Is the Fourier transform of a fast-vanishing function continuous?

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Can someone tell me if the continuous Fourier transform of a continuous (and vanishing fast enough ) function is also a continuous function?
 
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I can tell you more: in fact, if [itex]f \in L^{1}(\mathbb R)[/itex] then its Fourier Transform is uniformly continuous.
 
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Delta² said:
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

[tex]\int_{-\infty}^{+\infty} |f(x)|dx[/tex]

is finite (and the integral makes sense).