What is $X^*$, the topological dual of the space $X:=AC[0,1]$? Let $X = AC[0,1]$ be the space of all absolutely continuous functions from  $[0, 1]$ to $\mathbb{R}^n$, with $W^{1,1}$ norm that is 
$$   \|f\| = \int_{0}^{1} |f(t)| \; dt +   \int_{0}^{1} |f'(t) | \; dt        $$
My question is that: What is $X^*$, the dual of the space $AC[0,1]$?
I searched in the internet I only found the dual of $W^{p,1}$ with $p > 1$ !
 A: Let's first consider $n=1$.  Then $X = AC([0,1]; \mathbb{R}^1)$ is isomorphic as a normed space to $\mathbb{R} \oplus L^1([0,1])$; one possible isomorphism is $Tf = (f(0), f')$.  Therefore $X^*$ is isomorphic to $\mathbb{R} \oplus L^\infty([0,1])$ via the adjoint of this isomorphism.  In other words, every continuous linear functional $\ell$ on $X$ can be uniquely written in the form $\ell f = a f(0) + \int_0^1 h(t) f'(t)\,dt$ for some $a \in \mathbb{R}$, $h \in L^\infty([0,1])$.
(I mean "isomorphism" here in the sense of "linear homeomorphism".  My map $T$ is not an isometry.)
A few hints in verifying that $T$ is an isomorphism:


*

*$(T^{-1}(c,g))(x) = c + \int_0^x g(t)\,dt$

*To see that the map $X \ni f \mapsto f(0) \in \mathbb{R}$ is continuous, use the identity $f(0) = \int_0^1 f(t)\,dt - \int_0^1 \int_0^x f'(t)\,dt\,dx$.
For the general case, you can note that $AC([0,1]; \mathbb{R}^n)$ is isomorphic to the direct sum of $n$ copies of $AC([0,1]; \mathbb{R}^1)$ by taking components.  This leads to the conclusion that its dual is isomorphic to $\mathbb{R}^n \oplus L^\infty([0,1]; \mathbb{R}^n)$.
