UMD Analysis Qualifying Exam/Jan10 Real

Problem 1 edit

Assume that   is an integer, and let  . Prove that if  , then   a.e.

Solution 1 edit

We will show that  

Since the simple functions are dense in  , it suffices to show the result where  . Without loss of generality, we can assume that   is a disjoint family of measurable sets. Then,


We now wish to compute this limit.

An upper bound is:   since our function   is defined only on the interval  . The right hand side, goes to   as  .

A lower bound is:   since we assumed each of the   and   to be positive. This also tends to   as  .

Thus we have shown that   for simple  . By density of the simple functions in  , this shows the same result for general   functions.

So  . Now suppose  . Then for some   we have   for every  . Thus


Thus if we have any hope of   we must have   which implies that   a.e. on [0,1].

Problem 3 edit

Let  .

(a) Determine


(b) Determine


Solution 3 edit

For both parts (a) and (b), we have a clear upper bound of 2||f||1, by the triangle inequality. It remains to be shown that this is a tight upper bound in both cases.

The same approach can be used for both, with minor changes for the two different limits. Since step functions are dense in L1(R), pick epsilon e>0 and approximate f by some step function g, such that ||f-g||1<e. Let fx(t)=f(x+t), gx(t)=g(t+x).

Then ||fx+f|| = ||fx+f+(gx+g)-(gx+g)||. By the triangle inequality, this is greater than or equal to ||gx+g||-||fx+f-(gx+g)||. By yet another application of the triangle inequality, the second term is greater than or equal to -2e. The proof diverges at this point.

For part (a), for any particular t, x, |gx(t)+g(t)| is less than |gx(t)|+|g(t)| if and only if gx(t) and g(t) have opposite signs. Since g is a step function, this can clearly only happen when t and t+x are in intervals with different coefficients, and hence can happen at most for a distance of x per interval, across a finite number n of intervals. Since the difference for any particular step function g is bounded by M=max(g)-min(g), we get the following inequality:

||gx+g|| >= ||gx||+||g|| - x*n*M. Clearly, the limit of this as x goes to 0 is 2||g||, which is itself bounded below by 2||f||-2e. Adding up the two parts, we get a lower bound of the limit: limx->0||fx+f||>=2||f||-4e, for any positive epsilon. Thus the bound of 2||f|| is tight.

The justification in part (b) is simpler. Since g is a step function in L1, it has bounded support, so some value of x will be large enough that gx and g have disjoint supports. Hence we can just say that the limit is ||gx||+||g||>=2||f||-2e.

Problem 5 edit

Suppose that   is a sequence in   with   for all   and


(a) Prove that   and that  

(b) Show that


Solution 5 edit