The Real And Complex Number Systems

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|x r y r | 2 n r 1 n r r 1 n |2r 1| , asn since r 0, which is absurb with (*). Hence, we know that x r is not uniformly continuous on 0, 1, for r 0. Ps: The reader should try to realize why x r is not uniformly continuous on 0, 1, for r 0. The ruin of non-uniform continuity comes from that x is small enough. 4.54 Assume f : S T is uniformly continuous on S, whereS and T are metric spaces. If x n is any Cauchy sequence in S, prove that fx n is a Cauchy sequence in T. (Compare with Exercise 4.33.) Proof: Given 0, we want to find a positive integer N such that as n, m N, we have dfx n , fx m . For the same , sincef is uniformly continuous on S, then there is a 0 such that as dx, y , x, y S, wehave dfx, fy . For this , sincex n is a Cauchy sequence in S, then there is a positive integer N such that as n, m N, wehave dx n , x m . Hence, given 0, there is a postive integer N such that as n, m N, wehave dfx n , fx m . That is, fx n is a Cauchy sequence in T. Remark: The reader should compare with Exercise 4.33 and Exercise 4.55. 4.55 Let f : S T be a function from a metric space S to another metric space T. Assume that f is uniformly continuous on a subset A of S and let T is complete. Prove that there is a unique extension of f to clA which is uniformly continuous on clA. Proof: Since clA A A , it suffices to consider the case x A A. Since x A A, then there is a sequence x n A with x n x. Note that this sequence is a Cauchy sequence, so we have by Exercise 4.54, fx n is a Cauchy sequence in T since f is uniformly on A. In addition, since T is complete, we know that fx n is a convergent sequence, say its limit L. Note that if there is another sequence x n A with x n x, then fx n is also a convergent sequence, say its limit L . Note that x n x n is still a Cauchy sequence. So, we have dL, L dL, fx n dfx n , fx n dfx n , L 0asn . So, L L . That is, it is well-defined for g : clA T by the following gx fx if x A, lim n fx n if x A A, wherex n x. So, the function g is a extension of f to clA. Claim that this g is uniformly continuous on clA. That is, given 0, we want to find a 0 such that as dx, y , x, y clA, wehave dgx, gy . Since f is uniformly continuous on A, for /3, there is a 0 such that as
dx, y , x, y A, wehave dfx, fy . Let x, y clA, and thus we have x n A with x n x, andy n A with y n y. Choose /3, then we have dx n , x /3 and dy n , y /3 as n N 1 So, as dx, y /3, wehave(n N 1 ) dx n , y n dx n , x dx, y dy, y n /3 /3 /3 . Hence, we have as dx, y , (n N 1 ) dgx, gy dgx, fx n dfx n , fy n dfy n , fy * dgx, fx n dfy n , gy And since lim n fx n gx, and lim n fy n gy, we can choose N N 1 such that dgx, fx n and dfy n , gy . So, as dx, y , (n N) wehave dgx, gy 3 by (*). That is, g is uniformly on clA. It remains to show that g is a unique extension of f to clA which is uniformly continuous on clA. If there is another extension h of f to clA which is uniformly continuous on clA, thengivenx A A, we have, by continuity, (Say x n x) hx h lim n x n lim n hx n lim n fx n lim n gx n g lim n x n gx which implies that hx gx for all x A A. Hence, we have hx gx for all x clA. Thatis,g is a unique extension of f to clA which is uniformly continuous on clA. Remark: 1. We do not require that A is bounded, in fact, A is any non-empty set in a metric space. 2. The exercise is a criterion for us to check that a given function is NOT uniformly continuous. For example, let f : 0, 1 R by fx 1/x. Sincef0 does not exist, we know that f is not uniformly continuous. The reader should feel that a uniformly continuous is sometimes regarded as a smooth function. So, it is not surprising for us to know the exercise. Similarly to check fx x 2 , x R, and so on. 3. Here is an exercise to make us know that a uniformly continuous is a smooth function. Let f : R R be uniformly continuous, then there exist , 0 such that |fx| |x| . Proof: Since f is uniformly continuous on R, given 1, there is a 0 such that as |x y| , wehave |fx fy| 1. * Given any x R, then there is the positive integer N such that N |x| N 1. If x 0, we consider y 0 0, y 1 /2, y 2 ,...,y 2N1 N 2 , y 2N x. Then we have
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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
Page 75 and 76: 3.44 intM A M A . Proof: Let B
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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 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
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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: hx gfx if x S. Iff is uniformly c
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
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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 169 and 170: lim xa fx gx L. Remark: 1. The pr
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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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