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Unformatted text preview: MATH 202A HOMEWORK 14 Exercise 1 . (a) Proof. (i). R R K ( x ) dx = R R  d ( x/ ) dx = R R ( x ) d x = R R ( x ) dx = 1 (ii). R R  K ( x )  dx = R R  d  ( x/ )  dx = R R  ( x )  d x = R R  ( x )  dx < since is integrable on R . Hence,  K ( x )  is bounded. (iii). > 0, similarly, we know R  x   K ( x )  dx = R  t   ( t )  dt . > 0, Since is integrable on R , we know M > , s.t. R  t  M  ( t )  dt < . For M > 0, since as 0, we know > 0, s.t.   < , > M . Thus,   < , R  t   ( t )  dt R  t  M  ( t )  dt < , which implies R  x   K ( x )  dx = R  t   ( t )  dt 0 as 0. (b) Proof. (i). R R K ( x ) dx = 1 as the same argument above (ii). Since is bounded, c R s.t. ( x ) c . Hence,  K ( x )  c d . (iii). Since is bounded supported, X R s.t. ( x ) = 0 ,  x  > X . Hence  K ( x )  = 0 < A  x  d +1 ,  x  > X , where A is a constant number  x  X ,  K ( x )  < c d = c X d +1 X d +1 d +1 = A X d +1 A  x  d +1 , where A is a constant number. Hence  K ( x )  A  x  d +1 , x R , where A is a constant number. (c) Proof. Since R R K ( x ) dx = 1, we know f * K ( x ) f ( x ) = R R ( f ( x y ) f ( y )) K ( y ) dy . Hence,  f * K ( x ) f ( x )  R R  f ( x y ) f ( y )  K ( y )  dy By the measurability of f ( x y ) ,f ( y ) on R d R d , we can apply Fubini Theorem to the above to get k f * K ( x ) f ( x ) k R R k f ( x y ) f ( y ) k  K ( y )  dy . > 0, by the continuity with respect to the norm, we know > 0 s.t.  y  < , k f ( x y ) f ( y ) k < . Since K ( x ) is a family of good kernels, we know that > s.t. < R  y   K ( y )  dy < ....
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 Spring '10
 tao/analysis
 Math

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