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In Theorem 3.1 of this paper, the following result is formulated:

Suppose $f:S_1 \to \mathbb{R}$ is a continuous function where $S_1$ is a compact subset of $S(\mathcal{V}^p(J,E))$. Then for any $\epsilon > 0$, there exists a linear functional $L \in T((E))^*$ such that for every $a \in S_1$, $$ |f(a)-L(a)| \leq \epsilon. $$

Here $\mathcal{V}^p(J,E)$ denotes the space of paths of finite $p$-variation from an interval $J$ to a Banach space $E$ (assume finite dimensional), and $S$ is the signature map defined as a collection of iterated Young integrals.

My question is, what is the topology on $S(\mathcal{V}^p(J,E))$? The proof uses the Stone-Weierstrass theorem, and I think that even requires $S_1$ to be a metric space, but since $S(\mathcal{V}^p(J,E))$ is a subset of the full tensor algebra $T((E))$ and is sort of infinite dimensional (even assuming $E$ is finite dimensional), I don't see what topology or metric it should have.

Or maybe this theorem should be interpreted as: whatever suitable topology we give this space, we can approximate continuous, real-valued functions by linear functionals?

zhy
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2 Answers2

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Let $E=\mathbb{R}^d$. Each $E^{\otimes n}= (\mathbb{R}^d)^{\otimes n}$ can be endowed with the usual distance, e.g. when $a^2,b^2 \in (\mathbb{R}^d)^{\otimes 2} = \mathbb{R}^{d\times d}$, we can define $$ d_2(a^2,b^2) = \max_{1\le i,j \le d} |a^2_{i,j}-b^2_{i,j}|. $$ Then for $a = (1,a^1,a^2,\ldots),b=(1,b^1,b^2,\ldots) \in T((E)) = \bigoplus_{n=1}^{\infty} E^{\otimes n}, $ $$ d(a,b) = \sum_{n=1}^{\infty}\frac{d_n(a^n,b^n)}{2^n}$$ defines a natural distance between $a$ and $b$.

I'm not sure though if this is the topology the authors of the paper had in mind.

Sayantan
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    That's a good one. One small detail is that $T((E))$ is actually the closure of $\bigoplus_{n=0}^\infty E^{\otimes n}$, so in general its elements can grow increasingly large (don't need to have finite support). But for elements $a =S(X) \in S(\mathcal{V}^p(J,E)) \subset T((E))$ we have the bound $|a_n| \leq |X|_{1-var}^n/n!$, so on this space your metric is well-defined. – zhy Jun 12 '19 at 17:36
  • I know this is not the right place to post this, and this will get deleted soon. However, I am currently working on rough path theory; since this is a niche, if anybody wants to message me I would be very grateful: martin.geller@maths.ox.ac.uk – Martin Geller Oct 06 '21 at 11:06
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Usually in rough path theory, the norm on the space of full signature can be any thing, as long as it is "compatible". Lyons defined a set of properties for the norm to be qualified as "compatible". More details can be found in the book St Flour Lecture Notes

One example, which is quite standard, is demonstrated in the Definition 2.10 in this paper

quallenjäger
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