- #1

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[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.

- Thread starter pellman
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- #1

- 675

- 4

[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.

- #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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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.

[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.

For k = 0 you only know they have the same average value, not that they differ by a constant.By setting k = 0 you know they differ by a constant. Show that this constant is 0.

- #4

disregardthat

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My apologies, you are right of course.For k = 0 you only know they have the same average value, not that they differ by a constant.

- #5

mathman

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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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