# Generalized limits on $\ell^\infty(\mathbb{N})$

Let $\ell^\infty(\mathbb{N})$ denote the set of bounded real sequences $(a_n)_{n\in\mathbb{N}}$. The $\lim$ operator is a partial linear operator from $\ell^\infty(\mathbb{N})$ to $\mathbb{R}$. With the Hahn-Banach theorem, $\lim$ can be extended to a generalized limit functional. The Hahn-Banach theorem uses the Axiom of Choice.

Are there models of $\mathsf{ZF}$ in which no generalized limit functionals exist?

• Can we use the generalized limits of $0/1$-valued sequences to produce a nonprincipal ultrafilter on $\mathbb{N}$? If so, then the answer is yes, since if ZF is consistent, then there are models of ZF with no nonprincipal ultrafilters. – Joel David Hamkins May 17 '17 at 11:59
• – Martin Sleziak May 17 '17 at 13:29
• @JoelDavidHamkins All nonprincipal ultrafilters create generalised limits, but not all generalised limits come from ultrafilters. I myself was a little confused about this point, so I wrote a blog post recently in part to sort things out in my head once and for all: terrytao.wordpress.com/2017/05/11/… . There I also note that generalised limits can be used to create nonmeasurable sets and hence do not exist in Solovay models. – Terry Tao May 17 '17 at 17:13
• @TerryTao When you write "not all generalised limits come from ultrafilters" you mean that not every $f\in\ell_\infty^*$ which extends the usual limit, is of the form $f(x)=\operatorname{\mathscr{U}-lim} x_n$ for some ultrafilter $\mathscr U$, i.e., not all elements of $\ell_\infty^*$ are ultralimits, right? IIRC all such funtionals are in the closed convex hull of the set of ultralimits, which is also a possible interpretation of the phrase "come from ultrafilter". (Everything I write here is in ZFC, I guess that detailed analysis how much choice is needed is probably not simple.) – Martin Sleziak May 19 '17 at 8:11
• @JoelDavidHamkins I posted this question on math.SE, inspired basically by your comment about generalized limits and nonprincipal ultrafilters: Does existence of some (nice) non-trivial functionals in $\ell_\infty^*\setminus\ell_1$ give a free ultrafilter on $\omega$? – Martin Sleziak May 20 '17 at 10:27

Yes. In fact, if you work in $\mathsf{ZF}+\mathsf{DC}+$ "all sets of reals have the property of Baire" ($\mathsf{BP}$), say the Solovay or Shelah models, you can prove that $(\ell^\infty)^* = \ell^1$, so that the only continuous linear functionals on $\ell^\infty$ are those coming from $\ell^1$. The generalized limits are certainly not of this form, since they vanish on $c_0$ but not identically. Under these axioms, $\ell^1$ is reflexive.

You can find a proof in Schechter, Handbook of Analysis and its Foundations, sections 29.37–38.

Interestingly, under these axioms all linear functionals on any Banach space are continuous, so in fact the algebraic dual of $\ell^\infty$ also equals $\ell^1$. And the limit operator has no linear extension to $\ell^\infty$.

Digression: This is a fun example because there are several different ways to prove in $\mathsf{ZFC}$ that $(\ell^\infty)^* \ne \ell^1$. It is interesting to see how all of them fail in this model, and where they used choice. For instance:

1. Use Hahn-Banach as in your example. Hahn-Banach doesn't work without choice.

2. Use Alaoglu's theorem to produce a weak-* limit point of the usual basis $\{e_i\}$ of $\ell^1 \subset (\ell^\infty)^*$. This is another sort of generalized limit. So Alaoglu also must fail without choice; of course, the usual proof uses Tychonoff.

3. Recall the following exercise: "If $X^*$ is separable then so is $X$." Note that $\ell^1$ is separable and $\ell^\infty$ is not. But the "exercise" used Hahn-Banach, and isn't necessarily true without choice.

• About your digression: I cannot resist to mention another possibility, which is using limit along an ultrafilter. BTW perhaps you could post the proof using Banach-Alaoglu's theorem here: Dual of $l^\infty$ is not $l^1$. (It seems that such proof has not been posted there yet, Arguments based on separability and Hahn-Banach theorem are among the answers.) – Martin Sleziak May 20 '17 at 7:53