Fourier transformation of bump functions

Question

Suppose we have $\Omega\subset \mathbb{R} $ open and we are looking at the space $ C_c^\infty(\Omega):=\{\varphi\in C^\infty(\Omega) \ | \ \text{supp($\varphi$) is a compact subset of $\Omega$}\} $. Now let $\varphi\in C_c^\infty(\Omega)$. Is the Fouriertransformation $(\mathcal{F}\varphi)(\xi) $ in general an $L^1 $-function?

Answer 1

Any $f \in C^{\infty}_c(\Omega)$ can be completed to be $f \in C^{\infty}_c(\mathbb R)$ by just setting it to zero outside of $\Omega$, so the Fourier transform makes sense. Then $C^{\infty}_c(\mathbb R)$ is a subset of a more general space of Schwartz functions. This class of functions has the nice property that the Fourier transform of a Schwartz function is a Schwartz function. In particular all Schwartz functions are $L^1$ so the result you want does hold.

If you want to prove this yourself, a hint is that the Fourier transform exchanges decay and regularity. So to prove $\mathcal F f \in L^1$, you need to use smoothness of $f$ to show $\mathcal Ff$ has sufficient decay.