1458 CHAPTER 43. INTERPOLATION IN BANACH SPACE
Now
||a||qθ ,q,J ≤
∫∞
0
(t−θ J (t,u(t))
)q dtt
=∞
∑i=−∞
∫ 2i
2i−1
(t−θ J
(t,
ui
ln2
))q dtt
≤(
1ln2
)q ∞
∑i=−∞
∫ 2i
2i−1
(t−θ J
(2i,ui
))q dtt
≤(
1ln2
)q ∞
∑i=−∞
∫ 2i
2i−1
(t−θ 3(1+ ε)K
(2i,a
))q dtt
Using the above claim,K(2i,a)
2i ≤ K(2i−1,a)2i−1 and so K
(2i,a
)≤ 2K
(2i−1,a
). Therefore, the
above is no larger than
≤ 2(
1ln2
)q ∞
∑i=−∞
∫ 2i
2i−1
(t−θ 3(1+ ε)K
(2i−1,a
))q dtt
≤ 2(
1ln2
)q ∞
∑i=−∞
∫ 2i
2i−1
(t−θ 3(1+ ε)K (t,a)
)q dtt
= 2(
3(1+ ε)
ln2
)q ∫ ∞
0
(t−θ K (t,a)
)q dtt≡ 2
(3(1+ ε)
ln2
)q
||a||qθ ,q . (43.7.45)
This has shown that if a ∈ (A0,A1)θ ,q , then by 43.7.44 and 43.7.45, a ∈ (A0,A1)θ ,q,J and
||a||qθ ,q,J ≤ 2
(3(1+ ε)
ln2
)q
||a||qθ ,q . (43.7.46)
It remains to prove the other inclusion and norm inequality, both of which are mucheasier to obtain. Thus, let a ∈ (A0,A1)θ ,q,J with
a =∫
∞
0u(t)
dtt
(43.7.47)
where u is a strongly measurable function having values in A0∩A1 and for which∫∞
0
(t−θ J (t,u(t))
)qdt < ∞. (43.7.48)
K (t,a) = K(
t,∫
∞
0u(s)
dss
)≤∫
∞
0K (t,u(s))
dss. (43.7.49)
Now by Lemma 43.7.4, this is dominated by an expression of the form
≤∫
∞
0min
(1,
ts
)J (s,u(s))
dss
=∫
∞
0min
(1,
1s
)J (ts,u(ts))
dss
(43.7.50)