Instead of integrating with respect to a finite variation integrator function F, the function will
be of finite p variation. This is more general than finite variation.

Definition 10.1.1Define for a function F :

[0,T]

→ ℝ

α Holder continuous if

sup |F-(t)−-F-(s)| < C < ∞
0≤s<t≤T |t− s|α

Thus

|F (t) − F (s)| ≤ C|t− s|α

Finite p variation if for some p > 0,

( )1 ∕p
∑m p
∥F∥p,[0,T] ≡ suPp |F (ti+1)− F (ti)| < ∞
i=1

where P denotes a partition of

[0,T]

,P =

{t0,t1,⋅⋅⋅,tn}

for

0 = t0 < t1 < ⋅⋅⋅ < tn = T

also called a dissection.

|P|

denotes the largest length in any of the sub intervals. It willbe always assumed that actually p ≥ 1.

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

n
∑ |F (t )− F (t )|p instead of ∑ |F (t )− F (t)|p
P i+1 i i=1 i+1 i

Definition 10.1.2We use C^{α}

([0,T];ℝ)

to denote the α Holder functions andV^{p}

([0,T],ℝ)

to denote the continuous functions F which have finite p variation.

It is routine to verify that if α > 1, then any Holder continuous function is a constant. It is
also easy to see that any 1∕p Holder is p finite variation. To see this, note that you
have

1∕p
|F (t)− F (s)| ≤ C |t− s|

and so

( )1∕p ( )1 ∕p
∑m p m∑ p 1∕p
|F (ti+1)− F (ti)| ≤ C |ti+1 − ti| = CT
i=1 i=1

(∑ )1∕p ( ∑ )1 ∕p
∥F ∥p,[0,T ] ≡ sup |F (ti+1)− F (ti)|p = sup |F (ti+1) − F (ti)|p
P i P i

To save notation, it is customary to write

∥F∥∞ = sup |F (t)|
t∈[0,T]

Definition 10.1.3Suppose you have a set of functions V defined on someinterval I which satisfies cF ∈ V whenever c is a number and X ∈ V and thatX + Y ∈ V whenever X,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 10.1.4For each p ≥ 1,V^{p}is a set of continuous functions. Also

∥⋅∥

_{V p}is anorm. In addition, if 1 ≤ p < q,

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

The embeddings are continuous. Here, this means that

∥F ∥∞ ≤ ∥F∥Vq ≤ ∥F∥Vp

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, sup_{t∈[0,T]
}

|F (t)|

. It only remains to verify the
other axioms of a norm. It suffices to consider

∥⋅∥

_{p,[0,T]
}. Does it satisfy the triangle
inequality?

( ∑ )1∕p
∥Z + Y∥p,[0,T] ≡ sup |(Z + Y)(ti+1) − (Z + Y )(ti)|p
P i

(∑ )1∕p
≤ sup (|Z(ti+1) − Z (ti)|+ |Y (ti+1)− Y (ti)|)p
P ⌊ i ⌋
( ∑ )1∕p ( ∑ )1 ∕p
≤ sup ⌈ |Z (ti+1)− Z (ti)|p + |Y (ti+1)− Y (ti)|p ⌉
P i i

This is by

( ∑ )1∕p (∑ )1∕p
≤ sup |Z (ti+1)− Z (ti)|p +sup |Y (ti+1)− Y (ti)|p
P i P i
= ∥Z ∥p,[0,T] + ∥Y ∥p,[0,T]

Thus it clearly is a norm. It is easy to see that

∥Z + Y∥

_{∞}What about those inclusions. Let
F ∈ V^{p}. Is it also in V^{q}? Suppose F ∈ V^{p} and

∥F∥

_{p,[0,T]
} = 1. What about

∥F∥

_{q,[0,T]
}? For
any P

( ∑ )1∕p
|F (ti+1)− F (ti)|p ≤ 1
i

and so each

|F (ti+1)− F (ti)|

≤ 1. Hence

|F (ti+1) − F (ti)|

^{q}≤

|F (ti+1)− F (ti)|

^{p} and
so

( ∑ )1∕q
|F (ti+1)− F (ti)|q ≤ 1
i

also. Since this P is arbitrary, it follows that

∥F ∥

_{q,[0,T]
}≤ 1. Now let F ∈ V^{p} be arbitrary
then