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Thread: Total variation

  1. #1
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    Total variation

    Hello,

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

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

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

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

    Bye,
    Alexander
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  2. #2
    Super Member girdav's Avatar
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    Re: Total variation

    Here the dimension of $\displaystyle \Omega$ is 1, so $\displaystyle \operatorname{div}v$ has a simple expression. So you can compute $\displaystyle \int_{\Omega}u(x)\operatorname{div}v(x)dx$ in terms of $\displaystyle v$.
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  3. #3
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    Re: Total variation

    Hello,

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

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

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

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

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

    Is everything correct?

    Bye,
    Alexander
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  4. #4
    Super Member girdav's Avatar
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    Re: Total variation

    It seems correct.
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