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Let $p\ge1$ and $T>0$. Define $\mathscr D([0,T])$to be the space of partitions of $[0,T]$, where each partition is a finite collection of distinct points of $[0,T]$. Consider a continuous path $X:[0,T]\to V$, where $(V, |\cdot|)$ is a Banach space. The $p$-variation of $X$ is defined by:

$$\|X\|_{p,[0,T]}:= \left( \sup_{D\in \mathscr D([0,T])}\sum_{i=1}^{\text{Card}(D)-1}|X_{t_{i+1}}-X_{t_i}|^p\right)^{\frac{1}{p}}.$$

Consider the set of continuous paths of finite $p$-variation into $V$, $\mathcal V^p([0,T],V)$. It is easy to see that this is a vector space and that $\|\cdot\|_{\mathcal V^p([0,T],V)}:= \|\cdot\|_{\infty} + \|\cdot\|_{p,[0,T]} $ is a norm.

One can easily see that a path having finite $1$-variation is the same as a path having bounded variation in the more widespread sense. We will call continuous paths of bounded variation paths “BV”. One can then consider the closure of BV paths in the $\|\cdot\|_{\mathcal V^p([0,T],V)}$ norm. Analogously to the strict inclusion of so-called little Hölder spaces in Hölder spaces (see this other MO question), I am quite convinced one should find examples of paths in $\mathcal V^p([0,T],V)$ which are not inside the closure of BV paths in the $\|\cdot\|_{\mathcal V^p([0,T],V)}$ norm.

Could you please either tell me an example or point me towards the right direction?

UPDATE: this related post gets close to my question since it considers the closure of smooth functions in the $C^\infty$ sense, but I am interested in the closure of the paths which are continuous and of bounded variation.

Why I believe this is an interesting question: In the theory of rough paths (Lyons 1998; Friz & Hairer 2010), “geometric” rough paths are constructed as limits of (lifted) bounded variation paths, with the limit being taken in the (lifted) $p$-variation metric. “Weakly” geometric paths are those which are (lifted) continuous paths of finite $p$-variation. It turns out the space of weakly geometric rough paths is just “a little bit” larger than the soace of geometric rough paths. Maybe this behaviour can be observed at the basic level, without lifts.

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  • $\begingroup$ In the case where p = 1 and V is the real numbers, it seems to me that a sequence of BV functions which converge but are not uniformly bounded could provide you with a counter-example, i.e., do you need a condition corresponding to uniform boundedness in the more general case as part of your hypothesis? Or am I missing something? $\endgroup$
    – Derek
    Commented Mar 25, 2023 at 19:19
  • $\begingroup$ I am confused, as soon as your sequence converges in the $1$-variation norm, the limit will be in the space of BV functions so you will not have found a BV function that is not approximable by a sequence of BV functions (this is obviously not possible from the get-go) $\endgroup$
    – Martin Geller
    Commented Mar 27, 2023 at 15:05
  • $\begingroup$ I’m not too sure why my question got downvoted, it would be helpful if people posted their concerns $\endgroup$
    – Martin Geller
    Commented Mar 27, 2023 at 15:07

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