Timeline for What is a good reference that compact resolvent implies Fredholm operator?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2012 at 6:53 | comment | added | Amin | Oh right right, sorry about that, I recalled this morning about the usual case of elliptic operators with Sobolev as $E_1$ and figured out I was mistaken ;). | |
| Apr 13, 2012 at 1:03 | comment | added | Jeremy LeCrone | @Amin- It is bounded FROM $E_1$ TO $E_0$, not ON $E_1$. I.e. For every $x \in E_1$ we know that $\| Ax \|_0 \leq M \| x \|_1$, where $\| \cdot \|_i$ is the norm on $E_i$. This does not mean that we can extend $A$ to all of $E_0$. Specifically, you are likely thinking, let $x_0 \in E_0$ and take a sequence $(x_n) \subset E_1$, then define $A x_0$ by a limiting argument. BUT $\| Ax_0 - Ax_n \|_0 \leq M \|x_0 - x_n \|_1$ is the inequality you get and you only know that $\| x_0 - x_n \|_0 \rightarrow 0$. So, no it is not a misspell :) | |
| Apr 12, 2012 at 20:55 | comment | added | Amin | I may sound completely fool, but in your question, it's written bounded on $E_1$; is it then not directly extendable to a bounded op on $E_0$ ? Or it's a misspell and you meant unbounded? | |
| Apr 12, 2012 at 15:25 | comment | added | Jeremy LeCrone | Thank you Heiko, I will look into the reference in more detail, though at first glance it does appear to specifically concern bounded operators. | |
| Apr 12, 2012 at 11:58 | comment | added | Aaron Tikuisis | Doesn't this reference only concern bounded operators? | |
| Apr 12, 2012 at 10:45 | history | answered | Heiko | CC BY-SA 3.0 |