Ultrafilters in $\ell^\infty$ 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.
 A: 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 $C(\beta \Bbb N)$.
A: 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?
