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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.

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    $\begingroup$ 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. $\endgroup$ Commented Mar 29, 2022 at 21:39
  • $\begingroup$ 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? $\endgroup$
    – Arbiter
    Commented Mar 30, 2022 at 0:28

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