Timeline for Cesaro means and Banach limits
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Jan 11, 2013 at 1:03 | answer | added | Daniel Mansfield | timeline score: 4 | |
Apr 9, 2012 at 13:47 | comment | added | Martin Sleziak | Sequences with this property are called almost convergent and there exists a well-known characterization of such sequences due to Lorenz; which is described in the Wikipedia article. One possibility how to show this is using Hahn-Banach theorem, see this answer at Math.SE. Also the references from the Wikipedia article might be useful. | |
Apr 9, 2012 at 10:44 | vote | accept | kap44 | ||
Apr 9, 2012 at 6:10 | answer | added | Aaron Tikuisis | timeline score: 13 | |
Apr 8, 2012 at 21:21 | comment | added | Liviu Nicolaescu | Yes, thanks. $L$ is therefore an element of the topological dual of $\ell^\infty$. | |
Apr 8, 2012 at 20:40 | history | edited | Yemon Choi |
added FA tag
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Apr 8, 2012 at 20:34 | comment | added | kap44 | If $L$ is a Banach limit, then $|Lx|\le\|x\|_\infty$ for any sequence $x$. Does this answer your question? | |
Apr 8, 2012 at 20:29 | comment | added | Liviu Nicolaescu | I like the question, but I would like to make sure I understand correctly the notion of Banach limit. From your formulation I understand that a Banach limit is a linear map $L:\ell^infty(\mathbb{R})\to\mathbb{R}$ with the property that $L(\;(x_n)_{n\geq 0}\;)=L(\;(y_n)_{n\geq 0}\;) $ if $x_n=y_n$ for all $n$ sufficiently large. Is the functional $L$ continuous in the $\ell^\infty$ topology? | |
Apr 8, 2012 at 20:10 | history | edited | kap44 | CC BY-SA 3.0 |
added 24 characters in body
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Apr 8, 2012 at 19:58 | history | asked | kap44 | CC BY-SA 3.0 |