69.2. STANDARD TECHNIQUES IN EVOLUTION EQUATIONS 2361

Lemma 69.2.4 Let f be as defined in Definition 69.2.2. Then for f ∈ Lp (a,b;X) for p ∈[1,∞),

limδ→0

∫ b

a

∣∣∣∣ f (t−δ )− f (t)∣∣∣∣p

X dt = 0.

Proof: Regarding the measure space as (a,b) with Lebesgue measure, by regularity ofthe measure, there exists g∈Cc (a,b;X) such that || f −g||p < ε. Here the norm is the normin Lp (a,b;X) . Therefore,

|| fh− f ||p ≤ || fh−gh||p + ||gh−g||p + ||g− f ||p≤

(21/p +1

)|| f −g||p + ||gh−g||p

<(

21/p +1)

ε + ε

whenever h is sufficiently small. This is because of the uniform continuity of g. Therefore,since ε > 0 is arbitrary, this proves the lemma.

Definition 69.2.5 Let f ∈ L1 (a,b;X) . Then the distributional derivative in the sense of Xvalued distributions is given by

f ′ (φ)≡−∫ b

af (t)φ

′ (t)dt

Then f ′ ∈ L1 (a,b;X) if there exists h ∈ L1 (a,b;X) such that for all φ ∈C∞c (a,b) ,

f ′ (φ) =∫ b

ah(t)φ (t)dt.

Then f ′ is defined to equal h. Here f and f ′ are considered as vector valued distributionsin the same way as was done for scalar valued functions.

Lemma 69.2.6 The above definition is well defined.

Proof: Suppose both h and g work in the definition. Then for all φ ∈C∞c (a,b) ,∫ b

a(h(t)−g(t))φ (t)dt = 0.

Therefore, by Lemma 69.2.1, h(t)−g(t) = 0 a.e.The other thing to notice about this is the following lemma. It follows immediately

from the definition.

Lemma 69.2.7 Suppose f , f ′ ∈ L1 (a,b;X) . Then if [c,d]⊆ [a,b], it follows that(

f |[c,d])′=

f ′|[c,d]. This notation means the restriction to [c,d] .

Recall that in the case of scalar valued functions, if you had both f and its weak deriva-tive, f ′ in L1 (a,b) , then you were able to conclude that f is almost everywhere equal to acontinuous function, still denoted by f and

f (t) = f (a)+∫ t

af ′ (s)ds.