UMD Analysis Qualifying Exam/Aug08 Real

Problem 1 Suppose that is a sequence of absolutely continuous functions defined on such that for every and for every . Prove: the series converges for each pointwise to a function the function is absolutely continuous on Solution 1a Absolutely Continuous <==> Indefinite Integral is absolutely continuous if and only if can be written as an indefinite integral i.e. for all Apply Inequalities,Sum over n, and Use Hypothesis Let be given. Then, Hence Summing both sides of the inequality over and applying the hypothesis yields pointwise convergence of the series , Solution 1b Absolutely continuous <==> Indefinite Integral Let . We want to show: Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem Justification for Lebesgue Dominated Convergence Theorem Therefore is integrable The above inequality also implies a.e on . Therefore, a.e on to a finite value. Solution 1c Since ,   by the Fundamental Theorem of Calculus   a.e. Problem 3 Suppose that is a sequence of nonnegative integrable functions such that a.e., with integrable, and . Prove that Solution 3 Check Criteria for Lebesgue Dominated Convergence Theorem Define , . g_n dominates hat{f}_n Since is positive, then so is , i.e., and . Hence, g_n converges to g a.e. Let . Since , then , i.e., . integral of g_n converges to integral of g = Hence, hat{f_n} converges to hat{f} a.e. Note that is equivalent to i.e. Apply LDCT Since the criteria of the LDCT are fulfilled, we have that , i.e., Problem 5a Show that if is absolutely continuous on and , then is absolutely continuous on Solution 5a Show that g(x)=|x|^p is Lipschitz Consider some interval and let and be two points in the interval . Also let for all Therefore is Lipschitz in the interval Apply definitions to g(f(x)) Since is absolutely continuous on , given , there exists such that if is a finite collection of nonoverlapping intervals of such that then Consider . Since is Lipschitz Therefore is absolutely continuous. Problem 5b Let . Give an example of an absolutely continuous function on such that is not absolutely continuous Solution 5b f(x)= x^4sin^2(\frac{1}{x^2}) is Lipschitz (and then AC) Consider . The derivate of f is given by . The derivative is bounded (in fact, on any finite interval), so is Lipschitz. Hence, f is AC |f|^{1/2} is not of bounded variation (and then is not AC) Consider the partition . Then, Then, T(f) goes to as goes to . Then, is not of bounded variation and then is not AC