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,

Here's a sketch of what I think is an example of the sort you want. Consider a trapezoidal function T_δ, supported on [-1,1], which is 1 on [-1+δ , 1-δ ] and is defined on the remaining intervals by interpolation in the obvious way. Then as δ tends to zero, the Fourier transform of T_δ is going to tend to infinity in the L¹(R)-norm -- I can't remember the details of the proof, but since T_δ is a linear combination of Fourier transforms of Féjer kernels one can probably do a fairly direct computation.

Of course, the supremum norm of each T_δ 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 L¹-norm.

To be a little more precise: suppose that for each n we can find δ(n) such that T_δ(n) has a Fourier transform with L¹-norm equal to n² 3ⁿ.

Put S_m = Σ_j=1^m m^-2 T_δ(m) and note that the sequence (S_m) converges uniformly to a continuous function S which is supported on [-1,1]. The Fourier transform of S certainly makes sense as an L² function. On the other hand, the L¹-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 L¹-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 L¹, 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 L¹-norm at most M. But the functions T_δ show this is impossible.