I got the following the following idea in one of the articles that I'm reading. It goes this way. Let $X$ be a Hausdorff topological vector space and let $\mathcal{D}$ be the family of all divisions of the compact interval $[a,b]$. For function $g:[a,b]\to X$ and each $D\in \mathcal{D}$, we write $$V(g,D)=(D)\sum [g(v)-g(u)]$$ where $$D=\{[u,v]\}$$ and that $u$ and $v$ are the typical division points of $[a,b]$. We say that $g$ is of bounded variation if the set $$V_g=\{V(g,D):D\in\mathcal{D}\}$$ is a bounded subset of $X$. That is how the article introduced the concept of bounded variation in the setting of topologiacal vector spaces.
Question: Does the concept above coincides with the standard definition of bounded variation if the space $X$ is Banach?
First, let's state the definition given in the article by Weston J.D., Functions of Bounded Variation in Topological Vector Spaces, The Quarterly Journal of Mathematics, Vol 8, Issue 1, 1957, pp. 108-111:
The key point here is that $[a_i,b_i]$ do not have to be a partition of $[a,b]$: we can leave gaps between them. For example, if $g$ is real-valued, we would take only the intervals on which $g$ increases, or only those on which it decreases. It would not make sense to include intervals of both kinds, since it would decrease $V_S(g)$, creating cancellation in (1).
Now, you ask about
Well, is there the standard definition when $X$ is a Banach space? In Functional Analysis and Semi-Groups by E. Hille we find 3 definitions (see 3.2.4 on page 59):
According to Hille, the following results are due to Dunford and Gelfand:
The latter is shown by an example. Take $X=L^\infty[0,1]$ and $g(t)=\chi_{[0,t]}$ for $t\in [0,1]$. Since $\|g(t)-g(s)\|_{L^\infty}=1$ whenever $t\ne s$, it's clear that $g$ is not of strong bounded variation. On the other hand, for any $0\le t<s\le 1$ the difference $g_s-g_t$ is $\chi_{[s,t]}$; summing such differences over any collection of nonoverlapping intervals we get a function of norm $1$. Thus, $g$ is of bounded variation.