1990
DOI: 10.1016/0021-9045(90)90065-x
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Hilbert transforms and Lagrange interpolation

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Cited by 17 publications
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“…Thus, since G ∈ L ∞ and (4.10) holds, using a result in [39] we get Finally, combining (4.14)-(4.16) with (4.12), (4.11) follows. Now we can prove Theorem 2.1.…”
Section: Proofsmentioning
confidence: 70%
“…Thus, since G ∈ L ∞ and (4.10) holds, using a result in [39] we get Finally, combining (4.14)-(4.16) with (4.12), (4.11) follows. Now we can prove Theorem 2.1.…”
Section: Proofsmentioning
confidence: 70%
“…First of all we remark that under the assumptions ( 19) both ( 8) and Theorem 3.2 are true. Therefore, taking into account that, in view of (18), the sequence T 2 n m (f ) is obtained by alternating two subsequences of {I m (f )} m and {Σ 2m+1 (f )} m , then (20) follows by the first inequality in (8) and by (14), while (21) follows by the second inequality in (8) and by (16).…”
Section: The Proofsmentioning
confidence: 99%
“…In order to estimate the first summand, we note that the function under the sign of the Hilbert transform is bounded and the one outside is L(log + L) (see [8]). Therefore, with Γ = sgn H(p m (v α )v γ+r g) and ̺ = γ + r − α/2 − 1/4, we can write…”
Section: Proofsmentioning
confidence: 99%