Timeline for Is the Gelfand transform strictly continuous?
Current License: CC BY-SA 4.0
11 events
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| Jul 31, 2018 at 6:17 | history | edited | Jan_Ch. | CC BY-SA 4.0 |
deleted 95 characters in body; edited title
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| Jul 27, 2018 at 14:13 | comment | added | Nik Weaver | @Jan_Ch: it's considered bad form to change the title to a different question after the original question has been answered. Next time please ask the new question separately. | |
| Jul 27, 2018 at 11:21 | history | edited | Jan_Ch. | CC BY-SA 4.0 |
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| Jul 27, 2018 at 6:32 | history | edited | Jan_Ch. | CC BY-SA 4.0 |
edited title
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| Jul 26, 2018 at 7:23 | history | edited | Jan_Ch. | CC BY-SA 4.0 |
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| Jul 26, 2018 at 3:10 | comment | added | Yemon Choi | As @ChristianRemling's comment suggests, the answer is negative (if I have not made an error, just take a bounded net in $L^1$ that forms an approximate identity, e.g. Fejer kernels). I don't follow the "intuition" you mention in the last sentence; what is going on is that you have the Gelfand transform $L_1\to C_0$ and it is natural to guess that this would extend to a strictly-to-strictly continuous map $M(L_1) \to M(C_0) = C_b$, but this extension is unlikely to be strictly-to-norm continuous | |
| S Jul 25, 2018 at 22:39 | history | suggested | CommunityBot |
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| Jul 25, 2018 at 20:15 | comment | added | Christian Remling | So are you asking if $\| f*\mu_n\| \to 0$ for all $f\in L^1$ implies that $\|\widehat{\mu_n}\|_{\infty} \to 0$, with $\widehat{\mu}$ denoting the classical Fourier transform? | |
| Jul 25, 2018 at 19:41 | review | Suggested edits | |||
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| Jul 25, 2018 at 12:35 | review | First posts | |||
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| Jul 25, 2018 at 12:32 | history | asked | Jan_Ch. | CC BY-SA 4.0 |