let $X$ be a Banach space and suppose $T$ is an operator on $X$. If $T$ is bounded, then clearly $\sigma(T)$ is also bounded in the sense that $r(T) := \sup_{\lambda\in \sigma(T)}|\lambda| < \infty.$ Does this follow the other way, viz. if $r(T) < \infty$ does it follow that $T$ is bounded?

In the light of the counter-examples Tsemo and TrialAndError and Ben Wallis' comment, let us add the additional requirement that $r(T) > 0$ so operators with empty spectrum are not considered. I appreciate these answers and the people who wrote them, however we might agree that trivial cases are not generally the most interesting. The fault is mine for the imprecision.

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    $\begingroup$ Trivially not, since some unbounded operators have empty spectrum. Tsemo gave an example below of an unbounded operator on a Hilbert space with zero spectrum. I wonder if it is always possible to produce an example of a zero-spectrum unbounded operator on an arbitrary infinite-dimensional Banach space $X$. Here is an idea: Let $X=W\oplus\mathbb{K}$ for some 1-codimensional subspace $W$ of $X$. Does there always exist a unbounded operator $T_1:W\to W$ with empty spectrum? Now let $T=T_1\oplus 0$. Is it true that $\sigma(T)=\{0\}$? $\endgroup$ – Ben W Nov 2 '16 at 17:23

Let $X=C[0,1]$ and let $T : \mathcal{D}(T)\subset X \rightarrow X$ be defined on $$ \mathcal{D}(T) = \{ f\in C^1[0,1] : f(0) = 0 \}. $$ This is an unbounded operator because $T x^{n} = nx^{n-1}$. But it has bounded spectrum because its spectrum is empty. To see this, solve the resolvent equation $$ (T-\lambda I)f = g \\ f'-\lambda f = g, \;\;\; f(0)=0, $$ which is a simple ODE that can be solved with an integrating factor $e^{-\lambda x}$: $$ (e^{-\lambda x}f)'=e^{-\lambda x}g,\;\; f(0)=0, \\ f(x) = e^{\lambda x}\int_{0}^{x}e^{-\lambda t}g(t)dt. $$ You can check that the resolvent $R(\lambda)g=\int_{0}^{x}e^{\lambda(x-t)}g(t)dt$ is bounded on $X=C[0,1]$ for all $\lambda$.

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