Function $f:(a,b)\to \mathbb{R}$ is measurable and absolutely continuous if and only if there exists the weak derivative $\frac{df}{dx}\in L(a,b)$. 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?
Answer
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
$$
f(x)-f(x_{0})-\int_{x_{0}}^{x}f'(t)\,dt=0.
$$
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