Integral of a function with compact support












3












$begingroup$


I am reading a book with the following statement



Since $g$ has compact support $$sum_{i=1}^d int_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} dx = 0$$



where





  1. $frac{dx}{dt} = F(x)$ with $xin mathbb{R}^d$


  2. $g: mathbb{R}^d rightarrow mathbb{R}$ is continuously differentiable with compact support

  3. $f: mathbb{R}^d rightarrow mathbb{R}$


The compact support means that if it is zero outside of a compact set.



How can we say the sum of integral is zero?






integration derivatives partial-derivative







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$endgroup$








  • 2




    $begingroup$
    This is the volume integral of divergence of the vector field $fgF$. When it is written as a surface integral over a sufficiently large surface, which is beyond the support of $g$, it vanishes.
    $endgroup$
    – Amey Joshi
    Jan 1 at 7:31
















3












$begingroup$


I am reading a book with the following statement



Since $g$ has compact support $$sum_{i=1}^d int_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} dx = 0$$



where





  1. $frac{dx}{dt} = F(x)$ with $xin mathbb{R}^d$


  2. $g: mathbb{R}^d rightarrow mathbb{R}$ is continuously differentiable with compact support

  3. $f: mathbb{R}^d rightarrow mathbb{R}$


The compact support means that if it is zero outside of a compact set.



How can we say the sum of integral is zero?






integration derivatives partial-derivative







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This is the volume integral of divergence of the vector field $fgF$. When it is written as a surface integral over a sufficiently large surface, which is beyond the support of $g$, it vanishes.
    $endgroup$
    – Amey Joshi
    Jan 1 at 7:31














3












3








3


1



$begingroup$


I am reading a book with the following statement



Since $g$ has compact support $$sum_{i=1}^d int_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} dx = 0$$



where





  1. $frac{dx}{dt} = F(x)$ with $xin mathbb{R}^d$


  2. $g: mathbb{R}^d rightarrow mathbb{R}$ is continuously differentiable with compact support

  3. $f: mathbb{R}^d rightarrow mathbb{R}$


The compact support means that if it is zero outside of a compact set.



How can we say the sum of integral is zero?






integration derivatives partial-derivative







share|cite|improve this question











$endgroup$




I am reading a book with the following statement



Since $g$ has compact support $$sum_{i=1}^d int_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} dx = 0$$



where





  1. $frac{dx}{dt} = F(x)$ with $xin mathbb{R}^d$


  2. $g: mathbb{R}^d rightarrow mathbb{R}$ is continuously differentiable with compact support

  3. $f: mathbb{R}^d rightarrow mathbb{R}$


The compact support means that if it is zero outside of a compact set.



How can we say the sum of integral is zero?






integration derivatives partial-derivative




integration derivatives partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 1 at 6:23







sleeve chen

















asked Jan 1 at 6:09









sleeve chensleeve chen

3,14042053




3,14042053








  • 2




    $begingroup$
    This is the volume integral of divergence of the vector field $fgF$. When it is written as a surface integral over a sufficiently large surface, which is beyond the support of $g$, it vanishes.
    $endgroup$
    – Amey Joshi
    Jan 1 at 7:31














  • 2




    $begingroup$
    This is the volume integral of divergence of the vector field $fgF$. When it is written as a surface integral over a sufficiently large surface, which is beyond the support of $g$, it vanishes.
    $endgroup$
    – Amey Joshi
    Jan 1 at 7:31








2




2




$begingroup$
This is the volume integral of divergence of the vector field $fgF$. When it is written as a surface integral over a sufficiently large surface, which is beyond the support of $g$, it vanishes.
$endgroup$
– Amey Joshi
Jan 1 at 7:31




$begingroup$
This is the volume integral of divergence of the vector field $fgF$. When it is written as a surface integral over a sufficiently large surface, which is beyond the support of $g$, it vanishes.
$endgroup$
– Amey Joshi
Jan 1 at 7:31










1 Answer
1






active

oldest

votes


















4












$begingroup$

The key of the reasoning is that, defined
$$
V(x)=left(
begin{matrix}
fF_1 g\
fF_2 g\
vdots\
fF_{d-1} g\
fF_d g
end{matrix}
right) implies Vtext{ has compact support in }mathbb{R}^d
$$
Since
$$
sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x,
$$
then, by considering a ball $B(0,r)inmathbb{R}^d$ of radius $r>0$, we have that
$$
begin{split}
sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x & = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x\
&triangleqlim_{rtoinfty}intlimits_{B(0,r)} nablacdot V, mathrm{d}x\
&text{ by Gauss-Green}\
&=lim_{rtoinfty}intlimits_{partial B(0,r)} Vcdotnu_x, mathrm{d}sigma_x=0
end{split}
$$
where $nu_x$ is the normal unit vector to $partial B(0,r)$ in the point $xin partial B(0,r)$, since $V$ has compact support $iff$ $Vcdotnu_x=0$ for all $nu_x$ and all $r$ larger than a fixed finite value $r_0>0$.






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    4












    $begingroup$

    The key of the reasoning is that, defined
    $$
    V(x)=left(
    begin{matrix}
    fF_1 g\
    fF_2 g\
    vdots\
    fF_{d-1} g\
    fF_d g
    end{matrix}
    right) implies Vtext{ has compact support in }mathbb{R}^d
    $$
    Since
    $$
    sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x,
    $$
    then, by considering a ball $B(0,r)inmathbb{R}^d$ of radius $r>0$, we have that
    $$
    begin{split}
    sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x & = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x\
    &triangleqlim_{rtoinfty}intlimits_{B(0,r)} nablacdot V, mathrm{d}x\
    &text{ by Gauss-Green}\
    &=lim_{rtoinfty}intlimits_{partial B(0,r)} Vcdotnu_x, mathrm{d}sigma_x=0
    end{split}
    $$
    where $nu_x$ is the normal unit vector to $partial B(0,r)$ in the point $xin partial B(0,r)$, since $V$ has compact support $iff$ $Vcdotnu_x=0$ for all $nu_x$ and all $r$ larger than a fixed finite value $r_0>0$.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      The key of the reasoning is that, defined
      $$
      V(x)=left(
      begin{matrix}
      fF_1 g\
      fF_2 g\
      vdots\
      fF_{d-1} g\
      fF_d g
      end{matrix}
      right) implies Vtext{ has compact support in }mathbb{R}^d
      $$
      Since
      $$
      sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x,
      $$
      then, by considering a ball $B(0,r)inmathbb{R}^d$ of radius $r>0$, we have that
      $$
      begin{split}
      sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x & = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x\
      &triangleqlim_{rtoinfty}intlimits_{B(0,r)} nablacdot V, mathrm{d}x\
      &text{ by Gauss-Green}\
      &=lim_{rtoinfty}intlimits_{partial B(0,r)} Vcdotnu_x, mathrm{d}sigma_x=0
      end{split}
      $$
      where $nu_x$ is the normal unit vector to $partial B(0,r)$ in the point $xin partial B(0,r)$, since $V$ has compact support $iff$ $Vcdotnu_x=0$ for all $nu_x$ and all $r$ larger than a fixed finite value $r_0>0$.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        The key of the reasoning is that, defined
        $$
        V(x)=left(
        begin{matrix}
        fF_1 g\
        fF_2 g\
        vdots\
        fF_{d-1} g\
        fF_d g
        end{matrix}
        right) implies Vtext{ has compact support in }mathbb{R}^d
        $$
        Since
        $$
        sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x,
        $$
        then, by considering a ball $B(0,r)inmathbb{R}^d$ of radius $r>0$, we have that
        $$
        begin{split}
        sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x & = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x\
        &triangleqlim_{rtoinfty}intlimits_{B(0,r)} nablacdot V, mathrm{d}x\
        &text{ by Gauss-Green}\
        &=lim_{rtoinfty}intlimits_{partial B(0,r)} Vcdotnu_x, mathrm{d}sigma_x=0
        end{split}
        $$
        where $nu_x$ is the normal unit vector to $partial B(0,r)$ in the point $xin partial B(0,r)$, since $V$ has compact support $iff$ $Vcdotnu_x=0$ for all $nu_x$ and all $r$ larger than a fixed finite value $r_0>0$.






        share|cite|improve this answer











        $endgroup$



        The key of the reasoning is that, defined
        $$
        V(x)=left(
        begin{matrix}
        fF_1 g\
        fF_2 g\
        vdots\
        fF_{d-1} g\
        fF_d g
        end{matrix}
        right) implies Vtext{ has compact support in }mathbb{R}^d
        $$
        Since
        $$
        sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x,
        $$
        then, by considering a ball $B(0,r)inmathbb{R}^d$ of radius $r>0$, we have that
        $$
        begin{split}
        sum_{i=1}^d intlimits_{mathbb{R}^d} frac{partial (fF_i g)}{x_i} mathrm{d}x & = intlimits_{mathbb{R}^d} nablacdot V, mathrm{d}x\
        &triangleqlim_{rtoinfty}intlimits_{B(0,r)} nablacdot V, mathrm{d}x\
        &text{ by Gauss-Green}\
        &=lim_{rtoinfty}intlimits_{partial B(0,r)} Vcdotnu_x, mathrm{d}sigma_x=0
        end{split}
        $$
        where $nu_x$ is the normal unit vector to $partial B(0,r)$ in the point $xin partial B(0,r)$, since $V$ has compact support $iff$ $Vcdotnu_x=0$ for all $nu_x$ and all $r$ larger than a fixed finite value $r_0>0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 1 at 18:25

























        answered Jan 1 at 7:54









        Daniele TampieriDaniele Tampieri

        2,3272922




        2,3272922






























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