Fourier transforms of compactly supported functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:33:38Z http://mathoverflow.net/feeds/question/29991 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29991/fourier-transforms-of-compactly-supported-functions Fourier transforms of compactly supported functions Phil Isett 2010-06-30T00:43:22Z 2012-11-20T06:58:14Z <p>One manifestation of the uncertainty principle is the fact that a compactly supported function $f$ cannot have a Fourier transform which vanishes on an open set. As stated, this phenomenon applies when $f$ lives on the integers, or on Euclidean space, but it is false for, say, the p-adic rationals, on which the characteristic function of the p-adic integers provides a counterexample.</p> <p>One normally proves this phenomenon on Euclidean space through complex analysis, although it does follow from the real-variable argument on this post of Tao on <a href="http://terrytao.wordpress.com/2009/02/18/hardys-uncertainty-principle/" rel="nofollow">Hardy's Uncertainty Principle</a>.</p> <p>For $f$ on the integers, it's even easier because only the zero polynomial has infinitely many roots (and this appears to me to be more or less the kind of input which goes into the other proofs as well).</p> <p>What I want to know is whether there is a proof of this phenomenon -- namely, the refusal of a compactly supported functions F.T. to vanish on an open set -- which directly relates to the connectedness of the frequency space, if that is what is behind it.</p> <p>Here is a failed effort in this direction...</p> <p>You know that $f = f \chi$ for any function $\chi$ which is $1$ on the support of $f$. Then $\hat{f}$ should remain unchanged when convolved with the Fourier transform of $\chi$, but since $\hat{f}$ lives on the reals you would like to think that this convolution would partially fill up some open set connected to the boundary of the original support of $\hat{f}$ and thus enlarge the support of $\hat{f}$. This argument goes through as long as you don't get an absurd cancellation; but while $\hat{f}$ may be assumed positive, one has less freedom to renormalize $\chi$.</p> <p>Does anyone know if there is a version of this argument which actually goes through? As stated it's no different than saying "$\hat{f} \ast \hat{\chi} = 0$ whenever $\chi$ lives away from the support of $f$; how weird..." It's possible my intuition here is all wrong and it's really the regularity of $\hat{f}$ which is completely behind the phenomenon, but I am at a loss for other examples of connected, locally compact abelian groups (are there any?) and I don't know how connectedness behaves with respect to Pontrjagin duality.</p> http://mathoverflow.net/questions/29991/fourier-transforms-of-compactly-supported-functions/30924#30924 Answer by kh for Fourier transforms of compactly supported functions kh 2010-07-07T17:32:02Z 2012-11-20T06:58:14Z <p>For $\mathbb{R}$. Suppose f is our compactly supported function and g(x) is its Fourier transform. Since f is compactly supported, $\hat{f} = g$ is the restriction to $\mathbb{R}$ of an entire function g(z) by the Paley-Wiener theorems. Since g is entire and vanishes on an open set, $g \equiv 0$. The proof of this last fact (weakening the assumption to vanishing on a set with an accumulation point) uses that $\mathbb{C}$ is connected which is of course directly related to $\mathbb{R}$ being connected.</p> <p>I expect that you knew this proof, but maybe you accidentally overlooked where connectedness was used. Or more likely, this proof didn't explain what you had in mind and you want a more general proof for $\mathbb{R}^n$. I can't currently do that. Instead, I have another idea which focuses on a different aspect than connectedness, but seems to be related.</p> <p>In connection with the analogous statement for polynomials. A polynomial can only have finitely many zeroes over a field is proved via a complexity argument using that infinity > finite. Analytic functions, i.e. the completion of polynomials over $\mathbb{C}$ can have infinitely many zeroes, but uncountably many zeroes implies the analytic function is identically 0. So it seems that a set that has a limit point is more complex (in terms of complexity) than a countable set. I'm thinking the complexity argument should be interpreted in terms of density in topology - no finite subset of a $\mathbb{N}$ is dense in the discrete topology or any open subset of the co-finite topology on $\mathbb{N}$. Similarly for $\mathbb{R}$ and $\mathbb{C}$.</p> <p>I hope this is helpful. This is an interesting question and I'll think more about it.</p>