Eigenvalues of noncompact operators I am looking for an example of an (linear) operator $T:X\longrightarrow X$, $X$ being a Banch space, such that
(1) $T$ is non-compact, i.e., the closure of $T(B)$ is not comapct, where $B$ stands for the closed unit ball of $B$.
(2) $\sup \{ |\lambda|:\lambda \textrm{ is an eigenvalue of } T\}< \|T\|$ for each eigenvalue of $T$.
I don't find any example in the spae $\ell_{p},c_{0}$...Some suggestion?
Many thanks in advance for your comments.
 A: On $\ell^{2}$ define $T(x_n)=((1-\frac  1 n) x_n)$. Then the eigen values are $\{1-\frac  1 n: n \geq 1\}$ and $\|T\|=1$.
A: On $\ell^2(\mathbb N)$, consider the canonical basis $\{e_n\}$, and define $T$ to be the linear operator with
$$
Te_1=0,\qquad Te_2=2e_1,\qquad Te_{k+2}=e_{k+2}. 
$$
That is,
$$
T=\begin{bmatrix} 0&2\\0&0\end{bmatrix}\oplus I. 
$$
Then $\|T\|=2$, while the only eigenvalues of $T$ are $\{0,1\}$. And in this case the whole spectrum is $\{0,1\}$.
This idea can be used in other sequence spaces. And by using other operators instead of the identity, you can do pretty much whatever you want with the spectrum. In this sense:

Fix a compact subset $K\subset\mathbb C$, numbers $\{\lambda_1,\lambda_2,\ldots\}$ and $c>0$ with $c\geq \max\{|z|:\ z\in K\}$ and $c\geq \lambda_n$ for all $n$. Then there exists an operator $T$ with $\|T\|=c$, $\sigma(T)=K$, and $\lambda_n$ an eigenvalue of $T$ for all $n$.

Indeed, just do
$$
T=\begin{bmatrix} 0&c\\0&0\end{bmatrix}\oplus S_1\oplus S_2, 
$$
where $S_1$ is an operator with spectrum $K$ and $S_2$ an operator with eigenvalues $\{\lambda_n\}$.
