By Alessandra Lunardi

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**Extra info for An Introduction to Interpolation Theory**

**Example text**

We already know that D(A) ∈ J1/2 (X, D(A2 )) and that D(B) ∈ J1/2 (X, D(B 2 )). Therefore there is C > 0 such that x K 1 ≤ x D(A) + x D(B) ≤C x 1/2 ( x 1/2 D(A2 ) + x 1/2 ) D(B 2 ) ≤C x 1/2 x 1/2 , K2 which means that K 1 ∈ J1/2 (X, K 2 ). 3). Then v(λ) − x ≤ λR(λ, A)(λR(λ, B)x − x) + λR(λ, A)x − x = λR(λ, A)R(λ, B)Bx + R(λ, A)Ax ≤ λ−1 (M (M + 1) Bx + M Ax ), and v(λ) K2 = v(λ) + A2 v(λ) + ABv(λ) + B 2 v(λ) ≤ M 2 x + 2M (M + 1)λ( Ax + Bx ). Setting λ = t−1/2 we deduce that t−1/2 K(t, x, X, K 2 ) ≤ t−1/2 ( x − v(t−1/2 ) + t v(t−1/2 ) ≤ C( x K1 K2 ) + t1/2 x ), is bounded in (0, 1).

For every a ∈ K 1 , b ∈ K 2 such that x = a + b we have λθ BR(λ, B)Ax ≤ λθ BR(λ, B)Aa + λθ BR(λ, B)Ab ≤ λθ (M + 1) Aa + M λθ−1 BAb ≤ (M + 1)λθ ( a K1 + λ−1 b K2 ) 56 Chapter 3 so that λθ BR(λ, B)Ax ≤ (M + 1)λθ K(λ−1 , x, K 1 , K 2 ), λ > 0. It follows that λθ BR(λ, B)Ax ∈ Lp∗ (0, ∞) with norm not exceeding (M +1) x (K 1 ,K 2 )θ,p , and the embedding ⊂ is proved. 11, and is omitted. Let us prove that DA (θ+1, p)∩DB (θ+1, p) ⊂ {x ∈ K 1 : Ax, Bx ∈ DA (θ, p)∩DB (θ, p)}. We have only to show that if x ∈ DA (θ + 1, p) ∩ DB (θ + 1, p) then Ax ∈ DB (θ, p) and Bx ∈ DA (θ, p).

This is well known if X = C, and may be recovered for general X by the following argument. For every ζ ∈ Ω let x ∈ X be such that f (ζ) X = f (ζ), x and x X = 1. Applying the maximum principle to the complex function z → f (z), x we get f (ζ) X = | f (ζ), x | ≤ max{| f (z), x | : z ∈ ∂Ω} ≤ max{ f (z) X : z ∈ ∂Ω}. The maximum principle holds also for functions defined in strips. Dealing with complex interpolation, we shall consider the strip S = {z = x + iy ∈ C : 0 ≤ x ≤ 1}. If f : S → X is holomorphic in the interior of S, continuous and bounded in S, then for each ζ ∈ S f (ζ) X ≤ max{sup f (it) X , sup f (1 + it) X }.

### An Introduction to Interpolation Theory by Alessandra Lunardi

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