Does the derivative of a vector-valued BV function $f(x)$ equal to the norm of $f'(x)$? 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).

Today a friend of mine provided a brilliant example:
Let $E = L^1[0,1]$, $f(x) = \chi_{[0,x]}$. Then for $0\le s\le t\le 1$, $||f(t) - f(s)|| = ||\chi_{(s,t]}|| = |t-s|$, which means that $f: [0,1]\to E$ is Lipschitz. But it is nowhere differentiable.
