Timeline for Splines, harmonic analysis, singular integrals.
Current License: CC BY-SA 2.5
28 events
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Jul 23, 2010 at 12:48 | vote | accept | Olumide | ||
Jul 23, 2010 at 12:44 | answer | added | Olumide | timeline score: 1 | |
Jul 16, 2010 at 15:56 | comment | added | Zen Harper | Forget about the rigour as a first step! Understand the formal algebraic calculation fully, then try to make it rigorous! I haven't read the paper, but $$ \frac{|\omega|^{2p+2}}{(1+|\omega|^2)^{p+1}}f(\omega) $$ has the same asymptotic behaviour as $f$ for $|\omega| \to \infty$, but also no singularity at $0$ (using zeros to destroy poles by multiplication). Presumably the reason for choosing this particular function, rather than some other function with similar properties, is that the algebra is not too hard when you take the Fourier transform (or whatever else is being done in the paper). | |
Jul 15, 2010 at 13:13 | comment | added | Olumide | It was suggested by someone, who I discussed this paper with a while back, that the purpose of the multiplying factors $$\frac{|\omega|^{2p + 2}}{ (1 + |\omega|^2)^{p+1}}$$ in both expressions is to cause the either expression to "drop off sufficiently fast that it has a finite integral ... then take the limit as the drop-off is reduced so that the weighting as less and less effect." Is this so? | |
Jul 11, 2010 at 11:56 | comment | added | Willie Wong | One last remark, the whole point of the exercise is that suppose you are interested in $\int fg dx$. This integral makes sense if the product $fg$ is integrable. One way to guarantee this is to ask both $f$ and $g$ to be square integrable. But there's no reason why $f$ and $g$ have to have equal footing: you can make $f$ bad at some places and compensate by making $g$ good at those places. So if $f$ grows polynomially, the integral will make sense if $g$ decays polynomially. It is just a simple give and take. | |
Jul 11, 2010 at 11:51 | comment | added | Willie Wong | ... especially addition, subtraction, taking derivatives and Fourier transforms. So for the most part you can treat them like normal functions. Except that you can only evaluate them in an integral sense by testing against a Schwartz function. | |
Jul 11, 2010 at 11:48 | comment | added | Willie Wong | Let me put it this way: if you intend to make new theories and construct new theorems, then you do need to understand the intricacies of the details of generalized functions, especially what you are allowed to do and what you are not. If you intend to be a user of results, or are just surveying the literature, all that suffices you know is that for tempered distributions (a subset of generalized functions which grows at most polynomially near infinity; sort of opposite of Schwartz functions), many things that are defined for functions can be done for them... | |
Jul 11, 2010 at 11:44 | comment | added | Willie Wong | Now of course in general the statement won't make sense if you relax $f$. But if you are only interested in integrating it against functions of rapid decrease (Schwartz type), the polynomial blow-up at infinity (2.7) is no biggie. To deal with the inverse polynomial blow-up at the origin, one has to add vanishing moment conditions (in Fourier space) on the function you are testing against, so that the test function is $o(|x|^m)$ at the origin and cancels out the singularity. | |
Jul 11, 2010 at 11:36 | comment | added | Willie Wong | I am afraid I don't understand your question. Equations 2.5 and 2.7 in the linked paper are statements/assumptions for the paper. Basically the authors said: a certain expression (2.4) is trivial under Plancherel/Parsevel for functions $f$ that are sufficiently nice. Can we make $f$ not as nice? Before one of the conditions for $f$ to be nice is that it is $L^1(\mathbb{R})$. As it turns out that with a bit of work $f$ can be relaxed to have singularities at the origin (2.5) and at infinity (2.7). The mathematical justification for the two statements are thus: that is what they can prove. | |
Jul 11, 2010 at 5:03 | comment | added | Yemon Choi | I agree with Willie. I think that this question is really starting to drift unless a specific question can be extracted (such as "what is the justification for the following step?") from this. | |
Jul 11, 2010 at 4:01 | history | edited | Will Jagy | CC BY-SA 2.5 |
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Jul 11, 2010 at 1:59 | comment | added | Olumide | My primary objective is to understand the paper by Kent and Mardia. But in order to begin to do so, I think need to become familiar with the mathematics they use which I "assumed" was from the area of generalized functions and specifically Schwartz spaces, which is why I've been reading up on both subjects. But it would be helpful if someone would kindly confirm my "suspicions" and point me in the right direction. I apologize for any confusion I may have caused. | |
Jul 10, 2010 at 23:40 | history | edited | Will Jagy | CC BY-SA 2.5 |
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Jul 10, 2010 at 22:57 | comment | added | Willie Wong | Somewhere below in the comments ( mathoverflow.net/questions/31308/… ) the OP gave a link to the paper he is looking at. Somehow I feel like this rather open-ended question here is getting nowhere. I think it may be more productive if the OP points out (in perhaps a new question) the precise statement that is giving him/her trouble. That way he/she is likely to get a more focused and to the point response. Just my 2 pence. | |
Jul 10, 2010 at 21:45 | answer | added | Peter Luthy | timeline score: 3 | |
Jul 10, 2010 at 20:15 | history | edited | Will Jagy | CC BY-SA 2.5 |
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Jul 10, 2010 at 18:41 | answer | added | Zen Harper | timeline score: 5 | |
Jul 10, 2010 at 18:18 | answer | added | Helge | timeline score: 3 | |
Jul 10, 2010 at 18:08 | history | edited | Olumide | CC BY-SA 2.5 |
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Jul 10, 2010 at 18:07 | answer | added | Olumide | timeline score: 1 | |
Jul 10, 2010 at 18:01 | history | edited | Olumide | CC BY-SA 2.5 |
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Jul 10, 2010 at 17:48 | history | edited | Charles Matthews | CC BY-SA 2.5 |
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Jul 10, 2010 at 17:27 | answer | added | Will Jagy | timeline score: 4 | |
Jul 10, 2010 at 17:24 | answer | added | Willie Wong | timeline score: 8 | |
Jul 10, 2010 at 17:23 | comment | added | Mariano Suárez-Álvarez | I would say that «because with that condition things work as one wants them to» in some contexts. Notice that there are other spaces of test functions which are useful (and in fact, generalized functions are most generally introduced using not the ones you mention but $C^\infty$ functions of compact support) | |
Jul 10, 2010 at 17:20 | history | edited | mathphysicist |
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Jul 10, 2010 at 17:19 | comment | added | Qiaochu Yuan | The faster a function decays, the more functions you can integrate it against. | |
Jul 10, 2010 at 17:14 | history | asked | Olumide | CC BY-SA 2.5 |