$\liminf_{N\to\infty}\frac{1}{N}\sum_{n=1}^Na_nb_n\ge\liminf_{N\to\infty}\frac{1}{N}\sum_{n=1}^Na_n\liminf_{N\to\infty}\frac{1}{N}\sum_{n=1}^Nb_n$?

Question

Suppose $a_n$ and $b_n$ are uniformly bounded sequences of non-negative numbers. Is it true that $$ \liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N a_n b_n \ge \liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N a_n \liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N b_n $$

My attempt. The observation should be that $b_n\ge \liminf_{N} \frac{1}{N} \sum_{i=1}^N b_n$ for $n$ large enough. I'm not sure whether this is correct.

This question may be related to this one.

Answer 1

No.

Take $$a_n \stackrel{\rm def}{=} \begin{cases}1 & \text{ if } n \text{ even} \\ 0 & \text{ if } n \text{ odd}\end{cases}$$ and $b_n\stackrel{\rm def}{=} 1- a_n$, for $n\geq 0$.

Then, for any $N\geq 0$ $$ \sum_{n=0}^N a_nb_n = 0 \tag{1} $$ and so $\liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N a_n b_n = 0$. But

$$\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N a_n = \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N b_n = \frac{1}{2} \tag{2}$$ which implies $$ \liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N a_n \liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N b_n = \frac{1}{4}. $$