UMD Analysis Qualifying Exam/Aug08 Real

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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