B.1. FINITE p VARIATION 375

Note that when p = 1 having finite p variation is just the same as saying that it has finitetotal variation. Thus this is including more general considerations. Also, to simplify thenotation, for P such a dissection, write

∑P

|F (ti+1)−F (ti)|p instead ofn

∑i=1|F (ti+1)−F (ti)|p

Definition B.1.2 Let Cα ([0,T ] ;R) denote the α Holder functions and denote byV p ([0,T ] ,R) the continuous functions F which have finite p variation, V p for short.

It is routine to verify that if α > 1, then any Holder continuous function is a constant. Itis also easy to see that any 1/p Holder is p finite variation. To see this, note that you have|F (t)−F (s)| ≤C |t− s|1/p and so(

m

∑i=1|F (ti+1)−F (ti)|p

)1/p

≤

(m

∑i=1

Cp |ti+1− ti|)1/p

=CT 1/p

From now on p ≥ 1 and define ∥F∥V p([0,T ],R) ≡ ∥F∥p,[0,T ]+ supt∈[0,T ] |F (t)| where asdescribed above,

∥F∥p,[0,T ] ≡ supP

(∑

i|F (ti+1)−F (ti)|p

)1/p

=

(supP

∑i|F (ti+1)−F (ti)|p

)1/p

To save notation, it is customary to write ∥F∥∞= supt∈[0,T ] |F (t)|

Definition B.1.3 Suppose you have a set of functions V defined on some interval Iwhich satisfies cX ∈ V whenever c is a number and X ∈ V and that X +Y ∈ V wheneverX ,Y ∈V. Then ∥·∥ : V → [0,∞) is a norm if it satisfies.

∥X∥ ≥ 0,∥X∥= 0 if and only if X = 0

∥X +Y∥ ≤ ∥X∥+∥Y∥

For c a number, ∥cX∥= |c|∥X∥

Proposition B.1.4 For each p≥ 1,V p is a set of continuous functions. Also ∥·∥V p is anorm. In addition, if 1≤ p < q,

V p ([0,T ] ;R)⊆V q ([0,T ] ;R)⊆C0 ([0,T ] ;R)

The embeddings are continuous. In fact, ∥F∥∞≤ ∥F∥V q ≤ ∥F∥V p .

Note that, although ∥·∥V p is a norm, ∥·∥p,[0,T ] is not.Proof: It is clear that ∥F∥V p equals 0 if and only if F = 0. This follows from the

inclusion in the definition for the norm, supt∈[0,T ] |F (t)|. It only remains to verify the otheraxioms of a norm. It suffices to consider ∥·∥p,[0,T ]. Does it satisfy the triangle inequality?

∥Z +Y∥p,[0,T ] ≡ supP

(∑

i|(Z +Y )(ti+1)− (Z +Y )(ti)|p

)1/p