Show ${int_{0}^{1}|e^{-2picdot int} f(t)| dt} = {int_{0}^{1}|f(t)| dt}$












0














The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).



Some inequalities Fourier coefficient to function
For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?






fourier-analysis fourier-series







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    0














    The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).



    Some inequalities Fourier coefficient to function
    For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?






    fourier-analysis fourier-series







    share|cite|improve this question



























      0












      0








      0







      The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).



      Some inequalities Fourier coefficient to function
      For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?






      fourier-analysis fourier-series







      share|cite|improve this question















      The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).



      Some inequalities Fourier coefficient to function
      For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?






      fourier-analysis fourier-series




      fourier-analysis fourier-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 26 at 2:37









      Tianlalu

      3,06021038




      3,06021038










      asked Nov 26 at 1:59









      hgil

      226




      226






















          2 Answers
          2






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          1














          Note that



          $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



          It is because



          $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



          Thus,



          $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$






          share|cite|improve this answer





















          • Perfect! Thanks!
            – hgil
            Nov 26 at 2:13



















          1














          For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
          $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
          if $ntin mathbb{R}$.






          share|cite|improve this answer





















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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

            oldest

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            1














            Note that



            $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



            It is because



            $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



            Thus,



            $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$






            share|cite|improve this answer





















            • Perfect! Thanks!
              – hgil
              Nov 26 at 2:13
















            1














            Note that



            $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



            It is because



            $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



            Thus,



            $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$






            share|cite|improve this answer





















            • Perfect! Thanks!
              – hgil
              Nov 26 at 2:13














            1












            1








            1






            Note that



            $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



            It is because



            $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



            Thus,



            $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$






            share|cite|improve this answer












            Note that



            $$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$



            It is because



            $$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$



            Thus,



            $$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 at 2:11









            LeB

            991217




            991217












            • Perfect! Thanks!
              – hgil
              Nov 26 at 2:13


















            • Perfect! Thanks!
              – hgil
              Nov 26 at 2:13
















            Perfect! Thanks!
            – hgil
            Nov 26 at 2:13




            Perfect! Thanks!
            – hgil
            Nov 26 at 2:13











            1














            For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
            $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
            if $ntin mathbb{R}$.






            share|cite|improve this answer


























              1














              For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
              $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
              if $ntin mathbb{R}$.






              share|cite|improve this answer
























                1












                1








                1






                For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
                $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
                if $ntin mathbb{R}$.






                share|cite|improve this answer












                For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
                $$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
                if $ntin mathbb{R}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 2:13









                M1183

                943




                943






























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