= Spectral permanence for Hermitian Banach algebras
{title2=$\sigma_B(b)=\sigma_A(b)$}
Let $B$ be a <closed> <star-subalgebra> of a unital <Hermitian Banach algebra> $A$, containing the same identity. For Hermitian $h\in B$, the complement of $\sigma_A(h)\subseteq\mathbb R$ is <connected>. The resolvent belongs to $B$ at large modulus by the <Neumann series>; its membership in $B$ is both open and <closed> in that complement, so $\sigma_B(h)=\sigma_A(h)$. If $b$ is invertible in $A$, both $bb^*$ and $b^*b$ are invertible there and <Hermitian algebra elements>, hence invertible in $B$. Their inverses supply respectively a right and a left inverse for $b$ in $B$, proving invertibility there. Applying this to $b-\lambda1$ proves the spectrum equality.
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