1496 CHAPTER 45. TRACES OF SOBOLEV SPACES

= ||u||pLp(Rn)+

n

∑i=1

∫∞

0t(1−θ)p

∣∣∣∣∣∣∣∣u(·+ tei)−u(·)t

∣∣∣∣∣∣∣∣pLp

dtt

(45.3.13)

and u ∈W θ ,p (Rn) when this norm is finite. The only new detail is that in showing that foru ∈ T0 it follows it is in T, you use the function

f (t)≡ φ (t)1tn

∫ t

0· · ·∫ t

0G1 (τ1)G2 (τ2) · · ·Gn (τn)udτ1 · · ·dτn

and the fact that these semigroups commute. To get this started, note that

g(t)≡∫ t

0· · ·∫ t

0G1 (τ1)G2 (τ2) · · ·Gn (τn)udτ1 · · ·dτn ∈ D(Λi)

for each i. This follows from writing it as∫ t

0Gi (τ i)(wi)dτ i

for wi ∈ Lp coming from the other integrals and then repeating the earlier argument to get

Λig(t) = Gi (t)wi−wi

and then ∫∞

0t p(1−θ) ||Λi f ||pLp

dtt

≤∫

∞

0t p(1−θ)

∣∣∣∣∣∣∣∣Gi (t)wi−wi

t

∣∣∣∣∣∣∣∣pLp

dtt

≤ C∫

∞

0t p(1−θ)

∣∣∣∣∣∣∣∣Gi (t)u−ut

∣∣∣∣∣∣∣∣pLp

dtt≤C ||u||pT0

Thus all is well as far as f is concerned and the proof will work as it did earlier in Theorem45.3.6. What about f ′? As before, the only term which is problematic is

φ (t)(

1tn

∫ t

0· · ·∫ t

0G1 (τ1)G2 (τ2) · · ·Gn (τn)udτ1 · · ·dτn

)′After enough massaging, it becomes

n

∑i=1

∏j ̸=i

1t

∫ t

0G j (τ j)dτ j

1t2

∫ t

0(Gi (t)u−Gi (τ i)u)dτ i

where the operator ∑ni=1 ∏ j ̸=i

1t∫ t

0 G j (τ j)dτ j is bounded. Thus similar arguments to thoseof Theorem 45.3.6 will work, the only difference being a sum.