Let f be a complexvalued function of one real variable, continuous and compactly supported. Can it have a Fourier transform that is not Lebesgue integrable?
6 Answers
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 , 1δ ] 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 L^{1}(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 L^{1}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 L^{1}norm equal to n^{2} 3^{n}.
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 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^{m1} + ... + 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 normone function in C has L^{1}norm at most M. But the functions T_{δ} show this is impossible.

$\begingroup$ I like this answer. It almost surely leads to the right answer (haven't done the calculation myself), but leaves enough to be done that the querent actually learns something from carrying out the steps. And moreover, if the specific example is not quite correct, there is enough of the "this is how I thought of it" that a modicum of additional ponderance should be able to adapt it to a correct example. $\endgroup$ Nov 2, 2009 at 13:38

$\begingroup$ Thanks for the kind words, although I have to admit that the gaps were not so much left for pedagogic reasons, as because my brain was too fried to fill them all in. I should probably have linked to some stuff on BanachSteinhaus/Uniform Boundedness, since that's really what's going on here. $\endgroup$ Nov 2, 2009 at 21:39

$\begingroup$ Many thanks for your nice example and explanation. I haven't succeeded in proving that the Fourier transform of S has infinite L^1norm, though, so I don't have an explicit example yet. But your open mapping argument suffices to show the existence of this type of functions. $\endgroup$ Nov 15, 2009 at 16:43

$\begingroup$ In the answer of Yemon, the theoretical one, going by contradiction, how do you prove that the graph is closed in order to get the function continuous? $\endgroup$– jessicaApr 19, 2011 at 7:06

$\begingroup$ I think if you allow use of Distribution Theory, a.k.a. Generalised Functions, i.e. dirac deltas and all that stuff, using Fourier transforms of tempered distributions, then it's trivial to see it's closed. Convergence in $C$ (for $f$) or in $L^1$ (for the Fourier transform $\widehat{f}$) implies convergence as tempered distributions; and, the Fourier transform IS continuous on the tempered distributions; hence if $f_n \to 0$ in $C$ and $\widehat{f_n} \to g$ in $L^1$, then $g=0$ as a distribution, therefore also as an element of $L^1$. $\endgroup$ Apr 19, 2011 at 8:50
There's an explicit example in dimension 2 or more, i.e.: sqrt(1x^2) on the unit disc. This is the BochnerRiesz multiplier (for delta = 1/2). The divergence of the derivative on the boundary of the disc is exactly what was needed.

$\begingroup$ Is sqrt(1x^2) a counterexample in one dimension? (I tried computing it explicitly, but without success.) $\endgroup$ Nov 4, 2009 at 16:54

$\begingroup$ I don't know if you can compute it explicitly, but trying to substitute x=Sin(y) in the Fourier integral gives you the oscillatory integral \int Cos^2(y) e^{i\pi z Sin(y)}dy. The general principles of oscillatory integrals would say to look at the critical points of the phase Sin(y). Since there Sin vanishes to first order one would expect a decay z^{1/2}, but unfortunately the coefficient Cos^2(y) vanishes (to second order), accelerating the decay for large values of z. $\endgroup$ Nov 5, 2009 at 9:56

2$\begingroup$ The Fourier transform of (1x^2)_+^a (a>0) in one dimension is given by a 0F1 hypergeometric function (Mathematica computes it exactly). It's behaviour at ∞ is y^{(1+a)}sin(π a/2y), so that it is integrable. Maybe one should look at (1log(1x^2)_+)^{1}, or (1log(1+log(1x^2)))^{1}, or ... $\endgroup$ Nov 6, 2009 at 17:48
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.
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 $L^{1}$ .
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.
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.)

$\begingroup$ Hmm, hadn't thought of that approach. I've just bashed out a sketch of what I think was my intended argument, am leaving it as a comment to Jessica's question. $\endgroup$ Apr 20, 2011 at 8:23

$\begingroup$ @Yemon: Yeah, I think I agree with your argument, which is indeed easier than mine! $\endgroup$ Apr 20, 2011 at 9:43