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The Commutative Gelfand--Naimark theorem says that every commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C(\Delta (A))$, where $\Delta (A)$ is its compact character space and the map is the Gelfand transform

Indeed, maximal ideals give enough characters to identify $\sigma (a)$ with the range of $\hat{a}$. The C-star identity and the spectral radius formula giveso $\Gamma$ is isometric and has closed range. Characters send $a^{\ast }$ to $\overline{\chi (a)}$, so the range is self-conjugate; it contains constants and separates distinct characters. The complex Stone-Weierstrass theorem makes the range dense in $C(\Delta (A))$, and closedness makes it all of that algebra. This proves the theorem.

An element $x$ of a C-star algebra is positive when it is self-adjoint and $\sigma (x)\subseteq [0,\infty )$. Consider the commutative C-star subalgebra $C^{\ast }(1,x)$. Under its Gelfand–Naimark isomorphism, $x$ becomes a nonnegative continuous function $\hat{x}$. The function $\sqrt{\hat{x}}$ is continuous and nonnegative, so its inverse image $y$ is positive and satisfies

Mazur theorem says that the weak closure and norm closure of a convex subset of a real or complex normed space coincide. The norm closure is contained in the weak closure because every norm-continuous linear functional is norm-continuous. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and a real number $\alpha$ such thatThis weakly open separation shows that $x$ is outside the weak closure.

By definition,For each $n$, regard $x_n$ as the bounded functional $J{x}_n:X^{\ast }\to \mathbb{F}$ given by $J{x}_n(f)=f(x_n)$. Pointwise convergence makes the family $(J{x}_n)$ pointwise bounded on the Banach space $X^{\ast }$. The Uniform boundedness principle gives

Suppose first that $x_n\rightharpoonup 0$. Every subsequence indexed by an infinite set $M$ also converges weakly to zero. Thus zero belongs to the weak closure of the convex hull of $\{x_n:n\in M\}$, and Mazur's theorem puts it in its norm closure. This gives the required finite convex combination of norm below any prescribed $\epsilon$.

Conversely, if weak convergence fails, some $f\in X^{\ast }$ and an infinite subsequence satisfy either $\mathrm{Re}f(x_n)\ge \epsilon$ throughout or $\mathrm{Re}f(x_n)\le -\epsilon$ throughout. Every convex combination from that subsequence then has norm at least $\epsilon /\Vert f\Vert$, contradicting the stated property.

Now let the $x_n\in {\ell }^p$, $1<p<\infty$, have pairwise disjoint supports and satisfy $\Vert x_n{\Vert }_p\le C$. For $N$ distinct terms,The convex-combination criterion therefore proves $x_n\rightharpoonup 0$. This is weak convergence of bounded disjointly supported sequences in lp.

The final implication for a commutative unital C*-algebra is true. By the Commutative Gelfand--Naimark theorem, $A\cong C(K)$ and its characters are the point evaluations. The hypotheses say that the uniformly bounded functions $a_n$ converge pointwise to zero. Every functional on $C(K)$ is integration against a finite regular measure by the Riesz-Markov-Kakutani representation theorem; the dominated convergence theorem givesHence $a_n\rightharpoonup 0$.

A character of an algebra is a nonzero algebra homomorphism $\phi :A\to \mathbb{C}$. The character space isFor a unital algebra, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$, since applying $\phi$ shows that $a-\phi (a)1$ cannot be invertible. In a Banach algebra,Thus $\Vert \phi \Vert \le 1$, while $\phi (1)=1$ gives equality.

If $A$ is commutative and $\lambda \in {\sigma }_A(a)$, the proper ideal generated by $a-\lambda 1$ lies in a maximal ideal. The quotient by that maximal ideal is $\mathbb{C}$, and its quotient map is a character taking $a$ to $\lambda$. ThereforeSince every Banach-algebra element has nonempty spectrum, ${\Phi }_A$ is nonempty.

For $X=Y\oplus Z$, let $P,Q$ be the complementary coordinate projections. A character must send each idempotent to zero or one, and $P+Q=I$ forces their values to be different. Choose mutually inverse isomorphisms between $Y$ and $Z$ and regard them as off-diagonal operators $S,T$ on $X$. ThenMultiplicativity would give $\phi (P)=\phi (S)\phi (T)=\phi (T)\phi (S)=\phi (Q)$, a contradiction. Hence ${\Phi }_{\mathcal{B}(X)}$ is empty. This is the absence of characters on an operator algebra with isomorphic complementary summands.

If $a=a^{\ast }$ in a unital C*-algebra, then $e^{ita}$ is unitary for real $t$. If $\lambda \in \sigma (a)$, the spectral mapping theorem gives $e^{it\lambda }\in \sigma (e^{ita})$, whose modulus is one. Varying positive and negative $t$ forces $\mathrm{Im}\lambda =0$, so $\sigma (a)\subset \mathbb{R}$.

If $B\subseteq A$ is a unital C*-subalgebra and $x\in B$ is normal, spectral permanence holds. Indeed, when $x-\lambda$ is invertible in $A$, the positive normal elementhas spectrum bounded away from zero. Continuous functional calculus uniformly approximates its reciprocal by polynomials, placing the reciprocal in $B$. It follows that $(x-\lambda {)}^{-1}\in B$. Thus ${\sigma }_B(x)={\sigma }_A(x)$.

The Commutative Gelfand--Naimark theorem says that the Gelfand transform is an isometric unital star-isomorphismfor every commutative unital C*-algebra. If $a$ is positive, continuous functional calculus for the function $t\mapsto \sqrt{t}$ on $\sigma (a)\subset [0,\infty )$ gives a positive $b$ with $b^2=a$. Pointwise uniqueness in $C^{\ast }(1,a)\cong C(\sigma (a))$ gives the unique positive square root. This is the positive square root in a C*-algebra.