hw5 - Math 245C Homework 5 Brett Hemenway June 4 2007...

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Unformatted text preview: Math 245C Homework 5 Brett Hemenway June 4, 2007 Folland Chapter 9 29. (a) Suppose F = μ ∈ M ( R n ). Then by proposition 8.49, k μ * φ k p ≤ k φ k p k μ k , so F ∈ C 1 . On the other hand, if F ∈ C 1 , then let { φ n } be an approximate identity. Then sup n k F * φ n k p ≤ sup n C k φ n k p < ∞ . Thus the family { F * φ n } lies in a closed ball in C ( R n ) * . By Alaoglu’s Theorem, this ball is compact, and so the sequence { F * φ n } has a subsequence converging to μ ∈ C ( R n ) * , i.e. lim k →∞ h F * φ n k ,g i = h μ,g i∀ g ∈ C ( R n ) . It remains to show that F = μ as distributions. Since { φ n k } is an approximate identity, φ n k * g → g . Thus h μ,g i = lim k →∞ h F * φ n k ,g i = lim k →∞ h F,φ n k * g i = h F,g i . Thus F ∈ C ( R n ) * = M ( R n ). (b) Suppose F ∈ S , and ˆ F ∈ L ∞ . Then applying Plancherel’s The- orem twice, we have k F * φ k 2 = k [ F * φ k 2 = k ˆ F ˆ φ k 2 ≤ k ˆ F k ∞ k ˆ φ k 2 = k ˆ F k ∞ k φ k 2 . 1 Thus F ∈ C 2 . Conversely, assume F ∈ C 2 . Let g ∈ C ∞ c ( R n ). Then k g ˆ F k 2 = k [ ˇ g * F k 2 = k ˇ g * F k 2 ≤ C k ˇ g k 2 = C k g k 2 . Thus ≤ C 2 k g k 2 2- k g ˆ F k 2 2 = Z ( C 2- ˆ F 2 ) g 2 , for all g ∈ C ∞ c . Thus C 2- ˆ F 2 ≥ 0 almost everywhere. Which means k ˆ F k ∞ ≤ C...
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hw5 - Math 245C Homework 5 Brett Hemenway June 4 2007...

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