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Aug 12, 2016 at 9:07 comment added Nate Eldredge I'm skeptical. Morally speaking, if it's continuous then you ought to be able to define it explicitly, pathwise, on $W'$. I don't see how to do that. Remember that your net $S^\tau_\omega$ converges in probability but typically diverges almost surely.
Aug 12, 2016 at 8:54 comment added Matthias Ludewig Ok, but is there a subspace of continuous paths which is still large enough to support a Wiener measure, but small enough so that the enhancement map is everywhere defined and continuous? Would probably this intersection space that I wrote to the job?
Aug 12, 2016 at 8:46 comment added Nate Eldredge The map from the space of enhanced rough paths to the integral is continuous. The map from the space of actual paths to enhanced rough paths is only a.e. defined. What you want is their composition.
Aug 12, 2016 at 8:42 comment added Matthias Ludewig I meant the intersection instead of the union, sorry. However, you where saying that this solution map is continuous, and then again, that it is only defined almost everywhere?
Aug 12, 2016 at 8:40 history edited Matthias Ludewig CC BY-SA 3.0
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Aug 12, 2016 at 8:07 comment added Nate Eldredge The result I'm thinking of is that the rough path integral is supposed to be a continuous map on the space of enhanced rough paths (paths in the truncated tensor algebra) when it's equipped with the appropriate $p$-variation topology. But that's not what's going on here. Anyway, the enhancement that leads to the Stratonovich integral is something like Stratonovich stochastic area, and it is only defined almost everywhere.
Aug 12, 2016 at 7:06 comment added Matthias Ludewig Is this so? I am not an expert on the theory of rough paths and tried to find such a result in the literature, but I could not find the statement. Maybe this is just a language barrier, or maybe this way of thinking about matters is uncommon for probabilists.
Aug 11, 2016 at 21:54 comment added Nate Eldredge I'm fuzzy on the details, but since you mention rough paths, isn't this precisely what the $p$-variation topology accomplishes (with $p > 2$)?
Aug 11, 2016 at 14:45 history asked Matthias Ludewig CC BY-SA 3.0