Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 106 / 3

The Hahn-Banach separation theorem for two convex sets states that disjoint nonempty convex sets $A,B$ in a real locally convex space, with $A$ open, admit a nonzero continuous linear functional $\ell$ and $\alpha \in \mathbb{R}$ such thatThere need not be a positive uniform gap between the two sets.

For a proof, set $C=A-B$. It is open and convex, and $0\notin C$. Choose $c_0\in C$, let $V=C-c_0$, and let $p$ be the Minkowski functional of $V$. Because $V$ is open, convex and contains zero, $p$ is a finite sublinear functional, and $V=\{x:p(x)<1\}$. Put $z=-c_0$. Since $z\notin V$, $p(z)\ge 1$. On $\mathbb{R}z$ define ${\ell }_0(tz)=tp(z)$. For $t\ge 0$ this equals $p(tz)$; for $t<0$ it is nonpositive and hence at most $p(tz)$. The Hahn-Banach theorem extends it to a linear $\ell$ with $\ell \le p$. Both $p(x)$ and $p(-x)$ are bounded near zero, so this domination makes $\ell$ continuous; it is nonzero because $\ell (z)=p(z)\ge 1$. For $c\in C$,Thus $\ell (a)<\ell (b)$ for every $a\in A,b\in B$. Define $\alpha ={\sup }_{a\in A}\ell (a)$, which is finite because any fixed $b\in B$ bounds it above. Openness of $A$ and nonzeroness of $\ell$ ensure $\ell (a)<\alpha$ for every $a\in A$, and $\alpha \le \ell (b)$ for every $b\in B$. This proves the stated separation.

For the sequence criterion, suppose $f\in S_{X^{\ast }}$ and $f(x_i)\ge \varrho$. If each $t_i\ge 0$, thenConversely, the assumed inequality implies that every convex combination of the $x_i$ has norm at least $\varrho$. Hence $D=\{x:\Vert x\Vert <\varrho \}$ is disjoint from $C=\mathrm{conv}\{x_i:i\ge 1\}$. Apply the separation theorem with the open set $D$ to obtain $\ell \ne 0$ andThen $f=\ell /\Vert \ell \Vert$ has norm one and satisfies $f(x_i)\ge \varrho$ for every $i$. ThereforeThe PDF confirms the “if and only if” that is garbled in the TeX transcription.

By the Riesz-Markov-Kakutani representation theorem, the continuous dual space of $C(K)$ is isometrically the space of finite regular Borel measures, signed for real scalars and complex for complex scalars:Here $|\mu |$ is the variation measure, whose total mass is the total variation norm of a measure.

If $f_n\rightharpoonup 0$ in $C(K)$, evaluation at each $t\in K$ gives $f_n(t)\to 0$. The Uniform boundedness principle also gives $M={\sup }_n\Vert f_n{\Vert }_{\infty }<\infty$. For any $\mu \in M(K)$, the dominated convergence theorem with respect to the finite positive measure $|\mu |$ yieldssince $|f_n(t){|}^2\to 0$ pointwise and $|f_n(t){|}^2\le M^2$. ThusThe same proof works for complex squares, because the absolute-value domination remains valid.

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C({\Phi }_A)$ through its Gelfand transform, with ${\Phi }_A$ a compact Hausdorff space. The l-infinity sequence space is such an algebra under coordinatewise multiplication and conjugation, with the supremum norm. Thereforegives the required isometric isomorphism.

Uniqueness as a Banach space representation uses the Banach–Stone theorem, rather than just uniqueness as an algebra representation. To justify the relevant theorem, the Riesz-Markov-Kakutani representation theorem givesthe extreme points of the dual unit ball of C(K). A norm-one measure whose variation measure is not concentrated at one point splits into two normalized restrictions to disjoint sets of positive variation, and so is not extreme. Conversely, equality in the variation bound shows that any decomposition of $\alpha {\delta }_t$ into the average of two dual-unit-ball elements forces both to be $\alpha {\delta }_t$: after removing the phase, the two measures must be positive and their average is concentrated at $t$.

If $U:C(K)\to C(L)$ is a surjective linear isometry, its dual map preserves these extreme points and their scalar orbits. Hence $U^{\ast }{\delta }_s=u(s){\delta }_{h(s)}$ for a bijection $h:L\to K$ and $|u(s)|=1$. Evaluating at $1$ gives the continuous function $u=U1$, whileshows that $h$ is continuous, since continuous functions determine the topology of a compact Hausdorff space. It is consequently a homeomorphism. Thus any other compact space representing ${\ell }^{\infty }$ is homeomorphic to this $K$. The Banach–Stone theorem is also stated in Leonard Tomczak's notes on András Zsák's functional analysis lectures.

Embed $\mathbb{N}$ by the evaluation characters $\iota (n)(a)=a_n$. They are distinct because the coordinate indicator functions $e_n$ distinguish them. Moreover, $e_n^2=e_n$, so every character has $\chi (e_n)\in \{0,1\}$. If $\chi (e_n)=1$, thenand hence $\chi =\iota (n)$. It follows that $\{\iota (n)\}=\{\chi :|\chi (e_n)-1|<1/2\}$ is open in $K$. Thus $\iota$ is a homeomorphism from the discrete natural numbers onto its image.

To prove density, suppose $D=\overline{\iota (\mathbb{N})}\ne K$. The Urysohn lemma provides a nonzero $v\in C(K)$ vanishing on $D$. Write $v=\Gamma (a)$. Then $a_n=v(\iota (n))=0$ for all $n$, so $a=0$ and $v=0$, a contradiction. Every bounded function $b:\mathbb{N}\to \mathbb{C}$ is an element of ${\ell }^{\infty }$, and $\Gamma (b)$ is its continuous extension to $K$. Density makes this extension unique. ThereforeThis identifies $K$ with the Stone-Čech compactification of the natural numbers.