Timeline for What is a good reference that compact resolvent implies Fredholm operator?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| May 8, 2015 at 6:49 | comment | added | Liviu Nicolaescu | We are showing that if the resolvent $(T-\lambda)^{-1}$ is compact, then $T$ is Fredholm because $QT-1$ is compact. | |
| May 8, 2015 at 2:24 | comment | added | dh16 | Sorry for bringing up an old question. I do not understand the last sentence "If $Q=(T-\lambda)^{-1}$ is compact, ..... $QT-1=\lambda Q=$compact". Are we showing $T-\lambda$ is Fredholm, not $T$? Thanks | |
| Apr 13, 2012 at 10:56 | comment | added | Liviu Nicolaescu | @ Amin In these same notes I use Theorem 3.4.3 to prove that elliuptic pseudo-differential operators on compact manifolds are Fredholm. I do not regard them as unbounded operators. The unbounded operator point of view is disccussed in the notes below. nd.edu/~lnicolae/Lectures.pdf | |
| Apr 13, 2012 at 10:24 | comment | added | Amin | @Liviu : I remember that you showed that elliptic operators were unbounded Fredholm in other notes, and I don't know if the parallel can be done just based on the compact resolvent hypothesis (haven't checked at all). | |
| Apr 13, 2012 at 10:21 | comment | added | Amin | @Liviu : $E_1$ is a subset of $E_0$, so it makes sense. I was wondering if your proof goes well indeed in the unbounded case (thanks for all your notes btw). | |
| Apr 13, 2012 at 9:11 | comment | added | Liviu Nicolaescu | @Renato Thanks for pointing up that error. @Jeremy: Note that $E_0=E_1$ or else the operator $\lambda-T$ is meaningless. | |
| Apr 13, 2012 at 9:08 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
deleted 2 characters in body
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| Apr 13, 2012 at 1:10 | comment | added | Jeremy LeCrone | @Liviu- I had originally considered exactly this argument, but there seems to be a flaw in it. By assumption I only know that $Q = (\lambda - T)^{-1}$ is compact from $E_0$ into itself. However, when you are considering it in this context, you wind up with $Q$ as an operator from $E_0$ into $E_1$ (This is the only way that the equations $1 - QT$ and $1 - TQ$ make sense...) We, however, do not have compactness of $Q$ between the two spaces. I would love if I am wrong in this, but I think that this consideration hinders your argument. | |
| Apr 12, 2012 at 22:02 | comment | added | Renato G. Bettiol | I think you meant $Q=-(\lambda-T)^{-1}$, right? So that $QT=\lambda Q+1$. | |
| Apr 12, 2012 at 18:27 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
fixed typo
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| Apr 12, 2012 at 17:58 | history | answered | Liviu Nicolaescu | CC BY-SA 3.0 |