# Dominated convergence theorem for Banach limits

The notion of a Banach limit is usually defined for the space of bounded sequences, but one can define it for more general spaces (see "What is a generalized limit?" and "Do multiplicative Banach limits exist?" and references therein).

I am interested in the Banach space of Bounded functions $$f: \mathbb{R}_+ \to \mathbb{C}$$ with respect to the uniform norm $$\|f\| = \sup_{t \in \mathbb{R}_+} |f(t)|$$ and Banach limits on it. Consider a sequence $$f_n(t)$$ ,$$n\in \mathbb{N}$$, of such functions and a Banach limit denoted by $$L$$. Suppose that for all $$t\in \mathbb{R}_+$$: $$\sum_{n=1}^\infty f_n(t)$$ exists, and consider its Banach limit $$L\Big( \sum_{n=1}^{\infty} f_n(t) \Big)$$ Is it true that $$L\Big( \sum_{n=1}^{\infty} f_n(t) \Big) = \sum_{n=1}^{\infty}L\big( f_n(t) \big)$$

for all Banach limits $$L$$?

With usual limits, i.e. $$\lim_{t\to \infty} \lim_N \sum_{n=1}^N f_n(t)$$, it's not always the case that we can commute the two limits.

So, does the dominated convergence theorem extend to Banach limits, so that the Banach limit $$L$$ can be moved inside the summation/commute the limits? Under what assumptions?

I'd appreciate references.

• Doesn't that already fail for the usual Banach limit on $\ell_\infty$? Let $\mathbf{1}$ be the constant sequence with value $1$ and let $\mathbf{1}_n$ be the sequence with a $1$ in the $n$th place and all other entries $0$. Then $\mathbf{1}=\sum_{n=1}^\infty \mathbf{1}_n$. But $L(\mathbf{1})=1\neq 0=\sum_{n=1}^\infty L(\mathbf{1}_n)$. The point is that Banach limits coincide with the usual limits when they exist. Commented Mar 29, 2022 at 21:39
• Of course it wont hold for everything. We should ask for some form of uniform bound $g_n$ (independent of $t$ and summable) on the $|f_n(t)|$, like in the assumption of the dominated convergence theorem. That is my question, do we have an analogue of the dominated convergence theorem here? Commented Mar 30, 2022 at 0:28