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 $\limsup(x_n+y_n)=\limsup x_n + \lim y_n$."

$c$ 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.

Ok I see

Yes $\|\lim\|=1$ is exactly what I meant.

This is what we get as the operator norm if we consider $\lim$ as a linear functional on the space $c$.

So your proof of the Extension to the dimension one higher is another proof

not the same as the proof of the Hahn-Banach theorem?

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 $p(x+y)=f(x)=p(y)$.

Since $\limsup(x_n+y_n)=\limsup x_n + \lim y_n$, we have $p(x+cx_0)=f(x)+p(cx_0)$?

okay

For the proof of $-p(-x)\leq LIM x\leq p(x)$ Why I cannot get the result $LIM x\leq p(x)$? You say that is not right.

In this case we have the equality $p(x+y)=f(x)+p(y)$, therefore the range can be simplified.

@TYG There are examples of linear functionals $g: \ell_\infty \to\mathbb R$ which extend the limit and which do not fulfill $g(x)\le\limsup x$.

As I have mentioned in the footnote, you can look at this post: Continuous extension of the limit functional.

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

@MartinSleziak Okay, thanks.

In one of the answers you have a functional $f_1$ which maps $(1,-1,1,-1,\dots)$ to $2$.

There is also another functional which $f_2$ which maps this sequence to $-2$.

And both $f_1$ and $f_2$ extend limit.

@MartinSleziak I see

Clearly $2$ is bigger than $\limsup x$

@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.

But how to show every generalized limits takes same value? I mean if for a $(a_n)$ converges, then we have one generalized 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 $\|\LIM\|=1$.

If $a_n\to a$, then some entended limit denoted by $LIM_{\alpha}$ we have $LIM_{\alpha} a_n=a$. So how to show all $LIM_{\alpha}$ take same values?

By "$\LIM$ extends limit" I mean precisely this: If $a=(a_n)$ converges to $L$, i.e., $\lim a=L$, then also $\LIM(a)=L$.

I feel like there are many different generalized limits

In other words, $\LIM|_c=\lim$.

@TYG Yes, there are many of them. But what is the definition of the notion generalized limit.

For example, would you consider the functional defined as $g(x)=0$ to be a generalized limit?

I define each extension, we can have a new generalized limit. Isn't right?

I don't understand what you mean by this.

Again, is $\LIM x=0$ (i.e., we assign zero to every sequence) a generalized limit?

I mean for the sequence $(a_n)\to a$ we have an extension $LIM_{\alpha}$ such that $LIM_{\alpha}(a_n)=a

But we need to show that for every $\alpha$ this is true.

Ok, so the next question is, what yo mean by $\LIM_\alpha$? What is $\alpha$?

That is LIM_{\alpha}(a_n)=a and also maybe LIM_{\beta}(a_n)=b. Isn't right?

And what is $\beta$?

some extensions

Any extensions? Or do they fulfill some conditions?

We are still getting back to the question what you consider as the definition.

Because we can extend the classical limit to some subspaces of the the $X$?

Do you require for a generalized limit the condition $\LIM|_c=\lim$.

I.e., that you have $\LIM x=L$ for a convergent sequence $x$ which has limit $L$.

yes, we need that.

Or, if you prefer that notation, if $a_n\to a$, then for any two extensions $\LIM_\alpha$ and $\LIM_beta$ you have $\LIM_\alpha(a)=L=\LIM_\beta(a)$.

Why?$\LIM_\alpha(a)=L=\LIM_\beta(a)$? Just the definition?

@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.

Okay. I see. Thanks a lot.

@TYG Because $\LIM$ extends the usual limit.

So the $\LIM|_c=\lim$

Well, you agreed that this is part of the definition.

How to conclude $\LIM$ are same on the whole space?

@TYG I am not sure what you mean. I did not claim that $\LIM$ are the same on the whole space.

"this immediately tells you that for the every convergent sequence you have $\LIM(x)=\lim x$"

Why? I feel like $\LIM(x)|_C=\lim x$

There are many generalized limits. I only claim that for a convergent sequence $x$ each of them gives the same value, namely the usual limit $\lim x$.

okay. I see

thanks. I have no other questions.

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.

Okay, if I assume the $L(x)\le\limsup x$ as the third condition. So can I write the following proof about the $f(x)\leq \limsup x=p(x)$:

I know that $-p(-x)\geq f(x)\leq p(x)$

I ask a trivial question: why the norm of the limit operator is 1?

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.

Let $g\colon c\to\mathbb R$, $g(x)=\lim x$.

Clearly you have $|g(x)|\le \|x\|$, since $|\lim x_n| \le \sup |x_n|$.

So we know from this that $\|g\|\le 1$.

On the other hand, the constant sequence $\overline 1=(1,1,1,\dots)$ has norm equal to one and $g(1)=1$.

So for this sequence we have $|g(x)|/\|x\|=1$.

Hence $\|g\|=1$.

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 $f$.

@TYG I have tried to answer your two recent questions above.

In case it is useful, I have posted a separate question which should show why the additional condition is needed: Do we have $f(x)\le\limsup x_n$ for every functional extending limit?

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?

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)

Hence B(\{x_{n}\})=B(T^{k}\{x_{n}\})=B(\{x_{n+k\}).