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According to wikipedia, the total variation of the real-valued function $f$, defined on an interval $[a,b]\subset \mathbb{R}$, is the quantity $$V_b^a=\sup_{P\in\mathcal{P}}\sum_{i=0}^{n_P-1}\left | f(x_{i+1})-f({x_i)}\right |$$ where $\mathcal{P}= \left \{P=\{x_0,\ldots, x_{n_P}\} \mid P \text{ is a partition of } [a,b]\right \}$.

According to my professor, the total variation is the quantity $$V_b^a=\limsup_{\delta(P)\to 0}\sum_{i=0}^{n_P-1}\left | f(x_{i+1})-f({x_i)}\right |$$ where $\delta(P)=\max_k (x_k-x_{k-1})$.

Why are the two definitions equivalent?

user72870
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    this is because as you take more points, the variation is growing (triangle inequality). – mookid Mar 05 '14 at 22:47
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    I don't think that the second definition is correct. What does $\delta(P)\to 0$ mean? To give a precise meaning to this kind of things you should use the Moore-Smith theory of convergence. I would rather stick with the first definition. – Giuseppe Negro Mar 05 '14 at 23:47
  • Here is an example of what I mean. – Giuseppe Negro Mar 05 '14 at 23:49

2 Answers2

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Let $V_1$ be the first, $V_2$ be the second. Clearly $V_1 \ge V_2$ just because the $\sup$ is taken over all partitions, including those for which the mesh size goes to zero.

Suppose $P$ is a partition and $\sigma(P) = \sum_{i=0}^{n_P-1}\left | f(x_{i+1})-f({x_i)}\right| $. If $P'$ is a refinement of $P$ (that is, $P \subset P'$), we have $\sigma(P) \le \sigma(P')$. Hence we must have $V_2 \ge \sigma(P)$ for any $P$.

Now suppose $\epsilon>0$ and $P$ is a partition such that $\sigma(P) > V_1 -\epsilon$. Then we have $V_2 \ge \sigma(P) > V_1 -\epsilon$. Since $\epsilon >0$ was arbitrary, we have $V_2 \ge V_1$, and so $V_1 =V_2$.

copper.hat
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  • It seems like in the second paragraph you assumed that the sequence of partitions consists of successive refinements, but that was not assumed. And the proof does not use the continuity of $f$, which it should, since the result is otherwise not true – G. Chiusole Dec 03 '21 at 18:24
  • @G.Chiusole The variation is defined for any function, there is no continuity required, I am not sure where you get that from. The point of the 2nd paragraph is that $\sigma$ increases with refinement. So you can always choose a further refinement with smaller $\delta(P)$. It is just establishing that $V_2 \ge \sigma(P)$ for any partition $P$. – copper.hat Dec 03 '21 at 18:57
  • $\sigma$ increases with refinement, but not with making the mesh smaller. The second definition of OP considers a sequence of partitions where the mesh size goes to $0$, not a sequence of refinements i.e. we do not have $P_n \subseteq P_{n+1}$. My point about the continuity was that the two definitions are not in general equivalent, but only when the function is continuous (or when the sequence of meshes is a successive refinement). Your answer makes it sound like the two definitions are always equivalent. – G. Chiusole Dec 03 '21 at 19:02
  • @G.Chiusole They are equivalent. If there is a sequence of partitions that yields $V_1$ then you can refine said partitions so show that the sequence results in $V_1 \le V_2$. The $\limsup$ adds no restriction. – copper.hat Dec 03 '21 at 19:10
  • If the function is not continuous then they are not equivalent - consider the first example I gave in my answer. I agree that there always exists a sequence of partitions s.t. the sup is attained, and even that from any sequence one can build one which attains it (as you said, by $P_N := \cup_{n =1}^N P_n$). But not just any sequence of partitions with $\vert P \vert \rightarrow 0$ will attain the sup unless it is a successive refinement. – G. Chiusole Dec 03 '21 at 19:14
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    @G.Chiusole With $f=1_{{r}}$ I get the same answer. You don't get to pick a specific set of partitions, the $\limsup$ is taken over all partitions as the mesh goes to zero. Take the partitions $P_n$ formed by adjoining $r$ to the mesh formed from ${k \over n}$, then the $\limsup$ over this sequence is $2$. – copper.hat Dec 03 '21 at 19:18
  • Maybe there is a confusion about the notation of OP. I interpret $\limsup_{\delta(P) \rightarrow 0} ...$ as "for any sequence $(P_n){n = 1}^{\infty}$ with $\delta(P_n) \rightarrow 0$ we have $\limsup{n \rightarrow \infty} ...$" And with the $f = 1_{{r}}$ we can clearly find a sequence with $\delta(P) \rightarrow 0$ but which does not attain $V_1$. How do you interpret it? – G. Chiusole Dec 03 '21 at 19:23
  • @G.Chiusole I would interpret it as $\lim_{\epsilon \downarrow 0} \sup_{\delta(P)\le \epsilon} \sum_{i=0}^{n_P-1}\left | f(x_{i+1})-f({x_i)}\right |$. I am not sure how else one would interpret it, not do I see any advantage to the second definition. – copper.hat Dec 03 '21 at 19:34
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    Fair enough, given that interpretation, I agree with your answer. The advantage of the second definition is that it allows for a concrete computation of the variation (which is how I came about this question in the first place). – G. Chiusole Dec 03 '21 at 19:50
  • I'd like to add: You can only find such a $P$ such that $\sigma(P)>V_1-\epsilon$, if $V_1<\infty$. So I think your argument would not hold if $V_1=\infty$. It is however easy to fix, since if $V_1=\infty$, then $\forall k>0$ $\exists$ $P$ such that $\sigma(P)>k$ and with what you already proved, this would yield $V_2\geq \sigma(P)>k$, which essentially shows that $V_2$ can not be bounded, since $k>0$ is chosen arbitrarily. – SetofMeasureZero Apr 28 '23 at 08:41
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The two are equivalent when $f$ is continuous but not in general. To see this let $f(x) := 1_{\{x = r\}}$ where $r \in [0,1]$ is irrational i.e. the function which is $1$ at $r$ but $0$ elsewhere. Then by the sup definition of variation we have $\text{Var}^{\sup} (f) = 2$, but for the sequence of partitions $\{ \frac{i}{2^n}: 0 \leq i \leq n \}_{n \in \mathbb{N}}$ we have $\text{Var}^{\lim} = \limsup_{n \rightarrow \infty} 0 = 0$.

A proof in the continuous case is the following. Let $(Q_n)_{n = 1}^{\infty}$ be a sequence of partitions with $\vert Q_n\vert \rightarrow 0$ i.e. whose mesh size goes to $0$. For any finite partition, define

$$ \text{Var}^{P}(f) := \sum_{t_i \in P} \vert f(t_{i+1}) - f(t_i) \vert $$

for a finite partition $P$ of $[0,1]$, let $n_P$ denote the number of elements in $P$, and let

$$ \vert P \vert = \max_{P} \vert t_{i+1} - t_i \vert .$$

We will show that the limit of $\text{Var}^{Q_n}(f)$ exists and coincides with $\text{Var}^{\sup}(f)$.

Of course $\text{Var}^{\sup}(f) \geq \text{Var}^{\limsup}(f)$ (which is generally true).

For the other direction, we want to show that for every finite partition $P$ and any $\varepsilon > 0$ there exists an $n \geq 0$ s.t. for all $Q_m$ with $m \geq n$ we have

$$ \text{Var}^{P}(f) := \sum_{t_i \in P} \vert f(t_{i+1}) - f(t_i) \vert \leq \text{Var}^{Q_m}(f) + \varepsilon $$

and thus

$$ \text{Var}^{P}(f) := \sum_{t_i \in P} \vert f(t_{i+1}) - f(t_i) \vert \leq \inf_{m \geq n} \text{Var}^{Q_m}(f) + \varepsilon $$

and

$$ \text{Var}^{P}(f) := \sum_{t_i \in P} \vert f(t_{i+1}) - f(t_i) \vert \leq \lim_{n \rightarrow} \inf_{m \geq n} \text{Var}^{Q_m}(f) = \liminf_{n \rightarrow \infty} \text{Var}^{Q_n}(f)$$

Then taking the $\sup_{P}$ gives

$$ \text{Var}^{\sup}(f) \leq \text{Var}^{\liminf}(f) \leq \text{Var}^{\limsup}(f) \leq \text{Var}^{\sup}(f). $$

So let $P$ be an arbitrary finite partition of $[0,1]$ and let $\varepsilon > 0$ be arbitrary. Then choose $n \geq 0$ s.t. $\delta := \vert Q_n \vert$ small enough s.t. for any $x,y \in [0,1]$ with $\vert x - y \vert < \delta$ we have

$$ \vert f(x) - f(y) \vert < C := \frac{\varepsilon}{2 n_P}$$

Then

\begin{align} \text{Var}^{P}(f) &= \sum_{P} \vert f(t_{i+1}) - f(t_i) \vert \\ &\leq \sum_{P \cup Q_n} \vert f(t_{i+1}) - f(t_i) \vert \\ &\leq \sum_{Q_n} \vert f(t_{i+1}) - f(t_i) \vert + \sum_{t^P \in P} \vert f(q_{t-}) - f(t^P) \vert + \vert f(t^P) - f(q_{t+}) \vert \\ &\leq \text{Var}^{Q_n}(f) + n_P 2 C = \text{Var}^{Q_n}(f) + \varepsilon , \end{align}

where in line 3 for every $t^P$ in the partition $P$, the points $q_{t-}$ and $q_{t+}$ are the next points in $Q_n$ which lie to the left and right of $t^p$, respectively.

G. Chiusole
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