Download Solutions for Final Exam - Real Analysis I | MATH 515 and more Exams Mathematics in PDF only on Docsity! MATH 515 Fall 2007 Final Exam Explain your answers carefully! 1. If E ⊂ Rn has finite measure, and f is a bounded, measurable function on E, define the Lp integral average Ap = ( 1 m(E) ∫ E |f(x)|p dx ) 1 p for 1 ≤ p < ∞. Show that if p < q then Ap ≤ Aq. Solution: By Hölder’s inequality, using conjugate exponents q p , q q−p we have∫ E |f(x)|p dx ≤ (∫ E |f(x)|q dx )p/q (∫ E 1 dx )1−p/q or m(E)App ≤ (m(E)Aqq)p/qm(E)1−p/q Canceling powers of m(E) and taking the p’th root gives Ap ≤ Aq. 2. If X, Y are metric spaces, f : X → Y is continuous and K ⊂ X is compact, show that the image f(K) is compact in Y . Solution: Let {Oα}α∈A be an open cover of f(K) in Y . Since f is continuous it follows that f−1(Oα) is open in X for every α and so {f−1(Oα)}α∈A is an open cover of K. By compactness K ⊂ n⋃ j=1 f−1(Oαj) from which it follows that {Oα1, . . . Oαn} is a finite subcover of f(K). This can also be done by verifying that f(K) is sequentially compact. MATH 515 Fall 2007 Final Exam 3. Let X = (0, 1] and ρ(x, y) = ∣∣∣ 1x − 1y ∣∣∣. (a) Show that < X, ρ > is a metric space. (b) Is it a complete metric space? (c) Is the metric equivalent to the usual metric ρ0(x, y) = |x− y|? Solution: Clearly ρ : X → [0,∞), ρ(x, y) = 0 if and only if x = y, ρ(x, y) = ρ(y, x) and ρ(x, y) = ∣∣∣∣1x − 1y ∣∣∣∣ ≤ ∣∣∣∣1x − 1z ∣∣∣∣ + ∣∣∣∣1z − 1y ∣∣∣∣ = ρ(x, z) + ρ(z, y) so < X, ρ > is a metric space. < X, ρ > is complete: If {xn} is Cauchy in X, pick N such that ρ(xn, xm) < 1 for n, m ≥ N . Then in particular∣∣∣∣ 1xn − 1xN ∣∣∣∣ ≤ 1 n ≥ N so that xn ≥ δ > 0 for n ≥ N where δ = xN1+xN . Note that |x− y| = xyρ(x, y) ≤ ρ(x, y) so that {xn} is also Cauchy in the standard metric. It follows that there exists x ≥ δ such that xn → x in the standard metric, Therefore ρ(xn, x) ≤ |xn − x| δ2 n ≥ N which implies xn → x in < X, ρ >. The metrics are not equivalent: For example ρ(x, 1) |x− 1| = 1 x →∞ as x → 0. 4. Let E = {f ∈ C([0, 1]) : ∫ 1 0 f 2(x) dx < 1}. Show that E is an open, unbounded set in C([0, 1]). Page 2
