Our goal in the next two lectures will be to prove that rough differential equations driven by a Brownian motion seen as a -rough path, are nothing else but stochastic differential equations understood in the Stratonovitch sense. The proof of this fact requires an explicit approximation of the Brownian rough path in the rough path topology which is interesting in itself.

Let be a -dimensional Brownian motion and let us denote by

its lift in the free Carnot group of step 2 over .

Let us work on a fixed interval and consider a sequence of subdivisions of such that and whose mesh goes to 0 when . An example is given by the sequence of dyadic subdivisions. The family is then a filtration, that is an increasing family of -fields. We denote by the piecewise linear process which is obtained from by interpolation along the subdivision , that is for ,

The corresponding lifted process is then

The main result of the lecture is the following:

**Theorem:*** Let . When , almost surely, .*

We split the proof in two lemmas.

**Lemma:*** Let . When , almost surely, .*

**Proof:** We first observe that, due to the Markov property of Brownian motion, we have for ,

It is then an easy exercise to check that

As a conclusion, we get

It immediately follows that when . In the same way, we have

Indeed, for and small enough, we have by independence of and ,

and we conclude using the fact that Ito’s integral is a limit in of Riemann sums. It follows that, almost surely,

and we conclude that almost surely,

The second lemma is a uniform Holder estimate for .

**Lemma:*** For every , there exists a finite random variable that belongs to for every and such that for every , and every ,
*

**Proof:** By using the theorem of equivalence of norms, we see that there is a constant such that

From the Garsia-Rodemich-Rumsey inequality, we know that there is a finite random variable ( that belongs to for every ), such that for every ,

Since

we deduce that

where is a finite random variable that belongs to for every . Similarly, of course, we have

and this completes the proof

We are now in position to finish the proof that, almost surely, if . Indeed, if is a subdivision of , we have for ,

By using the second lemma, it is seen that is bounded when and by combining the first two lemmas we easily see that .

is this obvious that $K_2$ , which I believe equals $E[K_1|\mathcal F_n]$, does not depend on $n$.? Thanks!

yes, obvious, since a.s. convergent.