Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 (Introduction to Functional Analysis)

Choose $M$ with $|f|\le M$ on $U$, and fix $r=(1-\Vert x\Vert )/4>0$. We prove a local Lipschitz continuity bound, which gives more than continuity. Let $u,v\in B(x,r)$, put $d=\Vert v-u\Vert$, and suppose $d>0$. The point $w=u+(2r/d)(v-u)$ lies in $B(x,3r)\subset U$, while $d<2r$ andBy convexity, $f(v)-f(u)\le [d/(2r)](f(w)-f(u))\le Md/r$. Interchanging $u,v$ givesThe case $d=0$ is immediate. Every point has such a neighborhood, so $f$ is locally Lipschitz and therefore continuous throughout $U$. This is the bounded convex functions are locally Lipschitz on an open ball argument; no finite-dimensional assumption is needed.

Fix $y\in E$, and take $t>0$ sufficiently small that $x+ty\in U$. For $0<s<t$, convexity along the line segment givesThus the positive secant slopes are nondecreasing in their parameter, and have a limit as the parameter decreases to zero. The local Lipschitz continuity from part (a) bounds their absolute values by $L\Vert y\Vert$, where $L=M/r$. Hence the directional derivative exists as a finite real number andFor $y=0$ it is zero.

Write $p(y)=D_{y}f(x)$. Rescaling the parameter in the directional derivative gives $p(ay)=ap(y)$ for $a>0$, and $p(0)=0$ handles $a=0$. For sufficiently small $t>0$, convexity givesSubtract $f(x)$, divide by $t$, and let $t\downarrow 0$. The finite limits from part (b) giveThese are exactly the defining conditions of a sublinear functional. In particular, a directional derivative of a convex function is sublinear; it need not be linear.

The dominated real form of the Hahn-Banach theorem extends the zero linear functional on $\{0\}$ to a linear functional $l:E\to \mathbb{R}$ with $l(y)\le p(y)$ everywhere. If $x+z\in U$, convexity of the ball puts the whole segment in $U$. Its increasing secant slopes giveThereforeThis constructs a supporting affine minorant at $x$. Part (e) shows that it is a continuous supporting functional for a locally bounded convex function.

The bound in part (b) and domination $l\le p$ give $l(y)\le L\Vert y\Vert$. Applying the same bound to $-y$, and using linearity, gives $-l(y)=l(-y)\le L\Vert y\Vert$. ConsequentlyA bounded linear functional is continuous, so $l$ belongs to the continuous dual space $E^{\prime }$. The domination by a sublinear functional alone was algebraic; the local Lipschitz continuity supplies the needed boundedness.

By part (a), $f$ is a bounded continuous function, hence is Borel measurable and integrable for the given probability measure. By part (e), $l\in E^{\prime }$, and its restriction to the unit ball is also bounded and integrable. For every $u\in U$, the supporting inequality says $f(u)\ge f(x)+l(u)-l(x)$. Taking expectations and using the barycentre hypothesis for this particular $l$ givesThus $\boxed{\mathbb{E}f\ge f(x)}$, the Jensen inequality. This proof only uses the stated weak barycentre identity; a vector-valued integral for the identity map is not needed.

A completely regular Hausdorff space is a Hausdorff space in which each point $x$ outside a closed set $F$ is separated from $F$ by a continuous function $u:X\to [0,1]$ with $u(x)=1$ and $u{|}_F=0$. Put $E=C_b(X)$ and define the evaluation character by ${\delta }_x(g)=g(x)$. The weak-star topology on $E^{\prime }$ is the topology of pointwise convergence on $E$: its coordinate maps $p\mapsto p(g)$ are continuous, and generate the topology. It is Hausdorff, and the dual unit ball is compact in it by the Banach-Alaoglu theorem.

For nonempty $X$, $\Vert {\delta }_x\Vert =1$ by the supremum norm bound and the constant function $1$. All the coordinates of $x\mapsto {\delta }_x$ are continuous, so $\delta$ is continuous. Complete regularity separates distinct points, proving injectivity. To prove continuity of the inverse onto its image, let $x\in O$ with $O$ open in $X$. Separating $x$ from $X\setminus O$ gives a bounded continuous $u$ withSuch neighborhoods prove that $\delta$ is a homeomorphism onto $\delta (X)$. This qualification is essential: it is not onto the whole dual, since the zero functional is not an evaluation on a nonempty space.

Define the Stone-Čech compactification to beThe Banach-Alaoglu theorem makes this a compact Hausdorff space, with $X$ identified with its dense evaluation copy. For $g\in C_b(X)$, the coordinate function $\tilde{g}(p)=p(g)$ is a continuous extension. Density gives $\Vert \tilde{g}{\Vert }_{\infty }=\Vert g{\Vert }_{\infty }$. Conversely any $h\in C(\beta X)$ restricts to an element $g$ of $C_b(X)$, and $h=\tilde{g}$ because they agree on the dense subset $X$. Restriction and coordinate extension are inverse linear isometries, henceThey also preserve products: the corresponding continuous functions agree on $X$ and hence everywhere.

For the universal extension property, let $f:X\to K$ be continuous and $K$ a compact Hausdorff space. Pullback defines a bounded linear operator $T:C(K)\to C_b(X)$ by $Tg=g\circ f$. Its dual map $T^{\prime }:E^{\prime }\to C(K{)}^{\prime }$ is weak-star continuous, since $(T^{\prime }p)(g)=p(Tg)$. Evaluation ${\delta }_K:K\to C(K{)}^{\prime }$ is a homeomorphism onto a compact, thus closed, subset: a compact Hausdorff space is completely regular, by the Urysohn lemma. Now $T^{\prime }{\delta }_x={\delta }_K(f(x))$, soTherefore $\overline{f}={\delta }_K^{-1}\circ T^{\prime }{|}_{\beta X}$ is the required continuous extension. Two continuous maps into the Hausdorff space $K$ agreeing on the dense subset $X$ agree everywhere, proving uniqueness.

If $X$ is open in $\beta X$, then it is a locally compact space: around any $x$, regularity of the compact Hausdorff space $\beta X$ gives an open neighborhood whose closure is compact and lies in $X$. Conversely, suppose $X$ is locally compact. Choose an open neighborhood $U$ of $x$ in $X$ with compact closure $C\subseteq X$. There is an open $O\subseteq \beta X$ with $O\cap X=U$. Density of $X$ implies $O\subseteq {\overline{U}}^{\beta X}$, while compactness of $C$ makes it closed in $\beta X$, so ${\overline{U}}^{\beta X}\subseteq C$. Thus $x\in O\subseteq X$. This provesThe empty-space cases are immediate.

The final printed assertion, for an arbitrary subset of $\beta X$, is false. Take the infinite discrete space $X=\mathbb{N}$. It is not compact, so its compactification has a point $p\in \beta \mathbb{N}\setminus \mathbb{N}$. The singleton $\{p\}$ is closed, hence equals its own closure. It is not open, since every nonempty open subset of $\beta \mathbb{N}$ meets the dense copy of $\mathbb{N}$. This is a counterexample.

The valid result holds for subsets $D\subseteq X$, and more generally for open subsets of $\beta X$. For $D\subseteq X$, its indicator is bounded continuous because $X$ is discrete. Its extension $\widetilde{1_D}$ is still idempotent by density, so it takes only the values zero and one. The set $H=\{\widetilde{1_D}=1\}$ is clopen and has $H\cap X=D$. Density then gives $H={\overline{D}}^{\beta X}$. If $O$ is open in $\beta X$, the set $O\cap X$ is dense in $O$, so $\overline{O}=\overline{O\cap X}$ is clopen by the result just proved. Consequently $\beta X$ is extremally disconnected, which means the closure of every open set is open. This is the discrete Stone-Čech compactifications are extremally disconnected theorem, with the essential open-set qualification.

Positivity of the linear functional gives monotonicity. For any $g\in C(K)$, the pointwise bounds $-\Vert g{\Vert }_{\infty }1\le g\le \Vert g{\Vert }_{\infty }1$ therefore implyThus the positive linear functional is bounded, hence continuous, and its operator norm is at most $\varphi (1)$. Evaluating on the constant function $1$, which has supremum norm one when $K$ is nonempty, gives the reverse inequality. HenceFor empty $K$ the function space and functional are zero, so the same conclusion holds. This is the norm of a positive functional on C(K) identity.

Put $c=\Vert \varphi \Vert =\varphi (1)\ge 0$. If $c=0$, the linear functional is zero and is positive. Otherwise, for a nonzero $g\ge 0$ let $h=g/\Vert g{\Vert }_{\infty }$. Then $0\le h\le 1$ and $\Vert 1-h{\Vert }_{\infty }\le 1$, so boundedness givesPositive rescaling gives $\varphi (g)\ge 0$. Thus $\varphi$ is a positive linear functional. This is the scalar form of the unital contraction positivity criterion.

We construct the Jordan decomposition of a bounded functional on C(K) directly, without assuming a measure representation. For $f\ge 0$, defineThis is finite, nonnegative, and positively homogeneous, since $0$ is allowed and $|\varphi (g)|\le \Vert \varphi \Vert \Vert f{\Vert }_{\infty }$. If $f,h\ge 0$, separately chosen approximate maximizers add to an admissible function for $P(f+h)$, giving $P(f+h)\ge P(f)+P(h)$. For the converse, any $0\le g\le f+h$ splits into the continuous functionswith $0\le g_1\le f$ and $0\le g_2\le h$. Hence $\varphi (g)\le P(f)+P(h)$ and $P$ is additive on the positive cone.

Extend it to a linear functional by ${\varphi }^{+}(u-v)=P(u)-P(v)$ for $u,v\ge 0$. This is well defined: two decompositions satisfy $u+v^{\prime }=u^{\prime }+v$, and additivity gives the same difference. Every continuous real function has such a decomposition, for example into its pointwise positive and negative parts. The extension is positive. Define ${\varphi }^{-}={\varphi }^{+}-\varphi$; for $f\ge 0$, the choice $g=f$ shows $P(f)\ge \varphi (f)$, so ${\varphi }^{-}$ is also positive. Part (a) makes both positive linear functionals continuous.

Their operator norms satisfyThe last equality uses symmetry of the real unit ball. Therefore

Write $\varphi (g)=a+ib$ with $a,b\in \mathbb{R}$, where $g$ is real valued. The normalization and the operator norm bound imply, for every real $t$,The left side is $(1-tb{)}^2+t^2{a}^2$, soDivide by $t>0$ and let $t\downarrow 0$ to obtain $b\ge 0$. Divide by $t<0$, reversing the inequality, and let $t\uparrow 0$ to obtain $b\le 0$. Thus $\boxed{\mathrm{Im}\varphi (g)=0}$ for every real-valued $g$. The first-order term in $t$ is what forces reality; this is the complex-functional step in the unital contraction positivity criterion.

The two algebras have the same identity, so an inverse in $A$ is also an inverse in $B$. Therefore $G(A)\subseteq G_B(A)$.

For strictness, take $B=C_{\mathbb{C}}(\mathbb{T})$, where $\mathbb{T}$ is the unit circle, and let $A$ be the uniform closure of the polynomials in the coordinate function $z$. This is a closed unital subalgebra of the Banach algebra $B$. The coordinate $z$ has inverse $\overline{z}$ in $B$. But $\overline{z}$ cannot belong to $A$: for every polynomial $p$,and the same identity holds for every element of $A$ by uniform convergence. It fails for $\overline{z}$, since $z\overline{z}=1$. Thus $z\in G_B(A)\setminus G(A)$. This algebra is the boundary realization of the disk algebra.

Because $A$ is a closed subalgebra of a Banach algebra, it is itself complete. If $\Vert a-1\Vert <1$, the Neumann series converges in $A$:Multiplying the partial sums by $a$ on either side gives $1-(1-a{)}^{N+1}$, which tends to $1$. Thus every such $a$ is invertible in $A$, and $1$ is an interior point of $G(A)$.

For $a\in G(A)$, any $b\in A$ with $\Vert a^{-1}\Vert \Vert b-a\Vert <1$ has the factorizationThe second factor has a Neumann series inverse in $A$, so $b\in G(A)$. Hence the group of invertible elements of a Banach algebra $G(A)$ is open in $A$.

For $a\in G_B(A)$ use its inverse in $B$ instead. The same small-perturbation argument proves invertibility of $b$ in $B$ whenever $b\in A$ and $\Vert a^{-1}{\Vert }_B\Vert b-a{\Vert }_B<1$. Therefore both $G(A)$ and $G_B(A)$ are open subsets of $A$; in the second argument the inverse need not lie in $A$.

Let $c=(1-ab{)}^{-1}\in A$. From the two inverse identities we have $cab=c-1=abc$. The candidate inverse isIndeedandBoth products have been checked, so this is valid in a noncommutative Banach algebra. It is the identity underlying the nonzero spectra of products in opposite orders result.

Suppose the inverse norms did not tend to infinity. Some subsequence would satisfy $\Vert a_n^{-1}\Vert \le C$. Since $a_n\to a$, for its large indices $\Vert a_n^{-1}(a-a_n)\Vert <1$. Thenis invertible in $A$ by the Neumann series, contradicting the hypothesis. Hence $\boxed{\Vert a_n^{-1}\Vert \to \infty }$, the inverse norm divergence at noninvertible boundary points property.

If $a$ were invertible in $B$, then continuity of inversion in a Banach algebra would give $a_n^{-1}\to a^{-1}$ in $B$. Explicitly, the same factorization around $a$ gives a Neumann series and the bound $\Vert a_n^{-1}-a^{-1}\Vert \le \Vert a_n^{-1}\Vert \Vert a-a_n\Vert \Vert a^{-1}\Vert$, with the inverse norms locally bounded in $B$. Since every $a_n^{-1}$ belongs to the closed subalgebra $A$, so would $a^{-1}$. That would make $a\in G(A)$, again a contradiction. Therefore $a\notin G_B(A)$.

Part (c) makes $G(A)$ relatively open in $G_B(A)$. It is also relatively closed: if a sequence in $G(A)$ converges to $a\in G_B(A)$, part (e) forbids $a\notin G(A)$. Sequences detect closure here because the topology comes from a norm.

Intersecting this clopen subset with any connected component $C$ of $G_B(A)$ gives a clopen subset of the connected space $C$. It is therefore either empty or all of $C$. Thus $G(A)$ is the union of those connected components of $G_B(A)$ that it meets. This is connected-component permanence of subalgebra invertibility; it does not require connected components to be path connected.

The Gelfand-Mazur theorem says that a nonzero complex unital Banach algebra in which every nonzero element is invertible is isomorphic to $\mathbb{C}$ by the scalar map $\lambda \mapsto \lambda 1$. With the usual normalization $\Vert 1\Vert =1$, that isomorphism is isometric. Commutativity need not be assumed in the theorem; it follows from the conclusion.

Every complex Banach division algebra is the scalar field. We use complex scalars throughout this question, including for $C^1[0,1]$; the complex-algebra character and spectrum conclusions depend on that convention.

The assertion requires a proper ideal. If $I=A$, no proper ideal can contain it, so the literal statement without that restriction is false.

Let $I$ be proper, and consider the partially ordered collection of proper ideals containing $I$. It is nonempty. The union of a chain is an ideal, because any two of its elements belong together to some member of the chain. It is still proper: if it contained $1$, some member of the chain would contain $1$ and equal $A$. Thus every chain has an upper bound in the collection. The Zorn lemma gives a maximal member, and any proper ideal containing it still belongs to the collection. Hence this is a maximal ideal of $A$.

Therefore every proper ideal is contained in a maximal proper ideal. This algebraic argument only needs a nonzero unital commutative ring; the Banach algebra assumptions become important for closedness of maximal ideals in part (c).

A character of an algebra is a nonzero multiplicative complex linear functional $\varphi :A\to \mathbb{C}$. Nonzeroness and multiplicativity force $\varphi (1)=1$, so it is onto $\mathbb{C}$. Its kernel is an ideal, and the quotient is the field $\mathbb{C}$, making the kernel maximal.

Continuity is automatic. The element $a-\varphi (a)1$ cannot be invertible, since applying $\varphi$ to an inverse equation would give $0=1$. On the other hand $a-\lambda 1$ is invertible whenever $|\lambda |>\Vert a\Vert$, by the Neumann series. Hence $|\varphi (a)|\le \Vert a\Vert$, the automatic continuity of characters bound.

Conversely let $M$ be a maximal ideal. Its closure is an ideal, because multiplication is continuous. That closure is proper: otherwise there would be $m\in M$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Maximality gives $\overline{M}=M$. The quotient $A/M$ is consequently a Banach algebra. It is a division algebra: for any $a\notin M$, the larger ideal $M+Aa$ equals $A$, so the class of $a$ has an inverse. By the Gelfand-Mazur theorem, $A/M\cong \mathbb{C}$, and its quotient map gives a continuous character of an algebra with kernel $M$.

If two algebra characters have the same kernel, then $a-\varphi (a)1$ lies in the second kernel, giving $\psi (a)=\varphi (a)$ for every $a$. Thusis a bijection. This proves that maximal ideals of a commutative complex unital Banach algebra are character kernels.

For every character of an algebra $\varphi$, the element $a-\varphi (a)1$ cannot be invertible, so $\varphi (a)\in {\sigma }_A(a)$. Conversely, if $\lambda \in {\sigma }_A(a)$, the principal ideal $A(a-\lambda 1)$ is proper: otherwise $b(a-\lambda 1)=1$ for some $b$, which in this commutative algebra gives an inverse. Part (b) puts that ideal inside a maximal ideal $M$, and the corresponding algebra character from part (c) has $\varphi (a)=\lambda$. ThereforeHere $\hat{a}(\varphi )=\varphi (a)$ is the Gelfand transform. The spectrum equals character values in a commutative Banach algebra result does not require the Gelfand transform to be injective: the algebra need not be semisimple.

Take complex-valued functions. Pointwise multiplication is associative and commutative, with identity the constant function $1$, whose norm is one. The product rule givesTo prove completeness, let $(f_n)$ be a Cauchy sequence in this norm. The functions and their derivatives converge uniformly to continuous functions $f,g$. Passing to the limit in $f_n(t)-f_n(0)={\int }_0^t{f}_n^{\prime }(s)ds$ gives $f(t)-f(0)={\int }_0^{t}g(s)ds$. The fundamental theorem of calculus yields $f^{\prime }=g$, including the one-sided endpoint derivatives. Thus $f\in C^1[0,1]$ and $f_n\to f$ in the stated norm. Consequently $A$ is a unital commutative Banach algebra.

Let $u(t)=t$. If $\lambda \notin [0,1]$, then $(u-\lambda 1{)}^{-1}(t)=1/(t-\lambda )$ belongs to $C^1[0,1]$. If $\lambda \in [0,1]$, the function $u-\lambda 1$ vanishes somewhere and cannot have a pointwise inverse. Thus ${\sigma }_A(u)=[0,1]$. Every character of an algebra satisfies $\varphi (u)=t_0$ for some $t_0\in [0,1]$, by part (d), and therefore $\varphi (p(u))=p(t_0)$ for every polynomial $p$.

Polynomials are dense in this stronger norm, not just the supremum norm. By the Weierstrass approximation theorem, approximate the real and imaginary parts of $f^{\prime }$ uniformly by polynomials, combining them into $q_n\to f^{\prime }$. Put $p_n(t)=f(0)+{\int }_0^t{q}_n(s)ds$. Then $p_n$ is a polynomial andContinuity of the algebra character now gives $\varphi (f)=\lim p_n(t_0)=f(t_0)$. Conversely each point evaluation is a continuous algebra character. Hence the character space of C1 on a compact interval consists exactly of the evaluation functionals, and part (c) givesDistinct points give distinct maximal ideals, since the coordinate function separates them.