Lecture 5. Rough paths. Fall 2017

In this lecture we define the Young‘s integral $\int y dx$ when $x \in C^{p-var} ([0,T], \mathbb{R}^d)$ and $y \in C^{q-var} ([0,T], \mathbb{R}^{e \times d})$ with $\frac{1}{p}+\frac{1}{q} >1$ . The cornerstone is the following Young-Loeve estimate.

Theorem: Let $x \in C^{1-var} ([0,T], \mathbb{R}^d)$ and $y \in C^{1-var} ([0,T], \mathbb{R}^{e \times d})$ . Consider now $p,q \ge 1$ with $\theta=\frac{1}{p}+\frac{1}{q} > 1$ . The following estimate holds: for $0 \le s \le t \le T$ ,
$\left\| \int_s^t y(u)dx(u)-y(s)(x(t)-x(s)) \right\| \le \frac{1}{1-2^{1-\theta} }\| x \|_{p-var; [s,t]} \| y \|_{q-var; [s,t]}.$

Proof: For $0 \le s \le t \le T$ , let us define
$\Gamma_{s,t} =\int_s^t y(u)dx(u) -y(s)(x(t)-x(s)) .$
We have for $s < t < u$ ,
$\Gamma_{s,u}-\Gamma_{s,t}-\Gamma_{t,u} =-y(s)(x(u)-x(s))+y(s)(x(t)-x(s))+y(t)(x(u)-x(t))= (y(s)-y(t))(x(t)-x(u)).$
As a consequence, we get
$\| \Gamma_{s,u}\|\le \| \Gamma_{s,t} \|+\| \Gamma_{t,u}\| +\| x \|_{p-var; [s,t]} \| y \|_{q-var; [t,u]}.$
Let now $\omega(s,t)=\| x \|^{1/\theta}_{p-var; [s,t]} \| y \|^{1/\theta}_{q-var; [s,t]}$ . We claim that $\omega$ is a control. The continuity and the vanishing on the diagonal are obvious to check, so we just need to justify the superadditivity. Let $s < t < u$ , we have from Holder’s inequality,
$\omega(s,t)+\omega(t,u)$
$=\| x \|^{1/\theta}_{p-var; [s,t]} \| y \|^{1/\theta}_{q-var; [s,t]}+\| x \|^{1/\theta}_{p-var; [t,u]} \| y \|^{1/\theta}_{q-var; [t,u]}$
$\le (\| x \|^{p}_{p-var; [s,t]} + \| x \|^{p}_{p-var; [t,u]})^{\frac{1}{p\theta}}(\| y \|^{q}_{q-var; [s,t]} + \| y \|^{q}_{q-var; [t,u]})^{\frac{1}{q\theta}}$
$\le \| x \|^{1/\theta}_{p-var; [s,u]} \| y \|^{1/\theta}_{q-var; [s,u]}=\omega(s,u).$
We have then
$\| \Gamma_{s,u}\|\le \| \Gamma_{s,t} \|+\| \Gamma_{t,u}\| +\omega(s,u)^\theta.$
For $\varepsilon > 0$ , consider then the control
$\omega_\varepsilon (s,t)= \omega(s,t) +\varepsilon ( \| x \|_{1-var; [s,t]} + \| y \|_{1-var; [s,t]}).$
Define now
$\Psi(r)= \sup_{s,u, \omega_\varepsilon (s,u)\le r} \| \Gamma_{s,u}\|.$
If $s,u$ is such that $\omega_\varepsilon (s,u) \le r$ , we can find a $t$ such that $\omega_\varepsilon(s,t) \le \frac{1}{2} \omega_\varepsilon(s,u)$ , $\omega_\varepsilon(t,u) \le \frac{1}{2} \omega_\varepsilon(s,u)$ . Indeed, the continuity of $\omega_\varepsilon$ forces the existence of a $t$ such that $\omega_\varepsilon(s,t)=\omega_\varepsilon(t,u)$ . We obtain therefore
$\| \Gamma_{s,u}\|\le 2 \Psi(r/2) + r^\theta,$
which implies by maximization,
$\Psi(r)\le 2 \Psi(r/2) + r^\theta.$
By iterating $n$ times this inequality, we obtain
$\Psi(r)$
$\le 2^n \Psi\left(\frac{r}{2^n} \right) +\sum_{k=0}^{n-1} 2^{k(1-\theta)} r^\theta$
$\le 2^n \Psi\left(\frac{r}{2^n} \right) + \frac{1}{1-2^{1-\theta}} r^\theta.$
It is now clear that:
$\| \Gamma_{s,t} \|$
$\le \left\|\int_s^t (y(u)-y(s))dx(u) \right\|$
$\le \| x \|_{1-var; [s,t]} \| y-y(s) \|_{\infty; [s,t]}$
$\le ( \| x \|_{1-var; [s,t]} + \| y \|_{1-var; [s,t]})^2$
$\le \frac{1}{\varepsilon^2} \omega_\varepsilon (s,t)^2,$

Since
$\lim_{n \to \infty} 2^n \Psi\left(\frac{r}{2^n} \right) =0.$
We conclude
$\Psi(r) \le \frac{1}{1-2^{1-\theta}} r^\theta$
and thus
$\| \Gamma_{s,u}\| \le \frac{1}{1-2^{1-\theta}} \omega_\varepsilon(s,u) ^\theta$
Sending $\varepsilon \to 0$ , finishes the proof $\square$

It is remarkable that the Young-Loeve estimate only involves $\| x \|_{p-var; [s,t]}$ and $\| y \|_{q-var; [s,t]}$ . As a consequence, we obtain the following result whose proof is let to the reader:

Proposition: Let $x \in C^{p-var} ([0,T], \mathbb{R}^d)$ and $y \in C^{q-var} ([0,T], \mathbb{R}^{e \times d})$ with $\theta=\frac{1}{p}+\frac{1}{q} >1$ . Let us assume that there exists a sequence $x^n \in C^{1-var} ([0,T], \mathbb{R}^d)$ such that $x^n \to x$ in $C^{p-var} ([0,T], \mathbb{R}^d)$ and a sequence $y^n \in C^{1-var} ([0,T], \mathbb{R}^{e \times d})$ such that $y^n \to x$ in $C^{q-var} ([0,T], \mathbb{R}^d)$ , then for every $s < t$ , $\int_s^t y^n(u)dx^n(u)$ converges to a limit that we call the Young’s integral of $y$ against $x$ on the interval $[s,t]$ and denote $\int_s^t y(u)dx(u)$ .
The integral $\int_s^t y(u)dx(u)$ does not depend of the sequences $x^n$ and $y^n$ and the following estimate holds: for $0 \le s \le t \le T$ ,
$\left\| \int_s^t y(u)dx(u)-y(s)(x(t)-x(s)) \right\| \le \frac{1}{1-2^{1-\theta} }\| x \|_{p-var; [s,t]} \| y \|_{q-var; [s,t]}.$

The closure of $C^{1-var} ([0,T], \mathbb{R}^d)$ in $C^{p-var} ([0,T], \mathbb{R}^d)$ is $C^{0, p-var} ([0,T], \mathbb{R}^d)$ and we know that $C^{p+\varepsilon-var} ([0,T], \mathbb{R}^d) \subset C^{0, p-var} ([0,T], \mathbb{R}^d)$ . It is therefore obvious to extend the Young’s integral for every $x \in C^{p-var} ([0,T], \mathbb{R}^d)$ and $y \in C^{q-var} ([0,T], \mathbb{R}^{e \times d})$ with $\theta=\frac{1}{p}+\frac{1}{q} >1$ and the Young-Loeve estimate still holds
$\left\| \int_s^t y(u)dx(u)-y(s)(x(t)-x(s)) \right\| \le \frac{1}{1-2^{1-\theta} }\| x \|_{p-var; [s,t]} \| y \|_{q-var; [s,t]}.$
From this estimate, we easily see that for $x \in C^{p-var} ([0,T], \mathbb{R}^d)$ and $y \in C^{p-var} ([0,T], \mathbb{R}^{e \times d})$ with $\frac{1}{p}+\frac{1}{q} > 1$ the sequence of Riemann sums
$\sum_{k=0}^{n-1} y(t_i)( x_{t_{i+1}}-x_{t_i})$
will converge to $\int_s^t y(u)dx(u)$ when the mesh of the subdivision goes to 0. We record for later use the following estimate on the Young’s integral, which is also an easy consequence of the Young-Loeve estimate (see Theorem 6.8 in the book for further details).

Proposition: Let $x \in C^{p-var} ([0,T], \mathbb{R}^d)$ and $y \in C^{q-var} ([0,T], \mathbb{R}^{e \times d})$ with $\frac{1}{p}+\frac{1}{q} > 1$ . The integral path $t \to \int_0^t y(u)dx(u)$ is continuous with a finite $p$ -variation and we have
$\left\|\int_0^\cdot y(u) dx(u) \right\|_{p-var, [s,t] }$
$\le C \| x \|_{p-var; [s,t]} \left( \| y \|_{q-var; [s,t]} + \| y \|_{\infty; [s,t]} \right)$
$\le 2C \| x \|_{p-var; [s,t]} \left( \| y \|_{q-var; [s,t]} + \| y(0)\| \right)$