Let $f(x)$ be a real-valued function on $[a,b]$ of bounded variation. It is standard that $f(x)$ is almost everywhere differentiable, and that $\dfrac{{\rm d}}{{\rm d}x} V^x_a f = |f'(x)|$ for a.e. $x\in [a,b]$, where $V^x_a f$ is the total variation of $f$ on $[a,x]$. An outline of the proof for the equation can be found here (see Problem 3).
Now suppose that $(E,\lVert\cdot\rVert)$ is a real normed vector space. For $f: [a,b]\to E$, define $V^x_a f = \displaystyle{\sup_{a=t_0<t_1<\cdots<t_n=x}} \displaystyle{\sum^{n}_{i=1}} \lVert f(t_i) – f(t_{i-1}) \rVert$. Is it true that if $f$ is of bounded variation (meaning that $V^b_a f < +\infty$), then: (a) $f(x)$ is almost everywhere differentiable, and (b) $\dfrac{{\rm d}}{{\rm d}x} V^x_a f = \lVert f'(x)\rVert$ for a.e. $x\in [a,b]$? The only thing I can get now is that if both $f$ and $V^x_a f$ are differentiable at $x_0$, then $\left(\dfrac{{\rm d}}{{\rm d}x} V^x_a f\right)_{x=x_0} \ge \lVert f'(x_0)\rVert$.
Any help appreciated. Thank you in advance.
Note: This question states that if $E$ is complete, then $f(x)$ being of bounded variation implies that $f(x)$ has two one-side limits at every point. So I wonder if $f(x)$ can fail to be differentiable a.e. if $E$ is not complete.
I'd say that I'm most interested in the case where $E$ is finite-dimensional, but $f$ is not too good (for example, not absolutely continuous, in which case the fundamental theorem of calculus may fail).
Best Answer
If $E$ is supposed to be finite-dimensional of dimension $n \ge 1$, then all norms are equivalent. This implies that all coordinate maps $f_i; i=1, \dots,n$ are of bounded variation. Hence all the $f_i$ are a.e. differentiable with a.e. the relation $\frac{d}{dx} V_a^x f_i = \left\vert f_i^\prime \right\vert$.
For the $\left\Vert \cdot \right\Vert_1$ norm, we then get by additivity a.e.
$$\frac{d}{dx} V_a^x f = \left\Vert f^\prime \right\Vert_1.$$