43.8. DUALITY AND INTERPOLATION 1465
Now ∣∣∣∣∣∣∣∣∣∣∣∣ bi2iθ
max(||bi||A0
,2i ||bi||A1
)∣∣∣∣∣∣∣∣∣∣∣∣A0+A1
≤{
2iθ if i < 02−i(1−θ) if i≥ 0
. (43.8.63)
This is fairly routine to verify. Consider the case where i≥ 0. Then∣∣∣∣∣∣∣∣∣∣∣∣ bi2iθ
max(||bi||A0
,2i ||bi||A1
)∣∣∣∣∣∣∣∣∣∣∣∣A0+A1
≤
∣∣∣∣∣∣∣∣∣∣ bi2iθ
2i ||bi||A1
∣∣∣∣∣∣∣∣∣∣A0+A1
≤ 2−i(1−θ)
because ||bi||A1≥ ||bi||A0+A1
. Therefore,
M
∑i=0
∣∣∣∣∣∣∣∣∣∣∣∣ bi2iθ
max(||bi||A0
,2i ||bi||A1
)2−iθα i
∣∣∣∣∣∣∣∣∣∣∣∣A0+A1
≤
M
∑i=0
2−i(1−θ)2−iθα i ≤
(∞
∑i=0
2−i(1−θ)q′)1/q′(
∞
∑i=0
2−iqθα
qi
)1/q
< ∞
and similarly,0
∑i=−∞
∣∣∣∣∣∣∣∣∣∣∣∣ bi2iθ
max(||bi||A0
,2i ||bi||A1
)2−iθα i
∣∣∣∣∣∣∣∣∣∣∣∣A0+A1
converges. Therefore, a∞ makes sense in A0 +A1 and also from 43.8.63, we see that{||bi||A0+A1
2iθ
J (2i,bi)
}∈ λ
(1−θ)q′
Now letu(t)≡ α ibi
J (2i,bi) ln2on [2i−1,2i).
Then ∫∞
0u(t)
dtt
= ∑i
∫ 2i
2i−1
α ibi
J (2i,bi) ln2dtt
= ∑i
α ibi
J (2i,bi)= a∞.
Also ∫∞
0
(t−θ J (t,u(t))
)q dtt≤∑
i
∫ 2i
2i−1
(2(1−i)θ J
(2i,u
(2i−1))) dt
t
≤∑i
[2−(i−1)θ J
(2i,u
(2i−1))]q
ln2