45.3. INTRINSIC NORMS 1489

and so the left side equals ∫∞

−∞

eτ p(α−1)∣∣∣∣∫ τ

−∞

f (eσ )eσ dσ

∣∣∣∣p dτ

=∫

∞

−∞

∣∣∣∣∫ τ

−∞

eτα f (eσ )e−(τ−σ)dσ

∣∣∣∣p dτ

=∫

∞

−∞

∣∣∣∣∫ τ

−∞

e(τ−σ)α e−(τ−σ)eσα f (eσ )dσ

∣∣∣∣p dτ

=∫

∞

−∞

∣∣∣∣∫ ∞

−∞

X(−∞,0) (τ−σ)e(τ−σ)(α−1)eσα f (eσ )dσ

∣∣∣∣p dτ

and by Lemma 45.3.1,

≤(∫ 0

−∞

e(α−1)udu)p ∫ ∞

−∞

epσα | f (eσ )|p dσ

=

(1

1−α

)p ∫ ∞

−∞

epσα | f (eσ )|p dσ

which was to be shown. This proves the lemma.Next consider the case where G(t) , t > 0 is a continuous semigroup on A1 and A0 ≡

D(Λ) where Λ is the generator of this semigroup. Recall that from Proposition 19.14.5 onPage 577 Λ is a closed densely defined operator and so A0 is a Banach space if the norm isgiven by

||u||A0≡ ||u||A1

+ ||Λu||A1

Also assume ||G(t)|| is uniformly bounded for t ∈ [0,∞). I have in mind the case whereA1 = Lp (Rn) and G(t)u(x) = u(x+ tei) but it is notationally easier to discuss this in thegeneral case. First here is a simple lemma.

Lemma 45.3.3 Let A0 = D(Λ) as just described. Then for u ∈ A1

||u||A1+A0= ||u||A1

Proof: D(Λ)⊆ A1. Now let u ∈ A1.

||u||A0+A1≡ inf

{||u0||A1

+ ||Λu0||A1+ ||u1||A1

: u = u0 +u1

}To make this as small as possible you should clearly take u1 = u because

||u0||A1+ ||Λu0||A1

+ ||u1||A1≥ ||u0 +u1||A1

+ ||Λu0||= ||u||A1

+ ||Λu0||

Therefore, the result of the lemma follows.