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user12692
12:52 AM
Sorry for interrupting the long conversation.
 
user12692
I would just like to ask, any owner of a chat room can make others become an owner; any owner of a chat room can "cancel" other owners privilege as owners, is that right?
 
user12692
@MartinSleziak
 
1:19 AM
@Jack Yes. I just tested it on Gentle.
 
 
4 hours later…
5:49 AM
47
Q: Is there a list of SE chat privileges, and the minimum reputation required for those privileges?

kiamlalunoIs there a list of privileges that users have on any Stack Exchange chat rooms, and which lists the minimum reputation required for each privilege? If such list doesn't exist, what is the reputation required for the chat privileges?

I have just edited the answer so that it says more explicitly that RO can not only add but also remove the room owners.
 
 
1 hour later…
6:57 AM
@TYG Sorry, I missed your messages since I only saw the most recent ones by Jack and Zacky.
 
TYG
Thanks Martin
The lemma of Extension to the dimension one higher
Is the result \hat{f}(x)=θ and \hat{f}(x)\leq p(x)?
In your conclusion, why "Now if we apply Hahn-Banach theorem once again to fˆ, we get a functional LIM:ℓ∞→R which is a generalized limit and we also have"
Why we have LIM(x_0)=\hat{f}(x_0)=\theta?
Do you mean for each extension of \hat{f}, \hat{f}(x_0) always take same value \theta?
@MartinSleziak
Well, I am a little confused about your proof of question 1. I know your lemma shows that for the first extension(not the whole space) we still have the $\hat{f}$ is dominated and takes value \theta. But why we can apply H-B theorem again, state that $LIM$ still takes the same value $\theta$?
Use the fact $\liminf \leq LIM\leq \limsup$ then we need to let $\theta \in [\liminf, \limsup]$?
Anther question:
For "If x is a sequence such that for any generalized limit we get the same value L". Do we need to use the result in part (1)?
Another question, if I consider the Banach sequence which is defined by a generalized limit with shift-invariant. Is B(\{a_n\})=LIM(1/n \sum_{j=1}^n x_{j}) a Banach limit?
Here is my proof:
Do you think if it is right?
Here is the key point: LIM((\frac{x_1-x_{n+1}}{n})+B(x_2,x_3,x_4,\dots)=B(x_2,x_3,x_4,\dots) since $\Vert x_1-x_{n+1}\Vert \leq 2\Vert x\Vert_{\infty}$ which is bounded. Then LIM (\frac{x_1-x_{n+1}}{n})=0
Then I can show that $B(\{x_{n+1}\})=B(\{x_{n}\})$.
Remark: I use B(\{x_{n}\})=B(x_1,x_2,x_3,\dots)=B(x_1-x_2, x_2-x_3,x_3-x_4,\dots)+B(x_2,x_3,x_4,\dots)=LIM(\frac{1}{n}\sum_{i=1}^n(x_i-x_{i+1}))+B(x_2,x_3,x_4,\dots)
Denote the shift operator $T: l_{\infty}\to l_{\infty}$ by \[T: (x_1, x_2, x_3, \dots)\to (x_2, x_3, x_4, \dots).\] So we have $B(T\{x_{n}\})=B(\
 
Sorry, I won't be able to read all of this - I will have other duties today (and tomorrow).
Let me respond at least to this, since the answer there is simple:
8 hours ago, by TYG
Why we have LIM(x_0)=\hat{f}(x_0)=\theta?
We have $x_0\in\widehat M$. And $\operatorname{LIM}$ is an extension of $\widehat f$ from $\widehat M$ to the whole space. (Obtained using HBT.)
So we have $\operatorname{LIM} x= \widehat f(x)$ for every $x\in\widehat M$.
 
TYG
It's ok. I think I am right. I just want to make sure
 
In particular, we have $\operatorname{LIM} x_0 = \widehat f(x_0)=\theta$.
 
TYG
Just the Banach limit
 
7:03 AM
BTW there is also a room called Modern Abstract Analysis where one of the topics is functional analysis.
 
TYG
So $\operatorname{LIM} x= \widehat f(x)$ because \operatorname{LIM}| \widehat M=\widehat f(x)?
 
In case you have some other question from functional analysis. (Of course, we could discuss the stuff related to my answer here. But the question about Banach limits seems to be a new question.)
@TYG Yes. Because $\operatorname{LIM}|_{\widehat M}=\widehat f$ and $x\in\widehat M$.
 
TYG
I see your a paper
 
TYG
that states one proof of the ceraso limit is shift-invarinat
then the generalized limit is Banach limit
 
7:07 AM
I don't know what you mean. Do you have a link?
BTW you're probably not using bookmarklet for MathJax in chat, I guess. (Judging by the fact that you omit dollars in many places - which makes your messages a bit harder to read.)
 
TYG
Oh, sorry for that
In page 6
Your proof of Construction of Banach limit.
 
Ok, that's not really a paper, that's just a talk I gave at some students' seminar.
 
TYG
It only remains to prove the shift-invariance
Okay
"It only remains to prove the shift-invariance"
 
Yes, there is a proof that Banach limit exists using limit along ideal. (The same thing could be done using filters.)
 
TYG
That is the same of generalized limit ?
 
7:10 AM
> It only remains to prove the shift-invariance. To show this, simply observe that $\frac{x_1+\ldots+x_n}n - \frac{(Sx)_1+\ldots+(Sx)_n}n = \frac{x_1+\ldots+x_n}n - \frac{x_2+\ldots+x_{n+1}}n= \frac{x_1-x_{n+1}}n$.
 
TYG
If I want to construct a Banach limit from a generalized limit which is required that is shift-invariant
 
People use the term generalized iimit in many various meanings.
But generalized limit which you discussed is a different notion from Banach limit and a different notion from F-limit (where F is a filter).
 
TYG
Just the definition of generalized limit we talked yesterday
I define the Banach limit with generalized+shift-invariant property
 
It is still true that U-limit (where U is an ultrafilter) is special case of generalized limit. And U-limit is shift-invariant.
(Or, you can use maximal ideals instead of ultrafilters.)
 
TYG
I see.
 
7:13 AM
@TYG Yes, that's the same definition of Banach limit which I am using. (Or at least they are equivalent.)
 
TYG
I just follow your method to show that a generalized limit with cersaro sum (i.e. $\frac{1}{n}\sum x_j$) has shift-invariant property
 
Yes, that should work.
 
TYG
LIM $\frac{1}{n}\sum x_j$
Yes
 
As I said, I have some other stuff to do. And probably I'll be busy today and also tomorrow.
Sorry that I cannot have a more detailed look right now.
See you later!
 
TYG
Thank you very much
See you
 
 
7 hours later…
2:43 PM
in GENTLE, 46 mins ago, by Jack
The body of the answer you linked says (emphasis mine):

It depends; Look at the context of how they are asked.

You'll want to keep the question **if the wording provides another way for a search query to find the content.** It's a bit like a "see also …" entry in an index.

Delete it if the duplicate does not add terminology or alternate phrasing to find the question. It may not be worth cluttering up the system with this exact duplicate. There are several poorly phrased questions that will not direct traffic to us.
in GENTLE, 44 mins ago, by Jack
@MartinSleziak: do you know that why sometimes the formating **...** does not work in chat?
AFAIK many things do not work in chat if you have multiline messages. (Probably formatting. Links - at least in transcript.)
69
Q: Markdown in chat fails for multi-line messages

sbiWe've come to rename Markdown to Letdown in the C++ chat room because it lets you down so often. I've now just found a pattern. It seems markdown fails for multi-line messages. That is, this Letdown can't cope with multi-line comments. Let's see code? fails to display code marked as co...

Just a test. This is bold. Link to Google.
Just a test.
**This is bold.**
Link to [Google](https://www.google.com/).
@Jack See above.
 

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