Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 7 / 5 / Solution

An algebra involution is a map $x\mapsto x^{\ast }$ satisfyingThus it is a conjugate-linear map, reverses multiplication, and has square equal to the identity map. These are algebraic axioms; norm continuity, when included in a convention for a Banach star algebra, is an additional requirement and is not used here. A Hermitian element of a star algebra satisfies $h^{\ast }=h$.

Conjugate linearity gives $0^{\ast }=0$. The map $\ast$ is surjective, so every $u$ is $v^{\ast }$ for some $v$. The identities $(v1{)}^{\ast }=1^{\ast }{v}^{\ast }=v^{\ast }$ and $(1v{)}^{\ast }=v^{\ast }{1}^{\ast }=v^{\ast }$ show that $1^{\ast }$ is a two-sided identity; uniqueness of the identity gives $1^{\ast }=1$. Therefore both zero and the identity are Hermitian algebra elements.

For the final unheaded request, first prove equality of the two spectra for a Hermitian element of a star algebra $h\in B$. Write $D=\mathbb{C}\setminus {\sigma }_A(h)$. The spectrum is a compact subset of the real axis by the assumption that $A$ is a Hermitian Banach algebra, so $D$ is connected: points above and below the real axis can be joined by paths passing beyond the endpoints of an interval containing the spectrum; points of the real axis outside the spectrum can first move a short distance vertically. DefineFor large $|\lambda |$, the Neumann series ${\lambda }^{-1}{\sum }_{n\ge 0}(h/\lambda {)}^n$ converges in the closed unital star-subalgebra $B$, so $E$ is nonempty. It is relatively open in $D$ by the same series applied to a perturbation of an already invertible element of $B$. It is relatively closed in $D$, since the resolvent of an element is continuous in $A$ and $B$ is closed. Connectedness gives $E=D$. Conversely an inverse in $B$ is an inverse in $A$, soThis proves the needed spectrum in a closed unital subalgebra equality for Hermitian algebra elements without presupposing that their spectra in $B$ are real.

Now suppose $b\in B$ is invertible in $A$. Both $b{b}^{\ast }$ and $b^{\ast }b$ are Hermitian algebra elements, since reversing products and applying $\ast$ twice fixes them. They are invertible in $A$, by part (ii), so the preceding equality of spectra of Hermitian algebra elements makes them invertible in $B$. Part (ii), applied within $B$, then makes $b$ invertible in $B$. Applying this to $b-\lambda 1$ for every scalar $\lambda$, and using the automatic converse for an inverse already in $B$, proves spectral permanence for Hermitian Banach algebras:Only closedness, a common identity, closure under the algebra involution, and real ambient spectra of Hermitian algebra elements were used; the C-star identity was not assumed.