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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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  • $\begingroup$ I guess that a bounded spectrum always implies a bounded operator. Try looking for the keyword 'spectral radius'. $\endgroup$
    – Giuseppe Negro
    Commented Sep 12, 2012 at 13:21
  • $\begingroup$ but spectral radius is meaningful only for bounded linear operators,isn't it? $\endgroup$
    – Tomas
    Commented Sep 12, 2012 at 13:28
  • $\begingroup$ 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. $\endgroup$
    – Giuseppe Negro
    Commented Sep 12, 2012 at 13:39
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    $\begingroup$ @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. $\endgroup$
    – t.b.
    Commented Sep 12, 2012 at 13:45
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    $\begingroup$ What if you take the direct sum of an operator with empty spectrum and the zero operator? Would not the spectrum be $\{0\}$? $\endgroup$
    – user31373
    Commented Sep 12, 2012 at 18:14

2 Answers 2

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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 $\{0\}$.

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Weidmann gives a proposition to check for boundedness by investigating the numerical range (see proposition 2.51 in german version of Weidmann's 'Lineare Operatoren in Hilberträumen'). This makes sense since the numerical range is the natural object when studying boundedness rather than the spectrum...

It states that for either (a) arbitrary operators on complex Hilbert spaces or (b) not necessarily densely defined symmetric operator operators on arbitrary Hilbert spaces we have:

-->The operator is bounded iff its numerical range is bounded!

For not necessarily bounded normal densely defined operators we have:
(Relaxing to closed ones fails in general as seen in numerical range vs spectrum.)

-->The spectrum is contained in the closure of the numerical range!

So the scheme is as follows: $$A\text{ bounded}\iff\mathcal{W}(A)\text{ bounded}$$ $$N^*N=NN^*:\quad\sigma(N)\subseteq\overline{\mathcal{W}(N)}$$

But apart from that there's (unfortunately) nothing much to say I guess.

By the way if you're more concerned about little subtlties like denseness etc. Weidmann is a good book for. He tries to keep these aspects always (annoyingly or luckily) in discussion.

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