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True. The Banach-Alaoglu theorem makes $B^{\prime }$ compact for the weak-star topology. Its compactness can be seen by embedding it in the product ${\prod }_{x\in E}[-\Vert x\Vert ,\Vert x\Vert ]$: the closed conditions expressing linearity identify a closed subset with $B^{\prime }$, and the product is compact by the Tychonoff theorem. The product topology is exactly pointwise convergence on $E$.

Part (i) supplies a compatible metric. Every compact metric space has a finite $1/n$-net for each positive integer $n$. The countable union of these finite nets is dense. ThereforeThe argument concerns topological separability, with the relative weak-star topology specified in the question.

The Riesz-Markov-Kakutani representation theorem says that a positive linear functional on $C(K)$ has the form $f\mapsto {\int }_{K}fd\nu$ for a unique finite positive regular Borel measure $\nu$, with norm $\nu (K)$. Its complex form says that every bounded complex linear functional is represented by a unique finite regular complex measure $\mu$, andThe extension from positive to arbitrary linear functionals follows by positive/negative decomposition of real linear functionals and then real/imaginary decomposition. The last quantity is the total variation norm of a measure, so this identifies the dual isometrically.

An extreme point $e$ of a convex set $C$ cannot be written as $e=ta+(1-t)b$ with $0<t<1$ and distinct $a,b\in C$. The Krein-Milman theorem, applied with the underlying real locally convex weak-star topology, states

We next prove Milman's converse to the Krein-Milman theorem. Suppose an extreme point $e$ were outside the weak-star closure of $S$. A basic weak-star neighborhood of $e$ disjoint from $S$ uses finitely many real coordinates: real and imaginary parts of evaluations at elements of $X$. Its complement is the union of finitely many closed half-spacesIntersect these with $C$, discard empty intersections, and call the resulting compact convex sets $C_1,\dots ,C_m$. They cover $S$, and none contains $e$.

The convex hull of their union is compact. Every point in it can be written ${\sum }_{i=1}^m{t}_i{c}_i$ with $(t_i)$ in the finite simplex and $c_i\in C_i$, by combining terms from the same convex set. The map from the simplex times ${\prod }_i{C}_i$ to that sum is weak-star continuous, so its image is compact and closed. It therefore contains ${\overline{\mathrm{conv}S}}^{w^{\ast }}=C$, in particular $e$. But extremality forces every $c_i$ having a positive coefficient in a representation of $e$ to equal $e$, contradicting $e\notin C_i$. Hence

Assume $K\ne \varnothing$. We claim that the extreme points of the dual unit ball of C(K) areA measure of norm less than one is not extreme, since it admits a small nonzero perturbation $\pm \eta {\delta }_t$ within the ball. For a measure $\mu$ of norm one, suppose $|\mu |$ is not a point mass. There is a Borel set $E$ with $0<a=|\mu |(E)<1$: if the measure support has two points, choose disjoint neighborhoods of positive mass; a regular probability measure supported at just one point is the corresponding point mass. Thenis a convex combination of distinct norm-one measures. Thus an extreme measure must have variation concentrated at one point, and must be $\alpha {\delta }_t$ with $|\alpha |=1$.

Conversely, if $\alpha {\delta }_t=(\nu +\eta )/2$ with $\Vert \nu \Vert ,\Vert \eta \Vert \le 1$, thenEquality throughout forces both measures to have all their variation at $t$, and forces their phases to be $\alpha$. Hence $\nu =\eta =\alpha {\delta }_t$, proving extremality. If $K=\varnothing$, the dual ball is $\{0\}$ and its sole extreme point is $0$.

Every finite Borel measure on $[0,1]$ is regular, so the given $\mu$ belongs to the dual unit ball of $C[0,1]$. Apply Banach-Alaoglu theorem and Krein-Milman theorem to that ball. It is the weak-star closed convex hull of these phased point masses. The unit ball of the finite-dimensional vector subspace $F$ is norm compact. Choose a finite $\epsilon /4$-net $f_1,\dots ,f_m$ in it. There is a convex combinationwhose integrals differ from those of $\mu$ by less than $\epsilon /2$ on every $f_j$. Put $t_i=a_i{\alpha }_i$. Then ${\sum }_i|t_i|=1$ and $\Vert \nu \Vert \le 1$. For any $f$ in the unit ball of $F$, choose $f_j$ with $\Vert f-f_j\Vert <\epsilon /4$. The linear functional $L_{\mu }-L_{\nu }$ has norm at most two, soScaling yields the requested estimate for every $f\in F$. This is atomic approximation on finite-dimensional spaces of continuous functions. Repeated nodes are allowed: keeping their individual terms preserves the exact sum of coefficient magnitudes even if their phases cancel. If $F=\{0\}$, one point with coefficient one suffices.

A character of an algebra is a nonzero complex linear multiplicative map $\varphi :A\to \mathbb{C}$. It necessarily satisfies $\varphi (1)=1$. If $\lambda =\varphi (x)$, then $x-\lambda 1$ cannot be invertible, since applying $\varphi$ to an inverse identity would give $0=1$. Therefore $\varphi (x)\in \sigma (x)$. The Neumann series bound $\sigma (x)\subseteq \{z:|z|\le \Vert x\Vert \}$ givesproving automatic continuity of characters. In particular every algebra character belongs to the closed unit ball of $A^{\ast }$; no prior continuity was needed.

The character space of an algebra is ${\Phi }_A$, the set of all algebra characters. Its Gelfand topology is the weakest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous, equivalently the weak-star topology inherited from $A^{\ast }$. Within the weak-star compact ball $B_{A^{\ast }}$, it is the closed set specified byThese equations are closed conditions on finitely many evaluations at a time. By the Banach-Alaoglu theorem, ${\Phi }_A$ is compact, and it is Hausdorff because distinct characters differ on an evaluation.

To obtain every spectral value from a algebra character, take $\lambda \in \sigma (x)$. Commutativity makes the ideal $(x-\lambda 1)A$ proper, so it is contained in an algebraic maximal ideal $\mathfrak{m}$. Every such maximal ideal is norm closed: its closure is an ideal, and cannot be all of $A$, because an element of $\mathfrak{m}$ sufficiently close to $1$ would be invertible by the Neumann series. Maximality then makes its closure equal to itself.

The quotient $A/\mathfrak{m}$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. Composing the quotient map with this scalar identification gives a algebra character with $\varphi (x)=\lambda$. Together with the first inclusion,The same maximal-ideal argument shows that the algebra character space of a nonzero algebra is nonempty.

Now suppose $A$ is a commutative C-star algebra. All its elements are normal. The C-star identity givessince the self-adjoint element $h=x^{\ast }x$ satisfies $\Vert h^2\Vert =\Vert h{\Vert }^2$. Iterating yields $\Vert x^{2^n}\Vert =\Vert x{\Vert }^{2^n}$. The spectral radius formula along this subsequence provesThis is spectral radius norm equality for normal elements, applied to the commutative case.

For a self-adjoint $h$, the element $v_t=\exp (ith)$ is unitary for every real $t$, so $\Vert v_t\Vert =1$. Algebra character continuity and multiplicativity giveIts modulus is $\exp (-t\mathrm{Im}\varphi (h))$, forcing $\varphi (h)\in \mathbb{R}$. Write $x=h+ik$ with $h,k$ self-adjoint. ThenThis proves that characters of a C-star algebra respect the involution. The conjugation bar is present in the original PDF and must be retained; the TeX aid omits it.

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra is isometrically star-isomorphic to $C(K)$ for a compact Hausdorff space, canonically $K={\Phi }_A$. Its Gelfand transform isIt is a unital algebra homomorphism; the conjugation identity just proved makes it preserve the involution. The algebra character description of the spectrum and the spectral-radius equality giveThus it is injective and isometric, and its range is closed because $A$ is complete. The range contains constants, is closed under complex conjugation, and separates points of ${\Phi }_A$: two distinct algebra characters differ on some element of $A$.

The complex Stone-Weierstrass theorem says that a self-adjoint unital subalgebra of $C(K)$ separating points is uniformly dense. For completeness, its familiar lattice argument explains the last step. The real part of its uniform closure is closed under absolute value, by polynomial approximation of $|t|$ on bounded real intervals, and hence under finite maxima and minima. Constants and point separation permit a real function matching any given real continuous $g$ at any chosen pair of points. Fixing the first point, take a maximum of finitely many such functions to obtain a function above $g-\epsilon$ everywhere and equal to $g$ at that first point. It is below $g+\epsilon$ in a neighborhood of that point. A finite cover by these neighborhoods and the minimum of their associated functions then lies between $g-\epsilon$ and $g+\epsilon$ everywhere. Real and imaginary parts give density for complex functions.

Apply this to $\mathcal{G}(A)$. Its range is both dense and closed, so it is all of $C({\Phi }_A)$. ThereforeThis proves the commutative theorem in the setting of the question. For the zero algebra the corresponding compact space is empty and the representation is $C(\varnothing )=\{0\}$.

Use the standard setting of a compact metrizable convex set $K$ in a Hausdorff locally convex real topological vector space, with its metrizable topology. Write $A(K)$ for the continuous affine real functions on $K$. The affine upper envelope of a bounded real function isConstants make this infimum finite, and $f\le \overline{f}$. An infimum of affine continuous majorants is concave and upper semicontinuous, hence Borel. For continuous $f$ it is the upper concave envelope appropriate to barycentric measures; it is not the pointwise maximum of $f$ and a selected affine function.

For a fixed probability $\mu$, define $p(g)={\int }_K\overline{g}d\mu$ on real $C(K)$. Affine majorants showThus $p$ is sublinear and $p(a)=\int ad\mu$ for affine $a$. On the span of $f$, the linear functional taking $f$ to $p(f)$ is dominated by $p$: the negative-scalar condition follows from $-p(f)\le p(-f)$. For $f=0$ start from the zero subspace. The Hahn-Banach theorem extends it to a linear functional $L$ on $C(K)$ with $L\le p$ and $L(f)=p(f)$.

If $g\ge 0$, then $p(-g)\le 0$, so $L(g)\ge 0$. Also $p(1)=1$ and $p(-1)=-1$, forcing $L(1)=1$. Positivity gives $|L(g)|\le \Vert g{\Vert }_{\infty }$, and the Riesz-Markov-Kakutani representation theorem produces a Borel probability $\nu$ with $L(g)=\int gd\nu$. ThereforeFor affine $a$, testing both $a$ and $-a$ gives $\int ad\nu =\int ad\mu$: the two measures have the same barycenter. This proves the requested supporting measure lemma for affine upper envelopes.

Choquet's theorem: every $x\in K$ has a Borel probability measure $\lambda$ concentrated on the extreme points of $K$ whose barycenter is $x$; equivalently,Concentration is a measure-one assertion, not a claim that the topological support must be a closed subset of $\mathrm{Ex}K$.

To prove it, let ${\mathcal{M}}_x$ be the measures satisfying all the displayed affine equalities. It is nonempty because it contains ${\delta }_x$, and it is weakly closed in the compact space $P(K)$, hence compact. Let $h\in C(K)$ be strictly convex, as permitted. Choose $\lambda \in {\mathcal{M}}_x$ maximizing $\int hd\lambda$. Apply the supporting measure lemma with $\mu =\lambda$ and $f=h$. It gives $\nu$ with the same affine integrals, so $\nu \in {\mathcal{M}}_x$, andThus $\overline{h}-h\ge 0$ has integral zero. If $z$ is not extreme, write $z=ty+(1-t)w$ with $0<t<1$ and distinct $y,w\in K$. Strict convexity and every affine majorant giveTherefore the nonnegative gap is strictly positive at every nonextreme point. It is Borel, and $\mathrm{Ex}K$ is Borel by the allowed G-delta set assertion. Its zero integral forces $\lambda (K\setminus \mathrm{Ex}K)=0$, proving Choquet's theorem by strict convexity. No uniqueness is asserted for the representing measure.

For the real $L^{\infty }$ example, give its closed unit ball $K$ the weak-star topology $\sigma (L^{\infty },L^1)$, not the norm topology. The Banach-Alaoglu theorem makes it compact, and separability of $L^1[0,1]$ makes this ball metrizable. Its extreme points are exactly the classes $h$ satisfying $|h|=1$ almost everywhere. Indeed, if $|h|\le 1-\epsilon$ on a positive-measure set, adding and subtracting $\epsilon$ times its indicator decomposes $h$ nontrivially inside the ball. Conversely, if $|h|=1$ almost everywhere and $h=(u+v)/2$ with $|u|,|v|\le 1$, pointwise equality at the endpoints of $[-1,1]$ forces $u=v=h$ almost everywhere. This is the extreme-point criterion for the L-infinity unit ball.

For the hinted case $f=1_A-1_B$, put $C=[0,1]\setminus (A\cup B)$ and $h_{\pm }=1_A-1_B\pm 1_C$. Then $h_{\pm }$ are extreme andhas barycenter $f$. If $C$ is null, the two point masses coincide.

For a general real $f$, choose a measurable representative in $[-1,1]$ and set, for $0\le t\le 1$,Every $H_t$ is extreme. The map into the weak-star compact ball is Borel: for each $u\in L^1$, the function $t\mapsto \int u(s)H_t(s)ds$ is measurable by joint measurability and integration; a countable dense family of such tests generates the ball's topology and Borel sigma-algebra. The measure is independent of changes to $f$ on a null set. For each $s$,The integrand paired with $u$ is dominated by $|u|$, so Fubini's theorem yieldsThese continuous linear tests define the weak-star barycenter, hence that barycenter is $f$. Together with $\nu (\mathrm{Ex}K)=1$, this is the required threshold Choquet representation in L-infinity.

If $L^{\infty }$ is instead taken over complex scalars, its extreme points satisfy the same unit-modulus condition. Write $f(s)=r(s)\zeta (s)$, with $r=|f|$ and $|\zeta |=1$, choosing $\zeta =1$ where $f=0$. Replace the threshold family by $\zeta (s)(21_{\{t\le (1+r(s))/2\}}-1)$. Its members have unit modulus and its average is $f$, so the same pushforward and Fubini argument gives a representing measure in the real locally convex interpretation of the complex ball.

The weak topology on a normed vector space $X$ is $\sigma (X,X^{\ast })$, the coarsest topology making every bounded linear functional continuous. A neighbourhood base at $x$ is given by finitely many inequalities $|f_j(y-x)|<\epsilon$, with $f_j\in X^{\ast }$. In the complex case, separation uses real parts of bounded linear functionals.

Mazur theorem states that the weak closure of a convex set equals its closure in the norm topology. The weak topology is coarser than the norm topology, so the norm closure is contained in the weak closure. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and $a\in \mathbb{R}$ with $\mathrm{Re}f(x)>a\ge {\sup }_{c\in C}\mathrm{Re}f(c)$. The corresponding weak topology neighbourhood of $x$ misses $C$, so $x$ is outside its weak closure. This proves Mazur theorem, including the empty-set case. In particular, if $x_n$ converges weakly to $x$, then $x$ is in the weak closure of every tail and therefore in the norm closure of its convex hull. Choosing a finite convex combination of the $n$th tail within $1/n$ of $x$ proves the usual Mazur lemma formulation as well.

The weak-star topology on the continuous dual space $X^{\ast }$ is $\sigma (X^{\ast },X)$: convergence means pointwise convergence on $X$, and a neighbourhood base prescribes finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this weak-star topology, even if $X$ is incomplete. Embed this closed unit ball intowhere $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Each factor is compact, so $P$ is compact by the Tychonoff theorem. Inside $P$, the equations $a_{x+y}=a_x+a_y$ and $a_{\lambda x}=\lambda a_x$ define a closed set. Every such point defines a linear functional satisfying $|a_x|\le \Vert x\Vert$, hence belongs to the closed unit ball of $X^{\ast }$. Thus this image is closed in $P$. The product topology on it is exactly the weak-star topology, proving Banach-Alaoglu theorem. Evaluations also separate its points, so the weak-star topology is Hausdorff.

For the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, we have $(J_{X}x)(f)=f(x)$. Restricting all evaluations at $f\in X^{\ast }$ therefore gives exactly $\sigma (X,X^{\ast })$. Thus the induced subspace topology is the weak topology on $X$.

Now identify $X$ with $J_{X}X$. Let $C$ be a bounded convex set, write $D$ for its norm closure in $X$, and let $K$ be its weak-star topology closure in $X^{\ast \ast }$. Boundedness places $K$ in a multiple of the closed unit ball of $X^{\ast \ast }$, so $K$ is compact by Banach-Alaoglu theorem. The preceding subspace topology identification and Mazur theorem giveIf $D$ is a weakly compact set, its image in the Hausdorff weak-star topology is compact and therefore closed. It contains $C$, so $K\subseteq D\subseteq X$. Conversely, if $K\subseteq X$, the displayed identity gives $K=D$, and its compactness is precisely weak compactness in $X$. Hence $\boxed{D\text{ is weakly compact}⟺K\subseteq X}$. This argument also covers $C=\varnothing$.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint is $T^{\ast }:Y^{\ast }\to X^{\ast }$, defined by $(T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$. Evaluation at any fixed $x$ is thus evaluation at $Tx$ after applying $T^{\ast }$. Each is continuous in the relevant weak-star topology, proving that $T^{\ast }$ is weak-star continuous. Applying the same result to $T^{\ast }$ shows that $T^{\ast \ast }:X^{\ast \ast }\to Y^{\ast \ast }$ is weak-star continuous. Direct evaluation gives

Use $B_X$ for the closed unit ball; using the open ball gives the same norm closure of $T(B_X)$. Goldstine theorem says that $J_X{B}_X$ is weak-star dense in $B_{X^{\ast \ast }}$. Put $K=T^{\ast \ast }(B_{X^{\ast \ast }})$. It is compact and closed in the weak-star topology by Banach-Alaoglu theorem and the established continuity. It contains $J_{Y}T(B_X)$. Conversely, Goldstine theorem gives, for every $x^{\ast \ast }\in B_{X^{\ast \ast }}$, a net $(J_X{x}_{\alpha })$ from $J_X{B}_X$ converging weak-star to $x^{\ast \ast }$; its image converges weak-star to $T^{\ast \ast }{x}^{\ast \ast }$. ConsequentlyApply the preceding bounded convex set criterion in $Y$, and then scale the closed unit ball. We obtain the bidual characterization of weakly compact operators:The canonical embedding into the bidual on the right specifies exactly which copy of $Y$ is intended.

All vector spaces in this solution are complex, while convex combinations use real coefficients. An extreme point $x$ of a convex set $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.

The Krein-Milman theorem says that every nonempty compact convex set $K$ in a Hausdorff locally convex space is the closed convex hull of its extreme points. Here is a proof. A face of a convex set $F\subseteq K$ is a convex set such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contains distinct $x,y$, a continuous real linear functional $\ell$ on the underlying real locally convex space distinguishes them. Such a linear functional exists because the space is Hausdorff and locally convex, by the Hahn-Banach theorem. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an extreme point. The same argument inside any nonempty compact face of $K$ supplies an extreme point of $K$ lying in that face.

Let $D$ be the closed convex hull of the extreme points of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the Hahn-Banach separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an extreme point $e$ of $K$. But $e\in D$ and $\ell (e)\ge \ell (x_0)>{\sup }_D\ell$, a contradiction. Thus $\boxed{K=\overline{\mathrm{co}}(\mathrm{ext}K)}$, proving Krein-Milman theorem.

For a nonempty compact Hausdorff space $K$, the extreme points of the dual unit ball of C(K) areThis is the permitted description without proof, with $C(K)$ denoting the complex space of continuous functions on a compact space equipped with the supremum norm. If $K$ is empty, $C(K)=\{0\}$ and the closed unit ball of its continuous dual space has the single extreme point $0$ instead.

The complex Banach–Stone theorem states that a surjective complex-linear isometric isomorphism of normed spaces $T:C(K)\to C(L)$ has the formwhere $u\in C(L)$ and $\phi :L\to K$ is a homeomorphism. Conversely, every such map is a surjective complex-linear isometric isomorphism of normed spaces for the supremum norm.

To prove this, suppose first that $K,L$ are nonempty. The Banach-space adjoint $T^{\ast }$ is a bijective isometric isomorphism of normed spaces on the continuous dual spaces, so it bijects their closed unit balls and preserves extreme points. The displayed extreme point description gives uniquelyUniqueness follows by evaluating at the constant function $1$, and then using that continuous functions separate points of a compact Hausdorff space. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^{\ast }$ on extreme points proves surjectivity of $\phi$: the preimage of any ${\delta }_x$ is $\alpha {\delta }_y$ for some $y$. If $\phi (y_1)=\phi (y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\phi$ is bijective.

For every $f\in C(K)$, $f\circ \phi =(Tf)/u$ is continuous. The evaluation map $K\to {\mathbb{C}}^{C(K)}$, $x\mapsto (f(x){)}_f$, is a continuous injection of a compact Hausdorff space into a Hausdorff product topology, hence a homeomorphism onto its image. Continuity of every coordinate $f\circ \phi$ proves continuity of $\phi$. A continuous bijection between compact Hausdorff spaces is a homeomorphism. Conversely the weighted-composition formula plainly preserves the supremum norm, and its inverse isIf one compact space is empty, a surjective isometric isomorphism of normed spaces forces the other to be empty, and the empty homeomorphism gives the corresponding trivial case. This completes Banach–Stone theorem.

Neither the space of sequences converging to zero $c_0$ nor $L^1[0,1]$ can be isometrically a continuous dual space of a Banach space. Indeed, every nonzero continuous dual space has a nonempty weak-star compact closed unit ball by Banach-Alaoglu theorem, and Krein-Milman theorem then guarantees an extreme point. A bijective linear isometry preserves extreme points of closed unit balls.

The closed unit ball of $c_0$ has no extreme points. Given $x\in c_0$ with $\Vert x{\Vert }_{\infty }\le 1$, choose $n$ with $|x_n|<1$ and $0<\epsilon <1-|x_n|$. The two distinct elements $x\pm \epsilon e_n$ remain in that closed unit ball and have midpoint $x$.

The closed unit ball of the complex Lp space $L^1[0,1]$ also has no extreme points. An element of norm less than one can be perturbed by a sufficiently small nonzero Lp space element in both directions. If $\Vert f{\Vert }_1=1$, the non-atomic measure $|f(t)|dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto {\int }_0^s|f(t)|dt$ attains $1/2$. Put $h=f(1_A-1_{A^c})$. For $0<\epsilon <1$, the distinct functions $f\pm \epsilon h$ both have Lp normwith the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus neither proposed space is isometrically a Banach dual.

Finally, $[0,1]$ is a connected space, whereas $[0,1]\cup [2,3]$ is disconnected. They cannot be homeomorphic. By Banach–Stone theorem, $C[0,1]$ and $C([0,1]\cup [2,3])$ are not isometrically isomorphic as complex Banach spaces.

Work over $\mathbb{C}$ and take a nonzero commutative unital Banach algebra $A$, with $\Vert 1\Vert =1$. 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 $|\chi (a)|\le \Vert a\Vert$, proving automatic continuity of characters and $\Vert \chi \Vert =1$.

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:One inclusion was proved above. For the other, if $a-\lambda 1$ 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 $A^{\ast }$. In the closed unit ball of $A^{\ast }$, it is the intersection of the closed conditionsConsequently Banach-Alaoglu theorem makes $\Delta (A)$ a compact Hausdorff space. For every $a\in A$, define the Gelfand transform $\hat{a}(\chi )=\chi (a)$. 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 fieldMultiplicativity 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 isthe 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 $A/\mathrm{rad}A$ 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 $f_1,\dots ,f_n\in M$ 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 $\boxed{\hat{f}({\delta }_x)=f(x)}$, 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: 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 $h=h^{\ast }$, 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 $|e^{it\chi (h)}|\le 1$ for every real $t$, forcing $\chi (h)$ to be real. Writing $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint gives $\chi (a^{\ast })=\overline{\chi (a)}$. 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 $b^{\ast }b$, to obtainIts powers are also normal, so $\Vert a^{2^n}\Vert =\Vert a{\Vert }^{2^n}$. The spectral radius formula gives $r(a)=\Vert a\Vert$, hence $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$. 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 $\epsilon >0$, 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 $b_{x,y}>f-\epsilon$ cover $K$. Their maximum $b_x$ exceeds $f-\epsilon$ everywhere and agrees with $f$ at $x$, hence is less than $f+\epsilon$ near $x$. Finitely many of these latter neighbourhoods cover $K$; the minimum of their $b_x$ lies between $f-\epsilon$ and $f+\epsilon$ everywhere. Approximate real and imaginary parts separately. This proves the density used above and completes the C-star algebra conclusion.

For positive linear operators on continuous functions, a general Korovkin theorem, in its compact Korovkin test space form, can be stated using a test space $H\subset C(K,\mathbb{R})$, where $K$ is a compact Hausdorff space. Assume $H$ contains a strictly positive function $h_0$. Require that, for every $x\in K$, the only positive linear functional $L$ on $C(K)$ with $L(h)=h(x)$ for all $h\in H$ is evaluation at $x$. Then convergence on the test space implies convergence on every continuous function:By the Riesz-Markov-Kakutani representation theorem, the functional condition equivalently says that a positive measure with these test moments must be ${\delta }_x$. This permits arbitrary test families, not just the interval tests $1,x,x^2$.

For completeness, let $c={\min }_K{h}_0>0$. Positivity gives $U_{n}1\le c^{-1}{U}_n{h}_0$, so the operator norms are uniformly bounded. If the conclusion failed for some $f$, choose $x_n$ where its error is bounded away from zero. The evaluation functionals $L_n(g)=U_{n}g(x_n)$ have uniformly bounded norm. Compactness of $K$ and the Banach-Alaoglu theorem give a subnet on which $x_n\to x$ and $L_n$ converges in the weak-star topology to a positive $L$. For every $h\in H$, uniform test convergence gives $L(h)=h(x)$. The hypothesis forces $L(f)=f(x)$, contradicting the chosen errors. Complex-valued functions follow by applying the real result to their real and imaginary parts.

A useful concrete sufficient condition is that $1\in H$ and for each $x$ there is $g_x\in H$ with $g_x\ge 0$ and zero set exactly $\{x\}$. Its zero moment forces the representing measure to be supported at $x$, while the constant moment fixes its mass.

For the circle, take $H=\mathrm{span}\{1,\sin t,\cos t\}$. The nonnegative function $g_x(t)=1-\cos (t-x)$ belongs to $H$ and vanishes only at $x$ on the circle. Thus the periodic Korovkin test set isUniform convergence on these three functions is sufficient for uniform convergence on $C(\mathbb{T})$.

On a compact metric space $X$, a conservative Feller semigroup is a strongly continuous semigroup of positive linear maps on $C(X)$ satisfying $P_{t}1=1$. Positivity means $f\ge 0⟹P_{t}f\ge 0$. These conditions imply sup-norm contraction. Conversely, for a contraction semigroup with $1\in D(L)$ and $L1=0$, differentiation on the generator domain gives $\frac{d}{dt}{P}_{t}1=P_{t}L1=0$, so $P_{t}1=1$. On the real space $C(X)$, if $0\le f\le 1$, then $\Vert 1-P_{t}f{\Vert }_{\infty }\le \Vert 1-f{\Vert }_{\infty }\le 1$, implying $P_{t}f\ge 0$. Scaling proves positivity for every nonnegative $f$. This is the unital contraction positivity criterion. For complex $C(X)$ the same conclusion follows because each functional $f\mapsto P_{t}f(x)$ has norm one and value one on $1$, hence is a positive linear functional.

For each $t,x$, the Riesz-Markov-Kakutani representation theorem represents that functional by a unique Borel probability measure $p_t(x,dy)$:Continuity of $P_{t}f$ gives weak continuity of $x\mapsto p_t(x,\cdot )$. Approximating indicators of open sets increasingly by continuous functions proves Borel measurability in $x$; a monotone-class argument then gives a Markov kernel. The semigroup property and uniqueness of representing measures yield $p_0(x,\cdot )={\delta }_x$ and the Chapman-Kolmogorov equationThese are the required transition probabilities.

An invariant probability measure for a semigroup satisfies ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$ for every $t\ge 0$ and $f\in C(X)$. Equivalently its distribution is preserved by the Markov kernel. For $1\le p<\infty$, Jensen inequality gives $|P_{t}f{|}^p\le P_t(|f{|}^p)$ pointwise. Integrating and using invariance provesAlso $\Vert P_{t}f-f{\Vert }_{L^p(\mu )}\le \Vert P_{t}f-f{\Vert }_{\infty }\to 0$. Continuous functions are dense in these $L^p$ spaces for a finite Borel measure on a compact metric space, so the contraction extends to a strongly continuous semigroup on $L^p(\mu )$.

For real $f$ with $f,f^2\in D(L)$, the kernel form of Jensen inequality gives $P_t(f^2)-(P_{t}f{)}^2\ge 0$, with equality at $t=0$. Take its right derivative to obtain the generator square inequalityThis is nonnegativity of the associated carré du champ.

Each time-average functional ${\varphi }_n$ is positive with ${\varphi }_n(1)=1$, hence represents a Borel probability measure ${\mu }_n$. The compactness of probability measures on a compact metric space gives a subsequence converging weakly to a probability measure $\mu$. Explicitly, $C(X)$ is a separable Banach space, so Banach-Alaoglu theorem and a countable dense family give a subsequence converging on every continuous function, rather than merely a net. For fixed $t\ge 0$, the semigroup property gives the Krylov-Bogolyubov time-average argument:Its absolute value is at most $2t\Vert f{\Vert }_{\infty }/n$. Pass to the subsequential limit, observing that $P_{t}f\in C(X)$, to get ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$. Thus the limiting measure is invariant.

For a convex subset of a Banach space, weak closure equals norm closure. The Mazur theorem states that for every convex set $C\subseteq X$,Since every bounded linear functional is norm-continuous, the weak topology is weaker than the norm topology, giving ${\overline{C}}^{\Vert \cdot \Vert }\subseteq {\overline{C}}^w$.

For the reverse inclusion, use this form of the Hahn-Banach separation theorem: if $D$ is a nonempty norm-closed convex set in a real normed vector space and $x\notin D$, there is a continuous real-linear functional $F$ such thatApply it to $D={\overline{C}}^{\Vert \cdot \Vert }$. In a complex Banach space, use the underlying real space; every continuous real-linear $F$ is the real part of the complex bounded linear functional $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the weak topology of $x$ disjoint from $D$, so $x\notin {\overline{C}}^w$. The empty set case is immediate. This proves the Mazur theorem.

A weakly null sequence has disjoint convex blocks converging to zero in norm. If $x_n\stackrel{w}{\to }0$, then for every starting index $m$,The Mazur theorem places $0$ in the norm closure of that tail convex hull. Thus a finite convex combination of vectors from any tail can have norm less than any prescribed $\epsilon >0$.

Choose the convex blocks recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite convex combination from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yieldsTherefore $u_n\to 0$ in norm. The recursive tail selection is what makes these convex blocks, rather than merely unrelated convex combinations.

Next suppose $X^{\ast }$ is separable and $(x_n)$ is bounded. Use the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, where $J_{X}x(f)=f(x)$. If $\Vert x_n\Vert \le M$, then $J_X{x}_n\in M{B}_{X^{\ast \ast }}$. The Banach-Alaoglu theorem makes this ball compact in $\sigma (X^{\ast \ast },X^{\ast })$, and weak-star metrizability of the dual ball makes it metrizable because its predual $X^{\ast }$ is separable. A compact space that is a metric space is sequentially compact, so some subsequence $(y_n)$ satisfiesThe limit need not belong to $J_{X}X$; it is precisely the use of the bidual space that supplies compactness without assuming reflexivity.

For a convex block $u_n={\sum }_{i=p_n}^{q_n}{a}_i{y}_i$, every $f\in X^{\ast }$ satisfiesBoth $f(y_n)$ and $f(u_n)$ tend to $\varphi (f)$, soThis is the useful convex-block cancellation of a weak-star limit; it does not require the limit to be a vector in $X$.