= Solution
Work over $\mathbb C$ and take a nonzero commutative unital <Banach algebra> $A$, with $\|1\|=1$. A <character of an algebra> is a nonzero multiplicative complex <linear functional> $\chi:A\to\mathbb C$. It satisfies $\chi(1)=1$. Moreover $\chi(a)\in\sigma_A(a)$: otherwise $a-\chi(a)1$ would be invertible, although its image under $\chi$ is zero. The bound on the <spectrum of an element> therefore gives $|\chi(a)|\leq\|a\|$, proving <automatic continuity of characters> and $\|\chi\|=1$.
Every proper <maximal ideal> $M$ of $A$ is closed. Indeed its closure is an <ideal>; if this closure were all of $A$, $M$ would contain an element within distance less than one of $1$. Such an element is invertible by the Neumann <series>, forcing $1\in M$. Thus the closure is proper and maximality makes it equal to $M$. The <quotient Banach space> $A/M$, with its quotient <Banach algebra> structure, is a complex <normed division algebra>. By the <Gelfand-Mazur theorem>, it is $\mathbb C$, so the quotient map gives a <character of an algebra> with kernel $M$. Conversely, the kernel of every <character of an algebra> is a <maximal ideal>, since the character is onto $\mathbb C$. The <Zorn lemma> supplies a <maximal ideal> containing every proper <ideal>, so the <character space of an algebra> $\Delta(A)$ is nonempty.
These facts give the exact relation between the <character space> and the <spectrum of an element>:
$$
\boxed{\sigma_A(a)=\{\chi(a):\chi\in\Delta(A)\}.}
$$
One inclusion was proved above. For the other, if $a-\lambda1$ is noninvertible, the principal <ideal> it generates is proper because $A$ is commutative. Contain it in a <maximal ideal> and use its corresponding <character of an algebra> to obtain $\chi(a)=\lambda$.
Give $\Delta(A)$ the <Gelfand topology>, namely its <subspace topology> from the <weak-star topology> on $A^*$. In the <closed unit ball> of $A^*$, it is the intersection of the closed conditions
$$
\chi(1)=1,\qquad \chi(ab)=\chi(a)\chi(b)\quad(a,b\in A).
$$
Consequently <Banach-Alaoglu theorem> makes $\Delta(A)$ a compact <Hausdorff space>. For every $a\in A$, define the <Gelfand transform> $\widehat a(\chi)=\chi(a)$. This is a <continuous function> on $\Delta(A)$ by definition of the <Gelfand topology>. The <Gelfand representation theorem> gives a contractive unital <algebra homomorphism over a field>
$$
\Gamma:A\longrightarrow C(\Delta(A)),\qquad a\longmapsto\widehat a,
\qquad
\boxed{\|\widehat a\|_\infty=r(a)\leq\|a\|.}
$$
Multiplicativity and linearity follow by evaluating at each <character of an algebra>; the <supremum norm> equality follows from the preceding <spectrum of an element> identity. Its kernel is
$$
\ker\Gamma=\bigcap_{\chi\in\Delta(A)}\ker\chi
=\bigcap_{M\text{ maximal}}M=\operatorname{rad}A,
$$
the <Jacobson radical>. Equivalently, its elements have <spectrum of an element> $\{0\}$. Thus $\Gamma$ is injective precisely when $A$ is a <semisimple commutative Banach algebra>, and it gives a faithful continuous representation of $A/\operatorname{rad}A$ as a function algebra. Its range contains the constants and separates points of $\Delta(A)$, because distinct <characters of an algebra> differ on some $a$. An arbitrary <Banach algebra> need not have an isometric or surjective <Gelfand transform>, nor a uniformly dense range: those conclusions require further hypotheses.
For the <Banach algebra> $C(K)$ on a nonempty compact <Hausdorff space> $K$, all <characters of an algebra> are <evaluation characters>. To see this, let $M$ be a <maximal ideal>. If its elements had no common zero, compactness would supply $f_1,\ldots,f_n\in M$ with no common zero. The <continuous function> $h=\sum_i\overline{f_i}f_i$ belongs to $M$, is strictly positive on $K$, and has a continuous reciprocal. It is therefore invertible, a contradiction. Hence all elements of $M$ vanish at some $x\in K$, so $M\subseteq\ker\delta_x$ and maximality gives equality. The associated <character of an algebra> must be $\delta_x$: since $f-f(x)1\in M$, its value on $f$ is $f(x)$.
The map $x\mapsto\delta_x$ is a continuous bijection $K\to\Delta(C(K))$, using separation of points by <continuous functions>. Compactness and the Hausdorff property make it a <homeomorphism>. Under this identification the <Gelfand transform> is \b[$\boxed{\widehat f(\delta_x)=f(x)}$], so it is the identity representation of $C(K)$, in particular an isometric onto map. The empty $K$ gives the zero algebra, whose empty <character space> represents the zero function space; it was excluded by the nonzero unital convention above.
Now let $A$ be a commutative unital <C-star algebra>. The stronger conclusion is the <Commutative Gelfand--Naimark theorem>: \b[the <Gelfand transform> is an isometric onto <C-star homomorphism> $A\cong C(\Delta(A))$]. We prove the additional assertions without assuming this conclusion.
First every <character of an algebra> respects the <C-star algebra> involution. If $h=h^*$, the elements $e^{ith}$, $t\in\mathbb R$, are <unitary elements of a C-star algebra>, and their <norm> is one by the <C-star identity>. Continuity and multiplicativity give $\chi(e^{ith})=e^{it\chi(h)}$. Thus $|e^{it\chi(h)}|\leq1$ for every real $t$, forcing $\chi(h)$ to be real. Writing $a=h+ik$ with $h=(a+a^*)/2$ and $k=(a-a^*)/(2i)$ self-adjoint gives $\chi(a^*)=\overline{\chi(a)}$. Therefore the range of $\Gamma$ is closed under <complex conjugation>.
Every element of commutative $A$ is a <Normal element of a C-star algebra>. For a normal $b$, use the <C-star identity>, and then the same identity for the self-adjoint element $b^*b$, to obtain
$$
\|b^2\|^2=\|(b^2)^*b^2\|=\|(b^*b)^2\|
=\|b^*b\|^2=\|b\|^4.
$$
Its powers are also normal, so $\|a^{2^n}\|=\|a\|^{2^n}$. The <spectral radius formula> gives $r(a)=\|a\|$, hence $\|\widehat a\|_\infty=\|a\|$. The <Gelfand transform> is therefore an <isometry>, and its range is complete and closed in the <supremum norm>. It contains constants, separates points and is closed under <complex conjugation>. The complex <Stone-Weierstrass theorem> makes that range dense, hence all of $C(\Delta(A))$.
The approximation step in <Stone-Weierstrass theorem> can also be seen directly here. For a unital conjugation-closed point-separating subalgebra $B\subseteq C(K)$, the real-valued part of its uniform closure is closed under absolute values, by polynomial approximation to $|t|$ on bounded intervals, hence under pointwise maxima and minima. Its real-valued functions separate points. Given real $f\in C(K)$ and $\varepsilon>0$, for each $x,y$ an affine rescaling of a separating function produces $b_{x,y}$ agreeing with $f$ at $x,y$; take a constant when $x=y$. For fixed $x$, finitely many neighbourhoods of $y$ where $b_{x,y}>f-\varepsilon$ cover $K$. Their maximum $b_x$ exceeds $f-\varepsilon$ everywhere and agrees with $f$ at $x$, hence is less than $f+\varepsilon$ near $x$. Finitely many of these latter neighbourhoods cover $K$; the minimum of their $b_x$ lies between $f-\varepsilon$ and $f+\varepsilon$ everywhere. Approximate real and imaginary parts separately. This proves the density used above and completes the <C-star algebra> conclusion.
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