## Lecture 11. Estimating iterated integrals (Part 2)

Let $x \in C^{1-var}([0,T],\mathbb{R}^d)$. Since
$\omega(s,t)=\left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p$
is a control, the estimate
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k} \right\| \le \frac{C^k}{\left( \frac{k}{p}\right)!} \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^k, \quad 0 \le s \le t \le T.$
easily implies that for $k > p$,
$\left\| \int dx^{\otimes k} \right\|_{1-var, [s,t]} \le \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{k/p}.$
We stress that it does not imply a bound on the 1-variation of the path $t \to \int_{\Delta^k [0,t]} dx^{\otimes k}$. What we can get for this path, are bounds in $p$-variation:

Proposition: Let $p \ge 1$. There exists a constant $C \ge 0$, depending only on $p$, such that for every $x \in C^{1-var}([0,T],\mathbb{R}^d)$ and $k \ge 0$,
$\left\| \int_{\Delta^k [0,\cdot]} dx^{\otimes k} \right\|_{p-var, [s,t]} \le \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{1/p} \omega(0,T)^{\frac{k-1}{p}}$
where
$\omega(s,t)= \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p, \quad 0 \le s \le t \le T.$

Proof: This is an easy consequence of the Chen’s relations. Indeed,

$\left\| \int_{\Delta^k [0,t]} dx^{\otimes k} - \int_{\Delta^k [0,s]} dx^{\otimes k} \right\|$
$=\left\| \sum_{j=1}^k \int_{\Delta^j [s,t]} dx^{\otimes j} \int_{\Delta^{j-k} [0,s]} dx^{\otimes (k-j)} \right\|$
$\le \sum_{j=1}^k \left\| \int_{\Delta^j [s,t]} dx^{\otimes j} \right\| \left\| \int_{\Delta^{j-k} [0,s]} dx^{\otimes (k-j)} \right\|$
$\le C^k \sum_{j=1}^k \frac{1}{\left( \frac{j}{p}\right)!} \omega(s,t)^{j/p} \frac{1}{\left( \frac{k-j}{p}\right)!} \omega(s,t)^{(k-j)/p}$
$\le C^k \omega(s,t)^{1/p} \sum_{j=1}^k \frac{1}{\left( \frac{j}{p}\right)!} \omega(0,T)^{(j-1)/p} \frac{1}{\left( \frac{k-j}{p}\right)!} \omega(0,T)^{(k-j)/p}$
$\le C^k \omega(s,t)^{1/p} \omega(0,T)^{(k-1)/p}\sum_{j=1}^k \frac{1}{\left( \frac{j}{p}\right)!} \frac{1}{\left( \frac{k-j}{p}\right)!}.$
and we conclude with the binomial inequality $\square$

We are now ready for a second major estimate which is the key to define iterated integrals of a path with $p$-bounded variation when $p \ge 2$.

Theorem: Let $p \ge 1$, $K > 0$ and $x,y \in C^{1-var}([0,T],\mathbb{R}^d)$ such that
$\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \le 1,$
and
$\left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^p+ \left( \sum_{j=1}^{[p]} \left\| \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^p \le K.$
Then there exists a constant $C \ge 0$ depending only on $p$ and $K$ such that for $0\le s \le t \le T$ and $k \ge 1$
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}- \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{k/p} ,$
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}\right\| +\left\| \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le \frac{C^k}{\left( \frac{k}{p}\right)!} \omega(s,t)^{k/p}$
where $\omega$ is the control
$\omega(s,t)= \frac{ \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p+ \left( \sum_{j=1}^{[p]} \left\| \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} \right)^p } { \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^p+ \left( \sum_{j=1}^{[p]} \left\| \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^p }$
$+\left( \frac{\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j} - \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [s,t]} }{\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j} - \int dy^{\otimes j}\right\|^{1/j}_{\frac{p}{j}-var, [0,T]} } \right)^p$

Proof: We prove by induction on $k$ that for some constants $C,\beta$,
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}- \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \frac{C^k}{\beta \left( \frac{k}{p}\right)!} \omega(s,t)^{k/p},$
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}\right\| +\left\| \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le \frac{C^k}{\beta \left( \frac{k}{p}\right)!} \omega(s,t)^{k/p}$

For $k \le p$, we trivially have
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}- \int_{\Delta^k [s,t]} dy^{\otimes k} \right\|$ $\le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right)^k \omega(s,t)^{k/p}$
$\le \left( \sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} \right) \omega(s,t)^{k/p}.$
and
$\left\| \int_{\Delta^k [s,t]} dx^{\otimes k}\right\| +\left\| \int_{\Delta^k [s,t]} dy^{\otimes k} \right\| \le K^{k/p} \omega(s,t)^{k/p}$.
Not let us assume that the result is true for $0 \le j \le k$ with $k > p$. Let
$\Gamma_{s,t}=\int_{\Delta^k [s,t]} dx^{\otimes (k+1)}- \int_{\Delta^k [s,t]} dy^{\otimes (k+1)}$
From the Chen’s relations, for $0 \le s \le t \le u \le T$,
$\Gamma_{s,u}= \Gamma_{s,t}+ \Gamma_{t,u}$
$+\sum_{j=1}^{k} \int_{\Delta^j [s,t]} dx^{\otimes j }\int_{\Delta^{k+1-j} [t,u]} dx^{\otimes (k+1-j) }-\sum_{j=1}^{k} \int_{\Delta^j [s,t]} dy^{\otimes j }\int_{\Delta^{k+1-j} [t,u]} dy^{\otimes (k+1-j) }.$
Therefore, from the binomial inequality
$\| \Gamma_{s,u}\|$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\sum_{j=1}^{k} \left\| \int_{\Delta^j [s,t]} dx^{\otimes j }- \int_{\Delta^j [s,t]} dy^{\otimes j } \right\| \left\| \int_{\Delta^{k+1-j} [t,u]} dx^{\otimes (k+1-j) }\right\|$
$+\sum_{j=1}^{k} \left\| \int_{\Delta^{j} [s,t]} dy^{\otimes j }\right\| \left\| \int_{\Delta^{k+1-j} [t,u]} dx^{\otimes (k+1-j) }- \int_{\Delta^{k+1-j} [t,u]} dy^{\otimes (k+1-j) } \right\|$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\frac{1}{\beta^2}\tilde{\omega}(0,T) \sum_{j=1}^{k} \frac{C^j}{\left( \frac{j}{p}\right)!} \omega(s,t)^{j/p} \frac{C^{k+1-j}}{\left( \frac{k+1-j}{p}\right)!} \omega(t,u)^{(k+1-j)/p}$
$+\frac{1}{\beta^2}\tilde{\omega}(0,T) \sum_{j=1}^{k} \frac{C^j}{\left( \frac{j}{p}\right)!} \omega(s,t)^{j/p} \frac{C^{k+1-j}}{\left( \frac{k+1-j}{p}\right)!} \omega(t,u)^{(k+1-j)/p}$
$\le \| \Gamma_{s,t} \| + \| \Gamma_{t,u} \| +\frac{2p}{\beta^2} \tilde{\omega}(0,T) C^{k+1} \frac{ \omega(s,u)^{(k+1)/p}}{\left( \frac{k+1}{p}\right)! }$
where
$\tilde{\omega}(0,T)=\sum_{j=1}^{[p]} \left\| \int dx^{\otimes j}- \int dy^{\otimes j} \right\|^{1/j}_{\frac{p}{j}-var, [0,T]} .$
We deduce
$\| \Gamma_{s,t} \| \le \frac{2p}{\beta^2(1-2^{1-\theta})} \tilde{\omega}(0,T) C^{k+1} \frac{ \omega(s,t)^{(k+1)/p}}{\left( \frac{k+1}{p}\right)! }$
with $\theta= \frac{k+1}{p}$. A correct choice of $\beta$ finishes the induction argument $\square$

This entry was posted in Rough paths theory. Bookmark the permalink.