# Spectrum invariance under the passage to a sub Banach Algebra

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.

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 $$0\in\sigma_{\mathcal A}(A)$$.
• Can you please explain why are you changing your interval from $[0,2\pi]$ to $[0,1]$ and how $B$ is orthogonal to $A$ in $L^2$ as for $\text{A}=e^{it}\in A$, $B\text{A}\ne0$ Commented Mar 26, 2019 at 12:03
• Because typo. And "orthogonal to $A$ in $L^2$" does not mean $BA=0$. Commented Mar 26, 2019 at 12:24