The Real And Complex Number Systems

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fx x x i x i1 2 then f is a continuous function and is not bounded variation on 0, 1 since k1 , 1 2M diverges. In order to show that f satisfies uniform Lipschitz condition of order , we consider three cases. (1) If x, y x i , x i1 ,andx, y x i , x ix i1 2 or x, y x ix i1 2 , x i1 , then |fx fy| |x y | |x y| . (2) If x, y x i , x i1 ,andx x i , x ix i1 or y x ix i1 , x 2 2 i1 , then there is a z x i , x ix i1 such that fy fz. So, 2 |fx fy| |fx fz| |x z | |x z| |x y| . (3) If x x i , x i1 and y x j , x j1 ,wherei j. If x x i , x ix i1 , then there is a z x 2 i , x ix i1 such that fy fz. So, 2 |fx fy| |fx fz| |x z | |x z| |x y| . Similarly for x x ix i1 , x 2 i1 . Remark: Here is another example. Since it will use Fourier <strong>The</strong>ory, we do not give a proof. We just write it down as a reference. ft k1 cos3k t 3 k . (c) Give an example of a function f which is of bounded variation on a, b but which satisfies no uniform Lipschitz condition on a, b. Proof: Since a function satisfies uniform Lipschitz condition of order 0, it must be continuous. So, we consider x if x a, b fx b 1ifx b. Trivially, f is not continuous but increasing. So, the function is desired. Remark: Here is a good problem, we write it as follows. If f satisfies |fx fy| K|x y| 1/2 for x 0, 1, wheref0 0. define fx x 1/3 if x 0, 1 gx 0ifx 0. <strong>The</strong>n g satisfies uniform Lipschitz condition of order 1/6. Proof: Note that if one of x, andy is zero, the result is trivial. So, we may consider 0 y x 1 as follows. Consider |gx gy| fx x 1/3 fx x 1/3 fx x 1/3 fy y 1/3 fy x 1/3 fy x 1/3 fy fy x 1/3 x 1/3 fy y 1/3 1 k fy y 1/3 . *
For the part fx x fy 1 |fx fy| 1/3 x 1/3 x1/3 K x |x 1/3 y|1/2 by hypothesis K|x y| 1/2 |x y| 1/3 since x x y 0 K|x y| 1/6 . A fy For another part fy , we consider two cases. x 1/3 y 1/3 (1) x 2y which implies that x x y y 0, fy x 1/3 fy y 1/3 |fy| |fy| x 1/3 y 1/3 xy 1/3 x y 1/3 xy 1/3 since |x 1/3 y 1/3 | |x y| 1/3 for all x, y 0 |fy| x 1/3 xy 1/3 since x y 1/3 x 1/3 |fy| 1 y 1/3 K |y|1/2 |y| 1/3 K|y| 1/6 by hypothesis K|x y| 1/6 since y x y. B (2) x 2y which implies that x y x y 0, fy x 1/3 fy y 1/3 |fy| |fy| |fy| x 1/3 y 1/3 xy 1/3 x y 1/3 xy 1/3 since |x 1/3 y 1/3 | |x y| 1/3 for all x, y 0 x y 1/3 y 2/3 K|y| 1/2 x y 1/3 y 2/3 K|y| 1/6 |x y| 1/3 since x y by hypothesis K|x y| 1/6 |x y| 1/3 since y x y K|x y| 1/6 . C So, by (A)-(C), (*) tells that g satisfies uniform Lipschitz condition of order 1/6. Note: Hereisageneralresult.Let0 2. Iff satisfies |fx fy| K|x y| for x 0, 1, wheref0 0. define
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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
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
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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
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
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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
Page 181 and 182: ux, y ey e y cosx 2 and vx, y e
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Page 185: Functions of Bounded Variation and
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 197 and 198: n k1 n ||fb k | |fa k || |fb k
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
Page 205 and 206: which implies that c ≤ a b since
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 /
Page 225 and 226: then b k sin kx, a k1 − a k 1
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Page 229 and 230: Suppose NOT, it means that lim n→
Page 231 and 232: then since n lim n→ ∑ k1 n S n
Page 233 and 234: |sin k|r ∑ k So, ∑ |sink|r dive
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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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