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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).

Jianing Song
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  • Do you suppose that $E$ is finite dimensional? – mathcounterexamples.net Apr 11 '22 at 10:16
  • @mathcounterexamples.net Assume that if necessary :) – Jianing Song Apr 11 '22 at 11:08
  • @JianingSong: Constructing the integral in normed spaces that are not complete give rise to many complications (converge of Riemann sums for the LEbesgue type extension (Bochner integral) if the integral) Even for more general topological linear spaces $E$, some assumptions such as $E$ being a Frechet space (locally convex linear space with complete metric that is invariant under translations) some completeness may be required for the existence of the integral (Petits-integral). – Mittens Apr 12 '22 at 19:52

3 Answers3

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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.

$$\frac{d}{dx} V_a^x f = \left\Vert f^\prime \right\Vert_1.$$

mathcounterexamples.net
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  • Thanks! But what can be said for $\frac{{\rm d}}{{\rm d}x} V^x_a f$ using other norms ($\lVert\cdot\rVert_2$ norm for example) :) – Jianing Song Apr 12 '22 at 11:40
  • I think that this is true for whatever norm. But I don’t see a straightforward proof. By straightforward, I mean avoiding mimicking the exercise 3 of the link you gave. – mathcounterexamples.net Apr 12 '22 at 11:47
  • Mimicking the exercise 3 does not seem to work: given a.e. differentiable vector-valued function $f(x)$, I don't think we can easily construct a real-valued function $\tilde{f}(x)$ such that $|\tilde{f}'(x)| = \lVert f'(x)\rVert$. – Jianing Song Apr 12 '22 at 13:27
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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:

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

  2. 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.

Mittens
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  • Thanks! It seems that it is very difficult to generalize the result of a.e. differentiability of BV functions to the infinite-dimensional cases :) – Jianing Song Apr 13 '22 at 05:39
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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.

Jianing Song
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