Banach Alaoglu Goldstine and Krein Milman — Examples

If $K=\mathrm{\varnothing }$, no measure can have total mass one, so the probability space is empty and compact. Hence assume $K\mathrm{\ne }\mathrm{\varnothing }$. By [F5], compactness makes $K$ locally compact, and it is Hausdorff by hypothesis. Every continuous function on compact $K$ has compact support and compact closed superlevel sets, so [F6] gives $C_c(K)=C_0(K)=C(K)$.

A regular Borel probability $\mu$ defines $L_{\mu }(g)={\int }_{K}gd\mu$. For real or complex $g$, $\mathrm{\mid }{L}_{\mu }(g)\mathrm{\mid }\le \parallel g{\parallel }_{\mathrm{\infty }}\mu (K)=\parallel g{\parallel }_{\mathrm{\infty }}$, while $L_{\mu }(1)=1$; hence $L_{\mu }\in B_{C(K{)}^{\ast }}$ and $\parallel L_{\mu }\parallel =1$. It is positive on nonnegative real-valued $g$.

Conversely, let $L\in B_{C(K{)}^{\ast }}$ satisfy $L(1)=1$ and $L(g)\in [0,\mathrm{\infty })$ for every nonnegative real-valued $g\in C(K)$. In the real case [F2] represents $L$ by a finite regular measure $\mu$ in the sense of [F4], and $\mu (K)=L(1)=1$. In the complex case restrict $L$ to real-valued functions, apply [F2], and use complex linearity to recover $L(g+ih)=\int gd\mu +i\int hd\mu$; uniqueness in [F3] identifies this same positive probability measure.

Thus probabilities correspond exactly to the slice $S$ of $B_{C(K{)}^{\ast }}$ cut out by $L(1)=1$ and by $L(g)\in [0,\mathrm{\infty })$ for all nonnegative real-valued $g$. Each condition is weak-star closed because it is the inverse image of the closed set $\{1\}$ or $[0,\mathrm{\infty })$ under one evaluation; arbitrary intersections remain closed.

By [F1], the dual ball is weak-star compact under the ultrafilter lemma. Hence its closed subset $S$ is compact by [F7]. Under the two representation directions above, the weak-star topology is exactly convergence of $\int gd\mu$ for every continuous test function $g$, so the probability measures are compact in the asserted topology.

The initial reduction proves the empty case, and the closed-slice construction proves both scalar-field cases for nonempty compact Hausdorff $K$.

If $K=\mathrm{\varnothing }$, then $P(K)=\mathrm{\varnothing }$ and there are no Dirac measures, so the equality is empty on both sides. Hence assume $K\mathrm{\ne }\mathrm{\varnothing }$. By [F7], compactness makes $K$ locally compact, and it is Hausdorff by hypothesis, so the LCH regularity convention [F4] applies.

Let $\mu \in P(K)$ and let $A$ be Borel. The restriction ${\mu }_A(E)=\mu (E\cap A)$ is a measure by [F3]. It is regular: for Borel $E$, inner regularity of $\mu$ on $E\cap A$ gives ${\mu }_A(E)=\sup \{\mu (D):D\subseteq E\cap A,D\text{ compact}\}$; every such $D$ is also a compact subset of $E$ with ${\mu }_A(D)=\mu (D)$, while every compact $C\subseteq E$ satisfies ${\mu }_A(C)\le {\mu }_A(E)$. Thus the required supremum over compact $C\subseteq E$ equals ${\mu }_A(E)$.

Suppose that $\mu (E)\in \{0,1\}$ for every Borel $E$. Let $\mathcal{C}$ be all closed $C\subseteq K$ with $\mu (C)=1$. It contains $K$. A finite intersection of its members has measure one because the complement is a finite union of null sets, so $\mathcal{C}$ has the finite-intersection property. By [F5], choose $x\in \bigcap \mathcal{C}$.

Conversely, fix any $x\in K$. By [F2], ${\delta }_x$ is a probability measure. The Hausdorff hypothesis makes the compact singleton $\{x\}$ closed and hence Borel. Thus ${\delta }_x$ is regular: for a Borel set containing $x$, that singleton realizes mass one, while a set not containing $x$ has mass zero. If ${\delta }_x=t\mu +(1-t)\nu$ with $\mu ,\nu \in P(K)$ and $0<t<1$, evaluation on $K\setminus \{x\}$ gives $0=t\mu (K\setminus \{x\})+(1-t)\nu (K\setminus \{x\})$, so nonnegativity gives both masses zero. Thus $\mu (\{x\})=\nu (\{x\})=1$ and $\mu =\nu ={\delta }_x$; [F1] makes ${\delta }_x$ extreme.

If $0<t:=\mu (A)<1$, define ${\mu }_1=t^{-1}{\mu }_A$ and ${\mu }_2=(1-t{)}^{-1}{\mu }_{K\setminus A}$. Step 1.2 makes both regular probabilities, and $\mu =t{\mu }_1+(1-t){\mu }_2$. They are distinct because ${\mu }_1(A)=1$ and ${\mu }_2(A)=0$, so [F1] shows that $\mu$ is not extreme.

Every open neighborhood $U$ of the point $x$ from step 1.3 has measure one. Otherwise the zero-one hypothesis gives $\mu (U)=0$, so the closed complement $K\setminus U$ has measure one and belongs to $\mathcal{C}$, contradicting $x\in U\cap \bigcap \mathcal{C}$.

If $D\subseteq K\setminus \{x\}$ is compact, [F6] gives disjoint open sets $U,V$ with $x\in U$ and $D\subseteq V$. Step 2.2 gives $\mu (U)=1$, hence $\mu (V)=0$ and $\mu (D)=0$. Inner regularity [F4] on the Borel set $K\setminus \{x\}$ now gives $\mu (K\setminus \{x\})=0$, so $\mu ={\delta }_x$.

Let $\mu$ be extreme. Step 2.1 rules out every Borel set of intermediate mass, so $\mu$ is zero-one valued; steps 1.3, 2.2, and 3.1 then give $\mu ={\delta }_x$ for some $x\in K$. Step 1.4 proves the reverse implication, and step 1.1 covers the empty space.

Suppose $x$ lies in the closed unit ball and $\mathrm{\mid }{x}_N\mathrm{\mid }<1$ for some $N$. Over $\mathbb{R}$, choose $0<\epsilon \le 1-\mathrm{\mid }{x}_N\mathrm{\mid }$ and put $u_N=1$; over $\mathbb{C}$, put $u_N=1$ if $x_N=0$ and $u_N=i{x}_N\mathrm{/}\mathrm{\mid }{x}_N\mathrm{\mid }$ otherwise, and choose $0<\epsilon \le \sqrt{1-\mathrm{\mid }{x}_N{\mathrm{\mid }}^2}$. With $e$ supported at $N$ and equal there to $u_N$, both $x+\epsilon e$ and $x-\epsilon e$ have sup norm at most one, are distinct, and have midpoint $x$. Thus $x$ is not extreme by [F1].

Conversely suppose $\mathrm{\mid }{x}_n\mathrm{\mid }=1$ for all $n$ and $x=(1-t)y+tz$ with $y,z$ in the unit ball and $0<t<1$. For each $n$, the scalar identity $\mathrm{\mid }(1-t)y_n+t{z}_n{\mathrm{\mid }}^2=(1-t)\mathrm{\mid }{y}_n{\mathrm{\mid }}^2+t\mathrm{\mid }{z}_n{\mathrm{\mid }}^2-t(1-t)\mathrm{\mid }{y}_n-z_n{\mathrm{\mid }}^2$ gives $1\le 1-t(1-t)\mathrm{\mid }{y}_n-z_n{\mathrm{\mid }}^2$ by [F2]. Hence $y_n=z_n$, and their convex combination equals $x_n$, so $y_n=z_n=x_n$ for every $n$.

Step 1.1 excludes exactly the sequences with an interior coordinate, while step 1.2 and [F1] prove every sequence with all coordinates on the scalar unit circle is extreme.

Since $x_n\to 0$, there is an index $N$ with $\mathrm{\mid }{x}_N\mathrm{\mid }<1$; take the least such index if a canonical witness is desired. Put $\epsilon =(1-\mathrm{\mid }{x}_N\mathrm{\mid })\mathrm{/}2>0$ and let $e_N$ be the sequence equal to one at $N$ and zero elsewhere.

The sequences $y=x+\epsilon e_N$ and $z=x-\epsilon e_N$ still tend to zero and satisfy $\mathrm{\mid }{y}_N\mathrm{\mid },\mathrm{\mid }{z}_N\mathrm{\mid }\le \mathrm{\mid }{x}_N\mathrm{\mid }+\epsilon <1$, while every other coordinate is unchanged. Hence $y,z\in B_{c_0}$ by [F2]; they are distinct and $x=(y+z)\mathrm{/}2$.

By [F1], the nontrivial midpoint representation in step 2.1 shows that the arbitrary $x\in B_{c_0}$ is not extreme. Therefore the ball has no extreme points over either scalar field.

Assume for contradiction that $T:c_0\to Y^{\ast }$ is a surjective linear isometry. The space $Y$ cannot be zero, because then $Y^{\ast }=\{0\}$ whereas $c_0$ contains a nonzero coordinate vector.

The isometry maps $B_{c_0}$ bijectively onto $B_{Y^{\ast }}$. It preserves extreme points in both directions: applying $T$ or $T^{-1}$ to a strict convex representation preserves its coefficient and turns trivial endpoint equality into trivial endpoint equality.

By [F2], $B_{Y^{\ast }}$ has an extreme point, whose inverse image under $T$ is extreme in $B_{c_0}$ by step 2.1. This contradicts [F1], so the assumed surjective linear isometry does not exist.

For each $n\in \mathbb{N}$ define ${\mathrm{\Lambda }}_n\in ({\mathrm{\ell }}^{\mathrm{\infty }}{)}^{\ast }$ by ${\mathrm{\Lambda }}_n(x)=x_n$. Then $\mathrm{\mid }{\mathrm{\Lambda }}_n(x)\mathrm{\mid }\le \parallel x{\parallel }_{\mathrm{\infty }}$ and equality holds at the $n$th coordinate vector, so $\parallel {\mathrm{\Lambda }}_n\parallel =1$.

Consider any subsequence $({\mathrm{\Lambda }}_{n_k})$, with the indices $n_k$ strictly increasing. Define $x\in {\mathrm{\ell }}^{\mathrm{\infty }}$ by $x_{n_k}=(-1{)}^k$ and $x_j=0$ off the range of $(n_k)$. This is well defined because the indices are distinct and has $\parallel x{\parallel }_{\mathrm{\infty }}=1$ by [F3].

Its evaluations are ${\mathrm{\Lambda }}_{n_k}(x)=(-1{)}^k$, which do not converge in $\mathbb{R}$ or $\mathbb{C}$. By [F2], the chosen subsequence is not weak-star convergent. Since the subsequence was arbitrary, $({\mathrm{\Lambda }}_n)$ has no weak-star convergent subsequence.

Nevertheless [F1] makes the closed unit ball containing this sequence weak-star compact under the ultrafilter lemma. It is therefore a compact space with a sequence having no convergent subsequence, as claimed.