Timeline for Order between two completely monotone functions?
Current License: CC BY-SA 3.0
10 events
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| Sep 27, 2015 at 13:27 | comment | added | Christian Remling | @Alphonse: I can always add the same positive function to both $\phi_f$, $\phi_g$, so a compactly supported counterexample would suffice. (There is the issue of not being allowed singular measures now, but I don't think this can be a serious problem.) | |
| Sep 27, 2015 at 13:20 | comment | added | Alphonse | Yes that is what I thought at first, but then I realized that you really needed a compact support for your example to work. If you make the support of $\phi_f$ 'bigger and bigger', then you lose the $1/x$ part which makes it work. | |
| Sep 27, 2015 at 13:12 | comment | added | Christian Remling | @Alphonse: I think essentially the same counterexample will still work, if we just take $\phi_{f,g}$ that approximate my choices $\phi_f=\chi_{(\epsilon, 1+\epsilon)}$, $\phi_g=\epsilon\delta_{\epsilon} + (1-\epsilon)\delta_{\alpha}$. | |
| Sep 27, 2015 at 8:26 | comment | added | Alphonse | Thanks Christian, seems like a nice counterexample. Do you think it would still fail if I assume further that $f(x) = \int_0^\infty e^{-tx} \phi_f(t) dt$ and $g(x) = \int_0^\infty e^{-tx} \phi_g(t) dt$, where $\phi_f,\ \phi_g >0$ on $\mathbb{R}_+$? | |
| Sep 27, 2015 at 8:23 | vote | accept | Alphonse | ||
| Sep 26, 2015 at 23:54 | history | undeleted | Christian Remling | ||
| Sep 26, 2015 at 23:52 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Sep 26, 2015 at 19:21 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Sep 26, 2015 at 19:03 | history | deleted | Christian Remling | via Vote | |
| Sep 26, 2015 at 19:01 | history | answered | Christian Remling | CC BY-SA 3.0 |