Timeline for On a compact operator in the plane
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3 at 12:43 | vote | accept | Ali | ||
| May 1 at 14:36 | comment | added | Giorgio Metafune | I see. Writing $Tf-\lambda f=F$ and setting $u=Gf$, then $u=-\frac 1 \lambda e^{z/4 \lambda} G(F e^{-z/4\lambda})$ which shows that the resolvent satisfies an exponential estimate near zero (in 1d this is in fact the Volterra operator). Since 0 is not an eigenvalue ($Tf=0$ gives $f=\partial Tf=0$), 0 is not a pole of the resolvent, otherwise would be an eigenvalue. | |
| Apr 30 at 20:11 | history | answered | Ali | CC BY-SA 4.0 |