Timeline for Integrating the resolvent of a self-adjoint operator across a continuous part of the spectrum
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2016 at 9:12 | comment | added | oeiras | (ctd.) as can already be seen in the one dimensional case. | |
| Mar 16, 2016 at 9:11 | comment | added | oeiras | This is merely a coda to Nik Weaver's comments. We can assume as stated that we have multiplication by a real-valued measurable function $x$. If the set of values of $s$ for which $x(s)$ lies on the curve is negligible (e.g., if the curve crosses the real line tranversally at a finite, or even countably infinite, number of points in the continuous spectrum), then there is no problem. If it crosses it at an eigenvalue, then anything can happen. As already mentioned, you would have to specify in which sense the integral is to be taken and in any case it will depend on the function $f$ .... | |
| Feb 19, 2016 at 1:40 | comment | added | Nik Weaver | @IgorKhavkine: It really helps to assume $A$ is a multiplication operator. A crude intuition can be: $A$ is multiplication by a function $f(x)$ defined on a measure space $X$, and $\int_{C_\epsilon} R_z\, dz$ is multiplication by the function whose value at the point $x$ is $\int_{C_\epsilon} \frac{1}{z - f(x)}\, dz$. You only have a problem when $f(x) = a$ or $b$, but there the principal value integral converges nicely. | |
| Feb 18, 2016 at 20:58 | comment | added | Igor Khavkine | Although, perhaps self-adjointness (and thus a kind of symmetry between the upper and lower half planes) is precisely the condition that would make it work. | |
| Feb 18, 2016 at 20:57 | comment | added | Igor Khavkine | Ah, yes I see which norm you mean. I mistakenly read that you meant that the integrand is bounded in the operator norm over all of $C$, including the parts that approach the spectrum, which I think would not be true since $\|R_z\|$ grows as $1/\Im z$ when approaching the spectrum. So your suggestion essentially is to treat the resulting improper integral using the principal value regularization. I'm only a little bit concerned about the well-posedness of this regularization due to possibly different asymptotic behaviors as one approaches the spectrum from above and from below. | |
| Feb 18, 2016 at 19:37 | comment | added | Nik Weaver | I've never seen this idea before but I wouldn't be surprised if it's covered somewhere. Googling "principal value" "operator integral" didn't turn anything up. | |
| Feb 18, 2016 at 19:35 | comment | added | Nik Weaver | @IgorKhavkine: No, $\int f(z) R_z\, dz$ is an operator integral, and since the operators $R_z$ are uniformly bounded away from the spectrum of $A$ it defines a bounded operator. I mean operator norm, the norm of an operator in $B(H)$. | |
| Feb 18, 2016 at 19:26 | comment | added | Igor Khavkine | OK. By operator norm you mean the $\|x\|_A^2 = (x,x) + (Ax,Ax)$ norm? Do you think this topic is covered in some standard references on spectral theory? I couldn't find anything when I had a look. | |
| Feb 18, 2016 at 19:06 | history | answered | Nik Weaver | CC BY-SA 3.0 |