# Fourier transform of f \in C_c^\infty(R^n)

1. May 30, 2009

### chub

I have noticed that this result is hinted at in several books, but am having trouble proving it:

$$f, \hat{f} \in C_c^\infty(R^n) \Rightarrow f \equiv 0.$$

in other words, if both f and its fourier transform
are smooth, compactly supported functions on n-dimensional euclidean space
then f is identically zero.

any advice? i thought of using the fourier inversion theorem, which tells me that f agrees almost everywhere (Lebesgue) with the continuous function
$$f_0 = (\hat{f})\check{} = (\check{f})\hat{}$$
and then showing that one (or both) of those are zero; continuity of f would then take care of the "almost everywhere" part. but i'm not really sure what to do.

2. May 31, 2009

### maze

Ok here is one proof I found (g in C0): since exp(-i k x) is entire, the reimann sums approximating F[g] are entire. The reimann sums must converge uniformly on compact sets since g is in C0, so F[g] is entire. Since entire functions don't have compact support, F[g] can't have compact support.

I don't really like that proof very much.