6

Let $f:[a,b] \rightarrow \mathbb R$ be $C^1$. Is the length of $\{f(x): x \in [a,b] \}$ and the total variation of $f$ the same thing ? The definition are extremely similar to each others:

  • The total variation of a real-valued (or more generally complex-valued) function $f$, defined on an interval $[a, b] \subset \mathbb{R}$ is the quantity $$ V_{b}^{a}(f)=\sup _{\mathcal{P}} \sum_{i=0}^{n_{P}-1}\left|f\left(x_{i+1}\right)-f\left(x_{i}\right)\right| $$ where the supremum runs over the set of all partitions $\mathcal{P}=\left\{P=\left\{x_{0}, \ldots, x_{n_{P}}\right\} \mid P\right.$ is a partition of $\left.[a, b]\right\}$ of the given interval.

  • Let $f:[a, b] \rightarrow \mathbb{R}^{n}$ be a continuously differentiable function. The length of the curve defined by $f$ can be defined as the limit of the sum of line segment lengths for a regular partition of $[a, b]$ as the number of segments approaches infinity. This means $$ L(f)=\lim _{N \rightarrow \infty} \sum_{i=1}^{N}\left|f\left(t_{i}\right)-f\left(t_{i-1}\right)\right| $$ where $t_{i}=a+i(b-a) / N=a+i \Delta t$ for $i=0,1, \ldots, N .$ This defintiton is equivalent to the standard definition of arc length as an integral: $$ \lim _{N \rightarrow \infty} \sum_{i=1}^{N}\left|f\left(t_{i}\right)-f\left(t_{i-1}\right)\right|=\lim _{N \rightarrow \infty} \sum_{i=1}^{N}\left|\frac{f\left(t_{i}\right)-f\left(t_{i-1}\right)}{\Delta t}\right| \Delta t=\int_{a}^{b}\left|f^{\prime}(t)\right| d t $$

W. Volante
  • 2,244
  • 1
    Yes, the two notions are the same for $C^1$ functions. – zhw. Sep 07 '20 at 14:51
  • @zhw Formally your comment is correct, if you consider a curve in $\mathbb R$, which is not what we usually mean by a curve where $n \ge 2$. I think that the confusion is often coming from the fact that the graph of a real function can be seen as a curve in $\mathbb R^2$. See more detailed answer below. – mathcounterexamples.net Sep 07 '20 at 15:04

2 Answers2

6

No those are not the same things.

First, the total variation is defined for function defined on a closed segment of $\mathbb R$ and $\mathbb R$. While the length of a curve is defined for a curve defined between a closed segment of $\mathbb R$ and $\mathbb R^n$ where $n$ can be larger than $1$.

The confusion is often coming from the fact that a real function $f$ is usually represented by its graph which is the map $F: t \mapsto (t,f(t))$ taken its values in $\mathbb R^2$. This confusion is reinforced by the notations where $\vert \cdot \vert$ is used to denote both the real absolute value and the euclidean norm of $\mathbb R^n$.

For a $\mathcal C^1$ real function the total variation is $$\int_{a}^{b}\left|f^{\prime}(t)\right| d t$$ and the length of the curve representing its graph is

$$\int_{a}^{b}\left\Vert F^{\prime}(t)\right\Vert d t = \int_{a}^{b} \sqrt{1 + \left( f^\prime(x)^2\right)}d t.$$

mathcounterexamples.net
  • 70,018
0

check theorem 5.1 in https://www.whitman.edu/documents/Academics/Mathematics/grady.pdf, there is shown that the arclength $L_a^b(f)$ and the total variation $V_a^b(f)$ satisfies that:

  1. $V_a^b(f) \leq L_a^b(f)$
  2. $L_a^b(f) \leq b - a + V_a^b(f)$

Since $L_a^b(f) = \int_{a}^{b}{\sqrt{1+f'(x)^2}dx}$ and $V_a^b(f)= \int_{a}^{b}{|f'(x)|dx}$ are different, but similar, from point (2) you can see that they can be different for some functions $f(x)$ on the same interval $[a,b]$. I believe they are only equal if $f(x)=\pm ix/\sqrt{2} + constant$, so almost for reals functions they are different.

Joako
  • 1,380