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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.

Let $K$ be a compact Hausdorff space. A test subspace $H\subset C(K,\mathbb{R})$ containing a strictly positive function is sufficient for uniform convergence of positive linear operators on continuous functions if, for each $x\in K$, the only positive linear functional agreeing with point evaluation on $H$ is that evaluation itself. By the Riesz-Markov-Kakutani representation theorem, this is uniqueness of the positive measure with the prescribed moments. Boundedness from the positive test function and the Banach-Alaoglu theorem prove the assertion by compactness. When $1\in H$, nonnegative tests with a unique zero at each prescribed $x$ give a convenient sufficient condition.

The continuous functional calculus of a bounded normal operator $T$ is a representation of $C(K)$, with $K=\sigma (T)$. The Riesz-Markov-Kakutani representation theorem represents $f\mapsto ⟨f(T)v,v⟩$ by a positive regular measure ${\mu }_v$. The assignment $f\mapsto f(T)v$ is isometric for the $L^2({\mu }_v)$ norm, and extends to a unitary map onto the closed cyclic subspace it generates. Multiplication by $z$ and $\overline{z}$ represent $T$ and $T^{\ast }$. Orthogonal decomposition into cyclic reducing subspaces, allowing an uncountable family, constructs the Borel functional calculus for a normal operator without requiring separability.

On a compact metric space, a conservative Feller semigroup is a strongly continuous semigroup of positive unital operators on $C(X)$. The Riesz-Markov-Kakutani representation theorem associates a Markov kernel to each operator. On noncompact spaces the standard definition uses $C_0(X)$; conservativity then refers to the corresponding kernels having mass one, even when $1\notin C_0(X)$.

For a strongly continuous unitary representation, integration against $f\in L^1(G)$ defines an operator satisfying $\Vert \pi (f)\Vert \le \Vert f{\Vert }_1$. Fubini's theorem gives $\pi (f\ast g)=\pi (f)\pi (g)$, and the group-algebra involution gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. On an Abelian group, the operator-norm closure is a commutative C-star algebra. Its algebra characters produce continuous unitary group characters, allowing the Commutative Gelfand--Naimark theorem and Riesz-Markov-Kakutani representation theorem to turn a positive vector functional into a finite spectral measure.

For compact metric spaces and a continuous cost, define the marginal functional $\ell (f(x)+g(y))=\int fdP+\int gdQ$. Its majorant envelope $p(h)={\inf }_{s\ge h}\ell (s)$ is finite and sublinear. The dual value is $-p(-c)$. Assign that value to the cost direction and use the dominated Hahn-Banach theorem to obtain a positive linear functional on all continuous functions of two variables. The Riesz-Markov-Kakutani representation theorem turns it into an optimal transport plan with the prescribed marginals.

Let $K=\sigma (T)$ for a bounded normal operator on a nonzero complex Hilbert space. The zero Hilbert space has the trivial calculus. Take the inner product linear in its first argument. The continuous functional calculus is an isometric unital star-representation $\pi :C(K)\to \mathcal{B}(H)$ with $\pi (z)=T$. The Borel calculus will act on $B_b(K)$, the algebra of actual bounded Borel measurable functions. A quotient by null functions is introduced only after specifying the spectral measure; there is no preferred scalar measure such as plane area on $K$.

Here is a construction that also covers a nonseparable Hilbert space. For $v\in H$, the positive functional $f\mapsto ⟨\pi (f)v,v⟩$ has a finite regular Borel measure ${\mu }_v$ by the Riesz-Markov-Kakutani representation theorem. On continuous functions,Continuous functions are dense in $L^2(K,{\mu }_v)$. Hence $f\mapsto \pi (f)v$ extends to a unitary map from that space onto $H_v=\overline{\pi (C(K))v}$. This subspace reduces $T$ and $T^{\ast }$, which correspond to multiplication by $z$ and $\overline{z}$. Decompose $H$ as an orthogonal direct sum of such cyclic reducing subspaces, using maximality: a nonzero vector in the remaining orthogonal complement would generate another summand. The family need not be countable. This is the cyclic multiplication model for a normal operator.

On each summand define $f(T)$ to be multiplication by $f$, for $f\in B_b(K)$, and take their bounded direct sum. It defines a unital star-homomorphismThus $(fg)(T)=f(T)g(T)$, $\overline{f}(T)=f(T{)}^{\ast }$, and positivity of $f$ gives positivity of $f(T)$. The continuous functional calculus is recovered on $C(K)$.

Set $E(S)=\Phi (1_S)$ for a Borel subset $S$ of $K$. Then $E(S)$ is an orthogonal projection, $E(K)=I$, $E(S)E(R)=E(S\cap R)$, and disjoint countable unions add in the strong operator topology. Indeed on every multiplication summand the indicators add pointwise, and the tail squared norm is the measure of a decreasing sequence of sets tending to the empty set. Thus $E$ is a projection-valued measure, andFor $v,w\in H$, polarization gives scalar measures ${\mu }_{v,w}(S)=⟨E(S)v,w⟩$ satisfying $⟨f(T)v,w⟩=\int fd{\mu }_{v,w}$. Regularity of these finite scalar measures and uniqueness in the Riesz-Markov-Kakutani representation theorem make the projection-valued measure unique: its values on continuous functions determine all scalar measures and hence every projection. This also makes the calculus independent of the chosen cyclic decomposition.

The correct convergence assertion is bounded pointwise convergence in the strong operator topology. If $f_n\to f$ pointwise and ${\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$, thenby the dominated convergence theorem. This need not be convergence in operator norm. For multiplication by $t$ on $L^2[0,1]$, the functions $1_{[0,1/n]}$ converge pointwise to $1_{\{0\}}$, their operators converge strongly to zero, but their operator norms are all one.

A Borel measurable function can vanish spectrally without vanishing pointwise. Precisely, $f(T)=0$ if and only if $E(\{|f|>\epsilon \})=0$ for every $\epsilon >0$, andThe upper norm bound follows by integrating $|f{|}^2$ against vector measures; a unit vector in any nonzero projection $E(\{|f|>C\})$ gives $\Vert f(T)\Vert \ge C$. The induced map on the quotient by these spectral-null Borel measurable functions is isometric. Every nonempty relatively open subset of $K$ has nonzero spectral projection, since otherwise a nonzero continuous function supported there would contradict faithfulness of the continuous functional calculus. This full support gives the usual continuous spectral mapping theorem, but for a Borel measurable function the exact statement is insteadIf an indicated neighborhood has zero projection, define a bounded reciprocal of $f-\lambda$ off it and zero on it; the calculus supplies an inverse. Conversely unit vectors in the nonzero projections for arbitrarily small neighborhoods give $\Vert (f(T)-\lambda )v\Vert$ arbitrarily small, contradicting invertibility. This is spectral essential range in Borel functional calculus. For example $1_{\{0\}}(T)=0$ in the multiplication model on $[0,1]$, although the pointwise range of that function is $\{0,1\}$.

Spectral projections provide reducing subspaces, and $E(\{\lambda \})H=\ker (T-\lambda I)$: the squared norm of $(T-\lambda )v$ is $\int |z-\lambda {|}^{2}d{\mu }_v$. In finite dimension the calculus is simply $f(T)={\sum }_{j}f({\lambda }_j)P_j$, the sum over eigenspaces. In general it also permits discontinuous cuts and Borel choices of square roots, such as $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ with $s(0)=0$, giving a normal operator $s(T)$ with $s(T{)}^2=T$ even when a continuous scalar branch is unavailable.

Any bounded operator commuting with both $T$ and $T^{\ast }$ commutes with their continuous functional calculus and with all $E(S)$. To justify the latter, regularity approximates $1_S$ by uniformly bounded continuous functions in the $L^2$ norms of any finite list of vector measures; the resulting net converges strongly to $E(S)$. Thus the Borel calculus lies in the Von Neumann algebra generated by $T$. On a separable Hilbert space, a countable dense list of vectors gives a controlling measure $\mu ={\sum }_j{2}^{-j}{\mu }_{v_j}/(1+\Vert v_j{\Vert }^2)$; spectral-null sets are exactly its null sets, and the range is the represented $L^{\infty }(K,\mu )$, equal to that Von Neumann algebra. Without separability this last equality is not automatic: for the diagonal operator with one eigenvector for each $t\in [0,1]$, the generated Von Neumann algebra has arbitrary bounded diagonal functions, while the original Borel calculus has Borel diagonal functions. The spectral-measure construction and the strong convergence theorem themselves require no separability. This distinguishes the Borel functional calculus for a normal operator from its continuous predecessor without confusing pointwise functions, spectral equivalence classes and topological closure.

Put $K=\sigma (T)$ and use the continuous functional calculus $\pi :C(K)\to \mathcal{B}(H)$. It is a unital star-homomorphism taking the coordinate function to $T$. This follows from the commutative C-star theorem applied to $C^{\ast }(I,T,T^{\ast })$, together with spectral permanence for C-star subalgebras, which is among the permitted C-star results.

Take the Hilbert space inner product linear in its first variable. For a vector $\xi$, let $H_{\xi }=\overline{\{\pi (f)\xi :f\in C(K)\}}$. This cyclic subspace is invariant under every $\pi (f)$ and its adjoint, so it is reducing. The functional $f\mapsto ⟨\pi (f)\xi ,\xi ⟩$ is positive because $f\ge 0$ implies $\pi (f)=\pi (\sqrt{f}{)}^{\ast }\pi (\sqrt{f})$. It has, by the Riesz-Markov-Kakutani representation theorem, a finite regular positive Borel measure ${\mu }_{\xi }$ on $K$. The identitymakes $f\mapsto \pi (f)\xi$ an isometry on its quotient by the functions of zero $L^2$ norm. Density of continuous functions in $L^2({\mu }_{\xi })$ extends it to a unitary $U_{\xi }:L^2({\mu }_{\xi })\to H_{\xi }$. Multiplication by a continuous function intertwines with $\pi$ on this subspace.

Choose a maximal family of mutually orthogonal nonzero cyclic reducing subspaces. Their orthogonal sum is all of $H$: otherwise its reducing orthogonal complement contains a nonzero vector generating another such subspace. This argument does not require separability. For a bounded Borel measurable function $g$ on $K$, multiplication $M_g$ on every $L^2({\mu }_{\xi })$ is bounded with norm at most $\Vert g{\Vert }_{\infty }$. Define on that orthogonal sumMultiplication operators satisfy $M_{fg}=M_f{M}_g$, $M_{\overline{g}}=M_g^{\ast }$ and $M_1=I$. The same identities hold for their direct sums, and the norm bound is uniform in $\xi$. Thus ${\beta }_T$ is a norm-decreasing unital star-homomorphism on the algebra of actual bounded Borel measurable functions, and ${\beta }_T(Z)=T$ because it extends $\pi$. No quotient by a single unspecified measure is being substituted for that Borel algebra.

Choose the bounded Borel measurable square root $s(0)=0$ and $s(z)=|z{|}^{1/2}{e}^{i\mathrm{Arg}(z)/2}$ for $z\ne 0$, taking $\mathrm{Arg}\in [0,2\pi )$. Let $S={\beta }_T(s)$. ThenThus every bounded normal operator has a normal square root. The Borel branch is essential in general: on a circle winding about zero a continuous square-root branch need not exist.

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.

For a locally compact Hausdorff Abelian group $G$, Bochner's theorem states that a continuous positive-definite function $p$ has a unique representationwhere $\nu$ is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means ${\sum }_{i,j}{c}_i\overline{c_j}p(x_i{x}_j^{-1})\ge 0$ for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.

First prove existence without presupposing the representation. The elementary consequences of positive definiteness give $p(x^{-1})=\overline{p(x)}$, $p(e)\ge 0$ and $|p(x)|\le p(e)$. If $p(e)=0$, then $p=0$ and take $\nu =0$. Otherwise form finite formal linear combinations of symbols $[x]$, with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space $H$. Since $G$ is Abelian, translation $U_a[x]=[ax]$ preserves the form and extends to a unitary operator. Its inverse is $U_{a^{-1}}$, so $(U_a)$ is a unitary representation. With $v=[e]$,The translated copies of $v$ span a dense subspace. MoreoverFinite sums and density, together with $\Vert U_a\Vert =1$, give strong continuity on all of $H$. This constructs the cyclic unitary representation of a positive-definite function directly.

For $f\in L^1(G)$ define the integrated unitary representation $\pi (f)=\int f(x)U_{x}dx$, with the integral applied to each vector. Its norm is at most $\Vert f{\Vert }_1$. Fubini gives $\pi (f\ast g)=\pi (f)\pi (g)$, and inversion in the Abelian group gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions $u_V$ of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every $w\in H$. These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.

The Commutative Gelfand--Naimark theorem identifies $\mathcal{A}$ with $C_0(X)$, where $X$ is its character space. The positive vector functional $\omega (B)=⟨Bv,v⟩$ becomes a bounded positive functional on $C_0(X)$, so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure $\rho$ withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.

For each $\theta \in X$, the functional $f\mapsto \theta (\pi (f))$ is a nonzero algebra character of $L^1(G)$, since $\pi (L^1(G))$ is dense in $\mathcal{A}$. By Question 3 it has the form $\int f(x){\chi }_{\theta }(x)dx$. The resulting map $\iota :X\to \hat{G}$ is continuous: pointwise convergence of $\theta$ on $\mathcal{A}$ gives convergence on all $\pi (f)$, and the topological character identification in Question 3 gives compact-open convergence of ${\chi }_{\theta }$. Push $\rho$ forward under $\iota$ to obtain $\nu$. This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.

To recover $p(x)$, note that $U_x\pi (u_V)=\pi (T_x{u}_V)\in \mathcal{A}$. HenceThe left side tends to $p(x)$. On the right, $|{\Phi }_{\chi }(u_V)|\le 1$ and ${\Phi }_{\chi }(u_V)\to 1$ uniformly on compact subsets of $\hat{G}$. For completeness, the evaluation $(\chi ,x)\mapsto \chi (x)$ is jointly continuous: near $x_0$ restrict $x$ to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at $e$, proving the stated uniform convergence. Continuity of $\iota$ gives uniform convergence on compact subsets of $X$, and the finite Radon measure $\rho$ makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking $x=e$ also gives the required total mass $p(e)$.

Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in $x$ near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.

Finally suppose two finite Radon measures have the same transform. Multiply by $f\in L^1(G)$ and integrate over $G$. Fubini shows that the measures agree on every function $\chi \mapsto {\Phi }_{\chi }(f)$. Question 3 showed that these functions are uniformly dense in $C_0(\hat{G})$. Boundedness of both measure functionals extends equality to all of $C_0(\hat{G})$; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular $p(e)=1$ corresponds exactly to a probability measure on the dual group.

Give $C(G)$, the space of continuous functions on a compact space, its uniform norm. The left regular action is $l_{g}f(x)=f(g^{-1}x)$. The group law gives $l_g{l}_h=l_{gh}$ and $l_e=I$, and each translation is a linear isometry because $x\mapsto g^{-1}x$ is a bijection of $G$.

For fixed $f\in C(G)$, the map $(g,x)\mapsto f(g^{-1}x)$ is continuous. Compactness of the $x$-space makes this dependence uniform: for a given $g_0$ and $\epsilon >0$, continuity gives a neighbourhood of $g_0$ and a neighbourhood of each $x$ on which the difference from $f(g_0^{-1}x)$ is less than $\epsilon$; a finite cover of $G$ permits one common neighbourhood of $g_0$. Hence $\Vert l_{g}f-l_{g_0}f{\Vert }_{\infty }\to 0$. The estimatethen proves the joint continuity of the action on $G\times C(G)$.

For a left action on the continuous dual space, use the contragredient conventionIt satisfies $T_g{T}_h=T_{gh}$ and preserves the dual unit ball $B^{\prime }$. If the printed prime denotes the literal transpose $l_g^{\ast }\varphi =\varphi \circ l_g$, these maps instead satisfy $l_g^{\ast }{l}_h^{\ast }=l_{hg}^{\ast }$ and form a right action; replacing $g$ by $g^{-1}$ gives the left action above. Their invariant functionals are the same.

For $g_i\to g$ and ${\varphi }_i\to \varphi$ in the weak-star topology on $B^{\prime }$, every $f\in C(G)$ satisfiesThe first term uses strong continuity of translation and $\Vert {\varphi }_i\Vert \le 1$; the second uses weak-star convergence at one fixed function. This proves the requested joint weak-star topology continuity.

To obtain Haar measure, consider the set $K$ of positive linear functionals $\varphi$ on $C(G)$ with $\varphi (1)=1$. Such functionals have norm one: for complex $f$, choose a phase making $\varphi (f)$ real and nonnegative and use $\mathrm{Re}(e^{it}f)\le \Vert f{\Vert }_{\infty }$; this gives $|\varphi (f)|\le \Vert f{\Vert }_{\infty }$, with equality on the constant-one function. Thus $K\subseteq B^{\prime }$. Evaluation at the identity belongs to $K$, and positivity and normalization are weak-star closed conditions. The Banach-Alaoglu theorem makes $K$ a nonempty compact convex set. Each $T_g$ is a continuous affine self-map of it.

Here is the fixed-point step, including why no commutativity assumption on $G$ is needed. For any finite list $g_1,\dots ,g_n$, the Schauder-Tychonoff fixed-point theorem gives a fixed point $\varphi \in K$ of $n^{-1}{\sum }_j{T}_{g_j}$. Let $H$ be the closed subgroup they generate and, for real $f\in C(G)$, set $h(x)=\varphi (l_{x}f)$ on $H$. It is continuous and satisfiesAt a maximum point $x_0$, every summand must have the same maximum value. Iteration gives that value at all inverse-generator words times $x_0$. The closure of their positive-word monoid is a closed subsemigroup of a compact group, hence is the entire generated subgroup $H$. Density and continuity make $h$ constant on $H$. Therefore $\varphi$ is fixed by each of $T_{g_1},\dots ,T_{g_n}$, first on real functions and then on complex functions by linearity. The closed fixed-point sets consequently have the finite intersection property in $K$, so compactness gives a common fixed functional for every $g\in G$.

The Riesz-Markov-Kakutani representation theorem converts this positive functional into a regular Borel probability measure ${\mu }_l$ with $\varphi (f)=\int fd{\mu }_l$. Its invariance gives left translation invariance of the measure. It has full support: if a nonempty open set had measure zero, a finite collection of its left translates would cover the compact group and force ${\mu }_l(G)=0$, contradicting normalization. Thus it is a normalized left Haar measure. The right translation action gives a normalized right Haar measure ${\mu }_r$ by the same argument.

For uniqueness and equality, take any $f\in C(G)$ and apply the Fubini theorem toFor fixed $y$, left invariance of ${\mu }_l$ makes the inner integral $\int fd{\mu }_l$. Reversing the order, for fixed $x$, right invariance of ${\mu }_r$ makes the inner integral $\int fd{\mu }_r$. Since both total masses are one,Uniqueness in the Riesz-Markov-Kakutani representation theorem gives $\boxed{{\mu }_l={\mu }_r}$. Comparing any normalized left Haar measure to this fixed right Haar measure proves uniqueness as well. The conclusion is uniqueness of normalized Haar measure. Without normalization, the printed uniqueness claim is false: even on the one-point group, ${\delta }_e$ and $2{\delta }_e$ are distinct Haar measures. In general the result is uniqueness up to a positive scalar, and left and right probability normalizations coincide.

The discrete Bochner–Herglotz theorem says that a sequence $r:\mathbb{Z}\to \mathbb{C}$ is a positive-definite function, meaningif and only if there is a unique finite positive Borel measure $\sigma$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ such thatThe mass is $\sigma (\mathbb{T})=r(0)$. The necessity of positive definiteness follows by integrating $|{\sum }_j{c}_j{e}^{2\pi ijt}{|}^2$.

For sufficiency, positive definiteness makes the relevant Gram matrices Hermitian positive semidefinite, so $r(-n)=\overline{r(n)}$ and $|r(n)|\le r(0)$. DefineTaking $c_j=e^{-2\pi ijt}$ in the positive-definiteness condition gives $p_N(t)\ge 0$, while ${\int }_{\mathbb{T}}{p}_N(t)dt=r(0)$. Moreover,Thus the integrals against every trigonometric polynomial converge. These polynomials are uniformly dense in continuous functions on the circle: smooth periodic approximation followed by Fourier inversion for smooth functions gives this density. The uniform mass bound $r(0)$ then makes the integrals against every continuous $\phi$ converge as well. Their limit is a positive linear functional $L$ with $|L(\phi )|\le r(0)\Vert \phi {\Vert }_{\infty }$. The standard Riesz-Markov-Kakutani representation theorem gives a finite positive Borel measure $\sigma$ with $L(\phi )=\int \phi d\sigma$. The preceding coefficient computation proves its required Fourier coefficients. Finally, two such measures agree on all trigonometric polynomials, hence on all continuous functions by uniform density, and therefore are identical. This proves the Bochner–Herglotz theorem, including existence and uniqueness; if $r(0)=0$, the estimate $|r(n)|\le r(0)$ gives the zero measure directly.

Now let $U$ be the Koopman operator $Uf=f\circ T$ on the complex L2 space of the probability measure-preserving system. We use $⟨f,g⟩=\int f\overline{g}d\mu$, linear in the first entry. Since $U$ is an isometry, the sequence $r_f(n)=⟨U^{n}f,f⟩$ for $n\ge 0$, extended by $r_f(-n)=\overline{r_f(n)}$, is positive-definite: its matrix is the Gram matrix of $f,Uf,\dots$. The Bochner–Herglotz theorem gives its spectral measure of a Koopman observable ${\sigma }_f$.

We first prove exactly how an atom of a measure detects an eigenfunction. Fix $\zeta =e^{2\pi i{t}_0}$ and setThe mean ergodic theorem, applied to the isometry $V={\zeta }^{-1}U$, says these averages converge in L2 space to the orthogonal projection $P_{\zeta }f$ onto $\ker (U-\zeta I)$. Its usual Hilbert-space proof works without invertibility: the orthogonal complement of $\overline{\mathrm{ran}(I-V)}$ is $\ker (I-V^{\ast })=\ker (I-V)$, and the averages of $(I-V)h$ telescope to $(h-V^{N}h)/N\to 0$. The equality of the two kernels follows from $V$ being an isometry, by expanding $\Vert Vh-h{\Vert }^2$ when $V^{\ast }h=h$.

Expanding the averaged norm through the spectral measure of a Koopman observable givesThe integrand is at most one and converges to $1_{\{t_0\}}$. By the dominated convergence theorem,Thus a nonzero atom corresponds to a nonzero Koopman operator eigenfunction, with eigenvalue $\zeta$, in the cyclic spectral data of $f$.

We also need the Wiener Fourier coefficient lemma, which here has a short proof. For any finite positive circle measure $\sigma$ with coefficients $r(n)$,The bracket is bounded in absolute value by one and tends to the indicator of the diagonal $t=s$. Another application of the dominated convergence theorem givesIn particular the limit is zero for an atomless measure.

Suppose there are no nonconstant Koopman operator eigenfunctions. For every $f$ in the mean-zero L2 space, all its projections $P_{\zeta }f$ vanish: a nonzero projection would be a nonconstant eigenfunction, since it has mean zero. Hence ${\sigma }_f$ is atomless, and the Wiener Fourier coefficient lemma givesThe Cauchy-Schwarz inequality also gives vanishing averaged absolute autocorrelations. For two mean-zero observables $f,g$, the sesquilinear polarization identitythen gives vanishing averaged absolute cross-correlations. Taking $f=1_A-\mu (A)$ and $g=1_B-\mu (B)$ proveswhich is weak mixing.

Conversely, this set formulation of weak mixing extends to all pairs of mean-zero L2 space observables by finite-simple-function approximation and the uniform estimate $|⟨U^{n}f,g⟩|\le \Vert f{\Vert }_2\Vert g{\Vert }_2$. If $Uh=\zeta h$ and $h\ne 0$, the isometry property gives $|\zeta |=1$. If $\zeta \ne 1$, invariance of the integral gives $\int h=0$. If $\zeta =1$ and $h$ is nonconstant, replace it by $h-\int h$, a nonzero mean-zero eigenfunction. But in either casefor every $n$, contradicting weak mixing. We have proved weak mixing is equivalent to absence of nonconstant eigenfunctions, including for noninvertible transformations.

Finally, the doubling map $Tx=2x(\mathrm{mod}1)$ preserves normalized Lebesgue measure, as substitution on its two inverse branches shows. Let $f\in L^2(\mathbb{T})$ be an eigenfunction, $f(2x)=\zeta f(x)$, with Fourier coefficients $c_k$. The Fourier basis calculation givesSince $\zeta \ne 0$, all odd coefficients vanish. Every nonzero integer is $2^{r}m$ with $m$ odd, and repeated use of $\zeta c_{2^{r}m}=c_{2^{r-1}m}$ makes all nonzero coefficients vanish. Completeness of the Fourier basis then says $f$ is constant. Therefore the doubling map has no nonconstant eigenfunctions, and

We construct a normalized positive translation-invariant functional on $C(G)$ and then represent it by a measure. The main step is a finite matching in a graph argument, for which we establish all the covering-net properties.

Let $V$ be an open neighbourhood of the identity in the compact topological group $G$. The sets $xV$, $x\in G$, form an open cover. Compactness supplies a finite set $F$ with $G=FV$. Among all such finite sets choose one of smallest cardinality, denoted $m(V)$. This is a minimal left covering net of a compact group. If $g\in G$, then $G=gFV$, and $gF$ has the same cardinality. Hence $gF$ is also minimal for the same $V$.

Let $F,F^{\prime }$ be any two minimal left covering nets of a compact group for $V$. They both have cardinality $m(V)$. Form a bipartite graph by joining $x\in F$ to $y\in F^{\prime }$ whenever $xV\cap yV\ne \varnothing$. For a subset $S\subseteq F$, let $N(S)$ be its set of neighbours. Every point of $SV$ lies in some $yV$ with $y\in F^{\prime }$; that $y$ is a neighbour of a member of $S$. Thereforeand $(F\setminus S)\cup N(S)$ is still a left covering set. By minimality,Thus $|N(S)|\ge |S|$. Hall's marriage theorem gives a bijection $\pi :F\to F^{\prime }$ with $xV\cap \pi (x)V\ne \varnothing$ for every $x$. If $xv=\pi (x)v^{\prime }$ at an intersection point, thenThis proves the matching minimal covering nets of a compact group property, including the estimate needed below.

For a continuous complex-valued function $f$ on $G$, defineContinuity and compactness imply ${\omega }_f(W)\to 0$ as identity neighbourhoods $W$ shrink. Here is the uniformity argument: continuity of $(x,w)\mapsto f(xw)-f(x)$ at each $(x,e)$ gives a product neighbourhood where its absolute value is small; finitely many of the first-coordinate neighbourhoods cover $G$, and intersecting the corresponding second-coordinate neighbourhoods gives one $W$ that works for every $x$.

Matched pairs therefore giveIn particular, with $F^{\prime }=gF$,The estimate holds for every $g\in G$; no commutativity of $G$ is being assumed.

Direct the open identity neighbourhoods by reverse inclusion and choose a minimal left covering net of a compact group $F_V$ for each. The values $I_{F_V}(f)$ lie in the closed complex disk of radius $\Vert f{\Vert }_{\infty }$. The Tychonoff theorem makes the product of these disks over all $f\in C(G)$ compact. Consequently the net of vectors $(I_{F_V}(f){)}_{f\in C(G)}$ has a convergent subnet. Call its coordinatewise limit $I(f)$. Since each finite average is linear, positive on nonnegative real functions, and equal to one at the constant function one, the limit satisfiesand $|I(f)|\le \Vert f{\Vert }_{\infty }$. Thus $I$ is a bounded positive linear functional.

Given an identity neighbourhood $W$, continuity of multiplication and inversion supplies a sufficiently small $V$ with $V{V}^{-1}\subseteq W$. The translation estimate consequently tends to zero along the subnet. For every $f\in C(G)$ and every $g\in G$,The Riesz-Markov-Kakutani representation theorem now represents $I$ by a regular positive Borel measure $\mu$ with $\mu (G)=1$. The preceding equality and uniqueness in that representation imply that the pushforward under every left translation equals $\mu$. HenceThis is the required normalized Haar measure. It is nonzero because its total mass is one. In fact every nonempty open set has positive measure: its left translates cover $G$, a finite subcover exists, and a zero measure for that open set would force $\mu (G)=0$.

For completeness, the constructed probability measure is also right invariant. Apply Fubini's theorem to $f(x^{-1}y)$ with both variables distributed according to $\mu$. Integrating first in $y$ gives $\int fd\mu$ by left invariance. Integrating first in $x$ and substituting $x=yz$ gives $\int f(z^{-1})d\mu (z)$. Thus $\mu$ is invariant under inversion. Inversion turns a right translation into a left translation, so right invariance follows. This last observation is not needed for the existence of a left Haar measure, but confirms the usual two-sided normalization for compact groups.

We prove Kantorovich duality theorem here by a positive-functional extension argument; it produces the minimizing transport plan at the same time as equality of the values. Let $Y=X\times X$ and $c(x,y)=d(x,y)$. This is a bounded continuous function because $X$ is a compact metric space. Write $C(Y)$ for the real space of continuous functions on a compact space, and letOn this vector subspace define $\ell (s)=\int fd\mathbf{P}+\int gd\mathbf{Q}$. It is well-defined: two representations differ by a constant in the first variable and the opposite constant in the second, and both measures have mass one. It is positive, since $s\ge 0$ implies $\min f+\min g\ge 0$ and hence $\ell (s)\ge 0$. Also $\ell (1)=1$.