# Removing the interior of spectrums

Let $$A$$ be a Banach algebra. Is there a Banach algebra $$B$$ which contains $$A$$ but the spectrum of each elements of $$B$$ has empty interior(as a subset of $$\mathbb{C}$$)?

The motivation comes from the fact that the spectrum of elements in a smaller algebra possibly loses its interior when we compute its spectrum in a larger algebra.(Rudin, Functional analysis)

• In general, if you look for positive results (about shrinking the spectrum of certain elements by passing to a bigger algebra) then one place to start looking up some of the old results is this paper of C. J. Read: ams.org/journals/tran/1984-286-02/S0002-9947-1984-0760982-0 (Warning: I think that in this paper Charles uses the terminology "residual spectrum" in a way that is non-standard and conflicts with the accepted current terminology) – Yemon Choi Dec 16 '19 at 2:49

Let $$A$$ be the algebra of bounded linear operators on $$\ell^2(\mathbb{N})$$, and let $$a \in A$$ be the left shift on $$\ell^2(\mathbb{N})$$. Then the spectrum of $$a$$ is the closed unit disk $$\overline{D}$$, and the point spectrum of the operator $$a$$ is the open unit disk $$D$$.
Now we note that the notion "point spectrum" - which is defined for operators - can be translated into a notion that makes sense in Banach algebras: For each $$\lambda$$ in the point spectrum $$D$$ of $$a$$ there exists an element $$a_\lambda \in A \setminus \{0\}$$ such that $$(\lambda - a) a_\lambda = 0$$. Indeed, let $$x_\lambda \in \ell^2(\mathbb{N})$$ be an eigenvector of the operator $$a$$ and choose an arbitrary non-zero functional $$\varphi_\lambda$$ on $$\ell^2(\mathbb{N})$$. Then the operator $$a_\lambda := \varphi_\lambda \otimes x_\lambda \in A$$, given by $$a_\lambda x = \langle \varphi_\lambda, x \rangle x_\lambda \qquad \text{for } x \in \ell^2(\mathbb{N}),$$ satisfies $$(\lambda - a)a_\lambda = 0$$.
Hence, if $$B$$ is a unital Banach algebra that contains $$A$$ as a subalgebra, we still have $$(\lambda - a)a_\lambda = 0$$ in $$B$$, so $$\lambda - a$$ cannot be invertible in $$B$$ for any $$\lambda \in D$$.