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By Alessandra Lunardi

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6) to each of them. 10 For every θ ∈ (0, 1), [X, Y ]θ ∈ Kθ (X, Y ), that is [X, Y ]θ is continuously embedded in (X, Y )θ,∞ . Proof — Let a ∈ [X, Y ]θ . 6). Note that for z = x + iy ∈ S we have 0< eπ(y−t) sin(πx) 1 dt < 1, sin (πx) + (cos(πx) − exp(π(y − t)))2 eπ(y−t) sin(πx) 1 dt < 1. sin (πx) + (cos(πx) + exp(π(y − t)))2 R 0< R 2 2 Indeed, both kernels are positive so that both integrals are positive; moreover if f is holomorphic in S, continuous and bounded in S and f ≡ 1 on iR and on 1 + iR then f ≡ 1 in S, so that the sum of the integrals is 1.

Traditionally, a linear operator T : L1 (Ω) + L∞ (Ω) → L1 (Λ) + L∞ (Λ) is said to be of weak type (p, q) if there is M > 0 such that sup σ(ν{y ∈ Λ : |T f (y)| > σ})1/q ≤ M f Lp (Ω) , σ>0 Lp (Ω). for all f ∈ This is equivalent to say that the restriction of T to Lp (Ω) is a bounded operator from Lp (Ω) to Lq,∞ (Λ). Indeed, by the properties of the nonincreasing rearrangements, sup σ(ν{y ∈ Λ : |g(y)| > σ})1/q = sup t1/q g ∗ (t) = g σ>0 Lq,∞ . 28) t>0 T is said to be of strong type (p, q) if its restriction to Lp (Ω) is a bounded operator from Lp (Ω) to Lq (Λ).

Therefore ϕ(λ) = λ2θ (AR(λ, A))2 x is in Lp∗ (0, ∞), and x eθ,p ≤ 2−1/p (M + 1)2 x θ,p . (The formula is true also for p = ∞ if we set 1/∞ = 0). Conversely, if ϕ ∈ Lp∗ (0, ∞), from the obvious identity x = λ2 R(λ, A)2 x − 2λAR(λ, A)2 x + A2 R(λ, A)2 x, where λAR(λ, A)2 x = λ(λ − A)AR(λ, A)3 x = AR(λ, A)λ2 R(λ, A)2 x − λR(λ, A)A2 R(λ, A)2 x we get x = (I − 2AR(λ, A))λ2 R(λ, A)2 x + (2λR(λ, A) + I)A2 R(λ, A)2 x, λ ≥ 1, where (I − 2AR(λ, A))λ2 R(λ, A)2 x D(A2 ) = (I − 2AR(λ, A))λ2 R(λ, A)2 x + (I − 2AR(λ, A))λ2 A2 R(λ, A)2 x ≤ (2M + 3)M 2 x + (2M + 3)λ2 A2 R(λ, A)2 x and (2λR(λ, A) + I)A2 R(λ, A)2 x ≤ (2M + 1) A2 R(λ, A)2 x .

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An Introduction to Interpolation Theory by Alessandra Lunardi


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