Is the product of a Cesàro summable sequence of $0$s and $1$s Cesàro summable? 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?
 A: 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 $d(A \cap B) \notin \mathbb{R}$.
A: 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}
A: According to Cesàro's theorem, although the Cauchy product of a $(C,m)$-summable sequence and an $(C,n)$-summable sequence need not be Cesàro summable at the same level, it is $(C,m+n+1)$-summable.
