Total variation

**AlexanderW** · December 13th 2011, 12:32 AM

Hello,

can you please help me to calculate the total variation of the signum function $\mathrm{sgn}(x)$ on $\Omega=(-1,1)$ .

The total variation of $u \in L^1(\Omega)$ , whereas $\Omega \subseteq \mathbb R^n$ is defined by

$\left \| u \right \|_ {V(\Omega)}:=\sup\left \{ \int_\Omega u\ \mathrm{div}(v) dx : v \in C^1_0(\Omega)^n \text{ with }\left \| v \right \|_{\infty,\Omega} \leq 1 \right \}$ .

I don't know how to apply this abstract defintion.

Bye,
Alexander

**girdav** · December 13th 2011, 01:12 AM

Here the dimension of $\Omega$ is 1, so $\operatorname{div}v$ has a simple expression. So you can compute $\int_{\Omega}u(x)\operatorname{div}v(x)dx$ in terms of $v$ .

**AlexanderW** · December 13th 2011, 09:14 AM

Hello,

thanks for the hint.
Because of n=1, we can write $\text{div}(v)=Dv$ .

It is
$\int_\Omega u \ \text{div}(v)\ dx = \int_{-1}^1 \text{sign}(x)\ Dv\ dx$
$=\int_{-1}^0 -Dv \ dx + \int_0^1 Dv\ dx$
$=-[v(x)]_{-1}^0+[v(x)]_{0}^1 = -(v(0)-v(-1))+(v(1)-v(0))$
$=-v(0)+v(-1)+v(1)-v(0) = -2v(0)$

I used the Fundamental Theorem of Calculus and $v(-1)=v(1)=0$ ,
because $\text{supp}(v)$ is compact ( $v \in C^1_0(\Omega)$ ).

Now, by definition it is
$\left \| v \right \|_{\infty,\Omega} := \inf_{\mu(N)=0} \sup_{x\in (-1,1)\setminus N} \left | v(x) \right | \leq 1$
with $N$ is a zero set.

Because of $\left | v(x) \right | \leq 1$ , choose $v(0)=-1$ and then $\left \| \text{sign} \right \|_{V(\Omega)}=-2 \cdot (-1)=2$ .

Is everything correct?

Bye,
Alexander

**girdav** · December 13th 2011, 09:22 AM

It seems correct.

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