In the following definitions, we assumed that $(X,\left\|\cdot\right\|)$ is a Banach space.
Definition 1. $f:[a,b]\to X$ is of bounded variation on $[a,b]$ if $$\operatorname{Var}(f;[a,b])=\operatorname{sup}(D)\sum \left\|f(v)-f(u)\right\|<+\infty$$ where the supremum is taken over all divisions $D=\{[u,v]\}$ of $[a,b]$.
Definition 2. $f:[a,b]\to X$ is regulated if it has one-sided limits at every point of $[a,b]$, i.e. for every $c\in [a,b)$ there is a value $f(c+)\in X$ such that $$\lim_{t\to c^+}\left\|f(t)-f(c+)\right\|=0$$ and if for every $c\in (a,b]$ there is a value $f(c-)\in X$ such that $$\lim_{t\to c^-}\left\|f(t)-f(c-)\right\|=0.$$
Question. Using the preceding definitions, how do we show that every function of bounded variation is regulated.
I need some help on this and many thanks in advance...
Proceed by contradiction.
Assume for example that $f(c^-)$ does not exist for some $c\in (a,b]$. Then $f$ does not satisfies Cauchy's criterion at $c^-$. So one can find some $\varepsilon >0$ and two sequences $(u_n)$, $(v_n)$ such that $u_n<v_n<u_{n+1}<v_{n+1}<c$ for all $n$ and $\lim_n{u_n}=c=\lim_n v_n$ yet $\Vert f(v_n)-v(u_n)\Vert\geq \varepsilon$. For any $N\in\mathbb N$ we then have $\sum_{n=1}^N \Vert f(v_n)-f(u_n)\Vert\geq N\varepsilon$, which can be made arbitrarily large. Since $\{[u_1,v_1], \dots ,[u_N,v_N]\}$ can be "extended" into a division of $[a,b]$ (the intervals are pairwise disjoint), this shows that $f$ does not have bounded variation.