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 $V:= \{ f \mid \exists r=(r_n)_{n\in\mathbb{Z}} \in \ell^\infty(\mathbb{Z}) : f(x)= \sum\limits_{n\in\mathbb{Z}} r_n e^{2\gamma(xn/2)^2} \ \forall x\in\mathbb{R}\}$. Now suppose that for a countable set $\Lambda \subseteq \mathbb{R}$ there are constants $A,B>0$ s.th. for all $\textbf{nonnegative}$ $f \in V$ 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 nonnegative 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 $fg$ or $gf$ are not nonnegative, I don't know how to use the above statement.

1. If f and g are nonnegative, the absolute value is not needed.

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

If the stated result requires f to be nonnegative, then why the conclusion involves f? Are you sure that f has to be nonnegative 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.

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.

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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