hmwk14sol_a

hmwk14sol_a - MATH 202A HOMEWORK 14 Exercise 1 . (a) Proof....

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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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hmwk14sol_a - MATH 202A HOMEWORK 14 Exercise 1 . (a) Proof....

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