Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 106 (Functional Analysis)

Mazur theorem says that the weak closure and norm closure of a convex subset of a real or complex normed space coincide. The norm closure is contained in the weak closure because every norm-continuous linear functional is norm-continuous. 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 real number $\alpha$ such thatThis weakly open separation shows that $x$ is outside the weak closure.

By definition,For each $n$, regard $x_n$ as the bounded functional $J{x}_n:X^{\ast }\to \mathbb{F}$ given by $J{x}_n(f)=f(x_n)$. Pointwise convergence makes the family $(J{x}_n)$ pointwise bounded on the Banach space $X^{\ast }$. The Uniform boundedness principle gives

Suppose first that $x_n\rightharpoonup 0$. Every subsequence indexed by an infinite set $M$ also converges weakly to zero. Thus zero belongs to the weak closure of the convex hull of $\{x_n:n\in M\}$, and Mazur's theorem puts it in its norm closure. This gives the required finite convex combination of norm below any prescribed $\epsilon$.

Conversely, if weak convergence fails, some $f\in X^{\ast }$ and an infinite subsequence satisfy either $\mathrm{Re}f(x_n)\ge \epsilon$ throughout or $\mathrm{Re}f(x_n)\le -\epsilon$ throughout. Every convex combination from that subsequence then has norm at least $\epsilon /\Vert f\Vert$, contradicting the stated property.

Now let the $x_n\in {\ell }^p$, $1<p<\infty$, have pairwise disjoint supports and satisfy $\Vert x_n{\Vert }_p\le C$. For $N$ distinct terms,The convex-combination criterion therefore proves $x_n\rightharpoonup 0$. This is weak convergence of bounded disjointly supported sequences in lp.

The final implication for a commutative unital C*-algebra is true. By the Commutative Gelfand--Naimark theorem, $A\cong C(K)$ and its characters are the point evaluations. The hypotheses say that the uniformly bounded functions $a_n$ converge pointwise to zero. Every functional on $C(K)$ is integration against a finite regular measure by the Riesz-Markov-Kakutani representation theorem; the dominated convergence theorem givesHence $a_n\rightharpoonup 0$.

The holomorphic functional calculus assigns to $a$ in a unital complex Banach algebra and every function holomorphic near $\sigma (a)$ the elementwhere $\Gamma$ winds once around the spectrum. It is a unital algebra homomorphism, extends polynomial evaluation, is independent of the admissible contour, and obeys the spectral mapping theorem

Let $z_K\in \mathcal{R}(K)$ be the coordinate function. For a character $\phi$, put $\lambda =\phi (z_K)$. If $\lambda \notin K$, then $(z_K-\lambda {)}^{-1}$ is one of the admitted rational functions, contradicting the fact that $z_K-\lambda$ lies in the kernel of $\phi$. Thus $\lambda \in K$. For every rational function $r$ without poles on $K$,Continuity of characters and uniform density extend this identity to every member of $\mathcal{R}(K)$. Conversely evaluation at every $\lambda \in K$ is a character. Therefore the character space of R(K) is naturally $K$.

Runge approximation theorem says that if $K\subset \mathbb{C}$ is compact, $f$ is holomorphic on a neighbourhood of $K$, and one chooses one point in every bounded component of $\mathbb{C}\setminus K$, then $f$ can be approximated uniformly on $K$ by rational functions whose finite poles belong only to the chosen points.

To prove it, surround $K$ by finitely many small rectangles contained in the domain of $f$ and apply the Cauchy integral formula on their oriented boundaries:Riemann sums approximate this integral uniformly on $K$ by rational functions with poles on $\Gamma$. If a pole $b$ lies in a component containing the selected point $a$, choose a polygonal path from $b$ to $a$ inside that component and subdivide it finely. The resolvent identityallows the pole to be moved step by step along the path, with arbitrarily small uniform error on $K$. Moving every pole proves the theorem.

For an open set $U$, choose a compact exhaustion $K_1\subset K_2\subset \cdots$ such that every component of $\mathbb{C}\setminus K_n$ meets $\mathbb{C}\setminus U$. Runge's theorem approximates a holomorphic function on $K_n$ by a rational function with all poles outside $U$. A diagonal choice gives convergence uniformly on every compact subset, proving that such rational functions are dense in the space of holomorphic functions $\mathcal{O}(U)$ with its compact-open topology.

Finally, equality of the Gelfand transforms givesLet this common compact spectrum be $K$. The divided differenceis holomorphic near $K\times K$. The two-variable holomorphic functional calculus for the commuting pair $(x_1,x_2)$ gives an element $u=g(x_1,x_2)$ satisfyingFor every character,An element of a commutative unital Banach algebra is invertible exactly when its Gelfand transform has no zero. Thus $u$ is invertible, and $f(x_1)=f(x_2)$ implies $x_1=x_2$. This is injectivity through a holomorphic functional calculus with nonvanishing derivative.

Suppose first that $X$ is separable, and choose a norm-dense sequence $(x_n)$ in its unit ball. On the dual unit ball defineUniform boundedness on the unit ball and density of the $x_n$ show that this metric induces the weak-star topology. Conversely, if $B_{X^{\ast }}$ is weak-star metrizable, the Banach-Alaoglu theorem makes it a compact metric space. Hence $C(B_{X^{\ast }})$ is separable. The evaluation mapis an isometry by the Hahn-Banach theorem. A subspace of a separable metric space is separable, so $X$ is separable. This proves the weak-star metrizability criterion for a dual ball.

If $X$ has a countable weakly dense subset $D$, the rational linear span of $D$ is weakly dense. Its norm closure is a convex set, so Mazur theorem says that its weak and norm closures agree. Thus $X$ is norm separable. The weak-star compact metric ball $B_{X^{\ast }}$ consequently has a countable weak-star dense subset, and the union of its integer dilates is weak-star dense in $X^{\ast }$. Therefore $X^{\ast }$ is weak-star separable.

It need not be weakly separable. Take $X={\ell }^1$, whose dual is ${\ell }^{\infty }$. A weakly separable normed space is norm separable by the preceding convex-closure argument, whereas ${\ell }^{\infty }$ is not norm separable.

If the Banach space $X$ is reflexive, its closed unit ball identifies with the weak-star compact ball of $X^{\ast \ast }$, hence is weakly compact. Conversely, if $B_X$ is weakly compact, its canonical image $J(B_X)$ is weak-star compact and therefore weak-star closed in $X^{\ast \ast }$. Goldstine theorem says it is weak-star dense in $B_{X^{\ast \ast }}$, soScaling proves that $J$ is surjective and $X$ is reflexive. This is the weak compactness characterization of reflexivity.

The Krein-Milman theorem says that a nonempty compact convex subset of a locally convex space is the closed convex hull of its extreme points. For reflexive $X$, the ball $B_X$ is weakly compact, sowhere weak and norm closure agree for the convex hull by Mazur's theorem.

For the final claim, let $\mathcal{H}$ be the set of functions ${\mathbb{Z}}^2\to [0,1]$ with the mean-value property. It is a compact convex subset of the product $[0,1{]}^{{\mathbb{Z}}^2}$. If $f$ is extreme, its four unit translates also lie in $\mathcal{H}$, and the mean-value identity writes $f$ as their average. Extremality forces every translate to equal $f$, so $f$ is constant. Every extreme point is therefore constant. Krein--Milman now makes every member of $\mathcal{H}$ a limit of convex combinations of constant functions, and hence constant. This is the bounded harmonic function theorem on the integer lattice.

The useful forms of the Hahn-Banach separation theorem in a real locally convex space are:

The Banach-Alaoglu theorem says that the closed unit ball of $E^{\ast }$ is compact in $\sigma (E^{\ast },E)$. Goldstine theorem says that the canonical image of the closed unit ball of a normed space $E$ is weak-star dense in the closed unit ball of $E^{\ast \ast }$.

Give $Z$ its norm inherited from $X^{\ast }$ and defineIt is linear and contractive, and it is injective because $Z$ separates points. Goldstine's theorem followed by restriction from $X^{\ast }$ to $Z$ shows that $J(B_X)$ is weak-star dense in $B_{Z^{\ast }}$: a functional on $Z$ first extends norm-preservingly to $X^{\ast }$, and elements of $B_X$ approximate that extension on every finite subset of $Z$.

The topology induced by $\sigma (Z^{\ast },Z)$ on $J(B_X)$ is exactly the given topology $\sigma (X,Z)$. By hypothesis $B_X$ is compact, so $J(B_X)$ is weak-star compact and therefore closed in the Hausdorff space $B_{Z^{\ast }}$. Density now givesThus $J$ is surjective and maps closed unit ball onto closed unit ball, so it is an isometric isomorphism. Hence $X$ is a dual space. This is the compact norming dual-pair criterion.

The estimateshows $\Vert {\mu }_{x,y}\Vert \le \Vert x\Vert \Vert y\Vert$. For nonzero $x,y$, the rank-one operatorhas norm one and attains equality. The zero cases are immediate.

Embed the operator unit ball intoby $T\mapsto (⟨Tx,y⟩{)}_{x,y}$. The product is compact by Tychonoff theorem. A pointwise limit of these coordinates is a bilinear form $\theta$ satisfyingBy the stated representation theorem for bounded bilinear forms, $\theta (x,y)=⟨Tx,y⟩$ for a unique operator with $\Vert T\Vert \le 1$. The image is therefore closed and compact. Its product topology is precisely the weak operator topology $\sigma (\mathcal{B}(H),M)$.

The linear span $Z=\mathrm{span}M$ separates operators, and the preceding compactness lets part (a) identify $\mathcal{B}(H)$ isometrically with $Z^{\ast }$. Thus $\mathcal{B}(H)$ is a dual Banach space. This is the operator predual from matrix coefficients.

A character of an algebra is a nonzero algebra homomorphism $\phi :A\to \mathbb{C}$. The character space isFor a unital algebra, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$, since applying $\phi$ shows that $a-\phi (a)1$ cannot be invertible. In a Banach algebra,Thus $\Vert \phi \Vert \le 1$, while $\phi (1)=1$ gives equality.

If $A$ is commutative and $\lambda \in {\sigma }_A(a)$, the proper ideal generated by $a-\lambda 1$ lies in a maximal ideal. The quotient by that maximal ideal is $\mathbb{C}$, and its quotient map is a character taking $a$ to $\lambda$. ThereforeSince every Banach-algebra element has nonempty spectrum, ${\Phi }_A$ is nonempty.

For $X=Y\oplus Z$, let $P,Q$ be the complementary coordinate projections. A character must send each idempotent to zero or one, and $P+Q=I$ forces their values to be different. Choose mutually inverse isomorphisms between $Y$ and $Z$ and regard them as off-diagonal operators $S,T$ on $X$. ThenMultiplicativity would give $\phi (P)=\phi (S)\phi (T)=\phi (T)\phi (S)=\phi (Q)$, a contradiction. Hence ${\Phi }_{\mathcal{B}(X)}$ is empty. This is the absence of characters on an operator algebra with isomorphic complementary summands.

If $a=a^{\ast }$ in a unital C*-algebra, then $e^{ita}$ is unitary for real $t$. If $\lambda \in \sigma (a)$, the spectral mapping theorem gives $e^{it\lambda }\in \sigma (e^{ita})$, whose modulus is one. Varying positive and negative $t$ forces $\mathrm{Im}\lambda =0$, so $\sigma (a)\subset \mathbb{R}$.

If $B\subseteq A$ is a unital C*-subalgebra and $x\in B$ is normal, spectral permanence holds. Indeed, when $x-\lambda$ is invertible in $A$, the positive normal elementhas spectrum bounded away from zero. Continuous functional calculus uniformly approximates its reciprocal by polynomials, placing the reciprocal in $B$. It follows that $(x-\lambda {)}^{-1}\in B$. Thus ${\sigma }_B(x)={\sigma }_A(x)$.

The Commutative Gelfand--Naimark theorem says that the Gelfand transform is an isometric unital star-isomorphismfor every commutative unital C*-algebra. If $a$ is positive, continuous functional calculus for the function $t\mapsto \sqrt{t}$ on $\sigma (a)\subset [0,\infty )$ gives a positive $b$ with $b^2=a$. Pointwise uniqueness in $C^{\ast }(1,a)\cong C(\sigma (a))$ gives the unique positive square root. This is the positive square root in a C*-algebra.

SetBoth are hermitian and $x=h+ik$. If also $x=h^{\prime }+i{k}^{\prime }$ with hermitian $h^{\prime },k^{\prime }$, taking adjoints and then adding or subtracting the two equations gives $h=h^{\prime }$ and $k=k^{\prime }$.

For hermitian $x$ with $\Vert x\Vert \le 1$, its spectrum lies in $[-1,1]$. Continuous functional calculus therefore shows that $1-x^2$ is positive. LetIt commutes with $x$, andsatisfies $u^{\ast }u=u{u}^{\ast }=x^2+v^2=1$. Thus $u$ is unitary and

The hypothesis is possible only for $t\ge 0$. For every $\lambda \in \sigma (x)\subset \mathbb{R}$,so $0\le \lambda \le 2t$. The spectral characterization of a positive element in a C*-algebra gives $x\ge 0$.

Positivity and $\Vert x\Vert \le t$ give $\sigma (x)\subset [0,t]$. HenceThe element $t1-x$ is hermitian, so its norm equals its spectral radius and is at most $t$.

Put $s=\Vert x\Vert +\Vert y\Vert$. Part 4 givesThereforePart 3 now proves that $x+y$ is positive.