UMD Analysis Qualifying Exam/Aug08 Real
Contents
Problem 1[edit]
Suppose that is a sequence of absolutely continuous functions defined on such that for every and

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,
Hence
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
 a.e.
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]
Hence,
hat{f_n} converges to hat{f} a.e.[edit]
Note that is equivalent to
i.e.
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
then
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