Final Exam Solution | Fundamental Concepts of Analysis | MATH 425A, Exams of Mathematics

Download Final Exam Solution | Fundamental Concepts of Analysis | MATH 425A and more Exams Mathematics in PDF only on Docsity! FINAL EXAM SOLUTIONS, MATH 425A WEDNESDAY, DECEMBER 10, 1997 Problem 1. Suppose f : R→ R satisfies |f(x)− f(y)| ≤ (x− y)2 for all x, y ∈ R. Prove: f is a constant function. Proof. I hope you reviewed the results of the second midterm ;-) The condition implies that∣∣∣∣f(x+ h)− f(x)h ∣∣∣∣ ≤ |h| for all x ∈ R and h 6= 0; thus, given ε > 0, we can take δ = ε to conclude that if 0 < |h| < δ, then∣∣∣∣f(x+ h)− f(x)h − 0 ∣∣∣∣ < ε; that is, lim h→0 f(x+ h)− f(x) h = 0. Thus f ′ ≡ 0, and by a consequence of the Mean Value Theorem, this implies f is a constant. Problem 2. Suppose f : [a, b] → R is differentiable, and the derivative f ′ is bounded on [a, b] and Riemann-integrable (with respect to x) on [a, b]. Prove: ∫ b a f ′(x) dx = f(b)− f(a). Hint: for a partition P = {a = x0 < x1 < · · · < xn = b}, apply the mean-value theorem to f(xi)− f(xi−1); use this to show that L(P, f ′, x) ≤ f(b)− f(a) ≤ U(P, f ′, x).(1) Proof. We have f(b)− f(a) = n∑ i=1 (f(xi)− f(xi−1)) , since the sum telescopes. By the mean value theorem, there exists ξi ∈ (xi−1, xi) such that f(xi)−f(xi−1) = f ′(ξi)∆xi; letting Mi denote the sup of f ′ on [xi−1, xi] (and mi denote the inf), we therefore have mi∆xi ≤ f ′(ξi)∆xi = f(xi)− f(xi−1) ≤Mi∆xi, so that on adding, n∑ i=1 mi∆xi ≤ f(b)− f(a) ≤ n∑ i=1 Mi∆xi, i.e., we have proved the inequality (1). (So far we’ve only used the fact that f ′ is bounded.) Since f ′ is integrable, we also have L(P, f ′, x) ≤ ∫ b a f ′(x) dx ≤ U(P, f ′, x),(2) so putting inequality (1) together with inequality (2) we get∣∣∣∣f(b)− f(a)− ∫ b a f ′(x) dx ∣∣∣∣ ≤ U(P, f ′, x)− L(P, f ′, x). But since f ′ is given to be integrable, the right-hand side can be made as small as we wish; which means the left-hand side is zero, as was to be proved. Problem 3. Let f : [0, 1] → [0, 1] be a function. The graph of f is the set G(f) = {(x, f(x)) : x ∈ [0, 1]}. Prove: f is continuous if and only if the graph G(f) is compact. Proof. Suppose f is continuous. Then the function h : [0, 1]→ [0, 1]× [0, 1] given by h(x) = (x, f(x)) is also continuous (check: if xn → x, then (xn, f(xn))→ (x, f(x)) since the continuity of f implies f(xn)→ f(x)). And the graph G(f) is just the range of the function h. Since [0, 1] is compact and h is continuous, the range G(f) of h is compact. Conversely, suppose G(f) is compact; first I’ll give the clever proof. Consider the function π1 : R2 → R given by π(x, y) = x (“π” for “projection”). Clearly this is continuous. Let h be π1 restricted to G(f). Then h is also continuous, and is defined by h(x, f(x)) = x. Obviously h is also 1-1. (How could (x1, f(x1)) = (x2, f(x2))? Only if x1 = x2!) Since G(f) is compact, and h is a 1-1 continuous function 1 2 MATH 425A FINAL SOLUTIONS from G(f) onto [0, 1], by a theorem proved in class, h−1 : [0, 1] → G(f) is continuous. The second projection π2 : R2 → R, defined by π2(x, y) = y, is also continuous, and f = π2 ◦h−1 is the composition of two continuous functions, and is therefore continuous. If you didn’t have this idea—don’t worry, I’ve never met anybody who had this idea, who hadn’t seen the trick before—here’s another way. We want to prove f is continuous. So let {xn} be a sequence in [0, 1] which converges to a point x ∈ [0, 1]. Suppose f(xn) does not converge to f(x). Then there must exist ε > 0 such that |f(xn)− f(x)| ≥ ε for infinitely many n. Thus we can extract a subsequence xni such that |f(xni)− f(x)| ≥ ε.(3) Now {(xni , f(xni))} is a sequence in the supposedly compact graph G(f), and hence has a convergent subsequence; without loss of gen- erality we may suppose that is {(xni , f(xni))} itself (to avoid writing subscripts-of-subscripts-of-subscripts). This means that it converges to a point in G(f), which is therefore of the form (x, f(x)), i.e. xni → x, f(xni)→ f(x) But since xni → x, we must have x = x; and thus f(xni) → f(x), contradicting (3). Notice that a function can be discontinuous, yet its graph be closed. Example: f(x) = 1/x for 0 < x ≤ 1, but f(0) = 0. Problem 4. Let N denote the set of natural numbers, N = {1, 2, 3, . . . }. Define a function f on N× N by f(m,n) = 2m−1(2n− 1). Prove: f is a bijection between N× N and N. Proof. The problem is to show that this particular function f is a bi- jection, not that there exists a bijection. First, if m ≥ 1 and n ≥ 1, then f(m,n) is a positive integer; so f maps N× N into N. Second, f is 1-1, because if 2m−1(2n − 1) = 2m′−1(2n′ − 1) then the powers of 2 must be the same (m−1 = m′−1, hence m = m′), leaving the odd parts the same (2n − 1 = 2n′ − 1, hence n = n′). Why are the powers-of-2 the same? Well, if one were larger than the other—say m′ > m—we could divide both sides of the equality by 2m, leaving 2m ′−m(2m′ − 1) = 2n− 1, i.e. an even integer is equal to an odd integer. Don’t get this backward! You don’t want to show that m = m′ and n = n′ implies 2m−1(2n − 1) = 2m′−1(2n′ − 1); that’s trivial, and is only the statement that f is a function. 1-1 requires the reverse of this implication. Finally, we need to show that f is surjective. To do this we need to show that every positive integer can be written in the form 2m−1(2n− 1) for m,n ∈ N. This is a consequence of unique factorization, or even more easily, just take the given integer and keep dividing by 2 until the result isn’t divisible anymore; since dividing by 2 reduces the magnitude of the integer, this can only be done a finite number of times (perhaps 0 times); and you’re left with an odd integer, which is of the form 2n− 1 for some positive integer n. This is an arithmetic proof that N× N is equipotent with N. Problem 5. (This is a true story.) A student in Math 126 was asked whether the series ∑ n( √ n+ 1 − √n) converges. He replied that it does, and that ∞∑ n=0 ( √ n+ 1− √ n) = ( √ 1− √ 0) + ( √ 2− √ 1) + ( √ 3− √ 2) + . . . = 0, despite the fact that all the terms are positive, because the sum tele- scopes! The same student then stated that the series∑ n (−1)n( √ n+ 1− √ n) diverges (despite having been told, in an earlier problem, that √ n+ 1−√ n decreases to 0), because ∞∑ n=0 (−1)n( √ n+ 1− √ n) = ( √ 1− √ 0)− ( √ 2− √ 1) + ( √ 3− √ 2)− . . . = 2 √ 1− 2 √ 2 + 2 √ 3− 2 √ 4 + 2 √ 5− . . . , and the n-th term doesn’t go to zero! Explain the student’s errors.