Download Banach spaces of analytic functions by Kenneth Hoffman PDF

By Kenneth Hoffman

ISBN-10: 0486458741

ISBN-13: 9780486458748

This vintage of natural arithmetic bargains a rigorous research of Hardy areas and the invariant subspace challenge. Its hugely readable remedy of advanced features, harmonic research, and practical research is acceptable for complicated undergraduates and graduate scholars. The textual content gains a hundred tough routines. 1962 version.

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4. 20) where Ps is the Poisson kernel. The Fourier transform Φ is Φ (ξ ) = ∞ 1 ψ (s)Ps (ξ ) ds = ∞ 1 ψ (s)e−2π s|ξ | ds (cf. 11), which is easily seen to be rapidly decreasing as |ξ | → ∞. The same is true for all the derivatives of Φ . The function Φ is clearly smooth on Rn \ {0}. Moreover, 44 6 Smoothness and Function Spaces ∂ j Φ (ξ ) = L−1 ∑ (−2π )k+1 k=0 |ξ |k ξ j k! |ξ | ∞ 1 sk+1 ψ (s) ds + O(|ξ |L ) = O(|ξ |L ) as |ξ | → 0, which implies that the distributional derivative ∂ j Φ is continuous at the origin.

1. Fix k ∈ Z+ . Show that Dkh ( f )(x) = 0 for all x, h in Rn if and only if f is a polynomial of degree at most k − 1. Hint: One direction may be proved by direct verification. 1. 2. 1 to the case γ = 0 and show that for all continuous functions f we have f L∞ ≤ f Λ ≤ 3 f L∞ ; 0 can be identified with L∞ (Rn ) ∩C(Rn ). hence the space Λ0 (b) Given a measurable function f on Rn we define (Rn ) f L˙ ∞ = inf f +c L∞ : c∈C . Let L˙ ∞ (Rn ) be the space of equivalent classes of bounded functions whose difference is a constant, equipped with this norm.

6. 10). Fix s ∈ R and all 1 < p < ∞. Then there exists a constant C1 that depends only on n, s, p, Φ , and Ψ such that for all f ∈ Lsp we have ∞ S0 ( f ) Lp + ∑ (2 js |Δ j ( f )|)2 j=1 1 2 Lp ≤ C1 f p Ls . 11) 18 6 Smoothness and Function Spaces Conversely, there exists a constant C2 that depends on the parameters n, s, p, Φ , and Ψ such that every tempered distribution f that satisfies ∞ S0 ( f ) Lp 1 2 ∑ (2 js|Δ j ( f )|)2 + Lp j=1 <∞ is an element of the Sobolev space Lsp with norm f p Ls ∞ ≤ C2 S0 ( f ) 1 2 ∑ (2 js|Δ j ( f )|)2 + Lp Lp j=1 .

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