Showing that a $L^2$-norm is positive

Let $\lambda_n = n + \delta_n$ for all $n \in \mathbb{Z}$ where $\delta_n$ are a sequence of real numbers in $\ell^2(\mathbb{Z})$. I am looking to prove that the sequence $(x_n)_{n \in \mathbb{Z}} = (e^{i \lambda_n t})_{n \in \mathbb{Z}}$ is a minimal sequence 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}

I haven't really broken through aside from writing out the integral, so I would appreciate any input.

Edit: Assume moreover that $\delta_n \neq \delta_k$ for every $n \neq k$ (thanks to mechanodroid).

Edit: My question has gotten an answer at MO (https://mathoverflow.net/questions/289763/a-minimal-sequence-in-l2-pi-pi), but I leave it open if someone wants to add a suggestion.

• This may only be true if all $\lambda_n$ are different numbers. For $(\delta_n)_{n=1}^\infty$ defined as $\delta_{-1} = 1$, $\delta_1 = -1$, $\delta_n = 0$ for all $n \in \mathbb{Z}\setminus\{-1, 1\}$ we have: $$\lambda_1 = 1 + \delta_1 = 0 = (-1) + \delta_{-1} = \lambda_{-1}$$ so $e^{i\lambda_{-1}t} = 1 = e^{i\lambda_{1}t}$ so the sequence cannot be minimal. – mechanodroid Jan 2 '18 at 16:32