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Use the Lagrange multiplier with the sign conventionFor every $\lambda >0$, this is a strictly convex quadratic in $x$, with unique global minimizerThese formulas follow by setting the three partial derivatives of the optimization Lagrangian to zero; its Hessian is $\mathrm{diag}(2\lambda ,2\lambda ,4)$, which is positive-definite.

Choose $\lambda >0$ to make this minimizer feasible. Its constraint value is $5/(2{\lambda }^2)-\lambda /4$, so the scalar equation isThe left side of the first equation is strictly decreasing on $(0,\infty )$, with limits $+\infty$ and $-\infty$. Thus exactly one positive multiplier exists and can be obtained by a bracketed scalar root search. The Lagrangian sufficiency theorem proves that its associated feasible point is the global optimum; strict Lagrangian convexity makes that optimum unique.

Consequently the unique solution isIts objective value is $-5/\lambda +{\lambda }^2/8$. This scalar multiplier certificate for a quadratic equality constraint does not require finding the multiplier in radicals, and does not mistake the nonconvex equality surface for a convex set.

First, equality of two $U_0$ values forces equality of the corresponding $V_0$ values: neither strict comparison holds for $U_0$, and the same is then true for $V_0$ in both directions. Thus $V_0=g\circ U_0$ for a well-defined strictly increasing function $g$ on $J=U_0(\mathcal{P})$. Since $\mathcal{P}$ is a convex set and $U_0,V_0$ are affine functions, $J$ is a real interval andIf $J$ contains $u<v$, let $a=[g(v)-g(u)]/(v-u)>0$ and $b=g(u)-au$. The mixture identity gives $g(w)=aw+b$ for $w\in [u,v]$. For $w>v$, express $v$ as a convex combination of $u$ and $w$ and solve the same identity for $g(w)$; for $w<u$, express $u$ as a mixture of $w$ and $v$. Hence the formula holds on all of $J$, not merely between the chosen anchors. We obtain positive affine uniqueness of affine preference representations:If $U_0$ is constant, the same comparisons force $V_0$ to be constant; choose $a=1$ and the appropriate $b$. If the probability-measure set is empty, the assertion is vacuous. These degenerate cases do not require distinct anchors.

Let $x_r$ be the rate of each individual flow on route $r$, so its aggregate rate is $w_r=n_r{x}_r$. The feasible allocations satisfy ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. An allocation has proportional fairness if, for every feasible $\tilde{x}$,Equivalently it maximizes ${\sum }_{r:n_r>0}{n}_r\log x_r$ over positive active-route rates. This is the first-order optimality condition for a concave objective on a convex set. Rates for absent flows can be set to zero.

For the linear flow network, put $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. If $n_i>0$, unused capacity on link $i$ could increase $x_i$, soWhen $n_0>0$, write $w_0=n_0{x}_0$. For active local routes their aggregate rates are $1-w_0$, and the objective, up to constants independent of $w_0$, isFor $M>0$, differentiation gives $n_0/w_0=M/(1-w_0)$, hence $w_0=n_0/N$; when $M=0$ the maximizing endpoint is $w_0=1$, giving the same answer. Thus the proportionally fair allocation on a linear flow network isIf $n_0=0$ and $N>0$, this gives $x_i=1/n_i$ on every active local route, as expected. No departure occurs at the empty state.

Independent Poisson processes of flow arrivals and independent exponential distributions of document sizes make the count vector a continuous-time Markov chain. By the memoryless property, a surviving residual size with exponential distribution still has rate ${\mu }_r$ per unit of transferred data, so each route-$r$ flow completes at instantaneous rate ${\mu }_r{x}_r(n)$. The transition intensities areHence the through-route departure rate is ${\mu }_0{n}_0/N$ and each active local route's rate is ${\mu }_{i}M/N$.

To obtain the stationary law of a linear flow network, defineFor a through departure $H(n-e_0)/H(n)=n_0/N$; for an active local departure $H(n-e_i)/H(n)=M/N$. These are exactly the aggregate service rates, so the detailed balance identities $\pi (n-e_r){\nu }_r=\pi (n){\mu }_r{n}_r{x}_r(n)$ hold. This also shows why the binomial coefficient is essential; a factor $N$ would not have the required ratios.

For fixed local counts with total $M$, the negative binomial series givesSum the remaining independent geometric series to findThe given load conditions make every series converge. Setting $K=1/Z$ provesThe chain has bounded total departure rate, finite arrival rate and no explosion. With positive arrival rates it is irreducible, and this normalized invariant law gives the positive recurrent Markov chain property; zero arrival rates restrict the stationary support to the corresponding reachable class.

Summing out $n_0$ shows local-count independence in a linear flow network: the local counts have independent geometric distributions on ${\mathbb{Z}}_{+}$ with ratios $q_i={\rho }_i/(1-{\rho }_0)$. ThereforeThey are independent of one another under the stationary marginal, despite sharing the random through-flow population. The through count is generally dependent on them.

True, with the usual normalization $v(\varnothing )=0$. For a convex cooperative game, the supermodular inequality implies increasing marginal contributions: if $A\subseteq B$ and $i\notin B$, apply it to $A\cup \{i\}$ and $B$ to obtainFix an ordering $\pi$ and let $P_i$ be the set of players before $i$. Its marginal contribution vector is $m_i^{\pi }=v(P_i\cup \{i\})-v(P_i)$. Summing in order telescopes to ${\sum }_i{m}_i^{\pi }=v(N)$. For any coalition $S$, $S\cap P_i\subseteq P_i$, so increasing marginals giveThese are exactly the efficiency and coalition constraints of the core of a cooperative game. Thus every marginal contribution vector is in the core. The core is a convex set, being an intersection of linear half-spaces and an efficiency hyperplane. The Shapley value is the average of the marginal contribution vectors over all orderings, so it too lies in the core. This proves Shapley value belongs to the core of a convex game, without needing a separate existence theorem for the core.

The closest point theorem in a Hilbert space says that, for every nonempty closed convex set $C\subset H$ and $x\in H$, there is exactly one $p\in C$ minimizing $\Vert x-p\Vert$.

Put $d={\inf }_{y\in C}\Vert x-y\Vert$ and choose $y_n\in C$ with $\Vert x-y_n\Vert \to d$. The midpoint belongs to $C$ because it is a convex set. The parallelogram law givesThus $(y_n)$ is a Cauchy sequence. Completeness of the Hilbert space and closedness of $C$ give a limit $p\in C$, with $\Vert x-p\Vert =d$. Applying the same identity to two minimizers gives their squared distance at most zero, proving uniqueness.

The resulting projection is characterized byIndeed, differentiate $\Vert x-p-t(z-p){\Vert }^2$ at $t=0^{+}$; the minimum there gives the inequality. Conversely, expanding $\Vert x-z{\Vert }^2$ proves minimality from this inequality. For a closed linear subspace, both signs of each direction are allowed, so $x-p$ is orthogonal to that subspace: this recovers the orthogonal projection.

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

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

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

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

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

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

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

If $C={\mathbb{R}}^n$, the family of proper containing half-spaces is empty and its intersection is, by convention, ${\mathbb{R}}^n$. If $C=\varnothing$, every closed half-space contains it and their intersection is empty. Now suppose $C$ is a nonempty proper closed convex set. Take $x\notin C$ and let $z$ be its Euclidean projection onto a convex set. This projection exists: a minimizing sequence can be restricted to a bounded ball, and closedness gives attainment. It is unique by convexity and strict convexity of squared distance.

For $y\in C$, the segment $z+t(y-z)$ remains in $C$ for $0\le t\le 1$. Minimality at $t=0$ impliesThus the closed half-space $H_x=\{y:⟨x-z,y-z⟩\le 0\}$ contains $C$ but excludes $x$, since $⟨x-z,x-z⟩>0$. Every point outside $C$ is excluded by at least one containing half-space. The reverse inclusion is immediate, givingThis is the half-space representation of a closed convex set. The argument gives an explicit separating hyperplane rather than just citing a Hahn-Banach separation theorem.

Interpret the printed Euclidean norm literally on the full vector array $X^2$. It is a single global norm, not the sum of pixelwise gradient lengths in discrete isotropic total variation. Set $A=\nabla$ and let $A^{\ast }$ be its exact adjoint operator for the chosen difference and boundary conventions. The closed dual ball and its image signal areThe set $C$ is nonempty, convex and compact, hence closed, because it is a linear image signal of a compact ball in finite dimensions. Euclidean norm duality givesthe support function of $C$.

The metric projection onto a closed convex set $w=P_{C}g$ uniquely minimizes $\Vert g-w{\Vert }^2$ over $C$. Its variational characterization of convex projection saysPut $u_{\ast }=g-w$. This inequality says precisely that ${\sigma }_C(u_{\ast })=⟨u_{\ast },w⟩$. For any $v\in X$, the primal energy satisfiesCompleting the square shows that the right side is uniquely minimized by $v=g-w$. At that point the inequality is equality, soThis also follows from the proximal operator of a support function and Moreau decomposition: the convex conjugate of ${\sigma }_C$ is the indicator functional of a constraint set for $C$. The squared fidelity is strictly convex and coercive, so the primal minimizer exists and is unique.

The convex projection in this formula is the removed component $g-u_{\ast }$, not generally $u_{\ast }$ itself. A metric convex projection onto one fixed closed convex set is idempotent. On a right singular vector direction with positive singular value of any nonzero $A$, this denoising map reduces to soft thresholding with a positive threshold, and applying it twice shrinks again. It therefore cannot be such a convex projection for all data. This qualifies the printed convex projection wording while giving the required global gradient-norm projection residual.

Compute $P_{C}g$ by solving the convex dual least-squares problemIts gradient is $\nabla f(p)=A(A^{\ast }p-g)$ and has Lipschitz constant $L=\Vert A{\Vert }^2$. Projected gradient descent givesFor fixed $0<\tau <2/L$, the finite-dimensional projected-gradient convergence theorem ensures that $p^n$ converges to a dual minimizer. Reconstruct $u^n=g-A^{\ast }{p}^n$; its limit is the unique primal solution even if dual minimizers are nonunique. If $A=0$, the primal solution is simply $g$ and no iteration is needed. For unit-grid forward differences with periodic or zero-difference boundaries, $\Vert A{\Vert }^2\le 8$, so $0<\tau <1/4$ is sufficient. Grid-spacing factors or other boundary stencils change this bound.

The normalization here is global. If a pixelwise sum of gradient lengths had instead been intended, the dual feasible set would be a product of pixelwise balls and convex projection would normalize each block separately; it is a different regularizer and should not be silently substituted.

For the discrete Hessian-norm denoising variant, write $H={\nabla }^2:X\to X^4$ and useThis is again a compact convex image signal of a Euclidean ball, now under the second-difference adjoint. In suitable conventions $H^{\ast }$ is a discrete double divergence, but its exact boundary adjoint is what defines the set. It lies in $(\ker H{)}^{\perp }$, so components in $\ker H$ are preserved by denoising. Interior second derivatives annihilate affine image signals; whether all such image signals remain in the kernel depends on the boundary convention. With a pixelwise Hessian norm, the corresponding balls would instead be four-component blocks.

Let $R_s$ be the finite nonempty set of routes for source-sink pair $s$, with fixed nonnegative demand $f_s$. A Wardrop equilibrium is a feasible route-flow vector such that every route carrying positive flow has minimum total delay among the routes for its pair. Ifthis means that, for each $s$, some ${\lambda }_s$ satisfies $c_r(y)\ge {\lambda }_s$ for $r\in R_s$, with equality whenever $x_r>0$.

The link-route incidence matrix has $A_{jr}=1$ when route $r$ uses link $j$ and zero otherwise, or the traversal multiplicity if repeated traversals are permitted. The source-sink route incidence matrix has $H_{sr}=1$ when route $r$ serves pair $s$ and zero otherwise. Thus $Ax=y$ gives link throughputs, and $Hx=f$ gives the fixed demands.

The feasible route-flow set is a product of finite simplexes, hence nonempty and both a compact set and a convex set. The Beckmann potentialis continuous and convex, so it attains a minimum. Its route derivative is ${\partial }_r\Phi =c_r(Ax)$. At a minimum, moving a small amount of flow from any used route to any other route for the same pair cannot decrease the potential. Hence every used route has minimum route delay, exactly the Wardrop equilibrium condition. Conversely, at a Wardrop equilibrium $x$, for any feasible $\tilde{x}$,because the used flow has cost ${\lambda }_s$ and competing routes cost at least ${\lambda }_s$. The first-order inequality for a convex function makes $x$ a global minimum. This proves both existence and the specified optimization characterization.

Since each $D_j$ is strictly increasing, its primitive is a strictly convex function. Therefore the objective is strictly convex as a function of link loads $y$, and the equilibrium throughputs $y$ are unique. The mapping $x\mapsto Ax$ need not be injective on $Hx=f$, so route flows need not be unique.

For a concrete route-flow nonuniqueness at a Wardrop equilibrium example, use one unit-demand pair with two successive stages, each containing two parallel links. There are four routes, indexed by their two link choices. Give every link delay $D(u)=u$. For every $0\le t\le 1/2$,produces load $1/2$ on all four links. Every route has delay one, so each vector is a Wardrop equilibrium with the same unique link loads.

To implement minimum average delay, minimize the total delayThe average differs only by the fixed total demand ${\sum }_s{f}_s$, provided it is positive. Choose any global minimizer $x^{\ast }$, which exists by compactness, and write $y^{\ast }=A{x}^{\ast }$. Its first-order route-exchange conditions are the Wardrop equilibrium conditions for marginal link costsA genuinely traffic-dependent choice guaranteed under exactly the printed hypotheses is the calibrated toll functionHere $v_{+}=\max (v,0)$. The calibrated base charges are nonnegative because differentiable increasing delays have nonnegative derivatives, and the added penalties are nonnegative and nondecreasing in $u$. The perceived link costs $D_j(u)+T_j(u)$ remain strictly increasing, and at $y^{\ast }$ they equal $m_j^{\ast }$. Hence $x^{\ast }$ is a tolled Wardrop equilibrium. The same strictly convex Beckmann potential argument makes all its equilibrium link loads equal to $y^{\ast }$; every tolled equilibrium therefore has the globally minimum average delay. This is optimal-flow calibrated congestion tolling.

The familiar marginal external cost toll is $T_j(u)=u{D}_j^{\prime }(u)$, whose perceived-cost primitive is exactly $u{D}_j(u)$. It implements the optimum directly when the total-delay objective is convex, for example when each $u{D}_j(u)$ is convex. That extra hypothesis does not follow from the printed strict increase of $D_j$: for two identical parallel links with $D(u)=1-e^{-u}$ and total demand six, equal loads of three form a tolled Wardrop equilibrium, but the second derivative of $uD(u)$ at three is $-e^{-3}<0$, so an unequal nearby split lowers total delay. The calibrated traffic-dependent toll above establishes the requested existence without silently assuming this additional convexity.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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