By Alessandra Lunardi

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**Example text**

3) 40 Chapter 2 Proof — First let T L(X1 ,X2 ) = 0 and T be such that f (θ) = a. Set T T g(z) = = 0. If a ∈ [X1 , Y1 ]θ , let f ∈ F(X1 , Y1 ) L(Y1 ,Y2 ) L(X1 ,X2 ) z−θ T f (z), z ∈ S. L(Y1 ,Y2 ) Then g ∈ F(X2 , Y2 ), and g(it) X2 ≤( T 1−θ ( L(X1 ,X2 ) ) T θ L(Y1 ,Y2 ) ) f (it) g(1 + it) Y2 ≤( T 1−θ ( L(X1 ,X2 ) ) T θ L(Y1 ,Y2 ) ) f (1 + it) so that g F (X2 ,Y2 ) ≤ ( T [X2 , Y2 ]θ , and Ta [X2 ,Y2 ]θ ≤ g 1−θ ( L(X1 ,X2 ) ) ≤( T F (X2 ,Y2 ) θ L(Y1 ,Y2 ) ) T f F (X1 ,Y1 ) . 1−θ ( L(X1 ,X2 ) ) T X1 , Y1 , Therefore T a = g(θ) ∈ θ L(Y1 ,Y2 ) ) f F (X1 ,Y1 ) .

Since the set of such a’s is dense in Lpθ (Ω) the statement follows. 3 Exercises 1) The maximum principle for functions defined on a strip. Let f : S → X be holomorphic in the interior of S, continuous and bounded in S. Prove that for each ζ ∈ S f (ζ) ≤ max{sup f (it) , sup f (1 + it) }. t∈R t∈R (Hint: for each ε ∈ (0, 1) let z0 be such that f (z0 ) ≥ f ∞ (1 − ε); consider the functions fδ (z) = exp(δ(z − z0 )2 )f (z), and apply the maximum principle in the rectangle [0, 1] × [−M, M ] with M large).

1). This section is devoted to the study of the real interpolation spaces (X, D(A))θ,p , and, more generally, (X, D(Am ))θ,p . Since for every t, ω ∈ R the graph norm of D(Am ) is equivalent to the graph norm of D(B m ) with B = eit (A + ωI), the case of an operator B satisfying ρ(B) ⊃ {λeiθ : λ > λ0 }, ∃M : λR(λeiθ , B) L(X) ≤ M, λ > λ0 for some θ ∈ [0, 2π), λ0 ≥ 0, may be easily reduced to this one. The halfline r = {λeiθ : λ > λ0 } is said to be a ray of minimal growth of the resolvent of B.

### An Introduction to Interpolation Theory by Alessandra Lunardi

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