$Prove the uniqueness of a sequence using a norm inequality.$

Question

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 $A,B>0$ s.th. 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.

Real Analysis

Ichbinanonym

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Mathe

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

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

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.
Ichbinanonym

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

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

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

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.
Daniel90

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

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.

Mathe

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

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

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.
Ichbinanonym

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

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

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.
Daniel90

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

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.

Answer 1

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I donlt think the result you are trying to prove can be obtained from the stated result. Here I provide a solution using different ideas. I hope you find it helpful and instructional.

Here is an answer when $\Lambda$ has an accumulation point.

Let $f\in V.$ Then

\[f(x)=\sum_{-\infty}^\infty r_n(f)e^{-2\gamma(x-n/2)^2}=e^{-2\gamma x^2}\sum_{-\infty}^\infty r_n(f)\,e^{-\gamma n^2/2}\,e^{\gamma nx}\]

\[=e^{-2\gamma x^2}\sum_{-\infty}^\infty s_n(f)\,e^{\gamma nx},\quad \text{where} s_n(f)=r_n(f)\,e^{-\gamma n^2/2}.\]

For function $h\in V$ denote $M_f=\sup_n|r_n(h)|.$

Assume $f(x)=g(x)$ for $x\in \Lambda$. Then $e^{2\gamma x^2}f(x)-e^{2\gamma x^2}g(x)=0$ for $x\in \Lambda.$

Thus

$$\sum_{-\infty}^\infty [s_n(f)-s_n(g)]\,e^{\gamma nx}=0,\quad x\in \Lambda.$$
Consider the function

$$u(z)=\sum_{-\infty}^\infty [s_n(f)-s_n(g)]\,e^{\gamma nz},\quad z\in \mathbb{C}.$$

The series is uniformly convergent on bounded subsets of the complex plane due to

$$|s_n(f)-s_n(g)|\,|e^{\gamma nz}|\le [M(f)+M(g)]\,e^{-\gamma n^2/2}e^{\gamma n{\rm Re}(z)}.\quad (*)$$

Hence $u(z)$ is an entire function, which vanishes on $\Lambda$. If $\Lambda$ has an accumulation point, then $u=0,$ which implies $f=g.$

Nirenberg

$Prove the uniqueness of a sequence using a norm inequality.$

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