Proving that the point spectrum of the right shift operator on $\mathscr{l}^2(\mathbb{Z})$ is empty. 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.
Suppose there exists some $\lambda \in \mathbb{C}$ such that $\lambda \in \sigma_{pt} (R)$. Then, for some nonzero sequence $x \in l^2 (\mathbb{Z})$, $$Rx = \lambda x \leftrightarrow x_{n-1} = \lambda x_n \forall n \in \mathbb{Z} .$$

Let's then divide this into cases. First, suppose that $| \lambda | \geq 1$. In this case, the above equation can be rewritten as $$\frac{x_{n-1}}{x_n} = \lambda \rightarrow \left\vert \frac{x_{n-1}}{x_n} \right\vert = |\lambda| \geq 1.$$ However, this is a contradiction, since the terms of a sequence in $l^2(\mathbb{Z})$ must vanish as $n \rightarrow \pm \infty$. 

The second case follows a similar argument. Suppose $|\lambda| &lt; 1$. Then, $$\left\vert \frac{x_{n-1}}{x_n} \right\vert = |\lambda| \leftrightarrow \left\vert \frac{x_{n}}{x_{n-1}} \right\vert = \frac{1}{|\lambda|} &gt; 1.$$ The contradiction follows as before. 

Thus, there is no nonzero sequence $x \in l^2 (\mathbb{Z})$ such that $R x = \lambda x$ for some $\lambda \in \mathbb{C}$.
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
$$
                 0 = \|(M_{a}-\lambda I)f\|^{2}=\int_{-\pi}^{\pi}|a-\lambda|^{2}|f|^{2}dx \\
             \implies |a-\lambda||f| = 0\;\; a.e.
$$