Timeline for Is finding the CDF from the Laplace transform well-posed?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 27 at 9:54 | comment | added | Riemann | Thank you so much! | |
| Oct 27 at 9:53 | history | bounty ended | Riemann | ||
| Oct 27 at 9:53 | vote | accept | Riemann | ||
| Oct 27 at 9:26 | comment | added | Riemann | You're right, what I called the 'Laplace transform' is actually a restriction of the Laplace transform to a finite real interval. I should have emphasized this in my question. The reason for using this particular definition is that in Dynamic Light Scattering, we can only measure the Laplace transform for real inputs. | |
| Oct 27 at 5:13 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 27 at 5:02 | comment | added | Iosif Pinelis | @Riemann : I have now presented a different kind of argument, for the same "yes" answer. | |
| Oct 27 at 5:01 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 27 at 1:08 | comment | added | Iosif Pinelis | @Riemann : The standard definition of the Laplace transform is for all complex arguments $s$ with $\Re s>0$. Your definition, which I noticed only now, is very strange, for two additional reasons: (i) your "Laplace transform" is defined only on a finite interval and (ii) the co-domain must be the same as the domain. Do you want to say anything about this? | |
| Oct 25 at 18:11 | comment | added | Riemann | However, there is one thing remaining to verify. We know that $L(f_{n,\epsilon})(s)$ converges to $L(f_{\epsilon})(s)$ on real inputs $s\in [0,T]$. Why does it follow that the same holds for complex inputs $s+it$? | |
| Oct 25 at 18:09 | comment | added | Riemann | So we can use Mellin's inverse formula and Lebesgue's dominated convergence theorem to prove that $f_{n,\epsilon} \rightarrow f_{\epsilon}$. | |
| Oct 25 at 17:58 | comment | added | Riemann | Ok I checked and $|L(f_n)(s+it)|\leq1$ for all s and t, while $|L(\psi_{\epsilon})(s+it)|=e^{0.5s^2\epsilon^2-0.5t^2\epsilon^2}$ which is $O(e^{-0.5t^2\epsilon^2})$. | |
| Oct 25 at 17:50 | comment | added | Iosif Pinelis | @Riemann: Because $L(f_{n,\epsilon})(s+it)$ for real $s,t$ decreases (very) fast in $|t|$, whereas in general we cannot say this concerning $L(f_{n})(s+it)$. | |
| Oct 25 at 17:42 | comment | added | Riemann | Thanks for your answer! I get a bit lost where you say 'so, inversing the Laplace transform,'. Why can we do this for $f_{n,\epsilon}$ while it is not allowed for $f_n$? | |
| Oct 24 at 22:02 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |