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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.

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    $\begingroup$ Hopeless, take for example $q_n=1$ for the first $N$ indices, then $q_n=-1$ for the next $N$, and then $q_n=0$. $\endgroup$ Feb 4, 2022 at 18:49
  • $\begingroup$ confusing: "and then $q_n = 0$"??? $\endgroup$
    – Fei Cao
    Feb 4, 2022 at 19:12
  • $\begingroup$ "then" means "for $n>2N$" $\endgroup$ Feb 4, 2022 at 19:56
  • $\begingroup$ 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"? $\endgroup$ Feb 4, 2022 at 20:01
  • $\begingroup$ @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. $\endgroup$
    – Fei Cao
    Feb 4, 2022 at 20:14

1 Answer 1

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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.

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  • $\begingroup$ 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). $\endgroup$
    – Fei Cao
    Feb 5, 2022 at 16:58

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