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An extreme point $x$ of a convex set $K$ is one for which $x=(1-t)y+tz$, $y,z\in K$ and $0<t<1$ imply $y=z=x$. It is enough to test midpoint decompositions: a nontrivial interior point of a segment is itself the midpoint of a shorter nontrivial segment contained in that segment.

We prove the Krein-Milman theorem. Assume first that $K$ is nonempty. A face of a convex set $K$ means a convex subset $F$ with the property that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. The family of nonempty compact faces of $K$, including $K$ itself, has a minimal member by the Zorn lemma: every inclusion-decreasing chain has a nonempty intersection by compactness and the finite intersection property, and that intersection is again a compact face.

Let $M$ be such a minimal face. If it contains two distinct points, the continuous dual of a Hausdorff locally convex space separates them, by the permitted Hahn-Banach theorem. Choose a continuous real linear functional $\ell$ that is not constant on $M$. It attains a maximum on the compact set $M$, andis a nonempty proper compact face of $M$, and hence a face of $K$. Indeed, equality at the maximum in a convex combination forces both endpoint values to be maximal. This contradicts minimality. Thus $M$ is a singleton, and its point is extreme in $K$. The same argument applied within any nonempty compact face proves that every such face contains an extreme point of $K$.

Let $C=\overline{\mathrm{conv}}(\mathrm{ext}K)$. It is contained in $K$, since the latter is closed and convex, and extreme points exist when $K$ is nonempty. If $x_0\in K\setminus C$, the permitted separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_C\ell$. Let $F$ be its maximizer face on $K$. This nonempty compact face contains an extreme point $v$ of $K$, butcontradicting $v\in \mathrm{ext}K\subseteq C$. ThereforeFor empty $K$, the equality holds with the usual empty convex-hull convention. This proves both the extreme-point existence step and the closed-convex-hull step; no compactness of the set of extreme points itself is assumed.

The real absolutely summable sequence space is ${\ell }^1=\{x=(x_j):{\sum }_j|x_j|<\infty \}$ with norm $\Vert x{\Vert }_1={\sum }_j|x_j|$. The real l-infinity sequence space is ${\ell }^{\infty }=\{x=(x_j):{\sup }_j|x_j|<\infty \}$ with norm $\Vert x{\Vert }_{\infty }={\sup }_j|x_j|$. Their completeness is allowed as given; their extreme points still require proof.

For ${\ell }^1$, an interior point of the unit ball is not extreme because small opposite perturbations in one coordinate stay in the ball. At a boundary point with two nonzero coordinates $i,j$, choose $0<\epsilon <\min (|x_i|,|x_j|)$ and putThe two distinct vectors $x+z,x-z$ both have norm one and average to $x$, so $x$ is not extreme. A remaining boundary point must be $\pm e_j$. If $e_j=(y+z)/2$ with $\Vert y{\Vert }_1,\Vert z{\Vert }_1\le 1$, the $j$th coordinate forces $y_j=z_j=1$, exhausting both norm budgets and forcing all other coordinates to zero. The negative case is identical. ThusFor ${\ell }^{\infty }$, if $|x_j|<1$ in even one coordinate, small opposite changes to that coordinate give a nontrivial midpoint decomposition within the ball. If every $x_j=\pm 1$, a midpoint decomposition forces both summands to have that same value in every coordinate, so they equal $x$. HenceThese are precisely the extreme points of real sequence-space unit balls for the two spaces under consideration.

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.

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 linear operator between Banach spaces is a compact operator when it maps the unit ball into a relatively compact set, equivalently when every bounded sequence of inputs has an image subsequence converging in norm. Such an operator is bounded. For a bounded linear operator on a complex Banach space, the resolvent set consists of the complex $\lambda$ for which $T-\lambda I$ is bijective with bounded inverse; its complement is the spectrum. The point spectrum consists of those $\lambda$ for which $T-\lambda I$ has a nonzero kernel, namely the eigenvalues.

On a Hilbert space, $T$ is self-adjoint if $T=T^{\ast }$, or equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for all vectors. The given prescription extends uniquely toIt is bounded with norm one. The real diagonal entries prove self-adjointness by the inner-product identity. Define the finite-rank truncation $T_{m}x={\sum }_{j\le m}{x}_j{e}_j/j$. Then $\Vert T-T_m\Vert =1/(m+1)\to 0$. To see compactness directly, extract successively convergent subsequences of the finitely many first coordinates from a bounded input sequence; the uniform tail bound just displayed makes the resulting image subsequence Cauchy. Hence $T$ is compact and self-adjoint.

Each $1/j$ is an eigenvalue with eigenspace $\mathbb{C}{e}_j$. If $Tx=\lambda x$, each coordinate satisfies $(1/j-\lambda )x_j=0$, so there are no other eigenvalues; in particular zero is not an eigenvalue. But $\Vert T{e}_j\Vert =1/j\to 0$, which contradicts a bounded inverse at zero. Thus zero is in the spectrum.

For $\lambda \notin \{0,1,1/2,1/3,\dots \}$, its distance from this closed set is positive. The diagonal formulathen defines a bounded inverse. This proves the complete description

Two norms are Lipschitz equivalent norms if constants $c,C>0$ satisfy $c\Vert u{\Vert }_1\le \Vert u{\Vert }_2\le C\Vert u{\Vert }_1$ for every $u$. On the vector space of finitely supported sequences, the norms $\Vert x{\Vert }_1=\sum |x_i|$ and $\Vert x{\Vert }_{\infty }=\max |x_i|$ are not equivalent: a vector with $N$ entries equal to one has the two norms $N$ and one.

For dimension $d<\infty$, choose a basis and write $u=\sum x_i{e}_i$. Any norm satisfies $\Vert u\Vert \le \sum |x_i|\Vert e_i\Vert \le C|x{|}_2$. This also implies its continuity in the Euclidean coordinates, by $|\Vert u\Vert -\Vert v\Vert |\le \Vert u-v\Vert \le C|x-y{|}_2$. Its positive continuous values on the compact Euclidean unit sphere have a positive minimum $c$. Homogeneity gives $c|x{|}_2\le \Vert u\Vert$, proving equivalence with the Euclidean norm and hence between any two norms.

If a finite-dimensional subspace $U$ of a normed space contains a sequence converging in the ambient norm, its coefficient vectors are Cauchy by this equivalence. The coefficients converge in ${\mathbb{R}}^d$ or ${\mathbb{C}}^d$, and the reconstructed vector lies in $U$ and is the ambient limit. Thus finite-dimensional subspaces are closed.

A finite-dimensional normed space has nonempty open balls with compact closures by equivalence and Heine-Borel. Conversely let an open set $O$ have compact closure. Choose $w\in O$ and $r>0$ such that $B(w,r)\subset O$. Its closed ball of radius $r/2$ is a closed subset of the compact closure of $O$ and is therefore compact. Translation and scaling make the closed unit ball compact.

If the space were infinite-dimensional, construct unit vectors $x_j$ with $\mathrm{dist}(x_j,\mathrm{span}(x_1,\dots ,x_{j-1}))>1/2$. Here is the needed Riesz lemma: for a proper closed subspace $M$, choose $z\notin M$, put $d=\mathrm{dist}(z,M)>0$, and choose $m\in M$ with $\Vert z-m\Vert <2d$; then $(z-m)/\Vert z-m\Vert$ has distance greater than $1/2$ from $M$. The finite spans are closed by the earlier argument, so this constructs the sequence. Its members are separated by more than $1/2$, and have no convergent subsequence, contradicting compactness. Therefore the space is finite-dimensional exactly when it has a nonempty open set with compact closure.

A point $x$ of a set $S$ is an extreme point if $x=(1-t)y+tz$, with $y,z\in S$ and $0<t<1$, forces $y=z=x$. For a convex set it is enough to test midpoint decompositions: a nontrivial segment containing $x$ in its interior contains a smaller segment with midpoint $x$. The nonempty convex set ${\mathbb{R}}^2$ has no extreme points, since every $x$ is the midpoint of $x+e_1$ and $x-e_1$.

For the real l-infinity sequence space, a vector in the closed unit ball is extreme precisely when every coordinate is $1$ or $-1$. If $|x_j|<1$ for some $j$, choose $0<\delta \le 1-|x_j|$. The distinct vectors $x\pm \delta e_j$ remain in the unit ball and have midpoint $x$, so $x$ is not extreme. Conversely suppose all coordinates have modulus one and $x=(y+z)/2$, with $\Vert y{\Vert }_{\infty },\Vert z{\Vert }_{\infty }\le 1$. In each real interval $[-1,1]$, an endpoint can be a midpoint only if both entries equal that endpoint. Thus $y_j=z_j=x_j$ for every $j$ and $y=z=x$. Therefore

For the real absolutely summable sequence space, any point with $\Vert x{\Vert }_1<1$ is nonextreme: choose a nonzero coordinate perturbation of norm less than $1-\Vert x{\Vert }_1$. If $\Vert x{\Vert }_1=1$ and two coordinates $x_j,x_k$ are nonzero, take $0<\delta <\min (|x_j|,|x_k|)$ and putBoth perturbations preserve the signs of those two coordinates. One adds $\delta$ to one absolute value and subtracts it from the other, so $\Vert x+h{\Vert }_1=\Vert x-h{\Vert }_1=1$. Hence such an $x$ is not extreme. The only remaining candidates are $\pm e_j$. If $e_j=(y+z)/2$ with $\Vert y{\Vert }_1,\Vert z{\Vert }_1\le 1$, its $j$th coordinate forces $y_j=z_j=1$. That consumes the entire norm allowance, forcing all other coordinates to vanish. The argument for $-e_j$ is identical. Thus

For the real l2 sequence space, interior points of the unit ball again permit sufficiently small opposite perturbations and are not extreme. If $\Vert x{\Vert }_2=1$ and $x=(y+z)/2$, write $y=x+h$, $z=x-h$. The parallelogram law givesThe left side is at most one, so $h=0$ and $y=z=x$. ConsequentlyThese are the extreme points of real sequence-space unit balls: signs in ${\ell }^{\infty }$, coordinate spikes in ${\ell }^1$, and the whole sphere in ${\ell }^2$. The restriction to real scalars matters for the first description; the complex supremum ball has arbitrary unit-modulus coordinates instead.

For a $d$-dimensional normed vector space, with unit ball $K$ and John ellipsoid $\mathcal{E}$, its volume ratio isIt is at least $1$ and invariant under invertible linear coordinate changes.

For ${\ell }_1^n$, the unit ball has volume $2^n/n!$: its intersection with each orthant is the standard simplex of volume $1/n!$. Its John ellipsoid is $n^{-1/2}{B}_2^n$. To verify maximum volume, let $A{B}_2^n$ be any centered inscribed ellipsoid. For each sign vector $\epsilon \in \{-1,1{\}}^n$, containment gives $|A^T\epsilon |\le 1$. Averaging over independent Rademacher random variables yields $\mathrm{tr}(A{A}^T)\le 1$. The arithmetic-geometric mean inequality for its squared singular values gives $|\det A|\le n^{-n/2}$. The scalar matrix $n^{-1/2}I$ attains this bound because $\Vert x{\Vert }_1\le \sqrt{n}|x|$. Centered ellipsoids suffice: if $a+A{B}_2^n$ is contained in the symmetric convex unit ball, averaging it with $-a+A{B}_2^n$ gives the centered copy with the same volume.

Using the stated Euclidean ball volume and $n=2k$,since $(2k)!/k!={\prod }_{j=k+1}^{2k}j\ge k^k$. Therefore $\mathrm{vr}({\ell }_1^{2k})\le \sqrt{8/\pi }$ uniformly in $k$.

The subsequent assertion for an arbitrary $2k$-dimensional space is false as printed. Here is a counterexample independent of any asymptotic section theorem. Take $E={\ell }_{\infty }^{2k}$. Its John ellipsoid is $B_2^{2k}$: an inscribed $A{B}_2^{2k}$ has every row of $A$ of Euclidean length at most $1$, so the Hadamard inequality gives $|\det A|\le 1$, attained by $A=I$. Let $F$ be any $k$-dimensional subspace and $f_1,\dots ,f_k$ an orthonormal basis of it. The random unit vectorhas $|x|=1$. Since ${\sum }_j{f}_j(i{)}^2\le 1$ for each coordinate $i$, the moment-generating function bound for independent Rademacher random variables givesChoose $r=\sqrt{2\log (8k)/k}$; the latter probability is at most $1/2$, so some unit vector in $F$ has $\Vert x{\Vert }_{\infty }\le r$. The proposed section inequality would therefore forceThus even one half-dimensional section cannot have the printed universal constant.

The valid orthogonal splitting with bounded volume ratio is that $L$ may depend on $V=\mathrm{vr}(E)$, and $L=32V^2$ suffices. This supplies a dimension-independent constant for the preceding ${\ell }_1^{2k}$ case, or whenever $V$ is bounded uniformly. We prove the complete qualified assertion, including its epsilon-net ingredient.

In John position, $\Vert x{\Vert }_E\le |x|$. Polar integration of the radial function $\rho (u)=1/\Vert u{\Vert }_E$ giveswhere $\sigma$ is normalized spherical measure.

A maximal $\epsilon$-separated subset of $S^{k-1}$ is an epsilon-net. Balls of radius $\epsilon /2$ about its points have disjoint interiors and lie in the ball of radius $1+\epsilon /2$. Comparing volumes bounds its cardinality by $(1+2/\epsilon {)}^k$. Finite existence follows from this packing bound, and maximality proves the net property without assuming a separate covering theorem.

Choose two orthogonal $k$-dimensional subspaces $F_0,F_0^{\perp }$ and one such net on each unit sphere. Rotate both by a common uniformly distributed orthogonal matrix $U$. This distribution can be obtained from Question 1 applied to the compact orthogonal group with its Frobenius metric; left and right multiplication are transitive isometries. Consequently $Uv$ is uniformly distributed on $S^{2k-1}$ for every fixed unit vector $v$. The union bound shows that the probability that some rotated net point has norm less than $2\epsilon$ is at mostSet $\epsilon =(32V^2{)}^{-1}$. Because $V\ge 1$, the bracket is at most $1/4+1/256$, and twice its $k$th power is less than $1$ for every $k\ge 1$. Some rotation therefore makes every point of both nets have norm at least $2\epsilon$.

The triangle inequality and $\Vert x{\Vert }_E\le |x|$ imply that $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric. Every unit vector of $U{F}_0$ or $U{F}_0^{\perp }$ is within $\epsilon$ of a good net point and thus has norm at least $\epsilon$. By homogeneity, on both orthogonal subspaces,For ${\ell }_1^{2k}$ the first computation gives the absolute choice $L=256/\pi$. The missing bounded-volume ratio restriction is essential; the literal unrestricted assertion cannot be proved.

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.

For a discrete group, one standard definition of amenability is the fixed-point property: every continuous affine action on a nonempty compact convex set in a locally convex space has a fixed point. The equivalent state formulation, which may be stated without proof here, is that there is a left-invariant invariant mean on ${\ell }^{\infty }(\Gamma )$. Equivalently, every unital left-translation-invariant star-subalgebra of ${\ell }^{\infty }(\Gamma )$ admits a left-invariant state, with continuity for the supremum norm; one may equally pass to its norm closure. A state means a positive linear functional of norm one with value one on the constant function one. Restriction proves the subalgebra formulation from the full-algebra version, and taking the full algebra proves the converse. Such a mean is finitely additive on indicator functions, not necessarily countably additive.

Write $(\lambda (g)\mu )(h)=\mu (g^{-1}h)$ on ${\ell }^1(\Gamma )$ and $(L_{g}f)(h)=f(g^{-1}h)$ on bounded functions. We prove equivalence of this mean condition, the Reiter condition, and existence of a Følner sequence.

Assume first that an invariant mean $m$ exists. Fix a finite set $K\subseteq \Gamma$. In the real Banach space ${⨁}_{g\in K}{\ell }^1(\Gamma )$ with sum norm, consider the convex setIf zero were outside its norm closure, the Hahn-Banach separation theorem would give bounded real functions $f_g$ and $\delta >0$ such thatfor every such $\mu$. Testing point masses at $h$ showsApplying the positive invariant mean makes the left side zero and the right side at least $\delta$, a contradiction. Therefore zero is in the closure: for any $\epsilon >0$ a finitely supported probability measure makes the sum of the translation defects over $K$ less than $\epsilon$. Enumerate the countable group, take the first $n$ elements for $K$ and $\epsilon =1/n$, and obtain a sequence satisfying the Reiter condition. This proves (1) implies (2).

Next assume (2). An ${\ell }^1$ probability measure can be approximated in norm by finitely supported probability measures, by restricting to a large finite set and renormalizing. Replacing $\mu$ by $\nu$ changes each defect by at most $2\Vert \mu -\nu {\Vert }_1$. Thus simultaneous small defects for finitely many translations can be achieved with finite support.

For such $\nu$, set $F_t=\{h:\nu (h)>t\}$. The elementary identity $|r-s|={\int }_0^{\infty }|1_{r>t}-1_{s>t}|dt$ for $r,s\ge 0$ givesIf the sum of defects over $K$ is less than $\epsilon$, some $t$ with $F_t\ne \varnothing$ must satisfy ${\sum }_{g\in K}|F_t△g{F}_t|<\epsilon |F_t|$; otherwise integration would contradict that strict inequality. Applying this layer-cake extraction of Følner sets to successive finite sets of translations gives nonempty finite $F_n$ with each relative boundary tending to zero. Hence (2) implies (3).

Conversely, normalized counting measures ${\mu }_n=1_{F_n}/|F_n|$ satisfyso (3) implies (2). Finally, from (2) define states $m_n(f)={\sum }_h{\mu }_n(h)f(h)$. The dual unit ball of ${\ell }^{\infty }(\Gamma )$ is weak-star compact by the Banach-Alaoglu theorem; take a convergent subnet, since this compactness need not be metrizable. Its limit $m$ is positive with $m(1)=1$, andThus $m$ is invariant, proving (2) implies (1). Together these implications establish all three equivalences. Countability is used to obtain sequences from finite tests; it is not a justification for replacing the compactness subnet by a subsequence.

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.