Does there exist a continuous function of compact support with Fourier transform outside L^1? Let f be a complex-valued function of one real variable, continuous and compactly supported. Can it have a Fourier transform that is not Lebesgue integrable?
 A: There's an explicit example in dimension 2 or more, i.e.: sqrt(1-|x|^2) on the unit disc. 
This is the Bochner-Riesz multiplier (for delta = 1/2). The divergence of the derivative on the boundary of the disc is exactly what was needed.
A: A counter example of such a function is a sample path W of a Brownian motion on [0, 1]. It is continuous on [0, 1] and yet its Fourier tansform FW(u) decays at infinity with rate less that U^{-1/2} and hence, FW is not in L^1.  
A: Here's a sketch of what I think is an example of the sort you want. Consider a trapezoidal function Tδ, 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 Tδ is going to tend to infinity in the L1(R)-norm -- I can't remember the details of the proof, but since Tδ 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 Tδ 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 Tδ(n) has a Fourier transform with L1-norm equal to n2 3n.
Put Sm = Σj=1m  m-2 Tδ(m) and note that the sequence (Sm) 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 Sm is bounded below by
3m - (3m-1 + ... + 3 + 1) ~ 3m /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 Tδ show this is impossible.
A: I believe the following almost rigorous argument gives a positive answer.
Continuous compactly supported functions are in L^1 and so their Fourier Transform (FT) is bounded. So everything depends on the behaviour at infty. 
Observe that the characteristic function of a bounded interval is not (absolutely) integrable, being "almost" sen(y)/y. 
Since decay at infinity of the FT  reflects regularity of the function, if you strenghten continuity to C^k (k=2 is okay), you get immediately integrability fo the FT. 
The continuous case is intermediate between the discontinuous and regular one.
If you look to a counterexample, it's better looking at non differentiable functions. 
A good choice could be a function smooth apart from a single point x where the "tangent" to the graph is vertical and f(x) is not 0. I believe an application of the stationary phase principle should give a decay y^-1 and so the non integrability of FT(f).
I hope to provide more details later.
A: This is in answer to Jessica and Zen.  Yemon supposes that $\mathcal F$ restricts to a linear map $C_0(-1,1)\rightarrow L^1(\mathbb R)$.  We want this to have a closed graph-- so if $(f_n)\subseteq C_0(-1,1)$ with $f_n\rightarrow 0$ and $\mathcal F(f_n)\rightarrow g\in L^1(\mathbb R)$, why is $g=0$?
Well, I'd be tempted to use that $\mathcal F$ extends to a unitary $L^2(\mathbb R)\rightarrow L^2(\mathbb R)$ (which can be proved in a quite elementary way).  In particular, as $\|f_n\|_2\rightarrow 0$ we know that $\|\mathcal F(f_n)\|_2 \rightarrow 0$.  So if $h$ is a compactly support continuous function, then use the embedding $L^1(\mathbb R)\rightarrow
C_0(\mathbb R)^*$ to see that
$$\int_{\mathbb R} g(s) h(s) \ ds = \langle g, h\rangle_{(L^1(\mathbb R),C_0(\mathbb R))}
= \lim_n \langle \mathcal F(f_n), h\rangle_{(L^1(\mathbb R),C_0(\mathbb R))}
= \lim_n (\mathcal F(f_n)|\overline{h})_{L^2(\mathbb R)} = 0.$$
This shows that $g=0$ in $L^1(\mathbb R)$.
I'm not sure if that's what you're after, but it's the sort of "soft analysis" proof I'd use (that is, uses measure theory, Hilbert space theory, but no distributions etc.)
A: Choose
$$f(x)=
\begin{cases}
\dfrac{\frac12 -x}{\log(x)},&0<x\leq1/2\\
0,&\text{otherwise}
\end{cases}$$
Then $$\operatorname*{Im} \int_{0}^{\infty} dk \ e^{-\varepsilon k}
\int_{-\infty}^{\infty} dx \ f(x) e^{-ikx} = \int_{0}^{\frac12}
dx \ \frac{\frac12 -x}{\log(x)} \cdot
\frac{-x}{\varepsilon^{2}+x^{2}} \to \infty\text{ for }
\varepsilon \downarrow 0.$$
Therefore the Fourier transform of $f$ is not in $L^{1}$ .
