Proposition 30.4.3.label Let $\phi \in A(D)^{*}$ be a multiplicative linear functional, then there exists $z_{0} \in \ol D$ such that for every $f \in A(D)$, $\phi(f) = f(z_{0})$.
Proof. By Proposition 29.7.2, $\norm{\phi}_{A(D)^*}= 1$. Let $p$ be the identity polynomial, and $z_{0} = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(x_{0})$. By density of polynomials in $A(D)$, $\phi(f) = f(z_{0})$ for all $f \in A(D)$.$\square$
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