UMD Analysis Qualifying Exam/Aug06 Real

Problem 1a edit

Prove the following version of the Riemann-Lebesque Lemma: Let  . Prove in detail that


Here   denotes a positive integer. You may use any of a variety of techniques, but you cannot simply cite another version of the Riemann-Lebesque Lemma.

Solution 1a edit

Note that  .

Hence we can equivalently show


Claim edit

Let   be a step function.


Proof edit


Step functions approximate L^1 functions well edit

Since  , then  

Hence, given  , there exists   such that



Problem 1b edit

Let   be an increasing sequence of positive integers. Show that   has measure 0.

Notes: You may take it as granted that the above set is measurable.

Solution 1b edit

For the sake of brevity, let S be the set of such x. If S has positive measure, then it contains a subset of positive measure on which liminf[sin(nkx)] is bounded below by some positive constant; ie, the integral of liminf[sin(nkx)] over S will be positive. If S has zero measure, then the same integral will be zero. Thus, we must only compute an appropriate integral to show that m(S)=0.

Since we do not know that S has finite measure, take a sequence of functions fk(x)=2-|x|*sin(nkx), each of which is clearly integrable. By Fatou's lemma, the integral of liminf[fk(x)] over S <= the liminf of the integrals of fk(x) over S. However, each fk is the L1 function 2-|x|*sin(nkx). By the Riemann-Lebesgue Lemma, this goes to 0 as nk goes to infinity.

Hence our original integral of a strictly positive function over S is bounded above by 0, so m(S)=0.

Problem 3 edit

Suppose  , where  . Show that  .

Solution 3 edit

Let   then we can write


Hence  .

Problem 5 edit

Let  ,


(a) Show that   is differentiable a.e. and find  .

(b) Is   absolutely continuous on closed bounded intervals  ?

Solution 5 edit

Look at the difference quotient:


We can justify bringing the limit inside the integral. This is because for every  ,  . Hence, our integrand is bounded by   and hence is   for all  . Then by Lebesgue Dominated Convergence, we can take the pointwise limit of the integrand. to get


It is easy to show that   is bounded (specifically by  ) which implies that   is Lipschitz continuous which implies that it is absolutely continuous.