Timeline for Kolmogorov entropy of a subset of $L^1$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2020 at 4:45 | comment | added | Iosif Pinelis | I have now given a bound on the entropy assuming the domination. | |
| Oct 21, 2020 at 4:44 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 20, 2020 at 22:06 | comment | added | Riku | Thanks! Being dominated by a function in $L^1$ seems to be a nice assumption. What would the $\epsilon$-entropy be under this additional condition? | |
| Oct 20, 2020 at 19:42 | comment | added | Iosif Pinelis | For instance, if you add the condition that all the functions on $A$ are supported on one finite interval, that would be enough to have a finite $\epsilon$-entropy. More generally, it would be enough if you add the condition that all the functions on $A$ are dominated by one function in $L^1$. | |
| Oct 20, 2020 at 16:09 | comment | added | Riku | Then what could help to have finite $\epsilon$-entropy in this context? | |
| Oct 20, 2020 at 13:14 | comment | added | Iosif Pinelis | These conditions will not help. Indeed, in the above counterexample, the $f_n$'s have compact support and are in $L^1\cap L^\infty$. | |
| Oct 20, 2020 at 7:55 | comment | added | Riku | Thank you. So is compact support a necessary condition for the functions in $A$ for the set to have finite $\epsilon$-entropy? Or what if I replace $L^1$ with $L^1 \cap L^\infty$? | |
| Oct 20, 2020 at 0:44 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 20, 2020 at 0:07 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |