Ultrafilters in $\ell^\infty$

Question

Let $Ult(\mathbb{N})$ be the set of ultrafilters over $\mathbb{N}$ and $L=\{f\in(\ell^\infty(\mathbb{N}))':f(x)\in\overline{x(\mathbb{N})},\forall x\in\ell^\infty(\mathbb{N})\}$.

I want to prove that $$\Phi:Ult(\mathbb{N})\to L\\ \text{ }\text{ }\text{ }U\mapsto\lim_{U}$$ is well-defined. I know that for each $U\in Ult(\mathbb{N})$, $\lim_{U}$ is linear but I don't know if it's continuous.

I don't understand some of your notation..... Let $U \in Ult(\Bbb N).$ For each $x=(x_n)_{n\in \Bbb N}\in l^{\infty}$ let $f_U(x)=r$ iff ${n:x_n\in V}\in U$ for every nbhd $V$ of $ r.$ Then $ f_U$ is a well-defined continuous linear functional... Is this what you are trying to prove? — DanielWainfleet, Sep 11 '19 at 22:01
Yes, DanielWainfleet. I want to prove this. I forgot to specify. I'm a beginner in ultrafilters. — Meryjen Cacuango, Sep 16 '19 at 14:19
I was typing a thorough answer when the post became deleted. — DanielWainfleet, Sep 17 '19 at 06:54

s.harp · Answer 1 · 2019-09-12T11:49:30.053

Note that $\lim_U (x) \in\overline{x(\Bbb N)}$, hence $|\lim_U x|≤\|x\|$ for all $x\in\ell^\infty(\Bbb N)$ and $\lim_U$ is a bounded linear map, hence continuous.

You may alternatively view $\lim_U$ as a point evaluation map, which is also necessarily continuous.

First remember that if $x\in \ell^\infty(\Bbb N)$, then $x:\Bbb N\to \overline{x(\Bbb N)}\subset\Bbb C$ is valued in a compact set (as $x$ is bounded) and hence factors over the Stone-Cech compactification, that is there is a map $\beta x: \beta\Bbb N\to \Bbb C$ so that $x = \beta x\circ\iota$ where $\iota$ is the inclusion $\iota:\Bbb N\to \beta\Bbb N$ into the Stone-Cech compactification. (All this can be expressed in a commutative diagram, but I don't know how to do commutative diagrams in MathJax.)

We may thus identify $\ell^\infty(\Bbb N)$ with $C(\beta\Bbb N)$ (if you give the space on the right the uniform norm these are then the same as normed vector spaces). Now the thing is that $\beta\Bbb N$ is compact and hence every ultra-filter converges. Thus $\lim_U$ applied to a function of $C(\beta\Bbb N)$ is nothing other than the point evaluation of that function at the limit of $U$, which defines a continuous functional on $C(\beta \Bbb N)$.

Because $\overline{x(\Bbb N)}$ is closed, or does your problem lie at another part of that statement? — s.harp, Sep 16 '19 at 19:02
Yes. My question was wrong. I understand better now, thanks. — Meryjen Cacuango, Sep 16 '19 at 22:48
Why $\lim_{U}(x)\in \overline{x(\mathbb{N})}$ implies that $|\lim_{U}(x)|\leq \sup_{n\in\mathbb{N}}|x_n|$ ? — Meryjen Cacuango, Oct 27 '19 at 02:32

Martin Sleziak · Answer 2 · 2019-09-12T11:56:55.993

Another way to see that $\lim_U$ is continuous is to use monotonicity.

If you have any ultrafilter $U$ and two sequences such that $x\le y$ (i.e., $x_n\le y_n$ for each $n\in\mathbb N$), the you also have $\lim_U x\le\lim_U y$.

If you take as one of the sequences a constant sequence, you will get that $$\inf x \le \lim_U x \le \sup x.$$ And it's not difficult to prove this also directly, without using monotonicity of $\lim_U$.

And from this you get that $|\lim_U x| \le \sup\limits_{n\in\mathbb N} |x_n|$, i.e., $|\lim_U x| \le \|x\|$.

Mainly as a possible place where to find some references for limit along ultrafilter (filter, filter base), I'll add links to these two answers of mine: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter and Where has this common generalization of nets and filters been written down?

Ultrafilters in $\ell^\infty$

2 Answers2