On Riesz basis of exponentials

Question

Let $I$ and $I'$ be two intervals of $\mathbb{R}$ such that $I'\subset I$. If $\{{e^{{i\lambda_n}t}}\}_{n\in\mathbb{N}}$ is a Riesz basis in $L^{2}(I)$, can we confirm that $\{{e^{{i\lambda_n}t}}\}_{n\in\mathbb{N}}$ is a Riesz basis in $L^{2}(I')$?

Answer 1

No, it will never be a Riesz basis for $L^2(I')$: the lower bound fails. Consider the function $f=\chi_{I\setminus I'}$. As an element of $L^2(I)$, it can be represented in the Riesz basis as $$f(t) = \sum_{n\in\mathbb{N}} c_n e^{i\lambda_n t}$$ where the series converges in $L^2(I)$ and $\sum|c_n|^2$ is comparable to $\|f\|_{L^2}$. But this means that in the space $L^2(I')$, the sum $\sum_{n\in\mathbb{N}} c_n e^{i\lambda_n t}$ converges to $0$. So it's not even a Schauder basis for $L^2(I')$.