I want to proof that exists $f \in l_\infty '$ with $f(x) = \lim x_n, \forall x = (x_n) \in c$ and $f(x_1, x_2, x_3,...) = f(x_2,x_3,x_4,...)$

What I have been doing until now:

Consider the funtional $f_0: c \to \mathbb R $ defined by $f_0 (x) = \lim x_n, \forall x = (x_n) \in c$. It's well-defined because all sequence in $c$ converges and is linear

$$ f_0 (\lambda x + y) = \lim \lambda x_n + y_n = \lambda \lim x_n + \lim y_n = \lambda f_0(x) + f_0 (y)$$

And it's continuous. If $x = (x_n) \in c$, then it converges and it's bounded, so $\exists M = \sup |x_n| = ||x_n|| >0$ with $|x_n| < M$. Let $\bar x$ with $x_n \to \bar x$, so we can write for $n \geq n_0$, $|x_n - \bar x| < 1/ n$. Then

$$ |\bar x| \leq |\bar x - x_n| + |x_n| < 1/n + M, \quad n \geq n_0 $$

Well, if $n \to \infty$ we shall have $|\bar x | < M = ||x|| $, then $$ |f_0 (x)| = |\lim x_n| < ||x|| $$, so $f_0$ is continuous and $||f_0|| \leq 1$. Besides, we have $||f_0|| = 1$. Because,

$$ ||f_0|| = sup_{||x|| = 1} |f_0 (x)| \geq 1, \quad consider \quad x = (1,1,1,...) $$

Now, using Hahn-Banach Theorem, $\exists f \in l_\infty '$ with $f (x) = f_0 (x), \forall x \in c$ and $||f|| = ||f_0|| = 1$

I still need to show that $f(x_1, x_2, x_3,...) = f(x_2,x_3,x_4,...)$. Well, as $||f|| = 1$

$$||f(x_1 - x_2, x_2 - x_3, x_3 - x_4, ...)|| \leq || (x_1 - x_2, x_2 - x_3, x_3 - x_4,...) || = \sup_{n \geq 1} |x_n - x_{n+1}|$$

If I could conclude that $\sup_{n \geq 1} |x_n - x_{n+1}| = 0$, it's done. But I can't see that. Could someone keep this argument?

Thank You

  • $\begingroup$ With $c$ do you mean the linear subspace of $\mathscr l^\infty$ given by those sequences that converge? $\endgroup$ – s.harp Apr 24 '16 at 15:31
  • $\begingroup$ Yes, $c$ is the space of the sequences that converges. $\endgroup$ – user 242964 Apr 24 '16 at 15:32
  • 1
    $\begingroup$ The function $f$ you are looking for is called a Banach limit. A web search for "Banach limit shift invariance" will turn up several proofs of what you are looking for. $\endgroup$ – Umberto P. Apr 24 '16 at 15:41

There are versions of the Hahn-Banach theorem from which this follows more or less directly. We can get it from the Hahn-Banach that everybody knows by a little trick:

Say $D$ (for "difference") is the space of all $x\in\ell^\infty$ such that $$x=(y_1-y_2,y_2-y_3,\dots)$$for some $y\in\ell^\infty$.

Let $X$ be the span of $D$ and $c$: $$X=D+c.$$We want to define $f:X\to\Bbb C$ such that $fx=0$ for all $x\in D$, while $fx=\lim_nx_n$ for all $x\in c$. Does such a linear functional exist? Yes, if and only if the two definitions agree on $D\cap c$. And they do - it's easy to see that if $x\in D$ has a limit the limit must be $0$ (since the $y$ in the definition of $D$ lies in $\ell^\infty$).

Now Hahn-Banach extends $f$ to a map $f:\ell^\infty\to\Bbb C$ and we're done.

Oops. Was just about to hit the Post button when I realized we need to verify that $f:X\to\Bbb C$ is bounded (and in fact has norm $1$). This is not entirely clear at first, because it could happen that $||y||+||z||>||y+z||$ for $y\in D$ and $z\in c$.

Hmm. It's easy to see that $y\in D$ if and only if $|\sum_{j=1}^ny_j|$ is bounded. So in particular $$\limsup_{n\to\infty}\Re y_n\ge0\quad(y\in D).$$

Say $$x=y+z\in X,$$where $y\in D$ and $z\in c$. Say $\lim z_n=\alpha$. We need to show that $$\sup_n|y_n+z_n|\ge|\alpha|.$$ Assume WLOG that $\alpha\ge0$. Then $$\sup|y_n+z_n|\ge\limsup\Re(y_n+z_n)=(\limsup\Re y_n)+\alpha\ge\alpha.$$


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