Where F is Fourier Transform, prove the limit as |k| approaches infinity of Ff(k)=0 where f is differentiable...












0












$begingroup$


Prove $$lim_{|k|to infty}{mathscr F f(k)}=lim_{|k|to infty} int_{-infty}^infty f(x)e^{-ikx}dx=0$$ (imagine the low bound is negative infinity).



I tried integrating by parts... where the first term is evaluated at x from -infinity to +infinity $$lim_{|k|to infty} {left(frac{-1}{ik}right)f(x)e^{-ikx} - int_{-infty}^infty left(frac{-1}{ik}right)f'(x)e^{-ikx} dx}$$



This second integral can be substituted for the original integral (imagine the low bound is negative infinity), $$int_{-infty}^infty f(x)e^{-ikx}dx$$ by properties of Fourier Transforms, which would allow a cancellation however I still cannot prove the first part (the uv part of integration by parts) will go to zero.

The hint is to assume that since f'(x) is absolutely integrable that $$lim_{|x|to infty} f'(x) = 0$$, however I do not see how I can do this when f'(x) will always be inside of the integrand.

The next step is to broaden this argument to show that the $$lim_{|k|toinfty}{|k|^{n-1} mathscr F f(k)=0}$$ as k approaches infinity of |k|^(n-1)Ff(k)=0 when f is n times differentiable and the jth derivative of f is absolutely integrable where j=1,...,n. Any help is much appreciated!






integration fourier-analysis improper-integrals







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  • $begingroup$
    Is the integral wrt. $x$? I do not understand the first term. For fixed $x$ it certainly vanishes. Are you shure that $k$ goes to infinity and not $x$? Maybe it is better to start at the beginning. For now I do not see where the property for$f'(x)$ comes in.
    $endgroup$
    – Urgje
    Dec 2 '15 at 21:50










  • $begingroup$
    So, absolute integrability doesn't imply $lim_{|x|rightarrowinfty}f'(x)=0$, and I think you did the integration by parts wrong. See if you can derive the identity $F(f')(k)=ikF(f)(k)$.
    $endgroup$
    – charlestoncrabb
    Dec 2 '15 at 21:51












  • $begingroup$
    I apologize, part of the problem is that I am really not used to MathJax. The integral is in terms of x and yes the limit is as k goes to infinity, and I have derived that identity yes and used it in this problem.
    $endgroup$
    – mentorship
    Dec 2 '15 at 22:02
















0












$begingroup$


Prove $$lim_{|k|to infty}{mathscr F f(k)}=lim_{|k|to infty} int_{-infty}^infty f(x)e^{-ikx}dx=0$$ (imagine the low bound is negative infinity).



I tried integrating by parts... where the first term is evaluated at x from -infinity to +infinity $$lim_{|k|to infty} {left(frac{-1}{ik}right)f(x)e^{-ikx} - int_{-infty}^infty left(frac{-1}{ik}right)f'(x)e^{-ikx} dx}$$



This second integral can be substituted for the original integral (imagine the low bound is negative infinity), $$int_{-infty}^infty f(x)e^{-ikx}dx$$ by properties of Fourier Transforms, which would allow a cancellation however I still cannot prove the first part (the uv part of integration by parts) will go to zero.

The hint is to assume that since f'(x) is absolutely integrable that $$lim_{|x|to infty} f'(x) = 0$$, however I do not see how I can do this when f'(x) will always be inside of the integrand.

The next step is to broaden this argument to show that the $$lim_{|k|toinfty}{|k|^{n-1} mathscr F f(k)=0}$$ as k approaches infinity of |k|^(n-1)Ff(k)=0 when f is n times differentiable and the jth derivative of f is absolutely integrable where j=1,...,n. Any help is much appreciated!






integration fourier-analysis improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is the integral wrt. $x$? I do not understand the first term. For fixed $x$ it certainly vanishes. Are you shure that $k$ goes to infinity and not $x$? Maybe it is better to start at the beginning. For now I do not see where the property for$f'(x)$ comes in.
    $endgroup$
    – Urgje
    Dec 2 '15 at 21:50










  • $begingroup$
    So, absolute integrability doesn't imply $lim_{|x|rightarrowinfty}f'(x)=0$, and I think you did the integration by parts wrong. See if you can derive the identity $F(f')(k)=ikF(f)(k)$.
    $endgroup$
    – charlestoncrabb
    Dec 2 '15 at 21:51












  • $begingroup$
    I apologize, part of the problem is that I am really not used to MathJax. The integral is in terms of x and yes the limit is as k goes to infinity, and I have derived that identity yes and used it in this problem.
    $endgroup$
    – mentorship
    Dec 2 '15 at 22:02














0












0








0





$begingroup$


Prove $$lim_{|k|to infty}{mathscr F f(k)}=lim_{|k|to infty} int_{-infty}^infty f(x)e^{-ikx}dx=0$$ (imagine the low bound is negative infinity).



I tried integrating by parts... where the first term is evaluated at x from -infinity to +infinity $$lim_{|k|to infty} {left(frac{-1}{ik}right)f(x)e^{-ikx} - int_{-infty}^infty left(frac{-1}{ik}right)f'(x)e^{-ikx} dx}$$



This second integral can be substituted for the original integral (imagine the low bound is negative infinity), $$int_{-infty}^infty f(x)e^{-ikx}dx$$ by properties of Fourier Transforms, which would allow a cancellation however I still cannot prove the first part (the uv part of integration by parts) will go to zero.

The hint is to assume that since f'(x) is absolutely integrable that $$lim_{|x|to infty} f'(x) = 0$$, however I do not see how I can do this when f'(x) will always be inside of the integrand.

The next step is to broaden this argument to show that the $$lim_{|k|toinfty}{|k|^{n-1} mathscr F f(k)=0}$$ as k approaches infinity of |k|^(n-1)Ff(k)=0 when f is n times differentiable and the jth derivative of f is absolutely integrable where j=1,...,n. Any help is much appreciated!






integration fourier-analysis improper-integrals







share|cite|improve this question











$endgroup$




Prove $$lim_{|k|to infty}{mathscr F f(k)}=lim_{|k|to infty} int_{-infty}^infty f(x)e^{-ikx}dx=0$$ (imagine the low bound is negative infinity).



I tried integrating by parts... where the first term is evaluated at x from -infinity to +infinity $$lim_{|k|to infty} {left(frac{-1}{ik}right)f(x)e^{-ikx} - int_{-infty}^infty left(frac{-1}{ik}right)f'(x)e^{-ikx} dx}$$



This second integral can be substituted for the original integral (imagine the low bound is negative infinity), $$int_{-infty}^infty f(x)e^{-ikx}dx$$ by properties of Fourier Transforms, which would allow a cancellation however I still cannot prove the first part (the uv part of integration by parts) will go to zero.

The hint is to assume that since f'(x) is absolutely integrable that $$lim_{|x|to infty} f'(x) = 0$$, however I do not see how I can do this when f'(x) will always be inside of the integrand.

The next step is to broaden this argument to show that the $$lim_{|k|toinfty}{|k|^{n-1} mathscr F f(k)=0}$$ as k approaches infinity of |k|^(n-1)Ff(k)=0 when f is n times differentiable and the jth derivative of f is absolutely integrable where j=1,...,n. Any help is much appreciated!






integration fourier-analysis improper-integrals




integration fourier-analysis improper-integrals






share|cite|improve this question















share|cite|improve this question













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edited Jan 5 at 12:06









honey.mustard

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asked Dec 2 '15 at 21:18









mentorshipmentorship

1088




1088












  • $begingroup$
    Is the integral wrt. $x$? I do not understand the first term. For fixed $x$ it certainly vanishes. Are you shure that $k$ goes to infinity and not $x$? Maybe it is better to start at the beginning. For now I do not see where the property for$f'(x)$ comes in.
    $endgroup$
    – Urgje
    Dec 2 '15 at 21:50










  • $begingroup$
    So, absolute integrability doesn't imply $lim_{|x|rightarrowinfty}f'(x)=0$, and I think you did the integration by parts wrong. See if you can derive the identity $F(f')(k)=ikF(f)(k)$.
    $endgroup$
    – charlestoncrabb
    Dec 2 '15 at 21:51












  • $begingroup$
    I apologize, part of the problem is that I am really not used to MathJax. The integral is in terms of x and yes the limit is as k goes to infinity, and I have derived that identity yes and used it in this problem.
    $endgroup$
    – mentorship
    Dec 2 '15 at 22:02


















  • $begingroup$
    Is the integral wrt. $x$? I do not understand the first term. For fixed $x$ it certainly vanishes. Are you shure that $k$ goes to infinity and not $x$? Maybe it is better to start at the beginning. For now I do not see where the property for$f'(x)$ comes in.
    $endgroup$
    – Urgje
    Dec 2 '15 at 21:50










  • $begingroup$
    So, absolute integrability doesn't imply $lim_{|x|rightarrowinfty}f'(x)=0$, and I think you did the integration by parts wrong. See if you can derive the identity $F(f')(k)=ikF(f)(k)$.
    $endgroup$
    – charlestoncrabb
    Dec 2 '15 at 21:51












  • $begingroup$
    I apologize, part of the problem is that I am really not used to MathJax. The integral is in terms of x and yes the limit is as k goes to infinity, and I have derived that identity yes and used it in this problem.
    $endgroup$
    – mentorship
    Dec 2 '15 at 22:02
















$begingroup$
Is the integral wrt. $x$? I do not understand the first term. For fixed $x$ it certainly vanishes. Are you shure that $k$ goes to infinity and not $x$? Maybe it is better to start at the beginning. For now I do not see where the property for$f'(x)$ comes in.
$endgroup$
– Urgje
Dec 2 '15 at 21:50




$begingroup$
Is the integral wrt. $x$? I do not understand the first term. For fixed $x$ it certainly vanishes. Are you shure that $k$ goes to infinity and not $x$? Maybe it is better to start at the beginning. For now I do not see where the property for$f'(x)$ comes in.
$endgroup$
– Urgje
Dec 2 '15 at 21:50












$begingroup$
So, absolute integrability doesn't imply $lim_{|x|rightarrowinfty}f'(x)=0$, and I think you did the integration by parts wrong. See if you can derive the identity $F(f')(k)=ikF(f)(k)$.
$endgroup$
– charlestoncrabb
Dec 2 '15 at 21:51






$begingroup$
So, absolute integrability doesn't imply $lim_{|x|rightarrowinfty}f'(x)=0$, and I think you did the integration by parts wrong. See if you can derive the identity $F(f')(k)=ikF(f)(k)$.
$endgroup$
– charlestoncrabb
Dec 2 '15 at 21:51














$begingroup$
I apologize, part of the problem is that I am really not used to MathJax. The integral is in terms of x and yes the limit is as k goes to infinity, and I have derived that identity yes and used it in this problem.
$endgroup$
– mentorship
Dec 2 '15 at 22:02




$begingroup$
I apologize, part of the problem is that I am really not used to MathJax. The integral is in terms of x and yes the limit is as k goes to infinity, and I have derived that identity yes and used it in this problem.
$endgroup$
– mentorship
Dec 2 '15 at 22:02










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Actually the first term evaluated from - infinity to infinity goes to zero without application of the limit since f(x) is absolutely integrable and e^ix is bounded. After that the integral is bounded since f'(x) is absolutely integrable and e^ix is bounded again. Since 1/k will go to zero, the entire limit is 0






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    $begingroup$

    Actually the first term evaluated from - infinity to infinity goes to zero without application of the limit since f(x) is absolutely integrable and e^ix is bounded. After that the integral is bounded since f'(x) is absolutely integrable and e^ix is bounded again. Since 1/k will go to zero, the entire limit is 0






    share|cite|improve this answer









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      0












      $begingroup$

      Actually the first term evaluated from - infinity to infinity goes to zero without application of the limit since f(x) is absolutely integrable and e^ix is bounded. After that the integral is bounded since f'(x) is absolutely integrable and e^ix is bounded again. Since 1/k will go to zero, the entire limit is 0






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Actually the first term evaluated from - infinity to infinity goes to zero without application of the limit since f(x) is absolutely integrable and e^ix is bounded. After that the integral is bounded since f'(x) is absolutely integrable and e^ix is bounded again. Since 1/k will go to zero, the entire limit is 0






        share|cite|improve this answer









        $endgroup$



        Actually the first term evaluated from - infinity to infinity goes to zero without application of the limit since f(x) is absolutely integrable and e^ix is bounded. After that the integral is bounded since f'(x) is absolutely integrable and e^ix is bounded again. Since 1/k will go to zero, the entire limit is 0







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '15 at 23:52









        mentorshipmentorship

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