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Unformatted text preview: Math 245C Homework 5 Brett Hemenway June 4, 2007 Folland Chapter 9 29. (a) Suppose F = M ( R n ). Then by proposition 8.49, k * k p k k p k k , so F C 1 . On the other hand, if F C 1 , then let { n } be an approximate identity. Then sup n k F * n k p sup n C k n k p < . Thus the family { F * n } lies in a closed ball in C ( R n ) * . By Alaoglus Theorem, this ball is compact, and so the sequence { F * n } has a subsequence converging to C ( R n ) * , i.e. lim k h F * n k ,g i = h ,g i g C ( R n ) . It remains to show that F = as distributions. Since { n k } is an approximate identity, n k * g g . Thus h ,g i = lim k h F * n k ,g i = lim k h F, n k * g i = h F,g i . Thus F C ( R n ) * = M ( R n ). (b) Suppose F S , and F L . Then applying Plancherels The orem twice, we have k F * k 2 = k [ F * k 2 = k F k 2 k F k k k 2 = k F k k k 2 . 1 Thus F C 2 . Conversely, assume F C 2 . Let g C c ( R n ). Then k g F k 2 = k [ g * F k 2 = k g * F k 2 C k g k 2 = C k g k 2 . Thus C 2 k g k 2 2 k g F k 2 2 = Z ( C 2 F 2 ) g 2 , for all g C c . Thus C 2 F 2 0 almost everywhere. Which means k F k C...
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 Spring '10
 tao/analysis
 Math

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