Timeline for Does there exist a continuous function of compact support with Fourier transform outside L^1?
Current License: CC BY-SA 2.5
16 events
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| Apr 20, 2011 at 8:27 | comment | added | Yemon Choi | ... Now if we have a sequence in $C_b({\mathbb R})\cap L^1({\mathbb R})$, which converges to zero in the uniform norm and converges to $g$ in the $L^1$-norm, then by basic properties of the Lebesgue integral, $\int_E g = 0$ for any Lebesgue measurable set $E$ that has finite measure. This quickly implies that $g=0$ a.e. | |
| Apr 20, 2011 at 8:26 | comment | added | Yemon Choi | I think my original train of thought was something like the following: Zen, if you are reading, please feel free to correct things! Since $(f_n)$ is supported on $[-1,1]$ and converges uniformly to zero, ${\mathcal F}(f_n)$ converges uniformly to zero in $C_b({\mathbb R})$. (The Fourier transforms are all continuous, I think this can be shown directly, if one again uses the fact that the $(f_n)$ are all supported in $[-1,1]$.) ... | |
| Apr 19, 2011 at 9:08 | comment | added | Zen Harper | Ahh, stupid, me again! - of course $\int_0^t f(x) dx$ is continuous in $t$, using Lebesgue integrals, so a.e. $t$ is equivalent to all $t$. | |
| Apr 19, 2011 at 9:05 | comment | added | Zen Harper | I gave a lecture course years ago which almost reversed the process - one way to prove "classical" Fourier inversion, assuming that $f, \widehat{f} \in L^1$, uses a Lebesgue point of $f$ and Gaussian mollifiers! The inverse transform of $\widehat{f}$ is continuous, and agrees with $f$ at all Lebesgue points, so therefore a.e. I'm sure the experts know all this already, but I didn't and was very pleased with myself at working this all out in detail. But I suppose I should shut up now! | |
| Apr 19, 2011 at 8:55 | comment | added | Zen Harper | However, this is a good question, Jessica. Actually, things are not so simple; what's REALLY going on is that all the "hard" (i.e., non-soft) analysis and measure theory is encoded inside the statements that $C$ and $L^1$ are continuously embedded inside the distributions. For $C$, one easy way is to use $C^\infty$ mollifiers, with Gaussians. For $L^1$, this is not trivial - as far as I know, it needs the statement that $\int_0^t f(x) dx \equiv 0$ a.e. $\implies$ $f \equiv 0$ a.e., which is a special case of the Lebesgue differentiation theorem. | |
| Apr 19, 2011 at 8:50 | comment | added | Zen Harper | 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 $L^1$. | |
| Apr 19, 2011 at 7:06 | comment | added | jessica | 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? | |
| Dec 12, 2009 at 14:05 | vote | accept | Patrik Wahlberg | ||
| Nov 15, 2009 at 16:43 | comment | added | Patrik Wahlberg | Many thanks for your nice example and explanation. I haven't succeeded in proving that the Fourier transform of S has infinite L^1-norm, 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. | |
| Nov 5, 2009 at 7:55 | vote | accept | Patrik Wahlberg | ||
| Dec 12, 2009 at 14:05 | |||||
| Nov 2, 2009 at 21:39 | comment | added | Yemon Choi | 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 Banach-Steinhaus/Uniform Boundedness, since that's really what's going on here. | |
| Nov 2, 2009 at 20:53 | history | edited | Yemon Choi | CC BY-SA 2.5 |
fixed formatting and corrected a typo
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| Nov 2, 2009 at 14:38 | vote | accept | Patrik Wahlberg | ||
| Nov 2, 2009 at 14:38 | |||||
| Nov 2, 2009 at 14:38 | vote | accept | Patrik Wahlberg | ||
| Nov 2, 2009 at 14:38 | |||||
| Nov 2, 2009 at 13:38 | comment | added | Andrew Stacey | 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. | |
| Nov 2, 2009 at 12:36 | history | answered | Yemon Choi | CC BY-SA 2.5 |