Properties of Schwartz functions












0












$begingroup$


Let $f$ in Schwartz space, and $K, gamma >0$.
Then I guess the following:
begin{align}
int _{ -k }^{ -gamma }{ left| f(x) right| dx } +int _{ gamma }^{ k }{ left| f(x) right| dx } &=int _{ -k }^{ -gamma }{ frac { x }{ x } left| f(x) right| dx } +int _{ gamma }^{ k }{ frac { x }{ x } left| f(x) right| dx } \ &le { left| f right| }_{ 1,0 }int _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &={ -left| f right| }_{ 1,0 }int _{ -gamma }^{ -k }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &={ -left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &=0
end{align}



I did this calculation but I think that I am wrong or maybe not, Where am I wrong?






definite-integrals continuity schwartz-space







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$endgroup$

















    0












    $begingroup$


    Let $f$ in Schwartz space, and $K, gamma >0$.
    Then I guess the following:
    begin{align}
    int _{ -k }^{ -gamma }{ left| f(x) right| dx } +int _{ gamma }^{ k }{ left| f(x) right| dx } &=int _{ -k }^{ -gamma }{ frac { x }{ x } left| f(x) right| dx } +int _{ gamma }^{ k }{ frac { x }{ x } left| f(x) right| dx } \ &le { left| f right| }_{ 1,0 }int _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &={ -left| f right| }_{ 1,0 }int _{ -gamma }^{ -k }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &={ -left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &=0
    end{align}



    I did this calculation but I think that I am wrong or maybe not, Where am I wrong?






    definite-integrals continuity schwartz-space







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f$ in Schwartz space, and $K, gamma >0$.
      Then I guess the following:
      begin{align}
      int _{ -k }^{ -gamma }{ left| f(x) right| dx } +int _{ gamma }^{ k }{ left| f(x) right| dx } &=int _{ -k }^{ -gamma }{ frac { x }{ x } left| f(x) right| dx } +int _{ gamma }^{ k }{ frac { x }{ x } left| f(x) right| dx } \ &le { left| f right| }_{ 1,0 }int _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &={ -left| f right| }_{ 1,0 }int _{ -gamma }^{ -k }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &={ -left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &=0
      end{align}



      I did this calculation but I think that I am wrong or maybe not, Where am I wrong?






      definite-integrals continuity schwartz-space







      share|cite|improve this question











      $endgroup$




      Let $f$ in Schwartz space, and $K, gamma >0$.
      Then I guess the following:
      begin{align}
      int _{ -k }^{ -gamma }{ left| f(x) right| dx } +int _{ gamma }^{ k }{ left| f(x) right| dx } &=int _{ -k }^{ -gamma }{ frac { x }{ x } left| f(x) right| dx } +int _{ gamma }^{ k }{ frac { x }{ x } left| f(x) right| dx } \ &le { left| f right| }_{ 1,0 }int _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &={ -left| f right| }_{ 1,0 }int _{ -gamma }^{ -k }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &={ -left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } +{ left| f right| }_{ 1,0 }int _{ gamma }^{ k }{ frac { 1 }{ x } dx } \ &=0
      end{align}



      I did this calculation but I think that I am wrong or maybe not, Where am I wrong?






      definite-integrals continuity schwartz-space




      definite-integrals continuity schwartz-space






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 13 at 14:40









      Davide Giraudo

      128k17156268




      128k17156268










      asked Jan 10 at 20:57









      ManiMani

      92




      92






















          1 Answer
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          0












          $begingroup$

          Actually this would be correct if and only if $f$ vanishes on both intervals $(-k,-gamma)$ and $(gamma,k)$.



          The problem is in the first inequality: on $(-k,-gamma)$, $x$ is negative hence the integral $int_{-k}^{-gamma}1/xmathrm dx$ should be $-int_{-k}^{-gamma}1/xmathrm dx$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, can you explain me a litle bit more?. In this integral, I put x<0. $int _{ -k }^{ -gamma }{ left| f(x) right| dx }=int _{ -k }^{ -gamma }{ left| f(x) right| dx } le Cint _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } $ . if I set x<0. Why do you change the sign in the last integral?
            $endgroup$
            – Mani
            Jan 10 at 22:07










          • $begingroup$
            But we only know that $leftlvert xrightrvert leftlvert f(x)rightrvertleqslant C$ hence the bound should be $Cint _{ -k }^{ -gamma }{ frac { 1 }{ leftlvert xrightrvert } dx }$.
            $endgroup$
            – Davide Giraudo
            Jan 12 at 11:49












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          1 Answer
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          active

          oldest

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          0












          $begingroup$

          Actually this would be correct if and only if $f$ vanishes on both intervals $(-k,-gamma)$ and $(gamma,k)$.



          The problem is in the first inequality: on $(-k,-gamma)$, $x$ is negative hence the integral $int_{-k}^{-gamma}1/xmathrm dx$ should be $-int_{-k}^{-gamma}1/xmathrm dx$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, can you explain me a litle bit more?. In this integral, I put x<0. $int _{ -k }^{ -gamma }{ left| f(x) right| dx }=int _{ -k }^{ -gamma }{ left| f(x) right| dx } le Cint _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } $ . if I set x<0. Why do you change the sign in the last integral?
            $endgroup$
            – Mani
            Jan 10 at 22:07










          • $begingroup$
            But we only know that $leftlvert xrightrvert leftlvert f(x)rightrvertleqslant C$ hence the bound should be $Cint _{ -k }^{ -gamma }{ frac { 1 }{ leftlvert xrightrvert } dx }$.
            $endgroup$
            – Davide Giraudo
            Jan 12 at 11:49
















          0












          $begingroup$

          Actually this would be correct if and only if $f$ vanishes on both intervals $(-k,-gamma)$ and $(gamma,k)$.



          The problem is in the first inequality: on $(-k,-gamma)$, $x$ is negative hence the integral $int_{-k}^{-gamma}1/xmathrm dx$ should be $-int_{-k}^{-gamma}1/xmathrm dx$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, can you explain me a litle bit more?. In this integral, I put x<0. $int _{ -k }^{ -gamma }{ left| f(x) right| dx }=int _{ -k }^{ -gamma }{ left| f(x) right| dx } le Cint _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } $ . if I set x<0. Why do you change the sign in the last integral?
            $endgroup$
            – Mani
            Jan 10 at 22:07










          • $begingroup$
            But we only know that $leftlvert xrightrvert leftlvert f(x)rightrvertleqslant C$ hence the bound should be $Cint _{ -k }^{ -gamma }{ frac { 1 }{ leftlvert xrightrvert } dx }$.
            $endgroup$
            – Davide Giraudo
            Jan 12 at 11:49














          0












          0








          0





          $begingroup$

          Actually this would be correct if and only if $f$ vanishes on both intervals $(-k,-gamma)$ and $(gamma,k)$.



          The problem is in the first inequality: on $(-k,-gamma)$, $x$ is negative hence the integral $int_{-k}^{-gamma}1/xmathrm dx$ should be $-int_{-k}^{-gamma}1/xmathrm dx$.






          share|cite|improve this answer









          $endgroup$



          Actually this would be correct if and only if $f$ vanishes on both intervals $(-k,-gamma)$ and $(gamma,k)$.



          The problem is in the first inequality: on $(-k,-gamma)$, $x$ is negative hence the integral $int_{-k}^{-gamma}1/xmathrm dx$ should be $-int_{-k}^{-gamma}1/xmathrm dx$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 21:07









          Davide GiraudoDavide Giraudo

          128k17156268




          128k17156268












          • $begingroup$
            Sorry, can you explain me a litle bit more?. In this integral, I put x<0. $int _{ -k }^{ -gamma }{ left| f(x) right| dx }=int _{ -k }^{ -gamma }{ left| f(x) right| dx } le Cint _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } $ . if I set x<0. Why do you change the sign in the last integral?
            $endgroup$
            – Mani
            Jan 10 at 22:07










          • $begingroup$
            But we only know that $leftlvert xrightrvert leftlvert f(x)rightrvertleqslant C$ hence the bound should be $Cint _{ -k }^{ -gamma }{ frac { 1 }{ leftlvert xrightrvert } dx }$.
            $endgroup$
            – Davide Giraudo
            Jan 12 at 11:49


















          • $begingroup$
            Sorry, can you explain me a litle bit more?. In this integral, I put x<0. $int _{ -k }^{ -gamma }{ left| f(x) right| dx }=int _{ -k }^{ -gamma }{ left| f(x) right| dx } le Cint _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } $ . if I set x<0. Why do you change the sign in the last integral?
            $endgroup$
            – Mani
            Jan 10 at 22:07










          • $begingroup$
            But we only know that $leftlvert xrightrvert leftlvert f(x)rightrvertleqslant C$ hence the bound should be $Cint _{ -k }^{ -gamma }{ frac { 1 }{ leftlvert xrightrvert } dx }$.
            $endgroup$
            – Davide Giraudo
            Jan 12 at 11:49
















          $begingroup$
          Sorry, can you explain me a litle bit more?. In this integral, I put x<0. $int _{ -k }^{ -gamma }{ left| f(x) right| dx }=int _{ -k }^{ -gamma }{ left| f(x) right| dx } le Cint _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } $ . if I set x<0. Why do you change the sign in the last integral?
          $endgroup$
          – Mani
          Jan 10 at 22:07




          $begingroup$
          Sorry, can you explain me a litle bit more?. In this integral, I put x<0. $int _{ -k }^{ -gamma }{ left| f(x) right| dx }=int _{ -k }^{ -gamma }{ left| f(x) right| dx } le Cint _{ -k }^{ -gamma }{ frac { 1 }{ x } dx } $ . if I set x<0. Why do you change the sign in the last integral?
          $endgroup$
          – Mani
          Jan 10 at 22:07












          $begingroup$
          But we only know that $leftlvert xrightrvert leftlvert f(x)rightrvertleqslant C$ hence the bound should be $Cint _{ -k }^{ -gamma }{ frac { 1 }{ leftlvert xrightrvert } dx }$.
          $endgroup$
          – Davide Giraudo
          Jan 12 at 11:49




          $begingroup$
          But we only know that $leftlvert xrightrvert leftlvert f(x)rightrvertleqslant C$ hence the bound should be $Cint _{ -k }^{ -gamma }{ frac { 1 }{ leftlvert xrightrvert } dx }$.
          $endgroup$
          – Davide Giraudo
          Jan 12 at 11:49


















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