Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 7 (Aspects of Analysis)

A separated dual pair consists of real vector spaces $E,F$ and a bilinear form $⟨x,y⟩$ that separates both variables: each nonzero $x\in E$ pairs nontrivially with some $y\in F$, and conversely. Thus $F$ embeds in the algebraic dual of $E$, and $E$ embeds in the algebraic dual of $F$.

The weak topology $\sigma (E,F)$ is the coarsest topology making all maps $x\mapsto ⟨x,y⟩$, $y\in F$, continuous. It makes $E$ a Hausdorff locally convex space, with topology generated by the seminorms $p_y(x)=|⟨x,y⟩|$. A neighbourhood base at zero consists of finite intersectionsSeparation ensures that the common zero set of all these seminorms is just zero, giving Hausdorffness.

We determine the continuous dual of a weak topology. Suppose a linear functional $\ell$ is continuous. A basic neighbourhood as above is contained in $\{x:|\ell (x)|<1\}$. If $x$ is in the common kernel of the finitely many evaluations, every scalar multiple of $x$ lies in that neighbourhood, so $\ell (x)=0$. Hence $\ell$ factors through the image ofExtend the resulting linear functional on $T(E)$ to ${\mathbb{R}}^m$. It has the form $z\mapsto {\sum }_j{a}_j{z}_j$, giving $\ell (x)=⟨x,{\sum }_j{a}_j{y}_j⟩$. Conversely every such evaluation is continuous by the definition of the weak topology. Since the pairing separates $F$, this identification is injective, andFor the remaining parts every topological assertion about $E^{\prime }$ or its subsets uses the stipulated weak-star topology $\sigma (E^{\prime },E)$.

True. Choose a countable dense sequence $(x_j)$ in the separable Banach space $E$. For $\varphi ,\psi$ in its closed dual unit ball, putThis is a metric: positivity follows because continuous functionals agreeing on a dense subset agree everywhere; symmetry is immediate; and the triangle inequality follows from subadditivity of $t/(1+t)$ for $t\ge 0$. The tail of the series is uniformly small, so convergence on these coordinates implies convergence for this metric, and conversely metric convergence controls each individual coordinate.

The uniform bound $\Vert \varphi -\psi \Vert \le 2$ upgrades coordinate convergence to pointwise convergence on all of $E$. For any $x\in E$, choose $x_j$ close to $x$ and useFor neighbourhoods involving finitely many $x$, choose finitely many such approximations. This proves equality of the induced topologies, including for arbitrary nets. Thus the weak-star metrizability of the dual ball gives

True. The Banach-Alaoglu theorem makes $B^{\prime }$ compact for the weak-star topology. Its compactness can be seen by embedding it in the product ${\prod }_{x\in E}[-\Vert x\Vert ,\Vert x\Vert ]$: the closed conditions expressing linearity identify a closed subset with $B^{\prime }$, and the product is compact by the Tychonoff theorem. The product topology is exactly pointwise convergence on $E$.

Part (i) supplies a compatible metric. Every compact metric space has a finite $1/n$-net for each positive integer $n$. The countable union of these finite nets is dense. ThereforeThe argument concerns topological separability, with the relative weak-star topology specified in the question.

True in the specified topology. Let $D\subseteq B^{\prime }$ be the countable dense set obtained in part (ii). Scaling by a positive integer is a homeomorphism for the weak-star topology, so $nD$ is dense in $n{B}^{\prime }$. Sincethe countable set ${\bigcup }_{n}nD$ is dense in $E^{\prime }$: a nonempty open set contains a point of some $n{B}^{\prime }$, so its relative open intersection with that ball meets $nD$. This proves weak-star separability of the entire dual.

ThusThere is no assertion of norm separability here. For example, the dual of the separable Banach space ${\ell }^1$ is ${\ell }^{\infty }$, whose binary sequences form an uncountable set with pairwise norm distance one. That example is nevertheless separable in its weak-star topology.

True. We prove weak-star topology on an entire infinite-dimensional Banach dual is not metrizable. Suppose instead that $E^{\prime }$ has a countable local base $(W_n)$ at zero. Choose a basic weak-star topology neighbourhood $U_n\subseteq W_n$, with its conditions involving a finite set $S_n\subseteq E$. The $U_n$ still form a local base.

For any $x\in E$, the set $\{\varphi :|\varphi (x)|<1\}$ is a neighbourhood of zero, so some $U_n$ is contained in it. Every functional annihilating $S_n$, and every scalar multiple of that functional, belongs to $U_n$. Consequently every such functional also annihilates $x$. This forcesIndeed a finite-dimensional span is norm closed, and the Hahn-Banach theorem supplies a bounded linear functional vanishing on it and nonzero at any point outside it.

It follows that $E={\bigcup }_n\mathrm{span}{S}_n$. Each span is a proper finite-dimensional closed vector subspace and has empty interior in the infinite-dimensional Banach space $E$. This contradicts the Baire category theorem. HenceThe uniform norm bound that made the metric work on $B^{\prime }$ is absent on the whole dual. The answers to (i), (ii), (iii) and (iv) are therefore all true.

First we prove the Krein-Milman theorem in the required setting. If $K$ is empty the conclusion is immediate, so assume it is nonempty. A face of a convex set is a convex subset $F$ with the property that an interior point of a segment in $K$ belongs to $F$ only if both endpoints do. Consider all nonempty weakly compact faces of $K$, ordered by reverse inclusion. A chain has a nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma gives a minimal compact face $F$.

If $F$ contained distinct points, a bounded linear functional separating them would be nonconstant on $F$. Its maximizer set is a nonempty proper compact face of $F$, hence a face of $K$, contradicting minimality. Therefore $F$ is a singleton and its member is an extreme point of $K$. The same reasoning applies to each nonempty compact face of $K$, so each such face contains an extreme point of $K$.

Let $H$ be the norm-closed convex hull of the extreme points of $K$. A weakly compact subset of a Banach space is weakly closed, hence norm closed, so $H\subseteq K$. If $x\in K\setminus H$, the Hahn-Banach separation theorem gives $\ell \in E^{\prime }$ with $\ell (x)>{\sup }_H\ell$. The maximizer face of $\ell$ on $K$ contains an extreme point $e$ of $K$. Then $\ell (e)\ge \ell (x)>{\sup }_H\ell$, contradicting $e\in H$. We concludeNorm and weak closed convex hulls coincide by the same separation theorem. The printed phrase “closed convex cover” is understood in this standard closed-convex-hull sense.

Now work in the real space of continuous functions on a compact space $C(C)$ on the Cantor set. The extreme points of a real continuous-function unit ball are exactly the continuous sign functions:If $|f(x_0)|<1$, continuity supplies a neighbourhood where $|f|\le 1-\epsilon$. The Cantor cylinder sets form a clopen base, so choose a nonempty cylinder $D$ inside that neighbourhood. Its indicator function is continuous, and $f\pm \epsilon 1_D$ are distinct members of $B$ with midpoint $f$. Hence $f$ is not extreme. Conversely, if $f$ is pointwise sign-valued and $f=(g+h)/2$ with $g,h\in B$, equality at the endpoint of the scalar interval $[-1,1]$ forces $g(x)=h(x)=f(x)$ at every point.

To prove the closed-hull assertion without assuming weak compactness, take $f\in B$ and $\epsilon >0$. A sufficiently fine finite partition of $C$ into Cantor cylinders has oscillation of $f$ less than $\epsilon$ on each cell, by uniform continuity. Choose a value $a_j\in [-1,1]$ on each cell and let $h$ be the corresponding continuous step function. Then $\Vert f-h{\Vert }_{\infty }<\epsilon$. For each sign vector $s\in \{-1,1{\}}^m$, let $e_s$ take value $s_j$ on cell $j$, and assign the weightThese weights are nonnegative, sum to one, and satisfy ${\sum }_s{w}_s{s}_j=a_j$. Thus $h={\sum }_s{w}_s{e}_s$ is a convex combination of extreme points. This is clopen sign approximation in the real Cantor unit ball, provingNevertheless $B$ is not weakly compact. The suggested functions $f_n(x)=\min (3^{n}x,1)$ lie in $B$ and converge pointwise to the function equal to zero at $0$ and one at every other point of $C$. It is discontinuous at zero, since $2\cdot 3^{-j}\in C$ tends to zero. If $B$ were weakly compact, the sequence, viewed as a net, would have a weakly convergent subnet with limit in $C(C)$. Point evaluations are bounded linear functionals, so that subnet would have the same pointwise limit; its indices are cofinal in the original sequence. The limit would therefore be the discontinuous function just described, a contradiction. This discontinuous pointwise limit obstruction to weak compactness gives the required failure of weak compactness, despite the closed-hull equality.

The relevant version of Wiener covering lemma is the finite ball selection lemma: from a finite collection of open Euclidean balls one can select pairwise disjoint balls $B_j$ such that the original union is contained in ${\bigcup }_{j}3B_j$, where $3B_j$ has the same centre and three times the radius. To prove it, repeatedly retain a largest-radius remaining ball and discard every ball meeting it. If a discarded ball has radius $r\le R$ and meets a retained ball of radius $R$, every point of the former is at distance less than $r+(r+R)\le 3R$ from the latter's centre. This proves the containment and, by scaling Lebesgue measure,This is a covering result; no Fourier-algebra version of Wiener's lemma is involved.

We use it for a disjoint ball decomposition modulo null sets of $V$. The empty set needs only the empty family. Otherwise put $V_0=V$. For an open residual set $V_k$ of positive finite measure, inner regularity of Lebesgue measure supplies a compact $K_k\subseteq V_k$ with ${\lambda }_d(K_k)\ge {\lambda }_d(V_k)/2$. Cover $K_k$ by finitely many open balls whose closures lie in $V_k$, and use the Wiener covering lemma to select disjoint balls. Their total measure is at least ${\lambda }_d(V_k)/(2\cdot 3^d)$. Remove their closures to obtain the open residual set $V_{k+1}$. Ball boundaries are null sets, soAll retained balls, including those from different stages, are pairwise disjoint and contained in $V$. They form a countable family. The uncovered set is contained in ${\bigcap }_k{V}_k$ together with their countably many boundaries, and both have measure zero. Enumerating the retained balls gives

Next consider the uncentered maximal function of a finite measure. For every real $\alpha$, its strict superlevel set isIndeed, membership in the ball is exactly the strict condition $\Vert x-y\Vert <r$ in the supremum. This union is open, proving that $m_u$ is lower semicontinuous as an extended nonnegative function; it may take the value infinity.

For $\alpha >0$, take a compact subset $K$ of this superlevel set and choose a finite cover by balls satisfying the displayed density inequality. The Wiener covering lemma selects disjoint balls $B_j$ whose triples cover $K$. Since $\mu$ is a probability measure,Take the supremum over compact $K$ using inner regularity of Lebesgue measure. This works even if the open superlevel set was initially unbounded, and proves the uncentered maximal weak-type inequalityFor a finite positive measure of mass $M$, scaling gives $3^{d}M/\alpha$ instead.

One important use is the Lebesgue differentiation theorem. For $f\in L^1({\mathbb{R}}^d)$, approximate it in $L^1$ by a compactly supported continuous $g$, and set $h=|f-g|$. The limiting mean oscillation of $f$ at $x$ is at most $m_{h{\lambda }_d}(x)+h(x)$, since that of $g$ vanishes by continuity. The preceding weak-type estimate and the elementary integral bound on $\{h>t\}$ giveLet the approximation error tend to zero and then take countably many $t\downarrow 0$. The local averages of $f$ converge to $f(x)$ almost everywhere. Localizing extends this to locally integrable functions. Thus the maximal estimate turns norm approximation by continuous functions into almost-everywhere information about local averages.

We prove Kantorovich duality theorem here by a positive-functional extension argument; it produces the minimizing transport plan at the same time as equality of the values. Let $Y=X\times X$ and $c(x,y)=d(x,y)$. This is a bounded continuous function because $X$ is a compact metric space. Write $C(Y)$ for the real space of continuous functions on a compact space, and letOn this vector subspace define $\ell (s)=\int fd\mathbf{P}+\int gd\mathbf{Q}$. It is well-defined: two representations differ by a constant in the first variable and the opposite constant in the second, and both measures have mass one. It is positive, since $s\ge 0$ implies $\min f+\min g\ge 0$ and hence $\ell (s)\ge 0$. Also $\ell (1)=1$.

Define the majorant envelope, which is a sublinear functional:Constants majorize every $h$, and positivity gives $\min h\le p(h)\le \max h$, so this quantity is finite. Taking approximate minimizing majorants proves subadditivity; rescaling majorants proves positive homogeneity. Moreover $p(s)=\ell (s)$ for $s\in S$, and subtracting $s$ from a majorant provesThe dual value in the question is exactlySubadditivity at $c+(-c)=0$ gives $m_d\le p(c)$.

On $S+\mathbb{R}c$, assign the value $\ell (s)+t{m}_d$ to $s+tc$. If $c\notin S$ the representation is unique. If $c\in S$, then $m_d=\ell (c)$, so the assignment is still well-defined. It is dominated by $p$: for $t\ge 0$, use $t{m}_d\le tp(c)=p(tc)$; for $t<0$, positive homogeneity gives $p(tc)=(-t)p(-c)=t{m}_d$. Together with the translation identity, these verify domination in both cases.

The real Hahn-Banach theorem extends this functional to $T:C(Y)\to \mathbb{R}$ with $T\le p$. If $h\ge 0$, then $p(-h)\le \ell (0)=0$, so $T(h)\ge 0$. Thus $T$ is a positive linear functional, with $T(1)=1$ and $T(c)=m_d$. Positivity also gives $|T(h)|\le \Vert h{\Vert }_{\infty }$, so $T$ is continuous. The Riesz-Markov-Kakutani representation theorem supplies a Borel probability measure ${\pi }_0$ on the compact space $Y$, satisfying $T(h)={\int }_{Y}hd{\pi }_0$.

For every $f\in C(X)$, its pullback $f(x)$ belongs to $S$, so $\int f(x)d{\pi }_0=\int fd\mathbf{P}$. Likewise $\int g(y)d{\pi }_0=\int gd\mathbf{Q}$. Uniqueness in the Riesz-Markov-Kakutani representation theorem shows that these are exactly the two marginal distributions. Finally, every feasible pair $f+g\le c$ gives a lower bound for the cost of every transport plan, by integration. Our constructed plan achieves that bound:This Kantorovich duality by positive extension establishes the requested existence and equality without assuming an optimal plan in advance.

The metric structure additionally gives the Kantorovich–Rubinstein theorem formulation. For a feasible pair define $h(x)={\inf }_y\{d(x,y)-g(y)\}$. The triangle inequality for $d$ makes $h$ one-Lipschitz continuous, $h\ge f$, and $h(y)\le -g(y)$. Hence the dual objective is at most $\int hd\mathbf{P}-\int hd\mathbf{Q}$. Conversely $(h,-h)$ is feasible for every one-Lipschitz $h$. ThereforeNormalizing $h(x_0)=0$ for a fixed $x_0$ gives a uniformly bounded equicontinuous family; it is closed and compact in the uniform topology by the Arzela-Ascoli theorem. The objective is uniformly continuous on this family, so the supremum is attained as well. In particular, $m_d$ is the first Wasserstein distance between the two measures.