UMD Analysis Qualifying Exam/Aug05 Real

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Problem 1[edit | edit source]

Let be a bounded measurable function on for which there is a constant such that

Show that .

Solution 1[edit | edit source]

We will consider two different regions: {x:|f(x)|>1} and {x:|f(x)|<=1}, then show that the integrals of f over these regions are both finite. Since f is bounded by M, we can use the assumption that m(x:|f(x)|>1) < C/1, and the first integral is this bounded by MC.

For the second integral, consider the set En={x:|f(x)|<1/n}. By assumption, m(En)<C*n1/2. For n>m, En clearly contains Em, so we can take the sets Sn=En+1\En, which is then {x:1/(n+1)<|f(x)|<=1/n}. Using the set containment, we can also produce a bound on m(Sn). The measure of Sn can be larger than C*(n+1)1/2-C*n1/2 (up to C*(n+1)1/2), but this extra mass can only be acquired at the expense of m(En), and would thus reduce the integral of |f|. Since we are interested in maximizing this integral and proving that this maximum is finite, a bound of C*(n+1)1/2-C*n1/2 is justified (ie this is the worst case, with each subsequent En of maximum possible size as 1/n goes to zero).

This gives the integral an upper bound of the sum as n goes from 1 to infinity of (C*(n+1)1/2-C*n1/2)/n (by m(Sn)*(upper bound of |f| on Sn)). By comparing consecutive terms, we get the sum of C*n1/2*(1/(n-1) - 1/n), for n = 2 to infinity, with a finite term at n=1. After computing the subtraction, each term becomes C*n1/2/(n*(n-1)), which is of order C/n3/2. Hence the sum converges, and gives us an upper bound for the integral.

Note: the argument for the size of Sn is correct, but can definitely be expressed better.

Problem 3[edit | edit source]

Solution 3[edit | edit source]

Problem 5[edit | edit source]

Solution 5[edit | edit source]