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? 
 A: 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.
