Given a (bounded) sequence $\{q\_n\}\_{n\geq 0}$ such that $\lvert q\_n\rvert \leq 1$ for all $n \geq 0$ and $\sum\_{n\geq 0} q\_n = 0$. We can impose the condition that $\sum\_{n\geq 0} \lvert q\_n\rvert \leq 2$ as well. I am wondering whether there exists a fixed constant, independent of the sequence $\{q\_n\}$, such that $$\sum\_{n \geq 0} \lvert q\_n - q\_{n+1}\rvert^2 \geq C \sum\_{n \geq 0} \lvert q\_n\rvert^2$$ holds true?




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Edit: It seems that $\sum\_{n\geq 0} |q\_n| \leq 2$ might not be enough, I am wondering if higher order moment bound such as $\sum\_{n\geq 0} (1+n)|q\_n| < \infty$ will suffice.



A discrete version of Poincaré's inequality
You can duplicate the usual proof of Hardy type inequalities to the discrete case.


Suppose $\{q\_n\}$ is an eventually 0 sequence (you can weaken this to $\lim\_{n\to \infty} n^{1/2} q\_n = 0$). Then by telescoping you have (all sums are over $n\geq 0$)


$$ \sum (n+1) q\_{n+1}^2 - n q\_{n}^2 = 0 $$


Rewrite as


$$ \sum q\_n^2 = - \sum (n+1) (q\_{n+1} + q\_n) (q\_{n+1} - q\_n) $$


Take absolute values and apply Cauchy-Schwarz on the RHS


$$ \sum q\_n^2 \leq \left( \sum (n+1)^2 |q\_{n+1} - q\_n|^2 \right)^{1/2} \left( \sum (q\_{n+1} + q\_n)^2 \right)^{1/2} $$


the second factor can be bounded by $2 ( \sum q\_n^2)^{1/2}$. Cancelling and you get


$$ \sum q\_n^2 \leq C \sum (n+1)^2 |q\_{n+1} - q\_n|^2 $$




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Scott Armstrong's comment is similar. As long as $\lim\_{n\to\infty} q\_n = 0$ you have


$$ q\_n = \sum\_{k \geq n} q\_k - q\_{k+1} $$


then


$$ \sup\_{n\geq 0} |q\_n| \leq \sum\_{n\geq 0} |q\_n - q\_{n+1}| $$




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If you want to look at "Sobolev type" inequalities: they are all essentially based on the fundamental theorem of (discrete) calculus applied in various ways.




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Finally: note that you can also do a scaling argument.


Let $\lambda$ be a positive integer.


Let $q^{(\lambda)}\_{n}$ be such that
$$ q^{(\lambda)}\_m = q\_n \text{ if } m \in [\lambda n, \lambda (n+1))$$


This scaling preserves the $ \sum |q\_{n+1} - q\_n|^2$ semi-norm, but has $ \sum |q^{(\lambda)}\_n|^2 = \lambda \sum |q\_n|^2$, which immediately falsifies your proposed Poincaré inequality.


You see that the inclusion of weights in Hardy avoids this difficulty.


