UMD Analysis Qualifying Exam/Jan10 Real
Problem 1
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Assume that is an integer, and let . Prove that if , then a.e. 
Solution 1
editWe 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
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Let . (a) Determine
(b) Determine

Solution 3
editFor both parts (a) and (b), we have a clear upper bound of 2f_{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 L^{1}(R), pick epsilon e>0 and approximate f by some step function g, such that fg_{1}<e. Let f_{x}(t)=f(x+t), g_{x}(t)=g(t+x).
Then f_{x}+f = f_{x}+f+(g_{x}+g)(g_{x}+g). By the triangle inequality, this is greater than or equal to g_{x}+gf_{x}+f(g_{x}+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, g_{x}(t)+g(t) is less than g_{x}(t)+g(t) if and only if g_{x}(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:
g_{x}+g >= g_{x}+g  x*n*M. Clearly, the limit of this as x goes to 0 is 2g, which is itself bounded below by 2f2e. Adding up the two parts, we get a lower bound of the limit: lim_{x>0}f_{x}+f>=2f4e, for any positive epsilon. Thus the bound of 2f is tight.
The justification in part (b) is simpler. Since g is a step function in L^{1}, it has bounded support, so some value of x will be large enough that g_{x} and g have disjoint supports. Hence we can just say that the limit is g_{x}+g>=2f2e.
Problem 5
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Suppose that is a sequence in with for all and
(a) Prove that and that (b) Show that . 