Appendix B

Integration on Rough Paths∗The material on Stieltjes integrals has a very important generalization called integration onrough functions. This chapter gives an introduction to this topic. In order to show this, weneed a simple inequality called the triangle inequality. First here is a useful lemma.

As in the case of Stieltjes integrals all of this has generalizations to integrator functionswhich have values in various normed linear spaces but this is a book on single variableadvanced calculus and so this level of generality is avoided.

Lemma B.0.1 If a,b≥ 0, p > 1 and p′ is defined by 1p +

1p′ = 1, then

ab≤ ap

p+

bp′

p′.

Proof: Let p′ = q to save on notation and consider the following picture:

b

a

x

t

x = t p−1

t = xq−1

ab≤∫ a

0t p−1dt +

∫ b

0xq−1dx =

ap

p+

bq

q.

Note equality occurs when ap = bq.The following is a case of Holder’s inequality.

Lemma B.0.2 Let ai,bi ≥ 0. Then for p≥ 1,

∑i

aibi ≤

(∑

iap

i

)1/p(∑

ibp′

i

)1/p′

Proof: From the above inequality,

n

∑i=1

ai(∑i ap

i

)1/p

bi(∑i bp′

i

)1/p′ ≤n

∑i=1

1p

(ap

i

∑i api

)+

1p′

(bp′

i

∑i bp′i

)

=1p

(∑i ap

i

∑i api

)+

1p′

(∑i bp′

i

∑i bp′i

)=

1p+

1p′

= 1

Hence the inequality follows from multiplying both sides by(∑i ap

i

)1/p(

∑i bp′i

)1/p′

.

Then with this lemma, here is the triangle inequality.

373

Appendix BIntegration on Rough Paths*The material on Stieltjes integrals has a very important generalization called integration onrough functions. This chapter gives an introduction to this topic. In order to show this, weneed a simple inequality called the triangle inequality. First here is a useful lemma.As in the case of Stieltjes integrals all of this has generalizations to integrator functionswhich have values in various normed linear spaces but this is a book on single variableadvanced calculus and so this level of generality is avoided.Lemma B.0.1 /fa,b >0,p > 1 and p’ is defined by i+ ra = 1, then»" [x= telt=xI!taa b P bdab < | lars | odyJO JO P qNote equality occurs when a? = b17. JjThe following is a case of Holder’s inequality.Lemma B.0.2 Let a;,b; > 0. Then for p > 1,1/p ; \/pEan< (Ee) (EH)Proof: From the above inequality,/n a;Lai1/yial\ 1 (xo? \ 1 1(Ee,) +5 a = t+ 7 =1P\Liags / PP’ \y;b? P Pp. . La . p\1/p p\1/P"Hence the inequality follows from multiplying both sides by (D; a; ) (x; bi ) .-Then with this lemma, here is the triangle inequality.373