From: Thorsten Raasch Subject: Re: Fourier transform problem Date: 3 Aug 2001 13:30:28 +0200 Newsgroups: sci.math Summary: Comparing domain and range of Fourier Transform Whigdon2 wrote: >>From Thorsten Raasch >>Hi! >> >>I've got a simple problem concerning Fourier transformation: >>Is it true that the Fourier transform \widehat f of a continous >>L^2-integrable function f is already in L^1? >>If not, is there a simple counterexample? >> > Simple observe that there are many really nice L^2 functions which aren't in > L^1. Hope this helps. - Bill But that has nothing to do with my problem! The Fourier transform is an isometry of L^2(\mathbb R^n) onto itself. One way to construct it is setting \widehat f(\xi)=\int_{-\infty}^\infty f(t)e^{-i(x,\xi)}dt for functions f belonging to L^1. One can easily show that \widehat f is continuous. By restricting the transformation to L^1\cap L^2, one can prove it being an isometry. A density argument shows that the Fourier transform is an L^2-isometry. My problem is: Under what circumstances does \widehat f belong to L^1\cap L^2 ? One could consider the characteristic function \chi_{[0,1]} of the unit interval, surely belonging to every L^p. Its Fourier transform is \sin(\xi)/\xi, up to a constant factor. This function does not belong to L^1, but to L^2. Is this due to the fact that f lacks continuity? Best regards, -- Thorsten Raasch eMail: raasch@math.uni-siegen.de Homepage: http://www.math.uni-siegen.de/~raasch/ ============================================================================== From: israel@math.ubc.ca (Robert Israel) Subject: Re: Fourier transform problem Date: 3 Aug 2001 20:31:08 GMT Newsgroups: sci.math In article <3b6a8b54@si-nic.hrz.uni-siegen.de>, Thorsten Raasch wrote: >Whigdon2 wrote: >>>From Thorsten Raasch >>>I've got a simple problem concerning Fourier transformation: >>>Is it true that the Fourier transform \widehat f of a continous >>>L^2-integrable function f is already in L^1? >>>If not, is there a simple counterexample? >> Simple observe that there are many really nice L^2 functions which aren't in >> L^1. Hope this helps. - Bill >But that has nothing to do with my problem! Depends on your definition of "really nice" I guess. If f_n is the convolution of the indicator functions of the intervals [-n,n] and and [-1,1], then (f_n)^ is a constant times sin(x) sin(n x)/x^2. Note that f_n is continuous and uniformly bounded, but the L_1 norm of (f_n)^ goes to infinity as n -> infinity. In fact, if n >= 8 and (k+1/4)pi/n < x < (k+3/4)pi/n for 0 <= k < n/2 we have sin x > x/2, |sin(n x)| > 1/sqrt(2), |(f_n)^(x)| > const. n/(k+3/4), and thus ||(f_n)^||_1 >= c ln(n) for some positive constant c. To get your counterexample, it should suffice to take a series f = sum_j a_j f_(n_j) where sum_j |a_j| converges and n_j -> infinity rapidly enough. Then if (k+1/4)pi/n_j < x < (k+3/4)pi/n_j for 0 <= k <= sqrt(n_j) we have (as before) |(f_(n_j))^(x)| > const. sin(x)/x^2 > const. n_j/(k+3/4). for i < j, |(f_(n_i))^(x)| < const. sin(x) (n_i x)/x^2 < const. n_i. for i > j, |(f_(n_i))^(x)| < const. sin(x)/x^2 < const. n_j/(k+3/4). If a_j = 1/a^j where a is sufficiently large (depending on the constants), we will have sum_{i>j} |(f_(n_i))^(x)| < 1/3 |(f_(n_j))^(x)|. If n_j grows rapidly enough that a_j sqrt(n_j) > const. sum_{i= 1/3 |(f_(n_j))^(x)| > const. n_j/(k+3/4) on this interval, and we get ||f^||_1 >= const. sum_{k=0}^floor(sqrt(n_j)) 1/(k+3/4) >= const. ln(n_j) -> infinity Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ============================================================================== From: Stephen Montgomery-Smith Subject: Re: Fourier transform problem Date: Fri, 03 Aug 2001 13:05:55 GMT Newsgroups: sci.math Thorsten Raasch wrote: > > Hi! > > I've got a simple problem concerning Fourier transformation: > Is it true that the Fourier transform \widehat f of a continous > L^2-integrable function f is already in L^1? > If not, is there a simple counterexample? > Here is an idea. Consider the sequence of functions f_n: f_n(-1) = f(1) = 0 f_n(-1+1/n) = f_n(1-1/n) = 1 piecewise linear in between these points. Then ||hat f_n||_1 is probably something like log(n), or at least something that diverges. Let a_n be a sequence such that sum a_n converges, but sum a_n log(n) diverges. Then try f = sum a_n f_n Well maybe it will work. Or you could use some functional analysis and try to argue using closed graph theorem that if your assertion is true, then the map that takes f from L^2 cap C to L^1 would have to be a bounded map, and so the above sequence provides a counterexample. Well just ideas. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Fourier transform problem Date: Fri, 03 Aug 2001 14:59:37 GMT Newsgroups: sci.math On Fri, 03 Aug 2001 13:05:55 GMT, Stephen Montgomery-Smith wrote: >Thorsten Raasch wrote: >> >> Hi! >> >> I've got a simple problem concerning Fourier transformation: >> Is it true that the Fourier transform \widehat f of a continous >> L^2-integrable function f is already in L^1? >> If not, is there a simple counterexample? >> > >Here is an idea. Consider the sequence of functions f_n: >f_n(-1) = f(1) = 0 >f_n(-1+1/n) = f_n(1-1/n) = 1 >piecewise linear in between these points. > >Then ||hat f_n||_1 is probably something like log(n), or at least >something that diverges. Let a_n be a sequence such that sum a_n >converges, but sum a_n log(n) diverges. Then try > >f = sum a_n f_n > >Well maybe it will work. There's probably a very simple and very explicit counterexample, but I like this sort of thing better, because it explains more why the example exists. Whether the details above are exactly correct doesn't matter much; there are certainly continuous f_n which have L2 norm <= 1 and supremum <= 1, such that the L1 norm of the FT is unbounded; now the appropriate choice of a_n gives an example as above (forget the "log(n)" part and insert "||f_n^||_1" where it appears...) >Or you could use some functional analysis and try to argue using closed >graph theorem that if your assertion is true, then the map that takes f >from L^2 cap C to L^1 would have to be a bounded map, and so the above >sequence provides a counterexample. Or that. >Well just ideas. > >-- >Stephen Montgomery-Smith >stephen@math.missouri.edu >http://www.math.missouri.edu/~stephen David C. Ullrich ============================================================================== From: "David Petry" Subject: Re: Fourier transform problem Date: Fri, 3 Aug 2001 11:21:12 -0700 Newsgroups: sci.math >Thorsten Raasch wrote: >> Is it true that the Fourier transform \widehat f of a continous >> L^2-integrable function f is already in L^1? >> If not, is there a simple counterexample? The Fourier Transform of any L^1 function must be bounded, so if you have a continuous L^2 function which is not bounded, it's FT must not be in L^1. (Recall that the FT is almost its own inverse). So, a simple example might be: x^2 * |sin(x)|^exp(x^2) ============================================================================== From: artfldodgr@aol.com (ArtflDodgr) Subject: Re: Fourier transform problem Date: 3 Aug 2001 14:28:50 -0700 Newsgroups: sci.math Thorsten Raasch wrote in message news:<3b6a68a5@si-nic.hrz.uni-siegen.de>... > Hi! > > I've got a simple problem concerning Fourier transformation: > Is it true that the Fourier transform \widehat f of a continous > L^2-integrable function f is already in L^1? > If not, is there a simple counterexample? If \widehat f is integrable, then (by Fourier inversion) f is BOUNDED and continuous. So, for a counterexample, take f to be continuous and square integrable, but unbounded. For example, take f to be zero, except for tent shaped spikes at each positive integer, the spike at n being an isosceles triangle of height n and base n^{-4}. (The tents sit on the x-axis and point upward.) If you strengthen the requirement that f be continuous, by insisting that f be Holder continuous (or, more generally, possess a uniform modulus of continuity) then the square integrability of f implies that f must be bounded. So a counterexample with f Holder continuous does not come so cheaply. A quick calculation indicates that f(x) = min(1, |x|^{-p}) (where p is slightly larger than 1/2) is a Holder continuous square-integrable f with \widehat f not integrable. -- A.