Timeline for $C[0,1]$ is not a Grothendieck space
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| S Dec 14, 2020 at 8:57 | history | suggested | gmvh |
Added top-level tag
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| Dec 14, 2020 at 8:01 | review | Suggested edits | |||
| S Dec 14, 2020 at 8:57 | |||||
| Dec 14, 2020 at 0:10 | vote | accept | Dongyang Chen | ||
| Dec 13, 2020 at 22:51 | comment | added | Jochen Glueck | @DieterKadelka: Thank you for your response. I see. | |
| Dec 13, 2020 at 22:38 | comment | added | Dieter Kadelka | @Jochen Glueck: I was confused by the notation. I considered $h_n$ as a measure and then compared it with $\delta_0$ In principle it is a variant of the answer of Dirk Werner, I only had $n \cdot \lambda^1|{[0,1/n]}-\delta_0$ instead of $\delta_{1/n}-\delta_0$. Sorry for any confusion. | |
| Dec 13, 2020 at 22:16 | comment | added | Jochen Glueck | @DieterKadelka: Yes, indeed, we need a sequence that is weak${}^*$-convergent to $0$, but not weakly convergent; and the sequence $(h_n)$ suggested by Yemon Choi does the job. I was just wondering why you mentioned $\delta_0$ in your first comment, since the sequence $(h_n)$ has nothing to do with $\delta_0$. | |
| Dec 13, 2020 at 21:37 | history | edited | LSpice | CC BY-SA 4.0 |
$weak^*$ -> $\text{weak}^*$; deleted 'thank you'
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| Dec 13, 2020 at 19:34 | comment | added | Dirk Werner | @Dieter: If the sequence is weak$^*$ convergent to $0$ and weakly convergent to something, then something $=0$; so one just has to check weak convergence to $0$. | |
| Dec 13, 2020 at 19:31 | comment | added | Dieter Kadelka | @Jochen Glueck: Is this not what we need, to find a weak-*-0-convergent sequence which is not weakly convergent at all? | |
| Dec 13, 2020 at 19:31 | answer | added | Dirk Werner | timeline score: 12 | |
| Dec 13, 2020 at 19:08 | comment | added | Dieter Kadelka | @Jochen Glueck: You are right. The last sentence in the question irritated me, since it is not quite correct. Not beig weakly null doen't suffice. | |
| Dec 13, 2020 at 18:22 | comment | added | Jochen Glueck | @DieterKadelka: I'm not sure I follow. The sequence $(h_n)$ in $C([-1,1])^*$ does indeed weak${}^*$-converge to $0$, but it is not weakly convergent to any point. | |
| Dec 13, 2020 at 14:45 | answer | added | Dongyang Chen | timeline score: 4 | |
| Dec 13, 2020 at 11:40 | comment | added | Dieter Kadelka | @Dongyang Chen: The point is that this sequence converges to $\delta_0$. It is not a weakly null sequence. Try $\mu_n := n \cdot \lambda^1|_{[0,1/n]} - \delta_0$. I think it will do the job. | |
| Dec 13, 2020 at 8:57 | comment | added | Dongyang Chen | @YemonChoi Sorry. I can not prove that the sequence $(h_{n})_{n}$ is not weakly null as a sequence of measures on $[-1,1]$. | |
| Dec 13, 2020 at 8:39 | comment | added | Dongyang Chen | @YemonChoi I guess that mean zero means that $\int h_{n}=0$. I can prove that $\int f h_{n}\rightarrow 0$ for each $f\in C[-1.1]$. But I can not prove that the sequence $(h_{n})_{n}$ is weakly null as a sequence of measures on $[-1,1]$. | |
| Dec 13, 2020 at 7:44 | comment | added | Dongyang Chen | @YemonChoi Thanks, Yemon. Could you describe the function $h_{n}$ clearer? I do not know what mean zero is. | |
| Dec 13, 2020 at 3:59 | comment | added | Yemon Choi | Replacing $C[0,1]$ with $C[-1,1]$ for convenience, let $h_n$ be the "obvious" function in $L^1[-1,1]$ whichh as norm $1$ and mean zero and is supported on the interval $[-1/n, 1/n]$. If we view each $h_n$ as a measure on $[-1,1]$ then I think this should have the required properties - let me know if this doesn't work dor some reason | |
| Dec 13, 2020 at 3:33 | history | asked | Dongyang Chen | CC BY-SA 4.0 |