Timeline for Closure of the space of holomorphic functions on the open disk in $\mathbb{C}$ with respect to a Hardy-space-like semi-norm
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2019 at 20:38 | history | edited | MCS | CC BY-SA 4.0 |
added 361 characters in body
|
| Sep 4, 2019 at 20:32 | comment | added | MCS | @Yemon Choi: There exist a large family of functions for which the limit does exist. Those are the functions that I'm concerned with. For example, as a consequence of the Hardy-Littlewood Tauberian Theorem, all linear combinations of set-series of sets with well-defined natural density have finite semi-norm, as do all rational functions. | |
| Sep 4, 2019 at 20:29 | comment | added | MCS | @reuns: I know about semi-norms; I just goofed and failed to notice that this was only a semi-norm, not a norm. The counterexample didn't occur to me until just as I was going to sleep. xD | |
| Sep 4, 2019 at 20:28 | history | edited | MCS | CC BY-SA 4.0 |
deleted 99 characters in body; edited title
|
| Sep 4, 2019 at 0:21 | comment | added | Yemon Choi | I also don't understand why you get a well-defined seminorm, i.e. why should the limit in your definition be finite? | |
| Sep 3, 2019 at 23:49 | comment | added | reuns | To me this is not research level, why don't you ask on MSE. Your $\|.\|$ is only a semi-norm for entire function $\|f\|=0$ so the natural map $\to H$ isn't injective. For your question (1) sure the $L^2$ norm of $e^{1/(z+1)}$ on $|z|=r$ isn't $O( (1-r)^{-1})$. | |
| Sep 3, 2019 at 23:40 | history | asked | MCS | CC BY-SA 4.0 |