By Jordi Pau
We thoroughly describe the boundedness of the Volterra style operator Jg among Hardy areas within the unit ball of Cn. The facts of the only dimensional case used instruments, reminiscent of the powerful factorization for Hardy areas, that aren't on hand in larger dimensions, and accordingly different thoughts has to be used. specifically, a generalized model of the outline of Hardy areas by way of the realm functionality is required.
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Additional info for Integration operators between Hardy spaces on the unit ball of C^n
Put r = m p/(p − s) and set λQ = 2 s (r−1) if Q ∈ Em . Notice that μ G(Q) ≥ μ Q∈Em G(Q) = μ Em \ Em+1 . Q∈Em Then |λQ |s μ(G(Q)) = Q∈E 2m(r−1) m μ G(Q) Q∈Em J. Pau / Journal of Functional Analysis 270 (2016) 134–176 174 ≥ 2m(r−1) μ(Em ) − μ(Em+1 ) . m By a typical covering lemma of Vitali type (see [35, p. 9]), there is a sequence Fm of pairwise disjoint balls Q ∈ Em with σ(Em ) ≤ C Q∈Fm σ(Q) (here the constant C depends only on the dimension). This implies Q ≥μ μ(Em ) = μ Q∈Em Q = μ(Q) Q∈Fm Q∈Fm σ(Q) ≥ C2m σ(Em ).
1), Fubini’s theorem and Parseval’s identity, we have R−1,t Sek 2 A22t−1 |R−1,t Sek (z)|2 dv2t−1 (z) = k k B n Sek , Kzt = 2 H2 k B n SKzt = Bn 2 H2 T p Kzt , Kzt dv2t−1 (z) = Bn 2 H2 dv2t−1 (z) k H2 dv2t−1 (z) Bn T p kzt , kzt = ek , SKzt dv2t−1 (z) = H2 Kzt 2 H2 dv2t−1 (z). Bn Putting all together and taking into account that Kzt (1 − |z|2 )−(n+1) , we have that T is in Sp if and only if T p kzt , kzt H2 2 2 2t−1 H 2 (1 − |z| ) is comparable to dλn (z) < ∞. 31]) T p kzt , kzt H2 ≤ T kzt , kzt p H2 = [T t (z)]p , 0
Q∈Em Then |λQ |s μ(G(Q)) = Q∈E 2m(r−1) m μ G(Q) Q∈Em J. Pau / Journal of Functional Analysis 270 (2016) 134–176 174 ≥ 2m(r−1) μ(Em ) − μ(Em+1 ) . m By a typical covering lemma of Vitali type (see [35, p. 9]), there is a sequence Fm of pairwise disjoint balls Q ∈ Em with σ(Em ) ≤ C Q∈Fm σ(Q) (here the constant C depends only on the dimension). This implies Q ≥μ μ(Em ) = μ Q∈Em Q = μ(Q) Q∈Fm Q∈Fm σ(Q) ≥ C2m σ(Em ). 4) m where the last estimate is due to the fact that μ(ζ) 2m for ζ ∈ Em \ Em+1 . 2] together with the choice made on the points zQ , we see that zQ ∈ Γ(ζ) 1 implies that ζ ∈ Q, where Q = Q(ξ, 81 δ) if Q = Q(ξ, δ).