UMD Analysis Qualifying Exam/Jan08 Real

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Problem 1[edit]

Suppose that is a uniformly continous function. Show that

Solution 1[edit]

L^1 implies integral of tail end of function goes to zero[edit]

Assume Not[edit]

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[edit]

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[edit]

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)[edit]

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[edit]

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



then

Solution 3[edit]

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[edit]

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

Problem 5a[edit]

Show that it is a metric linear space with the metric



where .

Solution 5a[edit]

"One-half" triangle inequality[edit]

First, for all ,



Taking square roots of both sides of the inequality yields,


L^1/2 is Linear Space[edit]

Hence for all ,



Hence, is a linear space.


L^1/2 is Metric Space[edit]

Non-negativity[edit]

Since ,


Zero Distance[edit]


Triangle Inequality[edit]

Also, for all ,



From and , we conclude that is a metric space.

Problem 5b[edit]

Show that with this metric is complete.

Solution 5b[edit]

For ,



By induction, we then have for all and all


Work with Subsequence of Cauchy Sequence[edit]

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

Claim[edit]

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

Proof[edit]

Construct a subsequence[edit]

Choose such that for all ,



Setup telescoping sum[edit]

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[edit]

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[edit]

By the Monotone Convergence Theorem,



Hence,



Apply Lebesgue Dominated Convergence Theorem[edit]

From the Lebesgue dominated convergence theorem,



where the last step follows since


Hence,



i.e. is complete.