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.


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Erdos Erdos
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