Was there any relevant work to determine the dual (or more likely the predual) of the space of bounded variation functions $BV(\mathbb{R}^n)$ (I recall the definition : a function in $L^1(\mathbb{R}^n)$ such that the distributional gradient if a vector valued finite measure)?

2$\begingroup$ See Pełczyński, Aleksander; Wojciechowski, Michał Spaces of functions with bounded variation and Sobolev spaces without local unconditional structure. J. Reine Angew. Math. 558 (2003), 109–157. I cannot access the paper right now, but I think it contains the information you seek. $\endgroup$ – Bill Johnson Dec 23 '14 at 17:35

$\begingroup$ Thank you M. Johnson but I can't access the paper neither. $\endgroup$ – PaulBenjamin Dec 30 '14 at 9:55
Depending on the locally convex topology you take on $BV_c(\mathbb{R}^n)$ (BV functions with essentially compact support in $\mathbb{R}^n$), its dual is given by so called charges (or strong charges) as defined in "De Pauw, Pfeffer  Distributions for which div v = F has a continuous solution  Comm. Pure Appl. Math., 61(2), 2008. 230260". An electronic version of this paper is here:
http://webusers.imjprg.fr/~thierry.depauw/preprints/div.v.equals.F.pdf
A charge is a particular kind of distribution and its collection is given the structure of a Fréchet space. Moreover, a distribution in $\mathcal{D}'(\mathbb{R}^n)$ is a charge if and only if it can be realized as the distributional divergence of a continuous vector field, i.e. an element of $C(\mathbb{R}^n,\mathbb{R}^n)$. This result is also stated in Theorem 11.3.8 of the book "The Divergence Theorem and Sets of Finite Perimeter" by Pfeffer (some attention is needed since what is called a charge in this book is called a strong charge in the paper mentioned above). I hope this goes in the direction you are looking for...

$\begingroup$ Thank you very much rozu, in fact, I had found this article later (after a personal communication of a student of Prof. De Pauw actually), and forgot to update this question. But I really apreciate your concern (it's the second time you answer one of my personal "longstanding" question, thank you for that!). $\endgroup$ – PaulBenjamin Jun 4 '15 at 20:14