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.

  • 3
    $\begingroup$ 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.