Lecture 7. Young’s integral

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,
=\| 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
\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,
\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)

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