Download Composition operators on Hardy-Orlicz spaces by Pascal Lefevre, Daniel Li, Herve Queffelec, Luis PDF

By Pascal Lefevre, Daniel Li, Herve Queffelec, Luis Rodriguez-piazza

The authors examine composition operators on Hardy-Orlicz areas whilst the Orlicz functionality \Psi grows speedily: compactness, susceptible compactness, to be p-summing, order bounded, \ldots, and convey how those notions behave in response to the expansion of \Psi. They introduce an tailored model of Carleson degree. They build numerous examples exhibiting that their effects are basically sharp. within the final half, they learn the case of Bergman-Orlicz areas

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11), with Ψ1 = Ψ2 = Ψ. 3. 16. 1) μ is a Ψ-Carleson measure on D if and only if there exists some constant C ≥ 1 such that : uξ,1−h LΨ (μ) ≤ C Ψ−1 (1/h) , for every ξ ∈ T and every h ∈ (0, 1). 2) μ is a vanishing Ψ-Carleson measure on D if and only if lim sup Ψ−1 (1/h) uξ,1−h h→0 ξ∈T LΨ (μ) = 0. Proof. 12) in the preceding remark. The converse is an obvious consequence of the following lemma. 17. Suppose that there exist A > 0 and h0 ∈ (0, 1) such that: ρμ (h) ≤ 1 , for every h ∈ (0, h0 ). Ψ AΨ−1 (1/h) Then there exists h1 ∈ (0, 1) such that: uξ,1−h LΨ (μ) ≤ 24 , AΨ−1 (1/h) for every ξ ∈ T and every h ∈ (0, h1 ).

24 is not only a technical one, though it might perhaps be weakened. Proof. Let: Ψ(x) = exp (log x)2 e−1/4 x if if √ x ≥ e, √ 0 ≤ x ≤ e. It is plain that Ψ ∈ Δ1 ∩ ∇0 . Moreover, for every A > 0, one has, for h small enough: 1/h = exp − (log A)2 − 2(log A) Ψ AΨ−1 (1/h) log(1/h) . 1.

3 . Once again, the job is done by a discrete measure. (k + 1)! ; yk = ; rk = 1 − 1/3 Ψ(yk ) Ψ(xk ) k x2 < y2 < x3 < · · · . Let ν be the discrete measure defined by: ∞ νk , ν= k=2 where: νk = 1 Ψ (k + 1)! δr k a . k2 so that the series converges. 3 Ψ (k + 1)! that ν is supported in the union of the circles of radii rk and not in a subset of the segment [0, 1] as in the preceding counterexamples. In order to show that (R0 ) is satisfied, it is clearly sufficient to prove that, when 1 1 ≤h< (with k ≥ 3), we have: Ψ(yk ) Ψ(yk−1 ) Observe that νk ≤ 1 · k1/3 −1 Ψ (1/h) Ψ 2 1 1 , we have Ψ−1 (1/h) ≤ yk so ≤h< Supposing then Ψ(yk ) Ψ(yk−1 ) ρν (h) ≤ k1/3 −1 1 Ψ (1/h) ≤ Ψ (k + 1)!

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