Here's a sketch of what I think is an example of the sort you want. Consider a trapezoidal function 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 this function *T_\delta* is going to tend to infinity in the L^1(R)-norm -- I can't remember the details of the proof, but since *T_\delta* is a linear combination of Fourier transforms of Fejer kernels one can probably do a fairly direct computation.

Of course, the supremum norm of each *T_\delta* 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^1-norm.

To be a little more precise: suppose that for each n we can find \delta(n) such that *T_\delta(n)* has a Fourier transform with L^1-norm equal to n^2 3^n.

Put ***S_m = \sum_{j=1}^m m^{-2} T_\delta(m)*** and note that the sequence *(S_m)* converges uniformly to a continuous function S which is supported on [0,1]. The Fourier transform of S certainly makes sense as an L^2 function. On the other hand, the L^1-norm of the Fourier transform of S_m is bounded below by

3^m - (3^{m-1}+ ... + 3+ 1) ~ 3^m /2  

which suggests that the Fourier transform of S ought to have infinite L^1-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^1, 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>0, such that the Fourier transform of every norm-one function in C has L^1-norm at most M. But the functions *T_\delta* show this is impossible,