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).
 A: 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.
$$\frac{d}{dx} V_a^x f = \left\Vert f^\prime \right\Vert_1.$$
A: If $f:[a,b]\rightarrow E$, $E$ a real or complex Banach space, is continuously differentiable then the Bochner integral $\int_{(\alpha,\beta]}f'(t)\,dt$ exists and
$$f(\beta)-f(\alpha)=\int_{(\alpha,\beta]}f'(t)\,dt,\qquad a\leq \alpha\leq \beta\leq b$$
and $\|\int_{(\alpha,\beta]}f'\|_E\leq \int_{(\alpha,\beta]}\|f'\|$.
For any partition $a=t_0<t_1<\ldots<t_n=b$ we have
$$\begin{align}
\sum^n_{j=1}\|f(t_j)-f(t_{j-1})\|_E&=\sum^n_{j=1}\big\|\int_{(t_{j-1},t_j]}f'(t)\,dt\big\|_E\leq\sum^n_{j=1}\int_{(t_{j-1},t_j]}\|f'(t)\|_E\,dt\\
&=\int_{(a,b]}\|f'(t)\|_E\,dt
\end{align}$$
Hence $$V(f;a,b)\leq\int_{(a,b]}\|f'(t)\|_E\,dt$$
Conversely, for each $n\in\mathbb{N}$, consider the partition $\mathcal{P}_n:=\{t_{n,k}=a+2^{-n}k(b-a): 0\leq k\leq 2^n\}$ and define
\begin{align*} g_n(x)&:=\sum^{2^n-1}_{k=0}\frac{f(t_{n,k})-f(t_{n,k-1})}{t_{n,k}-t_{n,k-1}}\mathbb{1}_{[t_{n,k-1},t_{n,k})}(x)\\    &=\sum^{2^n-1}_{k=0}\Big(\frac{t_{n,k}-x}{t_{n,k}-t_{n,k-1}}\frac{f(t_{n,k})-f(x)}{t_{n,k}-x} +\frac{x-t_{n,k-1}}{t_{n,k}-t_{n,k-1}}\frac{f(x)-f(t_{n,k-1})}{x-t_{n,k-1}} \Big)\mathbb{1}_{[t_{n,k-1},t_{n,k})}(x)
 \end{align*}
Clearly $g_n\xrightarrow{n\rightarrow\infty} f'$ at any point $x\in[a,b]\setminus\{t_{n,k}:n\in\mathbb{N},\,0\leq k\leq 2^n\}$ at which $f$ is differentiable. By Fatou's lemma
\begin{align}
   \int^b_a \|f'(t)\|_E\,dt\leq\liminf_n\int^b_a\|g_n\|_E&=\liminf_n\sum^{2^n-1}_{k=0}\|f(t_{n,k})-f(t_{n,k-1})\|_E\leq V(f;a,b)
\end{align}
This all shows that
$$V(f;\alpha,\beta)=\int_{(\alpha,\beta]}\|f'(t)\|_E\,dt,\qquad a\leq\alpha\leq\beta\leq b$$
and so, if $V_f(x):=V(f;a,x)$, then $V'_f(x)=\|f'(x)\|_E$.

Comments:

*

*The result can be extended to functions $f:[a,b]\rightarrow E$ that are absolutely continuous.


*In the finite dimensional case, there is no need to invoke Bochner integration. The fundamental theorem of Calculus ( for either Lebesgue or Riemann integration) can be carried over without much complication.
A: 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.
