How can I prove that no series $x\in\mathscr{l}^2(\mathbb{Z})$ of the form $$\forall i\in\mathbb{Z}: x_{k-1}=\lambda x_k$$ exists other than the zero sequence?
In particular I want to prove that the point spectrum of the right shift operator $$R:\mathscr{l}^2(\mathbb{Z})\rightarrow\mathscr{l}^2(\mathbb{Z})$$ $$\forall i\in\mathbb{Z}:x_i\mapsto x_{i-1}$$ is empty. I did this for $\mathscr{l}^2(\mathbb{N})$ already but can't prove the first statement, which I think i need for the $\mathscr{l}^2(\mathbb{Z})$ case.
A representation of this operator is multiplication by $e^{ix}$ on $L^{2}[-\pi,\pi]$ with inner product $$ (f,g) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt. $$ This is because $\{ e^{inx} \}_{n=-\infty}^{\infty}$ is an orthonormal basis of $L^{2}[-\pi,\pi]$, and $$ e^{ix}\sum_{n=-\infty}^{\infty}f_n e^{inx}=\sum_{n=-\infty}^{\infty}f_{n-1}e^{inx}. $$ A multiplication operator $M_{a}f=a(x)f(x)$ on $L^{2}$ cannot have an eigenvalue unless what you're multiplying by ('a' in this case) is equal to a constant on a set of non-zero measure because $$ 0 = \|(M_{a}-\lambda I)f\|^{2}=\int_{-\pi}^{\pi}|a-\lambda|^{2}|f|^{2}dx \\ \implies |a-\lambda||f| = 0\;\; a.e. $$