Timeline for Discontinuous convolutions
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
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| Aug 31, 2010 at 20:14 | answer | added | Terry Tao | timeline score: 6 | |
| Aug 17, 2010 at 20:48 | comment | added | Ashutosh | Here's a sort of example I don't want: Let f_n be a smooth approximation to n times the characteristic function of [n, n + (1/n)^3]. Let f be the sum of f_n's where n runs over all positive integers. Let g be an even function which matches f on positive numbers. Then g is an infinitely differentiable L1 function whose convolution with itself at 0 is infinite and therefore discontinuous at 0. But this is not something I want. I want (f * g) to be finite everywhere and continuous nowhere and I don't know how to do this. | |
| Aug 17, 2010 at 20:18 | comment | added | Deane Yang | Don't know how to do the warmup, but here is a possibly silly suggestion. First, identify an $L_1$ smooth function $f$ and a sequence of real numbers $x_1, \dots \rightarrow \infty$ such that the function $\sum_i f(x - x_i)$ is discontinuous. Then $f \star g$ is discontinuous if $g(x) = \sum_i \delta(x - x_i)$. Now try replacing $g$ by a smooth approximation. | |
| Aug 17, 2010 at 18:30 | comment | added | Nate Eldredge | As a warmup, do you know an example where a convolution of integrable smooth functions is discontinuous even at one point? | |
| Aug 17, 2010 at 1:02 | answer | added | Vince | timeline score: 0 | |
| Aug 17, 2010 at 0:34 | comment | added | Yemon Choi | It might help if the original poster added a little more on what kind of examples he or she has already tried. One natural idea, if one is thinking of smooth, integrable but non-square-integrable functions, is to take a smooth compactly supported bump function, consider dilations of it to obtain bump functions which have small support and large height, and then stack together (translates of?) these bumps. But I haven't gone through actual calculations to determine what the convolution square of such an example would look like. | |
| Aug 17, 2010 at 0:30 | comment | added | Yemon Choi | Further to Nate's comment, it seems that it is precisely the bad behaviour at infinity which we either need to control (to prove no such pairs of functions exist) or to "exaggerate" (to give an example of two such functions). It may also be worth observing - apologies if this is known to everyone reading - that the convolution of two integrable and square-integrable functions is continuous and bounded on ${\mathbb R}$ (in fact it lies in $C_0({\mathbb R})$). | |
| Aug 17, 2010 at 0:16 | comment | added | Nate Eldredge | If the derivatives are integrable, the convolution will also be differentiable. So the derivatives would have to behave rather badly at infinity... | |
| Aug 16, 2010 at 23:56 | history | edited | Ashutosh | CC BY-SA 2.5 |
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| Aug 16, 2010 at 23:56 | comment | added | Ashutosh | I'm sorry. I meant nowhere continuous. | |
| Aug 16, 2010 at 23:51 | comment | added | Dylan Wilson | "can be nowhere continuous." I'm confused; are you asking for an examples of two smooth, integrable functions that have a continuous convolution? Because Nate gives an example. Sure you don't mean "nowhere continuous"? | |
| Aug 16, 2010 at 23:15 | history | edited | Ashutosh | CC BY-SA 2.5 |
added 22 characters in body
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| Aug 16, 2010 at 23:10 | comment | added | Ashutosh | The functions have to be in $L^1[\mathbf{R}]$ | |
| Aug 16, 2010 at 22:59 | history | edited | Yemon Choi |
added some tags which I think are relevant
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| Aug 16, 2010 at 22:58 | comment | added | Yemon Choi | First of all, if you are considering the convolution of two non-integrable functions, how are you defining this? | |
| Aug 16, 2010 at 22:24 | history | asked | Ashutosh | CC BY-SA 2.5 |