Spectrum invariance under the passage to a sub Banach Algebra

Question

Let $\mathcal{B}$ be a unital Banach Algebra, fix $A \in \mathcal{B}$. $\sigma_\mathcal{B}(A) = \{ \lambda \vert \lambda I - A \, not \,invertible \, in \, \mathcal{B}\}$ the specturm of $A$ in $\mathcal{B}$. Suppose $\mathcal{A} \subset \mathcal{B}$ is the closure of the algebra generated by $\{I, A \}$ in $\mathcal{B}$, which is a closed sub algebra, is it always true that $\sigma_\mathcal{B}(A) = \sigma_\mathcal{A}(A)$?

the inclusion $\sigma_\mathcal{B}(A) \subset \sigma_\mathcal{A}(A)$? is immediate from the definition, but can we expect equality? For the $C^*$ Algebra context it is true (see Rudin Functional Analysis p.296), but should it hold in general? If not, what are some counter examples?

any help is appreciated.

Answer 1

Let $\mathcal B=C[0,2\pi]$ with the uniform norm, $A$ the function $A(t)=e^{it}$. Note that $I$ is the function $I(t)=1$. Now $$ \mathcal A=\overline{\operatorname{span}}\{e^{ikt}:\ k=0,1,2,\ldots\}. $$ In $\mathcal B$, we have $0\not\in\sigma_{\mathcal B}(A)$ since $B(t)=e^{-it}$ is its inverse. But $B\not\in\mathcal A$; since uniform convergence on $[0,1]$ implies $L^2$ convergence, we would have $B\in\overline{\mathcal A}^{\|\cdot\|_2}$, the closure of $\mathcal A$ in the $2$-norm, inside $L^2[0,2\pi]$. But $B$ is orthogonal to $\mathcal A$ in $L^2$, a contradiction. Thus $A$ is not invertible in $\mathcal A$, and $0\in\sigma_{\mathcal A}(A)$.