43.8. DUALITY AND INTERPOLATION 1461

This proves the first part of the lemma.Claim: With this definition of the norm in 43.8.54, the operator norm of (a′0,a

′1) ∈

(A0×A1)′ = A′0×A′1 is ∣∣∣∣(a′0,a′1)∣∣∣∣(A0×A1)

′ =∣∣∣∣a′0∣∣∣∣A′0 + t

∣∣∣∣a′1∣∣∣∣A′1 . (43.8.55)

Proof of the claim:∣∣(a′0,a′1)(a0,a1)

∣∣ ≤ ∣∣∣∣a′0∣∣∣∣ ||a0||+ ||a′1|| ||a1|| . Now suppose that||a0||= max

(||a0|| , t−1 ||a1||

). Then this is no larger than(∣∣∣∣a′0∣∣∣∣+ t

∣∣∣∣a′1∣∣∣∣) ||a0||=(∣∣∣∣a′0∣∣∣∣+ t

∣∣∣∣a′1∣∣∣∣)max(||a0|| , t−1 ||a1||

).

The other case is that t−1 ||a1||= max(||a0|| , t−1 ||a1||

). In this case,∣∣(a′0,a′1)(a0,a1)

∣∣ ≤ ∣∣∣∣a′0∣∣∣∣ ||a0||+∣∣∣∣a′1∣∣∣∣ ||a1||

≤∣∣∣∣a′0∣∣∣∣ t−1 ||a1||+

∣∣∣∣a′1∣∣∣∣ ||a1||=

(∣∣∣∣a′0∣∣∣∣+ t∣∣∣∣a′1∣∣∣∣) t−1 ||a1||

=(∣∣∣∣a′0∣∣∣∣+ t

∣∣∣∣a′1∣∣∣∣)max(||a0|| , t−1 ||a1||

).

This shows∣∣∣∣(a′0,a′1)∣∣∣∣(A0×A1)

′ ≤(∣∣∣∣a′0∣∣∣∣+ t ||a′1||

). Is equality achieved? Let a0n and a1n

be points of A0 and A1 respectively such that ||a0n|| , ||a1n|| ≤ 1 and limn→∞ a′i (ain) = ||a′i|| .Then (

a′0,a′1)(a0n, ta1n)→

∣∣∣∣a′0∣∣∣∣+ t∣∣∣∣a′1∣∣∣∣

and also, ||(a0n, ta1n)||A0×A1= max

(||a0n|| , t−1t ||a1n||A1

)≤ 1. Therefore, equality is in-

deed achieved and this proves the claim.Consider 43.8.52. Take a′ ∈ A′0 +A′1 = (A0∩A1)

′ and let

E ≡ {(a,a) ∈ A0×A1 : a ∈ A0∩A1} .

Now define a linear map, λ on E as before.

λ ((a,a))≡ a′ (a) .

If a′ = ã′0 + ã′1,

|λ ((a,a))| ≤∣∣∣∣ã′0∣∣∣∣A′0 ||a||A0

+∣∣∣∣ã′1∣∣∣∣A′1 ||a||A1

=∣∣∣∣ã′0∣∣∣∣A′0 ||a||A0

+ t∣∣∣∣ã′1∣∣∣∣A′1 t−1 ||a||A1

≤(∣∣∣∣ã′0∣∣∣∣+ t

∣∣∣∣ã′1∣∣∣∣) ||(a,a)||A0×A1

so λ is continuous on the subspace, E of A0×A1 and

||λ ||E ′ ≤∣∣∣∣ã′0∣∣∣∣+ t

∣∣∣∣ã′1∣∣∣∣ . (43.8.56)

By the Hahn Banach theorem, there exists an extension of λ defined on all of A0×A1 withthe same norm. Thus, from 43.8.55, there exists (a′0,a

′1)∈ (A0×A1)

′ which is an extensionof λ such that ∣∣∣∣(a′0,a′1)∣∣∣∣(A0×A1)

′ =∣∣∣∣a′0∣∣∣∣A′0 + t

∣∣∣∣a′1∣∣∣∣A′1 = ||λ ||E ′