Is the convolution by BV function a BV function?











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Let $BV(mathbb R^n)$ be the set of functions $fin L^1(mathbb R^n)$ such that
$$supBig{int_U f(nablacdotphi),dx,|,phiin C_c^1(mathbb R^n;mathbb R^n), |phi|leq 1Big}<infty$$
Suppose $fin BV(mathbb R^n)$ and $gin L^1(mathbb R^n)$. Is it true that $fast gin BV(mathbb R^n)$?






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Let $BV(mathbb R^n)$ be the set of functions $fin L^1(mathbb R^n)$ such that
$$supBig{int_U f(nablacdotphi),dx,|,phiin C_c^1(mathbb R^n;mathbb R^n), |phi|leq 1Big}<infty$$
Suppose $fin BV(mathbb R^n)$ and $gin L^1(mathbb R^n)$. Is it true that $fast gin BV(mathbb R^n)$?






analysis geometric-measure-theory







share|cite|improve this question
























  • Added definition, thanks
    – user600464
    yesterday













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $BV(mathbb R^n)$ be the set of functions $fin L^1(mathbb R^n)$ such that
$$supBig{int_U f(nablacdotphi),dx,|,phiin C_c^1(mathbb R^n;mathbb R^n), |phi|leq 1Big}<infty$$
Suppose $fin BV(mathbb R^n)$ and $gin L^1(mathbb R^n)$. Is it true that $fast gin BV(mathbb R^n)$?






analysis geometric-measure-theory







share|cite|improve this question















Let $BV(mathbb R^n)$ be the set of functions $fin L^1(mathbb R^n)$ such that
$$supBig{int_U f(nablacdotphi),dx,|,phiin C_c^1(mathbb R^n;mathbb R^n), |phi|leq 1Big}<infty$$
Suppose $fin BV(mathbb R^n)$ and $gin L^1(mathbb R^n)$. Is it true that $fast gin BV(mathbb R^n)$?






analysis geometric-measure-theory




analysis geometric-measure-theory






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  • Added definition, thanks
    – user600464
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  • Added definition, thanks
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    yesterday
















Added definition, thanks
– user600464
yesterday




Added definition, thanks
– user600464
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Denote the supremum in the question as $V_f$.



You should be able to show that if $f in BV$ then the translate $f_y(x) = f(x-y)$ is also $BV$ and $V_{f_y} = V_f$.



Let $f in BV(mathbb R^n)$ and $g in L^1(mathbb R^n)$. It is a standard result that $f ast g in L^1(mathbb R^n)$. Assume that $g ge 0$.



Now let $phi in C_c^infty(mathbb R^n;mathbb R^n)$, $|phi| le 1$. Then begin{align*}int_{mathbb R^n} f ast g (nabla cdot phi) ,dx &= int_{mathbb R^n} int_{mathbb R^n} f(x-y) g(y) , dy (nabla cdot phi)(x) , dy dx \
&= int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy end{align*}
Now,
$$int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx = int_{mathbb R^n} f_y(x) (nabla cdot phi)(x) , dx le V_{f_y} = V_f$$
and since $g ge 0$
$$ int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy le V_f int_{mathbb R^n} g(y) , dy = V_f |g|_{L^1}.$$



Now take the sup over all admissible $phi$ to obtain $$V_{f ast g} le V_f |g|_{L^1}.$$



You can remove the requirement that $g ge 0$ by considering $g^+$ and $g^-$ separately.






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    Denote the supremum in the question as $V_f$.



    You should be able to show that if $f in BV$ then the translate $f_y(x) = f(x-y)$ is also $BV$ and $V_{f_y} = V_f$.



    Let $f in BV(mathbb R^n)$ and $g in L^1(mathbb R^n)$. It is a standard result that $f ast g in L^1(mathbb R^n)$. Assume that $g ge 0$.



    Now let $phi in C_c^infty(mathbb R^n;mathbb R^n)$, $|phi| le 1$. Then begin{align*}int_{mathbb R^n} f ast g (nabla cdot phi) ,dx &= int_{mathbb R^n} int_{mathbb R^n} f(x-y) g(y) , dy (nabla cdot phi)(x) , dy dx \
    &= int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy end{align*}
    Now,
    $$int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx = int_{mathbb R^n} f_y(x) (nabla cdot phi)(x) , dx le V_{f_y} = V_f$$
    and since $g ge 0$
    $$ int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy le V_f int_{mathbb R^n} g(y) , dy = V_f |g|_{L^1}.$$



    Now take the sup over all admissible $phi$ to obtain $$V_{f ast g} le V_f |g|_{L^1}.$$



    You can remove the requirement that $g ge 0$ by considering $g^+$ and $g^-$ separately.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      Denote the supremum in the question as $V_f$.



      You should be able to show that if $f in BV$ then the translate $f_y(x) = f(x-y)$ is also $BV$ and $V_{f_y} = V_f$.



      Let $f in BV(mathbb R^n)$ and $g in L^1(mathbb R^n)$. It is a standard result that $f ast g in L^1(mathbb R^n)$. Assume that $g ge 0$.



      Now let $phi in C_c^infty(mathbb R^n;mathbb R^n)$, $|phi| le 1$. Then begin{align*}int_{mathbb R^n} f ast g (nabla cdot phi) ,dx &= int_{mathbb R^n} int_{mathbb R^n} f(x-y) g(y) , dy (nabla cdot phi)(x) , dy dx \
      &= int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy end{align*}
      Now,
      $$int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx = int_{mathbb R^n} f_y(x) (nabla cdot phi)(x) , dx le V_{f_y} = V_f$$
      and since $g ge 0$
      $$ int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy le V_f int_{mathbb R^n} g(y) , dy = V_f |g|_{L^1}.$$



      Now take the sup over all admissible $phi$ to obtain $$V_{f ast g} le V_f |g|_{L^1}.$$



      You can remove the requirement that $g ge 0$ by considering $g^+$ and $g^-$ separately.






      share|cite|improve this answer























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        Denote the supremum in the question as $V_f$.



        You should be able to show that if $f in BV$ then the translate $f_y(x) = f(x-y)$ is also $BV$ and $V_{f_y} = V_f$.



        Let $f in BV(mathbb R^n)$ and $g in L^1(mathbb R^n)$. It is a standard result that $f ast g in L^1(mathbb R^n)$. Assume that $g ge 0$.



        Now let $phi in C_c^infty(mathbb R^n;mathbb R^n)$, $|phi| le 1$. Then begin{align*}int_{mathbb R^n} f ast g (nabla cdot phi) ,dx &= int_{mathbb R^n} int_{mathbb R^n} f(x-y) g(y) , dy (nabla cdot phi)(x) , dy dx \
        &= int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy end{align*}
        Now,
        $$int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx = int_{mathbb R^n} f_y(x) (nabla cdot phi)(x) , dx le V_{f_y} = V_f$$
        and since $g ge 0$
        $$ int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy le V_f int_{mathbb R^n} g(y) , dy = V_f |g|_{L^1}.$$



        Now take the sup over all admissible $phi$ to obtain $$V_{f ast g} le V_f |g|_{L^1}.$$



        You can remove the requirement that $g ge 0$ by considering $g^+$ and $g^-$ separately.






        share|cite|improve this answer












        Denote the supremum in the question as $V_f$.



        You should be able to show that if $f in BV$ then the translate $f_y(x) = f(x-y)$ is also $BV$ and $V_{f_y} = V_f$.



        Let $f in BV(mathbb R^n)$ and $g in L^1(mathbb R^n)$. It is a standard result that $f ast g in L^1(mathbb R^n)$. Assume that $g ge 0$.



        Now let $phi in C_c^infty(mathbb R^n;mathbb R^n)$, $|phi| le 1$. Then begin{align*}int_{mathbb R^n} f ast g (nabla cdot phi) ,dx &= int_{mathbb R^n} int_{mathbb R^n} f(x-y) g(y) , dy (nabla cdot phi)(x) , dy dx \
        &= int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy end{align*}
        Now,
        $$int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx = int_{mathbb R^n} f_y(x) (nabla cdot phi)(x) , dx le V_{f_y} = V_f$$
        and since $g ge 0$
        $$ int_{mathbb R^n} left( int_{mathbb R^n} f(x-y) (nabla cdot phi)(x) , dx right) g(y) , dy le V_f int_{mathbb R^n} g(y) , dy = V_f |g|_{L^1}.$$



        Now take the sup over all admissible $phi$ to obtain $$V_{f ast g} le V_f |g|_{L^1}.$$



        You can remove the requirement that $g ge 0$ by considering $g^+$ and $g^-$ separately.







        share|cite|improve this answer












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        Umberto P.

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