Timeline for What standard Banach space is isomorphic to the completion of this different normed structure on $\ell^1$?
Current License: CC BY-SA 4.0
17 events
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| Sep 28, 2023 at 23:48 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 |
Just correcting typos
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| Apr 12, 2023 at 18:27 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 |
added 15 characters in body
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| Apr 12, 2023 at 17:47 | comment | added | Lorenzo Pompili | @ChristianRemling thank you, mine was stupid :D | |
| Apr 12, 2023 at 17:46 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 |
I rewrote the proof of Lemma 1 (thanks C. Remling, mine was stupid)
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| Apr 12, 2023 at 17:40 | comment | added | Christian Remling | By the way, a simpler proof of your Lemma 1 is: $\int_0^1f(t)\, dt\ge\int_0^x f(t)\, dt\ge xf(x)$ | |
| Apr 12, 2023 at 12:16 | comment | added | Ali Taghavi | @BillJohnson Dear Prof. Johnson thank you very much for your comment and your attention to this post. | |
| Apr 12, 2023 at 12:15 | comment | added | Ali Taghavi | Thank you very much Lorenzo for your answer | |
| Apr 11, 2023 at 20:53 | comment | added | Bill Johnson | That this space is $\ell_1$ saturated follows from general results; e.g., the papers of Lindenstrauss-Tzafriri on Orlicz sequence spaces. Then it is also complementably $\ell_1$ saturated because the space has an unconditional basis. | |
| Apr 11, 2023 at 10:49 | comment | added | S Argyros | Nice result! I wonder if the alternative description of the norm could give a positive answer to the following: Consider a sequence $(x_n)_n $ in $l_1$ with each $x_n$ a finite linear combination of the basis, $|x_n| =1$ and $lim_n |x_n|_1 = \infty$. Then the norm of $x_n$ is asymptotically concentrated around one( in the same manner as happens for the normalized averages of the basis).If the answers is positive then it seems to me that the space is $l_1$ saturated. | |
| Apr 10, 2023 at 21:36 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 |
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| Apr 10, 2023 at 21:30 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 |
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| Apr 10, 2023 at 20:38 | comment | added | Lorenzo Pompili | @ChristianRemling thank you for the remark. I left it like that because $t/\ln t$ diverges at $t=1$, and also to make $\Phi$ convex (so that it plays well with the definition of Orlicz space), but it's right that the behaviour of phi as $t\to\infty$ is irrelevant concerning the characterization as long as $\Phi$ stays away from zero. | |
| Apr 10, 2023 at 17:28 | comment | added | Christian Remling | Since $x_j\to 0$, you could then also use $\Psi(t)=-|t|/\log |t|$. | |
| Apr 10, 2023 at 16:45 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 |
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| Apr 10, 2023 at 16:39 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 |
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| Apr 10, 2023 at 16:31 | history | edited | Lorenzo Pompili | CC BY-SA 4.0 |
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| Apr 10, 2023 at 16:21 | history | answered | Lorenzo Pompili | CC BY-SA 4.0 |