Carnot–Carathéodory norm and the inner product norm

Question

It is well-known that given the extended tensor algebra $T((\mathbb{R}^d))$ one may extract a separable Hilbert space by considering the subset $$T^1((\mathbb{R}^d)) := \left\{h \in T((\mathbb{R}^d)) : \sum_{n=0}^\infty \langle h_n, h_n \rangle_{(\mathbb{R}^d)^{\otimes n}}<\infty \right\},$$ and defining an inner product as $\langle h, h'\rangle_{T^1}:= \sum_{n=0}^\infty \langle h_n ,h_n' \rangle_{(\mathbb{R}^d)^{\otimes n}}$.

On another front, in the context of rough paths, one usually works with the Carnot–Carathéodory norm and its induced distance (refer to Theorem 7.32 in "Multidimensional stochastic processes as rough paths: theory and applications." by Friz, Peter K., and Nicolas B. Victoir).

Now, I am interested in how the Carnot–Carathéodory norm compares to the norm induced by the inner product above. Specifically, I'd like to see the norm induced by the inner product bounded from above by the CC norm. Do we have any results on this?

In the aforementioned book, the authors prove that all homogenous norms (in the sense of Definition 7.34) are equivalent. That said, the norm induced by the inner product, i.e. $\lVert\cdot\rVert_{T^1}$, is clearly not homogenous. They further compare homogenous norms with the maximum norm induced by the Euclidean norm (see Proposition 7.45, for instance). But I'm not seeing a way to leverage from this.

I am fairly new to rough paths, so perhaps there's already something on this that I haven't seen yet. Or maybe, there's a straightforward argument that uses equivalence of homogeneous norms and Proposition 7.45. Could you help me out on this? Any insights are appreciated.

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Just to provide more context, given a weakly geometric rough path $\mathbf{x} \in C^{\alpha\text{-Höl}}_o([0,T],G^{\lfloor 1/\alpha \rfloor}(\mathbb{R}^d))$, let $S_\infty(\mathbf{x})$ denote the unique Lyons' extension to the full signature. Using the notation introduced above, I'm looking for a bound of the form $$\lVert S_\infty(\mathbf{x})_T\rVert_{T^1} \leq C \lVert\mathbf{x}\rVert_{\alpha\text{-Höl};[0,T]},$$ where $S_\infty(\mathbf{x})_T$ denotes the path signature evaluated at $T$; $\lVert\cdot\rVert_{\alpha\text{-Höl};[0,T]}$ denotes the homogenous $\alpha$-Hölder norm (using the notation of the book above); and $C$ is a constant that possibly depends on $\alpha$.

Answer 1

I think the bound you are looking for cannot hold because of scaling. Here's an attempt at an argument: fix a path $\textbf{x}\in C^\alpha$ (for simplicity I assume $\alpha\in(1/2,1)$), and take $\lambda>0$. It is easy to see that at level $n$, $$S_n(\lambda\textbf{x})_T=\lambda^n S_n(\textbf{x})_T\in(\mathbb{R}^d)^{\otimes n}.$$ In particular, for $n=2$ we get $$S_2(\lambda\textbf{x})_T=\lambda^2 S_2(\mathbf{x})_T\in(\mathbb{R}^d)^{\otimes n}$$ so that

$$\Vert S_2(\lambda\textbf{x})\Vert_{(\mathbb{R}^d)^{\otimes 2}}=\lambda^2\Vert S_2(\textbf{x})\Vert_{(\mathbb{R}^d)^{\otimes 2}},$$

and if the inequality holds, then

$$\lambda^2\Vert S_2(\textbf{x})\Vert_{(\mathbb{R}^d)^{\otimes 2}}=\Vert S_2(\lambda\textbf{x})\Vert_{(\mathbb{R}^d)^{\otimes 2}}\le\Vert S_\infty(\lambda\textbf{x})\Vert_{T^1}\le C\lambda\Vert\textbf{x}\Vert_{\alpha;[0,T]}$$

and we get a contradiction letting $\lambda\to\infty$.

This is one of the reasons for considering homogeneous norms, in that for such a norm $|||S_\infty(\lambda\mathbf{x})||| = \lambda|||S_\infty(\mathbf{x})|||$.