Are Fourier transforms unique?

  • Thread starter pellman
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  • #1
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If

[tex]\int{g(x)e^{ikx}dx}=\int{h(x)e^{ikx}dx}[/tex]

for all real k, can I conclude that g(x) = h(x) for all real x?

If the answer is yes, please help me see it, if it is not too much trouble.
 

Answers and Replies

  • #2
disregardthat
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By setting k = 0 you know they differ by a constant. Show that this constant is 0.
 
  • #3
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If

[tex]\int{g(x)e^{ikx}dx}=\int{h(x)e^{ikx}dx}[/tex]

for all real k, can I conclude that g(x) = h(x) for all real x?

If the answer is yes, please help me see it, if it is not too much trouble.
You have to rely on Fourier inversion theorem, that you can find in any book that talks about Fourier transforms. In general, two functions (square integrable) that have the same Fourier transform are equal "almost everywhere", that is, everywhere except on a set of points with zero Lesbegue measure.

By setting k = 0 you know they differ by a constant. Show that this constant is 0.
For k = 0 you only know they have the same average value, not that they differ by a constant.
 
  • #4
disregardthat
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For k = 0 you only know they have the same average value, not that they differ by a constant.
My apologies, you are right of course.
 
  • #5
mathman
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Assuming the functions are nice, so that the inverse Fourier transform exists, then g(x)=f(x) almost everywhere. The expression for the inverse transform looks just like the expression for the transform except for a sign reversal in the exponent.

In other words, let H(t) be the Fourier transform of f(x) or g(x), then the inverse transform of H(t) (call it h(x)) equals f(x) (and g(x)) almost everywhere.
 
  • #6
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In general, two functions (square integrable) that have the same Fourier transform are equal "almost everywhere", that is, everywhere except on a set of points with zero Lesbegue measure

Thanks! I will have to read up on Lesbegue measure.
 

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