Download e-book for iPad: A Course of Mathematical Analysis (Vol. 2) by Nikolsky S.M.

By Nikolsky S.M.

A textbook for collage scholars (physicists and mathematicians) with exact supplementary fabric on mathematical physics. in response to the path learn by way of the writer on the Moscow Engineering Physics Institute. quantity 2 comprises a number of integrals, box conception, Fourier sequence and Fourier crucial, differential manifolds and differential varieties, and the Lebesgue indispensable.

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17. a) Let 1 < p < ∞, let E and F be Banach spaces of class HT and let T ⊂ L(E, F ) be R-bounded. 3) |ξ||γ| Dγ m(ξ); ξ ∈ Rn \ {0}, 0 ≤ γ ≤ 1 ⊂ T , then m defines a Fourier multiplier. 4) ξ γ Dγ m(ξ); ξ ∈ Rn \ {0}, 0 ≤ γ ≤ 1 ⊂ T 3 R-boundedness and operator-valued Fourier multiplier theorems 32 is sufficient. 4)} ⊂ L Lp (Rn , E), Lp (Rn , F ) is R-bounded again. The additional result on R-boundedness of operators associated with a family of multipliers in part b) is due to Girardi and Weis [GW03]. For further information on operator-valued Fourier multipliers we refer to [KW04] and [DHP03].

1]) α Hqα (R, X) := {f ∈ S (R, X); ∃fα ∈ Lq (R, X) : F fα (ξ) = (1 + |ξ|2 ) 2 Ff (ξ)} and f α,q := f α,q,R := fα q. For J = [0, T ] with T ∈ (0, ∞) we set Hqα (J, X) := {f |J ; f ∈ Hqα (R, X)} and f α,q := f α,q,J := inf gα g : g|J =f, g∈Hqα (R,X) q. 6. Let 1 < q < ∞, J = [0, T ] with T ∈ (0, ∞), and b ∈ L1loc (R+ ). 3) with b ∗ u ∈ Hqα (J, X) ∩ Lq (J, D(A)), and there is a constant C(T ) > 0 such that q q u q + b∗u α,q + Ab ∗ u q ≤ C(T ) f q. 3) based on R-sectoriality of A, some definitions of useful properties of b are in order.

The following lemma provides a condition on m which is equivalent for W ,p -regularity of Tm f . 7. Let 1 ≤ p < ∞, ∈ N0 and let m ∈ L∞ (Rn \ {0}, L(E, F )). Then the following assertions are equivalent: (i) Tm ∈ L(Lp (Rn , E), W (ii) For each |α| ≤ multiplier. ,p (Rn , F )). the function mα : ξ → ξ α m(ξ) defines a continuous Fourier Proof. (i) ⇒ (ii): Set κα (ξ) := ξ α . For arbitrary f ∈ S(Rn , E) we have F Dα Tm f = κα F F −1 mF f = mα Ff in S (Rn , F ). Hence, F −1 mα F f = Dα Tm f and there exists C > 0 such that F −1 mα F f p,F = D α Tm f p,F ≤C f p,E for all |α| ≤ and all f ∈ S(R , E).

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A Course of Mathematical Analysis (Vol. 2) by Nikolsky S.M.


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