Is $BV([0,1])$ a reflexive space?

Question

I was checking some properties of $BV([0,1])$ and I couldn't find this information anywhere. Below are some results that I know can help prove that this statement is true or false.

• If a vector space is reflexive, then its dual is also reflexive.
• Closed subspace of reflexive space is reflexive.
• A Banach space is reflexive if and only if its dual is reflexive. (I know that $BV([0,1])$ is Banach when equipped with $||f||=|f(0)|+Var(f)$.)
• A reflexive Banach space is separable if and only if its dual is separable. (I know that $BV([0,1])$ is not separable.)
• If a normed vector space, $X$, is reflexive and $\varphi\in X'$, then there is $x_0\in X$ such that $\varphi(x_0)=||\varphi||$.

Answer 1

The structure of the dual of $BV(\Omega)$ is still largely an open question. However, we do know that $BV(\Omega)$ is not reflexive. On the other hand, it is the dual of a separable space. A couple good resources for more about $BV$ are:

(1) Functions of Bounded Variation by Ambrosio, Fusco and Pallara (your question is addressed in Remark 3.12, but the authors don't prove it),

(2) Weakly Differentiable Functions by Ziemer (Theorem 5.12.4 may be of interest).

Though this is a bit beyond the scope of your question, an interesting article addressing the question of the dual of BV is "On SBV dual" by De Pauw, which considers the dual of $SBV(\mathbb{R}^d) \subset BV(\mathbb{R}^d)$. Another is "Integral Inequalities of Poincaré and Wirtinger Type for BV Functions" by Meyers and Ziemer. In this paper, the authors characterized the Radon measures in the dual of $BV(\mathbb{R}^d)$. I hope this helps.

Answer 2

$C[0,1]$ is separable. However, $C[0,1]^*$ consists of all finite Borel measures $\mu$ on $[0,1]$, which is definitely not separable because it contains the atomic probability measures $\mu_x$ for all $x\in [0,1]$, and $\|\mu_x-\mu_y\| \ne 0$ for all $x\ne y$. The new user NeoFanatic had this right except for a trivial normalization issue.