Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 106 (Functional Analysis)

A character of an algebra is a nonzero multiplicative complex-linear functional $\phi :A\to \mathbb{C}$, and the character space of an algebra is the set ${\Phi }_A$ of all such characters. Since $A$ is unital, $\phi (1)=1$. Moreover $\phi (a)\in {\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. The spectral radius estimate therefore yieldsThus every character is continuous and has norm one.

Let $M$ be a maximal ideal. Its norm closure is again an ideal. It cannot equal $A$, because then some $m\in M$ would satisfy $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Hence $M$ is closed. The quotient $A/M$ is a complex unital Banach division algebra, so the Gelfand-Mazur theorem identifies it with $\mathbb{C}$. Composing the quotient map with this isomorphism gives a character with kernel $M$. Conversely, a character kernel is maximal because its quotient is $\mathbb{C}$.

Now $\lambda \in {\sigma }_A(x)$ exactly when $x-\lambda 1$ is not invertible, equivalently when it lies in some maximal ideal. The preceding result turns that ideal into $\ker \phi$, giving $\phi (x)=\lambda$. The reverse implication follows from the first paragraph, so

The Gelfand topology is the weak-star topology on ${\Phi }_A\subseteq A^{\ast }$. The Gelfand transform isIts values are continuous by the definition of the topology, and multiplicativity and linearity of characters show that it is a unital algebra homomorphism. Finallyso it is continuous.

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The holomorphic functional calculus assigns to every function $f$ holomorphic on a neighbourhood of ${\sigma }_A(x)$ the elementwhere the oriented contour $\Gamma$ surrounds the spectrum inside that neighbourhood. The value is independent of the admissible contour, and $f\mapsto f(x)$ is a continuous unital algebra homomorphism sending the coordinate function $z$ to $x$.

Every $\phi \in {\Phi }_A$ commutes with the contour integral, so the Cauchy integral formula givesPart a applied to $f(x)$ now proves the spectral mapping theorem:

Let $\Omega$ be the unbounded component of $\mathbb{C}\setminus {\sigma }_A(x)$. On the spectrum, spectral mapping givesEvery remaining point of $\mathbb{C}\setminus \Omega$ lies in a bounded complementary component $D$. Its boundary is contained in ${\sigma }_A(x)$, and $f$ is holomorphic near $\overline{D}$. The maximum modulus principle therefore extends the same estimate from $\partial D$ to $D$. Hence

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For a closed unital subalgebra $A\subseteq B$ containing $x$, invertibility in $A$ implies invertibility in $B$, so ${\sigma }_B(x)\subseteq {\sigma }_A(x)$. On a connected component of the resolvent set of $x$ in $B$, the set of $\lambda$ for which $(x-\lambda 1{)}^{-1}\in A$ is both open, by a local Neumann series, and closed, by closedness of $A$. It contains all sufficiently large $\lambda$, hence the entire unbounded component. Thus spectrum in a closed unital subalgebra says that ${\sigma }_A(x)$ is ${\sigma }_B(x)$ with some bounded complementary components filled in.

Now let $A$ be the Banach subalgebra generated by one element $x$ and put $K={\sigma }_A(x)$. If $D$ were a bounded component of $\mathbb{C}\setminus K$, choose $\lambda \in D$. Since $(x-\lambda 1{)}^{-1}\in A$, polynomials $p_n(x)$ converge to it. The polynomialssatisfy $q_n(\lambda )=0$ and $q_n(x)\to 1$. Applying the contractive Gelfand transform gives $q_n\to 1$ uniformly on $K$, hence on $\partial D$. But the maximum modulus principle applied to $1-q_n$ givesa contradiction. Therefore $\mathbb{C}\setminus K$ is connected.

The mapis continuous and surjective by part a. It is injective because characters agreeing on $x$ agree on every polynomial in $x$, hence by continuity on their norm closure $A$. The character space is compact by the Banach-Alaoglu theorem, while $K$ is Hausdorff, so this continuous bijection is a homeomorphism.

Under this identification, the Gelfand transform $\theta :A\to C(K)$ obeysby the calculation in part b. Since $f(x)\in A$, choose polynomials $p_n$ with $p_n(x)\to f(x)$. Contractivity of $\theta$ givesThus every function holomorphic near $K$ is uniformly approximable there by polynomials.

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Choose a norm-dense sequence $(f_n)$ in the unit sphere of the separable dual $X^{\ast }$. For every $n$, choose $x_n\in B_X$ with $|f_n(x_n)|>1/2$. If a functional $f\in X^{\ast }$ vanished on the closed linear span of the $x_n$, normalize it and choose $f_n$ with $\Vert f-f_n\Vert <1/4$; then $|f(x_n)|>1/4$, a contradiction. The Hahn-Banach theorem therefore shows that the span of $(x_n)$ is dense, so $X$ is separable.

Choose a norm-dense sequence $(u_n)$ in $B_X$. On $B_{X^{\ast }}$ defineWeak-star convergence implies convergence in this metric. Conversely, metric convergence gives convergence on the dense set $(u_n)$, and the uniform norm bound on the dual ball extends it to every $u\in X$. Thus $d_{\ast }$ metrizes the weak-star topology, proving weak-star metrizability of the dual ball.

Similarly, for a norm-dense sequence $(f_n)$ in $B_{X^{\ast }}$,metrizes the weak topology on $B_X$. Indeed, convergence against the dense functionals extends to every $f\in X^{\ast }$ because $x-y$ remains norm bounded.

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By definition of the weak topology, $x_n\rightharpoonup x$ exactly when $f(x_n)\to f(x)$ for every $f\in X^{\ast }$.

Let $(x_n)$ be bounded and choose a norm-dense sequence $(f_j)$ in $X^{\ast }$. Successive subsequences make $f_1(x_n),f_2(x_n),\dots$ converge, and the diagonal argument produces a single subsequence $(y_n)$ on which every $f_j$ converges. Uniform boundedness of $(y_n)$ and norm approximation of an arbitrary $f$ by the $f_j$ show that $f(y_n)$ is Cauchy for every $f\in X^{\ast }$. Thus $(y_n)$ is weakly Cauchy, andfor every $f$, so its difference sequence is weakly null.

For the countable family $(y_{m,n}{)}_n$, use the weak metric from part a. Delete a finite initial segment from the $m$th sequence so that every remaining term has weak distance less than $1/m$ from zero, and relabel that tail. Enumerate all these tails while preserving the order within each one. For every weak neighbourhood of zero, all terms from sufficiently large $m$ lie inside it, and only finitely many terms from each of the finitely many remaining sequences lie outside it. The resulting enumeration $(z_n)$ is weakly null and contains the relabelled $m$th sequence as a subsequence for every $m$. Equivalently, without relabelling, it contains a tail-subsequence of every original sequence, which is the form used below.

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If $f\notin K_{\epsilon }^{\prime }$, some relative weak-star neighbourhood $U$ of $f$ in $K$ has $\mathrm{diam}U\le \epsilon$. Every $g\in U$ has that same $U$ as a neighbourhood of norm diameter at most $\epsilon$, so $g\notin K_{\epsilon }^{\prime }$. Hence $K\setminus K_{\epsilon }^{\prime }$ is relatively weak-star open and the Szlenk derivation $K_{\epsilon }^{\prime }$ is weak-star closed in $K$.

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The set $K$ is weak-star compact by the Banach-Alaoglu theorem and weak-star metrizable by part a. Choose a decreasing countable neighbourhood base $(U_n)$ at $f$. Since $f\in K_{\epsilon }^{\prime }$, the set $U_n\cap K$ cannot lie inside the closed norm ball of radius $\epsilon /2$ about $f$, for then its diameter would be at most $\epsilon$. ChooseThe neighbourhood-base property gives $g_n\stackrel{w^{\ast }}{\to }f$.

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Start with the sequence from part ii. For each $n$, the definition of the operator norm gives $x_n\in B_X$ withBecause $g_n\stackrel{w^{\ast }}{\to }f$, pass recursively to a subsequence so far out that its $n$th member also satisfies

Part b supplies a weakly Cauchy subsequence $(x_{k_j})$. PutThen $(y_j)$ is weakly null, $h_j\stackrel{w^{\ast }}{\to }f$, and the two preceding estimates giveSince $f(y_j)\to 0$, discard finitely many terms to obtain $|h_j(y_j)|>\epsilon /16$ for every remaining $j$. Relabelling proves the claim.

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Part i makes $K_{\epsilon }^{\prime }$ a closed subset of the compact metrizable space $K$, so it is compact metrizable and therefore separable. Choose the stated dense sequence $(f_m)$ and, using part iii, the sequences $(g_{m,n})$ and $(y_{m,n})$. In constructing the universal weakly null sequence $(z_n)$ from part b, retain a tail of each $y_m$ and write $z_{j(m,n)}=y_{m,n}$ along the retained subsequence, where $j(m,n)\to \infty$.

Eachis weak-star closed. Since $z_n\rightharpoonup 0$, every fixed $f\in K$ belongs to some $A_N$, so $K={\bigcup }_N{A}_N$. The compact Hausdorff space $K$ is a Baire space, and the Baire category theorem implies that some $A_N$ has nonempty relative weak-star interior.

Suppose $K_{\epsilon }^{\prime }=K$. Choose a nonempty relatively open $O\subseteq A_N$ and then $f_m\in O$ by density. Since $g_{m,n}\stackrel{w^{\ast }}{\to }{f}_m$, eventually $g_{m,n}\in O$. For a sufficiently late retained index, also $j(m,n)\ge N$, and hencecontradicting $|g_{m,n}(y_{m,n})|>\epsilon /16$. Therefore the One-step Szlenk derivation for a separable dual gives $K_{\epsilon }^{\prime }⊊K$ whenever $K$ is nonempty.

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An element $x$ of a unital C-star algebra is positive when $x=y^{\ast }y$ for some $y$, equivalently when $x=x^{\ast }$ and $\sigma (x)\subseteq [0,\infty )$. The continuous functional calculus for the nonnegative function $t\mapsto \sqrt{t}$ defines a positive element $x^{1/2}$ satisfying $(x^{1/2}{)}^2=x$. If a positive $z$ also satisfies $z^2=x$, functional calculus for $z$ gives $z=\sqrt{z^2}=\sqrt{x}$, proving uniqueness. For a positive operator $T=S^{\ast }S$ on ${\ell }^2$,

For arbitrary $T$, put $|T|=(T^{\ast }T{)}^{1/2}$. Thenso $\ker |T|=\ker T$. Defineon $\mathrm{im}|T|$. The kernel identity makes this well-defined, and the norm identity makes it an isometry. Extend it continuously toand set it equal to zero on $\ker T$. The resulting $U$ is a partial isometry, has $\ker U=\ker T$, and satisfies the polar decomposition of a bounded operator $T=U|T|$.

Finally $U^{\ast }U$ is the orthogonal projection onto $(\ker T{)}^{\perp }$, which contains the range of $|T|$. Therefore

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Twice applying Parseval identity for a Hilbertian basis to the matrix coefficients givesThus $\Vert T{\Vert }_2=\Vert T^{\ast }{\Vert }_2$.

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The operator norm givesUsing part i and the first inequality for the adjoints gives

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By the definition of the trace norm and positivity of $|T|$,

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Let $T=AB$ with $A,B$ Hilbert--Schmidt, and write $T=U|T|$. Since $|T|=U^{\ast }AB$, part iii and the Cauchy-Schwarz inequality for series giveThus $T$ is trace class with the required bound.

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For the polar decomposition $T=U|T|$, setThen $T=AB$. Part iii gives $\Vert B{\Vert }_2^2=\Vert T{\Vert }_1$. The range of $B$ lies in the initial space $(\ker T{)}^{\perp }$ on which $U$ is isometric, so $\Vert A{\Vert }_2=\Vert B{\Vert }_2$. Hence $A,B$ are Hilbert--Schmidt andproving the Hilbert-Schmidt factorization of a trace-class operator.

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Use the norm-attaining factorization $T=AB$ from part v. Then $ST=(SA)B$, parts ii and iv giveThus the trace-class operators form a left ideal. Applying the result to adjoints gives the corresponding right-ideal estimate as well.

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Factor $T=AB$ as in part b(v). Then Parseval identity for a Hilbertian basis and Cauchy-Schwarz inequality giveThe defining series for the operator trace is therefore absolutely convergent and satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

For unit vectors $x,y$, the adjoint operator of $R=x\otimes y$ is $R^{\ast }=y\otimes x$, andHence $|R|=P_{\mathbb{C}y}$ and part b(iii), followed by Parseval, givesScaling proves $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$. A second use of Parseval yieldswhich proves the asserted rank-one operator formulas.

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Part b(vi) and part c show that $\Vert T{\Vert }_1\le 1$ impliesConversely, choose a unit vector $x$ with $\Vert Sx\Vert$ arbitrarily close to $\Vert S\Vert$, put $y=Sx/\Vert Sx\Vert$, and take $T=x\otimes y$. Part c gives $\Vert T{\Vert }_1=1$ andTaking the supremum provesThus $S\mapsto [T\mapsto \mathrm{tr}(ST)]$ is an isometric embedding ${\mathcal{S}}_{\infty }\to {\mathcal{S}}_1^{\ast }$.

Suppose finite-rank operators are dense in ${\mathcal{S}}_1$ and let $L\in {\mathcal{S}}_1^{\ast }$. The sesquilinear formsatisfies $|b(x,y)|\le \Vert L\Vert \Vert x\Vert \Vert y\Vert$. The Riesz representation theorem gives $S\in {\mathcal{S}}_{\infty }$ with $b(x,y)=⟨Sx,y⟩$. Hence $L(x\otimes y)=\mathrm{tr}(S(x\otimes y))$, and linearity gives equality on every finite-rank operator. Density and continuity extend it to every $T\in {\mathcal{S}}_1$, proving surjectivity.

Conversely, if finite-rank operators were not dense, the Hahn-Banach theorem would give a nonzero $L\in {\mathcal{S}}_1^{\ast }$ vanishing on their closure. Surjectivity would represent it by some $S$, but thenfor all $x,y$, forcing $S=0$ and $L=0$, a contradiction. This proves the stated trace duality criterion.

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For a real locally convex space $(X,\mathcal{P})$, the continuous dual space $X^{\ast }$ consists of all continuous real-linear maps $X\to \mathbb{R}$. If $g\in Y^{\ast }$ on a subspace $Y$, continuity gives seminorms $p_1,\dots ,p_k\in \mathcal{P}$ and $C>0$ such thatThe right side is a continuous sublinear functional on $X$. The dominated Hahn-Banach theorem extends $g$ to a linear $f:X\to \mathbb{R}$ satisfying the same bound, so $f\in X^{\ast }$.

If $Y$ is closed and $x_0\notin Y$, the Hausdorff locally convex quotient $X/Y$ has a continuous seminorm $p$ with $p(x_0+Y)>0$. On $Y+\mathbb{R}{x}_0$, define $g(y+t{x}_0)=t$ and rescale $p$ so that $|g|\le p$. Hahn--Banach extends it to $f\in X^{\ast }$ with $f{|}_Y=0$ and $f(x_0)=1$.

The separation of a point and an open convex set says that if $C\subseteq X$ is nonempty, open, and convex and $x_0\notin C$, there is $f\in X^{\ast }$ such thatTo prove it, choose $a\in C$, put $U=C-a$, and let $p_U$ be its Minkowski functional. For $z=x_0-a\notin U$, $p_U(z)\ge 1$. Define $h(tz)=t{p}_U(z)$ on $\mathbb{R}z$; then $h\le p_U$. Hahn--Banach extends $h$ to $f\le p_U$. Since $p_U(u)<1$ for $u\in U$, one has $f(c-a)<1\le f(x_0-a)$, proving the claim.

If $K$ is closed and convex and $x_0\notin K$, a Hahn-Banach separation theorem gives $f\in X^{\ast }$ and a real number separating $x_0$ from $K$. The corresponding inverse image of an open interval is a weak neighbourhood of $x_0$ disjoint from $K$, so $K$ is closed in $\sigma (X,X^{\ast })$.

The unit sphere $S_X$ of a normed space is norm closed. If $X$ is infinite-dimensional, every basic weak neighbourhood of an interior point $x\in B_X$ constrains only finitely many functionals $f_1,\dots ,f_n$. Their common kernel contains a nonzero $y$. Continuity of $t\mapsto \Vert x+ty\Vert$, together with its value below one at $t=0$ and divergence as $|t|\to \infty$, supplies $t$ with $\Vert x+ty\Vert =1$. Since all $f_i$ have the same values at $x+ty$ and $x$, every weak neighbourhood of $x$ meets $S_X$. Points outside $B_X$ are separated from it by Hahn--Banach, so the weak closure of the unit sphere is exactly $B_X$. In particular, $S_X$ is not weakly closed.

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The seminorms in $\mathcal{R}$ have inverse images of intervals that constrain one coordinate at a time. Finite intersections of these sets are exactly the standard basic neighbourhoods for the product topology, so $(X\times Y,\mathcal{R})$ is the product of locally convex spaces. If $F\in (X\times Y{)}^{\ast }$, thenbelong to $X^{\ast }$ and $Y^{\ast }$ and $F(x,y)=f(x)+g(y)$. Conversely every such pair defines a continuous functional, so $(X\times Y{)}^{\ast }=X^{\ast }\oplus Y^{\ast }$.

For the open convex sets $K_1,\dots ,K_n$, considerThis set is open and convex, and $0\notin C$ because the $K_i$ have empty intersection. Separate $0$ from $C$ by a continuous linear functional on the product. By the dual description just proved, it has the formfor $f_j\in X^{\ast }$, and $L$ is nonzero with one strict sign on $C$. DefineIf some vector belonged to every $T(K_i)$, choose $x_i\in K_i$ with the same image. Then every $f_j(x_j-x_n)=0$, making $L(x_1-x_n,\dots ,x_{n-1}-x_n)=0$, a contradiction. Hence ${\bigcap }_{i}T(K_i)=\varnothing$, proving finite-dimensional separation of open convex sets.

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If $X$ is a reflexive Banach space, its closed unit ball is weakly compact. A bounded linear map is weak-to-weak continuous, so $T(B_X)$ is weakly compact in $Y$. Compact subsets of a Hausdorff space are closed, hence $T(B_X)$ is weakly closed and therefore norm closed.

Now suppose the two norms on $V$ have the same continuous dual as a set. Each $X_j^{\ast }$ is a Banach space in its dual norm. The identityhas closed graph: if $f_n\to f$ in the first dual norm and $f_n\to g$ in the second, evaluating at each $x\in V$ gives $f(x)=g(x)$. The closed graph theorem makes $I$ bounded, and the same argument for $I^{-1}$ makes the dual norms equivalent. Thus there are $c,C>0$ withThe dual formulafrom the Hahn--Banach theorem transfers these inequalities to the original norms. Hence $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ are equivalent.

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