# Integration of controlled rough paths via fractional calculus

Yu Ito
From the journal Forum Mathematicum

## Abstract

We develop a fractional calculus approach to rough path analysis, introduced by Y. Hu and D. Nualart [6], and show that our integration can be generalized so that it is consistent with the rough path integration introduced by M. Gubinelli [5].

MSC 2010: 26A42; 26A33; 60H05

Communicated by Jan Bruinier

Funding statement: This work was partially supported by JSPS Research Fellowships for Young Scientists.

## A Appendix

In this appendix, we prove some statements used in Sections 2 and 3. Note that almost all statements follow from slight modifications of the proofs in the preceding studies [6, 8, 9, 14].

First, Proposition A.2 shows that 𝒟t-γRZ is β-Hölder continuous on the interval [0,t]. The proposition is proved similarly to [6, Lemma 6.3]. The proof makes use of the following lemma, which is a slight reformulation of [6, Lemmas 6.1 and 6.2].

## Lemma A.1 ([8, Lemma 3.1]).

For 0<δ<ε1, there exists a positive constant C, depending only on δ and ε, such that

(A.1)yδ-xδCxδ-ε(y-x)εfor 0<x<y.

For δ,ε>0 with 0<ε-δ<1, there exists a positive constant C, depending only on δ and ε, such that

(A.2)01uε(u-δ-1-(u+z)-δ-1)𝑑uCzε-δfor 0z<.

## Proposition A.2.

Let (Z,Z)QXβ(Rn), α(0,β), and b(0,T]. Then, Db-αRZ is β-Hölder continuous on the interval [0,b].

## Proof.

Take the real numbers s and t such that 0s<tb and estimate 𝒟b-αRZ(t)-𝒟b-αRZ(s). First, with regard to the first term of 𝒟b-αRZ(t)-𝒟b-αRZ(s), we estimate the following:

|Rt,bZ(b-t)α-Rs,bZ(b-s)α||Rt,bZ(b-t)α-Rt,bZ(b-s)α|+|Rt,bZ(b-s)α-Rs,bZ(b-s)α|=:A1+A2.

Then, by using (A.1), we have

A1RZ2β(b-t)2β-α(b-s)-α((b-s)α-(b-t)α)
RZ2β(b-t)2β-α(b-s)-αC(b-t)α-(2β-α)((b-s)-(b-t))2β-α
CRZ2β(t-s)2β-α

and, from the equality Rs,bZ=Rs,tZ+Rt,bZ+δZs,tδXt,b, we have

A2(RZ2β(t-s)2β+ZβXβ(t-s)β(b-t)β)(b-s)-α
RZ2β(t-s)2β-α+ZβXβ(t-s)β(b-t)β-α.

Next, with regard to the second term of 𝒟b-αRZ(t)-𝒟b-αRZ(s), we estimate the following:

|tbRt,vZ(v-t)α+1dv-sbRs,vZ(v-s)α+1dv|st|Rs,vZ|(v-s)α+1dv+tb|Rt,vZ(v-t)α+1-Rs,vZ(v-s)α+1|dv=:B1+B2.

Then,

B1RZ2βst(v-s)2β-α-1𝑑v=RZ2β(2β-α)-1(t-s)2β-α

and

B2tb|Rt,vZ(v-t)α+1-Rt,vZ(v-s)α+1|dv+tb|Rt,vZ(v-s)α+1-Rs,vZ(v-s)α+1|dv=:B21+B22.

By using the change of variables u=(v-t)/(b-t) and (A.2) with z=(t-s)/(b-t), we have

B21RZ2βtb(v-t)2β((v-t)-α-1-(v-s)-α-1)𝑑v
=RZ2β(b-t)2β-α01u2β(u-α-1-(u+z)-α-1)𝑑u
RZ2β(b-t)2β-αCz2β-α=CRZ2β(t-s)2β-α

and, from the equality Rs,vZ=Rs,tZ+Rt,vZ+δZs,tδXt,v for v[t,b], we have

B22RZ2βtb(t-s)2β(v-s)-α-1𝑑v+ZβXβtb(t-s)β(v-t)β(v-s)-α-1𝑑v
RZ2βα-1(t-s)2β-α+ZβXβ(t-s)β(β-α)-1(b-s)β-α.

Therefore, by combining these estimates, we obtain the statement of the proposition. ∎

## Corollary A.3.

Let (X,X)Ωβ(Rd), α(0,β), and b(0,T]. Then, Db-αX is β-Hölder continuous on the interval [0,b].

## Proof.

We set (Z,Z)𝒬Xβ(dd) as Zt:=𝕏0,t and Zt(ξ):=(Xt-X0)ξ for t[0,T], where ξd. Then, the identity RZ=𝕏 holds from (2.1). Therefore, from Proposition A.2, we obtain the statement. ∎

We next prove the following proposition used in the proof of Theorem 2.3.

## Proposition A.4.

In the setting of Definition 2.1, for each (s,t) with s<t,

(A.3)lim|𝒫|0st|Ds+1-2γ(i=0N-1Yti1(ti,ti+1])(u)-Ds+1-2γY(u)|𝑑u=0,
(A.4)lim|𝒫|0st|Ds+1-γ(i=0N-1(Yti+YtiδXti,u)1(ti,ti+1])(u)-𝒟^s+1-γRY(u)|𝑑u=0,
(A.5)lim|𝒫|0st|Ds+1-γ(-i=0N-1YtiδXti,uδZti,u1(ti,ti+1])(u)-𝒟^s+1-γ(YδXδZ)(u)|𝑑u=0,
(A.6)lim|𝒫|0st|Ds+1-2γ(i=0N-1YtiZti1(ti,ti+1])(u)-Ds+1-2γ(YZ)(u)|𝑑u=0,

where the limits are taken over all finite partitions P={t0,t1,,tN} of the interval [s,t] such that s=t0<t1<<tN=t and |P|=max0iN-1|ti+1-ti|.

## Proof.

We prove only that (A.5) holds, since (A.3) and (A.6) follow from [14, Theorem 4.1.1], and (A.4) follows from [9, Proposition 2.4]. The proof of (A.5) here is based on the proofs of [14, Theorem 4.1.1] and [9, Proposition 2.4]. For u(s,t], we set

Γ(γ)(Ds+1-γ(-i=0N-1YtiδXti,uδZti,u1(ti,ti+1])(u)-𝒟^s+1-γ(YδXδZ)(u))
=-i=0N-1YtiδXti,uδZti,u1(ti,ti+1](u)(u-s)-(1-γ)+(1-γ)suΨv,u𝒫(u-v)(1-γ)+1𝑑v
=:S𝒫1(u)+S𝒫2(u),

where

Ψv,u𝒫:=-i=0N-1{YtiδXti,uδZti,u(1(ti,ti+1](u)-1(ti,ti+1](v))}-YvδXv,uδZv,u

for (v,u). It then suffices to show that

(A.7)lim|𝒫|0st|S𝒫k(u)|𝑑u=0

holds for k=1,2. First, by straightforward computation, we have

st|S𝒫1(u)|𝑑u=i=0N-1titi+1|YtiδXti,uδZti,u|(u-s)-(1-γ)𝑑u
YXβZβi=0N-1titi+1(u-ti)2β(u-s)-(1-γ)𝑑u
YXβZβ|𝒫|2βi=0N-1titi+1(u-s)-(1-γ)𝑑u
=YXβZβ|𝒫|2βγ-1(t-s)γ,

where Y denotes the supremum norm of Y on [s,t]. Thus, (A.7) holds for k=1. Next, by using the equalities

S𝒫2(u)=(1-γ)j=0i-1tjtj+1Ψv,u𝒫(u-v)(1-γ)+1𝑑v+(1-γ)tiuΨv,u𝒫(u-v)(1-γ)+1𝑑v

for u(ti,ti+1], we decompose the L1-norm of S𝒫2 on [s,t] as follows:

st|S𝒫2(u)|𝑑u=t0t1|S𝒫2(u)|𝑑u+i=1N-1titi+1|S𝒫2(u)|𝑑u
(1-γ)i=0N-1titi+1tiu|Ψv,u𝒫|(u-v)(1-γ)+1𝑑v𝑑u+(1-γ)i=1N-1titi+1j=0i-1tjtj+1|Ψv,u𝒫|(u-v)(1-γ)+1𝑑v𝑑u
=:A1+A2.

Then, from the equality Ψv,u𝒫=-YvδXv,uδZv,u for (v,u) with ti<vuti+1, we have

A1YXβZβ(1-γ)i=0N-1titi+1tiu(u-v)2β+γ-2𝑑v𝑑u
=YXβZβ(1-γ)(2β+γ-1)-1(2β+γ)-1i=0N-1(ti+1-ti)2β+γ
YXβZβ(1-γ)(2β+γ-1)-1(2β+γ)-1|𝒫|2β+γ-1(t-s),

where we use that 2β+γ>1 holds from (1-β)/2<γ<β. Also, for (v,u) with tj<vtj+1ti<uti+1, we have

Ψv,u𝒫=-YtiδXti,uδZti,u+YtjδXtj,uδZtj,u-YvδXv,uδZv,u
=-YtiδXti,uδZti,u+Ytj(δXtj,v+δXv,u)(δZtj,v+δZv,u)-YvδXv,uδZv,u
=-YtiδXti,uδZti,u+Ytj(δXtj,vδZtj,v+δXv,uδZtj,v+δXtj,vδZv,u)-(Yv-Ytj)δXv,uδZv,u

and so

|Ψv,u𝒫|max{Y,Yβ}XβZβ{(u-ti)2β+(v-tj)2β+2(u-v)β(v-tj)β+(v-tj)β(u-v)2β}.

Therefore, we have

A2max{Y,Yβ}XβZβ(1-γ)
×{i=1N-1j=0i-1titi+1(u-ti)2βtjtj+1(u-v)γ-2dvdu+i=1N-1j=0i-1titi+1tjtj+1(v-tj)2β(u-v)γ-2dvdu
+2i=1N-1j=0i-1titi+1tjtj+1(v-tj)β(u-v)β+γ-2dvdu+i=1N-1j=0i-1titi+1tjtj+1(v-tj)β(u-v)2β+γ-2dvdu}
C|𝒫|2β+γ-1,

where C is a positive constant that does not depend on 𝒫. The last inequality can be proved by straightforward computation, as in the proofs of [14, Theorem 4.1.1] and [9, Proposition 2.4]. Hence, it follows from the estimates of A1 and A2, and 2β+γ>1 that (A.7) holds for k=2. Thus, we obtain the statement of the proposition. ∎

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