Are these conditions regarding products of consecutive terms in a sequence of positive numbers equivalent?

Question

Assume $w_n$ is a bounded (weight) sequence of positive numbers. We want to consider products of consecutive terms in this sequence. For $i,j\in \mathbb{N}$, define $M_i^j = w_i w_{i+1}\cdots w_{i+j-1}$; that is to say, $M_i$ is the product of $j$ terms, starting with the $i$th term. Consider the following three statements concerning the sequence $w_n$.

\begin{align*} P: & \sup_{n\geq 1} M_1^n = \infty \hskip .2in \text{ and } \hskip .2in \inf_{n,i\geq 1} M_i^n >0 \\[0.2in] Q: & \inf_{n} w_n>0, \hskip .2in \text{ and } \hskip .2in \exists \text{ increasing } (n_k) \text{ s.t. } \lim_{k\to \infty} M_1^{n_k} = \infty \text{ and } \limsup_{k\to \infty} \left(\inf_{i\geq 1} M_i^{n_k}\right) >0\\[0.2in] R: & \inf_{n} w_n>0, \hskip .2in \text{ and } \hskip .2in \sup_{n\geq 1} M_1^n = \infty, \hskip .2in \text{ and } \hskip .2in \limsup_{n\to \infty} \left(\inf_{i\geq 1} M_i^n \right)>0\\[0.2in] \end{align*}

It's not terribly difficult to prove $P\implies Q \implies R$. My question is: must $R\implies Q$?

Some undergraduates and I proved $Q \nRightarrow P$ by coming up with a sequence $w_n$ which has property $Q$ but not property $P$. Here is such an example, minus a few details. Consider the sequence $w_n$ constructed in blocks $b_n$ given by \begin{align*} \underbrace{2}_{b_1} \underbrace{ \frac 12}_{b_2} \underbrace{22 }_{b_3} \underbrace{ \frac 12 1111 \frac 12}_{b_4} \underbrace{222}_{b_5} \underbrace{ \frac 12 1111111111 \frac 12 1111111111 \frac 12}_{b_6} \underbrace{2222}_{b_7} \underbrace{ \frac 12 1\cdots 1 \frac 12 1 \cdots}_{b_8} \end{align*}

Block $b_{2n-1}$ has $n$ 2's in it. Block $b_{2n}$ has $n$ $\frac 12$'s in it, and between any pair of $\frac 12$'s in $b_{2n}$, there are as many 1's as the total number of terms in all the previously appearing blocks. Let $|b_n|$ denote the length of block $b_n$. Then for $n_k=\sum_{i=1}^{k} |b_{2k-1}|$, the number $M_1^{n_k}$ represents the product of all the weights in the first $2k-1$ blocks, which we have arranged to equal $2^k$. One could then go on to show that there are enough 1's in the odd blocks to ensure $M_i^{n_k} \geq \frac 12$ for any $i$, and hence $Q$ is satisfied. But if we multiply all the weights in block $b_{2k}$, we get $2^{-k}$, so that $P$ is not satisfied.

The conditions $P, Q,$ and $R$ are related to (newly introduced) dynamical properties that a weighted backward shift operator on $c_0$ could have; the interested reader can check out this preprint, although this post is more related to ongoing work found here. I'm not confident those resources are useful, but they do provide the origin, context, and motivation for our question.

We are looking for any insight into whether or not $R\implies Q$ is true in general.

Answer 1

I think, $R$ does not yield $Q$.

Denote $f(n)=\log M_1^n$, then $R$ reads as

(i) $f(n+1)-f(n)$ is bounded both from below and from above (as $w_n$ is bounded and bounded away from zero);

(ii) $\sup f=+\infty$;

(iii) there exists a sequence $n_1<n_2<\ldots$ and a constant $C$ such that $f(x+n_i)-f(x)\geqslant -C$ for all positive integer $x$.

And $Q$ asks whether the sequence in (iii) may be chosen so that $\lim f(n_i)=+\infty$.

For a binary positive integer $x=\sum_{j=0}^\infty \varepsilon_j 2^j$, $\varepsilon_j\in \{0,1\}$ put $f(x)=\sum_{j=0}^\infty (-1)^j \varepsilon_j$.

Then (i) and (iii) for $n_i=2^i$ follows from the fact that when you add a power of 2 to a positive integer, several consecutive 1's in its binary expansion become equal to 0, and also one 0 becomes 1. This does not affect an alternating sum of binary digits too much. (ii) is obvious.

But since $f(2x)=-f(x)$, any sequence satisfying $f(n_i)\to +\infty$ also satisfy $f(2n_i)-f(n_i)=-2f(n_i)\to -\infty$, so such sequence does not satisfy (iii).