Prove the uniqueness of a sequence using a norm inequality.

There is a paper I am trying to understand and I am stuck on one thing, here is what I know:

Let $\gamma >0$. Consider the space of functions $V:= \{ f \mid \exists r=(r_n)_{n\in\mathbb{Z}} \in \ell^\infty(\mathbb{Z}) : f(x)= \sum\limits_{n\in\mathbb{Z}} r_n e^{-2\gamma(x-n/2)^2} \ \forall x\in\mathbb{R}\}$. Now suppose that for a countable set $\Lambda \subseteq \mathbb{R}$ there are constants $A,B>0$ for all $\textbf{non-negative}$ $f \in V$ and corresponding $r=(r_n)_{n\in\mathbb{Z}} \in \ell^\infty(\mathbb{Z})$ we have $A \sup_{\lambda \in\Lambda} |f(\lambda)| \leq ||r||_\infty \leq B \sup_{\lambda\in\Lambda} |f(\lambda)|$.

Now I want to use this to show that if $f,g \in V$ are non-negative and $|f(\lambda)|=|g(\lambda)|$ holds for all $\lambda \in\Lambda$, then the corresponding sequences in $\ell^\infty(\mathbb{Z})$ of $f$ and $g$ are the same. Since in general $f-g$ or $g-f$ are not non-negative, I don't know how to use the above statement.

  • Mathe Mathe

    1. If f and g are non-negative, the absolute value is not needed.

  • Mathe Mathe

    2. I believe the bounty is too low for an advanced question.

  • If the stated result requires f to be non-negative, then why the conclusion involves |f|? Are you sure that f has to be non-negative for the result to hold? Please double check your question.

  • yes I am sure, it was not necessary to involve |f| indeed

  • Then it is not straightforward to prove uniqueness using this result. This seems to be a very challenging problem.

  • M F H M F H

    by "for a countable set" do you mean "for any ..." ?

  • No, it is such that the above statement holds. To be specific, it is separated and has lower beurling density larger than two, but those are just tools to prove the above statements.

  • The bounty is too low for the level of the question.

  • Erdos Erdos

    Do you assume that \Lambda has an accumulation point? This can not be proved using the result you mentioned, but I can write a different solution which uses complex analysis.


Answers can be viewed only if
  1. The questioner was satisfied and accepted the answer, or
  2. The answer was disputed, but the judge evaluated it as 100% correct.
View the answer
Erdos Erdos
The answer is accepted.
Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.