Results 1 to 4 of 4

Math Help - Total variation

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    23

    Total variation

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    678
    Thanks
    32

    Re: Total variation

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    23

    Re: Total variation

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    678
    Thanks
    32

    Re: Total variation

    It seems correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. A problem about total variation of complex measure
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 6th 2011, 12:46 AM
  2. Total Differentiation; total mess.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 13th 2010, 01:14 PM
  3. variation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 26th 2008, 10:18 AM
  4. another variation
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 25th 2008, 11:05 AM
  5. Nim Variation
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: April 9th 2008, 09:22 AM

Search Tags


/mathhelpforum @mathhelpforum