I was looking at this superb question posted on MSE Spectrum can be an arbitrary subset.. And this question raised the following more elementary question
Find an example of bounded linear operator $T: X\to X$, defined in a normed vector space $X$, such that, $\sigma(T)$ is unbounded ($\sigma(T)$ is the spectrum of $T$).
I have tried to construct an example but I have failed miserably. Moreover, I searched online but I was not able to find anything related to the question.
Does anyone know an example?
Just a comment.
By Banach Completion we can always consider $X$ as a dense subspace of the Banach space $\widetilde X$. Using the following question extending a bounded linear operator there exists a unique bounded linear operator $S:\widetilde X\to \widetilde X$ such that $\|T\|=\|S\|$ and $\left.S\right|_{X}=T$.
It really seems like that $\sigma(T)\subset\sigma(S)$. If $\lambda \in \rho(S)$ (resolvent of $S$), then $(\lambda I-S)^{-1}$ is bounded. Therefore
$$(\lambda I-S)^{-1}(\lambda I-S)=\text{Id}_{\widetilde X}, $$
if we restric the above equation to the subspace $X$ we conclude that
$$(\lambda I-S)^{-1}(\lambda I-T)=\text{Id}_{X}. $$
Since $(\lambda I-S)^{-1}$ is bounded then $(\lambda I-T)^{-1}$ is bounded as well, therefore $\rho(S)\subset \rho(T)$, then $\sigma(T)\subset \sigma(S)$. Since $S$ is a bounded linear operator in a Banach space $\sigma(S)$ is bounded $\Rightarrow$ $\sigma(T)$ is bounded. Is this correct? (Maybe $(\lambda I-S)$ surjective will not imply $(\lambda I-T)$ surjective).
Best Answer
For a very simple example, let $X$ be the polynomial ring $\mathbb{C}[x]$ and let $T$ be multiplication by $x$. Then $\lambda I-T$ is not invertible for any $\lambda\in\mathbb{C}$ (it is never surjective), so the spectrum of $T$ is all of $\mathbb{C}$.
All that remains is to find a norm on $X$ for which $T$ is bounded. This is easy: for instance, you could consider $\mathbb{C}[x]$ as a subspace of $C[0,1]$ (the restrictions of polynomial functions to $[0,1]$), and then clearly $\|T\|\leq 1$.