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Let $(C, \|\cdot\|)$ be the Banach space of continuous paths $x: [0,1]\rightarrow\mathbb{R}^d$ starting at zero with sup-norm $\|\cdot\|$.

Let further $B\subset C$ be the subspace of $0$-started continuous paths of bounded variation, i.e.

$$B = \big\{x\in C \ \big| \ \|x\|_1:=\sup\nolimits_{(t_\nu)}{\textstyle\sum_\nu}|x_{t_\nu} - x_{t_{\nu-1}}|<\infty \ \text{ and } \ x(0)=0\big\}$$

where the above sup runs over all partitions $(t_\nu)$ of $[0,1]$.

The set $B$ can be endowed with both the (uniform) norm $\|\cdot\|$ and the (1-variation) norm $\|\cdot\|_1$.

Do you know if the Borel $\sigma$-algebras on $(B, \|\cdot\|)$ and $(B, \|\cdot\|_1)$ coincide?

Any references or hints are welcome.

Edit: As pointed out by Gerald Edgar, we narrowed $B$ down to paths started at zero to make the question more meaningful.

Edit 2: As Gerald Edgar pointed out in the comments to his answer below, the answer to this question is no, the Borel $\sigma$-algebras cannot coincide because there are more $\|\cdot\|_1$-open sets in $B$ than there are $\|\cdot\|$-Borel sets in $C$.

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I think there is a problem with a mere semi-norm. The constant functions have variation distance $0$ from each other. Any variation-open set contains either all the constants, or none of them. Therefore, any variation-Borel set contains either all the constants, or none of them.

Choose $\mathbf u \in \mathbb R^n \setminus \{0\}$. Take the subset $\{0,\mathbf{u}\} \subseteq B$ of two constant functions. It is sup-norm closed, so it is a uniform-Borel set. But it is not a variation-Borel set.

Maybe a better question would restrict to the subset of $B$ with $x(0) = 0$. Then at least we have a norm.

rmcerafl pointed out a problem with the following. For now, I will leave it here for reference.

Let $C_0 = \{x \in C : x(0)=0\}$ with the uniform norm. Let $B_0 = \{x \in B : x(0)=0\}$ with the variation norm. Both of these are complete separable metric spaces. The inclusion map $i : B_0 \to C_0$ is continuous and injective. Then citing a result or two from descriptive set theory*, we get: $i(B_0)$ is a Borel subset of $C_0$ and $i : B_0 \to i(B_0)$ is a Borel-isomorphic map. This shows, in particular, the variation-Borel subsets of $B_0$ coincide with the uniform-Borel subsets of $i(B_0)$.

${}^*$ Proposition 8.3.5 and Theorem 8.3.7 in:

Cohn, Donald L., Measure theory, Boston, Basel, Stuttgart: Birkhäuser. IX, 373 p. DM 42.00 (1980). ZBL0436.28001.

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  • $\begingroup$ Thank you, I edited the question accordingly. $\endgroup$
    – fsp-b
    Commented Jun 14, 2022 at 14:36
  • $\begingroup$ I don't think $(B_0, \|\cdot\|_1)$ is separable, see for instance Theorem 1.25 in: Friz and Victoir, Multidimensional Stochastic Processes as Rough Paths. $\endgroup$
    – fsp-b
    Commented Jun 14, 2022 at 15:19
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    $\begingroup$ I think you are right. References I consulted only discuss non-separability of BV and of NBV (=right-continuous BV), but not of continuous BV. If it is non-separable, I am inclinded to think that there are far too many Borel sets, then. $\endgroup$
    – Gerald Edgar
    Commented Jun 14, 2022 at 16:12
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    $\begingroup$ There are $2^{\mathfrak c}$ variation-open sets in $B_0$, but only $\mathfrak c$ uniform-Borel sets in $C_0$. $\endgroup$
    – Gerald Edgar
    Commented Jun 14, 2022 at 18:34
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    $\begingroup$ @GeraldEdgar. Nice! Except that your answer is in the comments while you've got a bunch of comments in your answer! $\endgroup$
    – Ruy
    Commented Jun 15, 2022 at 20:08

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