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The continuous dual space of ${\ell }^1$ is ${\ell }^{\infty }$ under the complex-linear coordinate pairing. Its subspace $c_0$ is norm closed: uniform approximation by a sequence with a small tail leaves a small tail. Nevertheless its preannihilator in ${\ell }^1$ is zero, because the coordinate unit vectors belong to $c_0$ and test every coordinate. Its double continuous annihilator is therefore all of ${\ell }^{\infty }$. It is weak-star dense directly as well: the finite truncations of any bounded sequence lie in $c_0$ and converge against every ${\ell }^1$ test by absolute summability. This demonstrates why norm closure cannot replace closure in the weak-star topology in the dual-side annihilator theorem.

The smallest closed convex set containing a set $C$, equivalently the closure of its convex hull in the specified topology. The topology must be stated: the norm topology, weak topology and weak-star topology need not give the same closure for an arbitrary set.

Even a bounded infinite input need not have a closed convex hull. In the real line, let $S=\{1/n:n\ge 1\}$. Every finite convex combination of these positive points is positive, so zero is absent. Conversely, any $x\in (0,1]$ lies between some $1/n$ and $1$, and is their convex combination. HenceThe closed convex hull adds the missing limit point zero. For the finite eight-point example in convex hull, the convex hull is already closed, so taking the closed convex hull adds nothing.

The closed unit ball of a normed vector space is the set of vectors of norm at most one. In a continuous dual space it is compact in the weak-star topology by the Banach-Alaoglu theorem.

For a compact group $G$, positive normalized linear functionals on $C(G)$ form a nonempty weak-star topology compact convex set. For finitely many translations, apply the Schauder-Tychonoff fixed-point theorem to the average of their induced affine dual maps. A fixed functional is stationary under this average. For each real continuous $f$, its orbit function $h(g)=\varphi (l_{g}f)$ is continuous on the compact subgroup generated by those translations and satisfies a finite averaging identity. At a maximum, every summand must attain that same maximum; iteration and density of the positive-word semigroup force $h$ to be constant. Thus the functional is fixed by each of the finitely many translations. Compactness and the finite intersection property now give a functional invariant under every translation. The Riesz-Markov-Kakutani representation theorem turns it into a normalized Haar measure, without assuming commutativity of the group.

A linear functional on $X^{\ast }$ is continuous for the weak-star topology exactly when it is evaluation at an element of $X$. Continuity makes it vanish on the common kernel of finitely many evaluations, so it is a linear combination of those evaluations. The canonical embedding into the bidual identifies this dual with $X$.

For finite complex measures on the additive circle group, convolution is the pushforward measure of their product under addition modulo $2\pi$. EquivalentlyBounded tests are integrable against the product variation measure, so Fubini's theorem applies. Testing the factorization $e^{-in(x+y)}=e^{-inx}{e}^{-iny}$ proves the displayed Fourier coefficient of a measure on the circle identity. The convolution of two probability measures is again a probability measure. On these measures convolution preserves simultaneous convergence in the weak-star topology: multiply the converging coefficients and use their common unit variation bound and dense trigonometric tests.

If $Z$ is a Banach space and $X$ is a norm-closed reflexive linear subspace of $Z^{\ast }$, evaluation gives the displayed surjection. Restriction $R:Z^{\ast \ast }\to X^{\ast }$ is weak-star continuous and maps the closed unit ball onto the target ball by the Hahn-Banach theorem. The Goldstine theorem then makes $T(B_Z)$ weak-star dense in $B_{X^{\ast }}$, where $T=R{J}_Z$. Reflexivity of $X$ makes this target weak-star topology equal to its weak topology. Convexity and the Hahn-Banach separation theorem convert weak density of this ball image to norm density.

To pass from density to surjectivity, approximate a desired $f$ by $T{z}_1$ with $\Vert z_1\Vert \le \Vert f\Vert$ and residual at most $\Vert f\Vert /2$. Repeat on each residual. The chosen vectors have summable norms, with sum at most $2\Vert f\Vert$, and the residuals tend to zero geometrically. Completeness of $Z$ produces $z={\sum }_n{z}_n$ with $Tz=f$. Using any smaller residual ratio proves the infimum lifting norm equals $\Vert f\Vert$, although that infimum need not be attained. This argument proves onto-ness, rather than merely density of the evaluation functionals.

A norm-closed finite-codimensional vector subspace $Y\subseteq X^{\ast }$ that is weak-star dense is a norming subspace. Its annihilator $F=Y^{\perp }\subseteq X^{\ast \ast }$ is finite-dimensional and disjoint from $JX$. The positive distance $d$ of the unit sphere from $F$ gives a triple-dual linear functional vanishing on $F$ and taking a value at least $d$ at a unit vector. Apply the finite-dimensional interpolation form of Goldstine's theorem to the Banach space $X^{\ast }$, then normalize and take a supremum to obtain norming constant $d$.

For a finite complex measure on the additive circle group $\mathbb{R}/(2\pi \mathbb{Z})$, its Fourier coefficients are integrals against the group characters $e^{-inx}$, $n\in \mathbb{Z}$. They obey $|\hat{\mu }(n)|\le \Vert \mu \Vert$, where the norm is the total variation norm of a measure. A probability measure has zeroth coefficient one. The normalized Haar measure $m(dx)=dx/(2\pi )$ has all nonzero-frequency coefficients zero. A bounded family of finite Radon measures converges in the weak-star topology exactly when its coefficients converge to those of the proposed limit: extend linear combinations to all continuous tests using the density of trigonometric polynomials and the common variation bound.

For a normed vector space $X$, take $K=B_{X^{\ast }}$ with the weak-star topology. Banach-Alaoglu theorem makes it compact Hausdorff, and each displayed evaluation function is continuous. Its supremum norm is $\Vert x\Vert$ by the Hahn-Banach theorem, giving a linear isometric embedding into $C(K)$.

A vector subspace $Z\subseteq X^{\ast }$ is norming if some $c>0$ satisfies the displayed inequality. It retains enough continuous linear functionals to control the norm of every vector. Every such vector subspace is dense for the weak-star topology: otherwise a nonzero evaluation linear functional would annihilate its weak-star closure, contradicting the inequality. In the real case its symmetric unit ball makes the absolute-value and signed supremum versions equivalent.

The weak-star topology on $X^{\ast }$ is the coarsest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous. Finite intersections of evaluation inequalities form a neighborhood basis. Distinct functionals differ on some $x$, so disjoint disks around their evaluation values give disjoint neighborhoods; thus the topology is Hausdorff. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this topology.

If $f\in X^{\ast \ast }$ is weak-star continuous, some $x_1,\dots ,x_n$ and $\epsilon >0$ make $|f(\varphi )|<1$ whenever $|\varphi (x_i)|<\epsilon$ for every $i$. If all evaluations vanish, every scalar multiple of $\varphi$ lies in this neighborhood, forcing $f(\varphi )=0$. Thus $f$ vanishes on the kernel of the map $E(\varphi )=(\varphi (x_1),\dots ,\varphi (x_n))$. It factors linearly through the finite-dimensional range of $E$. Extending that factor algebraically to ${\mathbb{K}}^n$ gives scalars $a_i$ withConsequently $f=J_{X}x$ for $x={\sum }_i{a}_i{x}_i$.

For the two induced topologies, let $B$ be the common norm-closed unit ball and write ${\tau }_j=\sigma (Y,X_j)$. Under either topology $B$ is compact by Banach-Alaoglu, and the ambient space is Hausdorff. If the two topologies are comparable, the identity map from $B$ with the finer topology to $B$ with the coarser one is a continuous bijection from a compact space to a Hausdorff space. Such a map is a homeomorphism, since it carries closed compact sets to closed sets. Thus in the comparable case the topologies agree on $B$, giving alternative (a).

An infinite-dimensional reflexive example is ${\ell }^2$, by the Riesz representation theorem.

The weak topology on $X$ is the coarsest topology making every $f\in X^{\ast }$ continuous. The weak-star topology on $X^{\ast }$ is the coarsest making every evaluation $f\mapsto f(x)$, $x\in X$, continuous. In each case basic neighborhoods impose finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of a dual space is compact for its weak-star topology. The separation theorem needed here says: if $K$ is a nonempty weak-star compact convex subset of a dual space $E^{\ast }$ and $z\notin K$, then some $e\in E$ strictly separates them,The real-part convention covers complex spaces; for real spaces omit it.

Let $K$ be the weak-star closure of $J_X(B_X)$ inside $B_{X^{\ast \ast }}$. This is compact by Banach-Alaoglu, convex, and nonempty. If $z\in B_{X^{\ast \ast }}\setminus K$, separation in the dual pair $(X^{\ast \ast },X^{\ast })$ gives $f\in X^{\ast }$ withBut $|z(f)|\le \Vert z\Vert \Vert f\Vert \le \Vert f\Vert$, a contradiction. Therefore $K=B_{X^{\ast \ast }}$, proving the required Goldstine theorem density.

If $X$ is reflexive, $J_X$ identifies its weak topology with the weak-star topology of $X^{\ast \ast }$ and carries $B_X$ onto $B_{X^{\ast \ast }}$, so Banach-Alaoglu gives weak compactness. Conversely, if $B_X$ is weakly compact, its image is weak-star compact, hence closed in the Hausdorff bidual topology. Density then forces $J_X(B_X)=B_{X^{\ast \ast }}$. Scaling gives surjectivity on the entire bidual. Thus

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every $f\in X^{\ast }$ continuous. Its basic neighbourhoods impose finitely many inequalities $|f_j(x-x_0)|<\epsilon$. The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology making each evaluation $f\mapsto f(x)$ continuous, for $x\in X$.

The canonical embedding into the bidual isIt is linear and $\Vert J_{X}x\Vert \le \Vert x\Vert$. For $x\ne 0$, the Hahn-Banach theorem supplies a norming functional $f$ with $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, proving equality. Thus $J_X$ is an isometric embedding. The coordinates defining the weak-star topology on $X^{\ast \ast }$, restricted to $J_X(X)$, are precisely $J_{X}x\mapsto f(x)$ for $f\in X^{\ast }$. These are exactly the coordinates defining the weak topology on $X$, so $J_X$ is also a topological embedding for these topologies.

A normed vector space is a reflexive space if $J_X$ is onto $X^{\ast \ast }$. Such a space is automatically a Banach space, since it is isometric to the complete continuous dual space $X^{\ast \ast }$. The Banach-Alaoglu theorem states that the closed unit ball of a continuous dual space is compact for its weak-star topology. The Goldstine theorem states that $J_X(B_X)$ is dense in the weak-star topology in $B_{X^{\ast \ast }}$. ConsequentlyIf $J_X$ is onto, its weak-to-weak-star homeomorphism carries $B_X$ onto the Banach-Alaoglu theorem's compact ball $B_{X^{\ast \ast }}$. Conversely, if $B_X$ is weakly compact, $J_X(B_X)$ is compact and hence closed in the Hausdorff weak-star topology. Its density from the Goldstine theorem makes it all of $B_{X^{\ast \ast }}$, and scaling gives surjectivity of $J_X$.

Now suppose $X$ is separable and nonzero. Choose a countable dense set $(u_n)$ in its unit sphere and, using the Hahn-Banach theorem, choose norming functionals $f_n$ with $\Vert f_n\Vert =1$ and $f_n(u_n)=1$. These continuous linear functionals separate points: for $x\ne 0$, choose $\Vert u_n-x/\Vert x\Vert \Vert <1/2$ to get $f_n(x)>\Vert x\Vert /2$. DefineThe series converges uniformly, separates distinct points, and obeys the triangle inequality because $\min (c,a+b)\le \min (c,a)+\min (c,b)$. Thus $d$ is a metric. Its topology is coarser than the weak topology: a finite set of coordinate controls makes the finite initial sum small, while the remaining sum is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$. On a weakly compact set $K$, the identity $(K,\text{weak})\to (K,d)$ is therefore a continuous bijection from a compact space into a Hausdorff space, hence a homeomorphism. This proves that $K$ is metrizable in its weak topology, an instance of countable separating family metrizes a weakly compact set.

If $X$ is also a reflexive space, its closed unit ball is weakly compact and therefore metrizable by this construction. Compactness in a metric space implies sequential compactness, so every sequence in $B_X$ has a weakly convergent subsequence. Scaling proves the claim for every bounded sequence.

For the converse, assume every bounded sequence in $X$ has a weakly convergent subsequence. Use the same metric $d$ on $B_X$. A weak limit of points of $B_X$ remains in $B_X$: $|f(x)|\le 1$ for every $\Vert f\Vert \le 1$, and the Hahn-Banach theorem recovers $\Vert x\Vert \le 1$. Thus every sequence in $B_X$ has a subsequence convergent in $d$ to a point of $B_X$, so $(B_X,d)$ is sequentially compact and hence compact.

We must still recover the entire weak topology, rather than just the selected coordinates. Suppose $x_n\to x$ in $d$ but not weakly. There is a neighbourhood in the weak topology $U$ of $x$ and a subsequence outside $U$. By hypothesis that subsequence has a further subsequence converging weakly to $y\in B_X$. It then converges in $d$ to $y$, so uniqueness of metric limits gives $y=x$, contradicting eventual membership in $U$. Hence convergence in $d$ implies weak convergence. Every $f\in X^{\ast }$ is therefore sequentially continuous on $(B_X,d)$ and hence continuous, since the domain is a metric space. The weak topology and $d$-topology on $B_X$ coincide. The ball is weakly compact, so the criterion above provesThe zero space satisfies all statements directly. The converse did not assume completeness, nor use the full Eberlein-Šmulian theorem.

A unital C-star algebra is a complex Banach algebra $A$ with a multiplicative identity $1_A$ and a conjugate-linear involution $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A C-star isomorphism $\Phi :A\to B$ is a bijective complex-linear map preserving products and the involution. It carries $1_A$ to $1_B$ and is automatically isometric.

For a commutative unital C-star algebra, let $\Omega$ be its maximal ideal space, equivalently the space of Gelfand characters, with the weak-star topology. A Gelfand character has $\chi (1)=1$ and $\Vert \chi \Vert =1$. The equations $\chi (ab)=\chi (a)\chi (b)$ and $\chi (1)=1$ define a closed in the weak-star topology subset of the dual unit ball. The Banach-Alaoglu theorem makes $\Omega$ a compact Hausdorff space. Maximal ideals give characters by the Gelfand-Mazur theorem, and the Banach algebra spectrum satisfies $\sigma (a)=\{\chi (a):\chi \in \Omega \}$.

The Gelfand transform is $\hat{a}(\chi )=\chi (a)$. It preserves products, constants and the involution; the latter uses the fact that characters of a C-star algebra respect the involution. Every element of a commutative C-star algebra is normal, so the C-star identity and the spectral radius formula giveThus its image is a closed self-adjoint subalgebra of $C(\Omega )$, contains the constants, and separates characters. The complex Stone-Weierstrass theorem makes that image dense and therefore all of $C(\Omega )$. This is the Commutative Gelfand--Naimark theorem.

For the convergent sequence space $c$, coordinatewise multiplication satisfies $\Vert \alpha \beta {\Vert }_{\infty }\le \Vert \alpha {\Vert }_{\infty }\Vert \beta {\Vert }_{\infty }$, and coordinatewise complex conjugation is an involution. The constant sequence $1$ is the identity, andIt is complete: a uniform limit of convergent sequences is convergent, so $c$ is a closed linear subspace of the Banach space ${\ell }_{\infty }$. Hence it is a commutative unital C-star algebra.

TakeEvery finite point is isolated, and the neighbourhoods of $\infty$ contain all but finitely many integers. This is a compact Hausdorff space: an open cover has a set containing $\infty$, leaving only finitely many points to cover. DefineContinuity at $\infty$ is exactly convergence of the sequence. Thus $\Phi$ is onto $C(\Omega )$, and pointwise operations show it is a C-star isomorphism with $\Vert \Phi \alpha {\Vert }_{\infty }=\Vert \alpha {\Vert }_{\infty }$. Explicitly, these are all Gelfand characters: each $\chi (e_n)$ is $0$ or $1$ and at most one is $1$. If $\chi (e_n)=1$, multiplication by $e_n$ gives $\chi (\alpha )={\alpha }_n$. If all vanish, density of finitely supported sequences in l-p in $c_0$ gives $\chi {|}_{c_0}=0$, so $\chi (\alpha )=\lim {\alpha }_n$.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ making every continuous linear functional $f\in X^{\ast }$ continuous. A basic neighborhood of $x_0$ is specified by finitely many inequalities $|f_j(x-x_0)|<ϵ$.

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology on $X^{\ast }$ making every evaluation $f\mapsto f(x)$, for $x\in X$, continuous. Its basic neighborhoods use finitely many such evaluations. Nets converge in these topologies exactly when all the corresponding scalar evaluations converge.

Let $J_X:X\to X^{\ast \ast }$ be the canonical isometric embedding, $J_{X}x(f)=f(x)$. Goldstine theorem statesHere the balls are closed unit balls. Thus any bidual unit vector can be approximated in finitely many dual evaluations by unit vectors from $X$; this does not claim norm density or sequential density in complete generality.

For Banach spaces $E,F$ and $1\le p<\infty$, define the weak $p$-norm of a finite family byA bounded linear operator $T:E\to F$ is an absolutely p-summing operator if $({\sum }_i\Vert T{x}_i{\Vert }^p{)}^{1/p}\le C{w}_p(x_1,\dots ,x_m)$ for every finite family. The least constant is ${\pi }_p(T)$.

The Pietsch factorization theorem states that such an operator admits a probability measure $\mu$ on $K=B_{E^{\ast }}$ with its weak-star topology such thatEquivalently, let $i:E\to C(K)$ be the evaluation embedding $i(x)(\varphi )=\varphi (x)$ and $j_p:C(K)\to L^p(\mu )$ the natural map. There is a bounded linear operator $V:H\to F$, where $H=\overline{j_{p}i(E)}\subset L^p(\mu )$, such that $T=V{j}_{p}i$ and $\Vert V\Vert \le {\pi }_p(T)$. Conversely, any such factorization makes $T$ an absolutely p-summing operator, with ${\pi }_p(T)\le \Vert V\Vert$. The final operator is required on this closed subspace $H$; no extension to all of $L^p(\mu )$ is assumed.

Here is the existence proof. If $T=0$ the conclusion is immediate. Otherwise write $C={\pi }_p(T)>0$. The Banach-Alaoglu theorem makes $K$ compact, and $\varphi \mapsto |\varphi (x){|}^p$ is continuous for each $x\in E$. For finitely many vectors $x_1,\dots ,x_m$, putWe claim the compact convex hull of $v(K)$ meets $a+[0,\infty {)}^m$. If it did not, finite-dimensional Hahn-Banach separation theorem would give weights ${\lambda }_i\ge 0$ withThe weights are nonnegative because the second convex set is unbounded in every positive coordinate direction. This strict inequality contradicts the defining summing inequality applied to ${\lambda }_i^{1/p}{x}_i$. Thus a finite convex combination of the vectors $v(\varphi )$ lies coordinatewise above $a$. Equivalently, a finitely supported probability measure satisfies all these finitely many domination constraints.

By the Riesz-Markov-Kakutani representation theorem, probability measures on compact $K$ are the positive norm-one functionals on $C(K)$ taking the constant function to one. They form a compact set in the weak-star topology, by the Banach-Alaoglu theorem. Each domination constraint $\int |\varphi (x){|}^{p}d\mu \ge C^{-p}\Vert Tx{\Vert }^p$ is closed. The finite feasibility just proved gives the finite intersection property, hence one measure $\mu$ satisfies every constraint.

Define $V(j_{p}i(x))=Tx$. The domination inequality makes this well-defined and bounded by $C$, and completeness of $F$ extends it to $H$. Conversely, any factorization satisfiesThis proves both directions of the Pietsch factorization theorem, including the optimal factorization norm.

For the multiplication operator, let $f\in L^p(0,1)$ and take $g_1,\dots ,g_m\in C([0,1])$. Point evaluations are norm-one linear functionals on $C([0,1])$, soThus $M_f$ is an absolutely p-summing operator and ${\pi }_p(M_f)\le \Vert f{\Vert }_p$. Taking the single function $g=1$ gives the reverse inequality, soFor $f\ne 0$, the explicit domination measure on $[0,1]$ is $|f(t){|}^{p}dt/\Vert f{\Vert }_p^p$, pushed into $K$ by $t\mapsto {\delta }_t$. For $f=0$ everything is zero; no essential boundedness of $f$ is needed.

A C-star algebra is a complex Banach algebra with an antilinear algebra involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. First take a nonzero unital algebra. Its identity has norm one, since $\Vert 1{\Vert }^2=\Vert 1\Vert$. Alsoand exchanging $a,a^{\ast }$ proves $\Vert a^{\ast }\Vert =\Vert a\Vert$. Thus the algebra involution is continuous. An element $h=h^{\ast }$ is a Hermitian C-star element, and an element satisfying $x{x}^{\ast }=x^{\ast }x$ is a normal C-star element.

We establish the spectral facts needed for the representation, to avoid assuming the conclusion. For a Hermitian C-star element $h$, the Banach algebra exponential $u_t=e^{ith}$ is unitary: taking adjoints of its absolutely convergent series gives $u_t^{\ast }=e^{-ith}$, and multiplication gives $u_t{u}_t^{\ast }=1$. Hence $\Vert u_t\Vert =1$. If $\lambda \in \sigma (h)$, then $e^{it\lambda }\in \sigma (u_t)$. Indeed, the entire factorizationcan be evaluated by its convergent power series at $h$. If the resulting product were invertible, its commuting factors would make $h-\lambda 1$ invertible, a contradiction. The spectral bound therefore gives $|e^{it\lambda }|\le 1$ for every real $t$. Both signs of $t$ force $\mathrm{Im}\lambda =0$. Thus Hermitian elements have real spectrum.

For completeness, the spectral radius formula needed below follows from the same resolvent theory. Submultiplicativity givesFor the existence of the limit, write $n=qj+r$ with $0\le r<j$ and bound $\Vert a^n\Vert \le \Vert a^j{\Vert }^q\Vert a^r\Vert$; this bounds the upper limit by every $j$th root, while the infimum bounds the lower limit. If a power vanishes, the conclusion follows directly. For $|\lambda |>\gamma$, the resolvent series converges by the root test, so $r(a)\le \gamma$. Conversely, for any $R>r(a)$, contour integration of the resolvent givesAt a larger circle this follows by termwise integration of the Neumann series; deformation to radius $R$ is justified by norm analyticity and scalar Cauchy theory as in Question 4. Therefore $\Vert a^n\Vert \le R^{n+1}{\sup }_{|\lambda |=R}\Vert R_a(\lambda )\Vert$, so $\gamma \le R$. Letting $R\downarrow r(a)$ proves $r(a)=\gamma$.

For a normal C-star element $a$, its adjoint commutes with it. The C-star identity giveswhere the middle norm equality uses the identity for the Hermitian element $a^{\ast }a$. Every power is again normal, so $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral-radius formula yieldsThis derives the spectral radius norm equality for normal elements from the axioms.

Now let $A$ be commutative and unital. Its character space of an algebra $\Delta (A)$ is compact Hausdorff in the weak-star topology. Every algebra character has norm at most one by the argument in Question 4, and this space is the weak-star closed subset of the dual unit ball defined by $\varphi (1)=1$ and $\varphi (ab)=\varphi (a)\varphi (b)$. Compactness follows from the Banach-Alaoglu theorem; weak-star separation makes it Hausdorff.

We also need the full character-spectrum correspondence, not just one inclusion. Every proper ideal is contained in a maximal ideal. Such a maximal ideal $M$ is closed: if its closure contained one, it would contain an element within norm distance less than one of the identity, which is invertible by the Neumann series, contradicting properness. Its closure is therefore proper and maximality gives equality. The quotient $A/M$ is a complex Banach division algebra. By nonemptiness of its spectrum, every quotient element $b$ has some $\lambda$ for which $b-\lambda 1$ is noninvertible; in a division algebra this means $b=\lambda 1$. Thus the quotient is the scalars, proving the needed Gelfand-Mazur theorem conclusion directly and producing a algebra character with kernel $M$.

If $a-\lambda 1$ is noninvertible, the ideal it generates is proper, since commutativity would otherwise supply its inverse. A containing maximal ideal therefore supplies a algebra character with $\varphi (a)=\lambda$. The converse inclusion was already proved. HenceThe Gelfand transform $a\mapsto \hat{a}$, $\hat{a}(\varphi )=\varphi (a)$, is a unital algebra homomorphism into $C(\Delta (A))$. If $h=h^{\ast }$, the reality of its spectrum makes $\varphi (h)$ real. Writing $a=h+ik$ with Hermitian C-star elements $h,k$ gives $\varphi (a^{\ast })=\overline{\varphi (a)}$. Thus the transform respects the algebra involution. Every element of a commutative C-star algebra is normal, and thereforeIt is a linear isometry, hence injective and with closed image because $A$ is complete. The image contains constants, separates algebra characters by their definition, and is closed under complex conjugation. The complex Stone-Weierstrass theorem makes that image dense in $C(\Delta (A))$, and closedness makes it all of that space. We have proved the Commutative Gelfand--Naimark theorem:These steps show exactly how completeness, the algebra involution and the C-star norm identity enter the theorem. A general Banach algebra need not have an injective or norm-preserving Gelfand transform.

The general commutative theorem also has a nonunital form. Here is an elementary construction of the necessary C-star unitization, without assuming the representation theorem. Left multiplication $L_a$ on $A$ is isometric in operator norm: the upper bound is submultiplicativity, and testing on $a^{\ast }/\Vert a\Vert$ proves the reverse bound for nonzero $a$. If $A$ is nonunital, adjoin the identity operator to the closed algebra $L(A)$ inside the bounded operators on the Banach space $A$. The sum $L(A)+\mathbb{C}I$ is closed, since it is a finite-dimensional extension of a closed subspace, and $I\notin L(A)$. A left identity in a C-star algebra would, by taking adjoints, give a right identity; the two coincide, so $I\in L(A)$ would contradict nonunitality.

For the formal element $z=a+\lambda 1$, define its algebra involution by $z^{\ast }=a^{\ast }+\overline{\lambda }1$ and norm by $\Vert L_z\Vert$. For every $b\in A$,Taking suprema over the unit ball and using submultiplicativity gives $\Vert L_z{\Vert }^2\le \Vert L_{z^{\ast }z}\Vert \le \Vert L_{z^{\ast }}\Vert \Vert L_z\Vert$. Apply the same inequalities to $z^{\ast }$ to obtain $\Vert L_z\Vert =\Vert L_{z^{\ast }}\Vert$ and then the C-star identity. This is the C-star unitization by left multiplication. The original algebra is a closed ideal in it.

For commutative nonunital $A$, apply the proved unital theorem to $\tilde{A}$. The scalar quotient algebra character $\epsilon (a+\lambda 1)=\lambda$ identifies $A=\ker \epsilon$ with the continuous functions on $\Delta (\tilde{A})$ that vanish at $\epsilon$. Removing this one algebra character gives a locally compact Hausdorff space; those functions are exactly its continuous functions vanishing at infinity. Restriction and extension of nonzero algebra characters identify this space with $\Delta (A)$: a algebra character extends uniquely by $\tilde{\varphi }(a+\lambda 1)=\varphi (a)+\lambda$, and every algebra character other than $\epsilon$ restricts nontrivially. Thus the full nonunital conclusion is $A\cong C_0(\Delta (A))$.

Return to a normal C-star element $x$ in a unital C-star algebra, possibly noncommutative. Let $B=C^{\ast }(1,x)$, the closed unital star algebra generated by $x$ and $x^{\ast }$. It is commutative because the two generators commute. We first prove the spectral permanence for C-star algebras needed to use the ambient spectrum rather than a potentially larger subalgebra spectrum.

If $h=h^{\ast }\in B$, both its ambient and subalgebra spectra are real. The complement of the compact real set ${\sigma }_A(h)$ is connected. Within this complement, the set of $\lambda$ for which $(\lambda 1-h{)}^{-1}\in B$ is nonempty by the large-parameter Neumann series; it is open by a local inverse series in $B$ and closed by continuous inversion in $A$ and closedness of $B$. It is therefore the entire complement, proving ${\sigma }_B(h)={\sigma }_A(h)$. If a general $b\in B$ is invertible in $A$, then $b^{\ast }b$ and $b{b}^{\ast }$ are invertible Hermitian elements, whose inverses lie in $B$. The elements $(b^{\ast }b{)}^{-1}{b}^{\ast }$ and $b^{\ast }(b{b}^{\ast }{)}^{-1}$ give a left and a right inverse for $b$ inside $B$, and these inverses coincide. Applying this to every $b-\lambda 1$ proves ${\sigma }_B(b)={\sigma }_A(b)$ for all $b\in B$.

By the proved commutative theorem, $B\cong C(\Delta (B))$. The mapis onto by the character-spectrum correspondence and spectral permanence. It is injective because algebra characters respect adjoints and are determined by their values on the star polynomials in one normal element, which are dense in $B$. It is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Consequently the inverse Gelfand transform, under this identification, suppliesIt is a continuous unital star homomorphism and a linear isometry. Its image is exactly $B$. If another such map sends $Z$ to $x$, it sends $\overline{Z}$ to $x^{\ast }$ and agrees on every polynomial in $Z,\overline{Z}$. These functions contain constants, separate points of the compact spectrum and are closed under conjugation; Stone-Weierstrass gives their uniform density. Continuity proves uniqueness on all of $C({\sigma }_A(x))$. This is the requested continuous functional calculus, including its isometry and image assertion.

Two conventions in the printed statement need to be made explicit. A unital map into $A$ requires $A$ to have an identity. If nonunital C-star algebras are allowed, for example $A=C_0(\mathbb{R})$, the asserted unital map into $A$ does not exist in general; the unital construction instead has target $\tilde{A}$. Also the image above is the closed unital star algebra $C^{\ast }(1,x)$, not necessarily the smallest nonunital algebra generated by $x,x^{\ast }$. For $x=0$ in $A=\mathbb{C}$, the former is $\mathbb{C}1$ and the latter is $\{0\}$. Thus the intended assertion is correct if “generated” includes the identity; without that convention, this is a counterexample. The distinction is the unital versus nonunital generation by a normal element.

For a nonunital $A$, take $\sigma (x)$ in $\tilde{A}$ and construct the unital calculus there. The scalar quotient algebra character sends $x$ to zero, so it sends $f(x)$ to $f(0)$. Functions with $f(0)=0$ therefore give elements of $A$. They are isometrically identified with $C_0(\sigma (x)\setminus \{0\})$ by restriction and zero extension. Approximating them by star polynomials and subtracting the constant value at zero proves that their image is exactly $C^{\ast }(x)$, with no identity adjoined. This proves the nonunital continuous functional calculus as well as specifying the corrected form of the original claim.

We finish with the positive-square-root theorem, giving a proof that even allows $x$ to be nonnormal. Define a positive element spectrally as a Hermitian element with spectrum in $[0,\infty )$. The calculus of a Hermitian element shows its square is positive. Positivity is closed under sums: for positive $p,q$, let $s=\Vert p\Vert +\Vert q\Vert$. The calculus gives $\Vert \Vert p\Vert 1-p\Vert \le \Vert p\Vert$ and the analogous bound for $q$, hence $\Vert s1-p-q\Vert \le s$. Since $p+q$ has real spectrum, the spectral norm bound forces it into $[0,2s]$. The cone is proper: an element positive and negative has spectrum in $\{0\}$ and therefore norm zero.

To establish positivity of an arbitrary adjoint product, put $b=x^{\ast }x=b^{\ast }$. Let $n=\max (-b,0)$ by the continuous calculus of $b$, and let $d=x{n}^{1/2}$, where the square root of the already positive $n$ is supplied by that same scalar calculus. Multiplicativity givesThus $d^{\ast }d$ has nonpositive spectrum. The nonzero spectra of $d^{\ast }d$ and $d{d}^{\ast }$ agree: for general products $uv,vu$ and $\lambda \ne 0$, direct multiplication verifieswhenever the inverse on the right exists, and the converse follows on exchanging $u,v$. Hence $d{d}^{\ast }$ is also nonpositive. Write $d=u+iv$ with Hermitian C-star elements $u,v$. Thenis positive, as a sum of Hermitian squares, and nonpositive, as a sum of the two nonpositive elements. Properness makes it zero. Therefore $d^{\ast }d=-d{d}^{\ast }$ is both positive and nonpositive, hence zero. The C-star identity gives $d=0$, and $n^2=0$ then implies $n=0$. Thus $b=x^{\ast }x$ is positive. This proof does not use positivity of adjoint products in establishing an earlier step.

Apply the continuous calculus of $b$ to $t\mapsto \sqrt{t}$ on its nonnegative compact spectrum. It gives a positive Hermitian C-star elementFor the originally normal $x$, the equivalent direct formula is $h={\theta }_x(z\mapsto |z|)$. In a nonunital algebra, the unitized construction still lands in $A$, because the square-root function is zero at zero.

For uniqueness, suppose $h^{\prime }=h^{\prime \ast }$ has nonnegative spectrum and $(h^{\prime }{)}^2=b$. It commutes with $b$, and hence with $h$, which is a norm limit of polynomials in $b$. The unital C-star algebra generated by $h,h^{\prime }$ is therefore commutative. Its Gelfand transform, together with spectral permanence, gives two nonnegative continuous functions whose squares are equal. Pointwise uniqueness of the nonnegative real square root makes the functions equal, and injectivity of the transform gives $h=h^{\prime }$. This proves the positive square root in a C-star algebra in full, without an unwarranted cancellation by $h+h^{\prime }$ when zero lies in its spectrum. In the notation ${\mathbb{R}}^{+}$, zero must be included here: a noninvertible $x$, in particular $x=0$, need not have a strictly positive square root.

A nontangential limit at $\xi \in \partial \mathbb{D}$ restricts approach to a fixed Stolz region ${\Gamma }_A(\xi )=\{z\in \mathbb{D}:|z-\xi |<A(1-|z|)\}$, where $A>1$. Radial approach is a special case. The Fatou theorem for bounded holomorphic functions says that every $f$ in H-infinity space has a finite boundary value $f^{\ast }(\xi )$ for almost every $\xi$, with convergence through every fixed Stolz region and $|f^{\ast }|\le \Vert f{\Vert }_{\infty }$ almost everywhere.

Here is a proof outline. Write $f=u+iv$ with bounded real harmonic functions $u,v$. Take a sequence $R_j\uparrow 1$. The bounded functions $u(R_j\xi )$ have a subsequence converging in the weak-star topology of $L^{\infty }(\partial \mathbb{D})$ by the Banach-Alaoglu theorem and weak-star metrizability of the dual ball; its $L^1$ predual is separable. Denote the limit by $U$. For $|z|<R_j$, the Poisson integral on the unit disk givesThe Poisson kernels converge uniformly for this fixed interior point and the boundary functions stay uniformly bounded, so the weak-star topology permits passage to the limit:Repeat for $v$, obtaining $V$. Thus $f$ is the Poisson integral of $F=U+iV\in L^{\infty }$.

The Lebesgue differentiation theorem makes almost every boundary point a Lebesgue point of $U$ and $V$. The lemma that a Poisson integral converges nontangentially at Lebesgue points then gives $f(z)\to F({\xi }_0)$ in every fixed Stolz region. To see why, put $\delta =1-|z|$ and let $d(\xi ,{\xi }_0)$ be the shorter angular distance. In ${\Gamma }_A({\xi }_0)$ the kernel satisfiesSplit the circle into a central arc of radius comparable to $\delta$ and annuli with radii $2^k\delta$. For a Lebesgue point, averages of $|F-F({\xi }_0)|$ on sufficiently small arcs are arbitrarily small. The annular contributions are bounded by these averages times a summable geometric sequence $C_A{2}^{-k}$. On the remaining fixed distant arc the kernel is $O(\delta )$, so its contribution vanishes. This proves the claimed nontangential limit. The same Lebesgue points work for every integer aperture $A\ge 2$, hence for all finite apertures.

These boundary values recover the interior by the Poisson integral, so they determine $f$ uniquely. They satisfy $\Vert f^{\ast }{\Vert }_{L^{\infty }}=\Vert f{\Vert }_{\infty }$: one inequality follows by taking limits of the bounded interior values, and the other from positivity and unit integral of the Poisson kernel. The dominated convergence theorem also gives $f(r\xi )\to f^{\ast }(\xi )$ in every finite $L^p$ norm. Its negative Fourier coefficients vanish, because the Taylor series of $f$ has only nonnegative powers. This explains the boundary realization of analytic Hardy spaces.

The exceptional set cannot generally be removed. For example, $f(z)=\exp (i\mathrm{Log}(1-z))$ is bounded and holomorphic on the unit disc, using the holomorphic logarithm on the right half-plane: $|f(z)|=e^{-\arg (1-z)}$ lies between $e^{-\pi /2}$ and $e^{\pi /2}$. At the boundary point one its radial values $e^{i\log (1-r)}$ oscillate without a limit. Nor does the theorem assert convergence along arbitrary tangential paths or a continuous boundary extension. Its strength is almost-everywhere control on every fixed approach cone.

The Tychonoff theorem states that an arbitrary product of compact spaces, with the product topology, is compact; no Hausdorff hypothesis is necessary. We prove it using Zorn's lemma and the axiom of choice.

First, every proper filter on a set extends to an ultrafilter. Order its proper extensions by inclusion. The union of a chain is again a proper filter: any finite collection of its members occurs in one member of the chain, and the empty set occurs in none. Zorn's lemma gives a maximal proper filter $\mathcal{U}$. If $A\notin \mathcal{U}$, adjoining $A$ must destroy properness, so some $F\in \mathcal{U}$ has $F\cap A=\varnothing$, and hence $X\setminus A\in \mathcal{U}$. This proves the decision property of an ultrafilter.

Next prove the ultrafilter characterization of compactness. In a compact space the closed sets $\overline{F}$, for $F\in \mathcal{U}$, have the finite intersection property, so their intersection contains a point $x$. Every open neighbourhood $O$ of $x$ belongs to $\mathcal{U}$: otherwise its closed complement belongs to $\mathcal{U}$ and contains $x$, a contradiction. Thus the ultrafilter converges to $x$. Conversely, if an open cover has no finite subcover, the closed complements have the finite intersection property and generate a proper filter. Extend it to an ultrafilter. A limit point lies in some member $O$ of the cover, so both $O$ and its complement would belong to the ultrafilter. This is impossible. These arguments also prove the closed-set form of compactness directly by taking complements.

Now let $X={\prod }_{j\in J}{X}_j$. If a factor is empty, the product is empty and compact. Otherwise the axiom of choice makes $X$ nonempty. For an ultrafilter $\mathcal{U}$ on $X$, its coordinate imageis an ultrafilter. Compactness of $X_j$ supplies at least one limit $x_j$; use the axiom of choice to choose one for every $j$. A basic open neighbourhood of $x=(x_j{)}_j$ restricts finitely many coordinates. Each corresponding inverse image belongs to $\mathcal{U}$, so their finite intersection also does. Hence $\mathcal{U}$ converges to $x$ in the product topology. The ultrafilter criterion proves the theorem. The explicit choices and the maximal-filter argument are exactly the places where choice enters.

The Banach-Alaoglu theorem states that the closed unit ball of the continuous dual space $V^{\ast }$ of a real or complex normed vector space $V$ is compact in the weak-star topology $\sigma (V^{\ast },V)$. Completeness of $V$ is not required.

For each $v\in V$, let $D_v=\{z\in \mathbb{K}:|z|\le \Vert v\Vert \}$, a compact scalar disk or interval. By the Tychonoff theorem, $P={\prod }_{v\in V}{D}_v$ is compact. Inside $P$, impose the equationsEach equation defines a closed subset, since it concerns continuous coordinate projections into a Hausdorff scalar field. Their intersection $K$ is therefore compact. A point in $K$ defines a linear functional $\ell (v)=z_v$, and the coordinate bounds give $|\ell (v)|\le \Vert v\Vert$. Conversely, every $\ell \in V^{\ast }$ with $\Vert \ell \Vert \le 1$ gives such a point. Thus $K$ is exactly the image of the dual unit ball under $\ell \mapsto (\ell (v){)}_{v\in V}$. The induced product topology is precisely pointwise convergence on $V$, which is the weak-star topology. This proves