# The spectrum of an unbounded operator

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.

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I guess that a bounded spectrum always implies a bounded operator. Try looking for the keyword 'spectral radius'. –  Giuseppe Negro Sep 12 '12 at 13:21
but spectral radius is meaningful only for bounded linear operators,isn't it? –  sun Sep 12 '12 at 13:28
I don't know. If $T$ is self-adjoint or normal in a Hilbert space, then surely the formula $\lVert T \rVert_{\mathrm{op}}=\rho(T)$ holds. This is a consequence of the spectral theorem, you probably already know those things. Unfortunately, you are asking for the general case and this I do not know. I guess that it is a general fact that $\lVert T \rVert_{\mathrm{op}}\le \rho(T)$, true for operators of all kinds, but I cannot prove this. –  Giuseppe Negro Sep 12 '12 at 13:39
@Giuseppe: your inequality goes in the wrong direction. It's always true that $\rho(T) \leq \lVert T \rVert$, but inequality can be strict. Recall the Volterra operator. –  t.b. Sep 12 '12 at 13:45
What if you take the direct sum of an operator with empty spectrum and the zero operator? Would not the spectrum be $\{0\}$? –  user31373 Sep 12 '12 at 18:14
$\displaystyle P = \frac{1}{2\pi i} \intop_\gamma (\lambda - T)^{-1} d\lambda$,
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 $\{0\}$.