Thanks for your nice answer. But how about the other direction? If all sequence $(a_n)$ converges, then evert generalised limit takes the same value? — TYG1 hour ago
Why we could assume $\Vert LIM \Vert$=1? — TYG57 mins ago
Also, I feel like $LIM(x)\leq p(x)$ is obvious since $LIM$ is the extension of $\lim$. By Hahn-Banach theorem, we have the extension $LIM x \leq p(x)$? Isn't it right? — TYG31 mins ago
@TYG It all depends on what you consider the definition of generalized limit. I suppose you're talking about the definition used in the linked blog post - I quoted the part which says, to my understanding, that the assumption $\|LIM\|\le1$ is part of the definition. (It is the part which says " without any increase in the operator norm.") — Martin Sleziak9 mins ago
From your comment it seem that you want to take just take this as the definition: $\operatorname{LIM}$ is a linear continuous functional which extends $\operatorname{LIM}$. Then it is no longer true that $\liminf x\le\operatorname{LIM}(x)\le\limsup x$. The footnote links to a post which has an example showing this. — Martin Sleziak9 mins ago
so I still do not understand why $\Vert LIM\Vert \leq 1$. What does "without any increase in the operator norm" mean? — TYG5 mins ago
I suppose that T. Tao say here that $\|\operatorname{LIM}\|$ is a functional which has the same norm as the operator norm of $\lim: c\to\mathbb R$. Which is $1$. — Martin Sleziak3 mins ago
Also. for your proof of the Extension to the dimension one higher, why $p(x+cx_0)=f(x)+p(cx_0)?$ I feel like $p(x+cx_0)\leq p(x)+p(cx_0)$. — TYG3 mins ago
Regarding $p(x+cx_0)\le f(x)+p(cx_0)$, this is explained in the post: " Moreover, for any convergent sequence $(y_n)$ we have $\limsup(x_n+y_n)=\limsup x_n + \lim y_n$."
What is the $c$? So the norm of all convergent sequences is 1?
$c$ is the space of all convergent sequences. If you look at the functional $\lim$ defined on this space, the norm of this functional is one, i.e., $\|\lim\|=1$. (We consider operator norm here.)
I'll just mention that a few comments related to extensions of limits can be found in the transcript of this room a few days ago: chat.stackexchange.com/transcript/19138/2019/11/29 In since we are looking at various rooms in chat, there is also room called Modern Abstract Analysis, where one of the topics is functional analysis.
Well, if you want avoid using the version of Hahn-Banach theorem I linked to, then I had to prove this fact separately.
@TYG This appears as a part of the usual proof of Hahn-Banach Banach theorem. However, in a slightly more complicated version. (Since in Hahn-Banach theorem you do not assume $p(x+y)=f(x)=p(y)$.
So if you look at the proof of Hahn-Banach theorem, you can find in there the proof of this lema, but $$\sup_{x\in M} [f(x)-p(x-{v})] \le \theta \le \inf_{y\in M} [p(y+{v})-f(y)]$$ rather that $$-p(-x) \le \theta \le p(x).$$
Also, for the statement:"(a_n) converges if and only if every generalized limit takes the same value." I know it is obvious that "if every generalized limit takes the same value" then $\limsup=\lininf$. So (a_n) converges. But how about the another direction
@TYG If $a=(a_n)$ is convergent, then for every generalized limit you have $\operatorname{LIM} a=\lim a$ simply because $\operatorname{LIM}$ extends limit.
This is part of the definition of generalized limit. At least the way I understand it.
> So let's take this as the definition of a generalized limit: $\LIM$ is a linear continuous functional defined on the space $\ell_\infty$ such that $\|\LIM\|=1$.
What do you consider as your definition of generalized limit.
Oh, I see that I have forgotten to include that $\LIM$ extends the usual limit in the definition.
I have now edited the post:
> So let's take this as the definition of a generalized limit: $\LIM$ is a linear continuous functional defined on the space $\ell_\infty$ such that $\LIM$ extends the usual limit and $\|\LIM\|=1$.
@TYG Well, if you take $\LIM|_c=\lim$ as a part of the definition of generalized limit, this immediately tells you that for the every convergent sequence you have $\LIM(x)=\lim x$. So for convergent sequences there is the only possible value.
I will just repeat that for me, this is definition of a generalized limit.
A function $f\colon \ell_\infty\to\mathbb R$ is called generalized limit if it fulfills these conditions:
1. $f$ is a linear continuous functional.
2. $f$ extends the usual limit.
3. An additional third condition which can be formulated in several equivalent ways.
We can take $\|f\|=1$ and the third condition.
We can take $f(x)\le\limsup x$ as the third condition.
We can take that $f$ is positive as the third condition.
It seems to me that one of the reasons that sometimes you and me did not understand what the other person is talking about is that you never clearly stated what you consider as your definition of a generalized limit.
So if I just want to use the Hahn-Banach theorem. I still feel like $f(x)\leq p(x)$ is the result of H-B theorem.
Well, it's true that HBT says that if you start for a with a functional majorized by $p(x)$ on the subspace, then you get a functional such that $f(x)\le p(x)$ is true for the whole space.
So if you take $f(x)\le\limsup x$ as a part of the definition of generalized limit, this is for free.
Of course, that isn't the only possible definition. If I understood it correctly, the blog post you linked uses a different definition of a generalized limit.
So if you work with a different definition, you need to prove the $f(x)\le\limsup x$ is true for any generalized limit $f$.
Let us consider functionals $f\colon\ell_\infty\to\mathbb R$ such that:
$f$ is linear and continuous,
$f$ extends the usual limit, i.e., if $x=(x_n)$ is convergent and $\lim\limits_{n\to\infty} x_n=a$, then also $f(x)=a$.
Question. Do these conditions already imply that $f(x)\le\limsup\limits...
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?
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 $\theta \in [\liminf, \limsup]$?
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 $B(T\{x_{n}\})=B(\{x_{n+1}\})$. Suppose for $k-1$ we have $B(T^{k-1}\{x_{n}\})=B(\{x_{n+k-1}\})=B(\{x_{n}\})$. Then for $k$, we have \[B(T^{k}\{x_{n}\})=B(T(T^{k-1}\{x_{n}\}))=B(T\{x_{n}\})=B(\{x_{n}\}). \]
Let us consider functionals $f\colon\ell_\infty\to\mathbb R$ such that:
$f$ is linear and continuous,
$f$ extends the usual limit, i.e., if $x=(x_n)$ is convergent and $\lim\limits_{n\to\infty} x_n=a$, then also $f(x)=a$.
Question. Do these conditions already imply that $f(x)\le\limsup\limits...
