Timeline for Absolutely summing operators from $l_{p}$ to $l_{q}$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15, 2023 at 17:04 | comment | added | Bill Johnson | @Dirk: Thanks. I deleted and made a correct comment. | |
| Jan 15, 2023 at 17:03 | comment | added | Bill Johnson | There is a compact operator from $\ell_p$ to $X$ that is not absolutely summing for all values of $1<p<\infty$ when $X$ is infinite dimensional. Otherwise, since $\ell_2^n$'s are uniformly complemented in $\ell_p$ and uniformly embed into $X$ by Dvoretzky's theorem, the Hilbert-Schmidt norm would be equivalent to the operator norm on $\ell_2$. | |
| Jan 15, 2023 at 14:27 | comment | added | Dirk Werner | @Bill: For the Pisier space $P$, all tensor norms on $P\otimes P$ are equivalent. So every bounded operator from $P$ to $P^*$ is integral, in particular 1-summing, with equivalent norms. [So there are infinite dimensional spaces between which the operator norm is equivalent to the 1-summing norm.] | |
| Jan 15, 2023 at 2:42 | vote | accept | Dongyang Chen | ||
| Jan 15, 2023 at 1:33 | comment | added | Bill Johnson | Absolutely summing operators factor through $\ell_2$ and operators from $\ell_2$ to $\ell_q$ are compact when $q<2$, so question 3 has a negative answer for $p=1$. | |
| Jan 15, 2023 at 0:50 | comment | added | Dongyang Chen | For $1<p,q<\infty$, is there a compact but not absolutely continuous operator from $l_{p}$ to $l_{q}$ ? | |
| Jan 15, 2023 at 0:41 | vote | accept | Dongyang Chen | ||
| Jan 15, 2023 at 0:53 | |||||
| Jan 15, 2023 at 0:40 | vote | accept | Dongyang Chen | ||
| Jan 15, 2023 at 0:40 | |||||
| Jan 15, 2023 at 0:38 | comment | added | Dongyang Chen | You are right, Dirk. What about Question 3 if $p=1$ and $q<2$. Obviously, Question 3 is true if $p=1$ and $q\geq 2$. | |
| Jan 14, 2023 at 20:24 | history | answered | Dirk Werner | CC BY-SA 4.0 |