# HW3 Solutions.pdf - Real Analysis HW3 solutions October 4th...

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This preview shows page 1 out of 9 pages. Unformatted text preview: Real Analysis HW3 solutions October 4th, 2019 Exercise 2.2.2 Prove part (ii) of Proposition 2.2.5. Proposition 2.2.5. Let {xn } and {yn } be convergent sequences. (ii) The sequence {zn }, where zn = xn − yn converges, and lim (xn − yn ) = lim zn = lim xn − lim yn n→∞ n→∞ n→∞ n→∞ Proof. Define x = limn→∞ xn and y = limn→∞ yn , and let  > 0 be arbitrary. Since {xn } converges to x, there exists an M1 such that for all n ≥ M1 ,  |xn − x| < , 2 and similarly there is an M2 such that for all n ≥ M2 ,  |yn − y| < , 2 Let M = max{M1 , M2 } , and let n ≥ M , then n ≥ M1 and n ≥ M2 , so using the triangle inequality , |zn − (x − y)| = |(xn − yn ) − (x − y)| = |(xn − x) − (yn − y)| ≤ |xn − x| + |yn − y|   < + 2 2 ≤ 1 Exercise 2.2.3 Prove that if {xn } is a convergent sequence, k ∈ N, then  k k lim xn = lim xn n→∞ n→∞ Proof. Let P (k) be the statement that limn→∞ xkn = (limn→∞ xn )k . We argue by induction: Base case (P (1)): P (1) simply states lim n→∞ x1n =  lim xn 1 n→∞ which is evidently true (both sides are equal to lim xn ). Inductive step (P (k) =⇒ P (k + 1)): Suppose P (k) holds. Then, by Prop. 2.2.5 part (iii), lim xk+1 = lim xkn xn n n→∞ n→∞ = lim xkn lim xn n→∞ n→∞  k = lim xn lim xn n→∞ n→∞  k+1 = lim xn n→∞ Exercise 2.2.9 Suppose {xn } is a sequence and suppose for some x ∈ R, the limit |xn+1 − x| n→∞ |xn − x| L := lim exists and L < 1. Show that {xn } converges to x. (It’s not given in the problem, but to make sense of L we will assume that xn 6= x for all n ∈ N). Proof. Let yn = xn − x. Then, yn 6= 0 for all n and |yn+1 | |xn+1 − x| = lim =L<1 n→∞ yn n→∞ |xn − x| lim so {yn } satisfies the hypotheses of Lemma 2.2.12 . Thus, yn → 0 . However, xn = yn + x. Thus, by limit arithmetic (Prop 2.2.5) and the fact that {x} → x, we have lim xn = lim yn + x = x. n→∞ n→∞ 2 Exercise 2.2.11 Let r > 0. Show that starting with any x1 6= 0, the sequence defined by xn+1 := xn − converges to √ x2n − r 2xn √ r if x1 > 0 and − r if x1 < 0. Proof. √ First, let us suppose that x1 > 0. We claim that the sequence {xn+1 } is bounded below by r . The key is observing that xn+1 = g(xn ) where g(x) = x − x2 −r 2x . Now, √ √ x2 − r 2x √ √ √ √ (x − r)(x + r) = r + (x − r) − 2x   √ √ √ x+ r = r + (x − r) 1 − 2x √ √ √ x− r = r + (x − r) √ 2 2x √ (x − r) = r+ 2x g(x) = x − r+ Base case: We must show that x2 ≥ r− √ r. In this case, we have that √ √ (x − r)2 √ x2 = g(x1 ) = r + ≥ r 2x since x1 > 0 . Inductive step: Suppose xn ≥ √ r. Then, √ √ (x − r)2 √ xn+1 = g(xn ) = r + ≥ r 2x √ √ since xn ≥ r > 0 . This proves {xn } is bounded below by r. Also, {xn+1 } is also monotone decreasing . For any n ≥ 2, xn+1 − xn = − √ √ x2n − r 1 = (xn − r) (xn + r) ≤ 0 2xn 2xn | {z } | {z } |{z} >0 ≤0 >0 3 Since {xn+1 } is monotone decreasing and and bounded below , it must converge to x = inf{xn+1 : n ∈ N} . By proposition 2.1.15, {xn } also converges to the same limit . Then, x2n − r n→∞ n→∞ 2x x2 − r x=x− 2x x2 − r = 0 x2 = r √ √ √ which implies that x = ± r. On √ the other hand, x ≥ r since r is a lower bound of {xn+1 : n ∈ N} , so lim xn = x = r. If only remains to prove the case when x1 < 0. Let yn = −xn . Then, y1 > 0 and for all n, lim xn+1 = lim xn − yn+1 = −xn+1   x2n − r = − xn − 2xn (−xn )2 − r = (−xn ) − 2(−xn ) 2 y −r = yn − n 2yn √ √ so, by the first part of the proof, yn → r . But then, lim xn = − lim yn = − r . Exercise 2.3.8 Let {xn } and {yn } be bounded sequences (from the previous exercise we know that {xn + yn } is bounded). a. Show that (lim sup xn ) + (lim sup yn ) ≥ lim sup(xn + yn ) n→∞ n→∞ n→∞ b. Find an explicit {xn } and {yn } such that (lim sup xn ) + (lim sup yn ) ≥ lim sup(xn + yn ) n→∞ n→∞ n∞ a. Proof. Let An = sup{xk : k ≥ n}, Bn = sup{yk : k ≥ n} Cn = {xk + yk : k ≥ n}, and define an = sup An bn = sup Bn cn = sup Cn 4 then, by definition, lim sup xn = lim an n→∞ n→∞ lim sup yn = lim bn n→∞ n→∞ lim sup xn + yn = lim cn n→∞ n→∞ By Proposition 2.2.5, this means that (lim sup xn ) + (lim sup yn ) = lim an + bn n→∞ n→∞ n→∞ On the other hand, Proposition 2.2.3 says that if an +bn ≥ cn for all n , then limn→∞ an + bn ≥ limn→∞ cn . Hence, we only need to show an + bn ≥ cn for all n . Since cn is the least upper bound of Cn , it is enough to show that an + bn is an upper bound of this set. Consider an element xk + yk ∈ Cn . Then, k ≥ n , so xk ∈ An . Since an is an upper bound of An , we conclude that an ≥ xk . Similar reasoning shows that bn ≥ yk . This shows that an + bn ≥ xk + yk , so an + bn is an upper bound of Cn . Since cn = sup Cn , this means an + bn ≥ cn . b. Let xn = (−1)n , yn = (−1)n+1 . Then, lim sup xn = limn→∞ sup{(−1)k : k ≥ n} = limn→∞ sup{−1, 1} = 1solim sup xn = 1 . Similar reasoning shows that lim sup yn = 1. However, xn + yn = 0 for all n , so lim sup(xn + yn ) lim sup{0 : k ≥ n} = 0. n→∞ Thus lim sup xn + lim sup yn > lim sup(xn + yn ) Exercise 2.3.15 Prove the following stronger version of Lemma 2.2.12, the ratio test. Suppose {xn } is a sequence such that xn 6= 0 for all n. a. Prove that if lim sup n→∞ |xn+1 | <1 |xn | then {xn } converges to 0. b. Prove that if lim sup n→∞ then {xn } is unbounded. 5 |xn+1 | >1 |xn | | a. Proof. Suppose that lim supn→∞ |x|xn+1 < 1. Let L = lim supn→∞ n| and define an = sup{rk : k ≥ N }. |xn+1 | . |xn | Define rn = |xn+1 | , |xn | Let  = 1−L . By the definition of lim sup, L = limn→∞ an , so there exists an M such 2 that for n ≥ M , an − L ≤ |an − L| <  which implies that aM ≤ L + Now, let A = max ( 1−L 1+L = < 1. 2 2 1+L 2 ) M −n |xn | : n ≤ M then, for n ≤ N ,  |xn | ≤ A 1+L 2 . n−M (1) On the other hand, if n > M , then rn−1 = |xn | 1+L < |xn−1 | 2 so |xn | |xn−1 | xn−1 = rn−1 |xn−1 | = rn−1 rn−2 |xn−2 | |xn | = .. . = rn−1 rn−2 · · · rM |xM |      1+L 1+L 1+L ≤ ··· |xM | 2 2 2 | {z } n−M times  ≤A 1+L 2 n−M so, again,  |xn | ≤ A 1+L 2 By limit arithmetic and Proposition 2.2.11, A tells us that xn → 0. n−M  1+L n−M 2 → 0. Thus, Proposition 2.2.10 b. Proof. Recall that to prove {xn } is unbounded, we must show that for every B > 0, | there exists an n so |xn | > B . Let B be arbitrary , and let L = lim infn→∞ |x|xn+1 . n| 6 | As in the previous problem, let rn = |x|xn+1 , and define bn = inf{rk : k ≥ N }. By the n| L−1 definition of lim inf, bn → L . Let  = 2 (notice  > 0). Then, there exists an M such that for all n ≥ M , L − bn ≥ |bn − L| <  which, after rearranging, gives L−1 1+L = ≥ bn . 2 2 n > 1 . By Proposition 2.2.11, 1+L is unbounded, so there exists an n 2 L− Notice that such that 1+L 2  1+L 2 n B > |xM |  1+L 2 M which rearranges to give  n−M  n  M 1+L 1+L B 1+L B< |xM | ≥ 2 2 |xM | 2 Thus, |xn | = rn−1 |xn−1 | = rn−1 rn−2 |xn−2 | .. . = rn−1 rn−2 · · · rM |xM |  n−M 1+L ≥ |xM | 2 >B so |xn | > B, as desired. Exercise 2.3.18 Suppose {xn } is a sequence such that both lim inf xn and lim sup xn are finite. Prove that {xn } is bounded. To prove B is bounded, we must show that there exists a B such that for all n, |xn | ≤ B. Proof. Let an = sup{xk : k ≥ n}, bn = inf{xk : x ≥ n}. Let U = lim supn→∞ xn and L = lim infn→∞ xn . Then, by definition, an → U and bn → L. Since an → U , there exists an MU such that for all n ≥ MU , |an − U | < 1 . Pick MU , and notice that for n ≥ MU , an is an upper bound of {xk : k ≥ n}, so in particular xn ≤ an ≤ |an − U | + U < U + 1. Similarly, bn → L, so there is an ML such that for all n ≥ ML , xn > L − 1 . Now, let M = max{MU , ML } . Define B = max{|U + 1|, |L − 1|, |xn | : n < M }. Now, let n ∈ N. There are two cases: either n < M for n ≥ M . 7 If n < M , then |xn | ∈ {|U + 1|, |L − 1|, |xn | : n < M } and so |xn | ≤ B. On the other hand, if n ≥ M , then n ≥ MU and n ≥ ML , so L − 1 < xn < U + 1 it follows that − max{|L − 1|, |U + 1|} < xn < max{|L − 1|, |U + 1|} so |xn | < max{|L − 1|, |U + 1|} and so |xn | ≤ B. In either case, |xn | ≤ B, so {xn } is bounded. Exercise 2.4.2 Let {xn } be a sequence such that there exists a 0 < C < 1 such that |xn+1 − xn | ≤ C|xn − xn−1 | Prove that {xn } is Cauchy. C Proof. Let  > 0 be arbitrary. Since 1−C |x2 − x1 | → 0, there exists an M > 0 such that for n C n ≥ M , 1−C |x2 − x1 | <  . Choose this M , and let n, m ≥ M . If n = m, then |xn −xm | = 0 <  and we are done, so assume n 6= m. Then, by interchanging the roles of n and m, we may assume without loss of generality that n > m . Then, n xn − xm = xn − xn−1 + xn−1 − xn−2 + · · · + xm+1 − xm so, by the iterated Triangle inequality (Corollary 1.3.4), |xn − xm | ≤ |xn − xn−1 | + |xn−1 − xn−2 | + · · · + |xm+1 − xm | On the other hand, by iterating the bound given in the hypothesis, we find that |xn − xn−1 | ≤ C|xn−1 − xn−2 | ≤ C 2 |xn−2 − xn−3 | .. . ≤ C n−1 |x2 − x1 | Thus,  |xn − xm | ≤ C n−1 + C n−2 + · · · + C m |x2 − x1 | Cn − Cm ≤ |x2 − x1 | 1−C Cn ≤ |x2 − x1 | 1−C < By the definition of M . 8 Exercise 2.4.6 Suppose that |xn − xk | ≤ n k2 for all n and k. Show that {xn } is Cauchy. Proof. Let  > 0 be arbitrary. By the Archimedean property, there is an M such that M1 < ; choose this M . Let n, m ≥ M . Since n and m play identical roles, we may assume without loss of generality that m ≥ n. Then, n m2 1 ≤ m 1 ≤ M < |xn − xm | ≤ Exercise 2.4.8 True or false, prove or find a counterexample: If {xn } is a Cauchy sequence then there exists an M such that for all n ≥ M we have |xn+1 − xn | ≤ |xn − xn−1 |. This is false. Let xn be defined by xn = ( 0 1 n n = 3k − 2, 3k − 1 n = 3k Then, |xn | ≤ n1 for all n, so by Proposition 2.2.10, xn → 0 . By Theorem 2.4.5, this also shows that xn is Cauchy. On the other hand, for any M , we have that 3M − 1 ≥ M , and |x3M − x3M −1 | = 1 > 0 = |x3M −1 − x3M −2 |. 3M 9 ...
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