43.7. THE J METHOD 1457

Proof: Begin with a ∈ (A0,A1)θ ,q . Thus

||a||qθ ,q =

∫∞

0

(t−θ K (t,a)

)q dtt< ∞ (43.7.43)

and it is necessary to produce u(t) as described above,

a =∫

∞

0u(t)

dtt

where∫

∞

0

(t−θ J (t,u(t))

)q dtt< ∞.

From 43.7.43, limt→0+ K (t,a) = 0 since t→ K (t,a) is nondecreasing and so if its limit ispositive, the integrand would have a non integrable singularity like t−θq−1. Next considerwhat happens to K(t,a)

t as t→ ∞.

Claim: t→ K(t,a)t is decreasing.

Proof of the claim: Choose a0 ∈ A0 and a1 ∈ A1 such that a0 +a1 = a and

K (t,a)+ εt > ||a0||A0+ t ||a1||A1

let s > t. Then

K (t,a)+ tεt

≥||a0||A0

+ t ||a1||A1

t≥||a0||A0

+ s ||a1||A1

s≥ K (s,a)

s.

Since ε is arbitrary, this proves the claim.Let r ≡ limt→∞

K(t,a)t . Is r = 0? Suppose to the contrary that r > 0. Then the integrand

of 43.7.43, is at least as large as

t−θqK (t,a)q−1 K (t,a)t≥ t−θqK (t,a)q−1 r

≥ t−θq (tr)q−1 r ≥ rqtq(1−θ)−1

whose integral is infinite. Therefore, r = 0.Lemma 43.7.3, implies there exist ui ∈ A0∩A1 such that a = ∑

∞i=−∞ ui, the convergence

taking place in A0 +A1with the inequality of that Lemma holding,

J(2i,ui

)≤ 3(1+ ε)K

(2i,a

).

For i an integer and t ∈ [2i−1,2i), let

u(t)≡ ui/ ln2.

Then

a =∞

∑i=−∞

ui =∫

∞

0u(t)

dtt. (43.7.44)