Exercise 3, § 5

Let $A$ be a commutative Banach algebra, $A'$ the dual Banach space of A. For every $x\in A$, let $\kappa (x)$ be the linear operator in $A'$ transposed from multiplication by $x$ in A. For $x\in A$ and $x'\in A'$, we denote $x ? x'=$ $\kappa (x)(x')\in A'$. A vector subspace of $A'$ is said to be invariant if it is stable under all the $\kappa (x)$.

a) For an element $x'$ of $A'$ to be proportional to a nonzero character of A, it is necessary and sufficient that $\mathbf{C}x'$ be invariant.

b) The mapping $V\mapsto V^{\circ}$ is a bijection of the set of closed ideals of A onto the set of weakly closed invariant vector subspaces of $A'$.

c) If W is an invariant vector subspace of $A'$, we denote by $\sigma (W)$ the set of characters belonging to W. If $W_1$ and $W_2$ are weakly closed invariant, and if W is the weakly closed vector subspace generated by $W_1$ and $W_2$, one has $\sigma (W) =\sigma (W_1)\cup \sigma (W_2)$. (Use b).)

d) Let W be a weakly closed invariant vector subspace of $A'$. Then $\sigma (W)$ is the set of the $\chi \in \widehat{A}$ such that the condition $x*x'= 0$ for every $x'\in W$ implies $\chi (x) = 0$.

e) If A admits a unit element, every weakly closed invariant vector subspace of $A'$ contains a nonzero character.

f) In the remainder of this exercise, we suppose that A is a regular algebra without radical and unital. If W is a weakly closed invariant vector subspace of $A'$ and U a neighbourhood of $\sigma (W)$ in $\mathsf{X}(A)$, then W is contained in the weakly closed vector subspace of $A'$ generated by U.

g) Let $x'\in A'$ and W the weakly closed invariant vector subspace of $A'$ generated by $x'$. We put $\sigma (x') =\sigma (W)$. If $x\in A$ is such that $\mathscr{G}x= 1$ on a neighbourhood of $\sigma (x')$, then $x ? x'=x'$. (Use f).)

h) If $\sigma (x')$ is the union of two disjoint closed sets $\sigma_1$ and $\sigma_2$, there exist $x'_1, x'_2\in A'$ such that $\sigma (x'_1) =\sigma_1,\sigma (x'_2) =\sigma_2$ and $x'=x'_1+x'_2$. (Put $x'_1=u_1?x'$, $x'_2=u_2? x'$, with $\mathscr{G}u_1= 1$ (resp. 0) in the neighbourhood of $\sigma_1$ (resp. $\sigma_2$), and $\mathscr{G}u_2= 1$ (resp. 0) in the neighbourhood of $\sigma_2$ (resp. $\sigma_1$)$.$)