Final Exam Solutions - Real Analysis | MA 452, Exams of Mathematical Methods for Numerical Analysis and Optimization

Download Final Exam Solutions - Real Analysis | MA 452 and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! MA 452/502: Introduction to Real Analysis Solutions to Final Exam Name: Score: /100 December 03, 2007 There are 7 problems in this exam for a maximum of 100 points. Please show your work and present your solutions in a well organized way. No work, no credit. 1. (25 points) Disprove the following statements by constructing counterexamples. Justify your answer please. (1). If a nonempty bounded set S ⊆ R contains its maximum and minimum, then S is compact. Counterexample: Let S = [0, 1) ∪ (2, 5]. S contains its min and max but it is not compact since it is not closed. (2). Every unbounded sequence has no convergent subsequences. Counterexample: Let sn = n + (−1)nn has a convergent subsequence s2k−1 = 0. (3). If limx→c f2(x) = L > 0, then limx→c f(x) exists and is finite. Counterexample: Let f(x) = 1 if x ∈ Q and f(x) = −1 if x ∈ Qc. Then f2(x) has limit 1 at every c ∈ R but f(x) doesn’t have a limit at any point c ∈ R. (4). Let f : I → R and g : I → R. If f and g are uniformly continuous on I, then the product function fg is uniformly continuous on I. Counterexample: Let f : R → R and g : R → R defined by f(x) = g(x) = x. By Example 23.2, they are uniformly continuous on R, but fg(x) = x2 is not uniformly continuous on R by Example 23.4. (5). Let f(x) and g(x) be integrable on [a, b], and let f(x) ≤ h(x) ≤ g(x) for all x ∈ [a, b], then h(x) is integrable on [a, b]. Counterexample: Let f(x) = 0, g(x) = 1 on [0, 1], and h(x) be the Dirichlet function restricted on [0,1]. Clearly, f and g are integrable on [0, 1] and f(x) ≤ h(x) ≤ g(x) for all x ∈ [0, 1], but h is not integrable on [0, 1]. 1 2. (9 points) Let S ⊆ R and s ∈ S′, prove that either s ∈ intS or s ∈ bdS. Proof. Suppose s /∈ intS, which implies that ∀ε > 0, N(s, ε) ∩ Sc 6= ∅. Since s ∈ S′, we have N∗(s, ε) ∩ S 6= ∅, which implies N(s, ε) ∩ S 6= ∅. Therefore, s ∈ bdS. 3. (16 points) Prove each of the following statements by using only the definition. (1). Let sn = √ n2 + 2− n, prove that ( sn ) converges to 0. Proof. Notice that | √ n2 + 2− n− 0| = 2√ n2 + 2 + n ≤ 1 n , ∀ε > 0, there exists N = ε−1 ∈ R such that for all n > N, we have | √ n2 + 2−n−0| < ε. This proves the conclusion. (2). Prove f(x) = |x− 2|+ x2 is continuous at 2. Proof. Notice that if |x− 2| < 1, then 1 < x < 3 and 3 < x + 2 < 5, which implies that for |x− 2| < 1, |f(x)− f(2)| = ||x− 2|+ (x− 2)(x + 2)| ≤ |x− 2|+ |x− 2||x + 2| ≤ 6|x− 2|. Since ∀ε > 0, there exists δ = min{1, ε/6} such that |f(x)−f(2)| < ε whenever |x−2| < δ, we have f(x) is continuous at 2. 4. (14 points) Let s1 = √ 2, and sn+1 = √ 2 + sn for n ≥ 1. Prove that ( sn ) is Cauchy, and find limn→∞ sn. Solution. We can follow the solution of Example 18.4 to prove that ( sn ) is convergent, and therefore it is Cauchy by Lemma 18.10. lim n→∞ sn = 2. 5. (10 points) Let f : [a, b] → [a, b] be a continuous function, prove that f(x) has a fixed point. That is, prove that ∃ c ∈ [a, b] such that f(c) = c. Proof. Let F : [a, b] → R defined by F (x) = x − f(x). Clearly, F (x) is continuous on [a, b], and F (a) = a − f(a) ≤ 0, F (b) = b − f(b) ≥ 0 (since f ( [a, b] ) ⊆ [a, b]). Therefore, by applying the Intermediate Value Theorem (Theorem 22.6) to F (x), ∃ c ∈ [a, b] such that F (c) = c− f(c) = 0, which proves that f(x) has a fixed point. 2