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I have a question regarding the definitions of functions of bounded variation.

I've been using the following definition, which I found in the book Variational Methods by Struwe:

$f \in BV(\Omega)$ if and only if $$ \int_{\Omega} |Df| = \sup\left\{\int_\Omega f \, \text{div}\, g \,\middle|\, g \in C^1_0(\Omega), \, \|g\|_\infty \leq 1 \right\} < \infty. $$

However, in some lecture notes, I saw that for the one-dimensional case, they define it as:

$f \in BV([a,b])$ if and only if $$ V(f,[a,b]) = \sup_{P \in \mathcal{P}} \sum_{i=0}^{n_P - 1} |f(x_{i+1}) - f(x_i)| < \infty, $$ where $\mathcal{P}$ is the set of all partitions of $[a,b]$.

I also came across this post, which explains that this pointwise definition does not extend to higher dimensions. But does it hold true in full generality for the case $n = 1$?

Daniele Tampieri
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Joaquín Ramírez
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1 Answers1

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Yes, the two definitions are not the same even in dimension $n=1$. To see this consider the following, very simple, counterexample: let $f:[0,1]\to [0,1]$ be defined as $$\DeclareMathOperator{\Dm}{d\!} f(x) = \begin{cases} 1 & \text{if } x= \frac{1}{n}, \; n\in \Bbb N_{>0}\\ 0 & \text{otherwise}. \end{cases} $$ We know that $f$ is $L^1$ as $$ \int\limits_{\Bbb R}|f(x)|\Dm x = \int\limits_0^1|f(x)|\Dm x = 0. $$ Moreover, let's define the indexed family $\{A_\epsilon\}_{\epsilon\in ]0,1[}$ of subsets of $[0,1]$ as follows: $$ A_\epsilon = [0, 1]\cap\bigcup_{n\in \Bbb N_{>0}} \big[\tfrac{1 -n\epsilon^2}{n}, \tfrac{1 + n\epsilon^2}{n}\big]. $$ Then for every $g\in C_0^1([0,1])$ such that $\|g\|_\infty=1$ we have $$ \bigg|\int\limits\limits_{\Bbb R} f(x) \frac{\Dm g(x)}{\Dm x}\Dm x \bigg| \le \int\limits_{A_\epsilon} \Big|\frac{\Dm g(x)}{\Dm x}\Big|\Dm x \le \|g^\prime\|_{\infty}\mu_{\mathrm L}({A_\epsilon})\underset{\epsilon \to 0}{\longrightarrow} 0 $$ where $\mu_{\mathrm L}$ is the Lebesgue measure on $\Bbb R$, and this in turn implies $$ \sup\left\{\int_{[0,1]} f {\Dm g}\,\Big |\, g \in C^1_0([0,1]), \, \|g\|_\infty \leq 1 \right\} = 0 \iff f\in BV([0,1]). $$ Therefore $f$ has a bounded Tonelli-Cesari variation.
On the other hand, let's consider the pointwise variation of $f$ and define the indexed family $\{P_k\}_{k\in \Bbb N_{>0}}\subsetneq \mathcal P$ of finite partitions of $[0, 1]$ as follows: $$ P_k= \{x_0, x_1, \ldots, x_{2k-1}\},\quad n_{P_k} = 2k, $$ where $$ x_i = \begin{cases} 0 & i= 0 \\ \text{any point in }\big]\frac{2}{2(k+1)-i}, \frac{2}{2k-i}\big[ & i= 0\mod 2 \wedge i \neq 0\\ \frac{2}{2k-i+1} & i= 1\mod 2\\ \end{cases} \quad\forall i= 0, \ldots, 2k-1. $$ It is simple to check that this partition, apart from the starting point $x_0=0$, includes all the first $k$ (for $k>1$) singular points of $f$ interleaved by points where $f=0$. This implies that each term $|f(x_{i+1} - f(x_i)|$ is 1 thus $$ \begin{split} V(f,[0,1]) & = \sup_{P \in \mathcal{P}} \sum_{i=0}^{n_P - 1} |f(x_{i+1}) - f(x_i)| \\ & > \sum_{P_k} |f(x_{i+1}) - f(x_i)| = 2k-1 \end {split} \quad \forall k\in \Bbb N_{>0} \iff V(f,[0,1])=\infty $$ therefore $f$ has not a bounded pointwise variation.

Notes

  • The above counterexample is possible because the value of Tonelli-Cesari variation depends only on the behaviour of the analysed function on sets whose Lebesgue measure is non zero, while the value pointwise variation depends on its behaviour at each single point.
  • The above counterexample shows also that functions whose Tonelli-Cesari variation is bounded are indeed equivalence classes of functions that are almost everywhere equal i.e. apart from null sets $A$ ($\mu_{\mathrm L} (A)=0$) as it happens for "functions" in $L^p$ spaces. This is obviously not true for functions with bounded pointwise variation, since in this case the total variation strictly depends on any single point value of the function.
Daniele Tampieri
  • 10,752