Lecture 26. Lyons’ continuity theorem: Proof

We now turn to the proof of Lyons’ continuity theorem.

Theorem: Let $\gamma > p \ge 1$ . Assume that $V_1, \cdots, V_d$ are $\gamma$ -Lipschitz vector fields in $\mathbb{R}^n$ . Let $x_1,x_2 \in C^{1-var}([0,T], \mathbb{R}^d)$ such that
$\| S_{[p]}(x_1) \|^p_{p-var,[0,T]}+\| S_{[p]}(x_2) \|^p_{p-var,[0,T]} \le K$
with $K \ge 0$ .
Let $y_1,y_2$ be the solutions of the equations
$y_i(t)=y(0)+\sum_{j=1}^d \int_0^t V_j(y_i(s)) dx_i^j(s), \quad 0 \le t \le T, \quad i=1,2$
There exists a constant $C$ depending only on $p,\gamma$ and $K$ such that for $0 \le s \le t \le T$ ,
$\|( y_2(t) -y_2(s)) - (y_1(t)-y_1(s))\| \le C \| V \|_{\text{Lip}^{\gamma}} e^{ C \| V \|^p_{\text{Lip}^{\gamma}}} d_{p-var,[0,T]} ( S_{[p]}(x_1), S_{[p]}(x_2)) \omega(s,t)^{1/p} ,$
where $\omega$ is the control
$\omega(s,t)=\left( \frac{d_{p-var,[s,t]} ( S_{[p]}(x_1), S_{[p]}(x_2)) }{d_{p-var; [0,T]}(S_{[p]}(x_1), S_{[p]}(x_2)) } \right)^p+\left( \frac{ \| S_{[p]}(x_1) \|_{p-var,[s,t]}}{\| S_{[p]}(x_1) \|_{p-var,[0,T]} } \right)^p+\left( \frac{ \| S_{[p]}(x_2) \|_{p-var,[s,t]}}{\| S_{[p]}(x_2) \|_{p-var,[0,T]} } \right)^p.$

Proof: We may assume $p < \gamma < [p]+1$ , and for conciseness of notations, we set $\varepsilon = d_{p-var,[0,T]} ( S_{[p]}(x_1), S_{[p]}(x_2))$ . Let
$g_i=\Delta_{\frac{1}{\omega(s,t)^{1/p}}} ( S_{[p]}(x_i)(s)^{-1} S_{[p]}(x_i)(t)), \quad i=1,2.$

We have,
$d(g_1,g_2) =\frac{1}{\omega(s,t)^{1/p}} d(S_{[p]}(x_1)(s)^{-1} S_{[p]}(x_1)(t) , S_{[p]}(x_2)(s)^{-1} S_{[p]}(x_2)(t))$
$\le \frac{1}{\omega(s,t)^{1/p}}d_{p-var,[s,t]} ( S_{[p]}(x_1), S_{[p]}(x_2))$
$\le \varepsilon$
and, in the same way,
$d(0,g_i) =\frac{1}{\omega(s,t)^{1/p}} d ( S_{[p]}(x_i)(s), S_{[p]}(x_i)(t))$
$=\frac{1}{\omega(s,t)^{1/p}} \| S_{[p]} (x_i) \|_{p-var,[s,t]} \le K.$

Therefore, there exist $x^{s,t}_1,x^{s,t}_2 \in C^{1-var}([s,t], \mathbb{R}^d)$ and a constant $C_1=C_1([p],K)$ such that
$S_{[p]}(x^{s,t}_i)(s)^{-1} S_{[p]}(x^{s,t}_i)(t) =S_{[p]}(x_i)(s)^{-1} S_{[p]}(x_i)(t) , i=1,2$
and
$\| x^{s,t}_1\|_{1-var,[s,t]} +\| x^{s,t}_2\|_{1-var,[s,t]} \le C_1\omega(s,t)^{1/p}$
and
$\| x^{s,t}_1-x^{s,t}_2 \|_{1-var,[s,t]} \le \varepsilon C_1\omega(s,t)^{1/p}.$
We define then $x_i^{s,t,u}$ as the concatenation of $x_i^{s,t}$ and $x_i^{t,u}$ . As in the proof of Davie’s lemma, we denote by $y_i^{s,t}$ the solution of the equation
$y^{s,t}_i(r)=y_i(s)+\sum_{j=1}^d \int_s^r V_j(y^{s,t}_i(v)) dx_i^j(v), \quad s \le r \le t, \quad i=1,2$
and consider the functionals
$\Gamma^i_{s,t}=(y_i(t)-y_i(s))-(y^{s,t}_i(t) -y^{s,t}_i(s))= y_i(t)-y^{s,t}_i(t),$
and
$\bar{\Gamma}_{s,t} =\Gamma^1_{s,t}-\Gamma^2_{s,t}$
From the proof of Davie’s estimate, it is seen that
$\| \Gamma^i_{s,t} \| \le \frac{1}{2} C_2 \left( \| V \|_{\text{Lip}^{\gamma}} \omega(s,t)^{1/p} \right)^{[p]+1},$
and thus
$\| \bar{\Gamma}_{s,t} \| \le C_2 \left( \| V \|_{\text{Lip}^{\gamma}} \omega(s,t)^{1/p} \right)^{[p]+1}.$
On the other hand, by estimating
$\bar{\Gamma}_{s,u}- \bar{\Gamma}_{s,t}- \bar{\Gamma}_{t,u},$
as in the proof of Davie’s lemma, that is by inserting $y_i^{s,t,u}$ which is the solution of the equation driven by the concatenation of $x_i^{s,t}$ and $x_i^{t,u}$ , and then by using the two lemmas of the previous lecture, we obtain the estimate
$\| \bar{\Gamma}_{s,u} \|$
$\le \| \bar{\Gamma}_{s,t} \| e^{C_3 \| V \|_{\text{Lip}^{\gamma}} \omega(s,u)^{1/p}} + \| \bar{\Gamma}_{t,u} \|+C_3( \| y_1-y_2\|_{\infty, [s,t]} +\varepsilon)\left( \| V \|_{\text{Lip}^{\gamma}} \omega(s,u)^{1/p} \right)^\gamma e^{C_3 \| V \|_{\text{Lip}^{\gamma}} \omega(s,u)^{1/p}}$
$\le \left( \| \bar{\Gamma}_{s,t} \| + \| \bar{\Gamma}_{t,u} \|+C_3( \| y_1-y_2\|_{\infty, [s,t]} +\varepsilon) \left( \| V \|_{\text{Lip}^{\gamma}} \omega(s,u)^{1/p} \right)^{\gamma} \right) e^{C_3 \| V \|_{\text{Lip}^{\gamma}} \omega(s,u)^{1/p}}.$
It remains to bound $\| y_1-y_2\|_{\infty, [s,t]}$ . For this let us observe that
$\| (y_1(t)-y_2(t))-(y_1(s)-y_2(s))-\bar{\Gamma}_{s,t} \| =\|( y^{s,t}_1(t) -y^{s,t}_2(t) )-( y^{s,t}_1(s) -y^{s,t}_2(s) )\|.$
$\|( y^{s,t}_1(t) -y^{s,t}_2(t) )-( y^{s,t}_1(s) -y^{s,t}_2(s) )\|$ can then be estimated by using classical estimates on differential equations driven by bounded variation paths. This gives,
$\|( y^{s,t}_1(t) -y^{s,t}_2(t) )-( y^{s,t}_1(s) -y^{s,t}_2(s) )\| \le C_4 \left( \| y_1(s) -y_2(s) \| +\varepsilon \right) \| V \|_{\text{Lip}^{\gamma}} \omega(s,t)^{1/p} e^{ C_4\| V \|_{\text{Lip}^{\gamma}} \omega(s,t)^{1/p}}.$
By denoting $z=y_1-y_2$ , we can summarize the two above estimates as follows:
$\| \bar{\Gamma}_{s,u} \| \le \left( \| \bar{\Gamma}_{s,t} \| + \| \bar{\Gamma}_{t,u} \|+C_3( \| z\|_{\infty, [s,t]} +\varepsilon) \left( \| V \|_{\text{Lip}^{\gamma}} \omega(s,u)^{1/p} \right)^{\gamma} \right) e^{C_3 \| V \|_{\text{Lip}^{\gamma}} \omega(s,u)^{1/p}}$
and
$\| z(t)-z(s) -\bar{\Gamma}_{s,t} \| \le C_4 \left( \| z \|_{\infty,[0,s]} +\varepsilon \right) \| V \|_{\text{Lip}^{\gamma}} \omega(s,t)^{1/p} e^{ C_4\| V \|_{\text{Lip}^{\gamma}} \omega(s,t)^{1/p}}.$
From a lemma already used in the proof of Davie’s estimate, the first estimate implies
$\| \bar{\Gamma}_{s,t} \| \le C_5 \left( \varepsilon + \| z \|_{\infty,[0,t]} \right) \left( \| V \|_{\text{Lip}^{\gamma}} \omega(s,t)^{1/p} \right)^{\gamma} e^{ C_5 \| V \|_{\text{Lip}^{\gamma}} \omega(s,t)^{1/p}}.$
Using now the second estimate we obtain that for any interval $[a,b]$ included in $[0,T]$ ,
$\sup_{s,t \in [a,b]} \| z(t)-z(s) \| \le C_6 (\varepsilon + \| z \|_{\infty, [0,b]} ) \| V \|_{\text{Lip}^{\gamma}} \omega(a,b)^{1/p} e^{C_6 \| V \|_{\text{Lip}^{\gamma}} \omega(a,b)^{1/p}}.$
Using the fact that $z(0)=0$ and picking a subdivision $0 = \tau_0 \le \tau_1 \le \cdots \le \tau_N \le T$ such that
$C_6 \| V \|_{\text{Lip}^{\gamma}} e^{C_6 \| V \|_{\text{Lip}^{\gamma}} } \omega(\tau_i , \tau_{i+1} )^{1/p} \le 1/2$
we see that it implies
$\| z \|_{\infty,[0,T]} \le C_7 \varepsilon e^{C_7 \| V \|^p_{\text{Lip}^{\gamma}} }.$
Coming back to the estimate
$\sup_{s,t \in [a,b]} \| z(t)-z(s) \| \le C_6 (\varepsilon + \| z \|_{\infty, [0,b]} ) \| V \|_{\text{Lip}^{\gamma}} \omega(a,b)^{1/p} e^{C_6 \| V \|_{\text{Lip}^{\gamma}} \omega(a,b)^{1/p}}.$
concludes the proof $\square$