Timeline for Invertible operator with countable spectrum
Current License: CC BY-SA 3.0
17 events
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| Aug 25, 2014 at 12:51 | comment | added | Ali Taghavi | @YemonChoi Yes By invertible I mean bijective. So as you said, the inverse is bounded, by open mapping theorem. | |
| Aug 25, 2014 at 9:28 | comment | added | Yemon Choi | @JoonasIlmavirta yes, my confusion was precisely this difference of terminology. But I am fairly sure the OP uses my definition of "invertible" | |
| Aug 25, 2014 at 3:38 | comment | added | Joonas Ilmavirta | @YemonChoi, a bounded invertible operator need not be surjective or have closed range. Consider for example the operator on $\ell^2(\mathbb N)$ that divides the $n$th coordinate by $n$. It is surely injective and so has an inverse, but the inverse is not continuous. Maybe the confusion is that for me invertible means injective and for you bijective, so my invertible is your left invertible. | |
| Aug 24, 2014 at 21:21 | comment | added | Yemon Choi | @JoonasIlmavirta My confusion is: you said above "The inverse of a bounded invertible operator need not be bounded" and the Banach isomorphism theorem tells us that actually, this is always the case | |
| Aug 24, 2014 at 19:39 | comment | added | Joonas Ilmavirta | @YemonChoi, I mean everywhere defined operators. Invertibility is not so important, spectrum is the main point. I was just wondering if one could use countable spectrum approximations for $B$ and $C$ to build one for $A$ and if this approximation would be invertible if the first ones were (since invertibility was asked for). | |
| Aug 24, 2014 at 19:25 | comment | added | Yemon Choi | @JoonasIlmavirta I don't understand your comment about inverses, unless you are dealing with densely-defined operators. In the case of bounded operators on a Banach space we have the Banach isomorphism theorem, unless I have misread or misunderstood something | |
| Aug 19, 2014 at 22:35 | answer | added | Leonel Robert | timeline score: 8 | |
| Aug 19, 2014 at 19:29 | vote | accept | Ali Taghavi | ||
| Aug 19, 2014 at 13:00 | answer | added | Mike Jury | timeline score: 6 | |
| Aug 19, 2014 at 11:03 | history | edited | Ali Taghavi |
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| Aug 19, 2014 at 9:55 | comment | added | Joonas Ilmavirta | Hmm... In a complex Hilbert space you can write $A=B+iC$ with $B$ and $C$ selfadjoint, and selfadjoint operators are easy to approximate since they diagonalize. I don't know what happens for invertibility and the spectrum of the sum though. | |
| Aug 19, 2014 at 9:53 | answer | added | Chandler | timeline score: 3 | |
| Aug 19, 2014 at 9:49 | comment | added | Joonas Ilmavirta | Ok. The inverse of a bounded invertible operator need not be bounded, so I wanted to check. | |
| Aug 19, 2014 at 9:43 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Aug 19, 2014 at 9:38 | comment | added | Ali Taghavi | @JoonasIlmavirta yes all things are bounded operators? | |
| Aug 19, 2014 at 9:16 | comment | added | Joonas Ilmavirta | Should the inverse also be bounded? | |
| Aug 19, 2014 at 9:12 | history | asked | Ali Taghavi | CC BY-SA 3.0 |