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9:37 AM
A reminder: For instructions how to render MathJax(TeX) in chat see this post on meta or go directly to robjohn's website
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A: Show that $\liminf_{n\to \infty}x_{n}\le\alpha(x)\le\limsup_{n\to\infty}x_{n}$ for $x=(x_{n})$ in $\ell^{\infty}$

Martin SleziakI suppose you speak about Banach limits (as was suggested in the comments). You wrote that you only want a hint, I tried to give a complete answer. So if you want to try it by yourself, do not scroll completely to the end. I will assume that your definition of Banach limit is that it is positiv...

@MartinSleziak Sorry, I do not know how you prove" we get that, for any given bounded sequence x, all values from the interval $[m(x),M(x)]$ can be attained by some Banach $\lim f$" — TYG 6 hours ago
@TYG Notice the remark in parenthesis which contains link to this post: Reference for the range of possible values in Hahn-Banach Theorem. From this version of HBT we get that each value $t$ between $-M(-x)$ and $M(x)$ there is a linear functional $f:\ell_\infty\to\mathbb R$ with $f(x)=t$ which extends limit and fulfills $f(x)\le M(x)$ for convergent sequences. — Martin Sleziak 5 hours ago
@MartinSleziak So if the Banach limit takes the same value. Since the value between $m(x)$ and $M(x)$. CAN I say $m(x)=M(x)$? Which theorem claims that? Then Banach limit is just the classcial limit. — TYG 5 hours ago
@TYG A sequence for which all Banach limit are the same is called almost convergent. The equality $m(x)=M(x)=L$ is basically a reformulation of Lorentz's characterization of almost convergent sequences; it says that $T_n(x)$ converges to $L$ uniformly in $n$. — Martin Sleziak 5 hours ago
Now I see that you have actually posted also a new question about this:
0
Q: How to show exists $\theta$ such that LIM $x_n=\theta$?

TYG Suppose $(x_n)$ is bounded sequence. For any number $\theta\in [\liminf a_n,\limsup a_n]$, there exists a generalised limit functional that assigns that number $\theta$ as the "limit" of the sequence $(x_n)$. By Hahn-Banach theorem, we can extend this limit. This claim can be seen by firs...

I would suggest the following:
We will use Hahn-Banach theorem with $p(x)=\limsup x_n$.
Using this we get a linear functional on $\ell_\infty$ such that $-p(-x)\le f(x) \le p(x)$.
This is precisely the functional such that $\liminf x_n \le f(x) \le \limsup x_n$.
If we use the version of HBT formulated in this question, we see that for a given $x$ it is possible to get any value between $-p(-x)=\liminf x_n$ and $p(x)=\limsup x_n$.
It is a bit unclear to me what you're actually asking. One thing is that the post doesn't define what you mean by generalized limit functional - but I assume that you mean a functional defined on $\ell_\infty$ which is linear, continuous and extends limit. Did you mean to ask how to show that for a given $x$ and $\theta$ between $\liminf x_n$ and $\limsup x_n$ there exist a functional $\operatorname{LIM}$ with $\operatorname{LIM} x=\theta$? — Martin Sleziak 2 mins ago
Since the way it is phrased - asking how to we know that there exists $\theta$ such that $\operatorname{LIM} x=\theta$ - the answer could simply be: Because $\operatorname{LIM}$ is defined on $\ell_\infty$, i.e., $x$ belongs into the domain. The question seems also related to our exchange in comments under another post - I left a few related comments in chat. — Martin Sleziak 43 secs ago
These posts seem to be related to extension of limit to the whole $\ell_\infty$.
4
Q: Continuous extension of the limit functional

Marco VerguraLet $(\ell^{\infty})'$ be the $\mathbb{F}$-vector space of linear and continuous (bounded) functionals $\ell^{\infty}\rightarrow \mathbb{F}$, where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$ (but we can assume $\mathbb{F}=\mathbb{R}$, if needed) and $\ell^{\infty}$ has the sup norm $\par...

4
Q: Linear functional and convergent series in $\ell^\infty$

JohanLet $\ell^\infty$ be the Banach space of bounded sequences with the usual norm and let $c,c_0$ be the subspaces of sequences that are convergent, resp. convergent to zero. Show that: The linear functional $\ell_0\colon c\rightarrow \mathbb{C}$ defined for $x = (x_n) \in c$ by $$ \ell_0(x) = \li...

3
Q: Hahn–Banach Theorem for Normed Spaces: not unique extension

MrReeseLet $\ell^{\infty}$ be the set of bounded sequences in $\mathbb{F}$, with the supremum norm. $c \subset \ell^{\infty}$ the sequences whose limit exists. Then there exists a $f \in (\ell^{\infty})'$, the dual, such that $f(x) = \lim_{n \to \infty} x(n)$ for all $x \in c$. Because we can define ...

@MartinSleziak Yes. you understand right. Just the Hahn-Banach theorem of the $l_{\infty}$. — TYG 3 mins ago
If you are online, why not try coming to chat, so that you can clarify what the question is about. (Or if you agree with my comments saying that the current wording is basically asking the question "the other way round" than intended, you could edit the question to get the formulation which you actually want to ask.) — Martin Sleziak 14 secs ago
 
 
2 hours later…
11:51 AM
@MartinSleziak But how about the generalized limit? without shift is invariant? — TYG 7 hours ago
@TYG I have posted an answer to one of your previous questions: A necessary and sufficient criterion for a bounded sequence ${(a_n)_{n=1}^\infty}$ to classically converge to a limit ${L}$. I suppose this question is related to your previous comment. — Martin Sleziak 26 secs ago
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A: A necessary and sufficient criterion for a bounded sequence ${(a_n)_{n=1}^\infty}$ to classically converge to a limit ${L}$

Martin SleziakWhat is a generalized limit?$\newcommand{\LIM}{\operatorname{LIM}}$ The question is about generalized limits $\LIM\colon\ell_\infty\to\mathbb R$. To deal with the question, we need to clearly define what we mean by a generalized limit. We want $\LIM$ to be a linear continuous functional on $\el...

 

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