Timeline for Spectrum of a differential operator on $L^2(0, \infty)$
Current License: CC BY-SA 4.0
10 events
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| Jul 30, 2019 at 5:59 | comment | added | Igor Khavkine | @Mainkit: the domain where the resolvent is defined depends on how it is constructed, and may be different from the range of $(A-\lambda I)$. For instance, in your case, the resolvent may be defined via a formula based on the Fourier transform or via the "variation of constants" formula. I presume that this is how you proved the existence of the resolvent in the first place. Well, these formulas lend themselves to checking boundedness on some dense domain. The relevant details are left to you as an exercise. | |
| Jul 29, 2019 at 16:00 | comment | added | Mike Van | @IgorKhavkine But I don't know if $R(A-\lambda I)$ is dense in $L^2(0, \infty)$ | |
| Jul 29, 2019 at 14:44 | comment | added | Igor Khavkine | @Mainkit: if the resolvent is bounded, but only defined on a dense domain, then it has a unique extension to all of $L^2$ by continuity. | |
| Jul 29, 2019 at 12:42 | comment | added | Mike Van | @JochenGlueck I have not tried. Maybe I can use Fourier Transform to give an expression for $u$. | |
| Jul 29, 2019 at 12:40 | comment | added | Mike Van | @IgorKhavkine Why you can say that if $(A-\lambda I)^{-1}$ exists and is bounded then $D((A-\lambda I)^{-1})=L^2(0, \infty)$?. I say that $(A-\lambda I)$ exists only on $R(A-\lambda I)$. | |
| Jul 29, 2019 at 12:17 | comment | added | Igor Khavkine | @JochenGlueck: I see, thanks! The notation was a bit confusing. But then, if one already knows that that $(A-\lambda I)^{-1}$ exists and is bounded, it's domain is already all of $L^2$. End of story. In a way, this is rephrasing of Jochen's last comment. | |
| Jul 29, 2019 at 4:54 | comment | added | Jochen Glueck | @Mainkit: Have you tried to explicitly solve the equation $Au-\lambda u =f$ (where $f\in L^2$), and to show that the solution $u$ is in $H^n$ if $\lambda\not\in S$? | |
| Jul 29, 2019 at 4:44 | comment | added | Jochen Glueck | @IgorKhavkine: I think that the notation $R(A-\lambda I)$ in the question means the range of $A-\lambda I$. The OP does not seem to use the letter $R$ to denote the resolvent. | |
| Jul 29, 2019 at 1:51 | comment | added | Igor Khavkine | When $\lambda \not\in \sigma(A)$, you expect the image of $A-\lambda I$ to be all of $L^2$. The image of the resolvent would then be all of the domain of $A$ (or more specifically the appropriate closure of $A$), which for unbounded operators is smaller than $L^2$ itself. I hope I didn't misunderstand what you wrote in the question. | |
| Jul 29, 2019 at 0:52 | history | asked | Mike Van | CC BY-SA 4.0 |