Can we identify absolutely continuous functions on $(a,b)$ with values in $X$ with $W^{1,1}((a,b);X)$? Function $f:(a,b)\to \mathbb{R}$ is measurable and absolutely continuous if and only if there exists the weak derivative $\frac{df}{dx}\in L^1(a,b)$. The weak derivative coincides with the classical derivative almost everywhere.
Can we prove the same theorem for $f:(a,b)\to X$, where $X$ is a Banach space?
In order to have differentiabilty almost everywhere of absolutely continuous function $f$ with the values in an abstract Banach space, we have to assume that $X$ is reflexive. I tried to copy the proof from the very first case but I stop when it comes to integrating by parts. Does the classical integrating by parts has the same form for absolutely continuous functions with values in Banach spaces?
 A: Yes it is still true. To have integration by parts you have to show that the fundamental theorem of calculus continues to hold. You first prove that for $T\in X^{\prime}$ the function $w:=T\circ f$ belongs to $AC([a,b])$. In turn,
$w$ is differentiable for $\mathcal{L}^{1}$-a.e. $x\in\lbrack
a,b]$. It follows from the differentiability of $f$ and the linearity of $T$ that for
$\mathcal{L}^{1}$-a.e. $x\in\lbrack a,b]$,$$
\frac{w(x+h)-w(x)}{h}=T\Bigl(\frac{f(x+h)-f(x)}{h}\Bigr)\rightarrow T(f'(x)),
$$
which shows that $w^{\prime}(x)=T(f'(x))$ for $\mathcal{L}^{1}$-a.e.
$x\in\lbrack a,b]$.
Hence, by the fundamental theorem of calculus applied to the function $w$, we
have that
\begin{align*}
T(f(x))-T(f(x_{0})) &  =w(x)-w(x_{0})=\int_{x_{0}}^{x}w^{\prime}%
(t)\,dt=\int_{x_{0}}^{x}T(f'(t))\,dt\\
&  =T{\Bigl(}\int_{x_{0}}^{x}f'(t)\,dt{\Bigr)},
\end{align*}
where you have to use the fact that the Bochner integral commutes with continuous linear functionals $T$. Hence,
$$
T{\Bigl(}f(x)-f(x_{0})-\int_{x_{0}}^{x}f'(t)\,dt{\Bigr)}=0.
$$
Taking the supremum over all $T$ with $\Vert T\Vert_{X^{\prime}}\leq1$ and
using the fact that for $x\in X$, $\Vert x\Vert=\sup_{\Vert T\Vert_{X^{\prime
}}\leq1}T(x)$, we get
$$
f(x)-f(x_{0})-\int_{x_{0}}^{x}f'(t)\,dt=0.
$$
