42.8 Duality And Interpolation
In this section it will be assumed that A0 ∩A1 is dense in Ai for i = 0,1. This is done so
and the inclusion map is continuous. Thus it makes sense to add
something in A0′
to something in A1′.
What is the dual space of
? The answer is based on the following lemma,
. Remember that
and this is a norm on A0 ∩ A1 and
As mentioned above, A0′ + A1′⊆
. In fact these two are equal. This is the first
part of the following lemma.
Lemma 42.8.1 Suppose A0 ∩ A1 is dense in Ai, i = 0,1. Then
and for a′∈ A0′ + A1′ =
is an equivalent norm to the usual operator norm on
′ taken with
respect to J
. If, in addition to this, Ai is reflexive, then for a′∈ A0′∩ A1′, and
a ∈ A0 ∩ A1,
Proof: First consider the claim that A0′ + A1′ =
. As noted above, ⊆
clear. Define a norm on A0 × A1
with the norm J
. Now define
, the subspace of
A0 × A1
Thus λ is a continuous linear map on E and in fact,
By the Hahn Banach theorem there exists an extension of λ to all of A0 × A1. This
extension is of the form
∈ A0′× A1′.
and therefore, a0′ + a1′ = a′ provided a0′ + a1′ is continuous. But
which shows that a0′
is continuous and in fact
This proves the first part of the lemma.
Claim: With this definition of the norm in 42.8.54, the operator norm of
= A0′× A1′
Proof of the claim:
Now suppose that
Then this is no larger than
The other case is that t−1
In this case,
Is equality achieved? Let a0n
be points of A0
respectively such that
Therefore, equality is
indeed achieved and this proves the claim.
Consider 42.8.52. Take a′∈ A0′ + A1′ =
Now define a linear map, λ on E as before.
If a′ =
is continuous on the subspace, E
of A0 × A1
By the Hahn Banach theorem, there exists an extension of λ defined on all of A0 × A1
with the same norm. Thus, from 42.8.55, there exists
which is an
extension of λ
and for all a ∈ A0 ∩ A1,
It follows that a0′ + a1′ = a′ in
. Therefore, from 42.8.56
because on E, J
which proves 42.8.52
To obtain 42.8.53 in the case that Ai is reflexive, apply 42.8.52 to the case where Ai′′
plays the role of Ai in 42.8.52. Thus, for a′′∈ A0′′ + A1′′,
Now a′′ = a1′′ + a0′′ = η1a1 + η0a0 where ηi is the map from Ai to Ai′′ which is onto and
preserves norms, given by ηa
Therefore, letting a1
Changing t → t−1,
which proves the lemma.
Definition 42.8.2 Let q ≥ 1. Then λθ,q will denote the sequences,
For α ∈ λθ,q,
Thus α ∈ λθ,q means
Lemma 42.8.3 Let f
0, and let f
αi for t ∈
) where α ∈ λθ,q. Then
there exists a constant, C, such that
Also, if whenever α ∈ λθ,q, and αi ≥ 0 for all i,
Proof: Consider 42.8.59.
42.8.61 is next. By 42.8.60, whenever α ∈ λθ,q,
It follows from the Riesz representation theorem that
This proves the lemma.
The dual space of
is discussed next.
Lemma 42.8.4 Let θ ∈
and let q ≥
and the inclusion map is continuous.
Proof: Let a′∈
then a has a representation of the form
∈ A0 ∩ A1.
by Lemma 42.7.2. Also