 Solution

A character of an algebra is a nonzero complex-linear multiplicative map $\phi :A\to \mathbb{C}$. The character space of an algebra ${\Phi }_A$ carries the Gelfand topology, namely pointwise convergence on $A$, equivalently the subspace weak-star topology in $A^{\ast }$.

For a unital complex Banach algebra, $\phi (1)=1$: multiplicativity and nonzeroness force this. If $a-\phi (a)1$ were invertible, applying $\phi$ to its product with its inverse would give $0=1$. Thus $\phi (a)\in {\sigma }_A(a)$. The Neumann series shows ${\sigma }_A(a)\subset \{|z|\le \Vert a\Vert \}$, so $|\phi (a)|\le \Vert a\Vert$ and the character of an algebra is automatically bounded. The same conclusion for a nonunital algebra follows by extending $\phi$ to its unitization of an algebra via $\tilde{\phi }(a,t)=\phi (a)+t$ and using the standard norm $\Vert (a,t)\Vert =\Vert a\Vert +|t|$.

In the algebra $R(K)$ let $u(z)=z$ and fix $\phi \in {\Phi }_{R(K)}$. Write $\lambda =\phi (u)$. If $\lambda \notin K$, the rational function $(u-\lambda {)}^{-1}$ belongs to $R(K)$, which is impossible because a character of an algebra cannot vanish on an invertible element. Thus $\lambda \in K$. For a rational function $r=p/q$ without poles in $K$, choose a reduced denominator having no zeros in $K$. Multiplicativity gives $\phi (r)=p(\lambda )/q(\lambda )=r(\lambda )$. Density of these rational functions and boundedness of $\phi$ then yield $\phi (h)=h(\lambda )$ for every $h\in R(K)$.

Conversely, evaluation at each $\lambda \in K$ is a character of an algebra. The map $\lambda \mapsto {\delta }_{\lambda }$ is continuous in the Gelfand topology because every $h\in R(K)$ is continuous; its inverse is $\phi \mapsto \phi (u)$, also continuous. Hence the character space of R(K) isThe identification includes the topology, not merely a bijection of sets.