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](https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_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?



Generalized limits on $\ell^\infty(\mathbb{N})$
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](https://en.wikipedia.org/wiki/Solovay_model) 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](https://books.google.com/books?id=eqUv3Bcd56EC&pg=PA810).


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.


