Download 3 Problems with Solutions on Real Analysis - Final Exam | MATH 245A and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity! Jeffrey Hellrung Friday, December 09, 2005 Math 245A, Sample Final Problems 1. Let f ∈ L1 loc (R) be such that f ( x + 1n ) ≥ f(x) for all n = 1, 2, . . . and a.e. x. Show that f(x+a) ≥ f(x) for all a > 0 and a.e. x. Solution We first notice that repeated application of the given inequality yields f(x + q) ≥ f(x) for q ∈ Q+ and a.e. x. Thus, continuity of f would then imply the claim by taking a sequence of {qn} ⊂ Q + such that qn → a > 0. For general f , we consider fs(x) = ∫ x+s x f(y)dy, s > 0. We note that fs is continuous, for given xn → x, fχ[xn,xn+s] → fχ[x,x+s] pointwise a.e., hence fs(xn) → fs(x) by the Dominated Convergence Theorem. Further, for q ∈ Q +, fs(x + q) = ∫ x+q+s x+q f(y)dy = ∫ x+s x f(y + q)dy ≤ ∫ x+s x f(y)dy = fs(x), so the previous comments above imply that fs(x + a) ≥ fs(x) for a > 0. Thus 0 ≤ fs(x + a) − fs(x) = ∫ x+a+s x+a f(y)dy − ∫ x+s x f(y)dy = ∫ x+s x (f(y + a) − f(y))dy. The above inequality holds for all s > 0, and the Lebesgue Differentiation Theorem tells us that lim s→0 1 s ∫ x+s x (f(y + a) − f(y))dy = f(x + a) − f(x) for a.e. x, hence we conclude that f(x + a) − f(x) ≥ 0 for a.e. x. 2. Let f ∈ L1(Rn). Show that lim |h|→∞ ∫ |f(x + h) − f(x)|dx = 2 ∫ |f(x)|dx. Solution Suppose first that f has compact support. Let M = supp f and τhM = M − h = supp f(· + h). Then for large enough |h|, M ∩ τhM = ∅, and in such case ∫ |f(x + h) − f(x)|dx = ∫ M |f(x)|dx + ∫ τhM |f(x + h)|dx = 2‖f‖1. Now take fn = fχ[−n,n]. Then by the above argument, the claim holds for each fn. Further, fntof pointwise, hence, by the Dominated Convergence Theorem, fn → f in L 1. Thus, 2‖f‖1 − ‖f(· + h) − f‖1 = 2‖f − fn + fn‖1 − ‖f(· + h) − fn(· + h) + fn(· + h) − fn + fn − f‖1 ≤ 2‖f − fn‖1 + 2‖fn‖1 − (‖fn(· + h) − fn‖1 − ‖f(· + h) − fn(· + h)‖1 − ‖fn − f‖1) . Now for |h| large, ‖fn(· + h) − fn‖1 = ‖fn‖1, hence for large enough n and |h| we have 2‖f‖1 − ‖f(· + h) − f‖1 < ǫ 1
