Show that $\liminf_{n\to \infty}x_{n}\le\alpha(x)\le\limsup_{n\to\infty}x_{n}$ for $x=(x_{n})$ in $\ell^{\infty}$

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Question:

Show that $\liminf_{n\to \infty}x_{n}\le\alpha(x)\le\limsup_{n\to\infty}x_{n}$ for $x=(x_{n})$ in $\ell^{\infty}$, where $\alpha$ is a bounded linear functional on $\ell^{\infty}$.

I try to solve this problem, but I do not know how to start, please I need a hint.

Are you asking about Banach-Limits? If so, you should probably provide a bit more context (the usual construction appeals to Hahn-Banach to extend the limit functional from the convergent sequences using the dominating sublinear functional $\limsup |x_n|$. Note that for a general bounded linear functional this is completely wrong (even in the case that it has norm $1$): take $\alpha$ to be $\alpha((x_n)) = x_1$, for example. — t.b., Nov 09 '11 at 16:28
That, or $\alpha((x_n))=0$ for every $(x_n)$ in $\ell^\infty$. — Did, Nov 09 '11 at 16:32
@HassanMuhammad: Please edit the question with your clarifications. It would help if you state exactly what you are assuming about $\alpha$. — Nate Eldredge, Nov 09 '11 at 16:54

score 8 · Accepted Answer · edited Apr 13 '17 at 12:20

8

I 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 positive, shift-invariant, linear function from $\ell_\infty$ to $\mathbb R$ which extends limit, i.e., $f: \ell_\infty \to \mathbb R$ is linear and

$x\ge 0$ $\Rightarrow$ $f(x)\ge 0$;
$(\forall x\in\ell_\infty) f(Tx)=f(x)$;
if $x$ is convergent then $f(x)=\lim x$.

Here $T:\ell_\infty\to\ell_\infty$ is shift-operator $T:{(x_n)}\mapsto{(x_{n+1})}$.

(In case you have a different definition, you should explain it in the question.)

Note that positivity and linearity imply that $$x\le y \qquad \Rightarrow \qquad f(x)\le f(y).$$

This gives $$\inf x_n \le f(x) \le \sup x_n$$ by comparing the sequence $x$ and constant sequences $(\inf x_n)$ and $(\sup x_n)$.

Now, since $f$ is shift-invariant, for any $k$ we get $f(x)=f(T^{k-1} x)$. But, using the above property: $$\inf_{n\ge k} x_n\le f(T^{k-1} x) \le \sup_{n\ge k} x_n$$ which means $$\inf_{n\ge k} x_n\le f(x) \le \sup_{n\ge k} x_n.$$

Taking limit $k\to\infty$ you get $$\liminf_{n\to\infty} x_n = \lim_{k\to\infty}\inf_{n\ge k} x_n\le f(x) \le \lim_{k\to\infty}\sup_{n\ge k} x_n = \limsup_{n\to\infty} x_n.$$

Without much work, this estimate can be improved in the following way.

For any bounded sequence $x$ we define $T_n(x)=\frac{x+Tx+\dots+T^{n-1}x}n$. I.e., $T_n(x)$ is the sequence $\left(\frac{x_k+x_{k+1}+\dots+x_{k+n-1}}n\right)_{k=1}^\infty$. Let us denote $$ \begin{gather*} M(x)=\lim_{n\to\infty} \limsup T_n(x),\\ m(x)=\lim_{n\to\infty} \liminf T_n(x). \end{gather*} $$ The existence of the above limits can be shown using Fekete's lemma - a proof of this lemma can be found in this answer.

We can show that for any Banach limit $m(x)\le f(x) \le M(x)$ (using the above estimates and linearity).

This result can be, in some sense, conversed: A linear function $f:\ell_\infty\to\mathbb R$ is a Banach limit if and only if $f(x)\le M(x)$ holds for each $x\in\ell_\infty$. Then, using Hahn-Banach theorem, we get that, for any given bounded sequence $x$, all values from the interval $[m(x),M(x)]$ can be attained by some Banach limit $f$. (Note that Hahn-Banach theorem gives in fact more than existence of a linear functional - we can get also the possible range of values of extensions, see this question.)

edited Apr 13 '17 at 12:20

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answered Nov 09 '11 at 16:51

Martin Sleziak

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Here in the definition of $M(x)$, does $\lim\limits_{n\to\infty}\lim\sup T_n(x)$ mean $\lim\sup\limits_{n\to\infty}T_n(x)$? – MAS Jul 15 '19 at 06:29
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@BijanDatta No, there are two variables involved. The sequence $T_n(x)$ has the $k$-th term $\frac{x_k+x_{k+1}+\dots+x_{k+n-1}}n$. So the more compact notation can be rewritten in more detail as $$\lim\limits_{n\to\infty} \limsup T_n(x) = \lim\limits_{n\to\infty} \limsup\limits_{k\to\infty} \frac{x_k+x_{k+1}+\dots+x_{k+n-1}}n.$$ – Martin Sleziak Jul 15 '19 at 06:38
I'm trying to understand about the existence of $M(x)$, but I can't. Here, which is subadditive sequence for applying Fekete's lemma? – MAS Jul 15 '19 at 09:47
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@BijanDatta To make use of the Fekete's lemma clearer let me rewrite this as $$\lim_{n\to\infty} \frac{\limsup_{k\to\infty} (x_k + x_{k+1} + \dots + x_{k+n-1})}n.$$ Now the denominator is $a_n= \limsup\limits_{k\to\infty} (x_k + x_{k+1} + \dots + x_{k+n-1})$ and we want to show (using Fekete's lemma) that the limit of $\frac{a_n}n$ exists. – Martin Sleziak Jul 15 '19 at 10:33
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@BijanDatta You might also check this answer: Help showing subadditivity of a map. In the last paragraph, this is explained in more detail. (Although the explanation is basically along the lines of what I wrote here in the comments.) – Martin Sleziak Jul 15 '19 at 10:37
For $$a_n=\limsup\limits_{k\to\infty}(x_k+x_{k+1}+\dots+x_{k+n-1})$$ Is $(a_n)_n$ a subadditive sequence? This is still not clear to me. – MAS Aug 08 '19 at 08:29
@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 $\lim f$" – Hermi Nov 29 '19 at 03:55
@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 Nov 29 '19 at 04:31
@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. – Hermi Nov 29 '19 at 04:57
@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 Nov 29 '19 at 05:04
This result has been mentioned in some other posts on the site: If $(x_n)$ is almost convergent to $\ell$, then $\lim\limits_{p\to\infty}\frac{x_n+x_{n+1}+x_{n+2}+\dots+x_{n+p-1}}{p}=\ell$ holds uniformly in $n$. or Can I define a bounded sequence whose Banach limit is not unique? – Martin Sleziak Nov 29 '19 at 05:04
@MartinSleziak But how about the generalized limit? without shift is invariant? – Hermi Nov 29 '19 at 05:05
@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 Nov 29 '19 at 11:51

Show that $\liminf_{n\to \infty}x_{n}\le\alpha(x)\le\limsup_{n\to\infty}x_{n}$ for $x=(x_{n})$ in $\ell^{\infty}$

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