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

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.

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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 9 '11 at 16:28
That, or $\alpha((x_n))=0$ for every $(x_n)$ in $\ell^\infty$. –  Did Nov 9 '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 9 '11 at 16:54
@NateEldredge $\alpha$ is the Banach Limit. –  Hassan Muhammad Nov 9 '11 at 20:26

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.)

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