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 $C$.



Borel $\sigma$-algebras on paths of bounded variation
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](https://zbmath.org/?q=an:0436.28001).


