## Lecture 6. Continuous paths with bounded p-variation

Our next goal in this course is to define an integral that can be used to integrate rougher paths than bounded variation. As we are going to see, Young’s integration theory allows to define $\int y dx$ as soon as $y$ has finite $q$-variation and $x$ and has a finite $p$-variation with $1/p+1/q>1$. This integral is simply is a limit of Riemann sums as for the Riemann-Stiletjes integral. In this lecture we present some basic properties of the space of continuous paths with a finite $p$-variation. We present these results for $\mathbb{R}^d$ valued paths but most of the results extend without difficulties to paths valued in metric spaces (see chapter 5 in the book by Friz-Victoir).

Definition. A path $x:[s,t] \to \mathbb{R}^d$ is said to be of finite $p$-variation, $p > 0$ if the $p$-variation of $x$ on $[s,t]$, which is defined as
$\| x \|_{p-var; [s,t]} :=\left( \sup_{ \Pi \in \Delta[s,t]} \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^p \right)^{1/p},$
is finite. The space of continuous paths $x : [s,t] \to \mathbb{R}^d$ with a finite $p$-variation will be denoted by $C^{p-var} ([s,t], \mathbb{R}^d)$.

The notion of $p$-variation is only interesting when $p \ge 1$.

Proposition: Let $x:[s,t] \to \mathbb{R}^d$ be a continuous path of finite $p$-variation with $p < 1$. Then, $x$ is constant.

Proof: We have for $s \le u \le t$,
$\| x(u)-x(s)\|$
$\le ( \max \| x(t_{k+1}) -x(t_k) \|^{1-p} ) \left( \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^p \right)$
$\le ( \max \| x(t_{k+1}) -x(t_k) \|^{1-p} ) \| x \|^p_{p-var; [s,t]}.$

Since $x$ is continuous, it is also uniformly continuous on $[s,t]$. By taking a sequence of subdivisions whose mesh tends to 0, we deduce then that
$\| x(u)-x(s)\|=0,$
so that $x$ is constant $\square$
The following proposition is immediate:

Proposition: Let $x:[s,t] \to \mathbb{R}^d$, be a continuous path. If $p \le p'$ then
$\| x \|_{p'-var; [s,t]} \le \| x \|_{p-var; [s,t]}.$
As a consequence $C^{p-var} ([s,t], \mathbb{R}^d) \subset C^{p'-var} ([s,t], \mathbb{R}^d).$

We remind that a continuous map $\omega: \{ 0 \le s \le t \le T \} \to [0,\infty)$ that vanishes on the diagonal is called a control f if for all $s \le t \le u$,
$\omega(s,t)+\omega(t,u) \le \omega (s,u).$

Proposition: Let $x \in C^{p-var} ([0,T], \mathbb{R}^d)$. Then $\omega(s,t)= \| x \|^p_{p-var; [s,t]}$ is a control such that for every $s \le t$,
$\| x(s) -x(t) \| \le \omega(s,t)^{1/p}.$

Proof: It is immediate that
$\| x(s) -x(t) \| \le \omega(s,t)^{1/p},$
so we focus on the proof that $\omega$ is a control. If $\Pi_1 \in \Delta [s,t]$ and $\Pi_2 \in \Delta [t,u]$, then $\Pi_1 \cup \Pi_2 \in \Delta [s,u]$. As a consequence, we obtain
$\sup_{ \Pi_1 \in \Delta[s,t]} \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^p +\sup_{ \Pi_2 \in \Delta[t,u]} \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^p \le \sup_{ \Pi \in \Delta[s,u]} \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^p,$
thus
$\| x \|^p_{p-var, [s,t]}+ \| x \|^p_{p-var, [t,u]} \le \| x \|^p_{p-var, [s,u]}.$
The proof of the continuity is left to the reader (see also Proposition 5.8 in the book by Friz-Victoir) $\square$

In the following sense, $\| x \|^p_{p-var; [s,t]}$ is the minimal control of a path $x$.

Proposition: Let $x \in C^{p-var} ([0,T], \mathbb{R}^d)$ and let $\omega: \{ 0 \le s \le t \le T \} \to [0,\infty)$ be a control such that for $0 \le s \le t \le T$,
$\| x(s)-x(t) \| \le C \omega (s,t)^{1/p},$
then
$\| x \|_{p-var; [s,t]} \le C \omega(s,t)^{1/p}.$

Proof: We have
$\| x \|_{p-var; [s,t]}$
$= \left( \sup_{ \Pi \in \Delta[s,t]} \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^p \right)^{1/p}$
$\le \left( \sup_{ \Pi \in \Delta[s,t]} \sum_{k=0}^{n-1} C^p \omega(t_{k}, t_{k+1}) \right)^{1/p}$
$\le C \omega(s,t)^{1/p}$
$\square$

The next result shows that the set of continuous paths with bounded $p$-variation is a Banach space.

Theorem: Let $p \ge 1$. The space $C^{p-var} ([0,T], \mathbb{R}^d)$ endowed with the norm $\| x(0) \|+ \| x \|_{p-var, [0,T]}$ is a Banach space.

Proof: The proof is identical to the case $p=1$, so we let the careful reader check the details $\square$

Again, the set of smooth paths is not dense in $C^{p-var} ([0,T], \mathbb{R}^d)$ for the $p$-variation convergence topology. The closure of the set of smooth paths in the $p$-variation norm shall be denoted by $C^{0,p-var} ([0,T], \mathbb{R}^d)$. We have the following characterization of paths in $C^{0,p-var} ([0,T], \mathbb{R}^d)$.

Proposition: Let $p \ge 1$. $x \in C^{0,p-var} ([0,T], \mathbb{R}^d)$ if and only if
$\lim_{\delta \to 0} \sup_{ \Pi \in \Delta[s,t], | \Pi | \le \delta } \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^p=0.$

Proof: See Theorem 5.31 in the book by Friz-Victoir $\square$

The following corollary shall often be used in the sequel:

Corollary: If $1 \le p< q$, then $C^{p-var} ([0,T], \mathbb{R}^d) \subset C^{0,q-var} ([0,T], \mathbb{R}^d)$.

Proof: Let $\Pi \in \Delta[s,t]$ whose mesh is less than $\delta > 0$. We have
$\sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^q$
$\le \left( \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^p\right) \max \| x(t_{k+1}) -x(t_k) \|^{p-q}$
$\le \| x \|^p_{p-var; [s,t]} \max \| x(t_{k+1}) -x(t_k) \|^{p-q}.$
As a consequence, we obtain
$\lim_{\delta \to 0} \sup_{ \Pi \in \Delta[s,t], | \Pi | \le \delta } \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|^q=0$ $\square$

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### 8 Responses to Lecture 6. Continuous paths with bounded p-variation

1. alabair says:

At a given moment we were fascinated by the Lebesgue integral and limitations of Riemann integral but now I think that it is Riemann to the honor.

• Well. The purpose of Lebesgue integration is to integrate functions against measures whereas the essence of Riemann-Stieltjes integral is to integrate functions against paths (functions).
Both theories are of course consistent in the sense that a bounded variation path induces a signed measure: Indeed, a bounded variation function is a difference of two increasing functions and an increasing function $f$ defines a measure by $\mu((a,b])=f(b)-f(a)$. The Riemann integral against $f$ coincides then with the Lebesgue integral against $\mu$.

2. alabair says:

By the way, I can not digest the fact that a function which $\alpha$-Hölderian became automatically to be finite $\frac{1}{\alpha}$-variation.

• If $f$ is $\alpha$-Holder, then $| f(t)-f(s) | \le C | t-s|^\alpha$. This implies
$\sum | f(t_{i+1})-f(t_i) |^{1/\alpha} \le C^{1/\alpha} \sum | t_{i+1}-t_i | \le C^{1/\alpha} T$, if $t_i$ is a subdivision of $[0,T]$.

3. Taras says:

it seems that the exponent $1/p$ is missing in the proposition of estimating the p-variation via the control function.

4. rong li says:

Merry Christmas !!!