Fourier transform of a convolution of measures












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Let $mathbb{T}$ be the 1-torus, let $mathcal{B}$ be the Borel subsets of $mathbb{T}$ and let $frak{M}(mathbb{T})$ be the set of the Borel complex measures of $mathbb{T}$. If $mu,tauinfrak{M}(mathbb{T})$, define:
$$mu*tau:mathcal{B}tomathbb{C}, Amapstofrac{1}{2pi}int_{mathbb{T}timesmathbb{T}}chi_{A}(x+y)operatorname{d}(muotimestau)(x,y),$$
so that $mu*tauinfrak{M}(mathbb{T}).$



Define the Fourier transform as:
$$mathcal{F}: {frak{M}} (mathbb{T}) to l^infty(mathbb{Z}), mu mapsto left(nmapstofrac{1}{2pi}int_{mathbb{T}}e^{-int}operatorname{d}mu(t)right).$$
How can we prove that:
$$forall mu,tauin {frak{M}}(mathbb{T}), mathcal{F}(mu*tau)=mathcal{F}(mu)mathcal{F}(tau)?$$
I know how to prove the corresponding result for $L^1(mathbb{T})$ functions, i.e. via Fubini theorem, but actually I've some trouble to prove the corresponding result for measure...



Any help?






measure-theory fourier-transform







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    0












    $begingroup$


    Let $mathbb{T}$ be the 1-torus, let $mathcal{B}$ be the Borel subsets of $mathbb{T}$ and let $frak{M}(mathbb{T})$ be the set of the Borel complex measures of $mathbb{T}$. If $mu,tauinfrak{M}(mathbb{T})$, define:
    $$mu*tau:mathcal{B}tomathbb{C}, Amapstofrac{1}{2pi}int_{mathbb{T}timesmathbb{T}}chi_{A}(x+y)operatorname{d}(muotimestau)(x,y),$$
    so that $mu*tauinfrak{M}(mathbb{T}).$



    Define the Fourier transform as:
    $$mathcal{F}: {frak{M}} (mathbb{T}) to l^infty(mathbb{Z}), mu mapsto left(nmapstofrac{1}{2pi}int_{mathbb{T}}e^{-int}operatorname{d}mu(t)right).$$
    How can we prove that:
    $$forall mu,tauin {frak{M}}(mathbb{T}), mathcal{F}(mu*tau)=mathcal{F}(mu)mathcal{F}(tau)?$$
    I know how to prove the corresponding result for $L^1(mathbb{T})$ functions, i.e. via Fubini theorem, but actually I've some trouble to prove the corresponding result for measure...



    Any help?






    measure-theory fourier-transform







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $mathbb{T}$ be the 1-torus, let $mathcal{B}$ be the Borel subsets of $mathbb{T}$ and let $frak{M}(mathbb{T})$ be the set of the Borel complex measures of $mathbb{T}$. If $mu,tauinfrak{M}(mathbb{T})$, define:
      $$mu*tau:mathcal{B}tomathbb{C}, Amapstofrac{1}{2pi}int_{mathbb{T}timesmathbb{T}}chi_{A}(x+y)operatorname{d}(muotimestau)(x,y),$$
      so that $mu*tauinfrak{M}(mathbb{T}).$



      Define the Fourier transform as:
      $$mathcal{F}: {frak{M}} (mathbb{T}) to l^infty(mathbb{Z}), mu mapsto left(nmapstofrac{1}{2pi}int_{mathbb{T}}e^{-int}operatorname{d}mu(t)right).$$
      How can we prove that:
      $$forall mu,tauin {frak{M}}(mathbb{T}), mathcal{F}(mu*tau)=mathcal{F}(mu)mathcal{F}(tau)?$$
      I know how to prove the corresponding result for $L^1(mathbb{T})$ functions, i.e. via Fubini theorem, but actually I've some trouble to prove the corresponding result for measure...



      Any help?






      measure-theory fourier-transform







      share|cite|improve this question









      $endgroup$




      Let $mathbb{T}$ be the 1-torus, let $mathcal{B}$ be the Borel subsets of $mathbb{T}$ and let $frak{M}(mathbb{T})$ be the set of the Borel complex measures of $mathbb{T}$. If $mu,tauinfrak{M}(mathbb{T})$, define:
      $$mu*tau:mathcal{B}tomathbb{C}, Amapstofrac{1}{2pi}int_{mathbb{T}timesmathbb{T}}chi_{A}(x+y)operatorname{d}(muotimestau)(x,y),$$
      so that $mu*tauinfrak{M}(mathbb{T}).$



      Define the Fourier transform as:
      $$mathcal{F}: {frak{M}} (mathbb{T}) to l^infty(mathbb{Z}), mu mapsto left(nmapstofrac{1}{2pi}int_{mathbb{T}}e^{-int}operatorname{d}mu(t)right).$$
      How can we prove that:
      $$forall mu,tauin {frak{M}}(mathbb{T}), mathcal{F}(mu*tau)=mathcal{F}(mu)mathcal{F}(tau)?$$
      I know how to prove the corresponding result for $L^1(mathbb{T})$ functions, i.e. via Fubini theorem, but actually I've some trouble to prove the corresponding result for measure...



      Any help?






      measure-theory fourier-transform




      measure-theory fourier-transform






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 10 at 20:53









      BobBob

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          $begingroup$

          Begin by noting that
          $$
          int_mathbb{T} fd(mu*tau) = int_{mathbb{T}timesmathbb{T}}f(x+y)d(muotimestau)(x,y)
          $$           for all bounded measurable $f$. Hence
          $$begin{eqnarray}
          mathcal{F}(mu*tau)_n &= &int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}d(muotimestau)(x,y)\
          &=&int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}dmu(x)dtau(y)\
          &=&int_{mathbb{T}}left[int_mathbb{T}e^{-inx}dmu(x)right]e^{-iny}dtau(y)\
          &=&mathcal{F}(mu)_nmathcal{F}(tau)_n.
          end{eqnarray}$$






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            $begingroup$

            Begin by noting that
            $$
            int_mathbb{T} fd(mu*tau) = int_{mathbb{T}timesmathbb{T}}f(x+y)d(muotimestau)(x,y)
            $$             for all bounded measurable $f$. Hence
            $$begin{eqnarray}
            mathcal{F}(mu*tau)_n &= &int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}d(muotimestau)(x,y)\
            &=&int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}dmu(x)dtau(y)\
            &=&int_{mathbb{T}}left[int_mathbb{T}e^{-inx}dmu(x)right]e^{-iny}dtau(y)\
            &=&mathcal{F}(mu)_nmathcal{F}(tau)_n.
            end{eqnarray}$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Begin by noting that
              $$
              int_mathbb{T} fd(mu*tau) = int_{mathbb{T}timesmathbb{T}}f(x+y)d(muotimestau)(x,y)
              $$               for all bounded measurable $f$. Hence
              $$begin{eqnarray}
              mathcal{F}(mu*tau)_n &= &int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}d(muotimestau)(x,y)\
              &=&int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}dmu(x)dtau(y)\
              &=&int_{mathbb{T}}left[int_mathbb{T}e^{-inx}dmu(x)right]e^{-iny}dtau(y)\
              &=&mathcal{F}(mu)_nmathcal{F}(tau)_n.
              end{eqnarray}$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Begin by noting that
                $$
                int_mathbb{T} fd(mu*tau) = int_{mathbb{T}timesmathbb{T}}f(x+y)d(muotimestau)(x,y)
                $$                 for all bounded measurable $f$. Hence
                $$begin{eqnarray}
                mathcal{F}(mu*tau)_n &= &int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}d(muotimestau)(x,y)\
                &=&int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}dmu(x)dtau(y)\
                &=&int_{mathbb{T}}left[int_mathbb{T}e^{-inx}dmu(x)right]e^{-iny}dtau(y)\
                &=&mathcal{F}(mu)_nmathcal{F}(tau)_n.
                end{eqnarray}$$






                share|cite|improve this answer









                $endgroup$



                Begin by noting that
                $$
                int_mathbb{T} fd(mu*tau) = int_{mathbb{T}timesmathbb{T}}f(x+y)d(muotimestau)(x,y)
                $$                 for all bounded measurable $f$. Hence
                $$begin{eqnarray}
                mathcal{F}(mu*tau)_n &= &int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}d(muotimestau)(x,y)\
                &=&int_{mathbb{T}timesmathbb{T}}e^{-in(x+y)}dmu(x)dtau(y)\
                &=&int_{mathbb{T}}left[int_mathbb{T}e^{-inx}dmu(x)right]e^{-iny}dtau(y)\
                &=&mathcal{F}(mu)_nmathcal{F}(tau)_n.
                end{eqnarray}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 21:08









                SongSong

                11.8k628




                11.8k628






























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