UMD Analysis Qualifying Exam/Aug08 Real

Problem 1 edit

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 edit

Absolutely Continuous <==> Indefinite Integral edit

  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 edit

Let   be given. Then,




Summing both sides of the inequality over   and applying the hypothesis yields pointwise convergence of the series  ,


Solution 1b edit

Absolutely continuous <==> Indefinite Integral edit

Let  .

We want to show:


Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem edit


Justification for Lebesgue Dominated Convergence Theorem edit


Therefore   is integrable

The above inequality also implies   a.e on  . Therefore,


a.e on   to a finite value.

Solution 1c edit

Since  ,   by the Fundamental Theorem of Calculus


Problem 3 edit

Suppose that   is a sequence of nonnegative integrable functions such that   a.e., with   integrable, and  . Prove that


Solution 3 edit

Check Criteria for Lebesgue Dominated Convergence Theorem edit

Define  ,  .

g_n dominates hat{f}_n edit

Since   is positive, then so is   , i.e.,   and  . Hence,


g_n converges to g a.e. edit

Let  . Since  , then

  , i.e.,


integral of g_n converges to integral of g = edit




hat{f_n} converges to hat{f} a.e. edit

Note that   is equivalent to




Apply LDCT edit

Since the criteria of the LDCT are fulfilled, we have that

  , i.e.,


Problem 5a edit

Show that if   is absolutely continuous on   and  , then   is absolutely continuous on  

Solution 5a edit

Show that g(x)=|x|^p is Lipschitz edit

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)) edit

Since   is absolutely continuous on  , given  , there exists   such that if   is a finite collection of nonoverlapping intervals of   such that




Consider  . Since   is Lipschitz


Therefore   is absolutely continuous.

Problem 5b edit

Let  . Give an example of an absolutely continuous function   on   such that   is not absolutely continuous

Solution 5b edit

f(x)= x^4sin^2(\frac{1}{x^2}) is Lipschitz (and then AC) edit

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) edit


Consider the partition  . Then,


Then, T(f) goes to   as   goes to  .

Then,   is not of bounded variation and then is not AC