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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

Assume for contradiction that any algebra norm exists. Part iii makes $K$ compact, so choose an interval disjoint from $K$ and a nonzero $a\in C(\mathbb{R})$ supported in that interval. Choose $b\in C(\mathbb{R})$ with $b=1$ on $K$ and $b=0$ on the support of $a$. Then $a,b$ are nonzero and $ab=0$.

In the completion $B$, every character is evaluation at a point of $K$, so ${\sigma }_B(b)=\{1\}$. Thus $q=1-b$ has spectral radius zero. The equality $ab=0$ givesfor every $n$. The spectral radius formula supplies an $n$ with $\Vert q^n\Vert <1$, and submultiplicativity then givesan impossibility because $a\ne 0$. Hence $C(\mathbb{R})$ admits no algebra norm at all.

In a complex unital Banach algebra, the spectral radius formula makes the root norms of $(x/t{)}^n$ tend to $r(x)/t$. If $t>r(x)$, those norms are eventually bounded above by a geometric sequence and the powers vanish. Conversely, vanishing powers eventually have norm less than one; the infimum version of the radius formula then forces $r(x)/t<1$. Consequently $r(x)=\inf \{t>0:(x/t{)}^n\to 0\}$, including the case $r(x)=0$.

For a complex unital Banach algebra and a circle enclosing its element's Banach algebra spectrum, the displayed identity holds for $n\ge 0$. On a sufficiently large circle, expand the Neumann series and integrate it term by term. At any resolvent point the local series $R({\lambda }_0+h)={\sum }_{j\ge 0}(-h{)}^{j}R({\lambda }_0{)}^{j+1}$ proves norm analyticity. Apply continuous complex linear functionals and the scalar Cauchy theorem to deform the large circle to any radius greater than the maximum spectral modulus; the Hahn-Banach theorem separates points and restores the vector integral identity. Its norm bound $\Vert x^n\Vert \le R^{n+1}{\max }_{|\lambda |=R}\Vert R(\lambda )\Vert$ gives the reverse inequality in the spectral radius formula.

For a finite commuting family in a unital Banach algebra, divide each $x_j$ by $r(x_j)+\epsilon /2$. The spectral radius formula makes each normalized element's powers bounded. Commutativity lets every product be reordered into one power of each generator, so the entire generated semigroup is bounded. Renorming a bounded multiplicative semigroup gives the displayed bounds simultaneously. Applying this to two commuting elements proves submultiplicativity and subadditivity of their spectral radius.

For a Banach algebra-valued holomorphic function, set $p_n(w)=\Vert f(w{)}^{2^n}{\Vert }^{1/2^n}$. These continuous functions decrease to the spectral radius by the spectral radius formula. If $M={\sup }_{\partial K}r(f(w))$, the continuous functions $\max \{p_n,M\}$ on $\partial K$ decrease to the continuous constant $M$. The Dini theorem gives uniform convergence. Apply the norm maximum principle for a Banach-space-valued holomorphic function to each power, and then let $n\to \infty$. This proof does not assume continuity of spectral radius.

For a Normal element of a C-star algebra, the C-star identity gives $\Vert a^2\Vert =\Vert a{\Vert }^2$, and iteration gives $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula then gives $r(a)=\Vert a\Vert$. This proves the isometry of the Gelfand transform for a commutative C-star algebra directly from its defining norm identity.

An algebra norm is submultiplicative when it satisfies the displayed product inequality. Induction gives $\Vert a_1\cdots a_n\Vert \le {\prod }_j\Vert a_j\Vert$, and in particular $\Vert a^n\Vert \le \Vert a{\Vert }^n$. The operator norm has this property because $\Vert STv\Vert \le \Vert S\Vert \Vert T\Vert \Vert v\Vert$. The supremum norm for bounded functions with pointwise multiplication is another example. This estimate supports the Neumann series and the spectral radius formula.