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Suppose $a_n$ and $b_n$ to be Cesàro summable sequences of zeros and ones, $a_n\in\{0,1\}$ and $b_n\in\{0,1\}$, i.e. the limits $$ \lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^{N}a_n, $$ and $$ \lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^{N}b_n, $$ do exist.

Is the product sequence $c_n=a_nb_n$ always Cesàro summable?

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No. Let $a_n = 1$ iff $n$ is even. Let $b_n = 1$ iff $n$ has an even number of digits and is even OR has an odd number of digits and is odd. (Write $n$ in base 10.)

Clearly $a_n$ is Cesaro summable with sum $\frac12$. $b_n$ is also Cesaro summable with sum $\frac12$; to see this, note that $\sum_{i=1}^N b_i = 1 + \sum_{i=1}^N a_i$ for all odd $N$.$^{1}$

However, we have $$ a_n b_n = 1 \iff n \text{ has an even number of digits and is even} $$ which is not Cesaro summable, since$^{2}$ \begin{align*} \frac{\sum_{i=1}^{10^{2k} - 1} a_i b_i}{10^{2k} - 1} &= \frac{9 \cdot 5 \cdot 10^{2k-2} + 9 \cdot 5 \cdot 10^{2k-4} + \cdots + 9 \cdot 5}{10^{2k} - 1} \\ &= \frac{45 \cdot (100^k - 1)/(100-1)}{100^k - 1} \\ &= \frac{45}{99} = \frac{10}{22} \end{align*} whereas \begin{align*} \frac{\sum_{i=1}^{10^{2k+1} - 1} a_i b_i}{10^{2k+1} - 1} &= \frac{9 \cdot 5 \cdot 10^{2k-2} + 9 \cdot 5 \cdot 10^{2k-4} + \cdots + 9 \cdot 5}{10^{2k+1} - 1} \\ &= \frac{45 \cdot (100^k - 1)/(100-1)}{10 \cdot 100^k - 1} \\ &< \frac{45 \cdot (100^k - 1)/(100-1)}{10 (100^k - 1)} \\ &= \frac{45}{990} = \frac{1}{22}. \end{align*}


$^{1}$: More detail on this: For every even $i \ge 2$, $i$ is even and $i+1$ is not even. So $a_i + a_{i+1} = 1 + 0 = 1$. Moreover, whether $i$ has an even or an odd number of digits, since $i$ is even $i+1$ has the same number of digits, which means that $b_i + b_{i+1} =$ either $0 + 1$ or $1 + 0$, but either way $= 1$. So when $N$ is odd, $a_i + a_{i+1} = b_i + b_{i+1}$ for each even $i$ from $2$ to $N-1$, so that $$ \sum_{i=2}^{N} b_i = \sum_{i=2}^{N} a_i. $$ Since $a_1 = 0$ and $b_1 = 1$, it follows that $$ \sum_{i=1}^{N} b_i = 1 + \sum_{i=1}^{N} a_i. $$

$^{2}$: More detail on this: The quantity $$ \sum_{i=1}^{10^{2k}-1} a_i b_i $$ is the number of integers with at most $2k$ digits which have an even number of digits and are even. For any $j$, $1 \le j \le k$ he number of even numbers with $2j$ digits is $9 \cdot 5 \cdot 10^{2j-2}$: there are $9$ choices for the first digit (so that the number of digits is exactly $2j$ and not less), $2$ choices for the middle $2j - 2$ digits, and $5$ choices for the last digit ($0, 2, 4, 6,$ or $8$, to make it even).

Therefore, this gives $$ \sum_{i=1}^{10^{2k} - 1} a_i b_i = 9 \cdot 5 \cdot 10^{2k-2} + 9 \cdot 5 \cdot 10^{2k-4} + \cdot + 9 \cdot 5 \cdot 10^{2} + 9 \cdot 5. $$ Likewise, since $a_i b_i = 0$ for a number with an odd number of digits (in particular for a number with $2k + 1$ digits), $$ \sum_{i=1}^{10^{2k+1} - 1} a_i b_i = \sum_{i=1}^{10^{2k} - 1} a_i b_i = 9 \cdot 5 \cdot 10^{2k-2} + 9 \cdot 5 \cdot 10^{2k-4} + \cdot + 9 \cdot 5 \cdot 10^{2} + 9 \cdot 5. $$

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  • $\begingroup$ thanks for answering. However I do not follow a couple of points, first one, could you give a bit more explanation of $\sum_{i=1}^{10^{2k+1}-1}a_ib_i=9\cdot5\cdot10^{2k-2}+9\cdot5\cdot2^{2k-4}+\dots+9\cdot5$? Secondly, I do not see $\sum_{i=1}^Na_i=\sum_{i=1}^Nb_i$ for even $N$. For example for $N=12$, we have $\sum_{i=1}^{12}a_i=0+1+0+1+0+1+0+1+0+1+0+1=6$ and $\sum_{i=1}^{12}b_i=1+0+1+0+1+0+1+0+1+1+0+1=7$. $\endgroup$ – NessunDorma Nov 5 '15 at 2:41
  • $\begingroup$ Thanks, I will appreciate if you can explain a bit more how to obtain $\sum_{i=1}^{10^{2k}-1}a_i b_i= 9\cdot 5 \cdot 10^{2k-2}+9\cdot 5 \cdot 2^{2k-4}+\dots+9\cdot 5$ $\endgroup$ – NessunDorma Nov 6 '15 at 0:56
  • $\begingroup$ @NessunDorma Sorry for all the errors, I realized my formula for $\sum_{i=1}^N b_i = 1 + \sum_{i=1}^N a_i$ is for ODD $N$ not for even $N$. You can check it still works in your example $N = 11$, but it also works for $N = 5, 7, 9, 13$, etc. (doesn't work for all even $N$ like $N = 8$). $\endgroup$ – 6005 Nov 6 '15 at 7:26
  • $\begingroup$ I added two footnotes with a lot more detail. First, the reason why $\sum_{i=1}^N b_i = 1 + \sum_{i=1}^N a_i$ which is that the terms of the each pair up and each pair equals $1$. Second, the justification for the formula you mention. (There was a typo--$2^{2k-4}$ was corrected to $10^{2k-4}$.) $\endgroup$ – 6005 Nov 6 '15 at 7:27
  • $\begingroup$ This answer should be correct now!! Please let me know if there are other errors or other things I can clarify. $\endgroup$ – 6005 Nov 6 '15 at 7:28
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REMARK

Let's be careful here... even though the product sum may not be Cesaro summable, they can still be bounded. How do we make sense of the Cauchy-Schwarz inequality here?

$$ \underbrace{\left( \frac{1}{N}\sum a_n b_n\right)^2 }_{C^2} \leq \left(\frac{1}{N}\sum a_n^2\right) \left(\frac{1}{N}\sum b_n^2\right) = \underbrace{\left(\frac{1}{N}\sum a_n\right)}_{A} \; \; \underbrace{\left(\frac{1}{N}\sum b_n\right)}_{B}$$

In some sense, the sum of the product can never get any larger than the square root of the product - or geometric mean.

$$ C \leq \sqrt{A \times B} $$

Remember that $a, b = 0$ or $1$ so that $a^2 = a$, since $0^2 = 0, 1^2 = 1$.


A FEW WAYS TO THINK ABOUT CESARO SUM

We can imagine the sequence $a_1, a_2, a_3 \dots $ as the outcomes of a random variable $A \in \{0,1\}$ (more correctly a dynamical system). Then we might try to define an expectation:

$$ \mathbb{E}(A) = \lim_{N \to \infty} \frac{1}{N} \sum a_n $$

Unfortunately this limit might not exist. At one scale the fraction of 0's might be $\frac{1}{2}$ on yet a larger scale that fraction might go down to $\frac{1}{3}$. One larger and larger scales that ratio could alternate between $\frac{1}{3}$ and $\frac{1}{2}$. So instead we might define two densities:

$$ \overline{\mathbb{E}(A)} = \limsup_{N \to \infty} \frac{\sum a_n}{N} = \overline{d(A)} \hspace{0.25in}\text{ vs }\hspace{0.25in} \underline{\mathbb{E}(A)} = \liminf_{N \to \infty} \frac{\sum a_n}{N}=\underline{d(A)}$$

Instead of a expectation of a random variable, we are taking the upper and lower density of the set $A = \{ n\in \mathbb{N}: a_n = 1\}$. Then we are asking:

$$ \overline{d(A)}=\underline{d(A)} \hspace{0.25in}\text{ and }\hspace{0.25in} \overline{d(B)}=\underline{d(B)} \hspace{0.25in}\longrightarrow?\hspace{0.25in} \overline{d(A\cap B)}=\underline{d(A\cap B)} $$

This notation is getting a little bit dense for me, so I am not happy with it but at least it's more standard.


CONSTRUCTION

The Cauchy Schwartz inequality basically says $\mathbb{E}(AB) \leq \mathbb{E}(A)\mathbb{E}(B)$. If we know that $\mathbb{E}(A) = \frac{1}{2}$ and $\mathbb{E}(B) = \frac{1}{2}$, do we know anything about the expectation $\mathbb{E}(AB)$ (or equivalently the density $d(A \cap B)$.

Is it even a number?

This is a statement about how the sets $A$ and $B$ are correlated. We can show that $A$ and $B$ can be correlated in different ways at different scales in such a way that $d(A \cap B)$ does not converge to any limit.

If $d(A) = d(B) = \frac{1}{2}$ then the density of the intersection could be any number $0 \leq d(A \cap B) \leq \frac{1}{2}$. In fact, over any finite interval we could certainly have $|A \cap B \cap [1, N]|$ be any number between $0$ and $\frac{N}{2}$.

On the interval $[0,N]$ let $A \cap B$ have density $0$ and on $[N, 2N]$ let $A \cap B$ have density $\frac{1}{2}$. More generally let $A \cap B$ have

  • density $0$ on any set $[2^{2k} N, 2^{2k+1}N]$
  • density $\frac{1}{2}$ on any set $[2^{2k+1} N, 2^{2(k+1)}N]$

This behavior should lead to conflicting behavior between the upper and lower densities $\overline{d(A\cap B)}$ and $\underline{d(A\cap B)}$.

  • $|A \cap B \cap [0, 2^{2k+1}]| < \frac{1}{2}2^{2k}$ so that $\underline{d(A \cap B)} < \frac{1}{2}\frac{2^{2k}}{2^{2k+1}} = \frac{1}{4}$

  • $|A \cap B \cap [0, 2^{2k}]| > \frac{1}{2}(2^{2k-1} + 2^{2k-3})$ so that $\overline{d(A \cap B)} > \frac{\frac{1}{2}(2^{2k-1} + 2^{2k-3})}{2^{2k}} = \frac{1}{4} + \frac{1}{8} = \frac{5}{16}$

Not as dramatic as I would like, but we have shown $\underline{d(A \cap B)} < \frac{1}{4}$ and yet $\overline{d(A \cap B)} > \frac{5}{16}$ therefore $d(A \cap B)$ cannot exist... rather $d(A \cap B) \notin \mathbb{R}$.

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A counter-example that may be simpler than 6005's is:

Define even-length frames of size $2^n$ for $n \in \{1, 2, 3, ...\}$. Define sequences $\{a_k\}$ and $\{b_k\}$ to periodically alternate between 0 and 1 over each frame. So the Cesaro average is 1/2 for both sequences. But on odd frames have them aligned (so $a_kb_k=a_k$ on odd frames). On even frames have them misaligned (so $a_kb_k=0$ on even frames).

Illustration (with odd frames labeled):

\begin{align} &\{a_k\}: \: [ 10 ] \: [1010] \: [10101010] \: [1010101010101010]...\\ &\{b_k\}: \underbrace{[10 ]}_{frame 1} [0101] \: \underbrace{[10101010]}_{frame 3} \: [0101010101010101]... \end{align}

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    $\begingroup$ But this is exactly my counterexample :) $\endgroup$ – 6005 Nov 7 '15 at 16:45
  • $\begingroup$ (In base 2 that is) $\endgroup$ – 6005 Nov 8 '15 at 0:50
  • $\begingroup$ @6005 : Maybe. Your answer looked complicated so I didn't really read it. Apologies. =) $\endgroup$ – Michael Nov 8 '15 at 21:41

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