# An integral by rough path.

If $(b, \mathbb{b})\in \mathcal{D}^{\alpha}[0,T],\ \alpha\in (\frac{1}{3}, \frac{1}{2})$. $\mathcal{D}^{\alpha}[0,T]$ is the space of those rough paths $(b,\mathbb{b})$ ￼￼such that $$\|b\|_\alpha=\sup_{s\neq t}\frac{|b_s-b_t|}{|s-t|^\alpha}<\infty; \quad \|\mathbb{b} \|_{2\alpha}= \sup_{s\neq t} \frac{|\mathbb{b}_{st}|}{|s-t|^{2\alpha}}<\infty.$$ (see Friz and Hairer's notes.)

Suppose $$f\in C^{1/2-}, B_t=\int_0^t b_s ds$$ Can we define the integral $$\int_0^tf(B_s)d b_s ?$$

Comments: if $b$ is an Brownian motion, then the above integral can by define in Ito sense or Stratonovich sense. But is it possible to define that in the sense of rough path?

• Is $(b,\mathbb{b})$ a controlled rough path? Are you allowing $\mathcal{D}^{\alpha}[0,T]$ to contain discontinuous paths too (discontinuity in $b$ if $(b,\mathbb{b})\in\mathcal{D}^{\alpha}$) ? Also, what does $\mathcal{C}^{1/2-}$ mean? Commented Nov 5, 2017 at 8:37
• @Sayantan $(b,\mathbb{b}) \in \mathcal{D}^\alpha[0,T]$ is continuous(see "A Short Course on Rough Paths"). And $f\in C^{1/2-}$ means $f\in C^\beta$ for any $\beta\in (0,1/2)$. Commented Nov 6, 2017 at 15:00

This is an old question but it's quite interesting. Here is an attempt with some extra assumptions because $$f$$ and $$b$$ are not concrete.

The reasonable approach is to try to find a Gubinelli derivative for $$Y_{t}:=f(B_{t})$$ (see Theorem 4.10 (Gubinelli) in "A course in rough paths"-CRP for short). To make some progress, lets assume that $$b$$ is truly rough (Definition 6.3 in CRP) which is also satisfied for Brownian motion (thm 6.6 CRP).

Then from Proposition 6.4 in CRP, we have uniqueness of the Gubinelli-derivative $$Y'$$. They have a formula for it:

$$Y'_{s}=\lim_{t_{n}\to s}\frac{Y_{t_{n}}-Y_{s}}{b_{t_{n}}-b_{s}}=\lim_{t_{n}\to s}\frac{f(B_{t_n})-f(B_{s})}{b_{t_{n}}-b_{s}}\label{1}\tag{1}$$

for a particular sequence $$t_{n}\to s$$.

So working in reverse, if you can show that the above expression $$Y'_{s}\in C^{\alpha}$$, then you got your Gubinelli-derivative and thus you got a rough-integral formulation (see Theorem 4.10 in CRP).

Heuristic: In \eqref{1}, let's say that the numerator term behaves like $$(|b_{s}| |t_{n}-s|)^{\ell}$$ for some $$\ell<\frac{1}{2}$$. The denominator term is unclear. In the case of Brownian motion the ratio diverges:

$$\frac{|t_{n}-s|^{\ell}}{|b_{t_{n}}-b_{s}|}\geq \frac{c}{|t_{n}-s|^{\frac{1}{2}-\ell}(\ln(1/|t_{n}-s|))^{1/2}}\to \infty.$$

If you want me to add more details, please let me know.

• But of course, unless $f$ relates to $b$ in a very very specific way, there is no reason whatsoever for (1) to converge. Under the assumptions given by the OP, $\int f(B)\,db$ typically just won't have any reasonable interpretation as a rough integral. Commented Nov 30, 2022 at 14:59