Here's a sketch of what I think is an example of the sort you want. Consider a trapezoidal function T<sub>&delta;</sub>, supported on [-1,1], which is 1 on [-1+&delta; , 1-&delta; ] and is defined on the remaining intervals by interpolation in the obvious way. Then as &delta; tends to zero, the Fourier transform of T<sub>&delta;</sub> is going to tend to infinity in the L<sup>1</sup>(R)-norm -- I can't remember the details of the proof, but since T<sub>&delta;</sub> is a linear combination of Fourier transforms of F&eacute;jer kernels one can probably do a fairly direct computation.

Of course, the supremum norm of each T<sub>&delta;</sub> is always 1. So the idea is to now stack scaled copies of these together, so as to obtain a function on [-1,1] which will be continuous (by uniform convergence) but whose Fourier transform is not integrable because its the limit of things with increasing L<sup>1</sup>-norm.

To be a little more precise: suppose that for each n we can find &delta;(n) such that T<sub>&delta;(n)</sub> has a Fourier transform with L<sup>1</sup>-norm equal to n<sup>2</sup> 3<sup>n</sup>.

Put S<sub>m</sub> = &Sigma;<sub>j=1</sub><sup>m</sup>  m<sup>-2</sup> T<sub>&delta;(m)</sub> and note that the sequence (S<sub>m</sub>) converges uniformly to a continuous function S which is supported on [-1,1]. The Fourier transform of S certainly makes sense as an L<sup>2</sup> function. On the other hand, the L<sup>1</sup>-norm of the Fourier transform of S<sub>m</sub> is bounded below by

3<sup>m</sup> - (3<sup>m-1</sup> + ... + 3 + 1) ~ 3<sup>m</sup> /2  

which suggests that the Fourier transform of S ought to have infinite L<sup>1</sup>-norm -- at the moment lack of sleep prevents me from remembering how to finish this off.


Alternatively, one could argue as follows. Consider the Banach space C of all continuous functions on [-1,1] which vanish at the endpoints, equipped with the supremum norm. If the Fourier transform mapped C into L<sup>1</sup>, then by an application of the closed graph theorem it would have to do so continuously, and hence boundedly. That means there would exists a constant M &gt;0, such that the Fourier transform of every norm-one function in C has L<sup>1</sup>-norm at most M. But the functions T<sub>&delta;</sub> show this is impossible.