Generalized limits

Question

Cross-posted to Mathoverflow.

$\DeclareMathOperator{\Lim}{Lim}$ $\DeclareMathOperator{\dom}{dom}$ $\DeclareMathOperator{\shift}{\sigma}$ $\DeclareMathOperator{\cesaro}{C}$

After reading Terry Tao's post on generalizations of the limit functional, I'm interested in the concept of a "generalized limit", which agrees with the ordinary limit when the latter exists, but also extends it to new sequences.

Let $X$ be a metrized module. A generalized limit is a partial function from sequences of elements of $X$ to $X$: $$\Lim : \bigcup_{S \subseteq X^\mathbb{N}} X^S$$

Let $\lim$ be the Cauchy limit. That is, $\lim x = a$ iff $$\forall \varepsilon \in \mathbb{R}^+ : \exists N \in \mathbb{N} : \forall n \in \mathbb{N} : n > N \rightarrow d(x(n),a) < \varepsilon$$

$\Lim_1$ is weaker than $\Lim_2$ iff $$\Lim_1 \subseteq \Lim_2$$

$\Lim$ is regular iff it is stronger than $\lim$. $\Lim_1$ is consistent with $\Lim_2$ iff $$\forall x \in (\dom \Lim_1 \cap \dom \Lim_2) : \Lim_1(x) = \Lim_2(x)$$

$\Lim$ is homogeneous iff $$\forall a \in X : \forall x \in \dom \Lim: \Lim(a x) = a \Lim(x)$$

$\Lim$ is additive iff $$\forall x, y \in \dom \Lim : \Lim(x + y) = \Lim(x) + \Lim(y)$$

$\Lim$ is linear iff it is homogeneous and additive. $\Lim$ is stable iff $$\Lim = \Lim \circ \shift$$

where $\shift : X^\mathbb{N} \rightarrow X^\mathbb{N}$ is the shift transform defined by $$\shift(a)(n) = a(n+1)$$

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If $a \neq 1$ and $\Lim$ is stable and homogeneous, then $\Lim (n \mapsto a^n) = 0$. Proof: \begin{align} \Lim (n \mapsto a^n) &= (\Lim \circ \shift) (n \mapsto a^n) & \text{(stability)} \\ &= \Lim(\shift(n \mapsto a^n)) \\ &= \Lim(n \mapsto a^{n+1}) \\ &= \Lim(a (n \mapsto a^n)) \\ &= a \Lim(n \mapsto a^n) & \text{(homogeneity)} \\ (1 - a) \Lim(n \mapsto a^n) &= 0 \\ \Lim(n \mapsto a^n) &= 0 \end{align}

Note that if $a$ is prime then this yields the correct $a$-adic limit. This also yields the expected generalized sum of a geometric series: \begin{align} \sum_{n=0}^\infty a^n &= \Lim\left(m \mapsto \sum_{n=0}^m a^n\right) \\ &= \Lim\left(m \mapsto \frac{1-a^{m+1}}{1-a}\right) \\ &= \frac{1 - \Lim(m \mapsto a^{m+1})}{1-a} \\ &= \frac{1}{1-a} \\ \end{align}

Consequently, if $\Lim$ is stable and linear then \begin{align} \Lim_{n \rightarrow \infty} \cos n &= \Lim_{n \rightarrow \infty} \frac{\mathrm{e}^{in} + \mathrm{e}^{-in}}{2} \\ &= \frac{\Lim_{n \rightarrow \infty} (\mathrm{e}^{i})^n + \Lim_{n \rightarrow \infty} (\mathrm{e}^{-i})^n}{2} \\ &= 0 \end{align}

and the same is true of $\sin$, we also have \begin{align} \Lim_{m \rightarrow \infty} \sum_{n=0}^m \cos n &= \Lim_{m \rightarrow \infty} \left(\frac{1}{2} + \frac{\cos(m)}{2} + \frac{1}{2} \cot \frac{1}{2} \sin m\right) \\ &= \frac{1}{2} \\ \Lim_{m \rightarrow \infty} \sum_{n=0}^m \sin n &= \Lim_{m \rightarrow \infty} \left(\frac{1}{2} \cot \frac{1}{2} + \frac{\sin m}{2} - \frac{1}{2} \cos m \cot \frac{1}{2}\right) \\ &= \frac{1}{2} \cot \frac{1}{2} \end{align}

More generally, if $a \neq 1$ and $\Lim$ is stable and linear then \begin{align} \Lim_{m \rightarrow \infty} \sum_{n=0}^m a^n n^b &= \Lim_{m \rightarrow \infty} (\mathrm{Li}_{-b}(a) - a^{m+1} \Phi(a,-b,m+1)) \\ &= \mathrm{Li}_{-b}(a) + \Lim_{m \rightarrow \infty} a^{m+1} \Phi(a,-b,m+1) \\ &= \mathrm{Li}_{-b}(a) \end{align} \begin{align} \Lim_{m \rightarrow \infty} \sum_{n=0}^m a^n n! &= \Lim_{m \rightarrow \infty} \left( -\frac{\mathrm{e}^{-1/a} \Gamma(0,-1/a)}{a} + (-1)^{m+1} \frac{\mathrm{e}^{-1/a} \Gamma(-m-1,-1/a) (m+1)!}{a} \right) \\ &= -\frac{\mathrm{e}^{-1/a} \Gamma(0,-1/a)}{a} + \Lim_{m \rightarrow \infty} (-1)^{m+1} \frac{\mathrm{e}^{-1/a} \Gamma(-m-1,-1/a) (m+1)!}{a} \\ &= -\frac{\mathrm{e}^{-1/a} \Gamma(0,-1/a)}{a} \end{align}

where $\mathrm{Li}$ is the polylogarithm and $\Phi$ is the Lerch transcendent. Thus the properties of stability and linearity alone make the generalized limit quite powerful.

Question: Is there a $\Lim$ such that

1. $\Lim$ can be explicitly defined in terms of the Cauchy limit, e.g. $\Lim = \lim \circ f$ where $f$ is some sequence transform.
2. $\Lim$ is stronger than (i.e. contains) the stable and linear closure of the Cauchy limit.

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The Cesàro transform $\cesaro : X^\mathbb{N} \rightarrow X^\mathbb{N}$ yields the sequence of partial averages: $$\cesaro(a)(n) = \frac{1}{n+1}\sum_{k=0}^n a(k)$$

For $n \in \mathbb{N}$, let the $n$-Cesàro limit be the Cauchy limit of the Cesàro transform iterated $n$ times: $$\lim \circ \cesaro^n$$

The $n$-Cesàro limit is regular, linear, and stable. Moreover, $$\lim \circ \cesaro^n \subsetneq \lim \circ \cesaro^{n+1}$$

The inclusion is proper as can be seen by taking the sequence $$k \mapsto (-1)^k k^n$$

which is in the domain of the latter but not the former. Let the $\omega$-Cesàro limit be the union of all such limits: $$\bigcup_{n < \omega} \lim \circ \cesaro^n$$

That is, the $\omega$-Cesàro limit is the closure of the Cauchy limit under Cesàro transforms.

My question is this: Are there any interesting examples of linear limits that are (non-strictly) stronger than the $\omega$-Cesàro limit? What about stronger than the closure of the $\omega$-Cesàro limit under stability (shift transforms)? This last property would allow it to regularize geometric limits. Note: It would not have to be stable itself, just stronger than the stable closure of the $\omega$-Cesàro limit.

I suspect sequence transformations used to accelerate the rate of convergence of sequences, like the Shanks transformation and Aitken's delta-squared process, might be useful here.