If $\{\lambda_n\}_n$ is a certain sequence of real constants such that $\{e^{i\lambda_n x}\}$ is complete in $L^2\left(-\gamma,\gamma\right)$, could I say that $\{e^{i\lambda_n x}\}$ is complete in $L^2\left(0,2\gamma\right)$?
I think that in general the answer is negative so I would like to know if there is any sufficient condition providing it.
I have also a doubt about the following theorem (from "Gap and Density theorems" by Levinson, page 3):
"The sequence $\{e^{i\lambda_n x}\}, \quad \lambda_n>0$, is closed over an interval of length L if $\liminf_{n\rightarrow \infty} \frac{n}{\lambda_n}>\frac{L}{2\pi}$."
The interval of length $L$ has to be symmetric with respect to $0$ or not?
I do not know what is meant with $\{\exp [i\lambda _{n}x]\}$ being complete. A basis? Anyway let us assume that the linear span of $\{\exp [i\lambda _{n}x]\}$ is dense in $L^{2}(-\gamma ,+\gamma )$. Then \begin{equation*} \int_{-\gamma }^{+\gamma }dx\exp [i\lambda _{n}x]f(x)=0,\;\forall n\;\Rightarrow \;f(x)=0 \end{equation*} Now consider \begin{equation*} \int_{0}^{2\gamma }dy\exp [i\lambda _{n}y]f(y)=\exp [i\lambda _{n}\gamma ]\int_{-\gamma }^{+\gamma }dx\exp [i\lambda _{n}x]f(x+\gamma )=0,\;\forall n \end{equation*} Then $f(x+\gamma )$ must vanish. Thus $\{\exp [i\lambda _{n}x]\}$ in $% L^{2}(0,2\gamma )$ is complete in your sense.
As to your second question what do you mean by closed?