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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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    I guess that a bounded spectrum always implies a bounded operator. Try looking for the keyword 'spectral radius'.2012-09-12
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    but spectral radius is meaningful only for bounded linear operators,isn't it?2012-09-12
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    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.2012-09-12
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    @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](http://en.wikipedia.org/wiki/Volterra_operator).2012-09-12
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    @t.b. : Of course you are right! That non-trivial operator has spectral radius zero. ShanLinHuang: I'm sorry, I was completely wrong.2012-09-12
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    What about the operator $Te_n = ne_1$ (where $e_n$ is chosen to have $\lvert\lvert e_n\rvert\rvert=1$ for all $n$)? This should be unbounded, has non-empty spectrum (because $e_1$ is eigenvector to eigenvalue $1$), but I don't think the spectrum should contain any other value (but I might be wrong in this respect).2012-09-12
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    @celtschk: the spectrum consists of those $\lambda$ such that $T-\lambda I$ has a **bounded** inverse. Since $(T-\lambda I)e_n=ne_1-\lambda e_n$, we would get $(T-\lambda I)^{-1}e_n=\lambda^{-1}(n(1-\lambda)e_1-e_n)$ and it wouldn't be bounded for any nonzero $\lambda\ne1$. That is, every nonzero scalar is in the spectrum.2012-09-12
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    What if you take the direct sum of an operator with empty spectrum and the zero operator? Would not the spectrum be $\{0\}$?2012-09-12
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    If we consider self-adjiont operator $T$ in a Hilbert space,then if it's spectrum is a bounded set of $\mathbb{R}$,then $T$ is automatically a bounded operator.2012-09-13

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