Canonical lift of a smooth path as a rough path

Question

Let $X_t$ be say, differentiable, importantly it is guaranteed to be at least more than $\frac{1}{2}$ Holder continuous. Then I can define the iterated integrals in the Riemann-Stieljes/Young sense as

$$\mathbb{X}^{ij}_{0,t} = \int_0^tX_r dX_r = \int_0^tX_r \frac{d}{dr}(X_r)dr$$

According to rough paths theory, this defines a rough path. The Chen identity holds. But I don't necessarily know how to show, or even believe, the continuity requirement:

$$ \|\mathbb{X}\|_{2 \alpha} \stackrel{\text { def }}{=} \sup _{s \neq t \in[0, T]} \frac{\left|\mathbb{X}_{s, t}\right|}{|t-s|^{2 \alpha}}<\infty $$

From Friz Hairer definition 2.1. Surely if $X_t$ is nice and smooth, then $\alpha > \frac{1}{2}$, but then $2\alpha > 1$, and wouldn't this degenerate into saying that $\mathbb{X}$ is constant? From the known result on Holder continuous greater than 1 $\implies$ constant?

How do I show that a smooth path is a particular case of rough paths with the obvious choice of signature?

Answer 1

Surely if $X_t$ is nice and smooth, then $\alpha > \frac{1}{2}$, but then $2\alpha > 1$, and wouldn't this degenerate into saying that $\mathbb{X}$ is constant? From the known result on Holder continuous greater than 1 $\implies$ constant?

As mentioned in the comments, the rough path lift $\mathbb{X}_{s,t}$ is not an increment. For example, in the case of Brownian motion we can take

$$\mathbb{B}_{s,t}^{Itô}:=\int_{s}^{t}(B_{r}-B_{s})\otimes dB_{r}.$$

This definition originates from the "Taylor"-like-expansion argument by repeatedly using FTC

$$\int^t_sF(X_u)dX_u = F(X_s)\int^t_sdX_u + \int^t_s(F(X_u)-F(X_s))\otimes dX_u$$

$$\stackrel{FTC}{=} F(X_s)\int^t_sdX_u + \int^t_s\int_{s}^{u}DF(X_v)dX_{v}\otimes dX_u$$

$$= F(X_s)\int^t_sdX_u + DF(X_s)\int^t_s\int_{s}^{u}dX_{v}\otimes dX_u+\int^t_s\int_{s}^{u}(DF(X_v)-DF(X_s))dX_{v}\otimes dX_u.$$

How do I show that a smooth path is a particular case of rough paths with the obvious choice of signature?

Indeed as mentioned in the comments, when we have two functions $Y\in C^{\alpha},X\in C^{\beta}$ with $\alpha+\beta>1$, then we can define their Young integral

$$\int YdX\approx \sum Y_{t_{i}} (X_{t_{i+1}}-X_{t_{i}})$$

and use Young's inequality as mentioned in "a course in rough paths" eq. (4.3)

$$|\int_{s}^{t} Y_{r}dX_{r}-Y_{s}X_{s,t}|\leq c \|Y\|_{\beta} \|X\|_{\alpha}|t-s|^{\alpha+\beta}.$$

So in the case of differentiable $X$ i.e. $\alpha\geq 1$,we can just take a rough path lift $\mathbb{X}$ to be

$$\mathbb{X}_{s,t}:=\int_{s}^{t} (X_{r}-X_{s})dX_{r}$$

and from above this indeed satisfies $\frac{|\mathbb{X}_{s,t}|}{|t-s|^{2\alpha}}<c\|X\|_{\alpha}^{2}$.