Banach Alaoglu Goldstine and Krein Milman

If $f\in B_{X^{\ast }}$, then $\mathrm{\mid }f(x)\mathrm{\mid }\le \parallel x\parallel$, so $E(f)$ belongs to the displayed product; evaluations separate functionals, so $E$ is injective. By the two initial-topology descriptions, the subspace topology pulled back by $E$ is exactly $\sigma (X^{\ast },X)$ on the ball.

Inside the product let $C$ be the set of all $z=(z_x{)}_{x\in X}$ satisfying

Every $E(f)$ lies in $C$. Conversely, if $z\in C$, then $f_z(x)=z_x$ is linear and its coordinate bound gives $\mathrm{\mid }{f}_z(x)\mathrm{\mid }\le \parallel x\parallel$ for every $x$. Thus $f_z$ is bounded with $\parallel f_z\parallel \le 1$, so $z=E(f_z)$. Consequently $E[B_{X^{\ast }}]=C$.

Steps 1.1 and 2.1 show that $E$ is a homeomorphism onto the closed subspace $C$. When $X=\{0\}$, both the ball and the product are one-point spaces and the same argument applies.

For each $x\in X$, the disk $D_x=\{z\in \mathbb{K}:\mathrm{\mid }z\mathrm{\mid }\le \parallel x\parallel \}$ is compact and Hausdorff: for $\mathbb{K}=\mathbb{R}$ it is a closed bounded interval, and for $\mathbb{K}=\mathbb{C}$ it is the closed Euclidean disk in ${\mathbb{R}}^2$. This finite-dimensional fact uses no choice; when $x=0$, $D_x=\{0\}$.

The product $P={\prod }_{x\in X}{D}_x$ is compact by compact-Hausdorff Tychonoff. This is the unique step that uses the assumed ultrafilter lemma.

By [F1], evaluation carries $B_{X^{\ast }}$ homeomorphically onto a closed subspace $C$ of $P$. The subspace $C$ is compact by [F4].

An open cover of $B_{X^{\ast }}$ transports under the homeomorphism to an open cover of $C$; a finite subcover of $C$ pulls back to a finite subcover of the ball. Therefore $B_{X^{\ast }}$ is weak-star compact. No step used completeness of $X$; if $X=0$, both spaces in [F1] are singletons.

Choose $\epsilon >0$ with $\{x:\parallel x\parallel <\epsilon \}\subseteq U$ and put $r=\epsilon \mathrm{/}2$. Then $r{B}_X\subseteq U$. If $f\in U^{\circ }$ and $\parallel x\parallel \le 1$, then $rx\in U$, whence $r\mathrm{\mid }f(x)\mathrm{\mid }\le 1$. Thus $\parallel f\parallel \le r^{-1}$ and $U^{\circ }\subseteq r^{-1}{B}_{X^{\ast }}$.

The polar is weak-star closed. Indeed, if $f_0\mathrm{\notin }{U}^{\circ }$, some $u\in U$ satisfies $\mathrm{\mid }{f}_0(u)\mathrm{\mid }>1$; the basic neighborhood $\{f:\mathrm{\mid }(f-f_0)(u)\mathrm{\mid }<\mathrm{\mid }{f}_0(u)\mathrm{\mid }-1\}$ misses $U^{\circ }$ by the reverse triangle inequality.

The map $f\mapsto r^{-1}f$ is a weak-star homeomorphism with inverse $g\mapsto rg$, by continuity of scalar multiplication. It carries $B_{X^{\ast }}$ onto $r^{-1}{B}_{X^{\ast }}$, so the latter is compact by [F1]. The ultrafilter lemma enters only through [F1].

By steps 1.1 and 1.2, $U^{\circ }$ is a closed subset of the compact space in step 1.3. Adding its open complement to any open cover of $U^{\circ }$ gives an open cover of that compact space, so deleting the complement from a finite subcover proves that $U^{\circ }$ is compact. For $X=0$ this says that a singleton is compact.

The series defining $d$ converges because its terms lie between $0$ and $2^{-n}$. Symmetry and the triangle inequality follow from those of the absolute value and from $\min (1,a+b)\le \min (1,a)+\min (1,b)$. If $d(f,g)=0$, then $(f-g)(x_n)=0$ for all $n$; for any $x\in X$, take for each $k\ge 1$ the least index $n_k$ with $\parallel x-x_{n_k}\parallel <1\mathrm{/}k$. Then $x_{n_k}\to x$ and $\mathrm{\mid }(f-g)(x)\mathrm{\mid }\le 2M\parallel x-x_{n_k}\parallel \to 0$. Thus $f=g$. This also covers $M=0$, when $A$ has at most one point.

Fix $f\in A$ and a basic weak-star neighborhood $U=\{g\in A:\mathrm{\mid }(g-f)(y_j)\mathrm{\mid }<\epsilon ,1\le j\le m\}$. If $m=0$, take any metric ball. Otherwise, when $M>0$, choose $n_j$ with $\parallel y_j-x_{n_j}\parallel <\epsilon \mathrm{/}(4M)$; when $M=0$ the assertion is immediate. Put $\delta ={\min }_j{2}^{-n_j}\min (1,\epsilon \mathrm{/}2)>0$. If $d(f,g)<\delta$, then $\mathrm{\mid }(g-f)(x_{n_j})\mathrm{\mid }<\epsilon \mathrm{/}2$, and hence $\mathrm{\mid }(g-f)(y_j)\mathrm{\mid }<2M\epsilon \mathrm{/}(4M)+\epsilon \mathrm{/}2=\epsilon$. Thus a $d$-ball about $f$ lies in $U$.

Conversely, given $\eta >0$, choose $N$ so that ${\sum }_{n>N}{2}^{-n}<\eta \mathrm{/}2$ and put $\rho =\min (1,\eta \mathrm{/}2)$. The weak-star neighborhood $V=\{g\in A:\mathrm{\mid }(g-f)(x_n)\mathrm{\mid }<\rho ,1\le n\le N\}$ satisfies $d(f,g)<\rho {\sum }_{n\le N}{2}^{-n}+\eta \mathrm{/}2<\eta$.

Step 2.1 makes every weak-star neighborhood contain a metric neighborhood, while step 2.2 makes every metric neighborhood contain a weak-star neighborhood. Hence the two relative topologies on $A$ agree.

Fix an at most countable dense set $D\subseteq X$. It is nonempty because $0\in X$ and the empty set is not dense in a nonempty space. If $D$ is countably infinite, a witnessing bijection $\mathbb{N}\to D$ is a dense sequence. If $D$ is finite, a witnessing finite list can be repeated periodically (and its first entry repeated after the list ends) to give a sequence with range $D$. Thus $X$ has a fixed dense sequence; no countable family of choices was made.

The same ball is weak-star compact by Banach–Alaoglu; the ultrafilter lemma is used at this step through [F3].

Applying [F2] to the norm-bounded set $B_{X^{\ast }}$ gives a metric inducing precisely its relative weak-star topology.

By steps 2.1 and 1.2 the ball is a compact metric space, hence countably compact by [F4] and sequentially compact by [F5]. Equivalently, every sequence in it has a weak-star convergent subsequence.

By [F1], $J_X(B_X)\subseteq B_{X^{\ast \ast }}$. This is where HB supplies the norm equality needed for the stated canonical isometric embedding.

Fix $x^{\ast \ast }\in B_{X^{\ast \ast }}$ and a basic weak-star neighborhood determined by $f_1,\dots ,f_m\in X^{\ast }$ and $\epsilon >0$. If $m=0$, it contains $J_X(0)$. Suppose $m\ge 1$, put $T(x)=(f_1(x),\dots ,f_m(x))$, $v=(x^{\ast \ast }(f_1),\dots ,x^{\ast \ast }(f_m))$, and let $C$ be the Euclidean closure of $T(B_X)$ in ${\mathbb{K}}^m$, viewed as ${\mathbb{R}}^m$ or ${\mathbb{R}}^{2m}$. The set $C$ is nonempty, closed and convex because $B_X$ is nonempty and convex and $T$ is real-linear.

If $v\mathrm{\notin }C$, [F2] gives a nonzero real-linear functional $\mathrm{\ell }$ and a real $b$ with $\mathrm{\ell }(z)\le b<\mathrm{\ell }(v)$ for every $z\in C$. Every real-linear functional on ${\mathbb{K}}^m$ has the form $\mathrm{\ell }(z)=\mathrm{Re}{\sum }_{j=1}^m{c}_j{z}_j$: in the complex case write its coefficients on real and imaginary coordinate vectors and take $c_j=a_j-i{b}_j$.

Put $f={\sum }_j{c}_j{f}_j\in X^{\ast }$. The separation inequalities give ${\sup }_{x\in B_X}\mathrm{Re}f(x)<\mathrm{Re}{x}^{\ast \ast }(f)$. Yet ${\sup }_{x\in B_X}\mathrm{Re}f(x)=\parallel f\parallel$: the inequality $\le$ is the norm bound, while for any $x$ rotate or change its sign so that $f(x)$ becomes the nonnegative real $\mathrm{\mid }f(x)\mathrm{\mid }$, and then take the supremum. Since $\parallel x^{\ast \ast }\parallel \le 1$, one also has $\mathrm{Re}{x}^{\ast \ast }(f)\le \mathrm{\mid }{x}^{\ast \ast }(f)\mathrm{\mid }\le \parallel f\parallel$.

The strict inequality in step 3.1 would therefore read $\parallel f\parallel <\mathrm{Re}{x}^{\ast \ast }(f)\le \parallel f\parallel$, which is impossible. Hence $v\in C$.

Because $v$ lies in the closure of $T(B_X)$, the open coordinate box $\{z:\mathrm{\mid }{z}_j-v_j\mathrm{\mid }<\epsilon ,1\le j\le m\}$ meets $T(B_X)$. Thus some $x\in B_X$ satisfies $\mathrm{\mid }{J}_X(x)(f_j)-x^{\ast \ast }(f_j)\mathrm{\mid }<\epsilon$ for every $j$, so $J_X(x)$ belongs to the chosen neighborhood.

Every basic weak-star neighborhood of every $x^{\ast \ast }\in B_{X^{\ast \ast }}$ therefore meets $J_X(B_X)$, including the empty-test and zero-space cases handled in step 1.2. This is exactly weak-star density.

Define $U=\{y^{\ast \ast }\in X^{\ast \ast }:\mathrm{\mid }{y}^{\ast \ast }(f_j)-x^{\ast \ast }(f_j)\mathrm{\mid }<\epsilon \text{ for }1\le j\le m\}$. It is a basic weak-star neighborhood of $x^{\ast \ast }$; when $m=0$, it is all of $X^{\ast \ast }$.

By Goldstine, $U$ meets $J_X(B_X)$, so there is $x\in B_X$ with $J_X(x)\in U$. This invocation carries the HB hypothesis; no sequence or family of approximants is selected.

Since $J_X(x)(f_j)=f_j(x)$, the witness $x$ from step 2.1 satisfies every displayed inequality, and hence is the required finite-data approximant.

If $E$ is weak-star closed, then so is $E\cap B_{X^{\ast }}$, because $B_{X^{\ast }}={\bigcap }_{x\in B_X}\{f:\mathrm{\mid }f(x)\mathrm{\mid }\le 1\}$ is an intersection of closed evaluation constraints.

For the reverse implication first suppose $\mathbb{K}=\mathbb{R}$ and $A:=E\cap B_{X^{\ast }}$ is weak-star closed. If $f_n\in E$ and $f_n\to f$ in norm, boundedness of the convergent sequence gives $R>0$ with $R^{-1}{f}_n\in A$; norm convergence implies weak-star convergence, so closedness of $A$ gives $R^{-1}f\in A$ and $f\in E$. If some $f_0\mathrm{\notin }E$ had ${\inf }_{f\in E}\parallel f-f_0\parallel =0$, DC could select $f_n\in E$ with $\parallel f_n-f_0\parallel <1\mathrm{/}(n+1)$, contradicting this sequential norm-closedness. Hence $d:=\mathrm{dist}(f_0,E)>0$; fix $0<\delta <d$.

For finite sets $S_k\subseteq B_X$, let $P_n(S_1,\dots ,S_{n-1})$ mean: every $f\in E$ with $\parallel f-f_0\parallel \le n\delta$ violates at least one earlier test, so $\mathrm{\mid }(f-f_0)(x)\mathrm{\mid }>k\delta$ for some $1\le k<n$ and $x\in S_k$. The assertion $P_1$ is vacuous because $\delta <d$.

Suppose $P_n$ holds. For finite $S\subseteq B_X$, let $E(S)$ consist of those $f\in E$ with $\parallel f-f_0\parallel \le (n+1)\delta$, all earlier tests at most $k\delta$, and the $S$-test at most $n\delta$. Put $R=\parallel f_0\parallel +(n+1)\delta$. The set $K=E\cap R{B}_{X^{\ast }}=RA$ is weak-star compact: $R{B}_{X^{\ast }}$ is compact by scaling [F1], $RA$ is weak-star closed, and [F1] uses the ultrafilter lemma through [F2]. Each $E(S)$ is weak-star closed in $K$, because each norm bound is the intersection over $x\in B_X$ of closed evaluation constraints. If every $E(S)$ were nonempty, the identity ${\bigcap }_{i=1}^{q}E(S_i)=E({\bigcup }_i{S}_i)$ would give the finite-intersection property; compactness would produce $f$ in every $E(S)$. Taking singleton $S=\{x\}$ for every $x\in B_X$ would give $\parallel f-f_0\parallel \le n\delta$, while all earlier tests hold, contradicting $P_n$. Thus some finite listed $S_n$ has $E(S_n)=\mathrm{\varnothing }$, and that emptiness is exactly $P_{n+1}$.

Apply DC to the relation that extends a finite list $(S_1,\dots ,S_{n-1})$ satisfying $P_n$ by a finite listed $S_n$ supplied in step 3.1. Starting from the empty list, it yields finite listed sets $S_n\subseteq B_X$ for all $n\ge 1$ with every $P_n$ true. Recording the finite listing as part of each state avoids a later countable choice of enumerations.

Concatenate, for $n=1,2,\dots$, the finite list $n^{-1}{S}_n$ followed by one zero padding term, obtaining a sequence $(x_i)$ in $B_X$. If a term lies in the $n$th block its norm is at most $1\mathrm{/}n$; because each block is finite and nonempty after padding, the block number tends to infinity with $i$. Hence $\parallel x_i\parallel \to 0$.

For every $f\in E$, choose an integer $n\ge \max (2,{\delta }^{-1}\parallel f-f_0\parallel )$. Property $P_n$ gives $k<n$ and $x\in S_k$ with $\mathrm{\mid }(f-f_0)(x)\mathrm{\mid }>k\delta$; the coordinate $x\mathrm{/}k$ occurs in $(x_i)$, so ${\sup }_i\mathrm{\mid }(f-f_0)(x_i)\mathrm{\mid }>\delta$.

Define $T:X^{\ast }\to c_0$ by $T(f)=(f(x_i){)}_i$. Step 5.1 makes every image a null sequence, and $\parallel T(f){\parallel }_{\mathrm{\infty }}\le \parallel f\parallel {\sup }_i\parallel x_i\parallel$, so $T$ is bounded and linear. With $y_0=T(f_0)$, step 6.1 gives $\parallel T(f)-y_0{\parallel }_{\mathrm{\infty }}>\delta$ for every $f\in E$; therefore the closed linear subspace $M=\overline{T(E)}$ has $\mathrm{dist}(y_0,M)\ge \delta$.

On $M+\mathbb{R}{y}_0$ define $g(m+a{y}_0)=a$. This is well defined because $y_0\mathrm{\notin }M$, and $\parallel m+a{y}_0\parallel \ge \mathrm{\mid }a\mathrm{\mid }\delta$ shows $\mathrm{\mid }g\mathrm{\mid }\le {\delta }^{-1}\parallel \cdot \parallel$. Applying HB to the sublinear function ${\delta }^{-1}\parallel \cdot \parallel$ extends $g$ to $\beta \in (c_0{)}^{\ast }$ with $\beta (y_0)=1$, $\beta {\mathrm{\mid }}_M=0$, and $\parallel \beta \parallel \le {\delta }^{-1}$. This is the sole HB use.

By [F5] there is $\alpha =({\alpha }_i)\in {\mathrm{\ell }}^1$ with $\beta (y)={\sum }_i{\alpha }_i{y}_i$ for $y\in c_0$ and ${\sum }_i\mathrm{\mid }{\alpha }_i\mathrm{\mid }=\parallel \beta \parallel \le {\delta }^{-1}$.

Since ${\sum }_i\parallel {\alpha }_i{x}_i\parallel \le {\sum }_i\mathrm{\mid }{\alpha }_i\mathrm{\mid }\le {\delta }^{-1}$ and $X$ is Banach, [F6] gives $x_0={\sum }_i{\alpha }_i{x}_i\in X$ with $\parallel x_0\parallel \le {\delta }^{-1}$.

Continuity of every $f\in X^{\ast }$ allows evaluation term by term: $f(x_0)={\sum }_i{\alpha }_{i}f(x_i)=\beta (Tf)$. Thus $f_0(x_0)=1$ and $f(x_0)=0$ for every $f\in E$. The weak-star neighborhood $\{h:\mathrm{\mid }(h-f_0)(x_0)\mathrm{\mid }<1\mathrm{/}2\}$ therefore misses $E$.

Every $f_0\mathrm{\notin }E$ has the weak-star neighborhood constructed in step 11.1 disjoint from $E$, so $E$ is weak-star closed in the real case. Together with step 1.1 this proves both directions there.

Now let $X$ be complex and write $X_{\mathbb{R}}$ for its realification. The map $\mathcal{R}:X^{\ast }\to (X_{\mathbb{R}}{)}^{\ast }$, $\mathcal{R}f=\mathrm{Re}f$, is a real-linear isometric bijection with inverse $g\mapsto (x\mapsto g(x)-ig(ix))$: complex linearity follows from the displayed formula, and rotating a vector shows norm equality. It is a weak-star homeomorphism because $g(x)=\mathrm{Re}f(x)$ and $\mathrm{Im}f(x)=-g(ix)$. For a complex-linear $E$, $\mathcal{R}(E)$ is real-linear and $\mathcal{R}(E\cap B_{X^{\ast }})=\mathcal{R}(E)\cap B_{(X_{\mathbb{R}}{)}^{\ast }}$. Hence closedness of the complex slice implies closedness of the real slice; step 12.1 makes $\mathcal{R}(E)$ real weak-star closed, and the homeomorphism makes $E$ complex weak-star closed.

Step 1.1 proves the forward implication over both scalar fields, step 12.1 proves the reverse implication over $\mathbb{R}$, and step 13.1 proves it over $\mathbb{C}$. Therefore the two weak-star closedness conditions are equivalent.

By [F1], some $x_0\in K$ satisfies $a(x_0)=m:={\min }_{x\in K}a(x)$, so $F$ is nonempty. Since $F=a^{-1}(\{m\})$ and $a$ is continuous, $F$ is closed in $K$; hence it is compact by [F2].

If $x,y\in F$ and $0\le t\le 1$, affinity gives $a((1-t)x+ty)=(1-t)m+tm=m$, so convexity of $K$ places the combination in $F$; thus $F$ is convex.

Suppose $x,y\in K$, $0<t<1$, and $(1-t)x+ty\in F$. Minimality gives $a(x),a(y)\ge m$, while affinity gives $(1-t)a(x)+ta(y)=m$; the two positive coefficients force $a(x)=a(y)=m$, so $x,y\in F$. Therefore $F$ is a face by [F3].

Let $G$ be a face of $F$, and suppose $x,y\in K$, $0<t<1$, and $(1-t)x+ty\in G$. Since $G\subseteq F$ and $F$ is a face of $K$, step 2.2 gives $x,y\in F$; the face condition for $G$ inside $F$ then gives $x,y\in G$. Since $G$ is already nonempty and convex, [F3] makes it a face of $K$.

Steps 1.1–2.2 prove that the minimum is attained and its level set is a nonempty compact face; step 3.1 proves that faces of faces are faces.

Let $\mathcal{P}$ be the set of nonempty faces of $K$ that are closed in $K$, ordered by $F\preccurlyeq G$ when $F\supseteq G$. It is a nonempty poset because $K\in \mathcal{P}$.

Let $\mathcal{C}$ be a chain in $\mathcal{P}$. If $\mathcal{C}=\mathrm{\varnothing }$, then $K$ is an upper bound. Otherwise every finite subfamily of $\mathcal{C}$ has intersection equal to its inclusion-smallest member and hence nonempty. Its members are closed in compact $K$, so [F3] gives a nonempty intersection $H:={\bigcap }_{F\in \mathcal{C}}F$, and $H$ is closed and convex.

If $x,y\in K$, $0<t<1$, and $(1-t)x+ty\in H$, then this combination lies in every $F\in \mathcal{C}$; since each $F$ is a face, $x,y$ lie in every $F$ and therefore in $H$. Thus $H$ is a face, hence belongs to $\mathcal{P}$, and $F\supseteq H$ for every $F\in \mathcal{C}$, so $H$ is an upper bound in the reverse-inclusion order.

By [F4], using AC as declared in [F5], $\mathcal{P}$ has a maximal element $M$; equivalently, $M$ is an inclusion-minimal nonempty closed face of $K$.

Suppose $p,q\in M$ are distinct. AC supplies HB by [F6], so [F2] gives $f\in X^{\mathrm{\prime }}$ with $u(p):=\mathrm{Re}f(p)\mathrm{\ne }\mathrm{Re}f(q)=:u(q)$. The restriction $u{\mathrm{\mid }}_M$ is continuous, real-valued, and affine.

By [F1], the minimizer set $N$ of $u{\mathrm{\mid }}_M$ is a nonempty compact face of $M$, hence a face of $K$. It is closed in $M$ and $M$ is closed in $K$, so it is closed in $K$ and lies in $\mathcal{P}$. Since $u(p)\mathrm{\ne }u(q)$, at least one of $p,q$ is not a minimizer, so $N⊊M$, contradicting the inclusion-minimality of $M$.

Hence $M$ is a singleton, say $M=\{x\}$. Since $M$ is a face of $K$, the singleton characterization in the face definition makes $x$ an extreme point of $K$.

The empty-chain case in step 2.1 and the nonempty-chain construction in steps 2.1–3.1 verify every chain hypothesis of Zorn; steps 4.1–7.1 then produce the required extreme point.

If $K=\mathrm{\varnothing }$, then $\mathrm{ext}K=\mathrm{\varnothing }$ and both sides are empty by the stated convention. Hence suppose $K\mathrm{\ne }\mathrm{\varnothing }$ and put $E=\mathrm{ext}K$ and $C=\stackrel{}{\mathrm{co}}(E)$. By [F1], $E$ and therefore $C$ are nonempty.

The set $K$ is closed by [F5] and convex by hypothesis, and it contains $E$; therefore it contains $\mathrm{co}(E)$ and its closure $C$. The set $C$ is closed by definition and convex by [F6].

Suppose for contradiction that $x\in K\setminus C$. Apply [F2] to the compact convex singleton $\{x\}$ and the nonempty closed convex set $C$, using HB supplied from AC by [F4]. After naming $u=\mathrm{Re}f$, the resulting inequalities give $u(x)<{\inf }_{c\in C}u(c)$.

By [F3], the minimizer set $M=\{y\in K:u(y)={\min }_{K}u\}$ is a nonempty compact face of $K$. By [F1], $M$ has an extreme point $e$. Then $\{e\}$ is a face of $M$, so face transitivity in [F3] makes $\{e\}$ a face of $K$; hence $e\in E\subseteq C$.

Since $e$ minimizes $u$ on $K$ and $x\in K$, one has $u(e)\le u(x)$; step 3.1 gives $u(x)<{\inf }_{C}u$, whereas $e\in C$ gives $u(e)\ge {\inf }_{C}u$, a contradiction. Thus no $x\in K\setminus C$ exists, so $K\subseteq C$.

Step 2.1 gives $C\subseteq K$ and step 5.1 gives the reverse inclusion; together with the empty case in step 1.1 this proves the asserted equality in every case.

For each $y\in K$ put $L_y=\{x\in K:f(x)\ge f(y)\}$, which is closed by [F1]. Any finite subfamily has nonempty intersection: choose from its finite list an index at which the finitely many real values $f(y)$ are largest, and the corresponding $y$ belongs to every listed $L_y$; the empty finite intersection is $K\mathrm{\ne }\mathrm{\varnothing }$. Thus [F4] supplies $z\in {\bigcap }_{y\in K}{L}_y$, so $m:=f(z)$ is the maximum of $f$ on $K$.

The maximizer set $M=\{x\in K:f(x)=m\}=\{x\in K:f(x)\ge m\}$ is nonempty and closed by [F1]. Call a subset $S\subseteq K$ $K$-extremal when $(1-t)x+ty\in S$, for $x,y\in K$ and $0<t<1$, implies $x,y\in S$.

The set $M$ is $K$-extremal. Indeed, if $w=(1-t)x+ty\in M$ with $0<t<1$, convexity and maximality give $m=f(w)\le (1-t)f(x)+tf(y)\le m$; positivity of both coefficients and $f(x),f(y)\le m$ force $f(x)=f(y)=m$.