= Solution
A <character of an algebra> is a nonzero multiplicative complex-linear functional $\varphi:A\to\mathbb C$, and the <character space of an algebra> is the set $\Phi_A$ of all such characters. Since $A$ is unital, $\varphi(1)=1$. Moreover $\varphi(a)\in\sigma_A(a)$: otherwise $a-\varphi(a)1$ would be invertible, while applying $\varphi$ to its inverse identity would give $0=1$. The <spectral radius> estimate therefore yields
$$
|\varphi(a)|\leq r(a)\leq\|a\|.
$$
Thus every character is continuous and has norm one.
Let $M$ be a <maximal ideal>. Its norm closure is again an ideal. It cannot equal $A$, because then some $m\in M$ would satisfy $\|1-m\|<1$, making $m$ invertible by the <Neumann series> and forcing $M=A$. Hence $M$ is closed. The quotient $A/M$ is a complex unital Banach division algebra, so the <Gelfand-Mazur theorem> identifies it with $\mathbb C$. Composing the quotient map with this isomorphism gives a character with kernel $M$. Conversely, a character kernel is maximal because its quotient is $\mathbb C$.
Now $\lambda\in\sigma_A(x)$ exactly when $x-\lambda1$ is not invertible, equivalently when it lies in some maximal ideal. The preceding result turns that ideal into $\ker\varphi$, giving $\varphi(x)=\lambda$. The reverse implication follows from the first paragraph, so
$$
\sigma_A(x)=\{\varphi(x):\varphi\in\Phi_A\}.
$$
The <Gelfand topology> is the <weak-star topology> on $\Phi_A\subseteq A^*$. The <Gelfand transform> is
$$
x\longmapsto\widehat x,
\qquad
\widehat x(\varphi)=\varphi(x).
$$
Its values are continuous by the definition of the topology, and multiplicativity and linearity of characters show that it is a unital algebra homomorphism. Finally
$$
\|\widehat x\|_\infty=\sup_{\varphi\in\Phi_A}|\varphi(x)|\leq\|x\|,
$$
so it is continuous.
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