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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).

Eric Wofsey
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Matheus Manzatto
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    It seems that Theorem 10.13a of Rudin's functional analysis says "no" when $X$ is a banach space (for those without the book, the theorem says that the spectrum of a member of a banach algebra is compact). –  Apr 13 '20 at 05:27
  • The map defined in the approved answer is not a well defined linear map. It would be nice not close this question lest this page would mislead others. – Kavi Rama Murthy Apr 13 '20 at 06:36
  • I have removed the green check. I am really sorry. My first impression was that the example was correct. – Matheus Manzatto Apr 13 '20 at 06:40
  • For a bounded smooth domain $\Omega$, the Laplace operator on $H^2(\Omega)$ is bounded, but it has unbounded spectrum. – Neal Apr 15 '20 at 16:27
  • @Neal As far as I remember, $H^{2}(\Omega)$ is a Banach space if Laplacian operator is bounded this would imply that the spectrum is compact. Am I missing something? – Matheus Manzatto Apr 15 '20 at 19:03

1 Answers1

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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 $\|T\|\leq 1$.

Eric Wofsey
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  • Wow, this example is everything that I was looking for it is really good. How did you came out with this example? Is it from some book? I am asking because I have been searching online / trying to construct one for a couple days. – Matheus Manzatto Apr 15 '20 at 05:55
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    From a purely algebraic perspective, a vector space with a linear operator is just a $\mathbb{C}[x]$-module, so you're looking for a $\mathbb{C}[x]$-module on which $x-\lambda$ acts non-invertibly for an unbounded set of $\lambda$. The obvious choice is $\mathbb{C}[x]$ itself, and then you just have to find a norm. – Eric Wofsey Apr 15 '20 at 05:56