Question about total variation When I was reading http://mathpost.la.asu.edu/~ylin/YLin_thesis.pdf , I didn't understand the following. Why is TV the sum of "jumps".
It seems to me that (1.24) is a formula of arc length. But why are $x_1, x_2$ and $x_4$ equal?
For example, consider $$x_1(t) = t^2, \quad x_2(t) = t^3$$
Then
$$ TV(x_1) = \int_0^1|2t|dt = 2 $$
$$ TV(x_2) = \int_0^1|3t^2|dt = 3 $$
Where am I wrong?
I am newbie for 'total variation'. So any recommendation for self-study material is also welcome.



 A: As mentioned, $ x_1, x_2$ and $x_4$ are monotonic and increasing. Another formulation of the Total Varation is:
$TV(x) = \sup_P{ \sum_{i=0}^{n_P-1} \lvert(f(x_{i+1})-f(x_i))\rvert}  $
Where the supremum is taken over all finite partitions of the domain of $f$. This may be a more intuitive definition.
A: First, if this was an arc length formula, then the length of $x^n$ on $[0, 1]$ would go to infinity with increasing n, while you want it to approach $2$ (two sides of a unit square). The $|\cdot|$ is actually a metric and your arcs are defined on a plane, i.e. $\mathbf{g}:[a,b] \rightarrow \mathbb{R}^2, \mathbf{g} = (g_1, g_2)$ where $g_1=t$ and $g_2=f(t)$, so the arc length is actually
$$ \int_a^b|\mathbf{g}'(t)| = \int_a^b\sqrt{g'_1(t)^2+g'_2(t)^2} = \int_a^b\sqrt{1+f'(t)^2} $$
More importantly, the way you compute the integrals is incorrect. The Newton integral is calculated by subtracting primitive functions at the boundary values, not the integrand directly.
The primitive function for $2t$ is $F(t)=t^2+C$ again, so
$$ \int_0^1|2t|dt = F(1) - F(0) = (1)^2 - (0)^2 = 1 $$
and similar for $3t^2$ it's $F(t)=t^3+C$:
$$ \int_0^1|3t^2|dt = F(1) - F(0) = (1)^3 - (0)^3 = 1 $$
