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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

A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

The Rosenthal l1 theorem states that every bounded sequence in a real or complex Banach space has either a weakly Cauchy subsequence or a subsequence equivalent to the unit vector basis of ${\ell }^1$. Equivalence means that there are $0<c\le C<\infty$ such that $c\sum |a_i|\le \Vert \sum a_i{x}_{n_i}\Vert \le C\sum |a_i|$ for every finite scalar combination.

We use the infinite Ramsey result that every closed subset of $[\mathbb{N}{]}^{\omega }$ is a Ramsey set of infinite subsets. It follows by taking complements in the open-set theorem proved in Question 6(a), and is also among the infinite Ramsey results permitted here.

Let $E$ be the closed span of the sequence and $K=B_{E^{\ast }}$ with its weak-star topology. It is compact by the Banach-Alaoglu theorem. The functions $f_n(\varphi )=\varphi (x_n)$ are continuous and uniformly bounded. Pointwise convergence of a subsequence on $K$ is precisely the weakly Cauchy conclusion.

Choose a countable collection of pairs $(Q_0,Q_1)$ of disjoint closed rational boxes in the scalar plane, or intervals for real scalars. Write $u,v$ for their centres, $d=|u-v|>0$, and require their diameters to be at most some $\eta <d/(4\pi )$. Include arbitrarily small such boxes, so any two distinct scalar cluster points lie in the interiors of one such pair. For a fixed pair, say an infinite set $M$ oscillates if some $\varphi \in K$ has $f_n(\varphi )\in Q_0$ and $f_n(\varphi )\in Q_1$ each for infinitely many $n\in M$.

There are two possibilities. If at some pair and some infinite $M$ every infinite subset oscillates, retain them. Otherwise enumerate the box pairs and successively thin so that no point oscillates for the current pair. A diagonal subsequence is eventually in each refined set, so has no point oscillating for any pair. A bounded scalar sequence with two cluster points would oscillate for a pair of boxes enclosing those points. Hence every scalar evaluation on that subsequence converges, giving the weakly Cauchy alternative.

In the remaining case put $A_n=f_n^{-1}(Q_0)$ and $B_n=f_n^{-1}(Q_1)$, closed subsets of $K$. ConsiderThis is closed. If the infinite intersection is empty, compactness gives an empty finite intersection, and every infinite sequence with the same sufficiently long initial segment also has an empty intersection. The complement is therefore open. Apply closed-set Ramsey to obtain an infinite $L\subseteq M$ homogeneous for $\mathcal{C}$.

It cannot be homogeneous outside $\mathcal{C}$: the retained oscillation property provides a point visiting both level sets infinitely often on $L$, from which we can select an alternating sequence belonging to $\mathcal{C}$. Thus every infinite subset of $L$ is in $\mathcal{C}$.

Pass to the even-indexed members $L^{\prime }=(l_2,l_4,\dots )$. The pairs $(A_n,B_n)$ for $n\in L^{\prime }$ are independent: every assignment of either $A$ or $B$ to finitely many chosen indices has a common point. Indeed insert zero or one unused members of $L$ before and between the specified indices to arrange their required odd or even positions, then complete to an infinite sequence. The gaps between even-indexed members supply these spare indices. Membership of that infinite sequence in $\mathcal{C}$ supplies the point realizing the finite assignment.

Take any finite complex coefficients $a_i$. Some unit scalar $\zeta$ satisfiesbecause the average over the unit circle is exactly the right side. Put ${\epsilon }_i=\mathrm{sign}\mathrm{Re}(\zeta a_i)$, choosing either sign at zero. Independence supplies functionals $\varphi ,\psi \in K$ such that $f_{n_i}(\varphi )$ lies near $u$ and $f_{n_i}(\psi )$ near $v$ when ${\epsilon }_i=1$, with the assignments reversed when ${\epsilon }_i=-1$. ThusAlso $|\sum {\epsilon }_i{a}_i|\ge \sum |\mathrm{Re}(\zeta a_i)|$. Since $\Vert \varphi -\psi \Vert \le 2$, we concludeThe coefficient is positive by the box choice, and boundedness of the original sequence gives the upper estimate. This proves the ${\ell }^1$ alternative over complex scalars. For real scalars simply partition by the signs of the coefficients; the same calculation works and gives the stronger lower constant $d/2-\eta$. This completes the proof.

LetThis scalar function is continuous. The first element is locally constant; deleting it is continuous on increasing sequences; and evaluation on a fixed $x_m$ is continuous in the weak-star topology. Thus $\mathcal{U}=\{M:|h(M)|<\delta \}$ is open.

By part (a), choose an infinite $L_0$ homogeneous for $\mathcal{U}$. If all its infinite subsets are in $\mathcal{U}$, the required conclusion follows. Suppose instead every one has $|h(M)|\ge \delta$. Choose finitely many unit scalars ${\zeta }_1,\dots ,{\zeta }_q$ such that every scalar $z$ with $|z|\ge \delta$ has $\mathrm{Re}({\zeta }_{j}z)>\delta /2$ for some $j$; for real scalars the two signs suffice. The open setscover $[L_0{]}^{\omega }$. Apply the open Ramsey set of infinite subsets result successively to them. After finitely many refinements, at least one, say ${\mathcal{U}}_j$, contains every infinite subset of a remaining infinite $L$; otherwise their cover would be violated.

For any finite $F\subseteq L$, choose an infinite $N\subseteq L$ after $\max F$. For each $m\in F$, apply this homogeneity to $M=\{m\}\cup N$. Its deleted tail is always the same $N$, so the same functional satisfiesBut $(x_m{)}_{m\in L}$ is a weakly null sequence. The Mazur lemma gives a finite convex combination $z={\sum }_{m\in F}{\lambda }_m{x}_m$ with $\Vert z\Vert <\delta /2$. The displayed inequalities give $\mathrm{Re}({\zeta }_j{x}_N^{\ast }(z))>\delta /2$, contradicting $\Vert x_N^{\ast }\Vert \le 1$. The bad homogeneous case is impossible. ThereforeContinuity is used to make the scalar tests open; weak nullity is used only for the finite convex-combination contradiction. The argument works over both real and complex scalars.

The Gelfand-Mazur theorem says that a nonzero complex unital Banach algebra in which every nonzero element is invertible is the complex field itself: $a\mapsto \lambda$ where $a=\lambda 1$. This is an isometric isomorphism under the usual normalization $\Vert 1\Vert =1$; without that normalization its norm is simply $\Vert \lambda 1\Vert =|\lambda |\Vert 1\Vert$. We first prove the required spectral and resolvent facts rather than assume them.

For any element $a$ of a complex unital Banach algebra $A$, define its spectrum of an element byIf $\Vert b\Vert <1$, the partial sums of ${\sum }_{n\ge 0}{b}^n$ converge in the complete algebra. Multiplying a partial sum by $1-b$ on either side gives $1-b^{N+1}\to 1$, proving the Neumann series inverse. For $|\lambda |>\Vert a\Vert$ this givesThus the spectrum is bounded. Under $\Vert 1\Vert =1$, $\Vert R(\lambda ,a)\Vert \le (|\lambda |-\Vert a\Vert {)}^{-1}$; for an unnormalized identity the same estimate with numerator $\Vert 1\Vert$ suffices. Also $\lambda R(\lambda ,a)\to 1$ as $|\lambda |\to \infty$.

If ${\lambda }_0$ is outside the spectrum and $R_0=R({\lambda }_0,a)$, then for $|h|\Vert R_0\Vert <1$,This follows by factoring $({\lambda }_0+h)1-a=({\lambda }_{0}1-a)(1+h{R}_0)$ and applying the just-proved series. The series converges locally in norm, proving that the resolvent set is open, the spectrum is closed, and the resolvent of an element is holomorphic. Subtracting inverse equations also gives the resolvent identityconsistent with the derivative $R^{\prime }(\lambda ,a)=-R(\lambda ,a{)}^2$ from the local series.

The spectrum is nonempty. If it were empty, the resolvent would be an entire algebra-valued function tending to zero at infinity. Choose a bounded linear functional $\ell :A\to \mathbb{C}$ with $\ell (1)=1$, extending the functional on $\mathbb{C}1$ by the complex Hahn-Banach theorem. Then $\ell (R(\lambda ,a))$ is an entire function, bounded outside a disk by the Neumann estimate and bounded inside by continuity and compactness. The Liouville theorem makes it constant, and its decay makes that constant zero. But $\lambda \ell (R(\lambda ,a))\to \ell (1)=1$, a contradiction. We have proved that every spectrum is a nonempty compact subset of the disk $|\lambda |\le \Vert a\Vert$.

Now assume every nonzero element of $A$ is invertible. For any $a$, choose $\lambda \in {\sigma }_A(a)$. The noninvertible element $a-\lambda 1$ must be zero. Thus every element is a scalar multiple of the identity, uniquely because $1\ne 0$, andNo commutativity assumption was needed for this part; it is forced by the conclusion. Complex scalars are essential to the spectral argument.

For the maximal ideal space and its correspondence with characters of an algebra, now take $A$ to be a commutative complex unital Banach algebra. Every proper ideal lies in an algebraic maximal ideal by Zorn's lemma: a chain's union is an ideal not containing $1$. Every proper ideal also has proper closure. Otherwise it contains an element $m$ with $\Vert 1-m\Vert <1$, which is invertible by the Neumann series, forcing the ideal to contain $1$. A maximal ideal $I$ is therefore closed, since its closure is a proper ideal containing it.

The quotient $A/I$, with its quotient norm, is a complex Banach algebra. Completeness follows from closedness of $I$, and submultiplicativity follows by taking infima over representatives. It is a division algebra: if $a\notin I$, maximality gives $I+Aa=A$, so $ba+i=1$ for some $b\in A$, $i\in I$. Hence $a+I$ is invertible. By the proved Gelfand-Mazur theorem, $A/I\cong \mathbb{C}$, with the identity mapping to $1$. Composing this isomorphism with the quotient map produces a nonzero multiplicative complex linear functional ${\varphi }_I$ with kernel $I$.

Conversely let $\varphi :A\to \mathbb{C}$ be a nonzero multiplicative complex linear functional, without assuming continuity. Nonzeroness and multiplicativity give $\varphi (1)=1$, hence $\varphi (\lambda 1)=\lambda$ and surjectivity onto $\mathbb{C}$. Its kernel is a maximal ideal. If $a-\varphi (a)1$ were invertible, applying $\varphi$ to its product with an inverse would give $0=1$. Therefore $\varphi (a)\in {\sigma }_A(a)$, soThis proves automatic continuity of characters; with normalized identity their operator norm is one. Finally a kernel determines its character uniquely: for every $a$, its class in $A/I=\mathbb{C}1$ is a unique scalar multiple of the unit, and every character with kernel $I$ must take that scalar value. We have obtained the bijectionThe commutativity in this correspondence cannot be dropped: ${\mathrm{Mat}}_2(\mathbb{C})$ has the maximal two-sided ideal $0$, but has no scalar-valued character, since such a character would have zero kernel and give an impossible injective linear map into $\mathbb{C}$.

For use in the next question, the spectrum equals character values in a commutative Banach algebra. We have proved one inclusion. For the other, if $a-\lambda 1$ is noninvertible, its generated ideal is proper, hence lies in a maximal ideal. The corresponding character takes $a$ to $\lambda$. ThusGive ${\mathcal{M}}_A$ the Gelfand topology, namely the weak-star topology of pointwise convergence on $A$ through these characters. It is Hausdorff and compact: the characters lie in the dual unit ball and form a weak-star closed subset, described by the closed equations $\varphi (1)=1$ and $\varphi (ab)=\varphi (a)\varphi (b)$. Apply the Banach-Alaoglu theorem. The Gelfand transform $a\mapsto \hat{a}$, where $\hat{a}(\varphi )=\varphi (a)$, is consequently a contractive unital algebra homomorphism into $C({\mathcal{M}}_A)$; it separates distinct characters. This argument proves compactness and continuity, but does not assert injectivity or isometry for an arbitrary Banach algebra.

Assume $X$ is nonempty, as is necessary for a probability measure to exist. Give its isometry group $I_X$ the uniform metricEvery distance-preserving self-map of a compact metric space is onto. Indeed, if $x_0\notin gX$, then $\delta =d(x_0,gX)>0$, and for $n>m$ the iterates satisfy $d(g^m{x}_0,g^n{x}_0)=d(x_0,g^{n-m}{x}_0)\ge \delta$. This infinite separated set contradicts compactness. Consequently $I_X$ is a group under composition, with the identity and inverses again isometries.

The metric $D$ is invariant under both left and right multiplication: left invariance uses preservation of distances, and right invariance uses surjectivity. Inversion is an isometry for $D$, and $D(gh,g^{\prime }{h}^{\prime })\le D(g,g^{\prime })+D(h,h^{\prime })$ proves continuity of multiplication. Moreover, $I_X$ is compact. A sequence of isometries has a subsequence converging at every point of a countable dense subset of $X$, by a diagonal argument. Their common Lipschitz constant $1$ and finite epsilon-nets of $X$ make this subsequence uniformly Cauchy. Its uniform limit preserves distances and is onto by the preceding argument. Thus $I_X$ is a compact metrizable topological group. A transitive group action means that for every $x,y\in X$ some $g\in I_X$ has $gx=y$.

We construct the invariant Borel probability measure directly. The Hall marriage theorem says that a finite bipartite graph with equal-sized vertex classes $A,B$ has a perfect matching precisely when every subset $S\subseteq A$ has at least $|S|$ neighbours in $B$. Its following consequence is useful. If $A,B$ are $\epsilon$-separated subsets of maximum cardinality in a compact metric space, join points whose distance is less than $\epsilon$. If some $S\subseteq A$ had fewer than $|S|$ neighbours, replacing those neighbours in $B$ by $S$ would produce a larger $\epsilon$-separated set. This contradicts maximal cardinality. Hence there is a bijection between $A$ and $B$ moving each point by less than $\epsilon$.

Choose ${\epsilon }_n\downarrow 0$ and maximum-cardinality ${\epsilon }_n$-separated sets $A_n\subset I_X$. Such sets exist: a finite covering by balls of radius less than ${\epsilon }_n/2$ bounds every separated set's cardinality. Fix $x_0\in X$ and define the empirical probability measuresFor $f\in C(X)$ let ${\omega }_f(r)={\sup }_{d(x,y)\le r}|f(x)-f(y)|$, which tends to zero with $r$ by uniform continuity. For every $h\in I_X$, left invariance of $D$ makes $h{A}_n$ another maximum-cardinality separated set. Match it to $A_n$ as above. ThenThe allowed compactness of probability measures on a compact metric space supplies a subsequence converging to $\mu$ in the weak-star topology. Passing to that limit for each fixed $h,f$ gives $\int fd\mu =\int f\circ hd\mu$. The same subsequence works for every $h,f$, since the displayed estimate held for all of them before taking the limit.

For uniqueness, put $F_n(x)=|A_n{|}^{-1}{\sum }_{a\in A_n}f(ax)$. For $x=h{x}_0$, right invariance of $D$ lets us match $A_{n}h$ with $A_n$. Transitivity therefore gives the uniform estimate $|F_n(x)-F_n(x_0)|\le {\omega }_f({\epsilon }_n)$ for every $x\in X$. If $\nu$ is any invariant Borel probability measure, thenTwo invariant probability measures consequently differ on the integral of $f$ by at most $2{\omega }_f({\epsilon }_n)$, hence agree on every continuous function. Continuous functions determine Borel probability measures on a compact metric space, so the invariant probability measure exists and is unique.

Fix $\epsilon >0$ and a finite epsilon-net $x_1,\dots ,x_N$ in $X$. Choose a Borel map $r:X\to \{x_1,\dots ,x_N\}$ with $d(x,r(x))\le \epsilon$, for example the nearest point with ties broken by index. Its cells $A_i$ give discrete probability measures with masses $p_i=P(A_i)$ and $q_i=Q(A_i)$. LetThe feasible set is nonempty and compact, so the minimum exists. Pushing any coupling forward by $(r,r)$ changes its cost by at most $2\epsilon$, hence $W_{\epsilon }\le W+2\epsilon$. Conversely, lift a discrete coupling by the measureTerms with zero row or column masses vanish. This lift has marginals $P,Q$, and its cost differs by at most $2\epsilon$. Therefore

Here is the finite linear programming duality proof. Write $d_{ij}=d(x_i,x_j)$ and consider the convex coneIt is closed and finitely generated. One elementary justification of closedness is to reduce each nonnegative representation to linearly independent generators, eliminating a coefficient along any linear dependence; after choosing a subsequence using the same subset of generators, their independent coordinates have bounded, convergent coefficients whenever the represented vectors converge. For $\eta >0$, the point $(p,q,W_{\epsilon }-\eta )$ lies outside $C$. Separation from this closed convex cone gives coefficients $(a,b,\tau )$ withIf $\tau =0$, summing the middle inequalities against any feasible $\pi$ contradicts the last one. Thus $\tau >0$, and $f_i=-a_i/\tau$, $g_j=-b_j/\tau$ obey $f_i+g_j\le d_{ij}$ with dual objective greater than $W_{\epsilon }-\eta$. The weak inequality already proved gives the reverse bound. Letting $\eta \downarrow 0$ proves finite strong duality.

Reduce this discrete dual to a single Lipschitz function. For a feasible pair put ${\varphi }_i={\min }_j(d_{ij}-g_j)$. The triangle inequality gives $|{\varphi }_i-{\varphi }_l|\le d_{il}$, while $f_i\le {\varphi }_i$ and $g_i\le -{\varphi }_i$. ConsequentlyConversely, any $1$-Lipschitz vector $\varphi$ gives a feasible pair $(\varphi ,-\varphi )$. Adding a constant leaves its objective unchanged, so impose ${\varphi }_1=0$. The resulting feasible set is compact, and its maximum is $W_{\epsilon }$.

For a maximizing vector, the McShane extension theorem has the following explicit proof in this situation. DefineThe triangle inequality proves that this extension is $1$-Lipschitz, and the discrete Lipschitz condition makes $\varphi (x_i)={\varphi }_i$. Thus $(\varphi ,-\varphi )$ is a continuous feasible pair. Because $|\varphi (x)-\varphi (r(x))|\le \epsilon$,Letting $\epsilon \downarrow 0$, together with part i, proves $m_d(P,Q)=W(P,Q)$. The infimum defining $W$ is also attained: the set of couplings is a closed subset of the compact space of probability measures on $X\times X$, and integrating the continuous cost is continuous for the weak-star topology.

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 topology needed here is the weak-star topology $\sigma (E^{\ast },E)$, meaning pointwise convergence on $E$. With the literal weak topology $\sigma (E^{\ast },E^{\ast \ast })$, the claimed compactness is false. For example, $E=c_0$ is separable and $E^{\ast }={\ell }^1$. Its unit vectors $e_n$ have no weak cluster point: every coordinate functional forces a cluster point to have all coordinates zero, while the bounded functional $x\mapsto {\sum }_j{x}_j$ has value one throughout. Thus the dual unit ball is not weakly compact. We prove the intended weak-star assertion in full.

Choose a countable dense set $(x_k)$ in the unit ball of $E$. On $E_1^{\ast }$ defineEach term is at most $2\cdot 2^{-k}$, so the series converges. The triangle inequality is immediate, and $d=0$ forces equality on the dense set and hence on all of $E$. It is therefore a metric. Convergence in this metric is equivalent to convergence on every $x_k$: one direction follows from any individual term, and the other by bounding the series tail uniformly. Because $\Vert \varphi \Vert ,\Vert \psi \Vert \le 1$, approximation of any unit vector by $x_k$ then gives convergence on all of $E$. The same finite-tail and approximation argument for neighborhoods proves equality with the weak-star topology, not just agreement of its convergent sequences.

Given a sequence $({\varphi }_n)$ in this ball, diagonal selection gives a subsequence converging at every $x_k$, since each coordinate lies in a compact disk. Uniform norm bounds then make it pointwise Cauchy at every $x\in E$. Define $\varphi (x)$ as the limit. Linearity passes to the limit, and $|\varphi (x)|\le \Vert x\Vert$, so $\varphi \in E_1^{\ast }$. The subsequence converges to it in the metric. Thus this metric space is sequentially compact, hence compact, proving

Use the same intended weak-star topology as in (b), and assume $C\ne \varnothing$. The induced operator is $T=S^{\ast }$, so $T\varphi (x)=\varphi (Sx)$; it is weak-star continuous because evaluation after applying $T$ is still evaluation at a vector of $E$.

Choose $\varphi \in C$ and form the averagesInvariance and convexity put every ${\varphi }_n$ in $C$. Weak-star closedness and (b) make $C$ compact, so some subsequence converges to $\psi \in C$. Telescoping givesbecause both orbit points lie in the dual unit ball. Weak-star continuity now gives $T\psi -\psi =0$. Hence $\boxed{T\text{ has a fixed point in }C.}$ This is the weak-star fixed point theorem for an adjoint operator; it does not require $T$ to be a contraction on the entire dual space.

If “weakly closed” is interpreted literally as closed for $\sigma (E^{\ast },E^{\ast \ast })$, the result can fail. Take $E=c_0$, let $S$ be the left shift and $T=S^{\ast }$ the right shift on ${\ell }^1$. The setis a nonempty weakly closed convex subset of the unit ball: its defining coordinate inequalities and sum equality are weakly closed. It is invariant under the right shift. But a fixed vector of that shift has first coordinate zero and then every coordinate zero, so none belongs to $C$. This supplies a counterexample to the literal topology and explains why the weak-star qualification is essential.

Fix Haar measure $m_G$ and use $\hat{f}(\chi )={\int }_{G}f(x)\overline{\chi (x)}d{m}_G(x)$. A positive-definite function $p$ satisfies ${\sum }_{i,j}{z}_i\overline{z_j}p(x_i-x_j)\ge 0$ for every finite choice of points and coefficients. The Bochner theorem states that a continuous positive-definite function on a locally compact Hausdorff Abelian group has a unique representationwhere ${\mu }_p$ is a Radon measure. Conversely, any finite positive Radon measure on the Pontryagin dual group has a continuous positive-definite inverse transform. The sign convention here is the one compatible with the displayed Fourier transform.

First note that the dual is a locally compact Hausdorff topological group. Its group operations are continuous in the compact-open topology proved in question 3. For local compactness, adjoin the zero functional to the character space of an algebra of $L^1(G)$. All its elements lie in the closed unit ball of $L^1(G{)}^{\ast }$, and the equations $\phi (a\ast b)=\phi (a)\phi (b)$ define a closed subset in the weak-star topology. The Banach-Alaoglu theorem makes that subset compact Hausdorff. Removing the zero functional gives an open, hence locally compact, subspace. This provides the setting for constructing the measure on the dual, without assuming a Fourier inversion theorem.

Let $u^{\ast }(x)=\overline{u(-x)}$ and let $\mathcal{P}$ be the finite sums of convolution squares $u\ast u^{\ast }$, $u\in C_c(G)$. Each such square is continuous with compact support, and is positive-definite becauseIts transform is $|\hat{u}{|}^2$, so every $p\in \mathcal{P}$ has $\hat{p}\ge 0$. Also $p\ast q\in \mathcal{P}$: convolution of two squares is $(u\ast v)\ast (u\ast v{)}^{\ast }$, and finite sums distribute. Insert the Bochner representation of $q$ into $p\ast q$ and use the Fubini theorem to obtainThe representing measure on the right is finite and positive. Uniqueness in the Bochner theorem and commutativity now give the Bochner consistency identity for convolution squares