Continuous Fourier Transform of Vanishing Fast Functions: Explained

[Delta2]

Can someone tell me if the continuous Fourier transform of a continuous (and vanishing fast enough ) function is also a continuous function?

 

[Xiuh]

I can tell you more: in fact, if $f \in L^{1}(\mathbb R)$ then its Fourier Transform is uniformly continuous.

 

[Delta2]

Thanks very much but can u ... remind me which functions belong to L1(R)?

 

[micromass]

Thanks very much but can u ... remind me which functions belong to L1(R)?

It are all the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ which are absolutely integrable. That is, for which

$$\int_{-\infty}^{+\infty} |f(x)|dx$$

is finite (and the integral makes sense).

 

[Delta2]

Thx again.