Fourier transform of $f^{n}$

Question

It is true that if a function $f$ has fourier transform $\hat{f}$ then, $(\hat{f})^n = FT(f^{n})$ (where $FT$ denotes the fourier transform). If so, why?

Answer 1

In general, no. The Fourier Transform of a product of functions is the convolution of their Fourier Transforms, e.g.

$$\widehat{(fg)}(x) = (\hat{f}* \hat{g})(x) = \int_{-\infty}^{\infty} \hat{f}(x-y) \hat{g}(y) \, \mathrm{d} y. $$

Showing this is a play with the definition of the Fourier Transform, and a good exercise to do to convince yourself.