45.3. INTRINSIC NORMS 1495

Now ∫∞

0X[0,1] (t) t pθ ||u||pA1

dt/t ≤C (p,θ) ||u||pT0

and using Lemma 45.3.2,∫∞

0X[0,1] (t)

∣∣∣∣ 1t2

∫ t

0||G(τ)u−u||dτ

∣∣∣∣p t pθ dtt

≤∫

∞

0

∣∣∣∣1t∫ t

0||G(τ)u−u||dτ

∣∣∣∣p t p(θ−1) dtt

≤ 1(1− (θ −1))p

∫∞

0||G(τ)u−u||p t p(θ−1) dt

t

=1

(2−θ)p

∫∞

0

∣∣∣∣∣∣∣∣G(τ)u−ut

∣∣∣∣∣∣∣∣p t pθ dtt≤C (θ , p) ||u||pT0

This proves the theorem.Of course the case of most interest here is where A1 = Lp (Rn) and

G(t)u(x)≡ u(x+ tei)

Thus Λu = ∂u/∂xi, the weak derivative. The trace space T (D(Λ) ,Lp (Rn) , p,1−θ) thenis a space of functions in Lp (Rn) which have a fractional order partial derivative withrespect to xi.

Recall from Definition 45.1.1 that for θ ∈ (0,1) ,

W θ ,p (Rn)≡ T(W 1,p (Rn) ,Lp (Rn) , p,1−θ

)Let f ∈W

(W 1,p (Rn) ,Lp (Rn) , p,1−θ

). Then

|| f ||W ≡max(∫

∞

0t(1−θ)p || f (t)||pW 1,p

dtt,∫

∞

0t(1−θ)p ∣∣∣∣ f ′ (t)∣∣∣∣pLp

dtt

)Letting Gi (t)u(x)≡ u(x+ tei) and Λi its generator,

W 1,p (Ω) = ∩ni=1DΛi∩Lp (Rn)

with the norm given by

||u||p = ||u||pLp +n

∑i=1||Λiu||pLp

which is equivalent to the norm

||u||p =n

∑i=1||u||pD(Λi)

.

Then by considering each of the Gi and repeating the above argument in Theorem 45.3.6,it follows an equivalent intrinsic norm is

||u||pW θ ,p(Rn)

= ||u||pLp(Rn)+

n

∑i=1

∫∞

0t(1−θ)p

∣∣∣∣∣∣∣∣Gi (t)u−ut

∣∣∣∣∣∣∣∣pLp

dtt