It's well known that the spectrum of a bounded operator on a Banach space is a closed bounded set (and non-empty)on the complex plane. And it's also not hard to find unbounded operators which their spectrum are empty or the whole complex plane.
Conversely, suppose $T$ is an unbounded operator on a Banach space $E$,and has non-empty spectrum, does this imply that the $\sigma(T)$ is unbounded on $\mathbb{C}$ ?
As far as I known, if $\sigma(T)$ is bounded,then it implied that $\infty$ is the essential singular point of the resolvent $(\lambda-T)^{-1}$, but I don't know how to form a contradiction.
Best Answer
If the spectrum is bounded, then by means of holomorphic functional calculus we can extract a projection:
$\displaystyle P = \frac{1}{2\pi i} \intop_\gamma (\lambda - T)^{-1} d\lambda$,
where $\gamma$ encloses the spectrum. Intuitively, this should be thought of as separating the "bounded part" $TP$ (which is indeed bounded, since $T\left(\lambda-T\right)^{-1}=\lambda\left(\lambda-T\right)^{-1}-1$) and the part $T-TP$, which has empty spectrum on $\mathbb{C}$ when restricted to $\ker P$, but should be thought of as having a point at infinity, since indeed its inverse is a bounded operator with spectrum $\{0\}$.