Approximate Identity in Schwartz Space












1












$begingroup$


Let ${({varphi}_{n}})_{n=1}^{infty}$ an Approximate Identity in Schwartz Space. Then,




  • for all $nin mathbb{Z}^+$ and for all $xin mathbb{R}$, $ varphi_n(x)ge 0$.


  • for all $nin mathbb{Z}^+$
    begin{equation}
    int _{ {mathbb{R} } }^{ }{ varphi _{ n }(x)dx } = 1,
    end{equation}


  • and for all $gamma >0$,
    begin{equation}
    int _{ left| x right| >gamma }^{ }{ varphi _{ n }(x)dx } {xrightarrow[ nlongrightarrow infty ]{}} 0.
    end{equation}
    EDIT:
    Let $alpha in mathbb{Z}^+$. Is it true or not the following statement?
    begin{equation}
    lim _{ nlongrightarrow infty }{ int _{ mathbb{R} }^{ }{ { left| y right| }^{ alpha }left| { varphi }_{ n }left( y right) right| } } dy=0.
    end{equation}






convolution schwartz-space approximate-integration







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$endgroup$












  • $begingroup$
    yes, I forgot to write that
    $endgroup$
    – Mani
    Jan 8 at 20:30










  • $begingroup$
    This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
    $endgroup$
    – Mani
    Jan 8 at 22:06










  • $begingroup$
    Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
    $endgroup$
    – Mani
    Jan 8 at 22:17
















1












$begingroup$


Let ${({varphi}_{n}})_{n=1}^{infty}$ an Approximate Identity in Schwartz Space. Then,




  • for all $nin mathbb{Z}^+$ and for all $xin mathbb{R}$, $ varphi_n(x)ge 0$.


  • for all $nin mathbb{Z}^+$
    begin{equation}
    int _{ {mathbb{R} } }^{ }{ varphi _{ n }(x)dx } = 1,
    end{equation}


  • and for all $gamma >0$,
    begin{equation}
    int _{ left| x right| >gamma }^{ }{ varphi _{ n }(x)dx } {xrightarrow[ nlongrightarrow infty ]{}} 0.
    end{equation}
    EDIT:
    Let $alpha in mathbb{Z}^+$. Is it true or not the following statement?
    begin{equation}
    lim _{ nlongrightarrow infty }{ int _{ mathbb{R} }^{ }{ { left| y right| }^{ alpha }left| { varphi }_{ n }left( y right) right| } } dy=0.
    end{equation}






convolution schwartz-space approximate-integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    yes, I forgot to write that
    $endgroup$
    – Mani
    Jan 8 at 20:30










  • $begingroup$
    This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
    $endgroup$
    – Mani
    Jan 8 at 22:06










  • $begingroup$
    Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
    $endgroup$
    – Mani
    Jan 8 at 22:17














1












1








1


1



$begingroup$


Let ${({varphi}_{n}})_{n=1}^{infty}$ an Approximate Identity in Schwartz Space. Then,




  • for all $nin mathbb{Z}^+$ and for all $xin mathbb{R}$, $ varphi_n(x)ge 0$.


  • for all $nin mathbb{Z}^+$
    begin{equation}
    int _{ {mathbb{R} } }^{ }{ varphi _{ n }(x)dx } = 1,
    end{equation}


  • and for all $gamma >0$,
    begin{equation}
    int _{ left| x right| >gamma }^{ }{ varphi _{ n }(x)dx } {xrightarrow[ nlongrightarrow infty ]{}} 0.
    end{equation}
    EDIT:
    Let $alpha in mathbb{Z}^+$. Is it true or not the following statement?
    begin{equation}
    lim _{ nlongrightarrow infty }{ int _{ mathbb{R} }^{ }{ { left| y right| }^{ alpha }left| { varphi }_{ n }left( y right) right| } } dy=0.
    end{equation}






convolution schwartz-space approximate-integration







share|cite|improve this question











$endgroup$




Let ${({varphi}_{n}})_{n=1}^{infty}$ an Approximate Identity in Schwartz Space. Then,




  • for all $nin mathbb{Z}^+$ and for all $xin mathbb{R}$, $ varphi_n(x)ge 0$.


  • for all $nin mathbb{Z}^+$
    begin{equation}
    int _{ {mathbb{R} } }^{ }{ varphi _{ n }(x)dx } = 1,
    end{equation}


  • and for all $gamma >0$,
    begin{equation}
    int _{ left| x right| >gamma }^{ }{ varphi _{ n }(x)dx } {xrightarrow[ nlongrightarrow infty ]{}} 0.
    end{equation}
    EDIT:
    Let $alpha in mathbb{Z}^+$. Is it true or not the following statement?
    begin{equation}
    lim _{ nlongrightarrow infty }{ int _{ mathbb{R} }^{ }{ { left| y right| }^{ alpha }left| { varphi }_{ n }left( y right) right| } } dy=0.
    end{equation}






convolution schwartz-space approximate-integration




convolution schwartz-space approximate-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Stephen Montgomery-Smith

17.7k12247




17.7k12247










asked Jan 8 at 19:45









ManiMani

52




52












  • $begingroup$
    yes, I forgot to write that
    $endgroup$
    – Mani
    Jan 8 at 20:30










  • $begingroup$
    This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
    $endgroup$
    – Mani
    Jan 8 at 22:06










  • $begingroup$
    Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
    $endgroup$
    – Mani
    Jan 8 at 22:17


















  • $begingroup$
    yes, I forgot to write that
    $endgroup$
    – Mani
    Jan 8 at 20:30










  • $begingroup$
    This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
    $endgroup$
    – Mani
    Jan 8 at 22:06










  • $begingroup$
    Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
    $endgroup$
    – Mani
    Jan 8 at 22:17
















$begingroup$
yes, I forgot to write that
$endgroup$
– Mani
Jan 8 at 20:30




$begingroup$
yes, I forgot to write that
$endgroup$
– Mani
Jan 8 at 20:30












$begingroup$
This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
$endgroup$
– Mani
Jan 8 at 22:06




$begingroup$
This is a more general question than the last that I made. I've done the proof with the classical examples of approximate identity but I cannot make a formal proof for whatever approximate identity.
$endgroup$
– Mani
Jan 8 at 22:06












$begingroup$
Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
$endgroup$
– Mani
Jan 8 at 22:17




$begingroup$
Oh I didn´t see that, now R is the whole space, I will immediately correct that.Thanks
$endgroup$
– Mani
Jan 8 at 22:17










1 Answer
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$begingroup$

It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
$$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
It is easily seen to satisfy the first two properties. But
$$ int_{mathbb R} |y|^alpha phi_x(y) , dy
ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
ge n^{-s} n^alpha ,$$
and this diverges if $alpha > s$.






share|cite|improve this answer











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    $begingroup$

    It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
    $$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
    It is easily seen to satisfy the first two properties. But
    $$ int_{mathbb R} |y|^alpha phi_x(y) , dy
    ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
    ge n^{-s} n^alpha ,$$
    and this diverges if $alpha > s$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
      $$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
      It is easily seen to satisfy the first two properties. But
      $$ int_{mathbb R} |y|^alpha phi_x(y) , dy
      ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
      ge n^{-s} n^alpha ,$$
      and this diverges if $alpha > s$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
        $$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
        It is easily seen to satisfy the first two properties. But
        $$ int_{mathbb R} |y|^alpha phi_x(y) , dy
        ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
        ge n^{-s} n^alpha ,$$
        and this diverges if $alpha > s$.






        share|cite|improve this answer











        $endgroup$



        It is false. Let $psi:mathbb R to [0,infty)$ be a smooth function supported on $[-1,1]$ such that $int_{mathbb R} psi = 1$, and suppose $s > 0$. Define
        $$ phi_n(x) = n (1-n^{-s}) psi(nx) + n^{-s} psi(x-n-1) .$$
        It is easily seen to satisfy the first two properties. But
        $$ int_{mathbb R} |y|^alpha phi_x(y) , dy
        ge int_{n}^{n+2} |y|^alpha n^{-s} psi(y-n-1) , dy
        ge n^{-s} n^alpha ,$$
        and this diverges if $alpha > s$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        Stephen Montgomery-SmithStephen Montgomery-Smith

        17.7k12247




        17.7k12247






























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