Timeline for Can we calculate the inner product of a semicontinous function with the Dirac delta function?
Current License: CC BY-SA 3.0
8 events
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| Aug 25, 2012 at 8:31 | comment | added | Anand | Thanks Wolfgang for your example and Timur for your update. :-) | |
| Aug 24, 2012 at 14:47 | history | edited | timur | CC BY-SA 3.0 |
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| Aug 24, 2012 at 14:44 | comment | added | timur | @Wolfgang: Thanks for your comment. So left and right limits exist for a smaller class of functions, e.g. for piecewise continuous functions. In any case, now the OP gave some concrete motivations, it appears that convolution with $\delta$ is what he is after, not $\delta$ itself. | |
| Aug 24, 2012 at 14:28 | comment | added | Wolfgang Loehr | Consider $f(x):=\sin(\frac1x)$ for $x\ne0$, $f(0):=-1$. Then clearly $f$ is lower semi-continuous but has neither a left nor a right limit at $0$. | |
| Aug 24, 2012 at 11:14 | comment | added | Anand | Dear Wolfgang Loehr, could you please clarify a bit your comments? So you don't agree what timur said? Thanks a lot. | |
| Aug 24, 2012 at 9:49 | comment | added | Wolfgang Loehr |
A semi-continuous function needs to have neither left nor right limits (though it is continuous on a dense $\mathcal{G}_\delta$-set).
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| Aug 23, 2012 at 21:45 | comment | added | Anand | Dear Timur, thanks for your answer. I have the same thinks as you. But do you know some references that I don't need to invent or care about everything? Thanks a lot! | |
| Aug 23, 2012 at 21:40 | history | answered | timur | CC BY-SA 3.0 |