UMD Analysis Qualifying Exam/Aug08 Real

Problem 1

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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

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Absolutely Continuous <==> Indefinite Integral

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  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

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Let   be given. Then,


 


Hence


 


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


 

Solution 1b

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Absolutely continuous <==> Indefinite Integral

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Let  .


We want to show:


 


Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem

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Justification for Lebesgue Dominated Convergence Theorem

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Therefore   is integrable


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


 


a.e on   to a finite value.

Solution 1c

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Since  ,   by the Fundamental Theorem of Calculus

    a.e.  

Problem 3

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


 

Solution 3

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Check Criteria for Lebesgue Dominated Convergence Theorem

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Define  ,  .

g_n dominates hat{f}_n

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

 

g_n converges to g a.e.

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Let  . Since  , then

  , i.e.,

 .

integral of g_n converges to integral of g =

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Hence,


 

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

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Note that   is equivalent to


 


i.e.


 

Apply LDCT

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

  , i.e.,

 

Problem 5a

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

Solution 5a

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

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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))

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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

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

Solution 5b

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

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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)

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Consider the partition  . Then,

 

Then, T(f) goes to   as   goes to  .

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