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A character of an algebra is a nonzero multiplicative complex-linear functional $\phi :A\to \mathbb{C}$, and the character space of an algebra is the set ${\Phi }_A$ of all such characters. Since $A$ is unital, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. The spectral radius estimate therefore yieldsThus 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 $\Vert 1-m\Vert <1$, 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 $x-\lambda 1$ is not invertible, equivalently when it lies in some maximal ideal. The preceding result turns that ideal into $\ker \phi$, giving $\phi (x)=\lambda$. The reverse implication follows from the first paragraph, so

The Gelfand topology is the weak-star topology on ${\Phi }_A\subseteq A^{\ast }$. The Gelfand transform isIts values are continuous by the definition of the topology, and multiplicativity and linearity of characters show that it is a unital algebra homomorphism. Finallyso it is continuous.

For a unital Banach algebra, $\phi (a)$ belongs to ${\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. Thereforeso every character of an algebra is continuous and has norm one. The nonunital case follows by extending the character to the unitization of an algebra.

The Gelfand topology on ${\Phi }_A$ is the weak-star topology inherited from $A^{\ast }$: a net ${\phi }_i$ converges to $\phi$ exactly when ${\phi }_i(a)\to \phi (a)$ for every $a\in A$. If $A$ is unital, ${\Phi }_A$ lies in the weak-star compact dual unit ball by the Banach-Alaoglu theorem. The equationsdefine a weak-star closed subset, so ${\Phi }_A$ is compact.

For a compact Hausdorff space $K$, every character of the Banach algebra $C(K)$ is an evaluation characterat a unique $x\in K$. The map $x\mapsto {\delta }_x$ is a homeomorphism from $K$ onto ${\Phi }_{C(K)}$ with its Gelfand topology.

Let $B$ be the completion under the assumed algebra norm. Restriction sends every $\psi \in {\Phi }_B$ to a character of the dense subalgebra $C(\mathbb{R})$, hence by part i to ${\delta }_x$ for some $x$. Continuity of the restriction says exactly that $x\in K$. Conversely, every ${\delta }_x$ continuous for the algebra norm extends uniquely through the completion, and continuity of multiplication makes the extension a character of $B$. Restriction and extension therefore give a bijection

For $b\in B$, choose $h_n\in C(\mathbb{R})$ converging to $b$ in the algebra norm. Characters on the unital Banach algebra $B$ are uniformly norm bounded, so $h_n{|}_K$ converges uniformly to the function $x\mapsto {\tilde{\delta }}_x(b)$. This function is continuous. Hence $x\mapsto {\tilde{\delta }}_x$ is continuous into the Gelfand topology. Its inverse is again $\psi \mapsto \psi (t)$, so the bijection is a homeomorphism. In particular, $K$ is compact.

Suppose an algebra norm made $C(\mathbb{R})$ a Banach algebra. Its unital character space of an algebra would be compact in the Gelfand topology. By part i it consists of the evaluations ${\delta }_x$. The mapis continuous from the usual topology because every $h\in C(\mathbb{R})$ is continuous, and its inverse is the continuous map $\phi \mapsto \phi (t)$ defined by the coordinate function $t$. Thus the character space is homeomorphic to the noncompact space $\mathbb{R}$, a contradiction. No complete algebra norm exists.