1470 CHAPTER 44. TRACE SPACES

Proof: Let 0 < s < t. Let ν + 1p = θ . Then for a generic g,

∫∞

0||τν g(τ)||p dτ =

∫∞

0

∣∣∣∣∣∣τθ g(τ)∣∣∣∣∣∣p dτ

τ

so that tν f ′ ∈ Lp (0,∞;A1) , the measure in this case being usual Lebesgue measure. Then

f (t)− f (s) =∫ t

sf ′ (τ)dτ =

∫ t

sτ

ν f ′ (τ)τ−ν dτ.

For 1p +

1p′ = 1, ν p′ =

(θ − 1

p

)p′ < 1 because θ < 1 = 1

p′ +1p . Therefore,

|| f (t)− f (s)||A0+A1

≤∫ t

s

∣∣∣∣ f ′ (τ)∣∣∣∣A0+A1dτ

≤∫ t

s

∣∣∣∣ f ′ (τ)∣∣∣∣A1dτ =

∫ t

s

∣∣∣∣τν f ′ (τ)∣∣∣∣

A1τ−ν dτ

≤(∫ t

s

∣∣∣∣τν f ′ (τ)∣∣∣∣p

A1dτ

)1/p(∫ t

sτ−ν p′dτ

)1/p′

≤ || f ||W

(t1−ν p′

1−ν p′− s1−ν p′

1−ν p′

)(44.1.3)

≤ || f ||Wt1−ν p′

1−ν p′.

which converges to 0 as t→ 0. This shows that limt→0+ f (t) exists in A0 +A1.Clearly Z is a subspace. Let fn→ f in W and suppose fn ∈ Z. Then since f ∈W, 44.1.3

implies f is continuous. Using 44.1.3 and replacing f with fn− fm and then taking a limitas s→ 0,

|| fn (t)− fm (t)||A0+A1≤ || fn− fm||W Cν t1−ν p′

Taking a subsequence, it can be assumed fn (t) converges to f (t) a.e. But the above inequal-ity shows that fn (t) is a Cauchy sequence in C ([0,β ] ;A0 +A1) for all β < ∞. Therefore,fn (t)→ f (t) for all t. Also,

|| fn (t)||A0+A1≤Cν || fn||W t1−ν p′ ≤ Kt1−ν p′

for some K depending on max{|| fn|| : n≥ 1} and so

|| f (t)||A0+A1≤ Kt1−ν p′

which implies f (0) = 0. Thus Z is closed.Consider the last claim. For a generic tθ g ∈ Lp

(0,∞, dt

t ;A), changing variables t = eτ ,∫

∞

0tθ p |g(t)|p dt

t=∫

∞

−∞

eτθ p |g(eτ)|p dτ