# Fractional Itô Calculus

Rama Cont and Ruhong Jin
Mathematical Institute, University of Oxford

(2021)

###### Abstract

We derive Itô-type change of variable formulas for smooth functionals of
irregular paths with non-zero $p-$th variation along a sequence of partitions
where $p\geq 1$ is arbitrary, in terms of fractional derivative operators,
extending the results of the Föllmer-Ito calculus to the general case of paths
with ’fractional’ regularity. In the case where $p$ is not an integer, we show
that the change of variable formula may sometimes contain a non-zero a
’fractional’ Itô remainder term and provide a representation for this
remainder term. These results are then extended to paths with non-zero
$\phi-$variation and multi-dimensional paths. Using these results, we derive
an isometry property for the pathwise Föllmer integral in terms of $\phi$
variation.

###### Contents

  1. 1 Preliminaries
    1. 1.1 Pathwise calculus for paths with finite p-th variation
    2. 1.2 Fractional derivatives and fractional Taylor expansions
    3. 1.3 A characterization of the local fractional derivative
  2. 2 Fractional Itô Calculus
    1. 2.1 A change of variable formula for time-independent functions
    2. 2.2 Change of variable formula for time-dependent functions
    3. 2.3 Extension to path-dependent functionals
  3. 3 Extension to $\phi$-variation
  4. 4 An isometry formula for the pathwise integral
  5. 5 Multi-dimensional extensions

Hans Föllmer derived in [13] a pathwise formulation of the Ito formula and
laid the grounds for the development of a pathwise approach to Ito calculus,
which has been developed in different directions in [1, 2, 5, 6, 7, 10, 8, 11,
19, 24].

Föllmer’s original approach focuses on functions of paths with finite
quadratic variation along a sequence of partitions. In a recent work [10],
Cont and Perkowski extended the Föllmer-Ito formula of [13] to function(al)s
of paths with variation of order $p\in 2\mathbb{N}$ along a sequence of
partitions and obtained functional change of variable formulas, applicable to
functionals of Fractional Brownian motion and other fractional processes with
arbitrarily low regularity (i.e. any Hurst exponent $H>0$). These results
involve pathwise integrals defined as limits of compensated left Riemann sums,
which are in turn related to rough integrals associated with a reduced order-p
rough path [10].

As the notion of $p$-the order variation may be defined for any $p>0$, an
interesting question is to investigate how the results in [10] extend to
’fractional’ case $p\notin\mathbb{N}$. In particular one may ask whether the
change of variable formula contains a fractional remainder term in this case
and whether the definition of the compensated integral needs to be adjusted.

We investigate these questions using the tools of fractional calculus [23].
Given that fractional derivative operators are (non-local) integral operators,
one challenge is to obtain non-anticipative, ’local’ formulas which have
similar properties to those obtained in the integer case [10]. We are able to
do so using a ’local’ notion of fractional derivative and exhibit conditions
under which these change of variable formulas contain (or not) a ’fractional
Itô remainder term’. It is shown that in most cases there is no remainder
term; we also discuss some cases where a non-zero remainder term appears and
give a representation for this term.

These results are first derived for smooth functions then extended to
functionals, using the concept of vertical derivative [9]. We extend these
results to the case of paths with finite $\phi$-variation [17] for a class of
functions $\phi$ and we obtain an isometry formula for the pathwise integral
in terms of $\phi$-variation, extending the results of [1, 10] to the
fractional case. Finally, we extend these results to the multi-dimensional
case.

Our change of variable formulas are purely analytical and pathwise in nature,
but applicable to functionals of fractional Brownian motions and other
fractional processes with arbitrary Hurst exponent, leading in this case to
non-anticipative ’Itô’ formulas for functionals of such processes. However, as
probabilistic assumptions play no role in the derivation of our results, we
have limited to a minimum the discussion of such examples.

#### Outline

Section 1 recalls some results on pathwise calculus for functions of irregular
paths (Sec. 1.1) and fractional derivative operators and associated fractional
Taylor expansions (Sec. 1.2).

Section 2 contains our main results on change of variable formulas for
function(al)s of paths with fractional regularity. We first give a change of
variable formula without remainder term for time-independent functions
(Theorem 2.7), followed by a discussion of an example where a remainder term
may appear (Example 2.9). We then provide a formula for computing this
fractional remainder term using an auxiliary space(Theorem 2.12). Sections 2.2
extends these results to the time-dependent case and Section 2.3 extends them
to the case of path-dependent functionals using the Dupire derivative.

In Section 3 we show that these results may be extended to the case where the
$p-$th variation is replaced by the more general concept of $\phi-$variation
[17]. In Section 4 we derive a pathwise isometry formula extending a result of
[1] to the case of $\phi-$variation. Finally, in Section 5 we discuss
extensions to the multidimensional case. These extensions are not immediate,
as the space $V_{p}(\pi)$ is not a vector space.

##  1 Preliminaries

###  1.1 Pathwise calculus for paths with finite p-th variation

We define, following [10, 13], the concept of $p$-th variation along a
sequence of partitions $\pi_{n}=\\{t_{0}^{n},\dots,t^{n}_{N(\pi_{n})}\\}$ with
$t_{0}^{n}=0<...<t^{n}_{k}<...<t^{n}_{N(\pi_{n})}=T$. Define the _oscillation_
of $S\in C([0,T],\mathbb{R})$ along $\pi_{n}$ as

$\operatorname{osc}(S,\pi_{n}):=\max_{[t_{j},t_{j+1}]\in\pi_{n}}\max_{r,s\in[t_{j},t_{j+1}]}|S(s)-S(r)|.$

Here and in the following we write $[t_{j},t_{j+1}]\in\pi_{n}$ to indicate
that $t_{j}$ and $t_{j+1}$ are both in $\pi_{n}$ and are immediate successors
(i.e. $t_{j}<t_{j+1}$ and $\pi_{n}\cap(t_{j},t_{j+1})=\emptyset$).

######  Definition 1.1 ($p$-th variation along a sequence of partitions).

Let $p>0$. A continuous path $S\in C([0,T],\mathbb{R})$ is said to have a
$p$-th variation along a sequence of partitions $\pi=(\pi_{n})_{n\geq 1}$ if
$\operatorname{osc}(S,\pi_{n})\to 0$ and the sequence of measures

$\mu^{n}:=\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\delta(\cdot-
t_{j})|S(t_{j+1})-S(t_{j})|^{p}$

converges weakly to a measure $\mu$ without atoms. In that case we write $S\in
V_{p}(\pi)$ and $[S]^{p}(t):=\mu([0,t])$ for $t\in[0,T]$, and we call
$[S]^{p}$ the _$p$ -th variation_ of $S$.

######  Remark 1.2.

  1. 1.

Functions in $V_{p}(\pi)$ do not necessarily have finite $p$-variation in the
usual sense. Recall that the $p$-variation of a function $f\in
C([0,T],\mathbb{R})$ is defined as [12]

$\|f\|_{\operatorname{p-var}}:=\Big{(}\sup_{\pi\in\Pi([0,T])}\sum_{[t_{j},t_{j+1}]\in\pi}|f(t_{j+1})-f(t_{j})|^{p}\Big{)}^{1/p},$

where the supremum is taken over the set $\Pi([0,T])$ of all partitions $\pi$
of $[0,T]$. A typical example is the Brownian motion $B$, which has quadratic
variation $[B]^{2}(t)=t$ along any refining sequence of partitions almost
surely while at the same time having infinite 2-variation almost surely [12,
25]:

$\mathbb{P}\left(\|B\|_{\operatorname{2-var}}=\infty\right)=1.$

  2. 2.

If $S\in V_{p}(\pi)$ and $q>p$, then $S\in V_{q}(\pi_{n})$ with $[S]^{q}\equiv
0$.

The following characterization is given in [10]:

######  Lemma 1.3.

Let $S\in C([0,T],\mathbb{R})$. $S\in V_{p}(\pi)$ if and only if there exists
a continuous function $[S]^{p}$ such that

$\forall
t\in[0,T],\qquad\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}:\\\
t_{j}\leq
t\end{subarray}}|S(t_{j+1})-S(t_{j})|^{p}\mathop{\longrightarrow}^{n\to\infty}[S]^{p}(t).$
(1)

If this property holds, then the convergence in (1) is uniform.

######  Example 1.4.

If $B$ is a fractional Brownian motion with Hurst index $H\in(0,1)$ and
$\pi_{n}=\\{kT/n:k\in\mathbb{N}_{0}\\}\cap[0,T]$, then $B\in V_{1/H}(\pi)$ and
$[B]^{1/H}(t)=t\mathbb{E}[|B_{1}|^{1/H}]$, see [21, 22].

Cont and Perkowski [10] obtained the following change of variable formula for
$p\in\mathbb{N}$, $S\in V_{p}(\pi)$ and $f\in C^{p}(\mathbb{R},\mathbb{R})$:

$f(S(t))-f(S(0))=\int_{0}^{t}f^{\prime}(S(s))\mathrm{d}S(s)+\frac{1}{p!}\int_{0}^{t}f^{(p)}(S(s))\mathrm{d}[S]^{p}(s),$

holds, where the integral

$\int_{0}^{t}f^{\prime}(S(s))\mathrm{d}S(s):=\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\sum_{k=1}^{p-1}\frac{f^{(k)}(S(t_{j}))}{k!}(S(t_{j+1}\wedge
t)-S(t_{j}\wedge t))^{k}$

is defined as a (pointwise) limit of compensated Riemann sums.

######  Remark 1.5 (Relation with Young integration and rough integration).

Note however that, given the assumptions on $S$, the pathwise integral
appearing in the formula cannot be defined as a Young integral. This relates
to the observation in Remark 1.2 that p-variation can be infinite for $S\in
V_{p}(\pi)$.

When $p=2$ it reduces to an ordinary (left) Riemann sum. For $p>2$ such
compensated Riemann sums appear in the construction of ‘rough path integrals’
[14, 16]. Let $X\in C^{\alpha}([0,T],\mathbb{R})$ be $\alpha$-Hölder
continuous for some $\alpha\in(0,1)$, and write $q=\lfloor\alpha^{-1}\rfloor$.
We can enhance $X$ uniquely into a (weakly) geometric rough path
$(\mathbb{X}^{1}_{s,t},\mathbb{X}^{2}_{s,t},\dots,\mathbb{X}^{q}_{s,t})_{0\leq
s\leq t\leq T}$, where $\mathbb{X}^{k}_{s,t}:=(X(t)-X(s))^{k}/k!$. Moreover,
for $g\in C^{q+1}(\mathbb{R},\mathbb{R})$ the function $g^{\prime}(X)$ is
controlled by $X$ with Gubinelli derivatives

$\displaystyle g^{\prime}(X(t))-g^{\prime}(X(s))$
$\displaystyle=\sum_{k=1}^{q-1}\frac{g^{(k+1)}(X(s))}{k!}(X(t)-X(s))^{k}+O(|t-s|^{q\alpha})$
$\displaystyle=\sum_{k=1}^{q-1}g^{(k+1)}(X(s))\mathbb{X}^{k}_{s,t}+O(|t-s|^{q\alpha}),$

and therefore the controlled rough path integral
$\int_{0}^{t}g^{\prime}(X(s))\mathrm{d}X(s)$ is given by

$\lim_{|\pi|\to
0}\sum_{[t_{j},t_{j+1}]\in\pi}\sum_{k=1}^{q}g^{(k)}(X(s))\mathbb{X}^{k}_{s,t}=\lim_{|\pi|\to
0}\sum_{[t_{j},t_{j+1}]\in\pi}\sum_{k=1}^{q}g^{(k)}(X(s))\frac{(X(t)-X(s))^{k}}{k!},$

where $|\pi|$ denotes the mesh size of the partition $\pi$, and which is the
type of compensated Riemann sum used to define the integral above, plus an
additional term (Ito term) absorbed into the sum).

###  1.2 Fractional derivatives and fractional Taylor expansions

We recall some definitions on fractional derivatives and their properties.
Several different notions of fractional derivatives exist in the literature
and it is not clear which ones are the right tools for a given context. Our
goal here is to shed some light on the advantages of different notions of
fractional derivative. Much of this material may be found in the literature
[23]. We have provided some detailed proofs for some useful properties whose
proof we have not been able to find in the literature.

######  Definition 1.6 (Riemann-Liouville fractional integration operator).

Suppose $f$ is a real function and $\alpha>0$, the left Riemann-Liouville
fractional integration operator of order $\alpha$ is defined by

$I_{a^{+}}^{\alpha}f(x):=\frac{1}{\Gamma(\alpha)}\int_{a}^{x}(x-t)^{\alpha-1}f(t)dt$

if integration exists for $x>a\in\mathbb{R}$. Similarly, the right Riemann-
Liouville fractional integration operator if given by

$I_{b^{-}}^{\alpha}f(x):=\frac{1}{\Gamma(\alpha)}\int_{x}^{b}(t-x)^{\alpha-1}f(t)dt$

for $x<b\in\mathbb{R}$.

This may be used to define a (non-local) fractional derivative of a real
function $f$ associated with some base point, as follows:

######  Definition 1.7 (Riemann-Liouville fractional derivative).

Suppose $f$ is a real function and $n\leq\alpha<n+1$ for some integer
$n\in\mathbb{N}$. Then left Riemann-Liouville fractional derivative of order
$\alpha$ with base point $a$ at $x>a$ is defined by

$D_{a^{+}}^{\alpha}f(x):=\left(\frac{d}{dx}\right)^{n+1}\left(I_{a^{+}}^{n+1-\alpha}f\right)(x)$

if it exists. Similarly, the right Riemann-Liouville fractional derivative of
order $\alpha$ with base point $b$ and $b>x$ is given by

$D_{b^{-}}^{\alpha}f(x):=\left(-\frac{d}{dx}\right)^{n+1}\left(I_{b^{-}}^{n+1-\alpha}f\right)(x)$

Some basic properties of the Riemann-Liouville fractional derivative are
described below.

######  Proposition 1.8.

Suppose $f$ is a real function, $\alpha,\beta$ are two real numbers. We will
also use the convention that $I_{a+}^{\alpha}=D_{a+}^{-\alpha}$. Then we have

  1. 1.

Fractional integration operators $\\{I_{a+}^{\alpha},\alpha\geq
0\\}$($\\{I_{b-}^{\alpha},\alpha\geq 0\\}$) form a semigroup in $L^{p}(a,b)$
for every $p\geq 1$. It is continuous in uniform topology for $\alpha>0$ and
strongly continuous for $\alpha\geq 0$.

  2. 2.

$I_{a+}^{\alpha}\circ
I_{a+}^{\beta}f=I_{a+}^{\alpha+\beta}f$($I_{b-}^{\alpha}\circ
I_{b-}^{\beta}f=I_{b-}^{\alpha+\beta}f$) is valid in the following cases:

     * •

$\beta\geq 0,\alpha+\beta\geq 0,f\in L^{1}([a,b])$

     * •

$\beta\leq 0,\alpha\geq 0,f\in I_{a+}^{-\beta}(L^{1})(I_{b-}^{-\beta}(L^{1}))$

     * •

$\alpha\leq 0,\alpha+\beta\leq 0,f\in
I_{a+}^{-\alpha-\beta}(L_{1})(I_{b-}^{-\alpha-\beta}(L_{1}))$

  3. 3.

If $n\leq\alpha<n+1$ and $D_{a^{+}}^{\alpha}f\in L^{1}([a,y])$ for some
$y>x\in[a,b]$ , then we have

$I_{a^{+}}^{\alpha}D_{a^{+}}^{\alpha}f(x)=f(x)-\sum_{j=1}^{n+1}\frac{D_{a^{+}}^{\alpha-j}f(a)}{\Gamma(\alpha-j+1)}\left(x-a\right)^{\alpha-j}$

and similarly,

$I_{b^{-}}^{\alpha}D_{b^{-}}^{\alpha}f(x)=f(x)-\sum_{j=1}^{n+1}\frac{(-1)^{j}D_{b^{-}}^{\alpha-j}f(b)}{\Gamma(\alpha-j+1)}\left(b-x\right)^{\alpha-j}$

Here $D^{\kappa}_{a^{+}}f(a)=\lim_{y\rightarrow a,y>a}D^{\kappa}_{a^{+}}f(y)$
for any $\kappa\in\mathbb{R}$ and similar for the right Riemann-Liouville
fractional derivative.

The Riemann-Liouville derivative has several shortcomings. One of them is that
the fractional derivative of a constant is not zero. To overcome this, we
introduce a modification of Riemann-Liouville fractional derivative which is
called the Caputo derivative.

######  Definition 1.9 (Caputo derivative).

Suppose $f$ is a real function and $n+1\geq\alpha>n$. We define the left and
right Caputo fractional derivatives of order $\alpha$ at $x\in(a,b)$ by

$C_{a^{+}}^{\alpha}f(x)=D_{a^{+}}^{\alpha}\left[f(t)-\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(t-a)^{k}\right](x)$
$C_{b^{-}}^{\alpha}f(x)=D_{b^{-}}^{\alpha}\left[f(t)-\sum_{k=0}^{n}\frac{(-1)^{k}f^{(k)}(b)}{k!}(b-t)^{k}\right](x)$

here $f^{(k)}$ denotes the $k$-th derivative of $f$.

We enumerate below some useful properties of the Caputo derivative:

######  Proposition 1.10.

Suppose $f$ is a real function and $\alpha,\beta>0$.

  1. 1.

If $\alpha=n\in\mathbb{N}$ and $f$ is $n-$th differentiable, then
$C_{a^{+}}^{\alpha}f=f^{(n)}$ and $C_{b^{-}}^{\alpha}f=(-1)^{n}f^{(n)}$

  2. 2.

Let $0<\alpha\leq\beta$ and $f\in L^{1}[a,b]$, then we have
$C_{a^{+}}^{\alpha}I_{a^{+}}^{\beta}f=I_{a+}^{\beta-\alpha}f$ and
$C_{b^{-}}^{\alpha}I_{b^{-}}^{\beta}f=I_{b-}^{\beta-\alpha}f$

  3. 3.

$C_{a^{+}}^{\alpha}C_{a^{+}}^{n}=C_{a^{+}}^{\alpha+n}$ and
$C_{b^{-}}^{\alpha}C_{b^{-}}^{n}=C_{b^{-}}^{\alpha+n}$ for all
$n\in\mathbbm{N}$.

######  Example 1.11.

We give some simple examples of Caputo fractional derivative here.

  1. 1.

Consider $f(x)=|x|^{\alpha}$ and $0<\beta\leq\alpha<1$. Then we have

$\displaystyle C_{a^{+}}^{\beta}f(x)$ $\displaystyle=$
$\displaystyle\frac{d}{dx}\left(\frac{1}{\Gamma(1-\beta)}\int_{a}^{x}\frac{|t|^{\alpha}-|a|^{\alpha}}{(x-t)^{\beta}}dt\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\Gamma(1-\beta)}\frac{d}{dx}\left(\int_{a}^{x}|t|^{\alpha}(x-t)^{-\beta}dt-|a|^{\alpha}\frac{(x-a)^{1-\beta}}{1-\beta}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\Gamma(1-\beta)}\frac{d}{dx}\left|\int_{\frac{a}{x}}^{1}|t|^{\alpha}|1-t|^{-\beta}dt|x|^{\alpha-\beta+1}\right|-\frac{1}{\Gamma(1-\beta)}|a|^{\alpha}(x-a)^{-\beta}$
$\displaystyle=$
$\displaystyle\frac{1}{\Gamma(1-\beta)}\left(|a|^{\alpha}(x-a)^{-\beta}\frac{a-x}{x}+(\alpha-\beta+1)|x|^{\alpha-\beta}\int_{\frac{a}{x}}^{1}|t|^{\alpha}|1-t|^{-\beta}dt\right)$

So we can see directly that $f^{(\beta+)}(a)=0$ for any $a\neq 0$ or
$\beta<\alpha$ but $f^{(\alpha+)}(0)=\Gamma(\alpha+1)$. It can be seen further
that for all $a\geq 0$, $C_{a^{+}}^{\alpha}f$ is continuous on $[a,\infty]$
but for $a<0$, $C_{a^{+}}^{\alpha}f$ has singularity at point $0$. In
particular, for $\beta=\alpha$, we have

$C_{a^{+}}^{\alpha}f(x)=\frac{1}{\Gamma(1-\beta)}\int_{\frac{a}{x}}^{1}g(t)-g\left(\frac{a}{x}\right)dt$

with $g(t)=\left|\frac{t}{1-t}\right|^{\alpha}$

Associated with the Caputo derivative is a fractional Taylor expansion:

######  Proposition 1.12.

Let $f\in C^{n}([a,b])$ and $n+1\geq\alpha>n$. If the Caputo fractional
derivative of order $\alpha$ of $f$ exists on $[a,b]$ and
$C_{a^{+}}^{\alpha}f\in L^{1}([a,b])$, then we have the Taylor expansion:

$f(x)=\sum_{k=0}^{n}f^{(k)}(a)\frac{(x-a)^{k}}{k!}+\frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{C_{a^{+}}^{\alpha}f(t)}{(x-t)^{1-\alpha}}dt$

and

$f(x)=\sum_{k=0}^{n}f^{(k)}(b)\frac{(x-b)^{k}}{k!}+\frac{1}{\Gamma(\alpha)}\int_{x}^{b}\frac{C_{b^{-}}^{\alpha}f(t)}{(t-x)^{1-\alpha}}dt$

As we were not able to find a proof of this expansion in the literature, we
provide a detailed proof below. Let us start with a simple but useful lemma,
which corresponds to Theorem 3.3 in [18] for the case $\rho=1$:

######  Lemma 1.13.

Suppose $f\in C^{n}([a,b])$, then for any $0<\alpha<n$ and
$\alpha\notin\mathbb{N}$, we have actually
$C_{a^{+}}^{\alpha}f(a)=C_{b^{-}}^{\alpha}f(b)=0$.

###### Proof of Proposition 1.12..

Define

$g(x)=f(x)-\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^{k}.$

From proposition 1.8, we have

$I_{a^{+}}^{\alpha}D_{a^{+}}^{\alpha}g(x)=g(x)-\sum_{j=1}^{n+1}\frac{D_{a^{+}}^{\alpha-j}g(a)}{\Gamma(\alpha-j+1)}\left(x-a\right)^{\alpha-j}$

Now we have by linearity of Riemann-Liouville fractional derivative operator
and definition of Caputo fractional derivative that

$D_{a^{+}}^{\alpha-j}g(a)=C_{a^{+}}^{\alpha-j}f(a)-\sum_{k=n-j+1}^{n}\frac{f^{(k)}(a)}{k!}D_{a^{+}}^{\alpha-j}((x-a)^{k})(a)$

Furthermore, for $k\geq n-j+1>\alpha-j$,
$D_{a^{+}}^{\alpha-j}((x-a)^{k})(a)=C_{a^{+}}^{\alpha-j}((x-a)^{k})(a)$ by
definition. Hence we have

$D_{a^{+}}^{\alpha-j}g(a)=C_{a^{+}}^{\alpha-j}f(a)-\sum_{k=n-j+1}^{n}\frac{f^{(k)}(a)}{k!}C_{a^{+}}^{\alpha-j}((x-a)^{k})(a)$

Now since $f\in C^{n}$ and $(\cdot-a)^{k}\in C^{\infty}$, we see that actually
$D_{a^{+}}^{\alpha-j}g(a)=0$ for every $j=1,\cdots,n$ by lemma 1.13. For the
case $j=n+1$, we actually have that

$\displaystyle\left|D_{a^{+}}^{\alpha-n-1}g(a)\right|$ $\displaystyle=$
$\displaystyle\lim_{y\rightarrow
a,y>a}\left|I_{a^{+}}^{n+1-\alpha}g(y)\right|$ $\displaystyle=$
$\displaystyle\lim_{y\rightarrow
a,y>a}\frac{1}{\Gamma(n+1-\alpha)}\left|\int_{a}^{y}\frac{g(t)}{(y-t)^{\alpha-n}}dt\right|$
$\displaystyle\leq$ $\displaystyle\lim_{y\rightarrow
a,y>a}\frac{1}{\Gamma(n+1-\alpha)}\left|\int_{a}^{y}\frac{\max_{t\in[a,y]}\\{|g(t)|\\}}{(y-t)^{\alpha-n}}dt\right|$
$\displaystyle=$ $\displaystyle\lim_{y\rightarrow
a,y>a}\frac{1}{\Gamma(n+1-\alpha)}\max_{t\in[a,y]}\\{|g(t)|\\}(y-a)^{n+1-\alpha}=0$

So actually we have

$g(x)=I_{a^{+}}^{\alpha}D_{a^{+}}^{\alpha}g(x)=I_{a^{+}}^{\alpha}C_{a^{+}}^{\alpha}f(x),$

hence the result. The derivation of the other formula is similar. ∎

The above derivative operators are non-local operators. We now introduce the
concept of local fractional derivative:

######  Definition 1.14 (Local fractional derivative).

Suppose $f$ is left(resp. right) fractional differentiable of order $\alpha$
on $[a,a+\delta]([a-\delta,a])$ for some positive $\delta$, then the
left(resp. right) local fractional derivative of order $\alpha$ of function
$f$ at point $a$ is given by

$f^{(\alpha+(resp.\alpha-))}(a)=\lim_{y\rightarrow
a,y\geq(\leq)a}C_{a+(a-)}^{\alpha}(y)$

######  Remark 1.15.

Using the integration by parts formula, we can see that we actually have

$D_{a^{+}}^{\alpha}f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(a)(x-a)^{k-\alpha-n}}{\Gamma(k+2-\alpha)}+D_{a^{+}}^{(\alpha-n)}f^{(n)}(x)$

Hence we will have furthermore that
$C_{a^{+}}^{\alpha}f(x)=C_{a^{+}}^{(\alpha-n)}f^{(n)}(x)$. This means the
existence of one side will imply the existence of the another side. And if we
take the limit we can see that the existences of $f^{(\alpha+)}(x)$ and
$(f^{(n)})^{(\alpha+)}(x)$ are equivalent and they are equal. This is very
important in the following proofs.

######  Corollary 1.16.

Let $f\in C^{n}([a,b])$ and $n+1\geq\alpha>n$. Suppose left(right) local
fractional derivative of order $\alpha$ of $f$ exists at $a$. This leads to
the following fractional Taylor formula with remainder term for $x\in[a,b]$

$\displaystyle f(x)$
$\displaystyle=\sum_{k=0}^{n}f^{(k)}(a)\frac{(x-a)^{k}}{k!}+\frac{1}{\Gamma(\alpha+1)}f^{(\alpha+)}(a)(x-a)^{\alpha}+o(|x-a|^{\alpha})$
(2) $\displaystyle f(x)$
$\displaystyle=\sum_{k=0}^{n}f^{(k)}(b)\frac{(x-b)^{k}}{k!}+\frac{1}{\Gamma(\alpha+1)}f^{(\alpha-)}(b)(b-x)^{\alpha}+o(|x-b|^{\alpha})$
(3)

###### Proof.

We only prove for the left Caputo derivative case.

$\displaystyle\frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{C_{a^{+}}^{\alpha}f(t)}{(x-t)^{1-\alpha}}dt$
$\displaystyle=$
$\displaystyle\frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{f^{(\alpha+)}(a)}{(x-t)^{1-\alpha}}dt+\frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{C_{a^{+}}^{\alpha}f(t)-f^{(\alpha+)}(a)}{(x-t)^{1-\alpha}}dt$
$\displaystyle=$
$\displaystyle\frac{f^{(\alpha+)}(a)}{\Gamma(\alpha+1)}(x-a)^{\alpha}+\frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{C_{a^{+}}^{\alpha}f(t)-f^{(\alpha+)}(a)}{(x-t)^{1-\alpha}}dt$

Furthermore,

$\displaystyle\left|\frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{C_{a^{+}}^{\alpha}f(t)-f^{(\alpha+)}(a)}{(x-t)^{1-\alpha}}dt\right|$
$\displaystyle\leq$
$\displaystyle\frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{\max_{t\in[a,x]}\\{|C_{a^{+}}^{\alpha}f(t)-f^{(\alpha+)}(a)|\\}}{(x-t)^{1-\alpha}}dt$
$\displaystyle=$
$\displaystyle\frac{\max_{t\in[a,x]}\\{|C_{a^{+}}^{\alpha}f(t)-f^{(\alpha+)}(a)|\\}}{\Gamma(\alpha+1)}(x-a)^{\alpha}$

Hence

$\lim_{x\rightarrow a,x\geq
a}\frac{\left|\frac{1}{\Gamma(\alpha)}\int_{a}^{x}\frac{C_{a^{+}}^{\alpha}f(t)-f^{(\alpha+)}(a)}{(x-t)^{1-\alpha}}dt\right|}{(x-a)^{\alpha}}\leq\lim_{x\rightarrow
a,x\geq
a}\frac{\max_{t\in[a,x]}\\{|C_{a^{+}}^{\alpha}f(t)-f^{(\alpha+)}(a)|\\}}{\Gamma(\alpha+1)}=0$

Thus we proved the result. ∎

A similar mean value theorem holds for the non-local fractional derivative.
The following result, which we state for completeness, is a direct consequence
of proposition 1.12:

######  Corollary 1.17.

Let $f\in C^{n}([a,b])$ and $n+1\geq\alpha>n$. Suppose Caputo fractional
derivative of order $\alpha$ of $f$ is continuous on $[a,b]$. Then we have the
for $x\in[a,b]$, there exists $\xi\in[a,x]$ such that

$f(x)=\sum_{k=0}^{n}f^{(k)}(a)\frac{(x-a)^{k}}{k!}+\frac{1}{\Gamma(\alpha+1)}C_{a^{+}}^{\alpha}f(\xi)(x-a)^{\alpha}$

A similar formula holds for the Caputo right derivative.

###  1.3 A characterization of the local fractional derivative

######  Proposition 1.18.

Let $p\notin\mathbb{N}$ and $f\in C^{p}([a,b])$. If $f^{(p+)}$ exists
everywhere on $[a,b]$ then $f^{(p+)}=0$ almost everywhere on $[a,b]$.

A rather complex proof of this proposition is given in [4, Corollary 3]. We
here give a simple proof of this property using only properties of monotone
functions.

###### Proof.

We first suppose $0<p<1$. And by Taylor expansion theorem for fractional
derivative, we actually have

$f(y)=f(x)+g(x)(y-x)^{p}+o(|y-x|^{p})$

Hence for any sequence $y_{n}\rightarrow x,y_{n}>x$, we have

$\lim_{n\rightarrow\infty}\frac{f(y_{n})-f(x)}{(y_{n}-x)^{p}}=g(x)$

Now let $E^{+}=\\{x\in\mathbb{R}:g(x)>0\\}$. If $\mathcal{L}(E^{+})>0$, then
$\exists K$ compact set such that $K\in E^{+}$ and $\mathcal{L}(K)>0$. Here
$\mathcal{L}$ denotes Lebesgue measure on the real line.

Let’s consider the open set $\cup_{x\in K}(x,x+\delta_{x})$ in $\mathbb{R}$,
where $\delta_{x}$ is the largest number so that $f(y)>f(x),\forall
y\in(x,x+\delta_{x})$. We can then write $\cup_{x\in
K}(x,x+\delta_{x})=\cup_{k=1}^{\infty}I_{k}$ for some open interval $I_{k}$.
By the additive property of measure, $\exists I_{r}$ such that
$\mathcal{L}(K\cap\bar{I_{r}})>0$.

For any $\bar{x}<\bar{y}\in K\cap\bar{I_{r}}$, there exists $x_{0}\in
K\cap[\bar{x},\bar{y}]$ such that

$x_{0}=\operatorname*{argmax}_{x\in K\cap[\bar{x},\bar{y}]}\\{f(x)\\}$

If $x_{0}\neq\bar{y}$, then there exists
$y_{0}\in(x_{0},x_{0}+\delta_{x_{0}})\cap K\cap[\bar{x},\bar{y}]$ such that
$f(y_{0})>f(x_{0})$, which is a contradiction. Hence $x_{0}=\bar{y}$ and we
have $f(\bar{y})>f(\bar{x}),\forall\bar{x}<\bar{y}\in K\cap\bar{I_{r}}$.

Now define $\bar{f}:\bar{I_{r}}\rightarrow\mathbb{R}$ such that $\bar{f}=f$ on
$K\cap\bar{I_{r}}$ and $\bar{f}$ is linear outside of $K\cap\bar{I_{r}}$. In
fact, let $\tilde{x}\in\bar{I_{r}}$, if $\tilde{x}\notin K$, then
$\tilde{x}\in(z_{0},z_{1})$ with $z_{0},z_{1}\in K$. By above reason,
$\bar{f}(z_{1})\geq\bar{f}(z_{0})$, and we perform linear interpolation to
define the value $\bar{f}(\tilde{x})$ here. It is then easy to conclude that
$\bar{f}$ is an increasing function. Hence $\bar{f}$ is differentiable almost
everywhere on $\bar{I_{r}}$.

Now we go back to the function $f$ on set $K$. It could be concluded that
almost surely on $K$, $\bar{f}$ is differentiable and we can simply suppose
here $\bar{f}$ is differentiable and $f=\bar{f}$ on $K$ whose Lebesgue measure
is positive. Now choose a point $x_{1}\in K$, either $x_{1}$ is isolated in
$K$ or it is an accumulated point in $K$. For the latter case, we would have

$g(x_{1})=\lim_{n\rightarrow\infty}\frac{f(y_{n})-f(x_{1})}{(y_{n}-x)^{p}}=\lim_{n\rightarrow\infty}\frac{\bar{f}(y_{n})-\bar{f}(x_{1})}{(y_{n}-x_{1})^{p}}=0$

Here $\\{y_{n}\\}$ is a sequence in $K$ approximating $x_{1}$. Since the set
of all isolated points in $K$ is of zero measure, we could get $g=0$ almost
surely on $K$, which is a contradiction. Hence $\mathcal{L}(E^{+})=0$.
Similarly, $\mathcal{L}(E^{-})=0$. Hence $f^{(p+)}(x)=0$ almost surely on
$\mathbb{R}$.

Now if $m<p<m+1$ for some integer $m>0$, since we have
$f^{(p+)}=(f^{(m)})^{(p-m+)}$, we could conclude again that $f^{(p+)}(x)=0$
almost surely on $\mathbb{R}$. ∎

######  Remark 1.19.

Here actually we only need a weaker condition on the function, namely that

$\lim_{y\rightarrow x,y>x}\frac{f^{(m)}(y)-f^{(m)}(x)}{|y-x|^{\alpha}}$

exists for $x\in\mathbb{R}$ with $m=\lfloor p\rfloor$ and $p=m+\alpha$, which
could be thought as classical definition of $\alpha$ order derivative. We call

$\lim_{x\rightarrow a,x>a}\frac{f^{(n)}(x)-f^{(n)}(a)}{|x-a|^{\alpha}}$

left ’classical’ fractional derivative of order $\alpha$ if it exists and
similar for right derivatives. However, when this limits exists, it is not
guaranteed that local fractional derivative of $f$ exists.

We can actually give a stronger result. Let
$E=\\{x\in\mathbb{R}:\lim_{y\rightarrow
x,y>x}\frac{f^{(m)}(y)-f^{(m)}(x)}{|y-x|^{\alpha}}\neq 0\\}$.

######  Proposition 1.20.

The Hausdorff dimension of the set $E$ is at most $\alpha$.

###### Proof.

For each $\epsilon,\delta$, we define the set $E_{\epsilon+}^{\delta}$ to be
the subset of $E$ such that for all $x\in E_{\epsilon+}^{\delta}$, we have

$f(y)>f(x)+\epsilon(y-x)^{\alpha},\forall x<y<x+\delta$

Then by the countably additive of measure, we only need to show the Hausdorff
dimension of the set $E_{\epsilon+}^{\delta}$ is zero. Furthermore, we can
restrict the set $E_{\epsilon+}^{\delta}$ on the interval $[0,\delta]$ due to
the countably additive of measure. Hence we will work on $[0,\delta]$ with
$E_{\epsilon+}^{\delta}$.

First of all, suppose the Hausdorff dimension of set $E_{\epsilon+}^{\delta}$
is larger than $\alpha$. In this case, we considers partitions
$\\{0,\frac{\delta}{k},\cdots,\frac{(k-1)\delta}{k},\delta\\}$ for
$k\in\mathbb{N}$. In each interval
$[\frac{i\delta}{k},\frac{(i+1)\delta}{k}]$, we choose the leftmost and
rightmost points of the set $E_{\epsilon+}^{\delta}$ and denote this interval
as $I_{k}^{i}$(if exists). Then $\cup_{i=0}^{k-1}I_{k}^{i}$ covers the set
$E_{\epsilon+}^{\delta}$ and by the definition of Hausdorff measure, we know
that $\sum_{i=0}^{k-1}|I_{k}^{i}|^{\alpha}\rightarrow\infty$ as
$k\rightarrow\infty$. Now let $a_{k}$ be the left boundary point of
$\cup_{i=0}^{k-1}I_{k}^{i}$ and $b_{k}$ be the right boundary point of
$\cup_{i=0}^{k-1}I_{k}^{i}$. We know

$f(b_{k})-f(a_{k})>\epsilon\sum_{i=0}^{k-1}|I_{k}^{i}|^{\alpha}\rightarrow\infty$

which gives a contradiction when $k\rightarrow\infty$. ∎

The following result is a consequence of the fractional Taylor expansion
(2)-(3):

######  Proposition 1.21.

Suppose $f\in C^{\alpha}([a,b])$, then for any $\beta<\alpha$ and
$\beta\notin\mathbb{N}$, we have $f^{(\beta+)}(x)=0$ if $f^{(\beta+)}(x)$
exists. And for $\beta>\alpha$ with same integer part, $f^{(\beta+)}(x)$ will
either be $\pm\infty$ or not exist.

##  2 Fractional Itô Calculus

In this section we derive Itô-type change of variable formulas for smooth
functions of paths with $p-$th variation along a sequence of partitions
$\\{\pi_{n}\\}_{n\in\mathbb{N}}$, first for functions (Sections 2.1 and 2.2)
then for path-dependent functionals (Section 2.3). In each case the focus is
the existence or not of a ’fractional’ Itô remainder term: we will see that
the existence of a non-zero Itô term depends on the fine structure of the
function and its fractional derivative.

###  2.1 A change of variable formula for time-independent functions

We now derive an Itô-type change of variable formula for smooth functions of
paths with $p-$th variation along a sequence of partitions
$\\{\pi_{n}\\}_{n\in\mathbb{N}}$.

Let $S\in V_{p}(\pi)$ be a path which admits $p-$th order variation along some
sequence of partitions $\\{\pi_{n}\\}_{n\in\mathbb{N}}$. We make the following
assumptions:

######  Assumption 2.1.

For any $k\in\mathbb{R}$, we have

$\int_{0}^{T}\mathbf{1}_{\\{S(t)=k\\}}d[S]_{\pi}^{p}=0$

Let $m=\lfloor p\rfloor$.

######  Assumption 2.2.

$f\in C^{m}(\mathbb{R})$ and admits a left local fractional derivative of
order $p$ everywhere.

######  Assumption 2.3.

The set

$\Gamma_{f}=\\{x\in\mathbb{R}:\exists U\ni x\text{ open,
}C_{a^{+}}^{p}f(b)\text{ is continuous on }\\{(a,b)\in U\times U:a\leq
b\\}\\}$

is locally finite, i.e. for any compact set $K\in\mathbb{R}$, the set
$\Gamma_{f}^{c}\cap K$ has only finite number of points.

Assumption 2.1 will be automatically satisfied if the occupation measure of
$S$ is atom-less. In fact,

$\Gamma_{S}(\\{k\\})=\int_{0}^{T}\mathbf{1}_{\\{S(t)=k\\}}d[S]_{\pi}(t)$

We first give a simple lemma regarding the set $\Gamma_{f}$:

######  Lemma 2.4.

$\forall x\in\Gamma_{f}$, the left local fractional derivative of order $p$ of
$f$ exists and equals to zero, i.e. $f^{(p+)}(x)=0$

###### Proof.

Let $x\in\Gamma_{f}$. There exists $x\in U$ open such that $C_{a+}^{p}f(b)$ is
continuous on $\\{(a,b)\in U\times U:a\leq b\\}$. Hence the function
$f^{(p+)}(x)$ is continuous on $U$. Due to the fact that $f^{(p+)}$ is zero
almost surely, we know $f^{(p+)}(x)=0$. ∎

Assumption 2.1 will be satisfied in particular if $S$ admits an occupation
density [15], then . However, as the following example shows, Assumption 2.1
may fail to be satisfied even if the path has a non-zero $p-$th order
variation.

######  Remark 2.5 (A counterexample).

We now give an example of path failing to satisfy Assumption 2.1, in the
spirit of [11, Example 3.6.].111 [11, Example 3.6] aims to construct a path
which admits quadratic variation but does not possess local time along some
sequence of partitions. In fact there seems to be an issue with the
construction in [11] but the underlying idea is still useful for our
construction.

Let $p>2$ and define the intervals

$\displaystyle I_{1}^{1}=\left(\frac{1}{3},\frac{2}{3}\right),\quad
I_{2}^{1}=\left(\frac{1}{9},\frac{2}{9}\right),\quad
I_{2}^{2}=\left(\frac{7}{9},\frac{8}{9}\right),\cdots,I_{j}^{i},j=1,\cdots,2^{i-1}$
(4)

and let $C=[0,1]/\cup_{i=1}^{\infty}\cup_{j=1}^{2^{i-1}}I^{i}_{j}$, which is
the Cantor ternary set [3]. Let $c:[0,1]\to\mathbb{R}_{+}$ be the associated
Cantor function, which is defined by

$c(x)=\left\\{\begin{array}[]{ll}\sum_{n=1}^{\infty}\frac{a_{n}}{2^{n}},&x=\sum_{n=1}^{\infty}\frac{2a_{n}}{3^{n}}\in
C,\quad a_{n}\in\\{0,1\\}\\\ \sup_{y\leq x,y\in
C}c(y),&x\in[0,1]/C\end{array}\right.$

We can see it is a non-decreasing function increasing only on Cantor set $C$.

Consider the function

$S(t)=|2^{\log_{3}(2\cdot\min_{u\in C}|t-u|)}|^{\frac{1}{p}}.$

We are going to construct a sequence of partitions such that the $p-$th
variation of $S$ along this sequence will be the Cantor function $c$. In this
case, we will see

$\int_{0}^{1}\mathbf{1}_{\\{S(t)=0\\}}d[S]^{p}=\int_{0}^{1}\mathbf{1}_{t\in
C}dc(t)=c(1)-c(0)=1$

which shows Assumption 2.1 is not satisfied.

We begin with the partition $\\{\pi_{j,n}^{i}\\}$, which denotes the $n-$th
partitions in the interval $I_{j}^{i}$. Define $t_{j,n}^{i,0}=\inf I_{j}^{i}$
and

$t_{j,n}^{i,k+1}=\inf\left\\{t>t_{j,n}^{i,k}:S(t)\in\left(\frac{1}{k_{n}}\sup_{t\in
I_{j}^{i}}S(t)\right)\mathbb{Z}\right\\}$

then $t_{j,n}^{i,2k_{n}}=\sup I_{j}^{i}$. $k_{n}$ is an integer to be
determined. We then do the calculation

$\sum_{k=0}^{2k_{n}-1}|S(t_{j,n}^{i,k+1})-S(t_{j,n}^{i,k})|^{p}=\sum_{k=0}^{2k_{n}-1}\left|\frac{1}{k_{n}}2^{-\frac{i}{p}}\right|^{p}=2^{1-i}k_{n}^{1-p}$

Then the sum for the $n-$th partition will be

$\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}2^{1-i}k_{n}^{1-p}=n\cdot k_{n}^{1-p}$

We choose $k_{n}=\lfloor n^{\frac{1}{p-1}}\rfloor$. Then

$nk_{n}^{1-p}\geq n\cdot(n^{\frac{1}{p-1}})^{1-p}=1$

and

$nk_{n}^{1-p}\leq
n\cdot(n^{\frac{1}{p-1}}-1)^{1-p}=\left(1-n^{\frac{1}{1-p}}\right)^{1-p}\rightarrow
1$

as $n\rightarrow\infty$. Hence we see

$[S]_{\pi}^{p}(1)=1$

Furthermore, since

$2^{1-i}k_{n}^{1-p}\rightarrow 0$

we see that $[S]_{\pi}^{p}$ will not change on the interval $I_{j}^{i}$ and
hence due to symmetry, we finally can show that $[S]^{p}(t)=c(t)$, the Cantor
function.

######  Remark 2.6.

Assumption 2.3 is necessary for Theorem 2.7. There are continuous function
with $f^{(p+)}=0$ everywhere but no joint continuity for the Caputo
derivative. We note here that the joint continuity can lead to the uniform
convergence of the equation

$\lim_{y\rightarrow x}\frac{f(y)-f(x)}{|y-x|^{\alpha}}=0$

for $x\in K$ with compact set $K$.

The following theorem extends the result of [10] to the case of functions with
fractional regularity:

######  Theorem 2.7.

Let $S\in V_{p}(\pi)$ satisfying assumptions 2.1. If $f$ is a continuous
function satisfying assumptions 2.2 and 2.3 for $m=\lfloor p\rfloor$, then

$\forall t\in[0,T],\qquad f(S(t))=f(S(0))+\int_{0}^{T}(f^{\prime}\circ S).dS$

where the last term is a limit of compensated Riemann sums of order $m=\lfloor
p\rfloor$:

$\int_{0}^{T}(f^{\prime}\circ
S).dS=\lim_{n\rightarrow\infty}\sum_{t_{i}\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(S(t\wedge
t_{i}))}{j!}(S(t\wedge t_{j+1})-S(t\wedge t_{j}))^{j}.$

###### Proof of Theorem 2.7.

We first suppose that for any $t_{i}\in\pi_{n}$ other than $0$ and $T$,
$f^{(p+)}(S_{t_{i}})=0$. By Taylor’s formula with integral remainder, we have

$\displaystyle f(S(t))-f(S(0))=\sum_{[t_{i},t_{i+1}]\in\pi_{n}}f(S(t\wedge
t_{i+1}))-f(S(t\wedge t_{i}))$ $\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\left(f(S(t\wedge
t_{i+1}))-f(S(t\wedge t_{i}))-\cdots-\frac{f^{(n)}(S(t\wedge
t_{i}))}{n!}(S(t\wedge t_{i+1})-S(t\wedge t_{i}))^{n}\right)$ $\displaystyle+$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(S(t\wedge
t_{i}))}{j!}(S(t\wedge t_{i+1})-S(t\wedge t_{i}))^{j}$ $\displaystyle=$
$\displaystyle
L^{n}+\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t\wedge
t_{i})}^{S(t\wedge t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t\wedge t_{i})))(S(t\wedge
t_{i+1})-r)^{m-1}dr$

where
$L^{n}=\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(S(t\wedge
t_{i}))}{j!}(S(t\wedge t_{i+1})-S(t\wedge t_{i}))^{j}$. It is easy to notice
that there exists a constant $M>0$ and

$\left|\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t\wedge t_{i})))(S(t\wedge
t_{i+1})-r)^{m-1}dr\right|\leq M|S(t\wedge t_{i+1})-S(t\wedge t_{i})|^{p}$

for example $M=\frac{\Gamma(p-m+1)||f||_{C^{p}}}{\Gamma(p+1)}$. We denote
$C^{n}_{\epsilon}=\\{[t_{i},t_{i+1}]\in\pi_{n}:|S(t_{i})-k|\leq\epsilon\text{
for some }k\in\Gamma_{f}^{c}\\}$. Since $S\in V^{p}(\pi)$, there exists
$N_{\epsilon}\in\mathbb{N}$ such that for all $n>N_{\epsilon}$, we have for
all $[t_{i},t_{i+1}]\in\pi_{n}$

$|S(t_{i+1})-S(t_{i})|\leq\frac{\epsilon}{2}$

Thus for $[t_{i},t_{i+1}]\notin C^{n}_{\epsilon}$, we have

$(S(t_{i+1})-k)(S(t_{i})-k)>0$

for all $k\in\Gamma_{f}^{c}$ and $S(t_{i}),S(t_{i+1})$ is away from
$\Gamma_{f}^{c}$ with at least $\frac{\epsilon}{2}$ distance, which means we
have the estimation

$|C_{a_{1}^{+}}^{p}f(b_{1})-C_{a_{2}^{+}}^{p}f(b_{2})|\leq\omega_{\epsilon}(\sqrt{(a_{1}-a_{2})^{2}+(b_{1}-b_{2})^{2}})$
(5)

for some modulus of continuity $\omega_{\epsilon}$,
$a_{1},a_{2},b_{1},b_{2}\in[S(t_{i}),S(t_{i+1})]$ or $[S(t_{i+1}),S(t_{i})]$
and $a_{1}\leq b_{1},a_{2}\leq b_{2}$ by assumption 2.2 and 2.3. Furthermore,
we denote
$C_{1,\epsilon}^{n}=\\{[t_{i},t_{i+1}]\in\pi_{n}/C_{\epsilon}^{n}:S(t_{i})\leq
S(t_{i+1}),t_{i}\neq 0,t_{i+1}\neq T\\}$ and
$C_{2,\epsilon}^{n}=\\{[t_{i},t_{i+1}]\in\pi_{n}/C_{\epsilon}^{n}:S(t_{i+1})<S(t_{i}),t_{i}\neq
0,t_{i+1}\neq T\\}$. Then we can calculate separately

$\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C^{n}_{\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t\wedge t_{i})))(S(t\wedge
t_{i+1})-r)^{m-1}dr\right|$ $\displaystyle\leq$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in C^{n}_{\epsilon}}M|S(t\wedge
t_{i+1})-S(t\wedge
t_{i})|^{p}\leq\sum_{[t_{i},t_{i+1}]\in\pi_{n}}Mg_{\epsilon}(S(t\wedge
t_{i}))|S(t\wedge t_{i+1})-S(t\wedge t_{i})|^{p}$

Here $g_{\epsilon}$ is a continuous taking value 1 on
$[k-\epsilon,k+\epsilon]$ for each $k\in\Gamma_{f}^{c}$ and value 0 outside of
$\cup_{k\in\Gamma_{f}^{c}}[k-2\epsilon,k+2\epsilon]$. Here we also take
$\epsilon$ small enough such that $[k-2\epsilon,k+2\epsilon]$ are disjoint and
$||g_{\epsilon}||_{\infty}\leq 1$. Then by definition of the $p-$th order
variation, we see

$\displaystyle\limsup_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in
C^{n}_{\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t\wedge t_{i})))(S(t\wedge
t_{i+1})-r)^{m-1}dr\right|$ $\displaystyle\leq$
$\displaystyle\int_{0}^{t}Mg_{\epsilon}(S(r))d[S]^{p}(r)$

now let $\epsilon\rightarrow 0$, we obtain that

$\displaystyle\lim_{\epsilon\rightarrow
0}\limsup_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in
C^{n}_{\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t\wedge t_{i})))(S(t\wedge
t_{i+1})-r)^{m-1}dr\right|$ $\displaystyle\leq$
$\displaystyle\sum_{k\in\Gamma_{f}^{c}}\int_{0}^{t}M\mathbf{1}_{\\{S(r)=k\\}}d[S]^{p}(r)$

By assumption 2.1 we then see that

$\lim_{\epsilon\rightarrow
0}\limsup_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in
C^{n}_{\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t\wedge t_{i})))(S(t\wedge
t_{i+1})-r)^{m-1}dr\right|=0$

Next for the set $C^{n}_{1,\epsilon}$ and $\alpha=p-m$, we have

$\displaystyle\sum_{[t_{i},t_{i+1}]\in
C^{n}_{1,\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(S(t\wedge t_{i+1})-r)^{m-1}\frac{1}{\Gamma(\alpha)}\int_{S_{t\wedge
t_{i}}}^{r}\frac{C_{S(t\wedge t_{i})^{+}}^{p}f(s)}{(r-s)^{1-\alpha}}dsdr$
$\displaystyle=$ $\displaystyle\sum_{[t_{i},t_{i+1}]\in
C^{n}_{1,\epsilon}}\frac{1}{\Gamma(m)\Gamma(\alpha)}\int_{S(t\wedge
t_{i})}^{S(t\wedge t_{i+1})}\int_{s}^{S_{t\wedge t_{i+1}}}(S(t\wedge
t_{i+1})-r)^{m-1}\frac{C_{S(t\wedge
t_{i})^{+}}^{p}f(s)}{(r-s)^{1-\alpha}}drds$ $\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in
C^{n}_{1,\epsilon}}\frac{1}{\Gamma(p)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(S(t\wedge t_{i+1})-s)^{p-1}C_{S(t\wedge t_{i})^{+}}^{p}f(s)ds$

Then by inequality (5) and the fact that
$f^{(p+)}(S(t_{i}))=C_{S(t_{i})^{+}}^{p}f(S(t_{i}))=0$(this is because
$S(t_{i})\in\Gamma_{f}$ and the lemma 2.4), we have

$\displaystyle\left|\int_{S(t\wedge t_{i})}^{S(t\wedge t_{i+1})}(S(t\wedge
t_{i+1})-s)^{p-1}C_{S(t\wedge t_{i})^{+}}^{p}f(s)ds\right|$
$\displaystyle\leq$ $\displaystyle\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(S(t\wedge t_{i+1})-s)^{p-1}\omega_{\epsilon}(s-S(t\wedge t_{i}))ds$
$\displaystyle\leq$
$\displaystyle\frac{1}{p}\omega_{\epsilon}(S(t_{i+1})-S(t_{i}))|S(t_{i+1})-S(t_{i})|^{p}$

Thus we obtain

$\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C^{n}_{1,\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}\frac{(S(t\wedge t_{i+1})-r)^{m-1}}{\Gamma(\alpha)}\int_{S_{t\wedge
t_{i}}}^{r}\frac{C_{S(t\wedge
t_{i})^{+}}^{p}f(s)}{(r-s)^{1-\alpha}}dsdr\right|$ $\displaystyle\leq$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{\omega_{\epsilon}(S(t_{i+1})-S(t_{i}))}{\Gamma(p+1)}|S(t_{i+1})-S(t_{i})|^{p}$

Since $osc(S,\pi_{n})\rightarrow 0$ as $n\rightarrow\infty$, we have

$\lim_{n\rightarrow\infty}\sum_{[t_{i},t_{i+1}]\in
C^{n}_{1,\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(S(t\wedge t_{i+1})-r)^{m-1}\frac{1}{\Gamma(\alpha)}\int_{S(t\wedge
t_{i})}^{r}\frac{C_{S(t\wedge t_{i})^{+}}^{p}f(s)}{(r-s)^{1-\alpha}}dsdr=0$

Now for the set $C_{2,\epsilon}^{n}$, we have

$\displaystyle\sum_{[t_{i},t_{i+1}]\in
C^{n}_{2,\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i+1})}^{S(t\wedge
t_{i})}(f^{(m)}(S(t\wedge t_{i}))-f^{(m)}(r))(S(t\wedge t_{i+1})-r)^{m-1}dr$
$\displaystyle=$ $\displaystyle\sum_{[t_{i},t_{i+1}]\in
C_{2,\epsilon}}\frac{1}{\Gamma(m)}\int_{SS(t\wedge t_{i+1})}^{S(t\wedge
t_{i})}\frac{1}{\Gamma(\alpha)}\int_{r}^{S(t\wedge
t_{i})}\frac{C_{r^{+}}^{p}f(s)}{(S(t\wedge t_{i})-s)^{1-\alpha}}ds(S(t\wedge
t_{i+1})-r)^{n-1}dr$ $\displaystyle=$ $\displaystyle\sum_{[t_{i},t_{i+1}]\in
C_{2,\epsilon}}\frac{1}{\Gamma(m)\Gamma(\alpha)}\int_{S(t\wedge
t_{i+1})}^{S(t\wedge t_{i})}\int_{S(t\wedge t_{i+1})}^{s}(S(t\wedge
t_{i+1})-r)^{m-1}\frac{C_{r^{+}}^{p}f(s)}{(S(t\wedge
t_{i})-s)^{1-\alpha}}drds$

Again by the inequality (5), we will have

$|C_{r+}^{p}f(s)|\leq\omega_{\epsilon}(\sqrt{(r-S(t\wedge
t_{i+1}))^{2}+(s-S(t\wedge
t_{i+1}))^{2}})\leq\omega_{\epsilon}(\sqrt{2}|S(t\wedge t_{i+1})-S(t\wedge
t_{i})|)$

Hence, we can obtain the estimation:

$\displaystyle\left|\int_{S(t\wedge t_{i+1})}^{S_{t\wedge
t_{i}}}\int_{S(t\wedge t_{i+1})}^{s}(S(t\wedge
t_{i+1})-r)^{m-1}\frac{C_{r^{+}}^{p}f(s)}{(S(t\wedge
t_{i})-s)^{1-\alpha}}drds\right|$ $\displaystyle\leq$
$\displaystyle\int_{S(t\wedge t_{i+1})}^{S_{t\wedge t_{i}}}\int_{S(t\wedge
t_{i+1})}^{s}(S(t\wedge
t_{i+1})-r)^{m-1}\frac{\omega_{\epsilon}(\sqrt{2}|S(t\wedge t_{i+1})-S(t\wedge
t_{i})|)}{(S(t\wedge t_{i})-s)^{1-\alpha}}drds$ $\displaystyle\leq$
$\displaystyle\frac{1}{m\alpha}\omega_{\epsilon}(\sqrt{2}|S(t\wedge
t_{i+1})-S(t\wedge t_{i})|)|S(t_{i+1})-S(t_{i})|^{p}$

Thus we could obtain that

$\lim_{n\rightarrow\infty}\sum_{[t_{i},t_{i+1}]\in
C^{n}_{2,\epsilon}}\frac{1}{\Gamma(m)}\int_{S(t\wedge t_{i+1})}^{S(t\wedge
t_{i})}(f^{(m)}(r)-f^{(m)}(S(t\wedge t_{i})))(S(t\wedge t_{i+1})-r)^{m-1}dr=0$

And finally for the case $t_{i}=0$ or $t_{i+1}=T$, the continuity of $f^{(m)}$
and $S$ will show that

$\lim_{n\rightarrow\infty}\int_{S(t\wedge t_{i})}^{S(t\wedge
t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t\wedge t_{i})))(S(t\wedge
t_{i+1})-r)^{m-1}dr=0$

Thus, by summing all these together and let $\epsilon$ tends to 0, we have

$\lim_{\epsilon\rightarrow
0}\lim_{n\rightarrow\infty}\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t\wedge
t_{i+1})}^{S(t\wedge t_{i})}(f^{(m)}(S(t\wedge t_{i}))-f^{(m)}(r))(S(t\wedge
t_{i+1})-r)^{m-1}dr=0$

Hence we see $\lim_{n\rightarrow\infty}L^{n}$ exists and we denote it as
$\int_{0}^{t}(f^{\prime}\circ S).dS$. And this leads to the Ito formula

$f(S(t))-f(S(0))=\int_{0}^{t}(f^{\prime}\circ S).dS$

∎

######  Remark 2.8.

We can not expect the Itô formula in the fractional case to have the same form
as in the integer case, i.e. there might be no Itô term in the fractional case
even if Assumption 2.1 does not hold. For the example in Remark 2.5, we can
show that for $2<p<3$ and $f(x)=|x|^{p}$, the Itô term still vanishes even
though we have

$\int_{0}^{1}f^{(p+)}(S(t))d[S]^{p}(t)=\frac{1}{\Gamma(p+1)}\int_{0}^{1}\mathbf{1}_{\\{S(t)=0\\}}dc(t)=\frac{c(1)-c(0)}{\Gamma(p+1)}=\frac{1}{\Gamma(p+1)},$

since $\\{S(t)=0\\}=C$ is the support of the function $c$. In this case, we
will show that

$0=f(S(1))-f(S(0))=\int_{0}^{1}(f^{\prime}\circ S).dS.$

In fact, we will need to calculate

$\lim_{n\rightarrow\infty}\sum_{[t_{i},t_{i+1}]\in\pi_{n}}f^{\prime}(S(t_{i}))(S(t_{i+1})-S(t_{i}))+\frac{1}{2}f^{\prime\prime}(S(t_{i}))(S(t_{i+1})-S(t_{i}))^{2}$

which is in fact the ’rough integral’ of $f^{\prime}$ along the reduced
order-p Ito rough path associated with $S$ [10]. Splitting the terms across
each $I_{j}^{i}$ we have

$\displaystyle\sum_{k=0}^{k_{n}-1}p(S(t_{j,n}^{i,k}))^{p-1}\frac{1}{k_{n}}2^{-\frac{i}{p}}-\sum_{k=k_{n}}^{2k_{n}-1}p(S(t_{j,n}^{i,k}))^{p-1}\frac{1}{k_{n}}2^{-\frac{i}{p}}$
$\displaystyle=$
$\displaystyle-p(S(t_{j,n}^{i,k_{n}}))^{p-1}\frac{1}{k_{n}}2^{-\frac{i}{p}}=-p\cdot(2^{-\frac{i}{p}})^{p-1}\cdot\frac{2^{-\frac{i}{p}}}{k_{n}}$

Thus

$\sum_{[t_{i},t_{i+1}]\in\pi_{n}}f^{\prime}(S(t_{i}))(S(t_{i+1})-S(t_{i}))=\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}-p\cdot\frac{2^{-i}}{k_{n}}=-\frac{np}{2k_{n}}$

Now we consider the term involving the second derivative. On $I_{j}^{i}$, we
have

$\displaystyle\sum_{[t_{j,n}^{i,k},t_{j,n}^{i,k+1}]\in\pi^{i}_{j,n}}\frac{1}{2}f^{\prime\prime}(S(t_{j,n}^{i,k}))(S(t_{j,n}^{i,k+1})-S(t_{j,n}^{i,k}))^{2}$
$\displaystyle=$
$\displaystyle\sum_{[t_{j,n}^{i,k},t_{j,n}^{i,k+1}]\in\pi_{j,n}^{i}}\frac{1}{2}p(p-1)(S(t_{j,n}^{i,k}))^{p-2}\frac{2^{-\frac{2i}{p}}}{k_{n}^{2}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\sum_{k=0}^{k_{n}-1}p(p-1)k^{p-2}\frac{2^{-i}}{k_{n}^{p}}+\sum_{k=1}^{k_{n}}p(p-1)k^{p-2}\frac{2^{-i}}{k_{n}^{p}}\right)$

Summing the second order terms over $\pi_{n}$ we obtain:

$\frac{np(p-1)}{4}\frac{2\sum_{k=1}^{k_{n}}k^{p-2}-k_{n}^{p-2}}{k_{n}^{p}}$

Adding together the first order term $-\frac{np}{2k_{n}}$, we need to
calculate

$\lim_{n\rightarrow\infty}\frac{np}{4}\left((p-1)\frac{2\sum_{k=1}^{k_{n}}k^{p-2}-k_{n}^{p-2}}{k_{n}^{p}}-\frac{2}{k_{n}}\right)$

Since it is obvious to see that

$\lim_{n\rightarrow\infty}\frac{np}{4}(p-1)\frac{1}{k_{n}^{2}}=\lim_{n\rightarrow\infty}\frac{np}{4}(p-1)\frac{1}{n^{\frac{2}{p-1}}}$

and $\frac{2}{p-1}>1$ means that

$\lim_{n\rightarrow\infty}\frac{np}{4}(p-1)\frac{1}{k_{n}^{2}}=0$

Hence it remains to consider

$\lim_{n\rightarrow\infty}\frac{np}{2}\left((p-1)\frac{\sum_{k=1}^{k_{n}}k^{p-2}}{k_{n}^{p}}-\frac{1}{k_{n}}\right)$

Use the inequality

$\frac{k_{n}^{p-1}}{p-1}=\int_{0}^{k_{n}}x^{p-2}dx\leq\sum_{k=1}^{k_{n}}k^{p-2}\leq\int_{1}^{k_{n}+1}x^{p-2}dx=\frac{(k_{n}+1)^{p-1}-1}{p-1}$

We then obtain first that

$\lim_{n\rightarrow\infty}\frac{np}{2}\left((p-1)\frac{\sum_{k=1}^{k_{n}}k^{p-2}}{k_{n}^{p}}-\frac{1}{k_{n}}\right)\geq\lim_{n\rightarrow\infty}\frac{np}{2}\left((p-1)\frac{k_{n}^{p-1}}{(p-1)k_{n}^{p}}-\frac{1}{k_{n}}\right)=0$

and secondly

$\lim_{n\rightarrow\infty}\frac{np}{2}\left((p-1)\frac{\sum_{k=1}^{k_{n}}k^{p-2}}{k_{n}^{p}}-\frac{1}{k_{n}}\right)\leq\lim_{n\rightarrow\infty}\frac{np}{2}\left(\frac{(k_{n}+1)^{p-1}-1-k_{n}^{p-1}}{k_{n}^{p}}\right)$
(6)

Notice that

$\lim_{n\rightarrow\infty}nk_{n}^{1-p}=1$

and $k_{n}\rightarrow\infty$, we see the right hand side of the inequality
$(\ref{ineqexample})$ equals to

$\lim_{k_{n}\rightarrow\infty}p\frac{k_{n}^{p-1}}{2}\left(\frac{(k_{n}+1)^{p-1}-1-k_{n}^{p-1}}{k_{n}^{p}}\right)=\frac{p}{2}\lim_{x\rightarrow\infty}\frac{(x+1)^{p-1}-1-x^{p-1}}{x}$

By l’Hopital’s rule, above limit equals to

$\displaystyle\lim_{x\rightarrow\infty}\frac{p(p-1)}{2}((x+1)^{p-2}-x^{p-2})$
$\displaystyle=$
$\displaystyle\frac{p(p-1)}{2}\lim_{x\rightarrow\infty}x^{p-2}\left(\left(1+\frac{1}{x}\right)^{p-2}-1\right)$
$\displaystyle=$
$\displaystyle\frac{p(p-1)}{2}\lim_{x\rightarrow\infty}x^{p-2}\frac{p-2}{x}=0$

since $p-2<1$. Thus in this case we see the remainder term is zero, which is
different from the integer case. Hence we see that the pathwise change of
variable formula in Theorem 2.7 may hold even if Assumption 2.1 does not hold.

We now give an example of path with the same $p-$th variation as above but
leading to a non-zero remainder term in the change of variable formula.

######  Example 2.9.

Define the intervals $I^{i}_{j}$ as in (4). Then we define $S|_{I^{i}_{j}}$ by
induction on $i$. Let $g(t)=2min\\{t,1-t\\}$ on $[0,1]$ and $0$ otherwise. For
$a<b$, we define $g(t,[a,b])=g\left(\frac{t-a}{b-a}\right)$.

First for $i=1$, we divide the interval $I^{i}_{j}$ into $r_{i}$ smaller
intervals $I^{i}_{j,k},k=1,\cdots,r_{i}$ such that
$|I^{i}_{j,k}|=\frac{|I^{i}_{j}|}{r_{i}}$ and two of each are non-
intersecting. On each interval $I^{i}_{j,k}$, we define
$S(t)=2^{-i}g(t,I^{i}_{j,k})$ for $k=1,\cdots,r_{i},j=1,\cdots,2^{i-1}$. In
other words, $S$ is defined as the limit

$S(t)=\sum_{i=1}^{\infty}\sum_{j=1}^{2^{i-1}}\sum_{k=1}^{r_{i}}2^{-i}g(t,I^{i}_{j,k})$

Let $\pi_{n}=(\tau^{n}_{l})$ be the dyadic Lebesgue partition associated with
$S$:

$\tau^{n}_{0}=0,\qquad\tau^{n}_{l+1}=\inf\\{t>\tau^{n}_{l},\
|S(t)-S(\tau^{n}_{l})|>2^{-n}\\}$
$[S]^{p}(1)=\lim_{n\rightarrow\infty}\sum_{\pi_{n}}|S(\tau^{n}_{l+1})-S(\tau^{n}_{l})|^{p}=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}\sum_{k=1}^{r_{i}}2\times
2^{-np}\times 2^{n-i}=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}r_{i}2^{-np+n}$

We can choose

$r_{i}=\lfloor 2^{(i-1)(p-1)}(2^{p-1}-1)\rfloor$

so that

$1=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}r_{i}2^{-np+n}.$

Then similar to what was discussed in the above remark, we have
$[S]^{p}(t)=c(t)$, the Cantor function. Let $f(x)=|x|^{p},2<p<\frac{3}{2}$ and
$T=1$, we are going to calculate the Itô remainder term for $f(S(1))-f(S(0))$.
We calculate the limit

$\displaystyle\lim_{n\rightarrow\infty}\sum_{[t_{i},t_{i+1}]\in\pi_{n}}p(S(t_{i}))^{p-1}(S(t_{i+1})-S(t_{i}))+\frac{p(p-1)}{2}(S(t_{i}))^{p-2}(S(t_{i+1})-S(t_{i}))^{2}$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}r_{i}\left(-p2^{-n}(2^{-i})^{p-1}+\frac{p(p-1)}{2}2^{-2n}c_{n}^{i}\right)$

Here

$c_{n}^{i}=2\sum_{k=1}^{2^{n-i}-1}2^{-n(p-2)}k^{p-2}+2^{-n(p-2)}(2^{n-i})^{p-2}$

We have

$c_{n}^{i}<2^{1-n(p-2)}\int_{0}^{2^{n-i}}x^{p-2}dx+2^{-n(p-2)}(2^{n-i})^{p-2}$

and hence

$\frac{p(p-1)}{2}2^{-2n}c_{n}^{i}<p2^{-n-ip+i}+\frac{p(p-1)}{2}2^{-np+(n-i)(p-2)}$

Hence

$\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}r_{i}\left(-p2^{-n}(2^{-i})^{p-1}+\frac{p(p-1)}{2}2^{-2n}c_{n}^{i}\right)$
$\displaystyle<$
$\displaystyle\sum_{i=1}^{n-1}2^{i-1}r_{i}\left(\frac{p(p-1)}{2}2^{-2n-ip+2i}\right)+2^{n-1}r_{n}\left(-p2^{-np}+\frac{p(p-1)}{2}2^{-np}\right)$
$\displaystyle<$
$\displaystyle\sum_{i=1}^{n-1}2^{-p}(2^{p-1}-1)\left(\frac{p(p-1)}{2}2^{-2n+2i}\right)+2^{-p}(2^{p-1}-1)\left(-p+\frac{p(p-1)}{2}\right)$
$\displaystyle=$ $\displaystyle
p2^{-p}(2^{p-1}-1)\left(\frac{p-1}{2}\frac{1-(\frac{1}{4})^{n-1}}{3}+-1+\frac{p-1}{2}\right)$

Now we let $n\rightarrow\infty$, we have

$\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}r_{i}\left(-p2^{-n}(2^{-i})^{p-1}+\frac{p(p-1)}{2}2^{-2n}c_{n}^{i}\right)\leq
p2^{-p}(2^{p-1}-1)\left(\frac{2(p-1)}{3}-1\right)$

So if $2<p<\frac{5}{2}$, then

$\int_{0}^{1}f^{\prime}(S(t))dS(t)<0$

but $f(S(1))-f(S(0))=0$, which means there will be a non-zero remainder. And
in this case, we also have

$\int_{0}^{1}\mathbf{1}_{\\{S(t)=0\\}}d[S]^{p}(t)=1$

so Assumption 2.1 is not satisfied.

In fact, we can provide a formula for the Itô remainder term for this path and
function $f=|x|^{p}$. Take $T=1$, $m=\lfloor p\rfloor$ and $\alpha=p-m$. Let

$G_{f}^{p}(a,b)=\frac{1}{(m-1)!|b-a|^{p}}\int_{a}^{b}(f^{(m)}(x)-f^{(m)}(a))(b-x)^{m-1}dx$

for $a\neq b$ and take the limit value when exists for $a=b$. For the function
$f(x)=|x|^{p}$, we can see $G_{f}^{p}$ is defined on
$\mathbb{R}^{2}/\\{(0,0)\\}$ and for $k>0$,

$G_{f}^{p}(ka,kb)=G_{f}^{p}(a,b)$

which indicate us to consider $G_{f}^{p}$ as a function on the unit circle
$S^{1}$. We define the projection map $P:\mathbb{R}^{2}/\\{(0,0)\\}\rightarrow
S^{1}$ by $p(x)=\frac{x}{||x||}$ and define a sequences of measures by

$\tilde{\nu}_{n}:=\frac{1}{N_{n}}\sum_{i=0}^{N_{n}-1}\delta_{P(S(t_{i}),S(t_{i+1}))}$

where $N_{n}$ is the number of intervals in the partition $\pi_{n}$.
Furthermore, we define

$\hat{G}_{f}^{p}(\theta)=G_{f}^{p}(\cos(\theta),\sin(\theta))$

for $\theta\in[0,2\pi)$. Since the distance between two successive points in
partition $\pi_{n}$ is $2^{-n}$, the remainder term can be rewritten as

$N_{n}\cdot 2^{-np}\int\hat{G}_{f}^{p}(x)\tilde{\nu}_{n}(dx)$

and we have
$N_{n}=\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}\sum_{k=1}^{r_{i}}2^{n-i+1}=2^{n}\sum_{i=1}^{n}r_{i}$.
Hence the remainder term is

$2^{n-np}\sum_{i=1}^{n}r_{i}\int\hat{G}_{f}^{p}(x)\tilde{\nu}_{n}(dx)$

By definition, we have
$\lim_{n\rightarrow\infty}2^{n-np}\sum_{i=1}^{n}r_{i}=1$. And if we have the
weak convergence of $\tilde{\nu}_{n}$, then we can obtain the limit
expression. Now let’s calculate the limit of the measures $\tilde{\nu}_{n}$.

Since $S(t_{i})=k2^{-n}$ for some positive number $k$ and $t_{i}\in\pi_{n}$,
the result $\tan\left(P(S(t_{i}),S(t_{i+1}))\right)$ will equal to either
$\frac{k}{k+1}$ or $\frac{k}{k-1}$, which means the final measure
$\tilde{\nu}$ will be supported on these points. And

$\displaystyle\tilde{\nu}\left(\arctan\left(\frac{k}{k+1}\right)\right)$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\tilde{\nu}_{n}\left(\arctan\left(\frac{k}{k+1}\right)\right)$
$\displaystyle=$ $\displaystyle\frac{1}{2^{\lceil\log_{2}(k+1)\rceil
p}}\frac{2^{p-1}-1}{2^{p}-1}$

By symmetry, we have

$\tilde{\nu}\left(\arctan\left(\frac{k+1}{k}\right)\right)=\tilde{\nu}\left(\arctan\left(\frac{k}{k+1}\right)\right)=\frac{1}{2^{\lceil\log_{2}(k+1)\rceil
p}}\frac{2^{p-1}-1}{2^{p}-1}$

Hence we have the Itô formula for the path $S$ defined in the example 2.9 and
$f(x)=|x|^{p}$:

$f(S(1))-f(S(0))=\int_{0}^{1}f^{\prime}(S(u))dS(u)+\int_{S^{1}}\hat{G}_{f}^{p}(x)\tilde{\nu}(dx)$
(7)

The above calculations show that the remainder term may be non-standard when
assumption 2.1 is not satisfied.

Furthermore, through the calculation of exact expression of Itô’s remainder
term for the path in example 2.9, we can give a general method to calculate
Itô’s remainder term for paths satisfying some assumptions.

For a function $f$ and a positive number $p=m+\alpha$, where $m=\lfloor
p\rfloor$, we define the map

$\displaystyle
G_{f}^{p}(a,b)=\frac{1}{(m-1)!|b-a|^{p}}\int_{a}^{b}(f^{(m)}(x)-f^{(m)}(a))(b-x)^{m-1}dx$
(8)

Then $G_{f}^{p}$ defines a map from $\\{(x,y)\in\mathbb{R}^{2},\ x\neq y\\}$.
If $f^{(m)}$ is continuous, it is easy to see $G_{f}^{p}$ is a continuous
function on $\\{(x,y)\in\mathbb{R}^{2},\ x\neq y\\}$.

Denote by $N_{n}^{t}$ the number of intervals in $\pi_{n}$ on $[0,t]$.

In order to compute the remainder term in the fractional Ito formula, we
’stratify’ the increments across the partition by the values of $G_{f}^{p}$,
to build an auxiliary quotient space $X$ which is required to have certain
properties:

######  Assumption 2.10 (Auxiliary quotient space).

There exists a space $X$ and a map

$P^{p}_{f}:\\{(x,y)\in\mathbb{R}^{2},\ x\neq y\\}\rightarrow X$

such that

  1. (i)

For all $x\in X$, the map $G^{p}_{f}$ defined by (8) is constant on
$(P^{p}_{f})^{-1}(x)$;

  2. (ii)

The sequence of measures on $[0,T]\times X$ defined by

$\nu_{n}(dt,dx):=\sum_{[t_{i},t_{i+1}]\in\pi_{n}}|S(t_{i+1})-S(t_{i})|^{p}\delta_{t_{i}}(t)\delta_{P^{p}_{f}(S(t_{i}),S(t_{i+1}))}(x)$

converges weakly to a measure $\nu$ on $[0,T]\times X$.

######  Remark 2.11.

The measure $\tilde{\nu}$ in the example 2.9 is actually the measure
$\nu([0,1],dx)$ defined in assumption 2.10 on the space $X$. In fact, due to
the fact that partitions in the example 2.9 are of Lebesgue type, we have

$\tilde{\nu}_{n}(dx)=\frac{1}{[S]^{p}_{n}(1)}\nu_{n}([0,1],dx)$

Here $[S]^{p}_{n}(1)=\sum_{[t_{i},t_{i+1}]\in\pi_{n},t_{i}\leq
1}|S(t_{i+1})-S(t_{i})|^{p}$. Let $n\rightarrow\infty$, we then obtain

$\tilde{\nu}(dx)=\frac{1}{[S]^{p}(1)}\nu([0,1],dx)=\nu([0,1],dx)$

Hence assumption 2.10 is a reasonable condition in order to calculate Ito’s
remainder term such as example 2.9.

Furthermore, since the first marginal measure of $\nu_{n}$ is

$\nu_{n}(dt,X)=\sum_{[t_{i},t_{i+1}]\in\pi_{n}}|S(t_{i+1})-S(t_{i})|^{p}\delta_{t_{i}}(t)$

we see $S\in V^{p}(\pi_{n})$ is equivalent to the weak convergence of
$\nu_{n}(dt,X)$ on the interval $[0,T]$. In this sense, we see assumption 2.10
actually requires other properties in addition to the existence of $p-$th
variation along the sequence $\\{\pi_{n}\\}$.

The choice of $X$ is crucial in making the calculation of the remainder term
tractable. One can always choose $X=\mathbb{R}$ but this choice may not lead
to easy calculations (see below).

######  Theorem 2.12.

Let $\\{\pi_{n}\\}$ be a sequence of partitions and $S\in V_{p}(\pi)$ such
that there exists $(X,P^{p}_{f})$ satisfying Assumption 2.10. We define the
map $\hat{G}_{f}^{p}:X\rightarrow\mathbb{R}$ such that $\hat{G}_{f}^{p}\circ
P_{f}^{p}=G_{f}^{p}$. Then

$f(S(t))-f(S(0))=\int_{0}^{t}f^{\prime}(S(u))dS(u)+\int_{0}^{t}\int_{X}\hat{G}_{f}^{p}(x)\nu(ds,dx)$

where

###### Proof.

Using a Taylor expansion, we have actually

$\displaystyle f(S(t))-f(S(0))$ $\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n},t_{i}\leq
t}\sum_{j=1}^{m}\frac{f^{(j)}(S(t\wedge t_{i}))}{j!}(S(t\wedge
t_{i+1})-S(t\wedge t_{i}))^{j}$ $\displaystyle+$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n},t_{i}\leq
t}G_{f}^{p}(S(t_{i}),S(t_{i+1}))|S(t_{i+1})-S(t_{i})|^{p}$

by the definition of measures $\nu_{n}$, we have

$\displaystyle f(S(t))-f(S(0))$ $\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n},t_{i}\leq
t}\sum_{j=1}^{m}\frac{f^{(j)}(S(t\wedge t_{i}))}{j!}(S(t\wedge
t_{i+1})-S(t\wedge t_{i}))^{j}$ $\displaystyle+$
$\displaystyle\int_{0}^{t}\int_{X}\hat{G}_{f}^{p}(x)\tilde{\nu}_{n}(dsdx)$

Using the weak convergence of $\nu_{n}$ in the assumption 2.10, so taking
limits yields the desired result. ∎

######  Remark 2.13.

We can perturb the function $f$ without changing the choice of the space $X$,
by working in a neighborhood of the singularity of $G_{f}^{p}$. Take $S^{1}$
as an example. Suppose $f\sim|x|^{p}$ near zero, which means that $G_{f}^{p}$
has a unique singularity at $(0,0)$. If furthermore, on each ray emanating
from the origin, the function $G_{f}^{p}$ converges as point tends to origin
’uniformly’. Here uniform means that for each $\epsilon>0$, there exists
$\delta>0$ such that on the band $\\{(a,b):d((a,b),y=x)<\delta\\}$, we have
$|G_{f}^{p}(a,b)-G_{f}^{p}(c,d)|\leq\epsilon$ if
$P_{f}^{p}(a,b)=P_{f}^{p}(c,d)$ and $(a,b),(c,d)$ are on the band. Above
assumption describe the behaviour of $G_{f}^{p}$ around the singular point
$(0,0)$ and can be used to calculate the Itô remainder term. For such function
$f$ and path $S$, we can derive the formula

$f(S(t))-f(S(0))=\int_{0}^{t}f^{\prime}(S(u))dS(u)+\int_{S^{1}}g(x)\nu([0,t],dx)$

Here $g(x)$ is defined as the radial limit at the origin of $G_{f}^{p}$.

Furthermore if $f$ admits a bounded $p-th$ order Caputo derivative, then the
uniform convergence can be checked by uniform convergence of the Caputo
derivative $C_{a}^{p}f(b)$.

We can see from the theorem that when Itô’s remainder term is non-zero, the
formula is not in the classical form. Hence we will need more information to
describe the Itô remainder term in the fractional case.

Let us go back to zero remainder term case. Consider the set
$(E^{p})^{\prime}$ of all functions satisfying assumptions 2.2 and 2.3. Then
$(E^{p})^{\prime}$ is a subset of $C^{p}(\mathbb{R})$. Let $E^{p}$ be the
closure of $(E^{p})^{\prime}$ in $C^{p}(\mathbb{R})$ equipped with the semi-
norms

$\sup_{x\in K}\\{f^{(k)}(x):k=0,\cdots,m\\}+\sup_{x,y\in
K}\left\\{\frac{|f^{(m)}(y)-f^{(m)}(x)|}{|y-x|^{\alpha}}\right\\}$

here $p=m+\alpha$ with $m<p<m+1$. Then we have the following theorem

######  Theorem 2.14.

Let $S\in V_{p}(\pi)$ satisfying Assumption 2.1 and $f\in E^{p}$. Then for any
$t\in[0,T]$

$f(S(t))=f(S(0))+\int f^{\prime}(S(u))dS(u)$

where the integral is a limit of compensated Riemann sums of order $m=\lfloor
p\rfloor$:

$\int
f^{\prime}(S(u))dS(u)=\lim_{n\rightarrow\infty}\sum_{t_{i}\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(S(t_{i}))}{j!}(S(t_{j+1})-S(t_{j}))^{j}.$

###### Proof.

Suppose $f_{k}\in(E^{p})^{\prime}$ and $f_{k}\rightarrow f$ in $C^{p}$. We
have

$f(S(t))-f(S(0))=L^{n}+\sum_{t_{i}\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t_{i})))(S(t_{i+1})-r)^{m-1}dr$

Let $K$ be a compact set containing the path of $S([0,T])$. Since
$f_{k}\rightarrow f$ in $C^{p}$, we have for every $\epsilon>0$, there exists
$N>0$ and for all $k>N$,

$\sup_{x,y\in
K}\frac{|(f^{(m)}-f_{k}^{(m)})(y)-(f^{(m)}-f_{k}^{(m)})(x)|}{|y-x|^{\alpha}}\leq\epsilon$

Hence

$\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}((f^{(m)}-f_{k}^{(m)})(r)-(f^{(m)}-f_{k}^{(m)})(S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|$
$\displaystyle\leq$
$\displaystyle\epsilon\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}|r-S(t_{i})|^{\alpha}|S(t_{i+1})-r|^{m-1}dr$
$\displaystyle\leq$
$\displaystyle\epsilon\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{m!}|S(t_{i+1})-S(t_{i})|^{p}$

Hence we have

$\displaystyle\limsup_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|$
$\displaystyle\leq$
$\displaystyle\limsup_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(f_{k}^{(m)}(r)-f_{k}^{(m)}(S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|+\frac{\epsilon}{m!}[S]^{p}(t)$

for all $k>N$. Since $f_{k}\in E$, we then have

$\limsup_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|\leq\frac{\epsilon}{m!}[S]^{p}(t)$

Finally let $\epsilon$ tends to $0$, we have

$\lim_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(f^{(m)}(r)-f^{(m)}(S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|=0$

Hence $L^{n}$ converges when $n$ tends to infinity and we have

$f(S(t))-f(S(0))=\int_{0}^{t}f^{\prime}(S(u))dS(u)$

∎

######  Remark 2.15.

It can be seen in the proof of the theorem 2.7 that we only need

$\int_{0}^{T}\mathbf{1}_{\\{S(t)=k\\}}d[S]^{p}(t)=0$ (9)

for $k\in\Gamma_{f}^{c}$. And we can denote $P_{f}$ as the set of paths which
satisfies equation (9) on $\Gamma_{f}^{c}$.

On the other hand, if we fix the path $S$, then by Fubini’s theorem, we
actually have that

$\int_{\mathbb{R}}\int_{0}^{T}\mathbf{1}_{\\{S(t)=k\\}}d[S]^{p}(t)dk=0$

which means equation (9) is satisfied by almost surely all $k\in\mathbb{R}$.
Hence we can consider the set $(E_{S}^{p})^{\prime}$ with all functions $f$
such that $\Gamma_{f}$ contains all points that break the equation (9). And
then we can obtain the closure $E_{S}^{p}$ for fixed path $S$. We can also
ignore $p$ in the subscript because $p$ is included in the information of $S$.
So we can just write $E_{S}$. And it is obvious that
$(E_{S})^{\prime}\subset(E^{p})^{\prime}$ and $E_{S}\subset E^{p}$.

Furthermore, by positivity of
$\int_{\mathbb{R}}\int_{0}^{T}\mathbf{1}_{\\{S(t)=k\\}}d[S]_{t}^{p}dk$, we can
see that the set of $k$ which violates equation (9) is actually countable.
Hence we could define set $(E_{x}^{p})^{\prime}$ to be all functions $f$ such
that $\Gamma_{f}$ contains point $x$ and $E_{x}^{p}$ to be the completion
under Hölder norm. If we furthermore define
$R_{S}=\left\\{k\in\mathbb{R}:\int_{0}^{T}\mathbf{1}_{\\{S(t)=k\\}}d[S]^{p}(t)\neq
0\right\\}$, then we have actually $E_{S}=\cap_{x\in R_{S}}E_{x}^{p}$.

Now let’s give some examples of functions belonging to $E^{p}$.

######  Example 2.16 (Examples of functions belonging to $E^{p}$.).

  1. 1.

All functions $f\in C^{m+1}(\mathbb{R})$.

  2. 2.

The function $f(x)=|x-k|^{p}$ for some $k\in\mathbb{R}$. It can be seen in the
example 1.11 that $C_{a^{+}}^{p}f(x)$ is continuous on $\\{(a,x)\in U\times
U:a\leq x\\}$ for any compact set $U$ contained in positive real line or
negative real line, which means $\Gamma_{f}=\mathbb{R}/\\{0\\}$ and hence is a
function in $E_{p}^{\prime}\subset E_{p}$.

  3. 3.

The linear combinations of $|x-k|^{p}$. For example

$f(x)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}|x-x_{n}|^{p}$

with $\\{x_{n}\\}$ an ordered set of rationals in $[0,1]$.

######  Remark 2.17.

Smooth ($C^{\infty}$) functions belong to $E_{p}$. Denote by
$C^{p+}(\mathbb{R})\subset E_{p}$ the completion of the smooth function under
Hölder norm i.e. the set of functions $f\in C^{p}$ such that on every compact
set $K$:

$\lim_{\delta\rightarrow 0}\sup_{x,y\in
K}\left\\{\frac{|f^{(m)}(x)-f^{(m)}(y)|}{|x-y|^{\alpha}}:|x-y|\leq\delta\right\\}=0$

Then we can easily see that for any $q>p$, we will have $E_{q}\subset
C^{q}\subset C^{p+}\subset E_{S}\subset E_{p}$ for any path $S$. Here $q$ is a
non-integer.

######  Example 2.18 (Examples for $P_{f}$ with $f(x)=|x|^{p}$).

  1. 1.

Fractional Brownian Motion: Let $p=1/H$. We have

$\mathbb{E}\left[\int_{0}^{T}\mathbf{1}_{\\{B^{H}(t)=0\\}}dt\right]=0$

which means that almost surely $B^{H}\in P^{p}_{f}$.

######  Remark 2.19.

We give here another remark on Assumption 2.3. In fact, as we see in the
remark 2.6, Assumption 2.3 means uniform convergence of the equation

$\lim_{y\rightarrow x}\frac{f^{(m)}(y)-f^{(m)}(x)}{|y-x|^{\alpha}}=0$

on every compact set $K$. Hence, according to the proof of the theorem 2.7, we
can change 3 assumptions to the following one

######  Assumption 2.20.

There exists a sequence of open set $U_{i}$ such that $\bar{U}_{i+1}\subset
U_{i}$ and

$\lim_{y\rightarrow x}\frac{f^{(m)}(y)-f^{(m)}(x)}{|y-x|^{\alpha}}=0$

locally uniformly outside of $U_{i}$. Furthermore, Let
$C=\cap_{i}^{\infty}\bar{U}_{i}$, we have

$\int_{o}^{T}\mathbf{1}_{\\{S(t)\in C\\}}d[S](t)=0$

If we replace the 3 assumptions in the theorem 2.7 by the single Assumption
2.20, then Theorem 2.7 still holds. And in this case we do not need to
calculate the Caputo derivative of the function $f$. The only thing that we
need to check is the uniform Hölder continuity. This means we have

######  Theorem 2.21.

Suppose $S\in V_{p}(\pi)$ and $f$ satisfies assumption 2.20 for $m=\lfloor
p\rfloor$. Then

$\forall t\in[0,T],\qquad f(S(t))=f(S(0))+\int_{0}^{T}(f^{\prime}\circ S).dS$

where the last term is a Föllmer integral, defined a limit of compensated
Riemann sums of order $m=\lfloor p\rfloor$:

$\int_{0}^{T}(f^{\prime}\circ
S).dS=\lim_{n\rightarrow\infty}\sum_{t_{i}\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(S(t\wedge
t_{i}))}{j!}(S(t\wedge t_{j+1})-S(t\wedge t_{j}))^{j}.$

This is a special case of Theorem 2.30 on the path-dependent Itô formula,
whose proof will be given in Sec. 2.3.

###  2.2 Change of variable formula for time-dependent functions

We now extend the change of variable formula to time dependent functions such
as $f(t,x)=|x-g(t)|^{p}$ where $g$ is a smooth function.

For $f\in C^{1,p}([0,T]\times\mathbb{R})$ right-differentiable in the time
variable, we define the open set

$\displaystyle\Gamma_{f}=$
$\displaystyle\\{(t,x)\in[0,T]\times\mathbb{R}:\exists\text{ open set
}U\ni(t,x)\text{ such that }C_{a+}^{p}f(s,b)\text{ is continuous on }$
$\displaystyle\\{(a,b)\in U_{s}\times U_{s}:a\leq b\\}\text{ uniformly on
}P_{t}(U)\\}$

Here $U_{s}$ means the intersection of $U$ and $t=s$ and $P_{t}(U)$ is the
projection of $U$ to $t-$axis. For the change of variable formula for time-
dependent functions, we need the following assumption:

######  Assumption 2.22.

$\int_{0}^{T}\mathbf{1}_{\\{(t,S(t))\notin\Gamma_{f}\\}}d[S]^{p}(t)=0$

######  Theorem 2.23.

Let $f\in C^{1,p}([0,T]\times\mathbb{R})$ with left local derivative on
$\mathbb{R}$ on $[0,T]$, $S\in V_{p}(\pi)$. Under Assumption 2.22, we have

$f(t,S(t))-f(0,S(0))=\int_{0}^{t}\partial_{t}f(u,S(u))du+\int_{0}^{t}D_{x}f(u,S(u))dS(u)$

where $m=\lfloor p\rfloor$ and

$\int_{0}^{t}D_{x}f(u,S(u))dS(u)=\lim_{n\rightarrow\infty}\sum_{t_{i}\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(t\wedge
t_{i},S(t\wedge t_{i}))}{j!}(S(t\wedge t_{i+1})-S(t\wedge t_{i}))^{j}.$

###### Proof.

For simplicity, we only prove the case $t=T$ here. As before we write

$\displaystyle f(T,S(T))-f(0,S_{0})$ $\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}(f(t_{i+1},S(t_{i+1}))-f(t_{i},S(t_{i})))$
$\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}(f(t_{i+1},S(t_{i+1}))-f(t_{i},S(t_{i+1})))$
$\displaystyle+$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}(f(t_{i},S(t_{i+1}))-f(t_{i},S(t_{i})))$

And for the first sum we have

$\sum_{[t_{i},t_{i+1}]\in\pi_{n}}(f(t_{i+1},S(t_{i+1}))-f(t_{i},S(t_{i+1})))=\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\int_{t_{i}}^{t_{i+1}}\partial_{t}f(u,S(t_{i+1}))du$

Let’s write
$S^{n}(t)=\sum_{t_{i}\in\pi_{n}}S(t_{i+1})\mathbf{1}_{[t_{i},t_{i+1})}(t)+S(T)\mathbf{1}_{\\{T\\}}(t)$.
Then above sum can be written as

$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}(f(t_{i+1},S(t_{i+1}))-f(t_{i},S(t_{i+1})))$
$\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\int_{t_{i}}^{t_{i+1}}\partial_{t}f(u,S^{n}(u))du=\int_{0}^{T}\partial_{t}f(u,S^{n}(u))du$

Now since $||S^{n}-S||_{\infty}\rightarrow 0$ as $n\rightarrow\infty$ due to
continuity of $S$, we see that

$\lim_{n\rightarrow\infty}\sum_{[t_{i},t_{i+1}]\in\pi_{n}}(f(t_{i+1},S(t_{i+1}))-f(t_{i},S(t_{i+1})))=\int_{0}^{T}\partial_{t}f(u,S(u))du$

Now let’s deal with the second sum

$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}f(t_{i},S(t_{i+1}))-f(t_{i},S(t_{i}))$
$\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\left(f(t_{i},S(t_{i+1}))-f(t_{i},S(t_{i}))-\cdots-\frac{D_{x}^{m}f(t_{i},S(t_{i}))}{n!}(S(t_{i+1})-S(t_{i}))^{n}\right)$
$\displaystyle+$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\sum_{j=1}^{m}\frac{D_{x}^{j}f(t_{i},S(t_{i}))}{j!}(S(t_{j+1})-S(t_{j}))^{j}$
$\displaystyle=$ $\displaystyle
L^{n}+\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(D_{x}^{m}f(t_{i},r)-D_{x}^{m}f(t_{i},S(t_{i})))(S(t_{i+1})-r)^{m-1}dr$

Here
$L^{n}=\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\sum_{j=1}^{m}\frac{D_{x}^{j}f(t_{i},S(t_{i}))}{j!}(S(t_{j+1})-S(t_{j}))^{j}$.
Now let’s define set
$C^{n}_{\epsilon}=\\{[t_{i},t_{i+1}]\in\pi_{n}:(t_{i},S_{t_{i}})\notin\Gamma_{f}^{\epsilon}\\}$.
Here
$\Gamma_{f}^{\epsilon}=\\{(t,x)\in\Gamma_{f}:d((t,x),\partial\Gamma_{f})>\epsilon\\}$.
Then
$(\Gamma_{f}^{\epsilon})^{c}:=[0,T]\times\mathbb{R}/\Gamma_{f}^{\epsilon}$ is
a closed set. We have for $\alpha=p-m$,

$\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon}^{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(D_{x}^{m}f(t_{i},r)-D_{x}^{m}f(t_{i},S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|$
$\displaystyle\leq$ $\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon}^{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}M|r-S(t_{i})|^{\alpha}|S(t_{i+1})-r|^{m-1}dr\right|$
$\displaystyle=$ $\displaystyle\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon}^{n}}\frac{M\Gamma(\alpha+1)}{\Gamma(p)}|S(t_{i+1})-S(t_{i})|^{p}$
$\displaystyle=$
$\displaystyle\frac{M\Gamma(\alpha+1)}{\Gamma(p)}\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\mathbf{1}_{\\{(t_{i},S_{t_{i}})\notin\Gamma_{f}^{\epsilon}\\}}|S(t_{i+1})-S(t_{i})|^{p}$
$\displaystyle=$
$\displaystyle\frac{M\Gamma(\alpha+1)}{\Gamma(p)}\mu_{n}(\\{t:(t,S_{t})\notin\Gamma_{f}^{\epsilon}\\})$

Since $(\Gamma_{f}^{\epsilon})^{c}$ is closed, the set
$\\{t:(t,S_{t})\notin\Gamma_{f}^{\epsilon}\\}=\\{t:(t,S_{t})\in(\Gamma_{f}^{\epsilon})^{c}\\}$
is also closed. Then by portmanteau theorem, we have

$\limsup_{n\rightarrow\infty}\mu_{n}(\\{t:(t,S_{t})\notin\Gamma_{f}^{\epsilon}\\})\leq\mu(\\{t:(t,S_{t})\notin\Gamma_{f}^{\epsilon}\\})$

Let $\epsilon\rightarrow 0$, since $\Gamma_{f}$ is open, we have

$\lim_{\epsilon\rightarrow
0}\mu(\\{t:(t,S_{t})\notin\Gamma_{f}^{\epsilon}\\})=\mu(\\{t:(t,S_{t})\notin\Gamma_{f}\\})=0$

By assumption 2.22. Hence

$\lim_{\epsilon\rightarrow
0}\lim_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon}^{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(D_{x}^{m}f(t_{i},r)-D_{x}^{m}f(t_{i},S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|=0$

Next we define two sets $C_{\epsilon,1}^{n}=\\{[t_{i},t_{i+1}]\notin
C_{\epsilon}^{n}:S(t_{i})\leq S(t_{i+1})\\}$ and
$C_{\epsilon,2}^{n}=\\{[t_{i},t_{i+1}]\notin
C_{\epsilon}^{n}:S(t_{i})>S(t_{i+1})\\}$. On the set $C_{\epsilon,1}^{n}$, we
have

$\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon,1}^{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(D_{x}^{m}f(t_{i},r)-D_{x}^{m}f(t_{i},S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|$
$\displaystyle=$ $\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C^{n}_{\epsilon,1}}\frac{1}{\Gamma(p)}\int_{S(t_{i})}^{S(t_{i+1})}(S(t_{i+1})-s)^{p-1}C_{S(t_{i})^{+}}^{p}f(t_{i},s)ds\right|$

and on the set $C_{\epsilon,2}^{n}$, we have

$\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon,2}^{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(D_{x}^{m}f(t_{i},r)-D_{x}^{m}f(t_{i},S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|$
$\displaystyle=$ $\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon,2}^{n}}\frac{1}{\Gamma(m)\Gamma(\alpha)}\int_{S(t_{i+1})}^{S(t_{i})}\int_{S(t_{i+1})}^{s}(S(t_{i+1})-r)^{m-1}\frac{C_{r^{+}}^{p}f(t_{i},s)}{(S(t_{i})-s)^{1-\alpha}}drds\right|$

We notice that if $(t_{i},S_{t_{i}})\notin C_{\epsilon}^{n}$ and $n$ is large
enough, then $(t_{i},S_{t_{i}}),(t_{i+1},S_{t_{i+1}})$ is in the closure of
$\Gamma_{f}^{\epsilon/2}\cup[0,T]\times K$, which is a compact set. Here $K$
is chosen so that ${S_{t}}_{t\in[0,T]}$ lies in $K$. This means that
$C_{x+}^{p}f(t_{i},x)$ is continuous in a neighborhood of $S(t_{i})$, hence
they are identically zero. Then by uniform continuous property, we can have a
modulus of continuity $\omega_{\epsilon}$ such that

$|C_{a_{1}^{+}}^{p}f(t,b_{1})-C_{a_{2}^{+}}^{p}f(t,b_{2})|\leq\omega_{\epsilon}(\sqrt{(a_{2}-a_{1})^{2}+(b_{2}-b_{1})^{2}})$

for all
$(t,a_{1}),(t,a_{2}),(t,b_{1}),(t,b_{2})\in\Gamma_{f}^{\frac{\epsilon}{2}}$.
Hence we have for $j=1,2$

$\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon,j}^{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(D_{x}^{m}f(t_{i},r)-D_{x}^{m}f(t_{i},S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|$
$\displaystyle\leq$ $\displaystyle\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon,j}^{n}}M\omega_{\epsilon}(S(t_{i+1})-S(t_{i}))|S(t_{i+1})-S(t_{i})|^{p}$
$\displaystyle\leq$ $\displaystyle
M\omega_{\epsilon}(\sup\\{|S(t_{i+1})-S(t_{i})|\\})\mu_{n}([0,T])$

for some constant $M$. And let $n\rightarrow\infty$, we see that

$\lim_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in
C_{\epsilon,j}^{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(D_{x}^{m}f(t_{i},r)-D_{x}^{m}f(t_{i},S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|=0$

Hence we finally obtain that

$\lim_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S(t_{i})}^{S(t_{i+1})}(D_{x}^{m}f(t_{i},r)-D_{x}^{m}f(t_{i},S(t_{i})))(S(t_{i+1})-r)^{m-1}dr\right|=0$

and hence the result. ∎

We can do as the time-independent case to extend the space of functions for
fixed path $S$ using approximation in the $C^{1,p}$ space and denote the
result space as $E_{S}$.

######  Remark 2.24.

As in the time-independent case, we can use a uniform convergence assumption
to replace Assumption 2.22:

######  Assumption 2.25.

There exists a sequence of open set $U_{i}\in\mathbb{R}^{+}\times\mathbb{R}$
such that $\bar{U}_{i+1}\subset U_{i}$ and $\lim_{y\rightarrow
x}\frac{f^{(m)}(t,y)-f^{(m)}(t,x)}{|y-x|^{\alpha}}=0$ locally uniformly
outside of $U_{i}$. Furthermore, Let $C=\cap_{i}^{\infty}\bar{U}_{i}$, we have

$\int_{o}^{T}\mathbf{1}_{\\{(t,S(t))\in C\\}}d[S](t)=0$

Under this assumption, Theorem 2.23 still hold.

######  Theorem 2.26.

Let $f\in C^{1,p}([0,T]\times\mathbb{R})$ and $S\in V_{p}(\pi)$. Under
Assumption 2.25, we have

$f(t,S(t))-f(0,S(0))=\int_{0}^{t}\partial_{t}f(u,S(u))du+\int_{0}^{t}D_{x}f(u,S(u))dS(u)$

where $m=\lfloor p\rfloor$ and

$\int_{0}^{t}D_{x}f(u,S(u))dS(u)=\lim_{n\rightarrow\infty}\sum_{t_{i}\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(t\wedge
t_{i},S(t\wedge t_{i}))}{j!}(S(t\wedge t_{i+1})-S(t\wedge t_{i}))^{j}.$

###  2.3 Extension to path-dependent functionals

In this subsection $S_{t}$ will designate the path stopped at $t$, i.e.
$S_{t}(s)=S(s\wedge t)$.

We consider the space $D([0,T],\mathbb{R})$ of càdlàg paths from $[0,T]$ to
$\mathbb{R}$. Let

$\Lambda_{T}=\\{(t,\omega_{t}):(t,\omega)\in[0,T]\times
D([0,T],\mathbb{R})\\}$

be the space of stopped paths. This is a complete metric space equipped with

$d_{\infty}((t,\omega),(t^{\prime},\omega^{\prime})):=\sup_{s\in[0,T]}|\omega(s\wedge
t)-\omega^{\prime}(s\wedge t^{\prime})|+|t-t^{\prime}|$

We will also need to stop paths ’right before’ a given time, and set for $t>0$

$\omega_{t-}(s):=\left\\{\begin{array}[]{cc}\omega(s),&s<t\\\ \lim_{r\uparrow
t},&s\geq t\end{array}\right.$

while $\omega_{0-}=\omega_{0}$. We recall here some concepts from the non-
anticipative functional calculus [9, 7].

######  Definition 2.27.

A non-anticipative functional is a map $F:\Lambda_{T}\rightarrow\mathbb{R}$.
Let $F$ be a non-anticipative functional.

  * •

We write $F\in\mathbb{C}_{l}^{0,0}$ if for all $t\in[0,T]$ the map
$F(t,\cdot):D([0,T],\mathbb{R})\rightarrow\mathbb{R}$ is continuous and if for
all $(t,\omega)\in\Lambda_{T}$ and all $\epsilon>0$, there exists $\delta>0$
such that for all $(t^{\prime},\omega^{\prime})\in\Lambda_{T}$ with
$t^{\prime}<t$ and
$d_{\infty}((t,\omega),(t^{\prime},\omega^{\prime}))<\delta$, we have
$|F(t,\omega)-F(t^{\prime},\omega^{\prime})|<\epsilon$

  * •

We write $F\in\mathbb{B}(\Lambda_{T})$ if for every $t_{0}\in[0,T)$ and every
$K>0$ there exists $C_{K,t_{0}}>0$ such that for all $t\in[0,t_{0}]$ and all
$\omega\in D([0,T],\mathbb{R})$ with $\sup_{s\in[0,t]}|\omega(s)|\leq K$, we
have $|F(t,\omega)|\leq C_{K,t_{0}}$

  * •

$F$ is horizontally differentiable at $(t,\omega)\in\Lambda_{T}$ if its
horizontal derivative

$\mathcal{D}F(t,\omega):=\lim_{h\downarrow
0}\frac{F(t+h,\omega_{t})-F(t,\omega_{t})}{h}$

exists. If it exists for all $(t,\omega)\in\Lambda_{T}$, then $\mathcal{D}F$
is a non-anticipative functional.

  * •

$F$ is vertically differentiable at $(t,\omega)\in\Lambda_{T}$ if its vertical
derivative

$\nabla_{\omega}F(t,\omega):=\lim_{h\rightarrow
0}\frac{F(t,\omega_{t}+h\mathbf{1}_{[t,T]})-F(t,\omega_{t})}{h}$

exists. If it exists for all $(t,\omega)\in\Lambda_{T}$, then
$\nabla_{\omega}F$ is a non-anticipative functional. In particular, we define
recursively $\nabla_{\omega}^{k+1}F:=\nabla\nabla_{\omega}^{k}F$ whenever this
is well defined.

  * •

For $p\in\mathbb{N}_{0}$, we say that $F\in\mathbb{C}_{b}^{1,p}(\Lambda_{T})$
if $F$ is horizontally differentiable and $p$ times vertically differentiable
in every $(t,\omega)\in\Lambda_{T}$, and if
$F,\mathcal{D}F,\nabla_{\omega}^{k}F\in\mathbb{C}_{l}^{0,0}(\Lambda_{T})\cap\mathbb{B}(\Lambda_{T})$
for $k=1,\cdots,p$.

In order to extend our result to the functional case, we need to give the
definition of fractional Caputo derivative of a non-anticipative functional.
But here we choose another approach using the uniform convergence of Hölder
semi-norm as stated in Remark 2.6.

######  Definition 2.28 (Vertical Hölder continuity).

Let $0<\alpha<1$.

  * •

$F$ is vertically $\alpha-$Hölder continuous at $(t,\omega)\in\Lambda_{T}$ if

$L(t,\omega):=\limsup_{h\rightarrow
0}\frac{|F(t,\omega_{t}+h\mathbf{1}_{[t,T]})-F(t,\omega_{t})|}{|h|^{\alpha}}<\infty$

and we call $L(t,\omega)$ the Hölder coefficient of $F$ at
$(t,\omega)\in\Lambda_{T}$

  * •

$F$ is vertically $\alpha-$Hölder continuous on $E\subset\Lambda_{T}$ if

$L:=\lim_{\epsilon\rightarrow 0}\sup_{(t,\omega)\in
E}\sup_{|h|\leq\epsilon}\frac{|F(t,\omega_{t}+h\mathbf{1}_{[t,T]})-F(t,\omega_{t})|}{|h|^{\alpha}}<\infty$

and we call $L$ the Hölder coefficient of $F$ over $E$ and we denote
$F\in\mathbb{C}^{0,\alpha}(E)$.

  * •

Let $E_{n}=\\{(t,\omega)\in\Lambda:d((t,\omega),(0,0))<n\\}$. $F$ is locally
vertically $\alpha-$Hölder continuous on $\Lambda_{T}$ if $F$ is vertically
$\alpha-$Hölder continuous on $E_{n}$ for each $n$. We denote
$F\in\mathbb{C}^{0,\alpha}_{loc}$.

  * •

For $m<p<m+1$, we say $F\in\mathbb{C}_{loc}^{0,p}$ if
$\nabla_{\omega}^{m}F\in\mathbb{C}_{loc}^{0,p-m}$. For $E\subset\Lambda_{T}$,
we say $F\in\mathbb{C}^{0,p}(E)$ if
$\nabla_{\omega}^{m}F\in\mathbb{C}^{0,p-m}(E)$.

We now give assumptions on the functional
$F\in\mathbb{C}_{b}^{1,m}\cap\mathbb{C}^{0,p}_{loc}$, where $m=\lfloor
p\rfloor$. Given a path $S$ which admits $p-$th order variation along a
sequence of partitions $\pi$.

######  Assumption 2.29.

There exists a sequence of continuous functions
$g_{k}\in\mathbb{C}(\Lambda_{T},\mathbb{R})$ with support $E_{k}:=supp(g_{k})$
such that

  1. 1.

$E_{k+1}\subset E_{k}$, $g_{k}|_{E_{k+1}}=1,g_{k}\leq 1$ and
$F\in\mathbb{C}_{loc}^{0,p}(\Lambda_{T}-E_{k})$ for each $k$ with Hölder
coefficient $L=0$.

  2. 2.

There exists a bounded function $g$ on $\Lambda_{T}$ such that
$g_{k}\rightarrow g$ pointwise and

$\int_{0}^{T}g((t,S_{t}))d[S]_{t}=0$

Define the piecewise-constant approximation $S^{n}$ to $S$ along the partition
$\pi_{n}$:

$S^{n}(t):=\sum_{[t_{i},t_{i+1}]\in\pi_{n}}S(t_{j+1})\mathbf{1}_{[t_{j},t_{j+1})}(t)+S(T)\mathbf{1}_{\\{T\\}}(t)$

Then $\lim_{n\rightarrow\infty}||S^{n}-S||_{\infty}=0$ whenever
$osc(S,\pi_{n})\rightarrow 0$.

######  Theorem 2.30.

Let $p\notin\mathbb{N}$, $m=\lfloor p\rfloor$ and $S\in V_{p}(\pi)$. If
$F\in\mathbb{C}_{b}^{1,m}\cap\mathbb{C}^{0,p}_{loc}$ satisfies Assumption 2.29
then the limit

$\int_{0}^{T}\nabla_{\omega}F(t,S_{t-})d^{\pi}S(t):=\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\pi_{n}}\sum_{k=1}^{m}\frac{1}{k!}\nabla_{\omega}^{k}F(t_{j},S_{t_{j}-}^{n})(S(t_{j+1}\wedge
t)-S(t_{j}\wedge t))^{k}$

exists and we have

$F(t,S_{t})=F(0,S_{0})+\int_{0}^{t}\mathcal{D}F(s,S_{s})ds+\int_{0}^{t}\nabla_{\omega}F(s,S_{s})dS(s)$

###### Proof.

We write

$\displaystyle F(t,S^{n}_{t})-F(0,S^{n}_{0})$ $\displaystyle=$
$\displaystyle\sum_{[t_{j},t_{j+1}]\in\pi_{n}}(F(t_{j+1}\wedge
t,S^{n}_{t_{j+1}\wedge t-})-F(t_{j}\wedge t,S^{n}_{t_{j}\wedge t-}))$
$\displaystyle+F(t,S^{n}_{t})-F(t,S^{n}_{t-})$ $\displaystyle=$
$\displaystyle\sum_{[t_{j},t_{j+1}]\in\pi_{n}}(F(t_{j+1}\wedge
t,S^{n}_{t_{j+1}\wedge t-})-F(t_{j}\wedge t,S^{n}_{t_{j}\wedge t-}))+o(1)$

We only need to consider $j$ with $t_{j+1}\leq t$ since the remainder is
another $o(1)$ as $n\rightarrow\infty$ by left continuity of functional $F$.
Now we split the difference into two parts:

$F(t_{j+1},S^{n}_{t_{j+1}-})-F(t_{j},S_{t_{j}-}^{n})=(F(t_{j+1},S^{n}_{t_{j+1}-})-F(t_{j},S_{t_{j}}^{n}))+(F(t_{j},S^{n}_{t_{j}})-F(t_{j},S_{t_{j}-}^{n}))$

Now since $S_{t_{j}}^{n}=S_{t_{j+1}-}^{n}$ on $[0,t_{j+1}]$, we have

$F(t_{j+1},S^{n}_{t_{j+1}-})-F(t_{j},S_{t_{j}}^{n})=\int_{t_{j}}^{t_{j+1}}\mathcal{D}F(u,S_{t_{j}}^{n})du=\int_{t_{j}}^{t_{j+1}}\mathcal{D}F(u,S^{n})du$

since $\mathcal{D}F$ is non-anticipative functional. Hence we have

$\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}(F(t_{j+1},S^{n}_{t_{j+1}-})-F(t_{j},S_{t_{j}}^{n}))=\int_{0}^{t}\mathcal{D}F(u,S_{u})du$

by continuity. It remains to consider the term

$F(t_{j},S_{t_{j}}^{n})-F(t_{j},S_{t_{j}-}^{n})=F(t_{j},S_{t_{j}-}^{n,S_{t_{j},t_{j+1}}})-F(t_{j},S_{t_{j}-}^{n})$

where $S_{t_{j},t_{j+1}}=S(t_{j+1})-S(t_{j})$ and
$S_{t_{j}-}^{n,x}(s):=S_{t_{j}-}^{n}(s)+\mathbf{1}_{[t_{j},T]}(s)x$. Now from
Taylor’s formula and definition of vertical derivative, we obtain

$\displaystyle F(t_{j},S_{t_{j}-}^{n,S_{t_{j},t_{j+1}}})$ $\displaystyle=$
$\displaystyle
F(t_{j},S_{t_{j}-}^{n})+\sum_{k=1}^{m}\frac{\nabla_{\omega}^{k}F(t_{j},S_{t_{j}-}^{n})}{k!}(S(t_{j+1})-S(t_{j}))^{k}$
$\displaystyle+$
$\displaystyle\frac{1}{(m-1)!}\int_{0}^{S_{t_{j},t_{j+1}}}(\nabla_{\omega}^{m}F(t_{j},S_{t_{j}-}^{n,u})-\nabla_{\omega}^{m}F(t_{j},S_{t_{j}-}^{n}))(S_{t_{j},t_{j+1}}-u)^{m-1}du$

From the definition of $\mathbb{C}^{0,p}$, we see that there exists $M>0$ such
that

$|\nabla_{\omega}^{m}F(t_{j},S_{t_{j}-}^{n,u})-\nabla_{\omega}^{m}F(t_{j},S_{t_{j}-}^{n})|\leq
M|u|^{p-m}$

for each $t_{j}$ when $n$ is large enough. This shows that

$\left|\frac{1}{(m-1)!}\int_{0}^{S_{t_{j},t_{j+1}}}(\nabla_{\omega}^{m}F(t_{j},S_{t_{j}-}^{n,u})-\nabla_{\omega}^{m}F(t_{j},S_{t_{j}-}^{n}))(S_{t_{j},t_{j+1}}-u)^{m-1}du\right|\leq\tilde{M}|S(t_{j+1})-S(t_{j})|^{p}$

for some constant $\tilde{M}$. Fix $k\in\mathbb{N}$, we now divide the
partitions into two parts. The first part $\pi_{n}^{1}$ contains
$[t_{j},t_{j+1}]$ such that $(t_{j},S_{t_{j}-}^{n})\in E_{k}$ and the second
part $\pi_{n}^{2}$ contains $[t_{j},t_{j+1}]$ such that
$(t_{j},S_{t_{j}-}^{n})\notin E_{k}$. Then we have

$\displaystyle\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\left|\frac{1}{(m-1)!}\int_{0}^{S_{t_{j},t_{j+1}}}(\nabla_{\omega}^{m}F(t_{j},S_{t_{j}-}^{n,u})-\nabla_{\omega}^{m}F(t_{j},S_{t_{j}-}^{n}))(S_{t_{j},t_{j+1}}-u)^{m-1}du\right|$
$\displaystyle\leq$
$\displaystyle\sum_{[t_{j},t_{j+1}]\in\pi_{n}^{1}}\tilde{M}|S(t_{j+1})-S(t_{j})|^{p}+\sum_{[t_{j},t_{j+1}]\in\pi_{n}^{2}}\tilde{M}L(S_{t_{j}-}^{n},osc(S,\pi_{n}))|S(t_{j+1})-S(t_{j})|^{p}$

where

$L(\omega,\epsilon):=\sup_{|h|\leq\epsilon}\frac{|\nabla_{\omega}^{m}F(t,\omega_{t}+h\mathbf{1}_{[t,T]})-\nabla_{\omega}^{m}F(t,\omega_{t})|}{|h|^{p-m}}$

and by the definition of vertically Hölder continuity, we have
$\lim_{\epsilon\rightarrow 0}\sup_{\omega\in E/E_{k}}L(\omega,\epsilon)=0$ for
each $k$ with $E$ a bounded subset of $\Lambda_{T}$ containing all stopped
paths occurred in this proof. Hence

$\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}^{2}}\tilde{M}L(S_{t_{j}-}^{n},osc(S,\pi_{n}))|S(t_{j+1})-S(t_{j})|^{p}=0$

As for the first part, we have

$\displaystyle\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}^{1}}\tilde{M}|S(t_{j+1})-S(t_{j})|^{p}$
$\displaystyle\leq$
$\displaystyle\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\tilde{M}g_{k}((t_{j},S_{t_{j}-}^{n})|S(t_{j+1})-S(t_{j})|^{p}$
$\displaystyle=$ $\displaystyle\int_{o}^{t}\tilde{M}g_{k}((t,S_{t}))d[S]_{t}$

by assumption 2.29. Let $k$ tends to infinity, we get the result. ∎

######  Remark 2.31.

One may derive theorems 2.23 and 2.26 from above theorem. Let $\Gamma_{f}$
defined as Assumption 2.20,
$E_{k}=\\{(t,x):d((t,x),\Gamma_{f}^{c})\leq\frac{1}{k}\\}$ and define

$g((t,\omega))=\mathbf{1}_{\\{(t,\omega(t))\notin\Gamma_{f}\\}}.$

Then one may choose smooth functions $y_{k}$ with support $E_{k-1}$ such that
$y_{k}|_{E_{k}}=1,0\leq y_{k}\leq 1$ and set

$g_{k}(t,\omega):=y_{k}(t,\omega(t)).$

$g_{k}$ and $g$ then satisfy 2) in Assumption 2.29.

##  3 Extension to $\phi$-variation

The concept of $\phi-$variation $\phi-$variation along a sequence of
partitions [17] extends the notion of p-th order variation to more general
functions:

######  Definition 3.1 ($\phi$-variation along a partition sequence).

A continuous path $S\in\mathbb{C}([0,T],\mathbb{R})$ is said to have a
$\phi-$variation along a sequence of partitions
$\pi=\left(\pi_{n}\right)_{n\geq 1}$ if $osc(S,\pi_{n})\rightarrow 0$ and the
sequence of measures

$\mu^{n}:=\sum_{t_{i}\in\pi_{n}}|\phi(S(t_{i+1})-S(t_{i}))|\delta(\cdot-
t_{i})$

converges weakly to a measure $\mu$ without atoms. In this case we write $S\in
V_{\phi}(\pi)$ and $[S]^{\phi}(t)=\mu([0,t])$ for $t\in[0,T]$, and we call
$[S]^{\phi}$ the $\phi-$variation of $S$.

The function $\phi$ we considered in this section satisfies some assumption
given here.

######  Assumption 3.2.

There exists $m\in\mathbb{N}$ such that $\phi\in C^{m}$ and

$\lim_{x\rightarrow 0}\frac{\phi(x)}{x^{m}}=0,\qquad,\lim_{x\rightarrow
0}\frac{\phi(x)}{x^{m+1}}=\infty$

It is obvious that in this case $\phi(0)=0$. Furthermore, we let $\phi$ map
$\mathbb{R}^{+}$ to $\mathbb{R}^{+}$ and $\phi$ is either odd or even
function.

We say a function $f$ is $\phi-$continuous if and only if for each compact set
$K$, there exists a constant $M$ such that

$\frac{|f^{(m)}(y)-f^{(m)}(x)|}{|\phi^{(m)}(y-x)|}\leq M,\qquad x,y\in K$

We denote the collection of these functions by $C^{\phi}$.

######  Remark 3.3.

As in the fractional case, if we consider the Taylor expansion of $f$ up to
’$\phi$’, we should write it as

$f(y)=f(x)+\sum_{k=1}^{m}\frac{f^{(k)}}{k!}(y-x)^{k}+g(x)\phi(y-x)+o(\phi(y-x))$

it is obvious that if $g$ is continuous, then we would have the remainder term
in the change of variable formula. However, for the function $\phi$ satisfying
assumption 3.2, the function $g$ in the Taylor expansion above would be almost
surely zero, the same as stated in proposition 1.18 for $|x|^{p}$. What we
need is just to change $(y-x)^{p}$ by $\phi(y-x)$ in the proof of proposition
1.18 and notice that if $f$ is differential at some point $x$, then for the
case $m=0$,

$\lim_{y_{n}\rightarrow
x}\frac{f(y_{n})-f(x)}{\phi(y_{n}-x)}=\lim_{y_{n}\rightarrow
x}\frac{f(y_{n})-f(x)}{y_{n}-x}\frac{y_{n}-x}{\phi(y_{n}-x)}=0$

Follow the steps in the proof of Proposition 1.18, we can show that $g=0$
almost everywhere. Hence a regular remainder should not appear in this case.

We will use the uniform convergence assumption here.

######  Assumption 3.4.

There exists a sequence of open sets $U_{i}$ such that $\bar{U}_{i+1}\subset
U_{i}$ for $i=1,2,\cdots$ and $\lim_{y\rightarrow
x}\frac{f^{(m)}(y)-f^{(m)}(x)}{|\phi^{(m)}(y-x)|}=0$ locally uniformly outside
of each $U_{i}$. Furthermore, Let $C=\cap_{i}^{\infty}\bar{U}_{i}$, we have

$\int_{0}^{T}\mathbf{1}_{\\{S(t)\in C\\}}d[S]^{\phi}(t)=0$

Under the assumption 3.4, we have the theorem

######  Theorem 3.5.

Let $\phi$ satisfy the assumption 3.2, $\\{\pi_{n}\\}$ be a given sequence of
partitions and $S\in V_{\phi}(\pi)$. Then for every function $f$ satisfies
assumption 3.4 we have the formula

$f(S(t))-f(S(0))=\int_{0}^{t}f^{\prime}(S(u))dS(u)$

holds, where the integral

$\int_{0}^{t}f^{\prime}(S(u))dS(u)=\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\sum_{k=1}^{m}\frac{f^{(k)}(S(t_{j}))}{k!}(S(t_{j+1}\wedge
t)-S(t_{j}\wedge t))^{k}$

is defined as a (pointwise) limit of compensated Riemann sums.

###### Proof.

We have

$\displaystyle f(S_{t})-f(S_{0})$ $\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}f(S_{t\wedge
t_{i+1}})-f(S_{t\wedge t_{i}})$ $\displaystyle=$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\left(f(S_{t\wedge
t_{i+1}})-f(S_{t\wedge t_{i}})-\cdots-\frac{f^{(m)}(S_{t\wedge
t_{i}})}{m!}(S_{t\wedge t_{i+1}}-S_{t\wedge t_{i}})^{n}\right)$
$\displaystyle+$
$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(S_{t\wedge
t_{i}})}{j!}(S_{t\wedge t_{j+1}}-S_{t\wedge t_{j}})^{j}$ $\displaystyle=$
$\displaystyle
L^{n}+\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S_{t\wedge
t_{i}}}^{S_{t\wedge t_{i+1}}}(f^{(m)}(r)-f^{(m)}(S_{t\wedge
t_{i}}))(S_{t\wedge t_{i+1}}-r)^{m-1}dr$

where $L^{n}=\sum_{t_{i}\in\pi_{n}}\sum_{j=1}^{m}\frac{f^{(j)}(S_{t\wedge
t_{i}})}{j!}(S_{t\wedge t_{j+1}}-S_{t\wedge t_{j}})^{j}$. It is easy to notice
that there exists a constant $M>0$ and

$\displaystyle\left|\frac{1}{\Gamma(m)}\int_{S_{t\wedge t_{i}}}^{S_{t\wedge
t_{i+1}}}(f^{(m)}(r)-f^{(m)}(S_{t\wedge t_{i}}))(S_{t\wedge
t_{i+1}}-r)^{m-1}dr\right|$ $\displaystyle\leq$
$\displaystyle\left|M\int_{S_{t\wedge t_{i}}}^{S_{t\wedge
t_{i+1}}}\phi^{(m)}(r-S(t_{i}\wedge t))(S_{t\wedge t_{i+1}}-r)^{m-1}dr\right|$
$\displaystyle=$ $\displaystyle\left|M(m-1)\int_{S_{t\wedge
t_{i}}}^{S_{t\wedge t_{i+1}}}\phi^{(m-1)}(r-S(t_{i}\wedge t))(S_{t\wedge
t_{i+1}}-r)^{m-2}dr\right|$ $\displaystyle=$
$\displaystyle\left|M(m-1)!\int_{S_{t\wedge t_{i}}}^{S_{t\wedge
t_{i+1}}}\phi^{\prime}(r-S(t_{i}\wedge
t))dr\right|=M(m-1)!|\phi(S(t_{i+1}\wedge t)-S(t_{i}\wedge t))|$

Now we write

$\displaystyle\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S_{t\wedge
t_{i}}}^{S_{t\wedge t_{i+1}}}(f^{(m)}(r)-f^{(m)}(S_{t\wedge
t_{i}}))(S_{t\wedge t_{i+1}}-r)^{m-1}dr$ $\displaystyle=$
$\displaystyle\sum_{S(t_{i}\wedge t)\in
U_{k},[t_{i},t_{i+1}]\in\pi_{n}}\cdots+\sum_{S(t_{i}\wedge t)\notin
U_{k},[t_{i},t_{i+1}]\in\pi_{n}}\cdots$

We denote first part by $L_{1}^{k}$ and second part by $L_{2}^{k}$. and by
locally uniform convergence property in assumption 3.4, we have for fixed
$\epsilon>0$, all $n$ large enough,

$\displaystyle|L_{2}^{k}|$ $\displaystyle=$
$\displaystyle\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n},S(t_{i}\wedge t)\notin
U_{k}}\frac{1}{\Gamma(m)}\int_{S_{t\wedge t_{i}}}^{S_{t\wedge
t_{i+1}}}(f^{(m)}(r)-f^{(m)}(S_{t\wedge t_{i}}))(S_{t\wedge
t_{i+1}}-r)^{m-1}dr\right|$ $\displaystyle\leq$
$\displaystyle\epsilon\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\int_{S_{t\wedge
t_{i}}}^{S_{t\wedge t_{i+1}}}\phi^{(m)}(r-S(t\wedge t_{i}))(S_{t\wedge
t_{i+1}}-r)^{m-1}dr\right|$ $\displaystyle\leq$
$\displaystyle\epsilon(m-1)!\sum_{[t_{i},t_{i+1}]\in\pi_{n}}|\phi(S(t\wedge
t_{i+1})-S(t\wedge t_{i}))|$

And as for $L_{1}^{k}$, we have

$\displaystyle|L_{1}^{k}|$ $\displaystyle\leq$ $\displaystyle
M(m-1)!\sum_{[t_{i},t_{i+1}]\in\pi_{n},S(t_{i}\wedge t)\in
U_{k}}|\phi(S(t_{i+1}\wedge t)-S(t_{i}\wedge t))|$ $\displaystyle\leq$
$\displaystyle
M(m-1)!\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\varphi_{k}(S(t_{i}\wedge
t))|\phi(S(t_{i+1}\wedge t)-S(t_{i}\wedge t))|$

here $\varphi_{k}$ is a sequence of smooth functions such that

$\varphi_{k}(x)=\left\\{\begin{array}[]{rcl}1&x\in\bar{U}_{k}\\\ 0&x\notin
U_{k-1}\end{array}\right.$

Then $\varphi_{k}(x)\rightarrow\mathbf{1}_{\\{x\in C\\}}$ point-wisely as
$k\rightarrow\infty$. Combine $L_{1}^{k}$ and $L_{2}^{k}$, we obtain that

$\displaystyle\lim_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S_{t\wedge
t_{i}}}^{S_{t\wedge t_{i+1}}}(f^{(m)}(r)-f^{(m)}(S_{t\wedge
t_{i}}))(S_{t\wedge t_{i+1}}-r)^{m-1}dr\right|$ $\displaystyle\leq$
$\displaystyle\epsilon(m-1)![S]^{\phi}(t)+M(m-1)!\int_{0}^{t}\varphi_{k}(S(u))d[S]^{\phi}(u)$

We then let $\epsilon\rightarrow 0$ and $k\rightarrow 0$, obtain that

$\displaystyle\lim_{n\rightarrow\infty}\left|\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\frac{1}{\Gamma(m)}\int_{S_{t\wedge
t_{i}}}^{S_{t\wedge t_{i+1}}}(f^{(m)}(r)-f^{(m)}(S_{t\wedge
t_{i}}))(S_{t\wedge t_{i+1}}-r)^{m-1}dr\right|$ $\displaystyle\leq$
$\displaystyle M(m-1)!\int_{0}^{t}\mathbf{1}_{\\{S(t)\in
C\\}}d[S]^{\phi}(u)=0$

Hence we get the result. ∎

If we equip space $C^{\phi}(\mathbb{R})$ with semi-norms

$||f||_{\phi,K}=||f||_{m,K}+\sup_{x,y\in
K}\frac{|f^{(m)}(y)-f^{(m)}(x)|}{|\phi^{(m)}(y-x)|}$

we can again extend the function space for which theorem holds. But since the
assumption of function involving the path $S$, we can again denote this space
by $E_{S}^{\phi}$.

######  Example 3.6.

Here we give an example of application of the above theorem following Han et
al [17]. We take

$\phi(x)=\frac{|x|}{\sqrt{-\log(|x|)}}$

and

$S(t)=\sum_{k=0}^{\infty}\alpha^{k}\varphi(b^{k}t),\quad t\in[0,1]$

where $|\alpha|=\frac{1}{b}$ and $\varphi(t)=\min_{z\in\mathbb{Z}}|z-t|$ or
$\varphi(t)=\nu\sin(2\pi t)+\rho\cos(2\pi t)$. The sequence of partitions
considered here is the $b-$adic partitions of $[0,1]$, i.e.

$\pi_{n}=\\{kb^{-n}:k=0,\cdots,b^{n}\\},\quad n\in\mathbb{N}$

It is shown in [17] that the $\phi-$variation of $S$ is non-trivial.
Furthermore, let $m=1$, we see that

$\lim_{x\rightarrow 0}\frac{\phi(x)}{x}=0,\quad\lim_{x\rightarrow
0}\frac{\phi(x)}{x^{2}}=\infty$

Hence theorem 3.5 applies to the path $S$ and partitions $\pi$ with
$\phi=\frac{|x|}{\sqrt{-\log(|x|)}}$. And here we can choose $f(x)=\phi(x)$
which satisfies the assumption 3.4 with $C=\\{0\\}$.

##  4 An isometry formula for the pathwise integral

Ananova and Cont [1] proved a pathwise analogue of the Itô isometry for the
Föllmer integral with respect to paths with finite quadratic variation. This
relation was extended to $p\geq 1$ in [10]. In the same flavor we derive here
an isometry relation for the pathwise integral in terms of the
$\phi-$variation, where $\phi$ is a more general function satisfying the
following assumption:

######  Assumption 4.1.

For $x\geq 0$, we have

  1. 1.

$\phi$ is strictly increasing, continuous and $\phi(0)=0$

  2. 2.

$\phi$ is a convex function on $[0,\infty)$ and $\log\phi(x)$ is a convex
function of $\log x$.

  3. 3.

For each $x>0$, the limit

$\varphi(x):=\lim_{y\rightarrow 0}\frac{\phi(xy)}{\phi(y)}$

exists and the convergene is uniform on bounded sets.

  4. 4.

$\infty>p(\phi)=\sup\left\\{p:\lim_{x\rightarrow
0}\frac{\phi(x)}{|x|^{p}}=0\right\\}>0$.

We first introduce a generalized version of Minkowski’s inequality here[20].

######  Lemma 4.2.

Suppose $\phi$ satisfies top 2 conditions in assumption 4.1, then we have
inequality

$\phi^{-1}\left(\sum\phi(a_{n}+b_{n})\right)\leq\phi^{-1}\left(\sum\phi(a_{n})\right)+\phi^{-1}\left(\sum\phi(b_{n})\right)$

for two positive sequences $\\{a_{n}\\}$ and $\\{b_{n}\\}$.

Using this lemma, we can show that

######  Theorem 4.3.

Let $\phi$ be an even function satisfying Assumption 4.1 and $\pi_{n}$ a
sequence of partitions of $[0,T]$ with vanishing mesh. Suppose $S\in
V_{\phi}(\pi)\cap C^{\alpha}([0,T],\mathbb{R})$ for
$\alpha>\frac{\sqrt{1+\frac{4}{p(\phi)}}-1}{2}$. Let
$F\in\mathbb{C}_{b}^{1,2}(\Lambda_{T})$ such that
$\nabla_{\omega}F\in\mathbb{C}_{b}^{1,1}(\Lambda_{T})$. Assume furthermore
that $F$ is Lipschitz-continuous with respect to $d_{\infty}$. Then
$F(\cdot,S)\in V_{\phi}(\pi)$ and

$[F(\cdot,S)]^{\phi}(t)=\int_{0}^{t}\varphi(|\nabla_{\omega}F(s,S_{s})|)d[S]^{\phi}(s)$

###### Proof.

We have the same

$R_{F}(s,t):=|F(t,S_{t})-F(s,S_{t})-\nabla_{\omega}F(s.S_{s})(S(t)-S(s))|\leq
C|t-s|^{\alpha+\alpha^{2}}$

with $\gamma_{F}(s,t):=\nabla_{\omega}F(s.S_{s})(S(t)-S(s))$, we have from
lemma 4.2 that

$\displaystyle\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq
t\end{subarray}}\phi(|F(t_{j+1},S_{t_{j+1}})-F(t_{j},S_{t_{j}})|)\right)$
$\displaystyle=$
$\displaystyle\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq
t\end{subarray}}\phi(|R_{F}(t_{j},t_{j+1})+\gamma_{F}(t_{j},t_{j+1})|)\right)$
$\displaystyle\leq$
$\displaystyle\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq
t\end{subarray}}\phi(|R_{F}(t_{j},t_{j+1})|)\right)+\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq t\end{subarray}}\phi(|\gamma_{F}(t_{j},t_{j+1})|)\right)$
$\displaystyle\leq$ $\displaystyle
2\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq
t\end{subarray}}\phi(|R_{F}(t_{j},t_{j+1})|)\right)+\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq
t\end{subarray}}\phi(|R_{F}(t_{j},t_{j+1})+\gamma_{F}(t_{j},t_{j+1})|)\right)$

and since $R_{F}(s,t)\leq C|t-s|^{\alpha^{2}+\alpha}$, we have

$\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq
t\end{subarray}}\phi(|R_{F}(t_{j},t_{j+1})|)\right)\leq\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq t\end{subarray}}\phi(C|t_{j+1}-t_{j}|^{\alpha+\alpha^{2}})\right)$

Furthermore due to $\alpha>\frac{\sqrt{1+\frac{4}{p(\phi)}}-1}{2}$, we know

$\alpha+\alpha^{2}>\frac{1}{p(\phi)}$

Hence there exist $\epsilon>0$ such that

$\phi(C|t_{j+1}-t_{j}|^{\alpha+\alpha^{2}})\leq\phi(C|t_{j+1}-t_{j}|^{\frac{1}{p(\phi)-\epsilon}})$

Then by the definition of $p(\phi)$, we see that

$\phi(C|t_{j+1}-t_{j}|^{\frac{1}{p(\phi)-\epsilon}})\leq\omega(|t_{j+1}-t_{j}|)C^{p(\phi)-\epsilon/2}|t_{j+1}-t_{j}|^{\frac{p(\phi)-\epsilon/2}{p(\phi)-\epsilon}}$

where $\omega$ is continuous on $\mathbb{R}^{+}$ and $\omega(0)=0$. Combine
all above, we get

$\lim_{n\rightarrow\infty}\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq t\end{subarray}}\phi(|R_{F}(t_{j},t_{j+1})|)\right)=0$

and we also have

$\lim_{n\rightarrow\infty}\phi^{-1}\left(\sum_{\begin{subarray}{c}[t_{j},t_{j+1}]\in\pi_{n}\\\
t_{j+1}\leq
t\end{subarray}}\phi(|\gamma_{F}(t_{j},t_{j+1})|)\right)=\phi^{-1}\left(\int_{0}^{t}\varphi(|\nabla_{\omega}F(s,S_{s})|)d[S]^{\phi}(s)\right)$

In fact, since $\nabla_{\omega}F(t,S_{t})\in\mathbb{B}(\Lambda_{T})$, there
exist $M>0$ such that $|\nabla_{\omega}F(u,S_{u})|\leq M$. For each
$\epsilon>0$, there exists $\delta>0$ such that

$\left|\frac{\phi(xy)}{\phi(y)}-\varphi(x)\right|\leq\epsilon$

for all $|x|\leq M,|y|\leq\delta$. Then we have for $n$ large enough(i.e.
$osc(S,\pi_{n})\leq\delta$)

$\displaystyle\sum_{[t_{j},t_{j+1}]\in\pi_{n},t_{j+1}\leq
t}\phi(|\nabla_{\omega}F(t_{j},S_{t_{j}})(S(t_{j+1})-S(t_{j}))|)$
$\displaystyle\leq$ $\displaystyle\sum_{[t_{j},t_{j+1}]\in\pi_{n},t_{j+1}\leq
t}\varphi(|\nabla_{\omega}F(t_{j},S_{t_{j}})|)\phi(|S(t_{j+1})-S(t_{j})|)+\epsilon\phi(|S(t_{j+1})-S(t_{j})|)$

Then let $n$ tends to infinity and $\epsilon$ tends to $0$, we obtain

$\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n},t_{j+1}\leq
t}\phi(|\nabla_{\omega}F(t_{j},S_{t_{j}})(S(t_{j+1})-S(t_{j}))|)=\int_{0}^{t}\varphi(|\nabla_{\omega}F(s,S_{s})|)d[S]^{\phi}(s)$

Hence we get

$[F(\cdot,S)]^{\phi}(t)=\int_{0}^{t}\varphi(|\nabla_{\omega}F(s,S_{s})|)d[S]^{\phi}(s)$

∎

In particular, if we set $\phi(x)=|x|^{p}$ for $p>0$, we have

$\varphi(x)=\lim_{y\rightarrow 0}\frac{|xy|^{p}}{|y|^{p}}=|x|^{p}$

Hence, we have the corollary

######  Corollary 4.4.

Let $p\in\mathbb{R}^{+}$, and $\alpha>((1+\frac{4}{p})^{\frac{1}{2}}-1)/2$,
let $\pi_{n}$ be a sequence of partitions with vanishing mesh, and let $S\in
V_{p}(\pi)\cap C^{\alpha}([0,T],\mathbb{R})$. Let
$F\in\mathbb{C}_{b}^{1,2}(\Lambda_{T})$ such that
$\nabla_{\omega}F\in\mathbb{C}_{b}^{1,1}(\Lambda_{T}))$. Assume furthermore
that $F$ is Lipschitz-continuous with respect to $d_{\infty}$. Then
$F(\cdot,S)\in V_{p}(\pi)$ and

$[F(\cdot,S)]^{p}(t)=\int_{0}^{t}|\nabla_{\omega}F(s,S_{s})|^{p}d[S]^{p}(s)$

######  Remark 4.5.

There exists a non empty set of $p$ and $\alpha$ such that assumption $S\in
V_{\phi}\cap^{\alpha}([0,T],\mathbb{R})$ for
$\alpha>\frac{\sqrt{1+\frac{4}{p(\phi)}}-1}{2}$ is valid. In fact, what we
need to verify is that

$\sum_{[t_{i},t_{i+1}]\in\pi_{n}}\phi(S(t_{i+1})-S(t_{i}))$

should not be identical zero for all
$\alpha>\frac{\sqrt{1+\frac{4}{p(\phi)}}-1}{2}$. Due to the definition of
$p(\phi)$, $\phi(S(t_{i+1})-S(t_{i}))$ should behave like
$C(\pi_{n})|S(t_{i+1})-S(t_{i})|^{p(\phi)}$ when the oscillation of $\pi_{n}$
is small enough. Here $C(\pi_{n})$ are uniformly bounded. Hence it remains to
check

$\sum_{[t_{i},t_{i+1}]\in\pi_{n}}|S(t_{i+1})-S(t_{i})|^{p(\phi)}$

can not be zero for all $S\in C^{\alpha}$. By the definition of Hölder
property, it remains to verify

$p(\phi)\frac{\sqrt{1+\frac{4}{p(\phi)}}-1}{2}<1$

which is equivalent to

$1+\frac{4}{p(\phi)}<\left(1+\frac{2}{p(\phi)}\right)^{2}$

which is obviously true.

######  Example 4.6.

We can see $\phi(x)=\frac{x}{\sqrt{-\log x}}$ satisfies Assumption (4.1). In
fact, it is strictly increasing and continuous and $\phi(0)=0$. Furthermore,

$\phi^{\prime}(x)=\frac{1}{\sqrt{-\log x}}+\frac{1}{2(\sqrt{-\log
x})^{3}}>0,\quad{\rm and}\quad\phi^{\prime\prime}(x)=\frac{1}{2x(\sqrt{-\log
x})^{3}}+\frac{3}{4x(\sqrt{-\log x})^{5}}>0$

Hence $\phi$ is convex. And we have

$\log\phi(x)=\log x-\frac{\log(-\log x)}{2}$

It is easy to check that $x\rightarrow x-\frac{\log(-x)}{2}$ is convex. And

$\displaystyle\varphi(x)$ $\displaystyle=$ $\displaystyle\lim_{y\rightarrow
0}\frac{\phi(|x||y|)}{\phi(|y|)}=\lim_{y\rightarrow 0}|x|\sqrt{\frac{-\log
y}{-\log x-\log y}}=|x|$

Hence we have

$[F(\cdot,S)]^{\phi}(t)=\int_{0}^{t}|\nabla_{\omega}F(s,S_{s})|d[S]^{\phi}(s)$

##  5 Multi-dimensional extensions

The extension of the above results to the multidimensional case is not
entirely straightforward, as the space $V_{p}(\pi)$ is not a vector space
[24].

In the case of integer $p$, some definitions may be extended to the vector
case by considering symmetric tensor-valued measures as in [10] but this is
not convenient for the fractional case. We use a definition which is
equivalent to the definition in the paper [10] when $p$ is an integer.

######  Definition 5.1.

Let $p\geq 1$ and $S=(S^{1},\cdots,S^{d})\in C([0,T],\mathbb{R}^{d})$ be a
continuous path. Let $\\{\pi_{n}\\}$ be a sequence of partition of $[0,T]$. We
say that $S$ has a $p-$the order variation along $\pi=\\{\pi_{n}\\}$ if
$osc(S,\pi_{n})\rightarrow 0$ and for any linear combination
$S_{a}:=\sum_{i=1}^{d}a_{i}S^{i}$, we have $S_{a}\in V_{p}(\pi)$. Furthermore,
we denote $\mu_{S_{a}}$ to be the weak limit of measures

$\mu_{S_{a}}^{n}:=\sum_{[t_{j},t_{j+1]}\in\pi_{n}}\delta(\cdot-
t_{j})|S_{a}(t_{j+1})-S_{a}(t_{j})|^{p}$

######  Remark 5.2.

Although we can define the situation when $S$ admits a $p$-th variation, we
can not give an object directly related to the $p$-th variation of the path
$S$ in this definition.

However, it can be seen that multi-dimension fractional Brownian motion will
definitely admit $p-$th variation along sequences of partitions with fine
enough mes.

We now state the assumptions on the function $f$ that will appear in the Itô
formula.

######  Assumption 5.3.

Let $p=m+\alpha$ with $m=\lfloor p\rfloor$. The symmetric tensor product
$\nabla^{m}f(x)\in C_{loc}^{\alpha}(Sym_{m}(\mathbb{R}^{d}))$. Furthermore,
there exists a sequence of open sets $U_{k}$ such that $\bar{U}_{k+1}\subset
U_{k}$ and

$\lim_{y\rightarrow
x}\frac{||\nabla^{m}f(y)-\nabla^{m}f(x)||}{||y-x||^{\alpha}}=0$

locally uniformly on $U_{k}^{c}$.

######  Assumption 5.4.

The sequence $U_{k}$ in assumption 5.3 converges to a set $C=\cap_{k}U_{k}$
and for all $a=(a_{1},\cdots,a_{d})\in\mathbb{R}^{d}$,

$\int_{0}^{T}\mathbf{1}_{\\{S(t)\in C\\}}d\mu_{S_{a}}=0$

where $\mu_{S_{a}}$ is defined as in Definition 5.1.

######  Theorem 5.5.

Suppose $S\in V_{p}(\pi)$ and $f$ satisfies assumptions 5.3 and 5.4. Then we
have the formula

$f(S(t))-f(S(0))=\int_{0}^{t}\nabla f(S(u))dS(u)$

where

$\int_{0}^{t}\nabla
f(S(u))dS(u):=\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\sum_{k=1}^{m}\frac{1}{k!}<\nabla^{k}f(S(t_{j})),(S(t_{j+1})-S(t_{j})^{\otimes
k}>$

Before we prove the theorem, we give a lemma here.

######  Lemma 5.6.

Let $\alpha_{1},\cdots,\alpha_{d}$ be positive numbers such that
$\alpha=\sum_{i=1}^{d}\alpha_{i}>1$. Suppose $S\in V^{\alpha}(\pi)$ in the
sense of definition 5.1. Then the limit of

$\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\Pi_{i=1}^{d}|S^{i}(t_{j+1})-S^{i}(t_{j})|^{\alpha_{i}}$

can be bounded by sum of $\alpha-$th variations of $2^{d}$ different linear
combinations of components of path $S$.

###### Proof.

For any positive number $\alpha_{1},\cdots\alpha_{d}$ such that
$\alpha=\sum_{i=1}^{d}\alpha_{i}>1$, Young’s inequality tells us that

$\Pi_{i=1}^{d}|S^{i}(t_{j+1})-S^{i}(t_{j})|^{\frac{\alpha_{i}}{\alpha}}\leq\sum_{i=1}^{d}\frac{\alpha_{i}}{\alpha}|S^{i}(t_{j+1})-S^{i}(t_{j})|$

which shows

$\Pi_{i=1}^{d}|S^{i}(t_{j+1})-S^{i}(t_{j})|^{\alpha_{i}}\leq\sum_{\epsilon_{i}=\pm
1}|\sum_{i=1}^{d}\epsilon_{i}\frac{\alpha_{i}}{\alpha}(S^{i}(t_{j+1})-S^{i}(t_{j}))|^{\alpha}$

By taking sum over the partition $\pi_{n}$ and taking the limit of $n$ we then
obtain desired result. ∎

###### Proof of theorem 5.5.

The technique for the proof is the same as the previous theorems. We only
consider the situation $t=T$, the case $t<T$ is similar with an $o(1)$
additional term. Using a Taylor expansion,

$\displaystyle f(S(T))-f(S(0))$ $\displaystyle=$
$\displaystyle\sum_{[t_{j},t_{j+1}]\in\pi_{n}}(f(S(t_{j+1}))-f(S(t_{j})))$
$\displaystyle=$
$\displaystyle\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\sum_{k=1}^{m}\frac{1}{k!}<\nabla^{m}f(S(t_{j})),(S(t_{j+1})-S(t_{j}))^{\otimes
k}>$ $\displaystyle+$
$\displaystyle\frac{1}{(m-1)!}\int_{0}^{1}<\nabla^{m}f(S(t_{j})+\lambda(S(t_{j+1})-S(t_{j})))-\nabla^{m}f(S(t_{j}))$
$\displaystyle,(S(t_{j+1})-S(t_{j}))^{\otimes m}>(1-\lambda)^{m-1}d\lambda$

We denote last integral by $R(t_{j},t_{j+1})$. By Hölder continuity of
$\nabla^{m}f$, there exists $M>0$ such that

$\frac{||\nabla^{m}f(S(t_{j})+\lambda(S(t_{j+1})-S(t_{j})))-\nabla^{m}f(S(t_{j}))||}{\lambda^{\alpha}||S(t_{j+1})-S(t_{j})||^{\alpha}}\leq
M$

Then we would get

$\displaystyle|R(t_{j},t_{j+1})|\leq\frac{M}{(m-1)!}\int_{0}^{1}\lambda^{\alpha}||S(t_{j+1})-S(t_{j})||^{\alpha}||(S(t_{j+1})-S(t_{j}))^{\otimes
m}||(1-\lambda)^{m-1}d\lambda$

Fixed $k\in\mathbb{N}$, We divide the partition $\pi_{n}$ into two parts
$\pi^{1}_{n}:=\\{[t_{j},t_{j+1}]\in\pi_{n}:S(t_{j})\in U_{k}\\}$ and
$\pi_{n}^{2}:=\pi_{n}/\pi_{n}^{1}$. Then we have on $\pi_{n}^{1}$,

$\sum_{[t_{j},t_{j+1}]\in\pi_{n}^{1}}|R(t_{j},t_{j+1})|\leq\sum_{[t_{j},t_{j+1}]\in\pi_{n}^{1}}\tilde{M}||S(t_{j+1})-S(t_{j})||^{\alpha}||(S(t_{j+1})-S(t_{j}))^{\otimes
m}||$

with
$\tilde{M}=\frac{M}{(m-1)!}\int_{0}^{1}\lambda^{\alpha}(1-\lambda)^{m-1}d\lambda$
and on $\pi_{n}^{2}$, for each $\epsilon>0$, when $n$ is large enough,

$\sum_{[t_{j},t_{j+1}]\in\pi_{n}^{2}}|R(t_{j},t_{j+1})|\leq\sum_{[t_{j},t_{j+1}]\in\pi_{n}^{2}}\epsilon||S(t_{j+1})-S(t_{j})||^{\alpha}||(S(t_{j+1})-S(t_{j}))^{\otimes
m}||$

In order to give our result, we need to give a bound for

$\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}||S(t_{j+1})-S(t_{j})||^{\alpha}||(S(t_{j+1})-S(t_{j}))^{\otimes
m}||$

First we have

$||S(t_{j+1})-S(t_{j})||^{\alpha}\leq
C\sum_{i=1}^{d}|S^{i}(t_{j+1})-S^{i}(t_{j})|^{\alpha}$

for some constant $C$. By Lemma 5.6, we know that each component of

$\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}||S(t_{j+1})-S(t_{j})||^{\alpha}(S(t_{j+1})-S(t_{j}))^{\otimes
m}$

can be bounded by the sum of $2^{d}$ different $p-$th varaitions of linear
combinations of path $S$. Hence

$\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}||S(t_{j+1})-S(t_{j})||^{\alpha}||(S(t_{j+1})-S(t_{j}))^{\otimes
m}||$

is bounded by some constant $C(S,\pi)$ related to path $S$. Furthermore, in
the sense of measure, it will be dominant by sum of measures $\mu_{r}$ with
enough numbers $r=1,\cdots,R$. Then we have

$\displaystyle\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}|R(t_{j},t_{j+1})|$
$\displaystyle\leq$
$\displaystyle\lim_{n\rightarrow\infty}\sum_{[t_{j},t_{j+1}]\in\pi_{n}}\tilde{M}g_{k}(S(t_{j}))||S(t_{j+1})-S(t_{j})||^{\alpha}||(S(t_{j+1})-S(t_{j}))^{\otimes
m}||+\epsilon C(S,\pi)$ $\displaystyle\leq$
$\displaystyle\sum_{r=1}^{R}\int_{0}^{T}g_{k}(S(t))d\mu_{r}+\epsilon C(S,\pi)$

Here $g_{k}$ is a positive smooth function with support on $U_{k-1}$,
$g_{k}|_{U_{k}}=1$ and $g_{k}\leq 1$. Letting $k\rightarrow\infty$ and
$\epsilon\rightarrow 0$, by assumption 5.4, we obtain the result. ∎

######  Remark 5.7.

This result may be extended to the time-dependent functions as discussed
above.

######  Remark 5.8.

It is hopeless to obtain a multi-variate Taylor expansion up to order $p$ in
general. The simplest function $f(x)=|x|^{\alpha}$ with $0<\alpha<1$ does not
admit a Taylor-type expansion up to order $\alpha$ at $0$ since the limit

$\lim_{x\rightarrow
0}\frac{||x||^{\alpha}}{|x_{1}|^{\alpha}+\cdots|x_{d}|^{\alpha}}$

does not exist.

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