Download Integration operators between Hardy spaces on the unit ball by Jordi Pau PDF

By Jordi Pau

http://www.sciencedirect.com/science/article/pii/S0022123615004176

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.

Show description

Read Online or Download Integration operators between Hardy spaces on the unit ball of C^n PDF

Similar functional analysis books

Functional Equations with Causal Operators

Written for technology and engineering scholars, this graduate textbook investigates practical differential equations related to causal operators, that are often referred to as non-anticipative or summary Volterra operators. Corduneanu (University of Texas, emeritus) develops the lifestyles and balance theories for useful equations with causal operators, and the theories at the back of either linear and impartial practical equations with causal operators.

Complex Variables: A Physical Approach with Applications and MATLAB (Textbooks in Mathematics)

From the algebraic houses of a whole quantity box, to the analytic homes imposed via the Cauchy quintessential formulation, to the geometric characteristics originating from conformality, complicated Variables: A actual process with functions and MATLAB explores all aspects of this topic, with specific emphasis on utilizing idea in perform.

Real Analysis (4th Edition)

Actual research, Fourth version, covers the elemental fabric that each reader may still comprehend within the classical conception of features of a true variable, degree and integration concept, and a few of the extra very important and simple subject matters typically topology and normed linear area thought. this article assumes a normal heritage in arithmetic and familiarity with the basic ideas of research.

Conformal mapping on Riemann surfaces

This lucid and insightful exploration reviews complicated research and introduces the Riemann manifold. It additionally exhibits how to find actual capabilities on manifolds analogously with algebraic and analytic issues of view. Richly endowed with greater than 340 workouts, this ebook is ideal for lecture room use or self sustaining learn.

Additional info for Integration operators between Hardy spaces on the unit ball of C^n

Example text

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).

Download PDF sample

Rated 4.33 of 5 – based on 35 votes