By Alessandra Lunardi

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As easily seen, Ff L2 = f L2 for every f ∈ C0∞ (Rn ), so that F is canonically extended to an isometry (still denoted by F) to L2 (Rn ). 13 If 1 < p ≤ 2, F is a bounded operator from Lp (Rn ) to Lp (Rn ), p = p/(p − 1), and 1 F L(Lp ,Lp ) ≤ . 1): pθ = 2/(1 + θ), qθ = 2/(1 − θ) = pθ . Moreover, F p L(Lpθ ,L θ) ≤ F θ L(L1 ,L∞ ) F 1−θ L(L2 ) ≤ 1 (2π)n/2 2/p−1 . 11, [L1 , L2 ]θ = Lpθ , pθ = 2/(1+θ), and [L∞ , L2 ]θ = Lqθ , qθ = 2/(1 − θ), with identical norms. 7) holds. When θ runs in (0, 1), pθ = 2/(1 + θ) runs in (1, 2), and the statement is proved.

Let a ∈ Lp0 (Ω) ∩ Lp1 (Ω). We may assume without loss of generality that a Lp = 1. For z ∈ S set a(x) p 1−z + z f (z)(x) = |a(x)| p0 p1 , if x ∈ Ω, a(x) = 0, |a(x)| f (z)(x) = 0, if x ∈ Ω, a(x) = 0. Then f is continuous in S and holomorphic in the interior of S with values in Lp0 (Ω) + Lp1 (Ω), and for each t ∈ R p |f (it)(x)| = |a(x)| p0 , f (it) Lp0 ≤ a p/p0 Lp p |f (1 + it)(x)| = |a(x)| p1 , f (1 + it) Lp1 ≤ a = 1, p/p1 Lp = 1. Moreover, t → f (it) is continuous with values in Lp0 (Ω) and t → f (1 + it) is continuous with values in Lp1 (Ω).

32 Chapter 1 Proof. For i = 1, 2, T is bounded from Lpi (Ω) to Lqi ,∞ (Λ), with norm not exceeding CMi . 6, T is bounded from (Lp0 (Ω), Lp1 (Ω))θ,p to (Lq0 ,∞ (Λ), Lq1 ,∞ (Λ))θ,p , and T (Lp0 (Ω),Lp1 (Ω))θ,p ,(Lq0 ,∞ (Λ),Lq1 ,∞ (Λ))θ,p ) ≤ CM01−θ M1θ . 10 that (Lp0 (Ω), Lp1 (Ω))θ,p = Lp,p (Ω) = Lp (Ω), and Lqi ,∞ (Λ) = (L1 (Λ), L∞ (Λ))1−1/qi ,∞ , i = 1, 2 (it is here that we need qi > 1: L1,∞ (Λ) is not a real interpolation space between L1 (Λ) and L∞ (Λ)), so that by the Reiteration Theorem (Lq0 ,∞ (Λ), Lq1 ,∞ (Λ))θ,p = (L1 (Λ), L∞ (Λ))(1−θ)(1−1/q0 )+θ(1−1/q1 ),p = (L1 (Λ), L∞ (Λ))1−1/q,p .

### An Introduction to Interpolation Theory by Alessandra Lunardi

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