UMD Analysis Qualifying Exam/Jan08 Real

Problem 1 Suppose that is a uniformly continous function. Show that Solution 1 L^1 implies integral of tail end of function goes to zero Assume Not Suppose . Then, or Without loss of generality, we can assume the first one, i.e., (see remark below to see why this) Note that can be written as Then, the negation of the above statement gives Apply Uniform Continuity Because of the uniform continuity, for the there is a such that , whenever Then, if , by Triangle Inequality, we have which implies , whenever Construct Contradiction Let be a number greater than . Note that and do not depend on . With this in mind, note that Then, which is a huge contradiction. Therefore, Remark If we choose to work with the assumption that , then in (*), we just need to work with instead of the original one Solution 1 (Alternate) By uniform continuity, for all , there exists such that for all , if Assume for the sake of contradiction there exists such that for all , there exists such that and . Let , then there exists such that and . Let , then there exists such that and . Let , then there exists such that and . So we have with if and for all and for all . In other words, we are choosing disjoint subintervals of the real line that are of length , centered around each for , and separated by at least . Hence, which contradicts the assumption that . Therefore, for all there exists such that for all , i.e. Problem 3 Suppose is absolutely continuous on , and . Show that if in addition then Solution 3 By absolute continuity, Fatou's Lemma, and hypothesis we have Hence a.e. From the fundamental theorem of calculus, for all , i.e. is a constant . Assume for the sake of contradiction that , then . which contradicts the hypothesis . Hence, i.e. for all Problem 5 Suppose that is the set of all equivalence classes of measurable functions for which Problem 5a Show that it is a metric linear space with the metric where . Solution 5a "One-half" triangle inequality First, for all , Taking square roots of both sides of the inequality yields, L^1/2 is Linear Space Hence for all , Hence, is a linear space. L^1/2 is Metric Space Non-negativity Since , Zero Distance Triangle Inequality Also, for all , From and , we conclude that is a metric space. Problem 5b Show that with this metric is complete. Solution 5b For , By induction, we then have for all and all Work with Subsequence of Cauchy Sequence We can equivalently prove completeness by showing that a subsequence of a Cauchy sequence converges. Claim If a subsequence of a Cauchy sequence converges, then the Cauchy sequence converges. Proof Construct a subsequence Choose such that for all , Setup telescoping sum Rewrite as a telescoping sum (successive terms cancel out) i.e. . The triangle inequality implies, which means the sequence is always dominated by the sequence on the right hand side of the inequality. Define a sequence {g}_m Let , then and . In other words, is a sequence of increasing, non-negative functions. Note that , the limit of as , exists since is increasing. ( is either a finite number or .) Also, Hence, for all Apply Monotone Convergence Theorem By the Monotone Convergence Theorem, Hence, Apply Lebesgue Dominated Convergence Theorem From the Lebesgue dominated convergence theorem, where the last step follows since Hence, i.e. is complete.