UMD Analysis Qualifying Exam/Aug06 Real

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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



Let be a step function.



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.