This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 < ....
View
Full
Document
 Spring '10
 tao/analysis
 Math, Mathematical constant, constant number, s.t.

Click to edit the document details