Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 106 (Functional Analysis)

The weak-star topology $\sigma (X^{\ast },X)$ is generated by the seminorms $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\stackrel{w^{\ast }}{\to }f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\epsilon$. Pointwise convergence eventually satisfies all of them simultaneously. HenceThe Uniform boundedness principle, applied on the Banach space $X$, gives $M={\sup }_n\Vert f_n\Vert <\infty$. If $x_n\to x$ in the norm topology, thenTherefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.

If $C\subseteq X^{\ast }$ is norm compact, the identity map from $(C,\Vert \cdot \Vert )$ to $(C,w^{\ast })$ is a continuous bijection. The target is Hausdorff, since distinct functionals differ at some $x\in X$. The compact-to-Hausdorff continuous bijection theorem gives

The Banach-Alaoglu theorem states that the closed dual unit ball $B_{X^{\ast }}$ is compact in the weak-star topology; in fact $X$ need only be a normed vector space. To prove it, work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and formEach $D_x$ is compact, so $Q$ is compact by the Tychonoff theorem. Embed $B_{X^{\ast }}$ by $g\mapsto (g(x){)}_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfyingThese are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\le \Vert x\Vert$, hence an element of $B_{X^{\ast }}$. The image is therefore closed and compact. The induced product topology is precisely the weak-star topology, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.

For the dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$, each $x\in X$ satisfiesEvaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^{\ast }\text{ is weak-star-to-weak-star continuous}}$.

Suppose $T$ is a compact operator. By the allowed Schauder theorem for compact operators, $T^{\ast }$ is compact. If $S\subseteq Y^{\ast }$ is bounded, the norm closure $C=\overline{T^{\ast }(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^{\ast }{|}_S$ into $C$ is consequently norm continuous. Conversely, if $T^{\ast }$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^{\ast }}$ maps a weak-star compact set, by Banach-Alaoglu theorem, to a norm compact set. Thus $T^{\ast }$ is a compact operator, and the Schauder theorem for compact operators gives

Use the usual unital-subalgebra convention $I_H\in A$. The spectral theorem for a commutative operator algebra states that the character space ${\Phi }_A$, equipped with the Gelfand topology, is a compact Hausdorff space, and there is a unique regular projection-valued measure $E$ on its Borel sigma-algebra satisfyingwhere $\hat{a}(\chi )=\chi (a)$ is the Gelfand transform. The map $a\mapsto \hat{a}$ is an isometric unital star-isomorphism from $A$ onto $C({\Phi }_A)$ by the Commutative Gelfand--Naimark theorem.

More explicitly, every $E(B)$ is an orthogonal projection, $E(\varnothing )=0$, $E(B\cap C)=E(B)E(C)$, and for disjoint Borel sets $B_n$,with convergence in the norm of the Hilbert space. Regularity means that each scalar spectral measure $B\mapsto ⟨E(B)v,w⟩$ is a regular Borel measure. The integral identity then implies $\Vert a\Vert ={\sup }_{\chi \in {\Phi }_A}|\hat{a}(\chi )|$. No separability assumption on $H$ is needed.

First construct the spectral measure of a normal operator and prove the open-set assertion needed for all three numbered conclusions. The algebra $A=C^{\ast }(I,T)$ generated by $T$ and $T^{\ast }$ is commutative because $T$ is a normal operator. The Gelfand transform identifies $A$ with $C({\Phi }_A)$. By spectral permanence for C-star algebras,is onto: its image is the spectrum of $T$ in $A$, which is also its spectrum in $\mathcal{B}(H)$. It is injective because a character is determined by its values on $T,T^{\ast }$ and $I$, and $\chi (T^{\ast })=\overline{\chi (T)}$. Thus $h$ is a homeomorphism. Apply part (a) and setThis is a projection-valued measure with $P(K)=I$ andThe associated continuous functional calculus is isometric: $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }$ for $u\in C(K)$.

If $U\subseteq K$ is nonempty and relatively open, choose $z_0\in U$ and a nonzero continuous $u$ supported inside $U$, for example a small continuous bump about $z_0$ in the metric of $K\subseteq \mathbb{C}$. If $P(U)=0$, then $u(T)={\int }_{K}udP=0$, contradicting $\Vert u(T)\Vert =\Vert u{\Vert }_{\infty }>0$. ThereforeThis proves the claim from faithfulness of the representation rather than assuming it as part of the spectral theorem.

Now take an isolated point $\lambda \in K$. Its singleton is relatively open, so $P(\{\lambda \})\ne 0$. Multiplication of spectral integrals givesThus its range is contained in the eigenspace $\ker (T-\lambda I)$. Conversely, for $v$ in that kernel, the positive scalar spectral measure ${\mu }_v(B)=⟨P(B)v,v⟩$ satisfiesIt follows that ${\mu }_v(K\setminus \{\lambda \})=0$, and hence $\Vert (I-P(\{\lambda \}))v{\Vert }^2=0$. ConsequentlyThe identification of this projection with the eigenspace holds even for a non-isolated $\lambda$; isolation guarantees that it is nonzero.

If $K=\{{\lambda }_0\}$, the projection-valued measure satisfies $P(\{{\lambda }_0\})=P(K)=I$. Its integral representation therefore reduces toThus a normal operator with a singleton spectrum is a scalar multiple of the identity operator. Normality is essential: a nonzero nilpotent matrix has spectrum $\{0\}$ but need not be zero.

If the spectrum has one point, the preceding part gives $T=\lambda I$, and the linear span of any nonzero vector is a closed proper invariant subspace because $\dim H>1$.

Otherwise choose distinct points ${\lambda }_1,{\lambda }_2\in K$ and disjoint nonempty relatively open neighborhoods $U,V\subseteq K$. The open-set result gives $P(U)\ne 0$ and $P(V)\ne 0$, whileThus $P(U)\ne I$. Set $L=\mathrm{ran}P(U)$. As the range of an orthogonal projection, $L$ is closed, and the preceding observations give $\{0\}\ne L\ne H$. The Borel functional calculus for a normal operator gives $TP(U)=P(U)T$, soIn fact both $L$ and its orthogonal complement are invariant.

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.

The Mazur theorem states that the weak closure and norm closure of a convex set $C$ in a normed vector space coincide:Norm closure is contained in weak closure because the weak topology is coarser. Conversely, if $x\notin {\overline{C}}^{\Vert \cdot \Vert }$, the Hahn-Banach separation theorem provides a continuous real linear functional strictly separating $x$ from that closed convex set. In a complex space this is the real part of a continuous complex linear functional. A weak neighborhood of $x$ then misses $C$, so $x\notin {\overline{C}}^w$. This proves the equality. It also gives the usual Mazur lemma: if $x_n\rightharpoonup x$, then $x$ lies in the norm closure of the convex hull of each tail, so one can choose tail convex combinations $y_n$ with $\Vert y_n-x\Vert <1/n$.

Now let $K$ be a weakly compact set in a normed space $X$. Each $\phi \in X^{\ast }$ is bounded on $K$ because it is weakly continuous. The family $\{Jx:x\in K\}$ in $\mathcal{B}(X^{\ast },\mathbb{F})$, where $J$ is the canonical embedding into the bidual, is therefore pointwise bounded. The space $X^{\ast }$ is Banach even if $X$ is not. Apply the Uniform boundedness principle and use $\Vert Jx\Vert =\Vert x\Vert$ to obtainThis proves that a weakly compact set is norm bounded without assuming completeness of the original space.

For the real-valued dual and integral formulas that follow, take $X$ to be real, as in the PDF. In a complex space the norming formula uses real parts, and the integral identities use complex-valued functionals instead.

If the separable Banach space $X$ is nonzero, choose a norm-dense sequence $(u_n)$ in its unit sphere. By the Hahn-Banach theorem, choose ${\phi }_n\in X^{\ast }$ with $\Vert {\phi }_n\Vert =1$ and ${\phi }_n(u_n)=1$. For any unit vector $u$, arbitrarily close $u_n$ satisfyScaling gives the countable norming family identityFor $X=\{0\}$ use the constant sequence of zero functionals. If $f:\Omega \to X$ is norm-Borel measurable, every ${\phi }_n\circ f$ is measurable, so its countable supremum $\Vert f(\cdot )\Vert$ is measurable. Equivalently, this also follows directly from continuity of the norm.

For any $\phi \in X^{\ast }$, continuity makes $\phi \circ f$ measurable, andThus the assumed integrability of the norm implies scalar integrability, anddefines a bounded linear functional on $X^{\ast }$. Use the granted weak-star continuity of $T_f$. By the continuous dual of a weak-star topology, $T_f$ is evaluation at a vector of $X$. Indeed, continuity gives finitely many $x_1,\dots ,x_m$ and $\epsilon >0$ such that $|T_f(\phi )|<1$ whenever $|\phi (x_j)|<\epsilon$ for all $j$. Scaling shows that $T_f$ vanishes on the common kernel of these evaluations. It therefore factors through their finite-dimensional coordinate map, so $T_f(\phi )={\sum }_j{a}_j\phi (x_j)=\phi ({\sum }_j{a}_j{x}_j)$. The Hahn-Banach theorem makes this representing vector unique. HenceIn this separable setting the vector is the Bochner integral.

Return to a weakly compact set $K\subseteq X$ and its inclusion $f:K\to X$. For each fixed $a\in X$, the identity $\Vert x-a\Vert ={\sup }_n{\phi }_n(x-a)$ makes $x\mapsto \Vert x-a\Vert$ weakly Borel measurable. Norm balls are consequently weakly Borel measurable. Separability gives a countable base of such balls, so every norm-open set is weakly Borel measurable. This proves measurability of $f$, and establishes the equality of the weak and norm Borel sigma-algebras in a separable Banach space.

Put $M={\sup }_{x\in K}\Vert x\Vert <\infty$. For every finite signed Borel measure $\mu$ on $K$,For positive measures this is the integral in the question. For signed measures, the correct integrability condition uses the variation measure; define the integral by taking the difference of the positive and negative integrals. The printed $\Omega$ in this clause should be $K$, the domain of the inclusion.

The Riesz-Markov-Kakutani representation theorem now defines the bounded linear mapFor each $\phi \in X^{\ast }$ the restriction $\phi {|}_K$ is in $C(K)$, andThe right side is weak-star continuous in $\mu$. The defining property of the weak topology therefore proves that $T$ is weak-star-to-weak continuous, for arbitrary nets. For a Dirac measure, $\boxed{T({\delta }_x)=x}$.

If $K\ne \varnothing$, let $\mathcal{P}(K)$ be its regular probability measures. This is a weak-star closed subset of $B_{C(K{)}^{\ast }}$: its conditions are $\mu (1)=1$ and $\mu (g)\ge 0$ for every nonnegative $g\in C(K)$. It is compact by Banach-Alaoglu theorem. Thus $T(\mathcal{P}(K))$ is weakly compact and convex, and contains $K$ because it contains all $T({\delta }_x)$. It is weakly closed, hence norm closed, so it contains $C={\overline{\mathrm{conv}K}}^{\Vert \cdot \Vert }$. By Mazur theorem, $C$ is weakly closed. Therefore it is a closed subset of the weakly compact set $T(\mathcal{P}(K))$, provingThe empty case is immediate. In fact $T(\mathcal{P}(K))=C$: a barycenter of a measure on a Banach space outside $C$ would be strictly separated by a functional $\phi$, contradicting $\phi (T(\mu ))={\int }_K\phi d\mu \le {\sup }_K\phi$.

For the first norm-one assertion, use the usual unital Banach algebra convention $\Vert 1\Vert =1$. A character of an algebra is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\phi (1)=1$, since otherwise $\phi (a)=\phi (a1)=0$ for every $a$. Moreover, $a-\phi (a)1$ cannot be invertible: applying $\phi$ to an inverse identity would give $0=1$. ThusThis proves automatic continuity of characters and $\Vert \phi \Vert \le 1$. Evaluation at the normalized identity gives the reverse inequality, soIf a convention permits $\Vert 1\Vert >1$, only $\Vert \phi \Vert \le 1$ follows in general: on $\mathbb{C}$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.

If $a$ is invertible, $\phi (a)\phi (a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a maximal ideal $M$ by Zorn lemma. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series, which is impossible for a proper ideal. Maximality therefore gives $\overline{M}=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb{C}$ by the Gelfand-Mazur theorem. The quotient map yields a character vanishing at $a$. Consequently

Now let $A=C(K)$. Its evaluation characters give the map $x\mapsto {\delta }_x\in {\Phi }_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\dots ,f_m$ in the kernel with no common zero. Then $h={\sum }_j\overline{f_j}{f}_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\phi (f)1$ belongs to the kernel gives $f(x)=\phi (f)$ for every $f$. Continuous functions distinguish points of the compact Hausdorff space $K$, so the map is bijective. It is continuous for the Gelfand topology, and the compact-to-Hausdorff continuous bijection theorem givesFor $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne \varnothing$.

Let $B$ be the completion for the other algebra norm $\Vert \cdot {\Vert }_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi \in {\Phi }_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original supremum norm. Thus $R(\psi )=\psi {|}_A$ maps into ${\Phi }_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi )(f)=\psi (f)$ is weak-star continuous.

The character space ${\Phi }_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^{\ast }}$, and the conditions $\psi (1)=1$, $\psi (bc)=\psi (b)\psi (c)$ define a weak-star closed set. Apply Banach-Alaoglu theorem. Consequently $R$ is a homeomorphism onto a closed subset $L\subseteq K$.

Suppose $x\in U=K\setminus L$. The compact Hausdorff space $K$ is normal, so choose an open $V$ with $x\in V\subseteq \overline{V}\subseteq U$. By the Urysohn lemma, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline{V}$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi (f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.

For each $x\in K$, choose a character ${\psi }_x\in {\Phi }_B$ extending ${\delta }_x$. Its spectral bound givesTaking the supremum over $x$ proves the minimality of the supremum norm on C(K):

The Beurling–Gelfand spectral radius formula states that for every element $a$ of a complex Banach algebra,For a Hermitian element of a C-star algebra $a=a^{\ast }$, the C-star identity gives $\Vert a^2\Vert =\Vert a{\Vert }^2$. Iterating,The subsequence $n=2^m$ in the spectral-radius limit therefore proves $\boxed{r(a)=\Vert a\Vert }$.

Let $\theta :A\to B$ be a unital C-star homomorphism. If $a-\lambda 1$ is invertible in $A$, its inverse is sent to an inverse of $\theta (a)-\lambda 1$ in $B$. Thus ${\sigma }_B(\theta (a))\subseteq {\sigma }_A(a)$ without assuming continuity of $\theta$. For Hermitian $a$, $\theta (a)$ is also Hermitian, soFor arbitrary $x\in A$, apply this inequality to $x^{\ast }x$ and the C-star identity in both algebras:Hence $\boxed{\Vert \theta (x)\Vert \le \Vert x\Vert }$ for every $x$.

Now assume $\theta$ is injective. For a fixed $x$, it is enough to prove norm preservation on $C=C^{\ast }(1,x^{\ast }x)$, a commutative unital C-star algebra: the preceding square-norm identity will then recover the norm of $x$. By the Commutative Gelfand--Naimark theorem, identify $C$ with $C(K)$. Let $D=\overline{\theta (C)}\subseteq B$. It is also a commutative unital C-star algebra, and its character space ${\Phi }_D$ is compact. Each $\chi \in {\Phi }_D$ restricts along $\theta$ to a character of $C(K)$, hence to evaluation at a point $q(\chi )\in K$. The map $q:{\Phi }_D\to K$ is continuous, so $L=q({\Phi }_D)$ is closed.

If $L\ne K$, the Urysohn lemma provides a nonzero $f\in C(K)$ vanishing on $L$. Then $\chi (\theta (f))=f(q(\chi ))=0$ for every $\chi \in {\Phi }_D$. Faithfulness of the Gelfand transform in $D$ gives $\theta (f)=0$, contradicting injectivity. Thus $q$ is onto. The isometric Gelfand transform of $D$ now givesFinally,This proves that an injective C-star homomorphism is isometric without assuming its image is closed before norm preservation has been established.