Analysis I Solutions to Practice Final Exam, Exams of Mathematical Analysis

Download Analysis I Solutions to Practice Final Exam and more Exams Mathematical Analysis in PDF only on Docsity! | � < + + = �, � � � 18.100B/C: Fall 2010 Solutions to Practice Final Exam 1. Suppose for sake of contradiction that x > 0. Then 1 x > 0 because the product of two positive 2 · quantities is positive. Thus x 2 + 0 < x 2 + x 2 (because y < z implies x + y < x + z for all x), i.e., x x x< x. Also, since � := > 0 we have by assumption that x � . However, for a strict order 2 2 2 at most one of x < x 2 and x 2 < x can be true. Hence we obtain a contradiction to the assumption x > 0. Thus x >→ 0. Since x � 0, this implies x = 0, as desired. 2.(a) We use 2xy � x2 + y2 (which follows from (x − y)2 � 0) and an � 0 to estimate 0 < � anan+1 � 1 ( � an 2 + � an+1 2) = 1 an + 1 an+1.2 2 2 ∑ ∑ Next, the partial sums of n � =1 an+1 are the same ones (shifted by one – see (b)) as for n � =1 an, and ∑ so by assumption both series converge. Hence by linearity for limits, the series � 1 + 1 ∑ n=1 2 an 2 an+1 also converges. Now convergence of � �anan+1 follows from the comparison criterion. n=1 ∑ (b) Since an+1 � an we obtain 0 � an+1 � � anan+1, so � converges by the comparison ∑k ∑k−1 n=1 an+1 ∑ktest. But now limk�� n=1 an+1 exists iff limk�� n=0 an+1 = limk�� exists; proving n=1 an convergence of the latter. 3.(a) Both f(x) = 4x(1 − x) and f(x) = 1 − |2x − 1| work nicely. (b) No function: continuous functions take connected sets to connected sets. (c) Define { f(x) = 0, x � 1, 1, x � 2. This function is continuous on [0, 1] ∈ [2, 3] and f([0, 1] ∈ [2, 3]) = {0, 1}. (d) No function: suppose such a function f exists. There exists x1 for which f(x1) = 1 and x2 for which f(x2) = 2, so by the Intermediate Value Theorem there is x between x1 and x2 for which f(x) = � 2, a contradiction. (Or, use connectedness again.) (e) No function: continuous functions take compact sets to compact sets. 4. Given � > 0, by uniform convergence of (fn), we can choose some N ⊂ N such that n � N implies |fn(x) − f(x) < � for all x ⊂ E. By uniform continuity of fN , we can choose some � such 3 that d(x, y) < � implies |fN (x) − fN (y)| < . Then for any x, y ⊂ E such that d(x, y) < � we have 3 |f(x) − f(y)| = |f(x) − fN (x) + fN (x) − fN (y) + fN (y) − f(y)| � |f(x) − fN (x)| + |fN (x) − fN (y)| + |fN (y) − f(y)| 3 3 3 so f is uniformly continuous. 5. (see Melrose Test 2) 6. (see Melrose Test 2) 1 ∣∣∣∣ ∣∣∣∣ ∣∣ ∣∣ ∣∣ ∣∣ � 7.(a) For all � > 0, there exists a (countable) collection {B(xi, ri)} of open balls such that N ≤ B(xi, ri) and i ri < �.i (b) We have {x f(x) = f(x)} = ∞ has measure 0, so f ⊆ f . The relation is symmetric since {x | g(x) = f(x)} | = {x | → g(x) = f(x)}. To check is transitivity assume f ⊆ g and g ⊆ h. Observe → → that if f(x) = h(x) then we must have either f(x) = g(x) or g(x) = h(x) (or both), so → → → {x | f(x) → | f(x) = g(x)} ∈ {x | f(x) = → g(x)}.= h(x)} � {x → So we must show that unions and subsets of measure-0 sets have measure 0. For subsets, just take a covering of the superset of measure 0 to cover its subset. For the union, take the union of two coverings of measure less than �/2 to cover the union with sets of total measure less than �. (c) Since f and g are both integrable, f − g is integrable as well, and we are asked to show that f − g = 0 given that f − g = 0 almost everywhere. Since f − g is integrable, the integral is equal to the infimum over all upper Riemann sums. Since f − g is zero almost everywhere, every interval contains a point at which f − g = 0, so the upper Riemann sum for any fixed partition is a sum of nonnegative numbers and thus nonnegative. The infimum of a set of nonnegative quantities ∫ 1must itself be nonnegative, so f − g � 0. However, we may apply identical reasoning to get that ∫ 1 0 g − f � 0. Since these two quantities are negatives of each other, they both must equal 0, as needed. 8.(a) Fix � > 0. For each fi, choose a �i such that d(x, y) < �i implies |fi(x) − fi(y)| < � for all x, y. Then let � = min{�i} > 0 and we have that for any fi ⊂ F and any x, y in the common domain that if d(x, y) < � then |fi(x) − fi(y) < �, so F is equicontinuous. (b) Let � = �/n. If |x − y| < � then | ∑∪ ∫ 1 0 0 |fn(x) − fn(y)| = x x + 1 n − y y + 1 n = |x−y| n |x−y| n 1 = n|x − y| < �, x + 1 y + 1 n 2·n n so fn is uniformly continuous for all n. (c) We have fn(0) = 0 for all n and fn(x) ≥ 1 as n ≥ ∼ for any fixed x ⊂ (0, 1], so (fn) converges pointwise to the function f(x) = { 0, x = 0 1, x ⊂ (0, 1]. However, fn( n 1 ) = 12 for all n, so for all n there exists x such that d(fn(x), f(x)) > 1 3 . Thus no subsequence of the (fn) can converge uniformly. (Alternatively, invoke problem 4 here: if convergence were uniform, the limit function would be uniformly continuous, when in fact it’s not even continuous.) In addition, we have 0 � fn(x) � 1 for all n ⊂ N and all x ⊂ [0, 1], so (fn) is uniformly bounded. By Arzelà-Ascoli, any equicontinuous pointwise bounded sequence of continuous functions has a uniformly convergent subsequence, so it follows that our sequence of functions is not equicontinuous. 9. see Melrose Test 1 .. hence no solution here 2