Timeline for What is a good reference that compact resolvent implies Fredholm operator?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 19, 2017 at 12:54 | comment | added | Chris Judge | Let $E_0=E_1$ be a Hilbert space, let $i: E_0 \to E_0$ be compact embedding with dense image. Let $B: E_1 \to E_0$ be the identity map. Define an unbounded operator $A:E_0 \to E_0$ with domain $i(E_0)$ by setting $A= B \circ i^{-1}$. Then $A^{-1}= i \circ B^{-1}= i$ is compact and $B-1 \cdot B$ is not Fredholm. | |
| Apr 15, 2012 at 3:14 | comment | added | Jeremy LeCrone | @jjcale - Yes. It should be a bijection when considering it's restriction to the residual space, where we need to be careful to only consider elements in $((1 - P_{\lambda})E_0) \cap E_1$, as you mentioned. This care is not necessary on $P_{\lambda}E_0$ however, since one can show that $P_{\lambda}E_0 \subset D(A^n)$ for any $n \in \mathbb{N}$. | |
| Apr 14, 2012 at 8:18 | comment | added | jjcale | You mean $\lambda I - A$ is bijective from $((1 - P_{\lambda})E_0) \cap E_1$ to $(1 - P_{\lambda})E_0$ , or ? | |
| Apr 12, 2012 at 17:58 | answer | added | Liviu Nicolaescu | timeline score: 3 | |
| Apr 12, 2012 at 10:45 | answer | added | Heiko | timeline score: 1 | |
| Apr 11, 2012 at 20:40 | history | edited | Yemon Choi |
added ref-req tag
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| Apr 11, 2012 at 20:04 | history | asked | Jeremy LeCrone | CC BY-SA 3.0 |