2
$\begingroup$

Let $\lambda_n = n + \delta_n $ for all $n \in \mathbb{Z}$ where $\delta_n$ are a sequence of real numbers in $\ell^2(\mathbb{Z})$. How can one show that the sequence $(x_n)_{n \in \mathbb{Z}} = (e^{i \lambda_n t})_{n \in \mathbb{Z}}$ is minimal in $L^2([-\pi, \pi])$, in the sense that \begin{equation} \forall n \in \mathbb{Z}, \quad \|x_n - x \|_{L^2([-\pi, \pi])} > 0 \quad \forall x \in \overline{\text{span} \{x_k, k \in \mathbb{Z} \setminus \{n \} \}}? \end{equation}

$\endgroup$
1
  • 4
    $\begingroup$ Considering that $\delta_n$ can be large for a few (finitely many) $n$ while still satisfying $(\delta_n)\in\ell^2(\mathbb Z)$, what prevents you from having $\lambda_1=\lambda_2$? That would violate your conclusion for $n=1$ and $x=x_2$. $\endgroup$ Commented Jan 2, 2018 at 16:58

1 Answer 1

4
$\begingroup$

First, you need to say what $\delta_n$'s are.

I am assuming you meant $0\le \delta_n\le 1$. This and somewhat more general statement follows from a formula of Carleman: if $\liminf \frac{n}{\lambda_n}>\frac{A}{\pi}$, then the system is complete in $L_2[-A,A]$. For detailed proof see the chapter 3 of Young's book

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.