Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 6 (Introduction to Functional Analysis)

First prove the Baire category theorem in its complete-metric form. Let $X$ be a nonempty complete metric space, let $G_n$ be open dense sets, and let $V$ be any nonempty open set. Choose a closed ball $\overline{B}(x_1,r_1)\subset V\cap G_1$ with $0<r_1<1/2$. Inductively, density and openness permit the choiceFor $m>n$, $x_m$ belongs to the $n$th closed ball, so $d(x_m,x_n)\le r_n$. The centers form a Cauchy sequence and converge to some $x\in X$ by completeness. Each closed ball contains all later centers and therefore contains their limit. Hence $x\in V\cap {\bigcap }_n{G}_n$. Since this holds for every $V$, every countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of closed sets with empty interior; this is the form needed below.

For the Banach-Steinhaus theorem, let $X$ be a Banach space, $Y$ a normed vector space, and $\mathcal{T}$ a family of bounded linear operators from $X$ to $Y$ such that ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x$. DefineEach $E_m$ is closed, as an intersection of inverse images of closed balls under continuous maps, and pointwise boundedness gives $X={\bigcup }_m{E}_m$. The Baire category theorem gives some $E_m$ with nonempty interior. Choose $B(x_0,r)\subset E_m$. If $\Vert h\Vert \le r/2$, both $x_0+h$ and $x_0$ lie in $E_m$, so $\Vert Th\Vert \le 2m$ for every $T$. Applying this to $h=(r/2)u$ with $\Vert u\Vert \le 1$ provesThis proves the Uniform boundedness principle; completeness is required for the domain, but not for $Y$.

Now fix the prescribed point ${\theta }_0$ on the unit circle, parametrized by angles modulo $2\pi$. On the real Banach space $C(\mathbb{T};\mathbb{R})$ with the supremum norm, consider the continuous linear functionals $T_{N}f=S_{N}f({\theta }_0)$, where $S_N$ is the symmetric Fourier partial sum. Summing the finite geometric series gives its Dirichlet kernel representationThe continuous extension at $t=0$ is $2N+1$. The kernel bound gives $\Vert T_N\Vert \le {\Lambda }_N=(2\pi {)}^{-1}\int |D_N|$. Equality holds: the continuous periodic functions defined byhave supremum norm at most one, and dominated convergence gives $T_N{f}_{\delta }\to {\Lambda }_N$ as $\delta \downarrow 0$.

These norms are unbounded. Indeed, evenness and $\sin (t/2)\le t/2$ for $0<t\le \pi$ giveFor the second inequality, integrate over $[(j-1)\pi ,j\pi ]$, where $1/u\ge 1/(j\pi )$ and the integral of $|\sin u|$ is two. The harmonic series diverges, proving the Dirichlet kernel harmonic lower bound. If every continuous $f$ had bounded partial sums at ${\theta }_0$, the Banach-Steinhaus theorem would bound all the operator norms, a contradiction. Consequently there exists a continuous real function $f$ with ${\sup }_N|S_{N}f({\theta }_0)|=\infty$, and its Fourier series diverges at the prescribed point. The function depends on the chosen point; the argument requires no claim of simultaneous divergence everywhere.

Use the dominated form of the real Hahn-Banach theorem. Let $p:V\to \mathbb{R}$ be a sublinear functional on a real vector space, meaning $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for $t\ge 0$. If a linear functional $f$ on a vector subspace $M$ satisfies $f\le p$ there, then it has a linear extension $F$ to $V$ with $F\le p$ everywhere.

Here is the one-dimensional extension step. For $z\notin M$, the value $a=F(z)$ must satisfyThis interval is nonempty: for any $m,n\in M$,so every candidate lower bound is at most every candidate upper bound. The candidates with $m=n=0$ also bound the endpoints by finite numbers, so a real $a$ can be chosen. Define $F(m+tz)=f(m)+ta$. For $t>0$, the upper bound, applied to $m/t$, gives $F(m+tz)\le p(m+tz)$ after multiplication by $t$. For $t<0$, the lower bound applied to $m/(-t)$ gives the same conclusion after multiplication by $-t$; $t=0$ is the original hypothesis. Thus the extension remains dominated.

Order all dominated extensions by extension of their domains and values. Every chain has the dominated linear union as an upper bound. The Zorn lemma supplies a maximal extension. If its domain were not all of $V$, the step just proved would enlarge it, a contradiction. This proves the real Hahn-Banach extension theorem. In particular, taking $p(x)=\Vert f\Vert \Vert x\Vert$ on a normed vector space and applying the domination to both $x$ and $-x$ yields the usual norm-preserving extension of a bounded linear functional.

For generalized limits, take $V={\ell }^{\infty }(\mathbb{N};\mathbb{R})$, the l-infinity sequence space, and $M=c$, the convergent sequence space. Let $f(x)={\lim }_n{x}_n$ on $c$ and $p(x)={\mathrm{limsup}}_n{x}_n$ on $V$. This $p$ is finite and sublinear, and agrees with $f$ on $c$. The dominated extension $L$ therefore satisfiesThus a generalized limit exists, withIt is positive, agrees with the ordinary limit on every convergent sequence, and satisfies $|L(x)|\le \Vert x{\Vert }_{\infty }$. Since $L(1)=1$, its operator norm is exactly one.

If generalized limits are additionally required to be shift invariant, the same theorem gives the stronger Banach limit. Replace the dominating functional byIt is again sublinear and agrees with $f$ on $c$, by the Cesàro theorem for convergent sequences. The resulting $L_C$ lies between the lower and upper limits of the Cesàro means. These in turn lie between the ordinary lower and upper limits of a bounded sequence, so $L_C$ is a positive, norm-one extension of the ordinary limit. For the left shift on bounded sequences $S$, the means of $Sx-x$ equal $(x_{N+1}-x_1)/N\to 0$. Applying domination to both signs forces $L_C(Sx-x)=0$. HenceThis Cesàro construction of a Banach limit proves existence even under the stronger convention, rather than leaving shift invariance implicit.

Work in a nonzero complex unital Banach algebra $B$, with identity $1$. Write $R(x)=\{\lambda \in \mathbb{C}:\lambda 1-x\text{ is invertible}\}$ for the resolvent set, $\sigma (x)=\mathbb{C}\setminus R(x)$ for the spectrum of an element, and $Q(\lambda )=(\lambda 1-x{)}^{-1}$ for its Banach algebra resolvent. Distinguishing the set from the inverse-valued function avoids a notational ambiguity.

The printed formula omits an essential modulus. Complex spectral values cannot be ordered to take the indicated supremum. Even for $x=-1$ in $B=\mathbb{C}$, the spectrum is $\{-1\}$ while the spectral radius is one. The correctly interpreted assertion, proved below, isAll required analytic facts about the Banach algebra resolvent will be derived from norm-convergent series and scalar complex analysis.

For $\Vert a\Vert <1$, the Neumann series ${\sum }_{n\ge 0}{a}^n$ converges in $B$ and is a two-sided inverse of $1-a$: multiplying its finite sums gives $1-a^{N+1}$ and passing to the limit gives $1$. Therefore for $|\lambda |>\Vert x\Vert$,The series converges locally uniformly on that exterior region and tends to zero in norm as $|\lambda |\to \infty$. This statement does not require the optional normalization $\Vert 1\Vert =1$; the zeroth term is $1/\lambda$, and the other terms have the usual geometric bound.

At any ${\lambda }_0\in R(x)$, factor ${\lambda }_{0}1-x+h1=({\lambda }_{0}1-x)(1+hQ({\lambda }_0))$. If $|h|\Vert Q({\lambda }_0)\Vert <1$, the Neumann series givesThus the resolvent set is open, $Q$ is norm-continuous there, and the series proves norm differentiability with $Q^{\prime }(\lambda )=-Q(\lambda {)}^2$. Repeated differentiation gives $Q^{(j)}(\lambda )=(-1{)}^{j}j!Q(\lambda {)}^{j+1}$. Direct subtraction of the inverse equations also yields the resolvent identityIn particular, every continuous linear functional $\varphi$ makes $\varphi (Q(\lambda ))$ a scalar holomorphic function; this has been proved, not assumed as an algebra-valued analytic theorem.

The spectrum of an element is closed, and is contained in $\{\lambda :|\lambda |\le \Vert x\Vert \}$ by the exterior series, so it is compact. It is nonempty. Otherwise $Q$ would be defined and norm-continuous on the entire plane, bounded on every fixed closed disk and tending to zero at infinity. Each $\varphi (Q)$ would be a bounded entire scalar function, hence identically zero by Liouville's theorem. Continuous complex linear functionals separate points of $B$, so $Q(\lambda )=0$, impossible for an inverse. This separation follows from the real Hahn-Banach theorem: a continuous real functional $h$ not vanishing on a given vector gives the complex linear functional $\varphi (v)=h(v)-ih(iv)$, which also does not vanish there. We have therefore proved thatis well-defined and finite.

Next establish the existence of the power-norm limit. If $x^m=0$ for some $m$, the limit is zero. Otherwise fix $m$ and write $n=qm+r$, $0\le r<m$. Submultiplicativity givesThe finitely many remainder factors have bounded size, and $q/n\to 1/m$. Hence ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$ for every $m$. The opposite lower bound by the infimum is automatic, soFor $|\lambda |>r_0$, the root test proves convergence of $\sum x^n/{\lambda }^{n+1}$ even if $|\lambda |\le \Vert x\Vert$. Its finite sums telescope against $\lambda 1-x$, and their remainder tends to zero, so this sum is an inverse. Thus $s\le r_0$.

For the reverse inequality, fix $R>s$ and first choose $T>\max (R,\Vert x\Vert )$. The exterior Neumann series converges uniformly on $|\lambda |=T$, so termwise vector integration givesOnly its $x^n{\lambda }^{-n-1}$ term contributes. The annulus between radii $R$ and $T$ is in the resolvent set. For every continuous complex linear $\varphi$, the scalar Cauchy integral theorem applied to ${\lambda }^n\varphi (Q(\lambda ))$ makes the two contour integrals equal. Continuous linear functionals commute with vector integration and separate points, so the vector integrals themselves agree. This proves the resolvent Cauchy coefficient formula without presupposing any Banach-algebra-valued analytic-function theorem:Let $M_R={\max }_{|\lambda |=R}\Vert Q(\lambda )\Vert$, finite by continuity. The contour-length estimate gives $\Vert x^n\Vert \le M_R{R}^{n+1}$. Taking $n$th roots gives $r_0\le R$, and letting $R\downarrow s$ gives $r_0\le s$, including $s=0$. Combined with the earlier inequality, this proves the spectral radius formula and the corrected printed identity.

Use the following general Banach algebra facts: in a complex commutative unital Banach algebra, maximal ideals are exactly the kernels of algebra characters, meaning nonzero complex multiplicative linear functionals. Such a character satisfies $\varphi (1)=1$ and $|\varphi (f)|\le \Vert f\Vert$, hence is continuous. The maximal ideal space is given the Gelfand topology, that is, pointwise convergence of characters on algebra elements. These facts follow from closedness of maximal ideals, the Gelfand-Mazur theorem for their quotients, and the spectral bound for character values. They do not identify any of the three spaces by themselves.

For the Wiener algebra, write $f={\sum }_{n\in \mathbb{Z}}{a}_n{u}^n$, where $u(e^{it})=e^{it}$ and $\Vert f{\Vert }_A=\sum |a_n|$. The Fourier series converges absolutely in the algebra norm. If $\varphi$ is a character, set $z=\varphi (u)$. Since $u$ is invertible and $\Vert u{\Vert }_A=\Vert u^{-1}{\Vert }_A=1$, multiplicativity and the norm bound give $|z|\le 1$, $|z^{-1}|\le 1$. Thus $|z|=1$. Continuity now givesConversely, for every $z\in \mathbb{T}$, evaluation is nonzero, linear and multiplicative, and $|f(z)|\le \sum |a_n|$ makes it continuous. Distinct $z$ give distinct characters because they have different values on $u$. Hence the character space of the Wiener algebra is the circle.

For the analytic Wiener algebra, the same expression uses only $n\ge 0$. A character is again determined by $z=\varphi (u)$, but $u^{-1}$ is no longer in this algebra, so only $|z|\le 1$ is forced. Continuity gives $\varphi (f)={\sum }_{n\ge 0}{a}_n{z}^n$. Conversely, every $z$ in the closed unit disk defines evaluation on this absolutely convergent power series. It is bounded and multiplicative, since multiplication of the series is absolutely convergent discrete convolution. Distinct $z$ again differ on $u$. Therefore the character space of the analytic Wiener algebra is the closed disk, including its interior and boundary, not just the original boundary circle.

For $C(\mathbb{T})$ with the supremum norm, let $M=\ker \varphi$ for a character. Suppose the functions in $M$ had no common zero. At each point choose some $f\in M$ nonzero there. Continuity supplies a neighborhood where it remains nonzero; compactness gives a finite subcover associated to $f_1,\dots ,f_k\in M$. Thenis continuous and strictly positive everywhere. It belongs to $M$ because $M$ is an ideal, but it is invertible in $C(\mathbb{T})$ with continuous inverse $1/g$. This would imply $1\in M$, a contradiction. There is therefore a common zero $z$ of $M$. The point-evaluation kernel $I_z=\{f:f(z)=0\}$ is maximal, and $M\subseteq I_z$ forces equality. Since $\varphi (1)=1$, writing $f=f(z)1+(f-f(z)1)$ then proves $\varphi (f)=f(z)$. Conversely every evaluation is a character, and $u$ distinguishes the points. Thus the character space of continuous functions on the circle is the circle.

These are topological identifications as well as bijections. The evaluation map $z\mapsto {\varphi }_z$ is continuous for the Gelfand topology, because each fixed algebra element is a continuous function of $z$: absolute convergence gives uniform convergence on the circle or closed disk in the first two cases. The parameter spaces are compact, while the character spaces are Hausdorff as subspaces of the pointwise topology. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Consequently the three maximal ideal spaces are, respectively,In each case the maximal ideal associated to a point is the kernel of the evaluation just described. The spelling “Weiner” in the source refers to the Wiener algebra.

A sufficient geometric form of the Hahn-Banach separation theorem is this: if $C$ is a nonempty closed convex set in a real normed vector space and $x\notin C$, there exists a continuous linear functional $\ell$ such thatWe use this form without proving it, as permitted. We will prove the Krein-Milman theorem for a nonempty compact convex set $K$: $K$ is the closed convex hull of its extreme points.

First prove the existence of extreme points by the minimal compact face argument. A face of a convex set $K$ is a convex subset $F$ such that, whenever $ta+(1-t)b\in F$ with $a,b\in K$ and $0<t<1$, both $a$ and $b$ belong to $F$. Consider all nonempty compact faces of $K$, ordered by reverse inclusion. The set $K$ itself is such a face. A chain has a nonempty compact intersection, by the finite intersection property and compactness of $K$. This intersection is convex and a face, since membership in every face forces the same membership of endpoints. It is an upper bound in the reverse order. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contained distinct $a,b$, geometric separation of one point from the other would give a continuous real linear $\ell$ with $\ell (a)\ne \ell (b)$. Compactness supplies a maximum $m={\max }_F\ell$. The set $F^{\prime }=\{v\in F:\ell (v)=m\}$ is a nonempty proper compact convex subset. It is also a face of $K$: if a strict convex combination belongs to $F^{\prime }$, first the face property of $F$ puts both endpoints in $F$, and then linearity and maximality of $m$ force both their values to equal $m$. Thus $F^{\prime }$ contradicts minimality of $F$. Hence $F=\{e\}$, whose face property says precisely that $e$ is an extreme point of $K$. The same argument starts with any nonempty compact face, so every such face contains an extreme point of $K$.

Let $E$ be the set of all extreme points of $K$ and $C=\overline{\mathrm{conv}}E$. It is nonempty, and $C\subseteq K$ because $K$ is closed and convex. Suppose $x\in K\setminus C$. Geometric separation gives a continuous real linear $\ell$ with $\ell (x)>{\sup }_C\ell$. Let $m={\max }_K\ell$ and consider the maximizer face $F=\{v\in K:\ell (v)=m\}$. It is nonempty, compact and a face of $K$, so the previous argument gives an extreme point $e\in F$. But $e\in E\subseteq C$, whilea contradiction. Therefore every point of $K$ belongs to the closed convex hull of its extreme points, provingNeither completeness of the ambient normed space nor finite dimensionality was used; compactness is the key hypothesis.