Timeline for When does $C_b(X)$ admit a Schauder Basis?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 28, 2022 at 18:16 | vote | accept | Carlos_Petterson | ||
| Jan 28, 2022 at 15:26 | comment | added | Carlos_Petterson | @TomaszKania Very nice point! | |
| Jan 28, 2022 at 14:14 | comment | added | Tomasz Kania | @Carl_Petterson, this is far from isometry, only isomorphism. Otherwise it would have contradicted the Banach-Stone theorem. | |
| Jan 28, 2022 at 13:45 | comment | added | bathalf15320 | I only know of an existence proof using the Pełczyński decomposition method so it won't give an explicit basis. For these, I would try Semadeni. | |
| Jan 28, 2022 at 11:52 | comment | added | Carlos_Petterson | @bathalf15320 Ah I wasn't aware of Milyutin's result (though I do know of the strict topology; as an example of colimits in the category of LCSs); is Milyutin's isometry explicit or is there currently only an existence version of the theorem known? | |
| Jan 28, 2022 at 11:02 | comment | added | bathalf15320 | The original Banach space basis was constructed naturally and explicitly in $C([0.1])$ by Schauder (hence the name). It is in Banach's book. This shows, by Milyutin's theorem, that there is a positive answer for any non countable compact metric soace. | |
| Jan 28, 2022 at 10:33 | comment | added | Michael Greinecker | @Carl_Petterson Take any element $f$ of $C_B(\mathbb{N})$. For each $n$, let $f(n\pm 1/2)=0$ and connect the values of $f(n\pm 1/2)$ and $f(n)$ by a straight line. | |
| Jan 28, 2022 at 10:32 | comment | added | bathalf15320 | Two remarks which might be of interest. 1. A standard reference on your topic is the Springer Lecture Notes "Schauder Bases in Banach Spaces of Continuous Functions" by Z. Semadeni. 2. As a general rule of thumb, if you want to extend results and concepts on spaces of bounded, continuous functions on compact spaces to the non-compact case (Gelfand-Naimark, duality for bounded Radon measures, tensor product representations,...), one way to go is to replace the norm by the strict topology (R.C. Buck), i.e., the finest l.c. topology which agrees with compact convergence on the unit ball. | |
| Jan 28, 2022 at 10:14 | answer | added | Tomasz Kania | timeline score: 2 | |
| Jan 28, 2022 at 10:10 | comment | added | Carlos_Petterson | @MichaelGreinecker What do you mean? | |
| Jan 28, 2022 at 10:06 | comment | added | Michael Greinecker | Isn't it straightforward to embed $C_B(\mathbb{N})$ in $C_B(\mathbb{R})$ via zigzag-curves? | |
| Jan 28, 2022 at 9:59 | history | edited | Carlos_Petterson | CC BY-SA 4.0 |
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| Jan 28, 2022 at 9:59 | comment | added | Carlos_Petterson | @GeraldEdgar Actually not (now that you mention in), so I refined my question (so that is directly to the point of the manner) I also just noticed the discrete counter-example (as usual these are pathological) so I added the connectedness condition. | |
| Jan 28, 2022 at 9:58 | comment | added | Gerald Edgar | Note: Consider $X = \mathbb N$ with the discrete metric. Then $C_b(X) = l^\infty$ is not separable. Thus no Schuder basis. Is there any non-compact $X$ where $C_b(X)$ is separable? | |
| Jan 28, 2022 at 9:58 | history | edited | Carlos_Petterson | CC BY-SA 4.0 |
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| Jan 28, 2022 at 9:56 | comment | added | Gerald Edgar | Do you know any example where $C_b(X)$ is separable, yet does not admit a Schauder basis? | |
| Jan 28, 2022 at 9:44 | history | asked | Carlos_Petterson | CC BY-SA 4.0 |