Timeline for A question about asymptotic affinity and strict convexity with unbounded means
Current License: CC BY-SA 4.0
14 events
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| Aug 25, 2020 at 8:54 | comment | added | Ron P | @AsafShachar, you're right. I've corrected the reduction from Lemma 3 to Lemma 4. Instead of using the continuity of $D_f$ in $a$, I now use monotonicity. | |
| Aug 25, 2020 at 8:50 | history | edited | Ron P | CC BY-SA 4.0 |
Correted the redudtion from Lemma 3 to Lemma 4.
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| Aug 24, 2020 at 11:39 | comment | added | Asaf Shachar | I have edited the question to make my misunderstanding clear. Thanks. | |
| Aug 24, 2020 at 11:39 | comment | added | Asaf Shachar | Thanks for the elaboration. I am still somewhat troubled tough. It seems to me that the most crucial non-trivial step is in passing from Lemma 3 to Lemma 4 (from $a_n$ to $\lim a_n$). I understand why $\liminf\lambda(a_n,c_n,b_n)=\liminf\lambda(\lim a_n,c_n,b_n)$ (because the difference between the 'lambdas' tend to zero.) However, I don't see why $\limsup D_f(a_n,c_n,b_n)= \limsup D_f(\lim a_n,c_n,b_n)$. I think it reduces to $\lim_{n \to \infty }( \lambda(a_n,c_n,b_n)-\lambda(\lim a_n,c_n,b_n))F(b_n)=0, $ but I don't see why this limit should be zero. | |
| Aug 23, 2020 at 10:01 | history | edited | Ron P | CC BY-SA 4.0 |
edited body
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| Aug 21, 2020 at 19:23 | history | edited | Ron P | CC BY-SA 4.0 |
added 612 characters in body
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| Aug 21, 2020 at 12:48 | history | edited | Ron P | CC BY-SA 4.0 |
added 1 character in body
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| Aug 21, 2020 at 12:42 | history | edited | Ron P | CC BY-SA 4.0 |
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| Aug 21, 2020 at 12:39 | comment | added | Ron P | @AsafShachar I've elaborated on "re-scaling." Note that the answer does not assume neither that $F$ is non-negative, nor that $c_n\to\infty$, rather just that it is bounded away from $a_n$. | |
| Aug 21, 2020 at 12:34 | history | edited | Ron P | CC BY-SA 4.0 |
Elaborated on what "rescaling" means
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| Aug 21, 2020 at 12:28 | history | edited | Ron P | CC BY-SA 4.0 |
Elaborated on what "rescaling" means
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| Aug 21, 2020 at 7:09 | comment | added | Ron P | @AsafShachar you're right. I should elaborate about this point. I will do it soon. The idea is to replace $a_n$ with its limit in the expression of $D_n$. By continuity, such a modification does not change whether it converges to zero or not. | |
| Aug 20, 2020 at 17:20 | history | edited | Ron P | CC BY-SA 4.0 |
Elaborating on the previous version of my proof
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| Aug 20, 2020 at 11:53 | history | answered | Ron P | CC BY-SA 4.0 |