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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)

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    $\begingroup$ 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) $\endgroup$ – Yemon Choi Dec 16 '19 at 2:49
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The answer is no, in general. Here is a counterexample:

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 $\lambda \in D$.

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  • $\begingroup$ Thank you very much for your very interesting answer. $\endgroup$ – Ali Taghavi Dec 15 '19 at 20:45
  • $\begingroup$ @YemonChoi: I'm not sure I follow: the right shift has empty point spectrum, so I indeed mean the left shift (as specified in the post). $\endgroup$ – Jochen Glueck Dec 16 '19 at 7:28
  • $\begingroup$ @JochenGlueck Sorry, I was reading the question while ill+tired and made a stupid misreading. Of course what you have written is fine $\endgroup$ – Yemon Choi Dec 16 '19 at 13:41

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