43.1. SOME STANDARD TECHNIQUES IN EVOLUTION EQUATIONS 1415

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 43.1.5 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 43.1.6 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 43.1.1, 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 43.1.7 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.

In particular, you can define f (a) to be the initial value of this continuous function. Itturns out that an identical theorem holds in this case. To begin with here is the same sortof lemma which was used earlier for the case of scalar valued functions. It says that iff ′ = 0 where the derivative is taken in the sense of X valued distributions, then f equals aconstant.

Lemma 43.1.8 Suppose f ∈ L1 (a,b;X) and for all φ ∈C∞c (a,b) ,∫ b

af (t)φ

′ (t)dt = 0.

Then there exists a constant, a ∈ X such that f (t) = a a.e.