Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 4

The finite-dimensional Krein-Milman theorem says that a nonempty compact convex set $K$ is the convex hull of its extreme points. A point is extreme if a proper convex combination of two points of $K$ equals it only when both points equal it.

Prove the theorem by induction on the dimension of the affine hull. Dimension zero is immediate. In positive dimension work inside that affine hull, where $K$ has interior. Every boundary point $x$ lies on a supporting hyperplane. To see this, choose exterior points $y_j\to x$ and nearest points $p_j\in K$. Convexity and the minimum-distance condition give $(y_j-p_j)\cdot (z-p_j)\le 0$ for $z\in K$. A subsequence of the normalized nonzero vectors $y_j-p_j$ converges to a unit vector $v$, while $p_j\to x$, giving $v\cdot (z-x)\le 0$. The resulting face $F=\{z\in K:v\cdot (z-x)=0\}$ has smaller affine dimension. Its extreme points are extreme in $K$, since equality in the supporting inequality forces both endpoints of any decomposition to lie in $F$. Induction expresses $x$ as a convex combination of extreme points of $K$.

Every interior point is a convex combination of the two endpoints of a line segment through it maximal in $K$; compactness gives the endpoints, and both are boundary points. Combining their convex decompositions completes the proof. In the generalization requested, a nonempty weak-star compact convex subset $K$ of the dual unit ball of a separable Banach space satisfiesThe closure is now essential; separability makes this compact weak-star setting metrizable but the general Krein-Milman theorem holds more broadly in locally convex spaces.

For a bounded self-adjoint $T$, nonnegative spectrum is equivalent to $⟨Tx,x⟩\ge 0$. Here is a useful elementary justification of the nontrivial direction. Let $a={\inf }_{\Vert x\Vert =1}⟨Tx,x⟩$ and $S=T-aI$, so $S$ has nonnegative quadratic form. Cauchy-Schwarz for that positive form gives $\Vert Sx{\Vert }^2\le \Vert S\Vert ⟨Sx,x⟩$. Taking unit vectors approaching the infimum makes $S{x}_j\to 0$, hence $a\in \sigma (T)$. Thus nonnegative spectrum forces $a\ge 0$. The reverse direction follows from the lower bound and adjoint argument for $T-\lambda$ when $\lambda <0$, together with reality of the self-adjoint spectrum.

If $T=0$ its positive square root is zero. Otherwise set $M=\Vert T\Vert$ and $C=I-T/M$, so $0\le C\le I$ and $\Vert C\Vert \le 1$. Write the scalar binomial seriesThe last equality follows by taking $z\uparrow 1$ in this positive-coefficient series. Consequentlyconverges absolutely in operator norm. Each partial sum is positive, because $0\le C^n\le I$ and the sum of its coefficients is at most one. The limit is positive and self-adjoint. Multiplying the absolutely convergent series gives $B^2=M(I-C)=T$.

For uniqueness let $R$ be another positive square root. It commutes with $T=R^2$, hence with the polynomial-norm-limit $B$. Thus $(R-B)(R+B)=0$. The kernel of $R+B$ is $\ker R\cap \ker B$: zero quadratic form for a positive operator implies it annihilates the vector, by positive-form Cauchy-Schwarz. The difference $R-B$ vanishes there and on $\mathrm{ran}(R+B)$; the latter is dense in the kernel complement because $R+B$ is self-adjoint. Therefore $R=B$. The positive square root of an operator exists and is unique.

Let $X$ be the character space of an algebra, consisting of nonzero multiplicative complex linear functionals on $A$. We construct the required Gelfand transform and prove its properties.

Characters exist. A maximal proper ideal exists by the axiom of choice and is closed: if it were dense it would contain an element within distance one of the identity, hence an invertible element, contradicting properness. The quotient is a complex unital Banach division algebra, and the Gelfand-Mazur theorem identifies it with $\mathbb{C}$, producing a character. The same argument applied to the proper ideal generated by $a-\lambda$ provesEvery character has norm one, since its value belongs to the spectrum and is bounded by $\Vert a\Vert$, while $\chi (1)=1$. The character equations are weak-star closed in the dual unit ball. Thus Banach-Alaoglu theorem makes $X$ compact Hausdorff; separability of $A$ makes it metrizable.

The algebra is inverse-closed in $B(H)$. If $a\in A$ is invertible there, $a^{\ast }a\ge cI$ for some $c>0$. For $M=\Vert a^{\ast }a\Vert$, the Neumann series for $(a^{\ast }a{)}^{-1}=M^{-1}[I-(I-a^{\ast }a/M){]}^{-1}$ converges in $A$, since the positive contraction in brackets has norm at most $1-c/M<1$. Therefore $a^{-1}=(a^{\ast }a{)}^{-1}{a}^{\ast }\in A$. This proves spectral permanence for C-star algebras in the present concrete setting. In particular a self-adjoint element has real spectrum in $A$, so characters take real values on it. Decomposing $a$ into self-adjoint real and imaginary parts gives $\chi (a^{\ast })=\overline{\chi (a)}$.

Define $\hat{a}(\chi )=\chi (a)$. This is a unital star-homomorphism to $C(X)$. Every element of the abelian star-algebra is normal, and the C-star identity gives $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral radius formula therefore gives $r_A(a)=\Vert a\Vert$. The displayed spectrum identity now yields $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$, proving isometry and injectivity. Its image is closed because $A$ is complete, contains the constants, is closed under conjugation, and separates distinct characters. The Stone-Weierstrass theorem makes that image dense in $C(X)$, hence equal to it. $\boxed{A\cong C(X)}$ isometrically as a star-algebra.

Apply part (c) to the commutative unital C-star algebra $A=C^{\ast }(I,T)$, commutative because $T$ is normal. A character is determined by $z=\chi (T)$, since $T,T^{\ast }$ generate the algebra and $\chi (T^{\ast })=\overline{z}$. The spectrum identity in part (c) shows that every $z\in {\sigma }_A(T)$ occurs. Thus $X\to {\sigma }_A(T)$ is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Spectral permanence identifies this spectrum with the operator spectrum.

For any complex polynomial $p$ in two commuting variables, evaluation gives $\chi (p(T,T^{\ast }))=p(z,\overline{z})$. Again using the spectrum identity and spectral permanence,This proves the requested spectral mapping theorem, including constants, and explains why conjugation in the second coordinate is required.