. Remark: 1. The pr
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every g k is never zero in c , c
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f x 1 ...x n n f x 1 ...x n1 n x
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f x 1 ...x n n fx 1 ...fx n n . C
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lx fc f cx c fx. (Exercise 2)
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So, we have Partial derivatives fb
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ux, y ey e y cosx 2 and vx, y e
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ux, y (1) arctany/x, ifx 0, y R
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Functions of Bounded Variation and
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For the part fx x fy 1 |fx fy| 1
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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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