A discrete version of Poincaré's inequality 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?

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: 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
$$ \sum q_n^2 \leq C \sum (n+1)^2 |q_{n+1} - q_n|^2 $$

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
$$ \sup_{n\geq 0} |q_n| \leq \sum_{n\geq 0} |q_n - q_{n+1}| $$

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.

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.
