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}
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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 $x=x_2$. $\endgroup$– Andreas BlassCommented Jan 2, 2018 at 16:58
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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
