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245B_W09_hwk1

# 245B_W09_hwk1 - 245B Winter 2009 Assignment 1 Notes and...

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245B, Winter 2009, Assignment 1: Notes and selected model answers (Model solutions follow question numbers in bold .) Folland Chapter 3 3. a. Our convention for a signed measure ν is simply to define L 1 ( ν ) := L 1 ( ν + ) L 1 ( ν - ) in terms of the Jordan Decomposition, and now f L 1 ( ν + ) L 1 ( ν - ) | f | d ν + < & | f | d ν - < | f | d ν + + | f | d ν - < ( both values 0 ) | f | d | ν | < f L 1 ( | ν | ) . [Note: although it looks tempting, one can’t really ‘deduce’ from | f | d ν < that | f | d ν + < and | f | d ν - < , because by definition for a signed measure we only allow ourselves to write ‘ | f | d ν ’ at all given these two separate finiteness conditions.] b. It suffices for f L 1 ( ν ) = L 1 ( | ν | ) to write in terms of the Jordan Decom- position ν = ν + - ν - that f d ν = f d ν + - f d ν - f d ν + + f d ν - | f | d ν + + | f | d ν - = def | f | d | ν | . c. On the one hand, if X = X + X - is a Hahn Decomposition for ν , then 1

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f := χ X + - χ X - satisfies | f | ≤ 1 and E f d ν = ν ( E X + ) - ν ( E X - ) = def ν + ( E ) + ν - ( E ) = | ν | ( E ) , so the supremum over all such f is certainly at least | ν | ( E ) . On the other hand, for any measurable f with | f | ≤ 1 part (b) and monotonicity for the (unsigned) measure | ν | give E f d ν E | f | d | ν | ≤ E χ X d | ν | = | ν | ( E ) , giving the reverse inequality. 7. Be careful for both parts (as in all other evaluations of a supremum) to prove carefully that (a) a claimed ‘supremum’ is an upper bound for the desired set of values, and (b) that no smaller number is an upper bound for that set of values, or, equivalently, that that set of values comes arbitrarily close to the ‘supremum’ from below; and similarly for infima. It turns out in these cases that you can achieve equality for some member of the set of values. 10. Whichever option you choose, be careful to prove both (a) that ν μ (i.e., μ ( E ) = 0 ν ( E ) = 0 ) and (b) that for some ε > 0 , for every δ > 0 there is a measurable set E with μ ( E ) < δ but ν ( E ) ε . 11. a. First recall that uniform integrability for a single function is the conclu- sion of Corollary 3.6 in Folland [note: or reprove it yourself, say from the mono- tone convergence theorem, if you like]. Hence given a finite collection { f α 1 , f α 2 , . . . , f α n } and ε > 0 we can choose for each i n some δ i > 0
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