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Oct 14, 2020 at 7:03 vote accept sharpe
Sep 9, 2020 at 16:46 answer added Mohan Ramachandran timeline score: 2
Sep 9, 2020 at 15:10 comment added Giorgio Metafune Yes, exactly. The same you wrote
Sep 9, 2020 at 14:35 comment added sharpe @GiorgioMetafune Thank you for letting me know. $f=\alpha^{-1}(1-|x|^2)^{\alpha}$ ? This seems to be the simplest.
Sep 9, 2020 at 13:17 comment added Giorgio Metafune I find $Lf=\left(4(1-c/2-\alpha)r^2+2cx\cdot \theta +2n (1-r^2)\right )(1-r^2)^{\alpha-1}$. Then $(4-4\alpha -2c)r^2+2cx\cdot \theta +2n (1-r^2) \ge (4-4\alpha -2c)r^2-2rc|\theta|+2n (1-r^2)$ and this last expressione is nonegative if $c(1+|\theta|) <2$, choosing $\alpha$ small as you did.
Sep 9, 2020 at 12:51 comment added sharpe @GiorgioMetafune Your result is consistent with the case of $\theta=0$.
Sep 9, 2020 at 11:31 comment added sharpe @GiorgioMetafune I don't find it. Could you tell me the reason?
Sep 9, 2020 at 7:12 comment added Giorgio Metafune It seems that your barrier works if $c(1+|\theta|) <2$; do you find the same?
Sep 9, 2020 at 1:51 history edited sharpe CC BY-SA 4.0
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Sep 8, 2020 at 13:10 history edited sharpe CC BY-SA 4.0
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Sep 8, 2020 at 12:56 comment added sharpe @GiorgioMetafune Thank you for your comment. There will be no function satisfying the conditions if $c \ge 2$ and $\theta=0$. This is based on probabilistic considerations.
Sep 8, 2020 at 12:35 comment added Giorgio Metafune Do you know what happens if $\theta=0$ and $c\geq 2$?
Sep 8, 2020 at 7:19 history edited sharpe CC BY-SA 4.0
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Sep 8, 2020 at 5:47 history edited sharpe CC BY-SA 4.0
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Sep 8, 2020 at 5:38 history edited sharpe CC BY-SA 4.0
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Sep 8, 2020 at 5:06 history asked sharpe CC BY-SA 4.0