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The sphere is the inverse image of $\{1\}$ under the continuous map $u\mapsto \Vert u{\Vert }^2$, so it is closed in ${\mathbb{R}}^3$. It is bounded. The Heine-Borel theorem therefore makes it compact. Every metric space is Hausdorff: distinct points have disjoint open balls of radii smaller than half their distance. The Euclidean subspace metric supplies that property here. Hence $S^2$ is compact and Hausdorff.

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.

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.

Two Lipschitz equivalent norms $N_1,N_2$ satisfy $c{N}_1(v)\le N_2(v)\le C{N}_1(v)$ for all $v$, for constants $0<c\le C<\infty$. These inequalities give the same convergent sequences in both norms. A subset of a metric space is closed exactly when it contains the limits of all convergent sequences from that subset. Therefore closedness is the same in the two norms.

For finite dimension, fix a basis $e_1,\dots ,e_d$ and let $x={\sum }_j{x}_j{e}_j$. For any norm $N$,The reverse triangle inequality then makes $N$ continuous in the Euclidean coordinate norm. If no positive lower bound existed on the Euclidean unit sphere, choose $x_m$ there with $N(x_m)\to 0$. By the Bolzano-Weierstrass theorem, a subsequence converges to $x$ on the unit sphere. Continuity would give $N(x)=0$, although $x\ne 0$, a contradiction. Thus $N$ is bounded below by a positive multiple of the Euclidean norm. Comparing any two norms through that coordinate norm proves all norms in finite dimension are Lipschitz equivalent.

On $C[0,1]$, nonnegativity, absolute homogeneity and the triangle inequality for $\Vert f{\Vert }_1={\int }_0^1|f|$ follow from the corresponding real-number properties and integration. If a continuous $f$ is nonzero at a point, it has absolute value bounded below by a positive constant on a nontrivial one-sided or two-sided interval around that point. Its integral is then positive. Hence $\Vert f{\Vert }_1=0$ implies $f=0$, completing the norm axioms.

For $n\ge 2$, put $g_n(x)=\max \{1-n|x-1/2|,0\}$ and $f_n=1-g_n$. Then $f_n(1/2)=0$, butThe limit function $1$ is not in the indicated point evaluation kernel. Consequently that set is not closed in the integral norm. This is the point-evaluation kernel is not closed in the integral norm phenomenon.

Finally $\Vert g_n{\Vert }_{\infty }=1$ while $\Vert g_n{\Vert }_1=1/n$. Thus no constant can bound the uniform norm by a multiple of the integral norm. Although $\Vert f{\Vert }_1\le \Vert f{\Vert }_{\infty }$ on this unit interval, the converse comparison fails: the two norms are not Lipschitz equivalent. The finite-dimensional theorem cannot be applied to this infinite-dimensional function space.

In $\mathbb{R}$, takeThese are nonempty descending sets of diameter $1/n$, with empty intersection. They are not closed sets, so there is no contradiction with the Cantor intersection theorem for the complete metric space $\mathbb{R}$.

Take the incomplete metric space $X=(0,1)$ with its usual distance, andEach is nonempty and closed relative to $X$: it is $X\cap (-\infty ,1/(n+1)]$. The sequence is strictly descending and its diameters are $1/(n+1)\to 0$. Its intersection is empty, since every positive point exceeds $1/(n+1)$ for sufficiently large $n$. The excluded limiting point is zero.

A metric space is complete if every Cauchy sequence in it converges to a point of that space. The stated shrinking-set characterization is the Cantor intersection theorem together with its converse.

First assume completeness, and choose $x_n\in A_n$. If $m,n\ge N$, then both points belong to $A_N$, so $d(x_m,x_n)\le \mathrm{diam}(A_N)$. The vanishing diameters imply that $(x_n)$ is a Cauchy sequence, hence it converges to some $x\in X$. For every fixed $N$, its tail belongs to the closed set $A_N$, so $x\in A_N$. Thus $x\in {\bigcap }_N{A}_N$. In fact the intersection contains only one point, since any two of its points have distance at most every diameter.

Conversely assume the stated intersection property and let $(x_n)$ be a Cauchy sequence. PutThese closures are nonempty, closed and nested. Taking a closure does not increase the diameter: approximate two points of the closure by points of the set and use continuity of distance. The Cauchy sequence therefore makes $\mathrm{diam}(A_N)\to 0$. Choose $x$ in the intersection. For $n\ge N$, both $x_n$ and $x$ belong to $A_N$, and $d(x_n,x)\le \mathrm{diam}(A_N)\to 0$. Every Cauchy sequence converges in $X$, proving completeness.

A contraction mapping $T$ on a metric space satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one constant $0\le q<1$. The contraction mapping theorem says that on a nonempty complete metric space such a map has a unique fixed point; successive iterates from any starting point converge to it. More precisely, if $x_n=T^n{x}_0$, then $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

The Volterra integration operator maps continuous functions to continuous functions. In the uniform norm,This constant cannot be decreased: for $f=1$ and $g=0$, $Af(x)=x$, and both uniform norms are one. Hence $A$ is not a contraction.

Changing the order of integration over the triangular region givesThus$A^2$ is a contraction, with sharp constant $1/2$. This is an example of an iterated contraction.