# Spectrum can be an arbitrary subset.

Given any subset $$E$$ of field $$\mathbb{F}$$ (real or complex), does there exist a normed linear space $$X$$ over $$\mathbb{F}$$ and a bounded linear operator $$A:X\rightarrow X$$ such that spectrum of $$A$$ is precisely the set $$E$$.

NOTE : It is known that this is true for compact sets as we can use their separability to construct such an operator.

• You really don't want any adjectives on the normed linear space? Not Banach? – Operatorerror Oct 18 '19 at 19:10
• No. For Banach spaces, we know that spectrum is a compact set. However, for normed spaces, this need not be true. – Akash Yadav Oct 18 '19 at 19:12
• Indeed. I am just asking for clarification. – Operatorerror Oct 18 '19 at 19:14

Yes. If $$E$$ is empty, let $$X=\{0\}$$. (By the Gelfand-Mazur theorem, this is the only possibility if $$\mathbb{F}=\mathbb{C}$$.)
If $$E$$ is not empty, we may assume without loss of generality that $$0\in E$$. Now take a Banach space $$Y$$ with a bounded operator $$T:Y\to Y$$ that is quasinilpotent (i.e., its spectrum is $$\{0\}$$) but not nilpotent. Let $$B(Y)$$ be the algebra of bounded operators on $$Y$$ and let $$X\subseteq B(Y)$$ be the subalgebra generated by $$T$$ and the elements $$(T-\lambda I)^{-1}$$ for all $$\lambda\in \mathbb{F}\setminus E$$ (here we use the assumption that $$0\in E$$, so that all these inverses must exist). Note that the assumption that $$T$$ is not nilpotent means that that no nontrivial polynomial in $$T$$ is zero, so $$X$$ is isomorphic as an $$\mathbb{F}$$-algebra to the subalgebra of the field of rational functions $$\mathbb{F}(x)$$ generated by $$x$$ and $$(x-\lambda)^{-1}$$ for $$\lambda\in\mathbb{F}\setminus E$$, by mapping $$T$$ to $$x$$. In particular, $$T-\lambda I$$ is not invertible in $$X$$ for any $$\lambda\in E$$.
Now let $$A:X\to X$$ be the operator given by multiplication by $$T$$. Then $$A$$ is bounded since $$X$$ is a normed algebra. Also, $$A-\lambda I$$ is invertible for any $$\lambda\in\mathbb{F}\setminus E$$ (the inverse is just multiplication by $$(T-\lambda I)^{-1}$$) but not for any $$\lambda\in E$$. Thus, the spectrum of $$A$$ is $$E$$.