# Do multiplicative Banach limits exist?

Let $$(D, \succeq)$$ be a directed set, and let $$B$$ be the space of real-valued bounded functions on $$D$$. A Banach limit $$\ell$$ on $$D$$ is a linear functional that satisfies $$\sup_{d \in D} \inf_{c \succeq d} f(c) \leq \ell(f) \leq \inf_{d \in D} \sup_{c \succeq d} f(c)$$ for all $$f \in B$$.

Banach limits exist by the Hahn-Banach theorem. In fact, if we say instead that our functions on $$D$$ take values in an abstract Dedekind complete ordered vector space, then the existence of Banach limits is equivalent to the Hahn-Banach theorem. But I am primarily interested in the real-valued case.

Question. Are there Banach limits that satisfy $$\ell(fg) = \ell(f)\ell(g)$$ for all $$f,g \in B$$? If so, does this require anything stronger than the Hahn-Banach theorem?

I do not think that this is the usual definition of Banach limit. (What I know under this name is linear functional on $$l_\infty$$ which is positive, shift-invariant and extends the usual limit, see the linked Wikipedia article. Of course, the terminology in various sources might differ.) In connection with the question it might be worth mentioning that Banach limit in this sense cannot be multiplicative.

EDIT: The OP mentioned in comments that the definition of Banach limit given in the question (i.e., with directed sets) can be found in Schechter's Handbook of Analysis and Its Foundations, see page 318 and also in Howard, Rubin: Consequences of the Axiom of Choice, see page 63, Form 372. (If we work on arbitrary directed sets, we do not have a natural way to talk about shift-invariance. So this is not a generalization of the notion of Banach limit on $$\mathbb N$$ mentioned in the first paragraph.)

Your conditions can be rewritten as1 $$\liminf_{c\in D} f(c) \le \ell(f) \le \limsup_{c\in D} f(c).$$

So you want a functional which is between $$\liminf$$ and $$\limsup$$ (and therefore extends the usual limit of a net) and is multiplicative.

You can simply take any ultrafilter $$\mathcal U$$ which contains all tail sets of the directed set $$D$$. (I.e., for any $$d\in D$$ you have $$d\uparrow=\{c\in D; c\ge D\}\in\mathcal U$$.) And then define $$\ell$$ using limit along this utlrafilter as: $$\ell(f) = \operatorname{{\mathcal U}-\lim} f(c).$$ This functional has the properties you want. (Boundedness of $$f$$ guarantees that the $$\mathcal U$$-limit exists.2 We get multiplicativity from the fact that $$\mathcal U$$-limit is multiplicative. And the fact that $$\mathcal U$$ contains the tail filter helps with the condition about limit inferior and limit superior.)

The same construction is mentioned in the answer to: What is a generalized limit?

In case it helps to find some references for $$\mathcal U$$-limit (limit along an ultrafilter or, more generally, limit along a filter or a filter base), I will mention my answers to these questions: Where has this common generalization of nets and filters been written down? and Basic facts about ultrafilters and convergence of a sequence along an ultrafilter.

Since you are interested in multiplicative functionals, this might be of interest, too: Every multiplicative linear functional on $$\ell^{\infty}$$ is the limit along an ultrafilter.

1For more on limit superior/inferior of a net, see: Limsups of nets and About the notion of limsup and liminf

2Limit along an ultrafilter with values in a compact space always exists. The proof is given, for example, in this answer: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter.

• Thanks. My definition comes from Howard & Rubin's Consequences of the Axiom of Choice and Schechter's Handbook of Analysis. It's just a bit more general than yours, so I don't think there are any issues there. – aduh Apr 7 at 7:41
• I'm not sure I see why the $\mathcal U$-limit exists. Can you explain? – aduh Apr 7 at 7:41
• Also, I wonder if, in general, the existence of multiplicative Banach limits requires the ultrafilter lemma (which is stronger than Hahn-Banach). – aduh Apr 7 at 7:43
• Re: Why $\mathcal U$-limit exists: Since $\mathcal U$ is an ultrafilter and the values are all on some bounded interval, which is a compact space. A proof that in compact space limit along an ultrafilter exists is given in this answer: Basic facts about ultrafilters and convergence of a sequence along an ultrafilter. (And probably in many other resources.) – Martin Sleziak Apr 7 at 7:51