# Are Fourier transforms unique?

If

$$\int{g(x)e^{ikx}dx}=\int{h(x)e^{ikx}dx}$$

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

disregardthat
By setting k = 0 you know they differ by a constant. Show that this constant is 0.

If

$$\int{g(x)e^{ikx}dx}=\int{h(x)e^{ikx}dx}$$

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

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.

disregardthat
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.

mathman