Timeline for A question about extension problem related to fractional laplacian
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2020 at 19:07 | comment | added | Giuseppe Negro | That's not a general identity which holds for all $u$. Take for example $u(x, y)=y^{1-a}$, you'll see that the LHS equals $(a-1)$ while the RHS equals $\frac{1}{1-a}$. For that identity to hold, $u$ must satisfy the given PDE. In any case, from the way the author phrase the result, I understand that they are going to prove the identity later in the paper. | |
| Nov 4, 2020 at 18:43 | comment | added | Iosif Pinelis | Yes, it appears that the constant factor in the equality in question in your post was computed incorrectly. | |
| Nov 4, 2020 at 17:46 | comment | added | inoc | If i apply l'Hospital's rule, i have that: $$ \frac{1}{1-a}\lim_{y\to0}\frac{u(x,y)-u(x,0)}{y^{1-a}}=\frac{1}{(1-a)^2}\lim_{y\to0}y^au_y(x,y),$$that is: $$ \lim_{y\to0}\frac{u(x,y)-u(x,0)}{y^{1-a}}=\lim_{y\to0}y^au_y(x,y),$$ or there is a mistake in may computation? Moreover doesn't appear the minus sign. Then i obtain $y\to0^+$ in the right limit by continuity? | |
| Nov 4, 2020 at 17:35 | comment | added | inoc | Sorry, i have dropped the hypothesis $a\in(-1,1)$. | |
| Nov 4, 2020 at 17:26 | comment | added | Iosif Pinelis | If $a<1$ and the limit of the left exists, then this follows by l'Hospital's rule. | |
| Nov 4, 2020 at 17:04 | history | asked | inoc | CC BY-SA 4.0 |