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hmwk14sol_a

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

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Unformatted text preview: MATH 202A HOMEWORK 14 Exercise 1 . (a) Proof. (i). R R K δ ( x ) dx = R R δ- d ϕ ( x/δ ) dx = R R ϕ ( x δ ) d x δ = R R ϕ ( x ) dx = 1 (ii). R R | K δ ( x ) | dx = R R δ- d | ϕ ( x/δ ) | dx = R R | ϕ ( x δ ) | d x δ = R R | ϕ ( x ) | dx < ∞ since ϕ is integrable on R . Hence, | K δ ( x ) | is bounded. (iii). ∀ η > 0, similarly, we know R | x |≥ η | K δ ( x ) | dx = R | t |≥ η δ | ϕ ( t ) | dt . ∀ > 0, Since ϕ is integrable on R , we know ∃ M > , s.t. R | t |≥ M | ϕ ( t ) | dt < . For M > 0, since η δ → ∞ as δ → 0, we know ∃ Δ > 0, s.t. ∀| δ | < Δ, η δ > M . Thus, ∀| δ | < Δ , R | t |≥ η δ | ϕ ( t ) | dt ≤ R | t |≥ M | ϕ ( t ) | dt < , which implies R | x |≥ η | K δ ( x ) | dx = R | t |≥ η δ | ϕ ( t ) | dt → 0 as δ → 0. (b) Proof. (i). R R K δ ( x ) dx = 1 as the same argument above (ii). Since ϕ is bounded, ∃ c ∈ R s.t. ϕ ( x ) ≤ c . Hence, | K δ ( x ) | ≤ c δ d . (iii). Since ϕ is bounded supported, ∃ X ∈ R s.t. ϕ ( x ) = 0 , ∀| x | > X . Hence | K δ ( x ) | = 0 < Aδ | x | d +1 , ∀| x | > X , where A is a constant number ∀| x | ≤ X , | K δ ( x ) | < cδ- d = cδ X d +1 X d +1 δ d +1 = Aδ X d +1 ≤ Aδ | x | d +1 , where A is a constant number. Hence | K δ ( x ) | ≤ Aδ | x | d +1 , ∀ x ∈ R , where A is a constant number. (c) Proof. Since R R K δ ( x ) dx = 1, we know f * K δ ( x )- f ( x ) = R R ( f ( x- y )- f ( y )) K δ ( y ) dy . Hence, | f * K δ ( x )- f ( x ) | ≤ R R | f ( x- y )- f ( y ) || K δ ( 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 δ ( x )- f ( x ) k≤ R R k f ( x- y )- f ( y ) k | K δ ( y ) | dy . ∀ > 0, by the continuity with respect to the norm, we know ∃ η > 0 s.t. ∀| y | < η , k f ( x- y )- f ( y ) k < . Since K δ ( x ) is a family of good kernels, we know that ∃ Δ > s.t. ∀ δ < Δ R | y |≥ η | K δ ( y ) | dy < ....
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hmwk14sol_a - MATH 202A HOMEWORK 14 Exercise 1(a Proof(i R...

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