Timeline for What standard Banach space is isomorphic to the completion of this different normed structure on $\ell^1$?
Current License: CC BY-SA 4.0
35 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 10, 2023 at 16:21 | answer | added | Lorenzo Pompili | timeline score: 14 | |
| Apr 8, 2023 at 12:13 | answer | added | S Argyros | timeline score: 8 | |
| Apr 6, 2023 at 17:05 | answer | added | Willie Wong | timeline score: 8 | |
| Apr 6, 2023 at 15:39 | comment | added | Jochen Glueck | @SArgyros: Argh, you're of course right - the norm that I suggested on the (purported) dual space was nonsense - what I wrote down is simply the $\ell^2$-norm. Don't know what I was thinking... | |
| Apr 6, 2023 at 15:29 | comment | added | S Argyros | @Jochen Glueck The description of the space you suggests seems to me quite promised. | |
| Apr 6, 2023 at 14:06 | comment | added | Jochen Glueck | If yes, then it would be natural to check next if the dual space is given by $\bigcup_{q \in [2,\infty)} \ell^q$ with the norm $\|x\| = \sup_{q \in [2,\infty)} \|x\|_q$. | |
| Apr 6, 2023 at 14:00 | comment | added | Jochen Glueck | Let me try to adapt @GeraldEdgar's first comment: Doesn't the completion consist of precisely those sequences in $\bigcap_{p \in (1,2]} \ell^p$ for which the integral is finite? | |
| Apr 6, 2023 at 13:07 | comment | added | S Argyros | @AliTaghavi Concerning the question the $lim$ concerns the natural numbers n. If the answer is positive then that means that $\epsilon $ mass of the integral quantity is asymptotically concentrated around 1. As I said before I made some experiments and here is one. Consider $ \int_1^{2 } 10^\frac{1000} {p}dp$ and $ \int_1^{1+\frac {1}{2^6 }} 10^\frac{1000} {p}dp$ if you calculate in an integral calculator these two integrals then you will see that are almost equal. This makes me to believe that $\epsilon = 1$. Concerning the measures $\mu_{x_n}$ are defined in one of my previous comments. | |
| Apr 6, 2023 at 12:21 | comment | added | Ali Taghavi | @SArgyros I have no idea. is n fix in your question? Do you mean that such a possible $\epsilon$ and $\delta$ work for all $n\in \mathbb{N}$? Could you plz elaborate the last part of your comment? (In particular the measur language you used in the last part). Is $\mu_{x_n}$ the dirac measure ? Or some thing else? | |
| Apr 6, 2023 at 12:09 | comment | added | S Argyros | @AliTaghavi I do not have to say something more related to the structure of the space. I would like only to know the answer of the question included in my previous comment. | |
| Apr 6, 2023 at 10:39 | comment | added | Ali Taghavi | @fedja I do not have your email address. I would appreciate if you send me email for more discussion. My email adreess is written in my profile. | |
| Apr 6, 2023 at 10:37 | comment | added | Ali Taghavi | @SArgyros I do not have your email address. I would appreciate if you send me email for more discussion. My email adreess is written in my profile. | |
| Apr 6, 2023 at 10:34 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 18 characters in body
|
| Apr 1, 2023 at 14:00 | comment | added | S Argyros | Can we prove that there exists an $\epsilon > 0$ such that for all $\delta>0$ $\lim \frac {\int_1^{1+\delta } n^{1/p} dp} { \int_1^2 n^{1/p} dp} \geq \epsilon $? I made some numerical experiments and I would not be surprised if $\epsilon = 1$.If this is true and $x_n$ is the $n $ normalized average of the basis then the sequence of the probability measures $(\mu_{x_n})$ is not uniformly integrable which yields that $(x_n)$ has a subsequence equivalent to $l_1$ basis.For me it is natural to expect that the space is $l_1$ saturated. | |
| Apr 1, 2023 at 8:53 | comment | added | Gerald Edgar | @AliTaghavi I agree with you, my first comment was wrong. | |
| Apr 1, 2023 at 8:50 | comment | added | Ali Taghavi | @GeraldEdgar I still do not understand your comment. I belive that the integral you mentioned is always finite if $x\in \ell_p$ for all $p\in [1,2]$. so the finiteness condition you mentioned is redundant. | |
| Apr 1, 2023 at 7:54 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 10 characters in body
|
| Mar 31, 2023 at 15:29 | comment | added | S Argyros | To each $x$ we assign the measure $\mu_x$ by the rule for $A$ Borel $\mu_x(A)=\int_A|x|_pdp$. I think that for a normalized block sequence $(x_n)$ the following holds. The sequence $(\mu_{x_n})$ is uniformly integrable iff the sequence $(x_n)$ is weakly null. Otherwise $(x_n)$ has a subsequence equivalent to $l_1$ basis. | |
| Mar 31, 2023 at 14:05 | comment | added | S Argyros | The averages of the basis with increasing size converge norm to zero. The norm of an n-average of the basis is equal to $\int_1^{2} n^{{1/p}-1}dp$ and the sequence of functions $ n^{{1/p}-1 }$ converges pointwise to the function which is zero everywhere except the one where its value is one. Also each one is dominated by the constant function one. Hence Lebesque dominated convergence theorem yields that the norm of the n-averages converges to zero.( My thanks to Alexandros Georgiou for pointing to me this proof) | |
| Mar 31, 2023 at 4:39 | comment | added | fedja | @OnurOktay Ermm... Not to the sum, but to the subspace of that sum cut by the diagonal $x^{1}=x^{2}=\dots$, which seems to change the game entirely. But yeah, an integral of a decreasing function is, indeed, equivalent to the sum you suggested. | |
| Mar 31, 2023 at 2:34 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 3 characters in body
|
| Mar 30, 2023 at 21:03 | comment | added | Onur Oktay | I'd like to write down a simple observation. The norm $\|x\|:= \sum_{n=0}^{\infty} 2^{-n} |x|_{p_n}$ where $p_n = 1+2^{-n}$ is an equivalent norm. So this space is isomorphic to $(\bigoplus \ell^{p_n})_{\ell^1}$ that is not reflexive. | |
| Mar 30, 2023 at 18:14 | comment | added | S Argyros | Is the following inequality true? $|x| \leq \int_1 ^{2} ( \int_1 ^{2} |x|_p ^{p }dp )^{1/p } dp $. If yes will be helpfull in understanding the space. | |
| Mar 30, 2023 at 16:50 | history | edited | Peter LeFanu Lumsdaine | CC BY-SA 4.0 |
replaced fragile link-only reference with a human-readable reference and archival-quality link
|
| Mar 30, 2023 at 16:37 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 219 characters in body
|
| Mar 30, 2023 at 15:46 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 161 characters in body
|
| Mar 30, 2023 at 14:46 | comment | added | terceira | There doesn't appear to be any substantial reason to suppose that this space is isomorphic to a classical one, rather it is natural to regard it as a space sui generis and to try to attack the second question by attempting to compute its dual. The natural conjecture is that it is the completion of the analogue space defined by the integral $\int_1^2|x|_q dp$ where $q$ is the conjugate of $p$, i.e., $\frac p {p-1}$. | |
| Mar 30, 2023 at 13:37 | comment | added | S Argyros | For $x$ in $l_1$ the function $|x|_p$ $1\leq p \leq 2$ is decreasing hence the norm is finite and roughly speaking is the average of $| x| _p$. The basis is symmetric and boundedly complete. Hence if the space does not contain $l_1$ it is reflexive. To show that this happens it is enough to prove that the norm of the averages of the normalized block sequences tend to zero when their size increases to infinity. | |
| Mar 30, 2023 at 13:04 | comment | added | Ali Taghavi | @GeraldEdgar This norm is already finite so I think the finitness condition you mention is redundant since $|x|_p\leq |x|_1$ | |
| Mar 30, 2023 at 12:46 | comment | added | Gerald Edgar | Can we show that it is the space of sequences $x$ such that $|x|=\int_1^2 |x|_p dp$ is finite? Is that complete in this norm? This space seems to be larger than $\bigcup_{p<2} \ell^p$ but smaller than $\ell^2$. | |
| Mar 30, 2023 at 10:05 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
added 1 character in body
|
| Mar 30, 2023 at 9:35 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
edited title
|
| Mar 30, 2023 at 9:27 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
edited title
|
| Mar 30, 2023 at 9:13 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
edited tags
|
| Mar 30, 2023 at 9:08 | history | asked | Ali Taghavi | CC BY-SA 4.0 |