Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
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
@t.b. Sure, I am asking about Banach-Limits. – Hassan Muhammad Nov 9 '11 at 16:31
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
up vote 4 down vote accepted

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.