Characters of continuous functions are evaluations

For $f\in C(K)$, continuity of $\mathrm{\mid }f\mathrm{\mid }$ and compactness give a finite maximum by [L5], so the supremum norm is well-defined. $C(K)$ is a nonzero commutative unital complex Banach algebra: pointwise operations give an associative commutative bilinear product and the constant function $\mathbf{1}$ is a unit with $\mathrm{\parallel }\mathbf{1}\mathrm{\parallel }=1$; the supremum norm is submultiplicative and satisfies the triangle inequality; and $C(K)$ is complete, because a Cauchy sequence $(f_n)$ in the supremum norm is uniformly Cauchy, so its pointwise limit $f$ exists by [L2] and is continuous by [L1], and $\mathrm{\parallel }{f}_n-f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$ by the definition of uniform Cauchyness. Nonzero: since $K\mathrm{\ne }\mathrm{\varnothing }$, the constant function $\mathbf{1}$ is not the zero function.

For every $x\in K$ the evaluation ${\mathrm{e}\mathrm{v}}_x(f):=f(x)$ is a character of $C(K)$: it is complex-linear, multiplicative, and nonzero since ${\mathrm{e}\mathrm{v}}_x(\mathbf{1})=1\mathrm{\ne }0$.

For any characters $\chi \mathrm{\ne }\psi$ of $C(K)$ there is $f$ with $\chi (f)\mathrm{\ne }\psi (f)$, so the evaluation-open sets $\{\phi :\mathrm{\mid }\phi (f)-\chi (f)\mathrm{\mid }<\epsilon \}$ and $\{\phi :\mathrm{\mid }\phi (f)-\psi (f)\mathrm{\mid }<\epsilon \}$ with $\epsilon =\mathrm{\mid }\chi (f)-\psi (f)\mathrm{\mid }\mathrm{/}2$ are disjoint; hence $\mathrm{\Delta }(C(K))$ is Hausdorff in the pointwise-evaluation topology.

The map $e:K\to \mathrm{\Delta }(C(K))$, $e(x)={\mathrm{e}\mathrm{v}}_x$, is continuous, because for each $f\in C(K)$ the composition $x\mapsto {\mathrm{e}\mathrm{v}}_x(f)=f(x)$ is continuous by the continuity of $f$.

The evaluations are pairwise distinct: if $x\mathrm{\ne }y$ in $K$, then $\{x\}$ and $\{y\}$ are disjoint closed subsets of the normal space $K$ by [L6], so by [L4] there is a continuous $f:K\to [0,1]$ with $f(x)=0$ and $f(y)=1$; then ${\mathrm{e}\mathrm{v}}_x(f)=0\mathrm{\ne }1={\mathrm{e}\mathrm{v}}_y(f)$.

Let $\chi$ be a character of $C(K)$ and suppose the common zero set of $\ker \chi$ is empty. The family $\{U_f:f\in \ker \chi \}$, where $U_f=\{x:f(x)\mathrm{\ne }0\}$, is an open cover of $K$ formed without choosing a function for each point. Compactness gives a finite subcover; choosing a witnessing function for each of its finitely many members gives $f_1,\dots ,f_n\in \ker \chi$ with no common zero. Here $n\ge 1$ since $K$ is nonempty. Then $g={\sum }_{i=1}^n\overline{f_i}{f}_i$ is in $\ker \chi$: each $\overline{f_i}$ is continuous and the kernel is an ideal, without any assumption that $\chi$ preserves conjugation. Also $g>0$ everywhere, so $1\mathrm{/}g$ is continuous and $1=g(1\mathrm{/}g)\in \ker \chi$. This contradicts $\chi (1)=1$ from [L3] applied using [step 1.1]. Thus there is $x\in K$ at which every member of $\ker \chi$ vanishes.

With $x$ as in [step 2.1], $\ker \chi \subseteq \ker {\mathrm{e}\mathrm{v}}_x$. Both are kernels of nonzero multiplicative linear functionals, hence both are maximal ideals: if $f\mathrm{\notin }\ker \chi$ then every $h\in C(K)$ has $h-(\chi (h)\mathrm{/}\chi (f))f\in \ker \chi$, so any ideal strictly containing $\ker \chi$ contains $f$ and hence equals $C(K)$. Therefore $\ker \chi =\ker {\mathrm{e}\mathrm{v}}_x$.

Hence for every $f\in C(K)$ one has $f-f(x)\mathbf{1}\in \ker {\mathrm{e}\mathrm{v}}_x=\ker \chi$, so $\chi (f)=f(x)\chi (\mathbf{1})=f(x)$, using $\chi (\mathbf{1})=1$ from [L3]; thus $\chi ={\mathrm{e}\mathrm{v}}_x$, and by [step 1.5] the point $x$ is unique.

By [step 1.2], [step 1.5] and [step 4.1] the map $e$ is a bijection from $K$ onto $\mathrm{\Delta }(C(K))$; by [step 1.4] it is continuous, $K$ is compact, and by [step 1.3] $\mathrm{\Delta }(C(K))$ is Hausdorff, so [L5] makes $e$ a homeomorphism.