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?
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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. |
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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. |
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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. |
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