# 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.

• Hopeless, take for example $q_n=1$ for the first $N$ indices, then $q_n=-1$ for the next $N$, and then $q_n=0$. Feb 4, 2022 at 18:49
• confusing: "and then $q_n = 0$"??? Feb 4, 2022 at 19:12
• "then" means "for $n>2N$" Feb 4, 2022 at 19:56
• This doesn't seem a straightforward discretization of Poincaré inequality, as the domain in this case is the non-compact set $\mathbb{N}_0$. There is a number of generalizations of P.i. to unbounded domains, which one you want to "discretize"? Feb 4, 2022 at 20:01
• @AlessandroDellaCorte I agree that the usual Poincare inequality is presented in the case of bounded domains of $\mathbb{R}^n$, so this conjecture is just a inequality of Poincare-type flavor. Feb 4, 2022 at 20:14

## 1 Answer

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.

• I like the comment "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." This reminds me of the proof of the usual Sobolev-type inequalities (based on the fundamental theorem of calculus and integration by parts). Feb 5, 2022 at 16:58