Where F is Fourier Transform, prove the limit as |k| approaches infinity of Ff(k)=0 where f is differentiable...
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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
edited Jan 5 at 12:06
honey.mustard
8619
asked Dec 2 '15 at 21:18
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1088
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add a comment |
$begingroup$
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
edited Jan 5 at 12:06
honey.mustard
8619
asked Dec 2 '15 at 21:18
mentorshipmentorship
1088
$endgroup$
$begingroup$
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.
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– 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
add a comment |
$begingroup$
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
edited Jan 5 at 12:06
honey.mustard
8619
asked Dec 2 '15 at 21:18
mentorshipmentorship
1088
$endgroup$
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
edited Jan 5 at 12:06
honey.mustard
8619
asked Dec 2 '15 at 21:18
mentorshipmentorship
1088
edited Jan 5 at 12:06
honey.mustard
8619
asked Dec 2 '15 at 21:18
mentorshipmentorship
1088
edited Jan 5 at 12:06
honey.mustard
8619
edited Jan 5 at 12:06
honey.mustard
8619
edited Jan 5 at 12:06
honey.mustard
8619
8619
asked Dec 2 '15 at 21:18
mentorshipmentorship
1088
asked Dec 2 '15 at 21:18
mentorshipmentorship
1088
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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
answered Dec 3 '15 at 23:52
mentorshipmentorship
1088
$endgroup$
add a comment |
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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
answered Dec 3 '15 at 23:52
mentorshipmentorship
1088
$endgroup$
add a comment |
$begingroup$
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
answered Dec 3 '15 at 23:52
mentorshipmentorship
1088
$endgroup$
add a comment |
$begingroup$
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
answered Dec 3 '15 at 23:52
mentorshipmentorship
1088
$endgroup$
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
answered Dec 3 '15 at 23:52
mentorshipmentorship
1088
answered Dec 3 '15 at 23:52
mentorshipmentorship
1088
answered Dec 3 '15 at 23:52
mentorshipmentorship
1088
answered Dec 3 '15 at 23:52
mentorshipmentorship
1088
1088
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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