Rough paths theory Fall 2017. Lecture 1

The first few lectures are essentially reminders of undergraduate real analysis materials. We will cover some aspects of the theory of differential equations driven by continuous paths with bounded variation. The point is to fix some notations that will be used throughout the course and to stress the importance of the topology of convergence in 1-variation if we are interested in stability results for solutions with respect to the driving signal.

If $s \le t$ , we will denote by $\Delta [s,t]$ , the set of subdivisions of the interval $[s,t]$ , that is $\Pi \in \Delta [s,t]$ can be written
$\Pi=\left\{ s= t_0 < t_1 < \cdots < t_n =t \right\}.$

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

$\| \cdot \|_{1-var; [s,t]}$ is not a norm, because constant functions have a zero 1-variation, but it is obviously a semi-norm. If $x$ is continuously differentiable on $[s,t]$ , it is easily seen (Exercise !) that
$\| x \|_{1-var, [s,t]}=\int_s^t \| x'(s) \| ds.$

Proposition: Let $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ . The function $(s,t)\to \| x \|_{1-var, [s,t]}$ is additive, i.e for $0 \le s \le t \le u \le T$ ,
$\| x \|_{1-var, [s,t]}+ \| x \|_{1-var, [t,u]}= \| x \|_{1-var, [s,u]},$
and controls $x$ in the sense that for $0 \le s \le t \le T$ ,
$\| x(s)-x(t) \| \le \| x \|_{1-var, [s,t]}.$
The function $s \to \| x \|_{1-var, [0,s]}$ is moreover continuous and non decreasing.

Proof: 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
$\sup_{ \Pi_1 \in \Delta[s,t]} \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \| +\sup_{ \Pi_2 \in \Delta[t,u]} \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \| \le \sup_{ \Pi \in \Delta[s,u]} \sum_{k=0}^{n-1} \| x(t_{k+1}) -x(t_k) \|,$
thus
$\| x \|_{1-var, [s,t]}+ \| x \|_{1-var, [t,u]} \le \| x \|_{1-var, [s,u]}.$
Let now $\Pi \in \Delta[s,u]$ :
$\Pi=\left\{ s= t_0 < t_1 < \cdots < t_n =u \right\}.$
Let $k=\max \{ j, t_j \le t\}$ . By the triangle inequality, we have
$\sum_{j=0}^{n-1} \| x(t_{j+1}) -x(t_j) \|$
$\le \sum_{j=0}^{k-1} \| x(t_{j+1}) -x(t_j) \| + \sum_{j=k}^{n-1} \| x(t_{j+1}) -x(t_j) \|$
$\le \| x \|_{1-var, [s,t]}+ \| x \|_{1-var, [t,u]}.$
Taking the $\sup$ of $\Pi \in \Delta[s,u]$ yields
$\| x \|_{1-var, [s,t]}+ \| x \|_{1-var, [t,u]} \ge \| x \|_{1-var, [s,u]},$
which completes the proof. The proof of the continuity and monoticity of $s \to \| x \|_{1-var, [0,s]}$ is let to the reader $\square$

This control of the path by the 1-variation norm is an illustration of the notion of controlled path which is very useful in rough paths theory.

Definition: A map $\omega: \{ 0 \le s \le t \le T \} \to [0,\infty)$ is called superadditive if for all $s \le t \le u$ ,
$\omega(s,t)+\omega(t,u) \le \omega (s,u).$
If, in adition, $\omega$ is continuous and $\omega(t,t)=0$ , we call $\omega$ a control. We say that a path $x:[0,T] \to \mathbb{R}$ is controlled by a control $\omega$ , if there exists a constant $C < 0$ , such that for every $0 \le s \le t \le T$ ,
$\| x(t) -x(s) \| \le C \omega(s,t).$

Obviously, Lipschitz functions have a bounded variation. The converse is of course not true: $t\to \sqrt{t}$ has a bounded variation on $[0,1]$ but is not Lipschitz. However, any continuous path with bounded variation is the reparametrization of a Lipschitz path in the following sense.

Proposition: Let $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ . There exist a Lipschitz function $y:[0,1] \to \mathbb{R}^d$ , and a continuous and non-decreasing function $\phi:[0,T]\to [0,1]$ such that $x=y\circ \phi$ .

Proof: We assume $\| x \|_{1-var, [0,T]} \neq 0$ and consider
$\phi(t)=\frac{ \| x \|_{1-var, [0,t]} }{ \| x \|_{1-var, [0,T]} }.$
It is continuous and non decreasing. There exists a function $y$ such that $x=y\circ \phi$ because $\phi(t_1)=\phi(t_2)$ implies $x(t_1)=x(t_2)$ . We have then, for $s \le t$ ,
$\| y( \phi(t)) -y ( \phi(s)) \|=\| x(t) -x (s) \| \le \| x \|_{1-var, [s,t]} =\| x \|_{1-var, [0,T]} (\phi(t)-\phi(s) )$ $\square$

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

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

Proof: Let $x^n \in C^{1-var} ([0,T], \mathbb{R}^d)$ be a Cauchy sequence. It is clear that
$\| x^n -x^m \|_\infty \le \| x^n(0)-x^m(0) \|+ \| x^n-x^m \|_{1-var, [0,T]}.$
Thus, $x^n$ converges uniformly to a continuous path $x :[0,T] \to \mathbb{R}$ . We need to prove that $x$ has a bounded variation. Let
$\Pi=\{ 0=t_0 <t_1 < \cdots <t_n=T \}$
be a a subdivision of $[0,T]$ . There is $m \ge 0$ , such that $\| x - x^m \|_\infty \le \frac{1}{2n}$ , thus
$\sum_{k=0}^{n-1} \|x(t_{k+1})-x(t_k) \|$
$\le \sum_{k=0}^{n-1} \|x(t_{k+1})-x^m(t_{k+1}) \| +\sum_{k=0}^{n-1} \|x^m(t_{k})-x(t_k) \| +\| x^m \|_{1-var,[0,T]}$
$\le 1+\sup_{n} \| x^n \|_{1-var,[0,T]}.$
Thus, we have
$\| x \|_{1-var,[0,T]} \le 1+\sup_{n} \| x^n \|_{1-var,[0,T]} < \infty$
$\square$

For approximations purposes, it is important to observe that the set of smooth paths is not dense in $C^{1-var} ([0,T], \mathbb{R}^d)$ for the 1-variation convergence topology. The closure of the set of smooth paths in the 1-variation norm, which shall be denoted by $C^{0,1-var} ([0,T], \mathbb{R}^d)$ is the set of absolutely continuous paths.

Proposition: Let $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ . Then, $x \in C^{0,1-var} ([0,T], \mathbb{R}^d)$ if and only if there exists $y \in L^1([0,T])$ such that,
$x(t)=x(0)+\int_0^t y(s) ds.$

Proof: First, let us assume that
$x(t)=x(0)+\int_0^t y(s) ds,$
for some $y \in L^1([0,T])$ . Since smooth paths are dense in $L^1([0,T])$ , we can find a sequence $y^n$ in $L^1([0,T])$ such that $\| y-y^n \|_1 \to 0$ . Define then,
$x^n(t)=x(0)+\int_0^t y^n(s) ds.$
We have
$\| x-x^n \|_{1-var,[0,T]}=\| y-y^n \|_1.$
This implies that $x \in C^{0,1-var} ([0,T], \mathbb{R}^d)$ . Conversely, if $x \in C^{0,1-var} ([0,T], \mathbb{R}^d)$ , there exists a sequence of smooth paths $x^n$ that converges in the 1-variation topology to $x$ . Each $x^n$ can be written as,
$x^n(t)=x^n(0)+\int_0^t y^n(s) ds.$
We still have
$\| x^m-x^n \|_{1-var,[0,T]}=\| y^m-y^n \|_1,$
so that $y^n$ converges to some $y$ in $L^1$ . It is then clear that
$x(t)=x(0)+\int_0^t y(s) ds$
$\square$

Exercise: Let $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ . Show that $x$ is the limit in 1-variation of piecewise linear interpolations if and only if $x \in C^{0,1-var} ([0,T], \mathbb{R}^d)$ .

Let $y:[0,T] \to \mathbb{R}^{e \times d}$ be a piecewise continuous path and $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ . It is well-known that we can integrate $y$ against $x$ by using the Riemann–Stieltjes integral which is a natural extension of the Riemann integral. The idea is to use the Riemann sums
$\sum_{k=0}^{n-1} y(t_k) (x(t_{k+1})-x(t_k)),$
where $\Pi=\{ 0 =t_0 < t_1 < \cdots < t_n =T \}$ . It is easy to prove that, when the mesh of the subdivision $\Pi$ goes to 0, the Riemann sums converge to a limit which is independent from the sequence of subdivisions that was chosen. The limit is then denoted $\int_0^T y(t) dx(t)$ and called the Riemann-Stieltjes integral of $y$ against $x$ . Since $x$ has a bounded variation, it is easy to see that, more generally,
$\sum_{k=0}^{n-1} y(\xi_k) (x(t_{k+1})-x(t_k)),$
with $t_k \le \xi_k \le t_{k+1}$ would also converge to $\int_0^T y(t) dx(t)$ . If
$x(t)=x(0)+\int_0^t g(s) ds$
is an absolutely continuous path, then it is not difficult to prove that we have
$\int_0^T y(t) dx(t) =\int_0^T y(t) g(t) dt,$
where the integral on the right hand side is understood in Riemann’s sense.

We have
$\left\| \sum_{k=0}^{n-1} y(t_k) (x(t_{k+1})-x(t_k))\right\|$
$\le \sum_{k=0}^{n-1} \| y(t_k)\| \| (x(t_{k+1})-x(t_k))\|$
$\le \sum_{k=0}^{n-1} \| y(t_k)\| \| (x(t_{k+1})-x(t_k))\|$
$\le \sum_{k=0}^{n-1} \| y(t_k)\| \| x \|_{1-var,[t_k,t_{k+1}]}.$
Thus, by taking the limit when the mesh of the subdivision goes to 0, we obtain the estimate
$\left\| \int_0^T y(t) dx(t) \right\| \le \int_0^T \| y(t) \| \| dx(t) \| \le \| y \|_{\infty, [0,T]} \| x \|_{1-var,[0,T]},$
where $\int_0^T \| y(t) \| \| dx(t) \|$ is the notation for the Riemann-Stieltjes integral of $\| y \|$ against the bounded variation path $l(t)= \| x \|_{1-var,[0,t]}$ . We can also estimate the Riemann-Stieltjes integral in the 1-variation distance. We collect the following estimate for later use:

Proposition: Let $y,y':[0,T] \to \mathbb{R}^{e \times d}$ be a piecewise continuous path and $x,x' \in C^{1-var} ([0,T], \mathbb{R}^d)$ . We have
$\left\| \int_0^{\cdot} y'(t) dx'(t)-\int_0^{\cdot} y(t) dx(t) \right\|_{1-var,[0,T]} \le \| x \|_{1-var,[0,T]} \| y-y' \|_{\infty, [0,T]} + \| y' \|_{\infty, [0,T]} \| x -x'\|_{1-var,[0,T]}.$

The Riemann-Stieltjes satisfies the usual rules of calculus, for instance the integration by parts formula takes the following form
Proposition: Let $y \in C^{1-var} ([0,T], \mathbb{R}^{e \times d} )$ and $x\in C^{1-var} ([0,T], \mathbb{R}^d)$ .
$\int_0^T y(t) dx(t)+\int_0^T dy(t) x(t)=y(T)x(T) -y(0)x(0).$

We also have the following change of variable formula:

Proposition: Let $x\in C^{1-var} ([0,T], \mathbb{R}^d)$ and let $\Phi: \mathbb{R}^d \to \mathbb{R}^e$ be a $C^1$ map. We have
$\Phi (x(T)) =\Phi (x(0)) + \int_0^T \Phi'(x(t)) dx(t).$

Proof: From the mean value theorem
$\Phi (x(T)) -\Phi (x(0))=\sum_{k=0}^{n-1} (\Phi (x(t_{k+1})) -\Phi (x(t_k)))=\sum_{k=0}^{n-1}\Phi'(x_{\xi_k}) (x(t_{k+1}) -x(t_k)),$
with $t_k \le \xi_k \le t_{k+1}$ . The result is then obtained by taking the limit when the mesh of the subdivision goes to 0 $\square$

We finally state a classical analysis lemma, Gronwall’s lemma, which provides a wonderful tool to estimate solutions of differential equations.

Proposition: Let $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ and let $\Phi: [0,T] \to [0,\infty)$ be a bounded measurable function. If,
$\Phi(t) \le A+B\int_0^t \Phi(s) \| d x(s)\|, \quad 0 \le t \le T,$
for some $A,B \ge 0$ , then
$\Phi(t) \le A \exp (B \| x \|_{1-var,[0,t]} )\quad 0 \le t \le T.$

Proof: Iterating the inequality
$\Phi(t) \le A+B\int_0^t \Phi(s) \| d x(s)\|$
$N$ times, we get
$\Phi(t) \le A+\sum_{k=1} ^n AB^{k} \int_0^ t \int_0^{t_1} \cdots \int_0^{t_{k-1}} \| d x(t_k)\| \cdots \| dx(t_1) \| +R_n(t),$
where $R_n(t)$ is a remainder term that goes to 0 when $n \to \infty$ . Observing that
$\int_0^ t \int_0^{t_1} \cdots \int_0^{t_{k-1}} \| d x(t_k)\| \cdots \| dx(t_1) \|=\frac{ \| x \|^k_{1-var,[0,t]} }{k!}$
and sending $n$ to $\infty$ finishes the proof $\square$

3 Responses to Rough paths theory Fall 2017. Lecture 1

marcogorelli says:

August 31, 2017 at 8:30 am

In the proof of the first proposition, should we not have
$$\Pi = \{s=t_0,ldots,t_n=u\}$$
rather than
$$\Pi = \{s=t_0,ldots,t_n=t\}$$
?

- Fabrice Baudoin says:
  
  August 31, 2017 at 2:14 pm
  
  Thanks, corrected.
  
  - marcogorelli says:
    
    August 31, 2017 at 9:03 pm
    
    Thank you, great lecture, I look forward to the rest!