Timeline for How to characterize a Banach space $X$ such that any operator from $X$ to $l_{p}$ is compact?
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 24, 2015 at 1:04 | comment | added | Dongyang Chen | @triple_sec the formal identity $i_{2,p}:l_{2}\rightarrow l_{p}(p>2)$ is non-compact and not surjective. There is no surjective linear continuous operator from $l_{2}$ to $l_{p}$. | |
| Sep 24, 2015 at 0:45 | comment | added | Dongyang Chen | Let us continue this discussion in chat. | |
| Sep 24, 2015 at 0:37 | comment | added | triple_sec | @DongyangChen I’m wondering whether this could help you sharpen the characterization of such an $X$ in any way. | |
| Sep 24, 2015 at 0:33 | comment | added | Dongyang Chen | @triple_sec You are right. If $X$ satisfies my condition, then there cannot exist a surjective linear continuous operator from $X$ to $l^{p}$. This is due to open mapping theorem. | |
| Sep 24, 2015 at 0:11 | comment | added | triple_sec | @DongyangChen I mean, if $X$ satisfies your condition that any bounded linear operator from $X$ to $\ell^p$ is a compact operator. Then, if there existed a surjective continuous linear operator $F:X\to\ell^p$, then $F$ would be an open mapping (given that both $X$ and $\ell^p$ are Banach spaces) and a compact operator by assumption. But, as I argued above, this would imply that the image of the open unit ball of $X$ under $F$ was a non-empty, precompact open subset of $\ell^p$, which is impossible (in an infinite-dimensional vector space—complete or not—any compact set is nowhere dense). | |
| Sep 23, 2015 at 23:59 | comment | added | Dongyang Chen | @triple_sec I think you are wrong. Obviously, there exists a surjective linear continuous operator $F: l_{1}\rightarrow l_{p}$. This is because each separable Banach space is isometrically isomorphic to $l_{1}$. | |
| Sep 23, 2015 at 23:10 | comment | added | triple_sec | @DongyangChen For what it’s worth, this argument shows that if $X$ is infinite-dimensional, then there cannot exist a surjective linear continuous operator $F:X\to\ell^p$. | |
| Sep 23, 2015 at 22:50 | comment | added | Dongyang Chen | @triple_sec Anyway, I thank you for your consideration. | |
| Sep 23, 2015 at 17:32 | comment | added | triple_sec | @DongyangChen In this case, I’m sorry my answer couldn’t help. Thank you for the clarification. | |
| Sep 23, 2015 at 15:39 | comment | added | Yemon Choi | I agree with @JochenWengenroth since the question is posed in the setting of Banach spaces by someone who works on Banach spaces. It is common practice for those who work in the areas of Banach spaces, Banach algebras, Cstar algebra, von Neumann algebras, etc to say "operator" as short-hand for "bounded linear map" | |
| Sep 23, 2015 at 13:05 | comment | added | Dongyang Chen | In my question, "operator"means linear bounded mapping. I am sorry. | |
| Sep 23, 2015 at 7:51 | comment | added | triple_sec | @JochenWengenroth At any rate, I added a warning in the beginning so that the OP can decide whether this interpretation is useful for him/her. | |
| Sep 23, 2015 at 7:50 | history | edited | triple_sec | CC BY-SA 3.0 |
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| Sep 23, 2015 at 7:48 | comment | added | triple_sec | @JochenWengenroth I beg to differ—interpretations vary. To the best of my knowledge, the most common definition of an operator is a linear function between vector spaces. | |
| Sep 23, 2015 at 7:43 | comment | added | Jochen Wengenroth | "Operator" means continuous linear map. | |
| Sep 23, 2015 at 7:31 | history | answered | triple_sec | CC BY-SA 3.0 |