UMD Analysis Qualifying Exam/Jan08 Real

Problem 1Edit

Suppose that   is a uniformly continous function. Show that


Solution 1Edit

L^1 implies integral of tail end of function goes to zeroEdit


Assume NotEdit

Suppose  . Then,




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 ContinuityEdit

Because of the uniform continuity, for the   there is a   such that



Then, if  , by Triangle Inequality, we have


which implies



Construct ContradictionEdit

Let   be a number greater than  . Note that   and   do not depend on  . With this in mind, note that




which is a huge contradiction.



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)Edit

By uniform continuity, for all  , there exists   such that for all  ,




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  .



which contradicts the assumption that  .

Therefore, for all   there exists   such that for all  ,




Problem 3Edit

Suppose   is absolutely continuous on  , and  . Show that if in addition



Solution 3Edit

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 5Edit

Suppose that   is the set of all equivalence classes of measurable functions for which


Problem 5aEdit

Show that it is a metric linear space with the metric


where  .

Solution 5aEdit

"One-half" triangle inequalityEdit

First, for all  ,


Taking square roots of both sides of the inequality yields,


L^1/2 is Linear SpaceEdit

Hence for all  ,


Hence,   is a linear space.

L^1/2 is Metric SpaceEdit


  Since  ,

Zero DistanceEdit


Triangle InequalityEdit

  Also, for all  ,


From   and   , we conclude that   is a metric space.

Problem 5bEdit

Show that with this metric   is complete.

Solution 5bEdit

For  ,


By induction, we then have for all   and all  


Work with Subsequence of Cauchy SequenceEdit

We can equivalently prove completeness by showing that a subsequence of a Cauchy sequence converges.


If a subsequence of a Cauchy sequence converges, then the Cauchy sequence converges.


Construct a subsequenceEdit

Choose   such that for all  ,


Setup telescoping sumEdit

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}_mEdit

Let  , then




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  .)



Hence, for all  


Apply Monotone Convergence TheoremEdit

By the Monotone Convergence Theorem,




Apply Lebesgue Dominated Convergence TheoremEdit

From the Lebesgue dominated convergence theorem,


where the last step follows since  



i.e.   is complete.