1460 CHAPTER 43. INTERPOLATION IN BANACH SPACE

As mentioned above, A′0+A′1 ⊆ (A0∩A1)′. In fact these two are equal. This is the first part

of the following lemma.

Lemma 43.8.1 Suppose A0∩A1 is dense in Ai, i = 0,1. Then

A′0 +A′1 = (A0∩A1)′ , (43.8.51)

and for a′ ∈ A′0 +A′1 = (A0∩A1)′ ,

K(t,a′)= sup

a∈A0∩A1

|a′ (a)|J (t−1,a)

. (43.8.52)

Thus K (t, ·) is an equivalent norm to the usual operator norm on (A0∩A1)′ taken with

respect to J(t−1, ·

). If, in addition to this, Ai is reflexive, then for a′ ∈ A′0 ∩A′1, and a ∈

A0∩A1,J(t,a′)

K(t−1,a

)≥∣∣a′ (a)∣∣ . (43.8.53)

Proof: First consider the claim that A′0 +A′1 = (A0∩A1)′. As noted above, ⊆ is clear.

Define a norm on A0×A1 as follows.

||(a0,a1)||A0×A1≡max

(||a0||A0

, t−1 ||a1||A1

). (43.8.54)

Let a′ ∈ (A0∩A1)′. Let

E ≡ {(a,a) : a ∈ A0∩A1}

with the norm J(t−1,a

)≡ max

(||a||A0

, t−1 ||a||A1

). Now define λ on E, the subspace of

A0×A1 byλ ((a,a))≡ a′ (a) .

Thus λ is a continuous linear map on E and in fact,

|λ ((a,a))|=∣∣a′ (a)∣∣≤ ∣∣∣∣a′∣∣∣∣J (t−1,a

).

By the Hahn Banach theorem there exists an extension of λ to all of A0×A1. This extensionis of the form (a′0,a

′1) ∈ A′0×A′1. Thus(

a′0,a′1)((a,a)) = a′0 (a)+a′1 (a) = a′ (a)

and therefore, a′0 +a′1 = a′ provided a′0 +a′1 is continuous. But∣∣(a′0 +a′1)(a)∣∣ =

∣∣a′0 (a)+a′1 (a)∣∣≤ ∣∣a′0 (a)∣∣+ ∣∣a′1 (a)∣∣

≤∣∣∣∣a′0∣∣∣∣ ||a||A0

+∣∣∣∣a′1∣∣∣∣ ||a||A1

≤∣∣∣∣a′0∣∣∣∣ ||a||A0

+ t∣∣∣∣a′1∣∣∣∣ t−1 ||a||A1

≤(∣∣∣∣a′0∣∣∣∣+ t

∣∣∣∣a′1∣∣∣∣)J(t−1,a

)which shows that a′0 +a′1 is continuous and in fact∣∣∣∣a′0 +a′1

∣∣∣∣(A0∩A1)

′ ≤(∣∣∣∣a′0∣∣∣∣+ t

∣∣∣∣a′1∣∣∣∣) .