Download Bochner-Riesz Means on Euclidean Spaces by Shanzhen Lu PDF

By Shanzhen Lu

This ebook in most cases offers with the Bochner-Riesz technique of a number of Fourier quintessential and sequence on Euclidean areas. It goals to provide a systematical creation to the basic theories of the Bochner-Riesz capability and demanding achievements attained within the final 50 years. For the Bochner-Riesz technique of a number of Fourier fundamental, it comprises the Fefferman theorem which negates the Disc multiplier conjecture, the well-known Carleson-Sjolin theorem, and Carbery-Rubio de Francia-Vega's paintings on nearly all over convergence of the Bochner-Riesz ability lower than the severe index. For the Bochner-Riesz technique of a number of Fourier sequence, it contains the idea and alertness of a category of functionality house generated via blocks, that is heavily on the topic of nearly all over the place convergence of the Bochner-Riesz ability. furthermore, the ebook additionally introduce a little analysis effects on approximation of features by way of the Bochner-Riesz capacity.

Readership: Graduate scholars and researchers in arithmetic.

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4 The disc conjecture and Fefferman theorem 51 Then we rewrite the above inequality as R TM h(ξ)γ(ξ)dξ ≤ Tm p h g Lp (R) Lp (Rn ) β Lq (Rn ) γ Lq (R) , which implies that TM h Lp (R) ≤ Tm p g Lp (Rn ) β Lq (Rn ) h Lp (R) . Hence, we have TM p ≤ Tm p g Lp (Rn ) β Lq (Rn ) . 3 and the above inequality, we obtain Rn m(ξ, η)ˆ g (η)β(η)dη ≤ Tm p g Lp (Rn ) β Lq (Rn ) , which can be rewritten as Rn Tm(ξ,η) g(η)β(η)dη ≤ Tm p g Lp (Rn ) β Lq (Rn ) . 1 that the conclusion of this theorem is true. 2, we have defined the Bochner-Riesz operator Tα by Tα f (x) = Φα (x)fˆ(x), for f ∈ S (Rn ).

1) m∈Zn which is an absolutely convergent series by the condition (i), which is called the Φ means of σ(f ). We consider the relation between σεΦ (f )(x) and f (x) as ε → 0+ . 1 Suppose f ∈ Lp (Q) (1 ≤ p ≤ ∞). 1) converges to f in the norm of Lp (Q) and converges to f at the Lebesgue points of f . Proof. 2) belongs to L(Rn ). It follows that ϕ = Φ. By letting ϕε (y) = ε−n ϕ(ε−1 y), we get ϕε (x) = Φ(εx). 2, we have Kε (x) := ϕ0ε (x + 2πm) = Φ(εm)eim·x ∈ C(Q). 3) Taking convolution on Q, we get that f ∗ Kε (x) := 1 (2π)n 1 (2π)n = m∈Zn f (y)e−im·y dyΦ(εm)eim·x Q Cm (f )Φ(εm)eim·x = = f (y)Kε (x − y)dy Q m∈Zn σεΦ (f )(x).

52 C2. 1 Let n ≥ 2. The operator T0 is bounded on Lp (Rn ) if and only if p = 2. For p = 2, by the definition of the multiplier T0 , it is obvious that T0 is bounded on L2 (Rn ). So the most essential part of the proof of the theorem lies in giving the conclusion that T0 is unbounded in Lp (Rn ), p = 2. Before proving, we need some lemmas. 1 Suppose that p > 2, vj , 1 ≤ j ≤ k, is the unit vector in R2 and Lj = x ∈ R2 : x · vj ≥ 0 is a half-plane. Define the operator Tj by Tj f (x) = χLj (x)fˆ(x), f ∈ S, where χE (x) is the characteristic function of the set E.

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