spectrum of a unilateral left weighted shift operator on c0 The spectrum of shift operators on $\ell_p$ is well-documented, however the spectrum of shift operators on $c_0$ does not seem to be nearly as commonly discussed.
Definitions.
Let $(e_n)_{n=1}^\infty$ denote the usual basis of $c_0$, and define the continuous linear operator $T:c_0\to c_0$ by the rules
\begin{equation}Te_{n+1}=w_ne_n\;\;\;(n\in\mathbb{N}),\;\;\;Te_1=0,\end{equation}
where $(w_n)_{n=1}^\infty$ is a strictly decreasing sequence of positive real numbers with $w_1=1$.  It is clear that
\begin{equation}\|T^n\|=\prod_{i=1}^nw_i\;\;\;(n\in\mathbb{N})\end{equation}
so that the spectral radius of $T$ is given by
\begin{equation}r:=r(T)=\lim_{n\to\infty}\left(\prod_{i=1}^nw_i\right)^{1/n}.\end{equation}
Conjecture.  Let $\sigma_p(T)$ denote the point spectrum of $T$ (defined as the set of all its eigenvalues).  Then
\begin{equation}\sigma_p(T)=D_r:=\left\{\lambda\in\mathbb{C}:|\lambda|\leq r\right\}.\end{equation}
Discussion.
The first answer below observes that
\begin{equation}D_r^o:=\{\lambda\in\mathbb{C}:|\lambda|<r\}\subseteq\sigma_p(T).\end{equation}
(From this it follows that $\sigma(T)=D_r$.)  However, I need to show that $\sigma_p(T)=D_r$.  As this is not the case when $w_n=1$ for all $n\in\mathbb{N}$ (see the comments below), I suspect the conjecture is not true even for strictly decreasing $(w_n)_{n=1}^\infty$.
If necessary, we could make additional assumptions about $(w_n)_{n=1}^\infty$ so that the conjecture is true.
Note.  I desire an operator with $0<r(T)<1$, so any additional assumptions we impose on $(w_n)_{n=1}^\infty$ should permit this.
Thank you!
 A: The point spectrum of your operator is empty.
Let first $0\neq \lambda \in \sigma_p(T).$ Then there exists an eigenvector $0\neq x=(x_i)_{i\in \mathbb{N}}\in c_0$ such that $T(x) = \lambda x$. This leads to the equations
$$ \lambda x_1 =0 \quad \text{and} \quad \lambda x_{i+1} = w_i x_i \quad \text{for } i\in \mathbb{N}.$$
As $\lambda\neq 0$, we get by induction that $x=0$, which is a contradiction.
Let now $\lambda=0$, then the equations become
$$ w_i x_i =0. $$
As all the $w_i$ are different from zero, we get again the contradiction $x=0$.
$\textbf{Edit}$: The above argument is wrong, I misread the definition of the operator. One can give a necessary and sufficient condition for the conjecture to be true.
The equations we really need to solve are
$$ \lambda x_i = w_i x_{i+1}.$$
Set $x_1=1$, then the solutions are
$$ x_{n+1} = \frac{\lambda^n}{\prod_{j=1}^n w_j} 
= \left( \frac{\lambda}{\left( \prod_{j=1}^n w_j \right)^{1/n}} \right)^n$$
For $\vert \lambda \vert < \lim_{n\rightarrow \infty} \left( \prod_{j=1}^n w_j \right)^{1/n}$ the solution defines a sequence converging to zero. Indeed, let $q:= \vert \frac{\lambda}{\lim_{n\rightarrow \infty} \left( \prod_{j=1}^n w_j \right)^{1/n}} \vert$ then
$$ \lim_{n\rightarrow \infty} \vert x_n \vert = \lim_{n\rightarrow \infty} \vert x_{n+1} \vert
\leq \lim_{n\rightarrow \infty} q^n =0.$$
We know that there cannot exist an eigenvalue with absolut value larger than the spectral radius. It remains to check, wheter all complex numbers of absolute value equals the spectral radius are also contained in the point spectrum.
One computes that $r=lim_{n\rightarrow \infty} w_n$. Let $N\in \mathbb{N}$, then as the $(w_n)_{n\in \mathbb{N}}$ is decreasing we get
$$ \lim_{k\rightarrow \infty} w_k \leq  \left(\prod_{i=1}^n w_i \right)^{1/n} 
= \left(\prod_{i=1}^N w_i \right)^{1/n} \cdot \left(\prod_{i=N+1}^n w_i \right)^{1/n}
\leq \left(\prod_{i=1}^N w_i \right)^{1/n} \cdot w_{N+1}^{(n-N)/n}$$
Taking the limit for $n\rightarrow \infty$
$$\lim_{k\rightarrow \infty} w_k  \leq r \leq  w_{N+1}.$$
Taking $N\rightarrow \infty$ yields $r=\lim_{k\rightarrow \infty} w_k$.
From this we get the necessary and sufficient condition for the conjecture to hold:
$$ \prod_{i=1}^n \frac{r}{w_i} \rightarrow 0.$$
An example of such a sequence would be $w_n=\frac{n}{n-1}$ for $n>2$.
