# Duality and Fourier Transforms [closed]

$(FF(f))(x)=2\pi f(-x)$, where $F$ is the Fourier transform

and $F(f(x-a))(k)=\exp(-ika) X(k)$ where $X(k)=F(f(x))$

implies $F(\exp(iax)f(x))(k)=X(k-a)$.

But I don't see how that is done... I am quite happy with getting $F^{-1}X(k-a)=\exp(iax)f(x)$ by brute force calculation. I would like to see how to use duality though.

-
math.stackexchange.com may be! –  MBN Nov 15 '11 at 11:58
@MBN: Okay, thanks. –  paul Nov 15 '11 at 12:10
Now posted at math.stackexchange.com/q/82339/11127 –  Qmechanic Nov 15 '11 at 18:11

## closed as off topic by Colin K, Qmechanic♦, David Z♦Nov 15 '11 at 21:02

Questions on Physics Stack Exchange are expected to relate to physics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

You need to know the basic Fourier transform delta-function identity

$$\int_{-\infty}^{\infty} e^{ikx} {dk\over 2\pi} = \delta(x)$$

Which implies Fourier inversion. Proving this identity is slightly subtle, because the right hand side is a distribution, but you can do the integral explicitly over a long interval from -M to M to get an object which has a unit integral and is shrinking in size with M as 1/M, so it must be a delta-function in any reasonable sense of limits.

The double fourier-transform is

$$FF(f) (x') = \int dk e^{ix'k} (\int dk e^{ikx} f(x))$$

And you can do the k integral using the identity to get the result.

-