INTEGRATION WITH RESPECT TO HOLDER ROUGH PATHS OFORDER GREATER THAN 1/4: AN APPROACH VIA FRACTIONAL

CALCULUS

YU ITO

Abstract. On the basis of fractional calculus, we introduce an integral of controlled

paths with respect to Holder rough paths of order β ∈ (1/4, 1/3]. Our definition of the

integral is given explicitly in terms of Lebesgue integrals for fractional derivatives, with-

out using any arguments from discrete approximation. We demonstrate that for suitable

classes of β-Holder rough paths and controlled paths, our definition of the integral is con-

sistent with the usual definition given by the limit of the compensated Riemann–Stieltjes

sum. The results of this paper also provide an approach to the integral of 1-forms against

geometric β-Holder rough paths.

1. Introduction

Let Y and Z be Holder continuous functions on the interval [0, T ] of orders λ ∈ (0, 1]

and µ ∈ (0, 1], respectively, with λ + µ > 1. Then there exists the Riemann–Stieltjes

integral of Y along Z: ∫ t

s

Yu dZu := lim|Ps,t|→0

m−1∑i=0

Yt∗i (Zti+1− Zti)

for 0 ≤ s < t ≤ T . Here, the limit is taken over all finite partitions Ps,t := t0, t1, . . . , tmof the interval [s, t] such that s = t0 < t1 < · · · < tm = t and all points t∗i ∈ [ti, ti+1] and

|Ps,t| := max0≤i≤m−1|ti+1−ti|. This follows from the classical results by Young [19]. Under

the above-mentioned assumptions, the Riemann–Stieltjes integral is called the Young

integral and has an explicit expression in terms of the Lebesgue integral for fractional

derivatives: ∫ t

s

Yu dZu = Ys(Zt − Zs) + (−1)α∫ t

s

Dαs+Ys+(u)D

1−αt− Zt−(u) du,

where 1 − µ < α < λ, (−1)α := eiπα, and Dαs+Ys+ and D1−α

t− Zt− are the Weyl–Marchaud

fractional derivatives of Ys+ := Y − Ys of order α and Zt− := Z − Zt of order 1 − α,

respectively. (See Eqs. (2.2) and (2.3) for the definition of the Weyl–Marchaud fractional

derivatives.) The preceding equality follows from the integration by Zahle [20] and can be

2010 Mathematics Subject Classification. Primary 26A42; Secondary 26A33, 60H05.Key words and phrases. Stieltjes integral, Fractional derivative, Rough path, Fractional Brownian

motion.

Supported by JSPS KAKENHI Grant Number JP18K13431.1

2 Y. ITO

regarded as an integration by parts formula of order α. The fractional calculus approach

to Young integration given above has provided many applications in stochastic calculus;

in particular, it has provided a useful method for studying stochastic differential equations

driven by fractional Brownian motion with the Hurst parameter H ∈ (1/2, 1); e.g., [17].

(Note that λ+ µ > 1 is equivalent to β > 1/2 when λ = µ =: β.)

Rough path theory by Lyons [14] and controlled path theory by Gubinelli [6, 7] have

provided valuable frameworks for integrations and differential equations with respect to

Holder continuous functions of arbitrary orders, known as rough integrations and rough

differential equations, respectively. These theories allow us to take pathwise approaches

to classical stochastic calculus and provide useful methods for studying stochastic ordi-

nary and partial differential equations. Conventionally, rough integral is defined as the

limit of the compensated Riemann–Stieltjes sum and is based on arguments from discrete

approximation by Young [19], and therefore, it can be regarded as a generalization of the

Young integral for Holder continuous functions of order β ∈ (0, 1/2].

A natural question to consider is whether the fractional calculus approach to Young in-

tegration mentioned above can be incorporated into rough integrations. There are several

affirmative answers to this question. First, Hu and Nualart [8] proposed an alternative

approach to the rough path theory on the basis of fractional calculus. They introduced an

integration and differential equation with respect to Holder continuous functions of order

β ∈ (1/3, 1/2) using the basic formulas of fractional calculus and the ideas from rough

path theory. Their integral is also defined explicitly in terms of the Lebesgue integrals for

fractional derivatives, without using any arguments from discrete approximation. Their

results have been applied to stochastic differential equations driven by Brownian motion

and fractional Brownian motion with the Hurst parameter H ∈ (1/3, 1/2); e.g., [1, 2].

From the viewpoint of rough path theory, their integral provided an alternative definition

of the first level of the integral of 1-forms against β-Holder rough paths by Lyons [14].

Next, the author’s previous studies [11,12] extended the integration introduced in [8] to the

integration of controlled paths with respect to Holder rough paths of order β ∈ (1/3, 1/2]

by Gubinelli [6]. The study [11] also provided an approach to the integration of 1-forms

against β-Holder rough paths, which means not only the first level, but also the second

level of the rough integral of 1-forms. It should be noted that the results of [11,12] allow

us to formulate differential equations driven by β-Holder rough paths with β ∈ (1/3, 1/2]

in the sense of Lyons [14], as well as Gubinelli [6], using such explicit definitions of rough

integrals. Although the fractional calculus approach to rough integrations described above

is substantially more complicated than that to Young integration, such explicit definitions

enable straightforward quantitative estimation of rough integrals and solutions to rough

differential equations. Therefore, we expect this approach to provide more direct ways to

the fundamental theories of rough paths and controlled paths and their applications.

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 3

The purpose of the present paper is to develop a fractional calculus approach to the

rough integrations mentioned above. We herein demonstrate that this approach is valid

for less regular functions; in particular, we introduce an approach to integrations with re-

spect to Holder rough paths of order β ∈ (1/4, 1/3] based on arguments from the previous

studies [20] and [11, 12]. This investigation is considered as a reasonable step to lay the

foundation for theoretical and practical developments; it is known that the class of geomet-

ric β-Holder rough paths with β ∈ (1/4, 1/3] contains an important example associated

with fractional Brownian motion with the Hurst parameter H ∈ (1/4, 1/3] (cf. [3]). The

case β ∈ (1/4, 1/3] is more complicated than the case β ∈ (1/3, 1/2], since the functions

being considered are more irregular and the third level of Holder rough paths, the sec-

ond derivative of controlled paths, and the second order compensations of the Riemann–

Stieltjes sum are required. As we will see in the following sections, these concepts lead to

complicated calculations. However, the present paper shows that the arguments from [20]

and [11,12] work in this case.

First, on the basis of fractional calculus, an integral of the controlled paths with respect

to Holder rough paths of the order β ∈ (1/4, 1/3] (Definition 2.2) is introduced. Our

definition of the integral is also given explicitly by the Lebesgue integrals for fractional

derivatives. Subsequently, as the main results of this paper, we show that for suitable

classes of β-Holder rough paths and controlled paths, our definition of the integral is

consistent with the usual definition given by the limit of the compensated Riemann–

Stieltjes sum (Theorems 2.8, 2.9, and Corollary 2.10). To prove this, we extend some

arguments from [20] and [11, 12] and provide several propositions (Propositions 2.6, 2.7,

2.11, 2.12, and 2.13), which play a fundamental role in our integration. In addition,

the main results of this paper provide an approach to the integration of 1-forms against

geometric β-Holder rough paths with β ∈ (1/4, 1/3], which yields not only the first level,

but also the second and third levels of the rough integral of 1-forms (Propositions 4.1,

4.2, and 4.3). This is described in the Appendix as the basic relationships between the

rough integral of 1-forms and the main results of this paper and is also an important

development in the fractional calculus approach to rough differential equations, because

it allows us to formulate a differential equation driven by the geometric β-Holder rough

paths with β ∈ (1/4, 1/3] in the sense of Lyons [14]. Notably, in the previous studies

of the author [9, 10, 13], approaches to rough integrations for arbitrary orders based on

fractional calculus were already proposed. In particular, the study [13] is based on the

arguments from [20] and [11,12] and allows us to formulate the differential equation driven

by geometric β-Holder rough paths with β ∈ (1/4, 1/3] on the basis of the formulation by

Gubinelli [6,7]. However, the studies [9,10,13] are concerned with the fractional calculus

approach to the first level of the rough integral of 1-forms and its improvements and

therefore differ from the results of the present paper; indeed, in the case β ∈ (1/4, 1/3],

4 Y. ITO

Definition 2.2 gives a generalization of the integrals in [9, 10, 13]. (See (5) of Remark 2.3

for relations to the integrals in [9, 10,13].)

The remainder of this paper is organized as follows. In Section 2, we introduce our

definition of the integral of controlled paths and describe the main results of this paper.

Section 3 is devoted to the proofs of some of the statements used in Section 2. Appendix

provides an approach to the integral of 1-forms against geometric Holder rough paths.

2. Definition of the integral and main results

In this section, we introduce our definition of the integral of controlled paths and

describe the main results of this paper. We also review some concepts, including Holder

rough paths, geometric Holder rough paths, controlled paths, and fractional integrals and

derivatives, as well as some basic formulas of fractional derivatives. We follow the standard

treatments for rough path analysis [4–7, 14–16] and the fractional calculus [18, 20] based

on the descriptions of [9–13] for consistency with the author’s previous studies.

2.1. Notation. Let V and W be finite-dimensional normed spaces with norms ∥ · ∥V and

∥ · ∥W , respectively. Although the fundamental theories of rough paths and controlled

paths are valid for suitable infinite-dimensional Banach spaces, we consider only finite-

dimensional cases in this paper to evade technical difficulties that are not relevant to our

theme. Let L(V,W ) denote the set of all linear maps from V toW . Let T denote a positive

constant, which will be fixed throughout this paper. The simplex (s, t) ∈ R2 : 0 ≤ s ≤t ≤ T is denoted by and is a closed subset of R2. Let C([0, T ], V ) and C(, V ) denote

the spaces of all V -valued continuous functions on the interval [0, T ] and , respectively.

For ψ ∈ C([0, T ], V ), we set ∥ψ∥∞ := sup0≤t≤T ∥ψt∥V . For ψ ∈ C([0, T ], V ), we define

δψ ∈ C(, V ) as δψs,t := ψt − ψs for (s, t) ∈ . Let λ ∈ (0,∞). We set

∥Ψ∥λ := sup0≤s<t≤T

∥Ψs,t∥V(t− s)λ

for Ψ ∈ C(, V ). We set Cλ2 (V ) := Ψ ∈ C(, V ) : ∥Ψ∥λ < ∞ and Cλ

1 (V ) := ψ ∈C([0, T ], V ) : ∥δψ∥λ <∞, and write ∥ψ∥λ instead of ∥δψ∥λ when there is no ambiguity.

Hereinafter, d1, d2, and d3 denote positive integers, E, F , and G the Euclidean spaces

Rd1 , Rd2 , and Rd3 , respectively, and | · | the Euclidean norms of E, F , G, and their tensor

spaces. Let k be a positive integer. For α ∈ R and a ∈ E⊗k, we set (−1)α := eiπα and

|(−1)αa| := |a|.

2.2. Rough paths and controlled paths. Let T (3)(E) denote⊕3

j=0E⊗j and we define

the norm on T (3)(E) as ∥a∥T (3)(E) :=∑3

j=0|aj| for a = (a0, a1, a2, a3) ∈ T (3)(E). The

set of all X = (X0, X1, X2, X3) ∈ C(, T (3)(E)) such that X0s,t = 1 for all (s, t) ∈ is

denoted by C0(, T (3)(E)). Let β denote a real number with 1/4 < β ≤ 1/3. We say

that X = (1, X1, X2, X3) ∈ C0(, T (3)(E)) is a β-Holder rough path in E if X satisfies

the following two properties:

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 5

(1) for n = 1, 2, 3, ∥Xn∥nβ <∞;

(2) for n = 1, 2, 3 and s, t, u ∈ [0, T ] with s ≤ u ≤ t,∑n

i=0Xis,u ⊗Xn−i

u,t = Xns,t.

The space of all β-Holder rough paths in E is denoted by Ωβ(E), which is a complete

metric space under the metric dβ(X, X) := max1≤n≤3 ∥Xn − Xn∥nβ for X, X ∈ Ωβ(E).

Let x ∈ C11(E). We set

X1s,t = xt − xs, X2

s,t =

∫ t

s

(xu − xs)⊗ dxu, and X3s,t =

∫ t

s

∫ u

s

(xv − xs)⊗ dxv ⊗ dxu

for (s, t) ∈ . Then X = (1, X1, X2, X3) is a β-Holder rough path in E and is called the

smooth rough path in E associated with x. The elements in the closure of the set of all

smooth rough paths in E with respect to the metric dβ are called the geometric β-Holder

rough paths in E. The spaces of all smooth rough paths and geometric β-Holder rough

paths in E are denoted by SΩβ(E) and GΩβ(E), respectively. If X ∈ GΩβ(E), then

Sym(Xns,t) =

1

n!(X1

s,t)⊗n(2.1)

holds for n = 1, 2, 3 and (s, t) ∈ . Here, Sym(Xns,t) denotes the symmetric part of Xn

s,t.

Let X ∈ Ωβ(E). We say that a 3-tuple Y = (Y (0), Y (1), Y (2)) is a path controlled by X

with values in F if Y satisfies the following two properties:

(1) Y (0) ∈ Cβ1 (F ), Y

(1) ∈ Cβ1 (L(E,F )), and Y

(2) ∈ Cβ1 (L(E

⊗2, F ));

(2) R20(Y ) ∈ C3β

2 (F ) and R11(Y ) ∈ C2β

2 (L(E,F )).

Here, R20(Y ) and R1

1(Y ) are defined by

R20(Y )s,t := Y

(0)t − Y (0)

s − Y (1)s X1

s,t − Y (2)s X2

s,t and R11(Y )s,t := Y

(1)t − Y (1)

s − Y (2)s X1

s,t

for (s, t) ∈ . Similarly, we set

R02(Y )s,t := Y

(2)t − Y (2)

s and R10(Y )s,t := Y

(0)t − Y (0)

s − Y (1)s X1

s,t

for (s, t) ∈ . Although X3 is not necessary for our definition of paths controlled by

X, we need it in the integration theory. The space of all paths controlled by X with

values in F is denoted by QβX(F ), which is a Banach space under the norm ∥Y ∥X,β :=∑2

l=0|Y(l)0 | +

∑2l=0 ∥R

2−ll (Y )∥(3−l)β for Y ∈ Qβ

X(F ). We also refer to Y ∈ QβX(F ) as a

controlled path for X with values in F . For further details and examples of Holder rough

paths and controlled paths, see, e.g., [4–7].

2.3. Fractional integrals and derivatives. Let a and b be real numbers with a < b.

For p ∈ [1,∞), let Lp(a, b) denote the complex Lp-space on the interval [a, b] with respect

to the Lebesgue measure. Let f ∈ L1(a, b) and α ∈ (0,∞). The left- and right-sided

Riemann–Liouville fractional integrals of f of order α are defined for almost all t ∈ (a, b)

by

Iαa+f(t) :=1

Γ(α)

∫ t

a

(t− s)α−1f(s) ds and Iαb−f(t) :=(−1)−α

Γ(α)

∫ b

t

(s− t)α−1f(s) ds,

6 Y. ITO

respectively, where Γ denotes the Euler gamma function, namely, Γ(α) :=∫∞0rα−1e−r dr.

For p ∈ [1,∞), let Iαa+(Lp) and Iαb−(L

p) denote the images of Lp(a, b) by the operators

Iαa+ and Iαb−, respectively. Letting f ∈ Iαa+(L1) (resp. Iαb−(L

1)) with 0 < α < 1, the

Weyl–Marchaud fractional derivative of f of order α is defined for almost all t ∈ (a, b) by

Dαa+f(t) :=

1

Γ(1− α)

(f(t)

(t− a)α+ α

∫ t

a

f(t)− f(s)

(t− s)α+1ds

)(2.2)

for the left-sided version, and

Dαb−f(t) :=

(−1)α

Γ(1− α)

(f(t)

(b− t)α+ α

∫ b

t

f(t)− f(s)

(s− t)α+1ds

)(2.3)

for the right-sided version. Here, the integrals above are well-defined for almost all t ∈(a, b). If f ∈ Iαa+(L

p) (resp. Iαb−(Lp)) with 0 < α < 1 and 1 ≤ p < ∞, then Dα

a+f ∈Lp(a, b) (resp. Dα

b−f ∈ Lp(a, b)) holds. The following two formulas are of particular

importance in this paper. The first is the composition formula:

Dαa+(D

βa+f) = Dα+β

a+ f (resp. Dαb−(D

βb−f) = Dα+β

b− f)(2.4)

for f ∈ Iα+βa+ (L1) (resp. Iα+β

b− (L1)), 0 < α < 1, and 0 < β < 1, with α+β < 1. The second

is the integration by parts formula of order α:

(−1)α∫ b

a

Dαa+f(t)g(t) dt =

∫ b

a

f(t)Dαb−g(t) dt(2.5)

for f ∈ Iαa+(Lp), g ∈ Iαb−(L

q), 0 < α < 1, 1 ≤ p < ∞, and 1 ≤ q < ∞, with 1/p + 1/q ≤1 + α. The following statements about Holder continuous functions are also important in

this paper. Let f be a real-valued Holder continuous function of order λ ∈ (0, 1] on the

interval [a, b]. Then f − f(a) ∈ Iαa+(Lp) and f − f(b) ∈ Iαb−(L

p) hold for α ∈ (0, λ) and

p ∈ [1,∞). The equality

f(y)− f(x) = (−1)α∫ b

a

Dαa+1(x,y](t)D

1−αb− (f − f(b))(t) dt(2.6)

holds for x, y ∈ [a, b] with x < y and α ∈ (1 − λ, 1). Here, 1(x,y] denotes the indicator

function of the interval (x, y], and the function (−1)αD1−αb− (f − f(b)) is real-valued. No-

tably, the right-hand side of Eq. (2.6) is well-defined from (1) 1(x,y] ∈ Iαa+(Lp) if and only

if αp < 1 and (2) D1−αb− (f − f(b)) is a bounded function on [a, b] (in fact, D1−α

b− (f − f(b))

is a Holder continuous function of order λ− (1− α) on [a, b]). Furthermore,

lim|Pa,b|→0

∫ b

a

|Dαa+(

m−1∑i=0

f(ti)1(ti,ti+1])(t)−Dαa+f(t)| dt = 0(2.7)

holds for α ∈ (0, λ). Here, the limit is taken over all finite partitions Pa,b = t0, t1, . . . , tmof the interval [a, b] such that a = t0 < t1 < · · · < tm = b and |Pa,b| := max0≤i≤m−1|ti+1 −ti|. For further details and proofs of Eqs. (2.6) and (2.7), see Proposition 2.2 and Theo-

rem 4.1.1 in [20], respectively.

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 7

2.4. Definition of the integral. For our definition of the integral of controlled paths,

we introduce the following functions. Let Ψ ∈ Cλ2 (V ) with λ > 0 and α ∈ (0,minλ, 1).

We define Dαs+Ψ with s ∈ [0, T ) and Dα

t−Ψ with t ∈ (0, T ] by

Dαs+Ψ(s) := 0,

Dαs+Ψ(u) :=

1

Γ(1− α)

(Ψs,u

(u− s)α+ α

∫ u

s

Ψv,u

(u− v)α+1dv

)for u ∈ (s, T ] and

Dαt−Ψ(t) := 0,

Dαt−Ψ(u) :=

(−1)1+α

Γ(1− α)

(Ψu,t

(t− u)α+ α

∫ t

u

Ψu,v

(v − u)α+1dv

)for u ∈ [0, t). For ψ ∈ C([0, T ], V ), we set ψs+(u) := ψu − ψs for u ∈ [s, T ] and ψt−(u) :=

ψu − ψt for u ∈ [0, t]. If Ψ ∈ Cλ2 (V ) is of the form Ψ = δψ for some ψ ∈ Cλ

1 (V ) with

0 < λ ≤ 1, then the identities Dαs+Ψ = Dα

s+ψs+ and Dαt−Ψ = Dα

t−ψt− hold, by definition,

for α ∈ (0, λ). The following estimates are used in some of the discussions in this paper:

∥Dαs+Ψ(u)∥V ≤ 1

Γ(1− α)

λ

λ− α∥Ψ∥λ(u− s)λ−α(2.8)

for u ∈ [s, T ] and

∥Dαt−Ψ(u)∥V ≤ 1

Γ(1− α)

λ

λ− α∥Ψ∥λ(t− u)λ−α(2.9)

for u ∈ [0, t]. These can be proved by a straightforward computation. For f ∈ C(,R)and g ∈ C([0, T ],C), we set fg ∈ C(,C) and gf ∈ C(,C) as (fg)s,t := fs,tgt and

(gf)s,t := gsfs,t for (s, t) ∈ . These symbols are used in some of the discussions on func-

tions Dαs+ and Dα

t−; e.g., Definition 2.1 below and (1) and (2) of Remark 2.3. Definition 2.1

is a slight reformulation of [9, Definition 2.2]. We recall that 1/4 < β ≤ 1/3.

Definition 2.1. Let X ∈ Ωβ(E), α ∈ (0, β), and t ∈ (0, T ]. We define functions

R(1,α)t− X, R(2,α)

t− X, and R(3,α)t− X on [0, t] inductively as follows: for u ∈ [0, t],

R(1,α)t− X(u) := Dα

t−X1(u),

R(2,α)t− X(u) := D2α

t−X2(u)−Dα

t−(X1 ⊗R(1,α)

t− X)(u),

R(3,α)t− X(u) := D3α

t−X3(u)−D2α

t−(X2 ⊗R(1,α)

t− X)(u)−Dαt−(X

1 ⊗R(2,α)t− X)(u).

We set S(a⊗b⊗c) := b⊗a⊗c for a, b, c ∈ E and R(3,α)t− X(u) := R(3,α)

t− X(u)+S(R(3,α)t− X(u))

for u ∈ [0, t]. Let Z ∈ QβX(G). Similarly, we define R(3,α)

t− Z by

R(3,α)t− Z(u) := D3α

t−R20(Z)(u)−D2α

t−(R11(Z)R

(1,α)t− X)(u)−Dα

t−(δZ(2)R(2,α)

t− X)(u)

for u ∈ [0, t].

8 Y. ITO

Notably, R(n,α)t− X(t) = 0 for n = 1, 2, 3 and R(3,α)

t− Z(t) = 0 hold, by definition, and

therefore

Djαt−(X

j ⊗R(1,α)t− X)(u) =

(−1)1+jαjα

Γ(1− jα)

∫ t

u

Xju,v ⊗R(1,α)

t− X(v)

(v − u)jα+1dv

for j = 1, 2 and

Dαt−(X

1 ⊗R(2,α)t− X)(u) =

(−1)1+αα

Γ(1− α)

∫ t

u

X1u,v ⊗R(2,α)

t− X(v)

(v − u)α+1dv

hold for u ∈ [0, t], and

D2αt−(R

11(Z)R

(1,α)t− X)(u) =

(−1)1+2α2α

Γ(1− 2α)

∫ t

u

R11(Z)u,vR

(1,α)t− X(v)

(v − u)2α+1dv

and

Dαt−(δZ

(2)R(2,α)t− X)(u) =

(−1)1+αα

Γ(1− α)

∫ t

u

δZ(2)u,vR(2,α)

t− X(v)

(v − u)α+1dv

hold for u ∈ [0, t]. It follows from Eq. (2.9) that R(n,α)t− X for n = 1, 2, 3 and R(3,α)

t− Z

are bounded functions on [0, t]. (See also Eqs. (2.12) and (2.13).) Furthermore, we can

verify that R(n,α)t− X for n = 1, 2, 3 and R(3,α)

t− Z are (β − α)-Holder continuous on [0, t],

from [13, Proposition 5.2 and Corollary 5.3]. Now, we are ready to define the integral of

Y along Z.

Definition 2.2. Let X ∈ Ωβ(E), Y ∈ QβX(L(G,F )), and Z ∈ Qβ

X(G). We set

Ξs,t := Y (0)s (Z

(0)t − Z(0)

s ) + Y (1)s Z(1)

s X2s,t + Y (2)

s Z(1)s X3

s,t + Y (1)s Z(2)

s (X3s,t + S(X3

s,t))

for (s, t) ∈ . Let γ be a real number such that (1 − β)/3 < γ < β. For (s, t) ∈ , we

define IX(Y, Z)s,t ∈ F by

IX(Y, Z)s,t := Ξs,t + (−1)1−γ

∫ t

s

D1−γs+ (Φ3

·,u)s+(u)R(1,γ)t− X(u) du

+ (−1)1−2γ

∫ t

s

D1−2γs+ (Φ2

·,u)s+(u)R(2,γ)t− X(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ Y

(0)s+ (u)R(3,γ)

t− Z(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ (Y (2)Z(1))s+(u)R(3,γ)

t− X(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ (Y (1)Z(2))s+(u)R(3,γ)

t− X(u) du.

Here, D1−γs+ (Φ3

·,u)s+(u) and D1−2γs+ (Φ2

·,u)s+(u) are defined as follows:

Φ3v,u := (Y (0)

v + Y (1)v X1

v,u + Y (2)v X2

v,u)Z(1)u − Y (2)

v X2v,uδZ

(1)v,u − Y (1)

v X1v,uR

11(Z)v,u

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 9

for (v, u) ∈ ,

D1−γs+ (Φ3

·,u)s+(u) :=1

Γ(1− (1− γ))

(Φ3

u,u − Φ3s,u

(u− s)1−γ+ (1− γ)

∫ u

s

Φ3u,u − Φ3

v,u

(u− v)(1−γ)+1dv

)for u ∈ [s, T ] and

Φ2v,u := (Y (0)

v + Y (1)v X1

v,u)Z(2)u + (Y (1)

v + Y (2)v X1

v,u)Z(1)u

− Y (2)v X1

v,uδZ(1)v,u − Y (1)

v R11(Z)v,u − Y (1)

v X1v,uδZ

(2)v,u

for (v, u) ∈ ,

D1−2γs+ (Φ2

·,u)s+(u) :=1

Γ(1− (1− 2γ))

(Φ2

u,u − Φ2s,u

(u− s)1−2γ+ (1− 2γ)

∫ u

s

Φ2u,u − Φ2

v,u

(u− v)(1−2γ)+1dv

)for u ∈ [s, T ].

Remark 2.3. Let us make some comments about Definition 2.2 mentioned above.

(1) We note that the equality

Φ3u,u − Φ3

v,u = R20(Y )v,uZ

(1)u + Y (2)

v X2v,uδZ

(1)v,u + Y (1)

v X1v,uR

11(Z)v,u

holds, by definition, for (v, u) ∈ , and thus

|Φ3u,u − Φ3

v,u| ≤ (∥R20(Y )∥3β∥Z(1)∥∞ + ∥Y (2)∥∞∥X2∥2β∥Z(1)∥β(2.10)

+ ∥Y (1)∥∞∥X1∥β∥R11(Z)∥2β)(u− v)3β.

Therefore, D1−γs+ (Φ3

·,u)s+(u) is well-defined for u ∈ [s, T ] because 1−γ < 3β follows

from (1− β)/3 < γ < β, and the equality

D1−γs+ (Φ3

·,u)s+(u) = D1−γs+ (R2

0(Y )Z(1) + Y (2)X2δZ(1) + Y (1)X1R11(Z))(u)

holds for u ∈ [s, T ].

(2) Similarly, the equality

Φ2u,u − Φ2

v,u = R10(Y )v,uZ

(2)u +R1

1(Y )v,uZ(1)u + Y (2)

v X1v,uδZ

(1)v,u

+ Y (1)v R1

1(Z)v,u + Y (1)v X1

v,uδZ(2)v,u

holds, by definition, for (v, u) ∈ , and therefore

|Φ2u,u − Φ2

v,u| ≤ (∥R10(Y )∥2β∥Z(2)∥∞ + ∥R1

1(Y )∥2β∥Z(1)∥∞ + ∥Y (2)∥∞∥X1∥β∥Z(1)∥β(2.11)

+ ∥Y (1)∥∞∥R11(Z)∥2β + ∥Y (1)∥∞∥X1∥β∥Z(2)∥β)(u− v)2β.

Therefore, D1−2γs+ (Φ2

·,u)s+(u) is well-defined for u ∈ [s, T ] because 1 − 2γ < 2β

follows from (1− β)/3 < γ < β, and the equality

D1−2γs+ (Φ2

·,u)s+(u) = D1−2γs+ (R1

0(Y )Z(2) +R11(Y )Z(1) + Y (2)X1δZ(1)

+ Y (1)R11(Z) + Y (1)X1δZ(2))(u)

10 Y. ITO

holds for u ∈ [s, T ].

(3) We recall that D1−3γs+ Y

(0)s+ , D1−3γ

s+ (Y (2)Z(1))s+, and D1−3γs+ (Y (1)Z(2))s+ are the left-

sided Weyl–Marchaud fractional derivatives of the β-Holder continuous functions

Y(0)s+ (u) = Y (0)

u − Y (0)s , (Y (2)Z(1))s+(u) = Y (2)

u Z(1)u − Y (2)

s Z(1)s ,

and (Y (1)Z(2))s+(u) = Y (1)u Z(2)

u − Y (1)s Z(2)

s ,

of order 1 − 3γ, respectively, and are well-defined on [s, T ] because 1 − 3γ < β

follows from (1− β)/3 < γ < β.

(4) We note that ∥IX(Y, Z) − Ξ∥4β < ∞ holds in the setting of Definition 2.2. This

follows from Eqs. (2.8), (2.9), (2.10), (2.11), (2.12), and (2.13) below. Eqs. (2.12)

and (2.13) are used in some of the discussions in this paper. Let X ∈ Ωβ(E).

Then, for n = 2, 3 and α ∈ (0, β), there exists a positive constant C, depending

only on n, β, and α, such that for t ∈ (0, T ] and u ∈ [0, t],

|R(n,α)t− X(u)| ≤ C(1 + max

1≤i≤n−1∥X i∥iβ)n−1 max

1≤i≤n∥X i∥iβ(t− u)n(β−α).(2.12)

Let Z ∈ QβX(G). Then, for α ∈ (0, β), there exists a positive constant C, depend-

ing only on β and α, such that for t ∈ (0, T ] and u ∈ [0, t],

|R(3,α)t− Z(u)| ≤ C(1 + max

i=1,2∥X i∥iβ)2 max

l=0,1,2∥R2−l

l (Z)∥(3−l)β(t− u)3(β−α).(2.13)

For a proof of Eq. (2.12), see [10, Lemma 4.1]. We omit a proof of Eq. (2.13)

because it is similar to a proof of (2.12) with n = 3.

(5) Relations to the integrals in the previous studies [9, 10, 13] are stated as follows.

Let G = E and set Z(0)t = X1

0,t, Z(1)t = I, and Z

(2)t = 0 for t ∈ [0, T ]. Here,

I ∈ L(E,E) denotes the identity map. Then Z = (Z(0), Z(1), Z(2)) ∈ QβX(E) holds

and

IX(Y, Z)s,t = Y (0)s X1

s,t + Y (1)s X2

s,t + Y (2)s X3

s,t

+ (−1)1−γ

∫ t

s

D1−γs+ R2

0(Y )(u)R(1,γ)t− X(u) du

+ (−1)1−2γ

∫ t

s

D1−2γs+ R1

1(Y )(u)R(2,γ)t− X(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ Y

(2)s+ (u)R(3,γ)

t− X(u) du

for (s, t) ∈ . The right-hand side of the preceding equality is the same as the

integral in [10, 13] when 1/4 < β ≤ 1/3. This corresponds to the integral of

Y along X. Furthermore, let f be a 3-times continuously Frechet differentiable

function from E to L(E,F ) such that f and ∇jf for j = 1, 2, 3 are bounded on

E and let X ∈ GΩβ(E) and ξ ∈ E and set Xt := ξ + X10,t for t ∈ [0, T ] and

Y(0)t = f(Xt), Y

(1)t = ∇f(Xt), and Y

(2)t = ∇2f(Xt) for t ∈ [0, T ]. Then Y =

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 11

(Y (0), Y (1), Y (2)) ∈ QβX(L(E,F )) holds, and IX(Y, Z) coincides with the integral

in [9] when 1/4 < β ≤ 1/3. This corresponds to the first level of the integral of f

against X ∈ GΩβ(E). (See also Proposition 4.1 in Appendix.)

2.5. Main results. Hereinafter, we always assume that γ is a real number with (1 −β)/3 < γ < β, as in Definition 2.2. We also recall that 1/4 < β ≤ 1/3. To describe

the main results of this paper, we introduce the following properties of X ∈ Ωβ(E) and

Z ∈ QβX(G).

Definition 2.4. Let X ∈ Ωβ(E). We say that X satisfies the property (Hrβ) if

Xna,b =

n∑k=1

(−1)1−kα

∫ t

s

D1−kαs+ 1(a,b](u)X

n−ka,u ⊗R(k,α)

t− X(u) du(2.14)

holds for n = 1, 2, 3, (s, t) ∈ with s < t, a, b ∈ [s, t] with a < b, and α ∈ (0, β).

Definition 2.5. Let X ∈ Ωβ(E) and Z ∈ QβX(G). We say that Z satisfies the prop-

erty (Hcβ) if

R20(Z)a,b = (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)R

11(Z)a,uR

(1,α)t− X(u) du(2.15)

+ (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)δZ

(2)a,uR

(2,α)t− X(u) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)R(3,α)

t− Z(u) du

holds for (s, t) ∈ with s < t, a, b ∈ [s, t] with a < b, and α ∈ (0, β).

The right-hand sides of Eqs. (2.14) and (2.15) are well-defined for the same reasons

as those of Eq. (2.6), and note that (−1)1−kαR(k,α)t− X(u) ∈ E⊗k for k = 1, 2, 3 and

(−1)1−3αR(3,α)t− Z(u) ∈ G for u ∈ [0, t]. The following propositions provide sufficient

conditions for X ∈ Ωβ(E) and Z ∈ QβX(G) to satisfy the properties (Hr

β) and (Hcβ),

respectively.

Proposition 2.6. Let X ∈ Ωβ(E). Suppose that there exists a sequence X(m) ∈SΩβ(E) such that limm→∞X(m)ns,t = Xn

s,t for (s, t) ∈ and supm ∥X(m)n∥nβ < ∞ for

n = 1, 2, 3. Then X satisfies the property (Hrβ). In particular, X ∈ GΩβ(E) satisfies the

property (Hrβ).

Proposition 2.7. Let X ∈ Ωβ(E) and Z ∈ QβX(G). Suppose that there exist sequences

X(m) ∈ SΩβ(E) and Z(m) ∈ QβX(m)(G) such that

(1) limm→∞X(m)ns,t = Xns,t for (s, t) ∈ and supm ∥X(m)n∥nβ <∞ for n = 1, 2,

(2) limm→∞R2−ll (Z(m))s,t = R2−l

l (Z)s,t for (s, t) ∈ and supm ∥R2−ll (Z(m))∥(3−l)β <

∞ for l = 0, 1, 2.

Then Z satisfies the property (Hcβ).

12 Y. ITO

Definition 2.4 and Proposition 2.6 are slight reformulations of Condition 3.3 and Propo-

sition 3.4 in [13], respectively. We prove Propositions 2.6 and 2.7 in Section 3. The

following theorem justifies the interpretation of IX(Y, Z) as an integral of Y along Z.

Theorem 2.8. Let X ∈ Ωβ(E), Y ∈ QβX(L(G,F )), and Z ∈ Qβ

X(G). Suppose that X

satisfies Eq. (2.1) with n = 2 and the property (Hrβ), and Z satisfies the property (Hc

β).

Then, for (s, t) ∈ ,

IX(Y, Z)s,t = lim|Ps,t|→0

m−1∑i=0

Ξti,ti+1,(2.16)

where the limit is taken over all finite partitions Ps,t = t0, t1, . . . , tm of the interval [s, t]

such that s = t0 < t1 < · · · < tm = t and |Ps,t| = max0≤i≤m−1|ti+1 − ti|.

Therefore, combining Theorem 2.8 with Propositions 2.6 and 2.7, we obtain the fol-

lowing theorem and corollary. Notably, under the assumption (1) of Proposition 2.7, X

satisfies Eq. (2.1) with n = 2.

Theorem 2.9. Let X ∈ Ωβ(E), Y ∈ QβX(L(G,F )), and Z ∈ Qβ

X(G). Suppose that

there exist sequences X(m) ∈ SΩβ(E) and Z(m) ∈ QβX(m)(G) such that

(1) limm→∞X(m)ns,t = Xns,t for (s, t) ∈ and supm ∥X(m)n∥nβ <∞ for n = 1, 2, 3,

(2) limm→∞R2−ll (Z(m))s,t = R2−l

l (Z)s,t for (s, t) ∈ and supm ∥R2−ll (Z(m))∥(3−l)β <

∞ for l = 0, 1, 2.

Then Eq. (2.16) holds for (s, t) ∈ .

Corollary 2.10. Let X ∈ GΩβ(E), Y ∈ QβX(L(G,F )), and Z ∈ Qβ

X(G). Suppose

that there exist sequences X(m) ∈ SΩβ(E) and Z(m) ∈ QβX(m)(G) such that

(1) limm→∞ dβ(X(m), X) = 0,

(2) limm→∞maxl=0,1,2 ∥R2−ll (Z(m))−R2−l

l (Z)∥(3−l)β = 0.

Then Eq. (2.16) holds for (s, t) ∈ .

Clearly, under the assumptions of Theorem 2.8, the equality

IX(Y, Z)s,u + IX(Y, Z)u,t = IX(Y, Z)s,t(2.17)

holds for s, t, u ∈ [0, T ] with s ≤ u ≤ t. However, it is unknown whether Eq. (2.17) holds

in the setting of Definition 2.2. Theorems 2.8, 2.9, and Eq. (2.17) will be used in Appendix

to describe the basic relations between the integral of 1-forms against X ∈ GΩβ(E) and

IX(Y, Z).

2.6. Proof of Theorem 2.8. In the remainder of this section, we prove Theorem 2.8

using Propositions 2.11, 2.12, and 2.13 below. We prove Propositions 2.11 and 2.13 in

Section 3.

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 13

Proposition 2.11. Under the assumptions of Theorem 2.8, for (s, t) ∈ with s < t,

a, b ∈ [s, t] with a < b, and α ∈ (0, β),

Ξa,b = (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)Φ

3a,uR

(1,α)t− X(u) du(2.18)

+ (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)Φ

2a,uR

(2,α)t− X(u) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)Y

(0)a R(3,α)

t− Z(u) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)Y

(2)a Z(1)

a R(3,α)t− X(u) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)Y

(1)a Z(2)

a R(3,α)t− X(u) du.

Proposition 2.12. Under the assumptions of Theorem 2.8, for (s, t) ∈ with s < t,

IX(Y, Z)s,t = (−1)1−γ

∫ t

s

D1−γs+ (Φ3

·,u)(u)R(1,γ)t− X(u) du

+ (−1)1−2γ

∫ t

s

D1−2γs+ (Φ2

·,u)(u)R(2,γ)t− X(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ Y (0)(u)R(3,γ)

t− Z(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ (Y (2)Z(1))(u)R(3,γ)

t− X(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ (Y (1)Z(2))(u)R(3,γ)

t− X(u) du,

where

D1−γs+ (Φ3

·,u)(u) :=1

Γ(1− (1− γ))

(Φ3

u,u

(u− s)1−γ+ (1− γ)

∫ u

s

Φ3u,u − Φ3

v,u

(u− v)(1−γ)+1dv

)

for u ∈ (s, T ] and

D1−2γs+ (Φ2

·,u)(u) :=1

Γ(1− (1− 2γ))

(Φ2

u,u

(u− s)1−2γ+ (1− 2γ)

∫ u

s

Φ2u,u − Φ2

v,u

(u− v)(1−2γ)+1dv

)

for u ∈ (s, T ].

14 Y. ITO

Proof. From Eq. (2.18) with a = s, b = t, and α = γ,

IX(Y, Z)s,t = (−1)1−γ

∫ t

s

(D1−γs+ (Φ3

·,u)s+(u) +D1−γs+ 1(s,t](u)Φ

3s,u)R

(1,γ)t− X(u) du

+ (−1)1−2γ

∫ t

s

(D1−2γs+ (Φ2

·,u)s+(u) +D1−2γs+ 1(s,t](u)Φ

2s,u)R

(2,γ)t− X(u) du

+ (−1)1−3γ

∫ t

s

(D1−3γs+ Y

(0)s+ (u) +D1−3γ

s+ 1(s,t](u)Y(0)s )R(3,γ)

t− Z(u) du

+ (−1)1−3γ

∫ t

s

(D1−3γs+ (Y (2)Z(1))s+(u) +D1−3γ

s+ 1(s,t](u)Y(2)s Z(1)

s )R(3,γ)t− X(u) du

+ (−1)1−3γ

∫ t

s

(D1−3γs+ (Y (1)Z(2))s+(u) +D1−3γ

s+ 1(s,t](u)Y(1)s Z(2)

s )R(3,γ)t− X(u) du.

From the equality

D1−nγs+ 1(s,t](u) =

1

Γ(1− (1− nγ))

1

(u− s)1−nγ

for n = 1, 2, 3 and u ∈ (s, t], we have

D1−γs+ (Φ3

·,u)s+(u) +D1−γs+ 1(s,t](u)Φ

3s,u = D1−γ

s+ (Φ3·,u)(u),

D1−2γs+ (Φ2

·,u)s+(u) +D1−2γs+ 1(s,t](u)Φ

2s,u = D1−2γ

s+ (Φ2·,u)(u),

D1−3γs+ Y

(0)s+ (u) +D1−3γ

s+ 1(s,t](u)Y(0)s = D1−3γ

s+ Y (0)(u),

D1−3γs+ (Y (2)Z(1))s+(u) +D1−3γ

s+ 1(s,t](u)Y(2)s Z(1)

s = D1−3γs+ (Y (2)Z(1))(u),

D1−3γs+ (Y (1)Z(2))s+(u) +D1−3γ

s+ 1(s,t](u)Y(1)s Z(2)

s = D1−3γs+ (Y (1)Z(2))(u)

for u ∈ (s, t]. Therefore, we obtain the statement of the proposition.

Proposition 2.13. In the setting of Definition 2.2, for (s, t) ∈ with s < t,

lim|Ps,t|→0

∫ t

s

|D1−γs+ (

m−1∑i=0

Φ3ti,u

1(ti,ti+1])(u)−D1−γs+ (Φ3

·,u)(u)| du = 0(2.19)

and

lim|Ps,t|→0

∫ t

s

|D1−2γs+ (

m−1∑i=0

Φ2ti,u

1(ti,ti+1])(u)−D1−2γs+ (Φ2

·,u)(u)| du = 0.(2.20)

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 15

Proof of Theorem 2.8. Fix (s, t) ∈ with s < t since Eq. (2.16) obviously holds when

s = t. From Eq. (2.18) with a = ti, b = ti+1, and α = γ,

m−1∑i=0

Ξti,ti+1= (−1)1−γ

∫ t

s

D1−γs+ (

m−1∑i=0

Φ3ti,u

1(ti,ti+1])(u)R(1,γ)t− X(u) du

+ (−1)1−2γ

∫ t

s

D1−2γs+ (

m−1∑i=0

Φ2ti,u

1(ti,ti+1])(u)R(2,γ)t− X(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ (

m−1∑i=0

Y(0)ti 1(ti,ti+1])(u)R

(3,γ)t− Z(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ (

m−1∑i=0

Y(2)ti Z

(1)ti 1(ti,ti+1])(u)R

(3,γ)t− X(u) du

+ (−1)1−3γ

∫ t

s

D1−3γs+ (

m−1∑i=0

Y(1)ti Z

(2)ti 1(ti,ti+1])(u)R

(3,γ)t− X(u) du.

Therefore, from Proposition 2.12,

|m−1∑i=0

Ξti,ti+1− IX(Y, Z)s,t|

≤∫ t

s

|D1−γs+ (

m−1∑i=0

Φ3ti,u

1(ti,ti+1])(u)−D1−γs+ (Φ3

·,u)(u)| du ∥R(1,γ)t− X∥∞

+

∫ t

s

|D1−2γs+ (

m−1∑i=0

Φ2ti,u

1(ti,ti+1])(u)−D1−2γs+ (Φ2

·,u)(u)| du ∥R(2,γ)t− X∥∞

+

∫ t

s

|D1−3γs+ (

m−1∑i=0

Y(0)ti 1(ti,ti+1])(u)−D1−3γ

s+ Y (0)(u)| du ∥R(3,γ)t− Z∥∞

+

∫ t

s

|D1−3γs+ (

m−1∑i=0

Y(2)ti Z

(1)ti 1(ti,ti+1])(u)−D1−3γ

s+ (Y (2)Z(1))(u)| du ∥R(3,γ)t− X∥∞

+

∫ t

s

|D1−3γs+ (

m−1∑i=0

Y(1)ti Z

(2)ti 1(ti,ti+1])(u)−D1−3γ

s+ (Y (1)Z(2))(u)| du ∥R(3,γ)t− X∥∞.

Therefore, from Eqs. (2.19), (2.20), and (2.7), we obtain the statement of the theorem.

3. Some proofs

In this section, we prove Propositions 2.6, 2.7, 2.11, and 2.13. We recall that 1/4 <

β ≤ 1/3 and (1− β)/3 < γ < β.

3.1. Proofs of Propositions 2.6 and 2.7. To prove Propositions 2.6 and 2.7, we first

introduce the following two lemmas.

16 Y. ITO

Lemma 3.1. Let Ψ ∈ Cλ2 (V ) with λ > 0. Suppose that there exists a sequence Ψ(m) ∈

Cλ2 (V ) such that limm→∞Ψ(m)s,t = Ψs,t for (s, t) ∈ and supm ∥Ψ(m)∥λ < ∞. Then,

for α ∈ (0,minλ, 1) and t ∈ (0, T ], limm→∞ Dαt−Ψ(m)(u) = Dα

t−Ψ(u) for u ∈ [0, t].

Proof. Fix u ∈ [0, t) since the equality obviously holds when u = t. According to the

definition of Dαt−, it suffices to show that

limm→∞

∫ t

u

∥Ψ(m)u,v −Ψu,v∥V(v − u)α+1

dv = 0.

This immediately follows from the Lebesgue’s dominated convergence theorem.

Lemma 3.2. The following statements (1) and (2) hold under the assumptions of Propo-

sitions 2.6 and 2.7, respectively: for α ∈ (0, β) and t ∈ (0, T ],

(1) limm→∞R(n,α)t− X(m)(u) = R(n,α)

t− X(u) for n = 2, 3 and u ∈ [0, t],

(2) limm→∞R(3,α)t− Z(m)(u) = R(3,α)

t− Z(u) for u ∈ [0, t].

Proof. We show statement (1) and omit (2) because statement (2) is similar to (1)

with n = 3. Fix u ∈ [0, t) since the equalities obviously hold when u = t. It follows

from Lemma 3.1 that limm→∞Dnαt−X(m)n(u) = Dnα

t−Xn(u) holds for n = 2, 3. Therefore,

according to the definition of R(n,α)t− , it suffices to show that

limm→∞

∫ t

u

|X(m)ju,v ⊗R(1,α)t− X(m)(v)−Xj

u,v ⊗R(1,α)t− X(v)|

(v − u)jα+1dv = 0(3.1)

for j = 1, 2 and

limm→∞

∫ t

u

|X(m)1u,v ⊗R(2,α)t− X(m)(v)−X1

u,v ⊗R(2,α)t− X(v)|

(v − u)α+1dv = 0.(3.2)

Eq. (3.1) follows from Lemma 3.1, Eq. (2.9), and the Lebesgue’s dominated convergence

theorem. In particular, we obtain statement (1) with n = 2. Similarly, Eq. (3.2) follows

from statement (1) with n = 2, Eq. (2.12) with n = 2, and the Lebesgue’s dominated

convergence theorem. Therefore, we obtain statement (1) with n = 3.

Proof of Proposition 2.6. Fix n with 2 ≤ n ≤ 3 because Eq. (2.14) obviously holds when

n = 1, from Eq. (2.6) with f = X1a,·. It follows from [13, Proposition 3.4] that Eq. (2.14)

holds when X ∈ SΩβ(E). Therefore, from Eq. (2.14) for X(m) ∈ SΩβ(E), the triangle

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 17

inequality, and the Lebesgue’s dominated convergence theorem,

|Xna,b −

n∑k=1

(−1)1−kα

∫ t

s

D1−kαs+ 1(a,b](u)X

n−ka,u ⊗R(k,α)

t− X(u) du|

≤ |Xna,b −X(m)na,b|

+n∑

k=1

∫ t

s

|D1−kαs+ 1(a,b](u)(X

n−ka,u ⊗R(k,α)

t− X(u)−X(m)n−ka,u ⊗R(k,α)

t− X(m)(u))| du

→ 0 as m→ ∞,

where we have used Lemmas 3.1, 3.2, Eqs. (2.9), and (2.12) to apply the Lebesgue’s

dominated convergence theorem. Therefore, we obtain the statement of the proposition.

Next, we introduce the following three lemmas to prove Proposition 2.7.

Lemma 3.3. Let X ∈ SΩβ(E) and Z ∈ QβX(G). Then the following statements (1) and

(2) hold:

(1) for α ∈ (0, β) and t ∈ (0, T ], the functions R(1,α)t− X and R(2,α)

t− X are Holder

continuous of orders (1− α) and (1− 2α), respectively on [0, t],

(2) for a ∈ [0, T ), the functions u 7→ R20(Z)a,u and u 7→ R1

1(Z)a,u are Holder continu-

ous of orders 3β and 2β, respectively on [a, T ].

Proof. Statement (1) follows from [13, Proposition 5.2 and Corollary 5.3]. State-

ment (2) follows easily from

R20(Z)a,v −R2

0(Z)a,u −R20(Z)u,v = R1

1(Z)a,uX1u,v + δZ(2)

a,uX2u,v,(3.3)

R11(Z)a,v −R1

1(Z)a,u −R11(Z)u,v = δZ(2)

a,uX1u,v(3.4)

for a ≤ u ≤ v ≤ t and ∥R2−ll (Z)∥(3−l)β <∞ for l = 0, 1, 2, ∥Xn∥n <∞ for n = 1, 2.

Statements (1) and (2) of Lemma 3.3 are used frequently in the proofs of Lemma 3.5

and Proposition 2.7 without being explicitly noted.

Lemma 3.4 (cf. Lemma 3.3 in [12]). Let f ∈ Cλ1 (L(V,W )) and g ∈ Cλ

1 (V ) with λ ∈ (0, 1].

Then, for α ∈ (0, λ) and (u, t) ∈ with u < t,

Dαt−(fg)(u) = f(u)Dα

t−g(u) +(−1)1+αα

Γ(1− α)

∫ t

u

(f(v)− f(u))g(v)

(v − u)α+1dv.(3.5)

Lemma 3.5. Let X ∈ SΩβ(E) and Z ∈ QβX(G). Then, for α ∈ (0, β) and a, t, u ∈ [0, T ]

with a < u < t,

R(3,α)t− Z(u) = D3α

t−(R20(Z)a,· −R2

0(Z)a,t)(u)(3.6)

−D2αt−(R

11(Z)a,·R

(1,α)t− X)(u)−Dα

t−(δZ(2)a,· R

(2,α)t− X)(u).

18 Y. ITO

Proof. We note that Eq. (3.6) is equivalent to the following equality:

D3αt−(R

20(Z)a,· −R2

0(Z)a,t)(u)−D3αt−R

20(Z)(u)(3.7)

= D2αt−(R

11(Z)a,·R

(1,α)t− X)(u)−D2α

t−(R11(Z)R

(1,α)t− X)(u)

+Dαt−(δZ

(2)a,· R

(2,α)t− X)(u)−Dα

t−(δZ(2)R(2,α)

t− X)(u).

Let us now prove the preceding equality (Eq. (3.7)). From Eq. (3.3),

D3αt−(R

20(Z)a,· −R2

0(Z)a,t)(u) = D3αt−R

20(Z)(u) +R1

1(Z)a,uD3αt−X

1(u) + δZ(2)a,uD3α

t−X2(u),

and therefore

(left-hand side of (3.7)) = R11(Z)a,uD3α

t−X1(u) + δZ(2)

a,uD3αt−X

2(u).

Subsequently, we calculate the right-hand side of Eq. (3.7). From Eqs. (3.5), (2.4), and

(3.4),

D2αt−(R

11(Z)a,·R

(1,α)t− X)(u)

(3.8)

= R11(Z)a,uD

2αt−(R

(1,α)t− X)(u) +

(−1)1+2α2α

Γ(1− 2α)

∫ t

u

(R11(Z)a,v −R1

1(Z)a,u)R(1,α)t− X(v)

(v − u)2α+1dv

= R11(Z)a,uD

3αt−(X

10,· −X1

0,t)(u) +(−1)1+2α2α

Γ(1− 2α)

∫ t

u

(R11(Z)u,v + δZ

(2)a,uX1

u,v)R(1,α)t− X(v)

(v − u)2α+1dv

= R11(Z)a,uD3α

t−X1(u) +D2α

t−(R11(Z)R

(1,α)t− X)(u) + δZ(2)

a,uD2αt−(X

1 ⊗R(1,α)t− X)(u).

Similarly,

Dαt−(δZ

(2)a,· R

(2,α)t− X)(u)(3.9)

= δZ(2)a,uD

αt−(R

(2,α)t− X)(u) +

(−1)1+αα

Γ(1− α)

∫ t

u

(δZ(2)a,v − δZ

(2)a,u)R(2,α)

t− X(v)

(v − u)α+1dv

= δZ(2)a,u(D3α

t−X2(u)−D2α

t−(X1 ⊗R(1,α)

t− X)(u)) +Dαt−(δZ

(2)R(2,α)t− X)(u),

where in the last equality, we have used

Dαt−(R

(2,α)t− X)(u) = D3α

t−X2(u)−D2α

t−(X1 ⊗R(1,α)

t− X)(u).

For a proof, see, e.g., [9, Proposition 3.3]. Therefore, from Eqs. (3.8) and (3.9),

(right-hand side of (3.7)) = R11(Z)a,uD3α

t−X1(u) + δZ(2)

a,uD2αt−(X

1 ⊗R(1,α)t− X)(u)

+ δZ(2)a,u(D3α

t−X2(u)−D2α

t−(X1 ⊗R(1,α)

t− X)(u))

= R11(Z)a,uD3α

t−X1(u) + δZ(2)

a,uD3αt−X

2(u).

Hence, Eq. (3.7) holds. Subsequently, we obtain the statement of the lemma.

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 19

Proof of Proposition 2.7. Let I1, I2, and I3 denote the first, second, and third terms of

the right-hand side of Eq. (2.15), namely,

I1 := (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)R

11(Z)a,uR

(1,α)t− X(u) du,

I2 := (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)δZ

(2)a,uR

(2,α)t− X(u) du,

I3 := (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)R(3,α)

t− Z(u) du.

We first show that Eq. (2.15) holds when X ∈ SΩβ(E). Then, from Eqs. (2.6), (2.4),

(2.5), and (3.6),

R20(Z)a,b = R2

0(Z)a,b −R20(Z)a,a

= (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)D

αt−(R

20(Z)a,· −R2

0(Z)a,t)(u) du

= I1 + (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)(D

αt−(R

20(Z)a,· −R2

0(Z)a,t)(u)

−R11(Z)a,uR

(1,α)t− X(u)) du

= I1 + (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)(D

2αt−(R

20(Z)a,· −R2

0(Z)a,t)(u)

−Dαt−(R

11(Z)a,·R

(1,α)t− X)(u)) du

= I1 + I2 + (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)(D

2αt−(R

20(Z)a,· −R2

0(Z)a,t)(u)

−Dαt−(R

11(Z)a,·R

(1,α)t− X)(u)− δZ(2)

a,uR(2,α)t− X(u)) du

= I1 + I2 + (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)(D

3αt−(R

20(Z)a,· −R2

0(Z)a,t)(u)

−D2αt−(R

11(Z)a,·R

(1,α)t− X)(u)−Dα

t−(δZ(2)a,· R

(2,α)t− X)(u)) du

= I1 + I2 + I3.

Hence, Eq. (2.15) holds when X ∈ SΩβ(E). Next, we show that Eq. (2.15) holds when

X ∈ Ωβ(E) and Z ∈ QβX(G) satisfy the assumptions of Proposition 2.7. Then, from

Eq. (2.15) for X(m) ∈ SΩβ(E) and Z(m) ∈ QβX(m)(G), the triangle inequality, and the

Lebesgue’s dominated convergence theorem,

|R20(Z)a,b − (I1 + I2 + I3)|

≤ |R20(Z)a,b −R2

0(Z(m))a,b|

+

∫ t

s

|D1−αs+ 1(a,b](u)(R

11(Z)a,uR

(1,α)t− X(u)−R1

1(Z(m))a,uR(1,α)t− X(m)(u))| du

20 Y. ITO

+

∫ t

s

|D1−2αs+ 1(a,b](u)(δZ

(2)a,uR

(2,α)t− X(u)− δZ(m)(2)a,uR

(2,α)t− X(m)(u))| du

+

∫ t

s

|D1−3αs+ 1(a,b](u)(R(3,α)

t− Z(u)−R(3,α)t− Z(m)(u))| du

→ 0 as m→ ∞,

where we have used Lemmas 3.1, 3.2, Eqs. (2.9), (2.12), and (2.13) to apply the Lebesgue’s

dominated convergence theorem. Therefore, we obtain the statement of the proposition.

3.2. Proof of Proposition 2.11. Proposition 2.11 follows straightforwardly from Defi-

nitions 2.4 and 2.5.

Lemma 3.6. Let X ∈ Ωβ(E) and Z ∈ QβX(G). Suppose that X and Z satisfy the

properties (Hrβ) and (Hc

β), respectively. Then, for (s, t) ∈ with s < t, a, b ∈ [s, t] with

a < b, and α ∈ (0, β),

Z(0)b − Z(0)

a = (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)Z

(1)u R(1,α)

t− X(u) du(3.10)

+ (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)Z

(2)u R(2,α)

t− X(u) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)R(3,α)

t− Z(u) du.

Proof. From Eqs. (2.14) with n = 1, 2 and (2.15),

Z(0)b − Z(0)

a = Z(1)a X1

a,b + Z(2)a X2

a,b +R20(Z)a,b

= (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)(Z

(1)a + Z(2)

a X1a,u +R1

1(Z)a,u)R(1,α)t− X(u) du

+ (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)(Z

(2)a + δZ(2)

a,u)R(2,α)t− X(u) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)R(3,α)

t− Z(u) du.

Therefore, we obtain the statement of the lemma.

Proof of Proposition 2.11. From Eq. (2.14) with n = 3,

X3a,b + S(X3

a,b)(3.11)

= (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)2Sym(X2

a,u)⊗R(1,α)t− X(u) du

+ (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)(X

1a,u ⊗R(2,α)

t− X(u) + S(X1a,u ⊗R(2,α)

t− X(u))) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)(R(3,α)

t− X(u) + S(R(3,α)t− X(u))) du.

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 21

Subsequently, from Eqs. (3.10), (2.14), (3.11), and (2.1) with n = 2, we have

Ξa,b = Y (0)a (Z

(0)b − Z(0)

a ) + Y (1)a Z(1)

a X2a,b + Y (2)

a Z(1)a X3

a,b + Y (1)a Z(2)

a (X3a,b + S(X3

a,b))

= (−1)1−α

∫ t

s

D1−αs+ 1(a,b](u)

× (Y (0)a Z(1)

u + Y (1)a X1

a,uZ(1)a + Y (2)

a X2a,uZ

(1)a + Y (1)

a X1a,uZ

(2)a X1

a,u)R(1,α)t− X(u) du

+ (−1)1−2α

∫ t

s

D1−2αs+ 1(a,b](u)

× (Y (0)a Z(2)

u + Y (1)a Z(1)

a + Y (2)a X1

a,uZ(1)a )R(2,α)

t− X(u)

+ Y (1)a Z(2)

a (X1a,u ⊗R(2,α)

t− X(u) + S(X1a,u ⊗R(2,α)

t− X(u))) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)Y

(0)a R(3,α)

t− Z(u) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)Y

(2)a Z(1)

a R(3,α)t− X(u) du

+ (−1)1−3α

∫ t

s

D1−3αs+ 1(a,b](u)Y

(1)a Z(2)

a (R(3,α)t− X(u) + S(R(3,α)

t− X(u))) du.

Therefore, we calculate the first and second terms of the right-hand side of the preceding

equality. Fix u ∈ (a, t] because D1−nαs+ 1(a,b](u) = 0 for u ∈ (s, a] and n = 1, 2, 3. For the

first term, we have

Y (0)a Z(1)

u + Y (1)a X1

a,uZ(1)a + Y (2)

a X2a,uZ

(1)a + Y (1)

a X1a,uZ

(2)a X1

a,u

= (Y (0)a + Y (1)

a X1a,u + Y (2)

a X2a,u)Z

(1)u

− (Y (1)a X1

a,u + Y (2)a X2

a,u)(Z(1)u − Z(1)

a ) + Y (1)a X1

a,uZ(2)a X1

a,u

= (Y (0)a + Y (1)

a X1a,u + Y (2)

a X2a,u)Z

(1)u − Y (2)

a X2a,u(Z

(1)u − Z(1)

a )

− Y (1)a X1

a,u(Z(2)a X1

a,u +R11(Z)a,u) + Y (1)

a X1a,uZ

(2)a X1

a,u

= (Y (0)a + Y (1)

a X1a,u + Y (2)

a X2a,u)Z

(1)u − Y (2)

a X2a,u(Z

(1)u − Z(1)

a )− Y (1)a X1

a,uR11(Z)a,u

= Φ3a,u.

For the second term, we show

(Y (0)a Z(2)

u + Y (1)a Z(1)

a + Y (2)a X1

a,uZ(1)a )R(2,α)

t− X(u)(3.12)

+ Y (1)a Z(2)

a (X1a,u ⊗R(2,α)

t− X(u) + S(X1a,u ⊗R(2,α)

t− X(u))) = Φ2a,uR

(2,α)t− X(u).

We subsequently calculate the first term of the left-hand side of Eq. (3.12):

Y (0)a Z(2)

u + Y (1)a Z(1)

a + Y (2)a X1

a,uZ(1)a

= (Y (0)a + Y (1)

a X1a,u)Z

(2)u − Y (1)

a X1a,uZ

(2)u − Y (1)

a (Z(1)u − Z(1)

a ) + Y (1)a Z(1)

u

− Y (2)a X1

a,u(Z(1)u − Z(1)

a ) + Y (2)a X1

a,uZ(1)u

22 Y. ITO

= (Y (0)a + Y (1)

a X1a,u)Z

(2)u + (Y (1)

a + Y (2)a X1

a,u)Z(1)u − Y (2)

a X1a,u(Z

(1)u − Z(1)

a )

− Y (1)a X1

a,uZ(2)u − Y (1)

a (Z(2)a X1

a,u +R11(Z)a,u)

= (Y (0)a + Y (1)

a X1a,u)Z

(2)u + (Y (1)

a + Y (2)a X1

a,u)Z(1)u − Y (2)

a X1a,u(Z

(1)u − Z(1)

a )− Y (1)a R1

1(Z)a,u

− Y (1)a X1

a,uZ(2)u − Y (1)

a (Z(2)a X1

a,u)

= Φ2a,u + Y (1)

a X1a,uδZ

(2)a,u − Y (1)

a X1a,uZ

(2)u − Y (1)

a (Z(2)a X1

a,u)

= Φ2a,u − Y (1)

a X1a,uZ

(2)a − Y (1)

a (Z(2)a X1

a,u).

Therefore,

(left-hand side of (3.12)) = (Φ2a,u − Y (1)

a X1a,uZ

(2)a − Y (1)

a (Z(2)a X1

a,u))R(2,α)t− X(u)

+ Y (1)a Z(2)

a (X1a,u ⊗R(2,α)

t− X(u) + S(X1a,u ⊗R(2,α)

t− X(u)))

= Φ2a,uR

(2,α)t− X(u).

Hence, Eq. (3.12) holds. Therefore, we obtain the statement of the proposition.

3.3. Proof of Proposition 2.13. Using Lemmas 3.7 and 3.8 below, we prove Proposi-

tion 2.13. Our proof of Lemma 3.8 is based on the proof of [12, Proposition 2.4].

Lemma 3.7 (cf. Lemma 3.9 in [12]). Let λ ∈ (0,∞) and µ ∈ (0, 1) with λ + µ > 1.

Take (s, t) ∈ with s < t. Then, for the partition t0, t1, . . . , tm of the interval [s, t]

such that s = t0 < t1 < · · · < tm = t,

m−1∑i=1

i−1∑j=0

(tj+1 − tj)λ

∫ ti+1

ti

∫ tj+1

tj

(u− v)µ−2 dv du

≤ (1− µ)−1µ−1 max0≤j≤m−2

|tj+1 − tj|λ+µ−1(tm−1 − s).

Lemma 3.8. Fix an integer n with 2 ≤ n ≤ 3. Let Ψ ∈ C(, V ) such that

∥Ψw,u −Ψv,u∥V ≤ C

n−1∑k=0

(w − v)(n−k)β(u− w)kβ

for u, v, w ∈ [0, T ] with v ≤ w ≤ u. Here, C is a positive constant that does not depend

on u, v, w. Set α := 1− (4− n)γ and

Dαs+(Ψ·,u)(u) :=

1

Γ(1− α)

(Ψu,u

(u− s)α+ α

∫ u

s

Ψu,u −Ψv,u

(u− v)α+1dv

)for u ∈ (s, T ]. Then, for (s, t) ∈ with s < t,

lim|Ps,t|→0

∫ t

s

∥Dαs+(

m−1∑i=0

Ψti,u1(ti,ti+1])(u)−Dαs+(Ψ·,u)(u)∥V du = 0.

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 23

Proof. By the definition of Ψ, ∥Ψu,u −Ψv,u∥V ≤ C(u− v)nβ holds for (v, u) ∈ , and

thus Dαs+(Ψ·,u)(u) is well-defined for u ∈ (s, T ] because α < nβ follows from (1− β)/3 <

γ < β. For u ∈ (s, t], we set

Γ(1− α)(Dαs+(

m−1∑i=0

Ψti,u1(ti,ti+1])(u)−Dαs+(Ψ·,u)(u))

= (m−1∑i=0

Ψti,u1(ti,ti+1](u)−Ψu,u)(u− s)−α + α

∫ u

s

ΨPv,u

(u− v)α+1dv =: S1

P(u) + S2P(u).

Here,

ΨPv,u :=

m−1∑i=0

Ψti,u(1(ti,ti+1](u)− 1(ti,ti+1](v))− (Ψu,u −Ψv,u)

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

lim|Ps,t|→0

∫ t

s

∥SlP(u)∥V du = 0(3.13)

holds for l = 1, 2. First, from the equality

Ψti,u1(ti,ti+1](u)−Ψu,u = Ψti,u −Ψu,u

for u ∈ (ti, ti+1], we have

∫ t

s

∥S1P(u)∥V du =

m−1∑i=0

∫ ti+1

ti

∥S1P(u)∥V du

=m−1∑i=0

∫ ti+1

ti

∥Ψti,u −Ψu,u∥V (u− s)−α du

≤ C

m−1∑i=0

∫ ti+1

ti

(u− ti)nβ(u− s)−α du

≤ C|Ps,t|nβm−1∑i=0

∫ ti+1

ti

(u− s)−α du

= C|Ps,t|nβ(1− α)−1(t− s)1−α.

Therefore, Eq. (3.13) holds for l = 1. Next, using the equality

S2P(u) = α

i−1∑j=0

∫ tj+1

tj

ΨPv,u

(u− v)α+1dv + α

∫ u

ti

ΨPv,u

(u− v)α+1dv

24 Y. ITO

for u ∈ (ti, ti+1], we have∫ t

s

∥S2P(u)∥V du =

∫ t1

t0

∥S2P(u)∥V du+

m−1∑i=1

∫ ti+1

ti

∥S2P(u)∥V du

≤ α

m−1∑i=0

∫ ti+1

ti

∫ u

ti

∥ΨPv,u∥V

(u− v)α+1dv du

+ αm−1∑i=1

∫ ti+1

ti

i−1∑j=0

∫ tj+1

tj

∥ΨPv,u∥V

(u− v)α+1dv du

=: A1 + A2.

When (v, u) ∈ with ti < v ≤ u ≤ ti+1, we have ΨPv,u = −(Ψu,u −Ψv,u). Hence,

A1 ≤ Cαm−1∑i=0

∫ ti+1

ti

∫ u

ti

(u− v)nβ−α−1 dv du

= Cα(nβ − α)−1(nβ − α + 1)−1

m−1∑i=0

(ti+1 − ti)nβ−α+1

≤ Cα(nβ − α)−1(nβ − α + 1)−1|Ps,t|nβ−α(t− s),

where we have used α < nβ. When (v, u) ∈ with tj < v ≤ tj+1 ≤ ti < u ≤ ti+1,

ΨPv,u = Ψti,u −Ψtj ,u − (Ψu,u −Ψv,u) = −(Ψu,u −Ψti,u) + (Ψv,u −Ψtj ,u),

and thus

∥ΨPv,u∥V ≤ C(u− ti)

nβ +n−1∑k=0

(v − tj)(n−k)β(u− v)kβ.

Therefore, by a straightforward computation,

A2 ≤ Cαm−1∑i=1

∫ ti+1

ti

(u− ti)nβ

∫ ti

s

(u− v)−α−1 dv du

+ Cα

n−1∑k=0

m−1∑i=1

i−1∑j=0

∫ ti+1

ti

∫ tj+1

tj

(v − tj)(n−k)β(u− v)kβ−α−1 dv du

≤ Cm−1∑i=1

∫ ti+1

ti

(u− ti)nβ−α du

+ Cα

n−1∑k=0

m−1∑i=1

i−1∑j=0

(tj+1 − tj)(n−k)β

∫ ti+1

ti

∫ tj+1

tj

(u− v)kβ−α−1 dv du

≤ C(nβ − α + 1)−1|Ps,t|nβ−α(t− s)

+ Cα

n−1∑k=0

(1− (kβ − α + 1))−1(kβ − α + 1)−1|Ps,t|nβ−α(t− s),

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 25

where we have used α < nβ. In the last inequality, we have used Lemma 3.7 with

λ = (n − k)β and µ = kβ − α + 1. Hence, from the estimates of A1 and A2, Eq. (3.13)

holds for l = 2. Therefore, we obtain the statement of the lemma.

Proof of Proposition 2.13. Fix u, v, w ∈ [0, T ] with v ≤ w ≤ u. Throughout this proof, C

denotes a positive constant that does not depend on u, v, w and may change line-by-line.

First, using Lemma 3.8 with n = 3, we prove Eq. (2.19). By the definition of Φ3,

Φ3w,u − Φ3

v,u = ((Y (0)w + Y (1)

w X1w,u + Y (2)

w X2w,u)− (Y (0)

v + Y (1)v X1

v,u + Y (2)v X2

v,u))Z(1)u

+ (−Y (2)w X2

w,uδZ(1)w,u + Y (2)

v X2v,uδZ

(1)v,u)

+ (−Y (1)w X1

w,uR11(Z)w,u + Y (1)

v X1v,uR

11(Z)v,u)

=: A1 + A2 + A3.

We decompose A1, A2, and A3 as follows:

A1 = (−R20(Y )w,u +R2

0(Y )v,u)Z(1)u = (R2

0(Y )v,w +R11(Y )v,wX

1w,u + δZ(2)

v,wX2w,u)Z

(1)u

and

A2 = −Y (2)w X2

w,uδZ(1)w,u + Y (2)

v (X2v,w +X2

w,u +X1v,w ⊗X1

w,u)(δZ(1)v,w + δZ(1)

w,u)

= (−Y (2)w + Y (2)

v )X2w,uδZ

(1)w,u

+ Y (2)v (X2

v,w(δZ(1)v,w + δZ(1)

w,u) +X2w,uδZ

(1)v,w + (X1

v,w ⊗X1w,u)(δZ

(1)v,w + δZ(1)

w,u)).

Similarly,

A3 = −Y (1)w X1

w,uR11(Z)w,u + Y (1)

v (X1v,w +X1

w,u)(R11(Z)v,w +R1

1(Z)w,u + δZ(2)v,wX

1w,u)

= (−Y (1)w + Y (1)

v )X1w,uR

11(Z)w,u

+ Y (1)v (X1

v,w(R11(Z)v,w +R1

1(Z)w,u + δZ(2)v,wX

1w,u) +X1

w,u(R11(Z)v,w + δZ(2)

v,wX1w,u)).

Therefore,

|Ai| ≤ C(w − v)3β + (w − v)2β(u− w)β + (w − v)β(u− w)2β

for i = 1, 2, 3, and thus

|Φ3w,u − Φ3

v,u| ≤ C(w − v)3β + (w − v)2β(u− w)β + (w − v)β(u− w)2β.

Hence, Eq. (2.19) holds from Lemma 3.8 with n = 3.

26 Y. ITO

Next, using Lemma 3.8 with n = 2, we prove Eq. (2.20). By the definition of Φ2,

Φ2w,u − Φ2

v,u = ((Y (0)w + Y (1)

w X1w,u)− (Y (0)

v + Y (1)v X1

v,u))Z(2)u

+ ((Y (1)w + Y (2)

w X1w,u)− (Y (1)

v + Y (2)v X1

v,u))Z(1)u

+ (−Y (2)w X1

w,uδZ(1)w,u + Y (2)

v X1v,uδZ

(1)v,u)

+ (−Y (1)w R1

1(Z)w,u + Y (1)v R1

1(Z)v,u)

+ (−Y (1)w X1

w,uδZ(2)w,u + Y (1)

v X1v,uδZ

(2)v,u)

=: B1 +B2 +B3 +B4 +B5.

We decompose B1, B2, B3, B4, and B5 as follows:

B1 = (−R10(Y )w,u +R1

0(Y )v,u)Z(2)u = (R1

0(Y )v,w + δY (1)v,wX

1w,u)Z

(2)u

and

B2 = (−R11(Y )w,u +R1

1(Y )v,u)Z(1)u = (R1

1(Y )v,w + δY (2)v,wX

1w,u)Z

(1)u .

In addition,

B3 = −Y (2)w X1

w,uδZ(1)w,u + Y (2)

v (X1v,w +X1

w,u)(δZ(1)v,w + δZ(1)

w,u)

= (−Y (2)w + Y (2)

v )X1w,uδZ

(1)w,u + Y (2)

v (X1v,w(δZ

(1)v,w + δZ(1)

w,u) +X1w,uδZ

(1)v,w)

and

B5 = −Y (1)w X1

w,uδZ(2)w,u + Y (1)

v (X1v,w +X1

w,u)(δZ(2)v,w + δZ(2)

w,u)

= (−Y (1)w + Y (1)

v )X1w,uδZ

(2)w,u + Y (1)

v (X1v,w(δZ

(2)v,w + δZ(2)

w,u) +X1w,uδZ

(2)v,w).

Furthermore,

B4 = −Y (1)w R1

1(Z)w,u + Y (1)v (R1

1(Z)v,w +R11(Z)w,u + δZ(2)

v,wX1w,u)

= (−Y (1)w + Y (1)

v )R11(Z)w,u + Y (1)

v (R11(Z)v,w + δZ(2)

v,wX1w,u).

Therefore,

|Bi| ≤ C(w − v)2β + (w − v)β(u− w)β

for i = 1, 2, 3, 4, 5, and thus

|Φ2w,u − Φ2

v,u| ≤ C(w − v)2β + (w − v)β(u− w)β.

Hence, Eq. (2.20) holds from Lemma 3.8 with n = 2. Therefore, we obtain the statement

of the proposition.

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 27

4. Appendix

The aim of this appendix is to provide basic relations between the integral of 1-forms

against geometric Holder rough paths and the main results of this paper. First, we briefly

review the definition of the rough integral of 1-forms and its fundamental properties in the

context of this paper. For more details on the rough integral of 1-forms, see, e.g., [14–16].

Let f be a 3-times continuously Frechet differentiable function from E to L(E,F ) such

that f and ∇jf for j = 1, 2, 3 are bounded on E. Let X ∈ GΩβ(E) and ξ ∈ E. We recall

that 1/4 < β ≤ 1/3. We set Xt := ξ +X10,t for t ∈ [0, T ] and

Y 1s,t := f(Xs)X

1s,t +∇f(Xs)X

2s,t +∇2f(Xs)X

3s,t,

Y 2s,t := f(Xs)⊗ f(Xs)X

2s,t +∇f(Xs)⊗ f(Xs)X

3s,t + f(Xs)⊗∇f(Xs)(X

3s,t + S(X3

s,t)),

Y 3s,t := f(Xs)⊗ f(Xs)⊗ f(Xs)X

3s,t

for (s, t) ∈ . We recall that S(a⊗ b⊗ c) = b⊗ a⊗ c for a, b, c ∈ E. Then the limits

Y 1s,t := lim

|Ps,t|→0

m−1∑i=0

Y 1ti,ti+1

,

Y 2s,t := lim

|Ps,t|→0

m−1∑i=0

Y 2ti,ti+1

+ Y 1s,ti

⊗ Y 1ti,ti+1

,

Y 3s,t := lim

|Ps,t|→0

m−1∑i=0

Y 3ti,ti+1

+ Y 1s,ti

⊗ Y 2ti,ti+1

+ Y 2s,ti

⊗ Y 1ti,ti+1

exist for (s, t) ∈ . We set Ys,t := (1, Y 1s,t, Y

2s,t, Y

3s,t) ∈ T (3)(F ) for (s, t) ∈ and call

Y = (1, Y 1, Y 2, Y 3) the integral of f againstX ∈ GΩβ(E) with ξ. The following properties

of Y are fundamental: (1) Y ∈ GΩβ(F ), (2) the map X 7→ Y is continuous from GΩβ(E)

to GΩβ(F ), and (3) ∥Y n − Y n∥4β <∞ for n = 1, 2, 3.

Next, under the above-mentioned notation and assumptions, we provide Propositions 4.1,

4.2, and 4.3 below as the basic relations between Y and IX(Y, Z). These propositions

straightforwardly follow from the definitions of Y and IX(Y, Z), the above-mentioned

properties of Y , and Theorems 2.8 and 2.9. For this reason, we omit the details of the

proofs of several statements in Propositions 4.1, 4.2, and 4.3 when the statements appear

to be well-known or can be verified by a straightforward computation.

Proposition 4.1. Set Y = (Y (0), Y (1), Y (2)) ∈ QβX(L(E,F )) as Y

(0)t = f(Xt), Y

(1)t =

∇f(Xt), and Y(2)t = ∇2f(Xt) for t ∈ [0, T ] and Z = (Z(0), Z(1), Z(2)) ∈ Qβ

X(E) as

Z(0)t = X1

0,t, Z(1)t = I, and Z

(2)t = 0 for t ∈ [0, T ]. Here, I ∈ L(E,E) denotes the identity

map. Then Y 1s,t = IX(Y, Z)s,t for (s, t) ∈ .

Proof. It is clear that Y ∈ QβX(L(E,F )) and Z ∈ Qβ

X(E) hold, and R2−ll (Z)s,t = 0

for l = 0, 1, 2 and (s, t) ∈ . In particular, Z satisfies the property (Hcβ). It is also clear

28 Y. ITO

that Ξs,t = Y 1s,t for (s, t) ∈ . Since X ∈ GΩβ(E) satisfies Eq. (2.1) with n = 2 and the

property (Hrβ), Theorem 2.8 gives Y 1

s,t = IX(Y, Z)s,t for (s, t) ∈ .

Proposition 4.2. Set Y = (Y (0), Y (1), Y (2)) ∈ QβX(L(F, F

⊗2)) as Y(0)t (η) = Y 1

0,t ⊗ η,

(Y(1)t (x))(η) = f(Xt)x⊗ η, and (Y

(2)t (x⊗ y))(η) = ∇f(Xt)x⊗ y ⊗ η for η ∈ F , x, y ∈ E,

and t ∈ [0, T ] and Z = (Z(0), Z(1), Z(2)) ∈ QβX(F ) as Z

(0)t = Y 1

0,t, Z(1)t = f(Xt), and

Z(2)t = ∇f(Xt) for t ∈ [0, T ]. Then Y 2

0,t = IX(Y, Z)0,t for t ∈ [0, T ] and Y 2s,t = Y 2

s,t +

(IX(Y, Z)s,t − Ξs,t) for (s, t) ∈ .

Proof. It is clear that Y ∈ QβX(L(F, F

⊗2)) and Z ∈ QβX(F ) hold. By the definition of

X ∈ GΩβ(E), there exists a sequence X(m) ∈ SΩβ(E) such that limm→∞ dβ(X(m), X) =

0. Subsequently, set Z(m) = (Z(m)(0), Z(m)(1), Z(m)(2)) ∈ QβX(m)(F ) as Z(m)

(0)t =

Y (m)10,t, Z(m)(1)t = f(X(m)t), and Z(m)

(2)t = ∇f(X(m)t) for t ∈ [0, T ]. Here, X(m)t :=

ξ + X(m)10,t for t ∈ [0, T ] and Y (m) denotes the integral of f against X(m) ∈ SΩβ(E)

with ξ. Then the assumption (2) of Theorem 2.9 is fulfilled. This can be proved by a

straightforward computation. In addition, from the definitions of Y and Z, for (s, t) ∈ ,

Ξs,t = Y 10,s ⊗ (Y 1

0,t − Y 10,s) + f(Xs)⊗ f(Xs)X

2s,t(4.1)

+∇f(Xs)⊗ f(Xs)X3s,t + f(Xs)⊗∇f(Xs)(X

3s,t + S(X3

s,t))

= Y 10,s ⊗ Y 1

s,t + Y 2s,t.

Therefore, from Theorem 2.9, for t ∈ [0, T ],

IX(Y, Z)0,t = lim|P0,t|→0

m−1∑i=0

Y 10,ti

⊗ Y 1ti,ti+1

+ Y 2ti,ti+1

= Y 20,t.

Hence, from Eqs. (2.17) and (4.1), for (s, t) ∈ ,

Y 2s,t = Y 2

0,t − Y 20,s − Y 1

0,s ⊗ Y 1s,t = IX(Y, Z)s,t − Y 1

0,s ⊗ Y 1s,t = Y 2

s,t + (IX(Y, Z)s,t − Ξs,t).

Thus, we obtain the statement of the proposition.

Proposition 4.3. Set Y = (Y (0), Y (1), Y (2)) ∈ QβX(L(F, F

⊗3)) as Y(0)t (η) = Y 2

0,t ⊗ η,

(Y(1)t (x))(η) = Y 1

0,t ⊗ f(Xt)x ⊗ η, and (Y(2)t (x ⊗ y))(η) = f(Xt)x ⊗ f(Xt)y ⊗ η + Y 1

0,t ⊗∇f(Xt)x⊗ y ⊗ η for η ∈ F , x, y ∈ E, and t ∈ [0, T ] and Z = (Z(0), Z(1), Z(2)) ∈ Qβ

X(F ),

as in Proposition 4.2. Then Y 30,t = IX(Y, Z)0,t for t ∈ [0, T ] and Y 3

s,t = Y 3s,t − Y 1

0,s ⊗ (Y 2s,t −

Y 2s,t) + (IX(Y, Z)s,t − Ξs,t) for (s, t) ∈ .

Proof. First, we show that Y belongs to QβX(L(F, F

⊗3)). It is clear that Y (l) ∈Cβ1 (L(E

⊗l, L(F, F⊗3))) holds for l = 0, 1, 2. For (s, t) ∈ ,

R20(Y )s,t = Y 2

0,t − Y 20,s − Y 1

0,s ⊗ f(Xs)X1s,t − f(Xs)⊗ f(Xs)X

2s,t − Y 1

0,s ⊗∇f(Xs)X2s,t

= (Y 2s,t + Y 1

0,s ⊗ Y 1s,t)− f(Xs)⊗ f(Xs)X

2s,t − Y 1

0,s ⊗ (f(Xs)X1s,t +∇f(Xs)X

2s,t)

= (Y 2s,t − f(Xs)⊗ f(Xs)X

2s,t) + Y 1

0,s ⊗ (Y 1s,t − f(Xs)X

1s,t −∇f(Xs)X

2s,t).

ROUGH INTEGRATION VIA FRACTIONAL CALCULUS 29

Hence, R20(Y ) ∈ C3β

2 (L(F, F⊗3)) holds. For (s, t) ∈ ,

R11(Y )s,t = Y 1

0,t ⊗ f(Xt)− Y 10,s ⊗ f(Xs)− f(Xs)X

1s,t ⊗ f(Xs)− Y 1

0,s ⊗∇f(Xs)X1s,t

= (Y 10,s + Y 1

s,t)⊗ f(Xt)− f(Xs)X1s,t ⊗ f(Xs)− Y 1

0,s ⊗ (f(Xs) +∇f(Xs)X1s,t)

= Y 1s,t ⊗ f(Xt)− f(Xs)X

1s,t ⊗ f(Xs) + Y 1

0,s ⊗ (f(Xt)− f(Xs)−∇f(Xs)X1s,t)

= Y 1s,t ⊗ (f(Xt)− f(Xs)) + (Y 1

s,t − f(Xs)X1s,t)⊗ f(Xs)

+ Y 10,s ⊗ (f(Xt)− f(Xs)−∇f(Xs)X

1s,t).

Hence, R11(Y ) ∈ C2β

2 (L(E,L(F, F⊗3))) holds, and thus Y ∈ QβX(L(F, F

⊗3)).

Next, we show the desired equalities. From the definitions of Y and Z, for (s, t) ∈ ,

Ξs,t = Y 20,s ⊗ (Y 1

0,t − Y 10,s) + Y 1

0,s ⊗ f(Xs)⊗ f(Xs)X2s,t(4.2)

+ f(Xs)⊗ f(Xs)⊗ f(Xs)X3s,t + Y 1

0,s ⊗∇f(Xs)⊗ f(Xs)X3s,t

+ Y 10,s ⊗ f(Xs)⊗∇f(Xs)(X

3s,t + S(X3

s,t))

= Y 3s,t + Y 2

0,s ⊗ Y 1s,t + Y 1

0,s ⊗ Y 2s,t.

Since Z is the same as in Proposition 4.2, we can use Theorem 2.9. Therefore, for t ∈ [0, T ],

IX(Y, Z)0,t = lim|P0,t|→0

m−1∑i=0

Y 3ti,ti+1

+ Y 20,ti

⊗ Y 1ti,ti+1

+ Y 10,ti

⊗ Y 2ti,ti+1

= Y 30,t.

In the last equality, we have used lim|P0,t|→0

∑m−1i=0 Y 1

0,ti⊗ (Y 2

ti,ti+1− Y 2

ti,ti+1) = 0 from the

property (3) of Y with n = 2 (or from Proposition 4.2 and (4) of Remark 2.3). Hence,

from Eqs. (2.17) and (4.2), for (s, t) ∈ ,

Y 3s,t = Y 3

0,t − Y 30,s − Y 1

0,s ⊗ Y 2s,t − Y 2

0,s ⊗ Y 1s,t

= IX(Y, Z)s,t − Y 10,s ⊗ Y 2

s,t − Y 20,s ⊗ Y 1

s,t

= Y 3s,t − Y 1

0,s ⊗ (Y 2s,t − Y 2

s,t) + (IX(Y, Z)s,t − Ξs,t).

Thus, we obtain the statement of the proposition.

Acknowledgements. The author thanks Professor Masanori Hino for his helpful sug-

gestions and comments on this work.

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Department of Mathematics

Faculty of Science

Kyoto Sangyo University

Kyoto 603-8555

Japan

Email address: itoyu@cc.kyoto-su.ac.jp