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 $$\shift(a)(n) = a(n+1)$$

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.

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.

  • $\begingroup$ Not sure why you give special names to the invariance properties you are assuming for your $Lim$ operator, also if $Lim(a^n)$ exists and $a \ne 0$ then $Lim(a^n) =0$ (with the usual $\lim$ then $\lim 2^n$ doesn't exist and $\lim 1^n= 1$). Also you didn't define the "iterated Cesaro limit". Do you mean $\lim_{n \to \infty} \frac{\sum_{k=0}^n (n-k)^n a_k}{\sum_{k=0}^n k^n}$ or $Lim a_n=\lim_{n \to \infty} (C^m a)_n$ if it exists for some $m$ ? $\endgroup$
    – reuns
    Jul 21, 2019 at 11:19
  • $\begingroup$ @reuns I don’t understand your first sentence regarding naming. Regarding your second sentence, did you mean $a \neq 1$? I proved this in my question. Regarding your third sentence, I did define the iterated Cesàro mean right below the definition of the Cesàro transform (it’s the same as your second expression). $\endgroup$
    – user76284
    Jul 21, 2019 at 17:29
  • 1
    $\begingroup$ this can be rephrased in terms of "divergent series" since any sequence $a_n$ can be written as a partial sum of the differences $a_{n+1} - a_n$. If the $\omega$-Cesaro limit is nothing but "find an $m$ where $C^m$ works, then use it", then I believe the Abel regularisation $$\lim_A : a \mapsto a_0 + \lim_{x\to 1} \sum_{i=1}^\infty (a_{n} - a_{n-1} ) x^n = \lim_{x\to 1} (1-x) \sum_{i=0}^\infty a_n x^n $$ is regular, homogeneous, additive, stable, and stronger than $\omega$-Cesaro. There are yet others in Hardy's book on Divergent Series $\endgroup$ Oct 14, 2019 at 5:52
  • 1
    $\begingroup$ @BigbearZzz G. H. Hardy's Divergent Series is considered the standard reference for this subject. I don't remember the exact reference for this particular property, but Cesàro summation is a special case of Nørlund means, which as stated here are stable. $\endgroup$
    – user76284
    Feb 25, 2020 at 2:00
  • 1
    $\begingroup$ @BigbearZzz Theorems 5.33 and 5.34 of The Nörlund and the weighted mean methods by P. N. Natarajan characterize the conditions for being translative (i.e. stable), citing Translativity of weighted means in non-archimedean fields, also by P. N. Natarajan. $\endgroup$
    – user76284
    Feb 26, 2020 at 2:03


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