- Suppose μ is a vector measure having values in ℝ
^{n}or ℂ^{n}. Can you show thatmust be finite? Hint: You might define for each e_{i}, one of the standard basis vectors, the real or complex measure, μ_{ei}given by μ_{ei}≡ e_{i}⋅ μ. Why would this approach not yield anything for an infinite dimensional normed linear space in place of ℝ^{n}? - The Riesz representation theorem of the L
^{p}spaces can be used to prove a very interesting inequality. Let r,p,q ∈satisfyThen

and so r > q. Let θ ∈

be chosen so that θr = q. Then alsoand so

which implies p

^{′}= q. Now let f ∈ L^{p},g ∈ L^{q},f,g ≥ 0. Justify the steps in the following argument using what was just shown that θr = q and p^{′}= q. Let(23.27) (23.28) Therefore

_{r}≤_{q}_{p}. How does this inequality follow from the above computation? Does 23.27 continue to hold if r,p,q are only assumed to be in? Explain. Does 23.28 hold even if r,p, and q are only assumed to lie in? - Suppose is a finite measure space and thatis a sequence of functions which converge weakly to 0 in L
^{p}. This means thatfor every g ∈ L

^{p′ }. Suppose also that f_{n}→ 0 a.e. Show that then f_{n}→ 0 in L^{p−ε}for every p > ε > 0. - Give an example of a sequence of functions in L
^{∞}which converges weak ∗ to zero but which does not converge pointwise a.e. to zero. Convergence weak ∗ to 0 means that for every g ∈ L^{1},∫_{−π}^{π}gf_{n}dt → 0. Hint: First consider g ∈ C_{c}^{∞}and maybe try something like f_{n}= sin. Do integration by parts. - Let λ be a real vector measure on the measure space . That is λ has values in ℝ. The Hahn decomposition says there exist measurable sets P,N such that
and for each F ⊆ P,λ

≥ 0 and for each F ⊆ N,λ≤ 0. These sets P,N are called the positive set and the negative set respectively. Show the existence of the Hahn decomposition. Also explain how this decomposition is unique in the sense that if P^{′},N^{′}is another Hahn decomposition, then∪has measure zero, a similar formula holding for N,N^{′}. When you have the Hahn decomposition, as just described, you define λ^{+}≡ λ,λ^{−}≡ λ. This is sometimes called the Hahn Jordan decomposition. Hint: This is pretty easy if you use the polar decomposition above. - The Hahn decomposition holds for measures which have values in (−∞,∞]. Let λ be such a
measure which is defined on a σ algebra of sets ℱ. This is not a vector measure because
the set on which it has values is not a vector space. Thus this case is not included in
the above discussion. N ∈ℱ is called a negative set if λ≤ 0 for all B ⊆ N. P ∈ℱ is called a positive set if for all F ⊆ P,λ≥ 0. (Here it is always assumed you are only considering sets of ℱ.) Show that if λ≤ 0, then there exists N ⊆ A such that N is a negative set and λ≤ λ. Hint: This is done by subtracting off disjoint sets having positive meaure. Let A ≡ N
_{0}and suppose N_{n}⊆ A has been obtained. Tell whyLet B

_{n}⊆ N_{n}such thatThen N

_{n+1}≡ N_{n}∖ B_{n}. Thus the N_{n}are decreasing in n and the B_{n}are disjoint. Explain why λ≤ λ. Let N = ∩N_{n}. Argue t_{n}must converge to 0 since otherwise λ= −∞. Explain why this requires N to be a negative set in A which has measure no larger than that of A. - Using Problem 6 complete the Hahn decomposition for λ having values in (−∞,∞]. Now the Hahn
Jordan decomposition for the measure λ is
Explain why λ

^{−}is a finite measure. Hint: Let N_{0}= ∅. For N_{n}a given negative set, letExplain why you can assume that for all n, t

_{n}< 0. Let E_{n}⊆ N_{n}^{C}such thatand from Problem 6 let A

_{n}⊆ E_{n}be a negative set such that λ≤ λ. Then N_{n+1}≡ N_{n}∪ A_{n}. If t_{n}does not converge to 0 explain why there exists a set having measure −∞ which is not allowed. Thus t_{n}→ 0. Let N = ∪_{n=1}^{∞}N_{n}and explain why P ≡ N^{C}must be positive due to t_{n}→ 0. - What if λ has values in [−∞,∞). Prove there exists a Hahn decomposition for λ as in the above
problem. Why do we not allow λ to have values in ? Hint: You might want to consider −λ.
- Suppose X is a Banach space and let X
^{′}denote its dual space. A sequence_{n=1}^{∞}in X^{′}is said to converge weak ∗ to x^{∗}∈ X^{′}if for every x ∈ X,Let

be a mollifier. Also let δ be the measure defined byExplain how ϕ

_{n}→ δ weak ∗. - A Banach space X is called strictly convex if <+. A Banach space is called uniformly convex if whenever→ 2, it follows that→ 0. Show that uniform convexity implies strict convexity. It was not done here, but it can be proved using something called Clarkson’s inequalities that the L
^{p}spaces for p > 1 are uniformly convex.

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