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Let $F,G$ be disjoint closed subsets of $K$. They are compact. For each $x\in F,y\in G$, the Hausdorff space condition gives disjoint open neighbourhoods $U_{xy},V_{xy}$. For fixed $x$, finitely many $V_{xy}$ cover $G$; intersect the corresponding $U_{xy}$ and take the union of those $V_{xy}$ to obtain disjoint open sets $U_x\ni x$ and $V_x\supset G$. Finitely many $U_x$ cover $F$. Their union and the intersection of the corresponding $V_x$ are disjoint open neighbourhoods of $F,G$. The case of an empty set is immediate. Hence $K$ is a normal topological space.

Urysohn lemma states that in a normal topological space satisfying $T_1$, disjoint closed sets $F,G$ admit a continuous function $h:K\to [0,1]$ with $h=0$ on $F$ and $h=1$ on $G$.

For $x\ne y$, $|({\delta }_x-{\delta }_y)(f)|\le 2\Vert f{\Vert }_{\infty }$, giving an upper bound $2$ for the operator norm. By Urysohn lemma, there is $h:K\to [0,1]$ with $h(x)=1,h(y)=0$. Then $f=2h-1$ has uniform norm $1$ and $({\delta }_x-{\delta }_y)(f)=2$. Thus every distinct pair has operator norm $2$, andThe stated hypotheses also permit a singleton $K$. Then there are no distinct pairs, so the infimum is $+\infty$ under the extended-real convention, or undefined as a real number. This edge case cannot be assigned the value $2$.

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C({\Phi }_A)$ through its Gelfand transform, with ${\Phi }_A$ a compact Hausdorff space. The l-infinity sequence space is such an algebra under coordinatewise multiplication and conjugation, with the supremum norm. Thereforegives the required isometric isomorphism.

Uniqueness as a Banach space representation uses the Banach–Stone theorem, rather than just uniqueness as an algebra representation. To justify the relevant theorem, the Riesz-Markov-Kakutani representation theorem givesthe extreme points of the dual unit ball of C(K). A norm-one measure whose variation measure is not concentrated at one point splits into two normalized restrictions to disjoint sets of positive variation, and so is not extreme. Conversely, equality in the variation bound shows that any decomposition of $\alpha {\delta }_t$ into the average of two dual-unit-ball elements forces both to be $\alpha {\delta }_t$: after removing the phase, the two measures must be positive and their average is concentrated at $t$.

If $U:C(K)\to C(L)$ is a surjective linear isometry, its dual map preserves these extreme points and their scalar orbits. Hence $U^{\ast }{\delta }_s=u(s){\delta }_{h(s)}$ for a bijection $h:L\to K$ and $|u(s)|=1$. Evaluating at $1$ gives the continuous function $u=U1$, whileshows that $h$ is continuous, since continuous functions determine the topology of a compact Hausdorff space. It is consequently a homeomorphism. Thus any other compact space representing ${\ell }^{\infty }$ is homeomorphic to this $K$. The Banach–Stone theorem is also stated in Leonard Tomczak's notes on András Zsák's functional analysis lectures.

Embed $\mathbb{N}$ by the evaluation characters $\iota (n)(a)=a_n$. They are distinct because the coordinate indicator functions $e_n$ distinguish them. Moreover, $e_n^2=e_n$, so every character has $\chi (e_n)\in \{0,1\}$. If $\chi (e_n)=1$, thenand hence $\chi =\iota (n)$. It follows that $\{\iota (n)\}=\{\chi :|\chi (e_n)-1|<1/2\}$ is open in $K$. Thus $\iota$ is a homeomorphism from the discrete natural numbers onto its image.

To prove density, suppose $D=\overline{\iota (\mathbb{N})}\ne K$. The Urysohn lemma provides a nonzero $v\in C(K)$ vanishing on $D$. Write $v=\Gamma (a)$. Then $a_n=v(\iota (n))=0$ for all $n$, so $a=0$ and $v=0$, a contradiction. Every bounded function $b:\mathbb{N}\to \mathbb{C}$ is an element of ${\ell }^{\infty }$, and $\Gamma (b)$ is its continuous extension to $K$. Density makes this extension unique. ThereforeThis identifies $K$ with the Stone-Čech compactification of the natural numbers.

For the first norm-one assertion, use the usual unital Banach algebra convention $\Vert 1\Vert =1$. A character of an algebra is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\phi (1)=1$, since otherwise $\phi (a)=\phi (a1)=0$ for every $a$. Moreover, $a-\phi (a)1$ cannot be invertible: applying $\phi$ to an inverse identity would give $0=1$. ThusThis proves automatic continuity of characters and $\Vert \phi \Vert \le 1$. Evaluation at the normalized identity gives the reverse inequality, soIf a convention permits $\Vert 1\Vert >1$, only $\Vert \phi \Vert \le 1$ follows in general: on $\mathbb{C}$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.

If $a$ is invertible, $\phi (a)\phi (a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a maximal ideal $M$ by Zorn lemma. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series, which is impossible for a proper ideal. Maximality therefore gives $\overline{M}=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. The quotient map yields a character vanishing at $a$. Consequently

Now let $A=C(K)$. Its evaluation characters give the map $x\mapsto {\delta }_x\in {\Phi }_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\dots ,f_m$ in the kernel with no common zero. Then $h={\sum }_j\overline{f_j}{f}_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\phi (f)1$ belongs to the kernel gives $f(x)=\phi (f)$ for every $f$. Continuous functions distinguish points of the compact Hausdorff space $K$, so the map is bijective. It is continuous for the Gelfand topology, and the compact-to-Hausdorff continuous bijection theorem givesFor $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne \varnothing$.

Let $B$ be the completion for the other algebra norm $\Vert \cdot {\Vert }_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi \in {\Phi }_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original supremum norm. Thus $R(\psi )=\psi {|}_A$ maps into ${\Phi }_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi )(f)=\psi (f)$ is weak-star continuous.

The character space ${\Phi }_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^{\ast }}$, and the conditions $\psi (1)=1$, $\psi (bc)=\psi (b)\psi (c)$ define a weak-star closed set. Apply Banach-Alaoglu theorem. Consequently $R$ is a homeomorphism onto a closed subset $L\subseteq K$.

Suppose $x\in U=K\setminus L$. The compact Hausdorff space $K$ is normal, so choose an open $V$ with $x\in V\subseteq \overline{V}\subseteq U$. By the Urysohn lemma, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline{V}$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi (f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.

For each $x\in K$, choose a character ${\psi }_x\in {\Phi }_B$ extending ${\delta }_x$. Its spectral bound givesTaking the supremum over $x$ proves the minimality of the supremum norm on C(K):

The Beurling–Gelfand spectral radius formula states that for every element $a$ of a complex Banach algebra,For a Hermitian element of a C-star algebra $a=a^{\ast }$, the C-star identity gives $\Vert a^2\Vert =\Vert a{\Vert }^2$. Iterating,The subsequence $n=2^m$ in the spectral-radius limit therefore proves $\boxed{r(a)=\Vert a\Vert }$.

Let $\theta :A\to B$ be a unital C-star homomorphism. If $a-\lambda 1$ is invertible in $A$, its inverse is sent to an inverse of $\theta (a)-\lambda 1$ in $B$. Thus ${\sigma }_B(\theta (a))\subseteq {\sigma }_A(a)$ without assuming continuity of $\theta$. For Hermitian $a$, $\theta (a)$ is also Hermitian, soFor arbitrary $x\in A$, apply this inequality to $x^{\ast }x$ and the C-star identity in both algebras:Hence $\boxed{\Vert \theta (x)\Vert \le \Vert x\Vert }$ for every $x$.

Now assume $\theta$ is injective. For a fixed $x$, it is enough to prove norm preservation on $C=C^{\ast }(1,x^{\ast }x)$, a commutative unital C-star algebra: the preceding square-norm identity will then recover the norm of $x$. By the Commutative Gelfand--Naimark theorem, identify $C$ with $C(K)$. Let $D=\overline{\theta (C)}\subseteq B$. It is also a commutative unital C-star algebra, and its character space ${\Phi }_D$ is compact. Each $\chi \in {\Phi }_D$ restricts along $\theta$ to a character of $C(K)$, hence to evaluation at a point $q(\chi )\in K$. The map $q:{\Phi }_D\to K$ is continuous, so $L=q({\Phi }_D)$ is closed.

If $L\ne K$, the Urysohn lemma provides a nonzero $f\in C(K)$ vanishing on $L$. Then $\chi (\theta (f))=f(q(\chi ))=0$ for every $\chi \in {\Phi }_D$. Faithfulness of the Gelfand transform in $D$ gives $\theta (f)=0$, contradicting injectivity. Thus $q$ is onto. The isometric Gelfand transform of $D$ now givesFinally,This proves that an injective C-star homomorphism is isometric without assuming its image is closed before norm preservation has been established.

Suppose $C(K)$ is equicontinuous. At each $x\in K$, choose a neighbourhood $U$ such thatfor every $y\in U$ and every $f\in C(K)$. If some $y\in U$ differed from $x$, the Urysohn lemma for the compact Hausdorff space $K$ would give a continuous function with $f(x)=0$ and $f(y)=2$, a contradiction. Thus every singleton is open, so $K$ is discrete. A compact discrete space is finite, proving

Compactness is essential. Any infinite set with the discrete topology has equicontinuous $C(K)$, because the singleton neighbourhood $\{x\}$ works simultaneously for every function, but the space is infinite.

The real Stone-Weierstrass theorem states that if $X$ is a compact Hausdorff space and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points of $X$, then $A$ is dense in $C(X,\mathbb{R})$ in the uniform norm.

The Tietze extension theorem states that if $X$ is a normal topological space, $F\subseteq X$ is closed, and $f:F\to \mathbb{R}$ is continuous, then $f$ extends to a continuous function on $X$. In its bounded form, an $f$ taking values in $[-M,M]$ has an extension with the same range bound. The underlying separation step is the Urysohn lemma.

Write $C(K)={\bigcup }_{n\ge 1}{\mathcal{S}}_n$ with every ${\mathcal{S}}_n$ equicontinuous. Its closure $E_n=\overline{{\mathcal{S}}_n}$ is still equicontinuous, because uniform limits preserve the same oscillation inequalities. The Baire category theorem applied to the Banach space $C(K)$ shows that some $E_n$ has nonempty interior. Hence there are $f_0\in C(K)$ and $r>0$ such that the open ball $B(f_0,r)$ lies in $E_n$.

Fix $x\in K$. Equicontinuity of $E_n$ gives a neighborhood $U$ of $x$ on which every member of $E_n$ changes by less than $r/4$. If $y\in U$ were distinct from $x$, the normality of a compact Hausdorff space and the Urysohn lemma would give $h\in C(K)$ with $h(x)=0$, $h(y)=r/2$, and $\Vert h{\Vert }_{\infty }=r/2$. Both $f_0$ and $f_0+h$ belong to $E_n$, yeta contradiction. Thus $U=\{x\}$, so every point is isolated and $K$ is a discrete space. A compact discrete space is finite, since its cover by singleton open sets has a finite subcover.

A locally compact Hausdorff space is regular. Cover $K$ by finitely many open sets whose compact closures lie in $U$; their union $V$ satisfies $K\subset V\subset \overline{V}\subset U$, with $\overline{V}$ compact. The compact Hausdorff space $\overline{V}$ is normal, so the Urysohn lemma gives a continuous function $g:\overline{V}\to [0,1]$ equal to $1$ on $K$ and $0$ on $\overline{V}\setminus V$. Extend $g$ by zero outside $\overline{V}$. The boundary values make the extension continuous, and its support is a compact subset of $U$.

The Urysohn lemma states that if $A$ and $B$ are disjoint closed subsets of a normal space $X$, there is a continuous $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

The Tietze extension theorem states that if $A$ is a closed subset of a normal space $X$, then every continuous $f:A\to \mathbb{R}$ has a continuous extension $F:X\to \mathbb{R}$. In the bounded form, if $f(A)\subseteq [a,b]$, the extension can be chosen with $F(X)\subseteq [a,b]$.

The Tietze extension theorem states: if $X$ is a normal topological space, $A\subseteq X$ is closed, and $f:A\to [-M,M]$ is continuous, then there is a continuous $F:X\to [-M,M]$ with $F{|}_A=f$.

We first prove an approximation lemma. Given continuous $h:A\to [-r,r]$, the closed subsetsare disjoint and closed in $X$. By the Urysohn lemma, there is a continuous $g:X\to [-r/3,r/3]$ equal to $-r/3$ on $A_{-}$ and $r/3$ on $A_{+}$. Then

Starting with $h_0=f$ and $r_0=M$, apply the lemma recursively to obtain continuous $g_j:X\to \mathbb{R}$ such thatwhere $r_j=M(2/3{)}^j$. The seriesconverges uniformly by the Weierstrass test, so its sum is continuous. On $A$, its remainder after $j$ terms is $h_{j+1}$, which tends uniformly to zero; hence $F{|}_A=f$. The bounds also keep $F$ in $[-M,M]$ after the standard endpoint-preserving version of the construction, proving the theorem.

Writewith each $U_n$ open. The closed sets $A$ and $X\setminus U_n$ are disjoint. By the Urysohn lemma, choose a continuous $f_n:X\to [0,1]$ withThe uniformly convergent seriesdefines a continuous map to $[0,1]$. It vanishes on $A$. If $x\notin A$, then $x\notin U_n$ for some $n$, so $f_n(x)=1$ and $f(x)>0$. Thuswhich is the closed G-delta set as a zero set construction.