I recently learned that a Cesaro summation extends the usual summation in the following way: Given a sequence $a_1, a_2, \ldots $ we construct the Cesaro sequence by defining $$\sigma_n = \frac{1}{n}\sum_{j=1}^n a_j$$ Then we say that $(a_j)$ is Cesaro summable if $\sigma_n$ coverages to some point.

Now let's relabel the sequence $(\sigma_j)$ as $(\sigma^{1}_j)$, and say $(a_j)$ is $1$-Cesaro summable if it is Cesaro summable. We could, obviously, construct a new Cesaro sequence, call it $(\sigma^2_j)$, which is the Cesaro sequence of the original Cesaro sequence. Then we say the sequence $(a_j)$ is $n$-Cesaro summable if the sequence $(\sigma^{n-1}_j)$ is Cesaro summable. Clearly if a sequence is $n$-Cesaro summable then it is $j$-Cesaro summable for all $j\ge n$.

Question: Is there a proper way to define $\infty$-Cesaro summability? Are there sequences which are not $n$-Cesaro summable for any finite $n$ yet are $\infty$-Cesaro summable? Finally, a good definition would require the existence of sequences which are not $\infty$-Cesaro.

  • $\begingroup$ Probably what you are looking for is en.wikipedia.org/wiki/…. Unfortunately that page is a bit misleading (at best); the Cesaro sum of the series $\sum_{j=1}^\infty a_j$ is $\lim_{n\to\infty}\frac1n\sum_{j=1}^ns_j$ where $s_j=\sum_{k=1}^j a_k$. $\endgroup$
    – stewbasic
    Commented Aug 22, 2017 at 4:10
  • $\begingroup$ Your response seems to assume that I have a basic misunderstanding of what a Cesaro sum is. I already stated in the body of my question precisely the definition of a Cesaro sum. This is not my question and this wiki page does not help, the page only appears to offer different naming to what I call n-Cesaro. However, the page does NOT discuss $\infty$-Cesaro. Also your definition of the Cesaro sum is incorrect. $\endgroup$
    – user351797
    Commented Aug 22, 2017 at 4:17
  • $\begingroup$ The answer appears to be yes. mathoverflow.net/questions/84772/… ... although I have little direct experience with this construction. Expressing Cesaro summation as multiplication by an (infinite) matrix seems to be a smart way to think of it for your purposes. $\endgroup$ Commented Aug 22, 2017 at 4:26
  • $\begingroup$ @EricTowers sorry I do not see the connection between my question and the math overflow post, nor do I understand what you mean by "Expressing Cesaro summation as multiplication by an (infinite) matrix ....". My gut tells me that some sequences that may be good candidates for testing would include $1-2+3-4+\cdots $ and $1+2+3+4+\cdots$. $\endgroup$
    – user351797
    Commented Aug 22, 2017 at 4:36
  • 5
    $\begingroup$ @Clclstdnt You should probably be nicer to people trying to help. The term Cesaro sum (of a series) typically refers to stewbasic's definition. Yours is typically called the Cesaro mean (of a sequence). $\endgroup$ Commented Aug 22, 2017 at 4:36

1 Answer 1

  1. Let $T = (T_{i,j})_{i,j \geq 1}$ be an infinite matrix (i.e. double sequence) given by

    $$ T_{i,j} = \begin{cases} \frac{1}{i} & \text{if } j \leq i \\ 0, & \text{otherwise} \end{cases} $$

    Then we easily check that $\sigma_i = [Ta]_i := \sum_{j=1}^{\infty} T_{i,j}a_j$. So if we regard a sequence as an infinite column vector, then $\sigma = Ta$ is simply a result of matrix multiplication, and $\sigma^n = T^n a$. This provides a convenient language for theoretical approach to Cesàro means.

  2. Unfortunately, we can prove that the limit of each entry of $T^n$ is rather trivial:

    $$ \lim_{n\to\infty} [T^n]_{i,j} = \begin{cases} 1 & \text{if } j = 1 \\ 0 & \text{if } j \geq 2\end{cases}. $$

    In other words, we have $\sigma^n_i = [T^n a]_i \to a_1$ as $n\to\infty$ for each fixed $i$.

  3. On the other hand, if $\sigma^n_i = [T^n a]_i \to \ell$ as $i\to\infty$ for some $n \geq 0$, then we can prove that the following convergence occurs as well:

    $$ \lim_{x \to 1^-} (1-x) \sum_{i=1}^{\infty} a_i x^i = \ell. $$

    In other words, if $(a_i)$ has the $n$-Cesàro mean, then it also has the Abel mean which is defined as the above limit (if exists, of course). So the notion of Abel mean can handle more sequences than any of $n$-Cesàro means.

  4. It is not hard to see that if $(\sigma^n_i)_{i\geq 1}$ is bounded, than we should have $a_i = \mathcal{O}(i^n)$. However, we can find a sequence $(a_i)$ which has Abel mean but satisfies $\lim_{i\to\infty} |a_i| / i^n = \infty$ for all $n$. So the Abel mean is strictly stronger than any $n$-Cesàro means.

Now here are my thought about extending the Cesàro mean in some sense. Let us focus on the case where $a_i$ are real and define

$$ \sigma^{n,+}_{\infty} = \limsup_{i\to\infty} \sigma^n_i, \qquad \sigma^{n,-}_{\infty} = \liminf_{i\to\infty} \sigma^n_i .$$

Now Cesàro-Stolz theorem tells that $(\sigma^{n,+}_{\infty})_{n\geq 0}$ is decreasing and $(\sigma^{n,-}_{\infty})_{n\geq 0}$ is increasing. And by the monotonicity, both sequences always have limit in $[-\infty, \infty]$ as $n\to\infty$. So if they happen to coincide, it would be no harm to declare this common value as the $\infty$-Cesàro mean of $(a_n)$.

Definition. If $\lim_{n\to\infty} \sigma^{n,+}_{\infty}$ and $\lim_{n\to\infty} \sigma^{n,-}_{\infty}$ coincide with value $\ell$ in $\mathbb{R}$, we say that $(a_n)$ has $\infty$-Cesàro mean with value $\ell$.

Also, we introduce the following symbols for convenience:

  • $\mathsf{Cesaro}(n)$ : the set of all sequences in $\mathbb{R}$ that have $n$-Cesàro mean, where $0 \leq n \leq \infty$.

  • $\mathsf{Abel}$ : the set of all sequences in $\mathbb{R}$ that have Abel mean.

Then the followings summarize what I know so far:

  1. $\mathsf{Cesaro}(n) \subsetneq \mathsf{Cesaro}(n+1)$ for all $n \geq 0$. (For instance, consider $a_i = (-1)^i i^{n+1}$.)

  2. $\mathsf{Cesaro}(n) \subsetneq \mathsf{Abel}$ for all $n \geq 0$. (This is what we discussed in the previous remark.)

  3. $\mathsf{Abel} \setminus \mathsf{Cesaro}(\infty)$ is non-empty.

    Indeed, if $(a_i) \in \mathsf{Abel}$ is chosen so that $\lim_{i\to\infty} |a_i|/i^n = \infty$ for all $n \geq 0$, then for each $n$ we must have either $\sigma^{n,+}_{\infty} = +\infty$ or $\sigma^{n,-}_{\infty} = -\infty$. So we must have $(a_i) \notin \mathsf{Cesaro}(\infty)$.

Remaining questions. I am not sure any of these have already answered, thought a quick googling showed no useful answers.

  1. Is $\bigcup_{n=0}^{\infty} \mathsf{Cesaro}(n) \subsetneq \mathsf{Cesaro}(\infty)$?

    If OP accepts the above definition, then this is exactly the second part of OP's question. I gut is telling that the sequence $a_i = (-1)^{\lfloor \log_2 i \rfloor}$ would provides one such example, though I have no proof for this.

    One supporting argument is that the continuous analogue of Cesàro-mean clearly shows this behavior: Let

    $$ Tf(x) = \frac{1}{x}\int_{0}^{x} f(t) \, dt $$

    The following figure is the plot of $T^1 f, \cdots, T^{20}f$ for the function $f(x) = \sin(\log x)\mathbf{1}_{\{x \geq 1\}}$:

    graph of iterates of Hardy operator for a log-oscillating function

    Here the $x$-axis uses logarithmic scale. Notice that the averaging fails to kill the logarithmic oscillation of $f$, but it does reduce the amplitude. To be precise, a direct computation shows that

    $$ T^{n+1}f(e^x) = \int_{0}^{x} \frac{u^n e^{-u}}{n!} \sin(x-u) \, du, $$

    from which we obtain a quantitative bound $\limsup_{x\to\infty} |T^n f(x)| \leq C 2^{-n/2}$ for some absolute constant $C> 0$. So it will have zero $\infty$-Cesàro mean.

  2. Another obvious question would be whether $\mathsf{Cesaro}(\infty) \subseteq \mathsf{Abel}$ is true or not. I also believe that this is true, but I have no good reason why I believe so.

Finally, my guess is that we have the following hierarchy

$$ \mathsf{Cesaro}(0) \subsetneq \mathsf{Cesaro}(1) \subsetneq \cdots \subsetneq \bigcup_{n=0}^{\infty} \mathsf{Cesaro}(n) \subsetneq \mathsf{Cesaro}(\infty) \subsetneq \mathsf{Abel}. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .