1474 CHAPTER 44. TRACE SPACES

Then as before,∣∣∣∣∣∣tθ fλ

∣∣∣∣∣∣Lp(0,∞, dt

t ;A0)= λ

−θ R,∣∣∣∣∣∣tθ ( fλ )

′∣∣∣∣∣∣

Lp(0,∞, dtt ;A1)

= λ1−θ S. (44.1.9)

so that || f ||W = max(R,S) . Then, changing the variables, letting λ = R/S,∣∣∣∣∣∣tθ fλ

∣∣∣∣∣∣Lp(0,∞, dt

t ;A0)=∣∣∣∣∣∣tθ ( fλ )

′∣∣∣∣∣∣

Lp(0,∞, dtt ;A1)

= R1−θ Sθ (44.1.10)

Since fλ (0) = a, fλ ∈W, and it is always the case that for positive R,S,

R1−θ Sθ ≤max(R,S) ,

this shows that

||a||T ≤ max(∣∣∣∣∣∣tθ fλ

∣∣∣∣∣∣Lp(0,∞, dt

t ;A0),∣∣∣∣∣∣tθ ( fλ )

′∣∣∣∣∣∣

Lp(0,∞, dtt ;A1)

)= R1−θ Sθ ≤max(R,S) = || f ||W < ||a||T + ε,

the first inequality holding because ||a||T is the infimum of such things on the right. Thisshows 44.1.6.

It remains to verify 44.1.7. To do this, let ψ ∈C∞ ([0,∞)) , with ψ (0) = 1 and ψ (t) = 0for all t > 1. Then consider the special f ∈W which is given by f (t) ≡ aψ (t) wherea ∈ A0 ∩A1. Thus f ∈W and f (0) = a so a ∈ T (A0,A1, p,θ) . From the first part, thereexists a constant, K such that

||a||T ≤∣∣∣∣∣∣tθ f

∣∣∣∣∣∣1−θ

Lp(0,∞, dtt ;A0)

∣∣∣∣∣∣tθ f ′∣∣∣∣∣∣θ

Lp(0,∞, dtt ;A1)

≤ K ||a||1−θ

A0||a||θA1

This shows 44.1.7 and the first inclusion in 44.1.8. From the inequality just obtained,

||a||T ≤ K((1−θ) ||a||A0

+θ ||a||A1

)≤ K ||a||A0∩A1

.

This shows the first inclusion map of 44.1.8 is continuous.Now take a ∈ T. Let f ∈W be such that a = f (0) and

||a||T + ε > || f ||W ≥ ||a||T .

By 44.1.3,||a− f (t)||A0+A1

≤Cν t1−ν p′ || f ||W

where 1p +ν = θ , and so

||a||A0+A1≤ || f (t)||A0+A1

+Cν t1−ν p′ || f ||W .