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A basis of a topology is a family $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. Equivalently, whenever $x\in U$ and $U$ is open, there is $B\in \mathcal{B}$ with $x\in B\subseteq U$. For a metric space, all open balls form a basis. For the product topology on $X\times Y$, the rectangles $U\times V$ with $U,V$ open in their respective factors form a basis; rectangles made from factor bases also suffice.

If $C\subseteq X$ and $D\subseteq Y$ are countable dense subsets, then $C\times D$ is countable. Every nonempty basic rectangle meets it, since $U\cap C$ and $V\cap D$ are nonempty. Every nonempty open set contains such a rectangle, proving that a product of two separable spaces is separable.

Suppose a metric space has a countable dense set $C$. The balls $B(c,r)$ with $c\in C$ and positive rational $r$ form a countable basis. Indeed, given $x\in U$ open, choose $ϵ>0$ with $B(x,ϵ)\subseteq U$, then choose $c\in C$ with $d(c,x)<ϵ/3$ and rational $r$ satisfyingThe triangle inequality gives $x\in B(c,r)\subseteq B(x,ϵ)\subseteq U$. Conversely, from any countable basis select one point of each nonempty member. These points form a countable dense subset, because every nonempty open set contains a basis member. This direction holds for arbitrary topological spaces. Thus a metric space is separable exactly when it is a second-countable space. Intersecting a countable basis with a subspace gives a countable basis for its subspace topology; selecting points as above shows that every subspace of a separable metric space is separable.

For the Sorgenfrey line, $\mathbb{Q}$ meets every nonempty interval $[a,b)$, so it is a countable dense subset. In the Sorgenfrey plane, however, for any $x\in \mathbb{R}$ and $ϵ>0$,Indeed, $(t,-t)$ in that rectangle requires both $t\ge x$ and $t\le x$. Every singleton is therefore open in the subspace topology on $Y$, which is the discrete topology. A dense subset of a discrete space must contain every point. Since $Y$ is in bijection with the uncountable set $\mathbb{R}$, it has no countable dense subset. Hence $X$ is separable and $Y$ is not separable, illustrating why the metric assumption in the inheritance statement matters.

The topology needed here is the weak-star topology $\sigma (E^{\ast },E)$, meaning pointwise convergence on $E$. With the literal weak topology $\sigma (E^{\ast },E^{\ast \ast })$, the claimed compactness is false. For example, $E=c_0$ is separable and $E^{\ast }={\ell }^1$. Its unit vectors $e_n$ have no weak cluster point: every coordinate functional forces a cluster point to have all coordinates zero, while the bounded functional $x\mapsto {\sum }_j{x}_j$ has value one throughout. Thus the dual unit ball is not weakly compact. We prove the intended weak-star assertion in full.

Choose a countable dense set $(x_k)$ in the unit ball of $E$. On $E_1^{\ast }$ defineEach term is at most $2\cdot 2^{-k}$, so the series converges. The triangle inequality is immediate, and $d=0$ forces equality on the dense set and hence on all of $E$. It is therefore a metric. Convergence in this metric is equivalent to convergence on every $x_k$: one direction follows from any individual term, and the other by bounding the series tail uniformly. Because $\Vert \varphi \Vert ,\Vert \psi \Vert \le 1$, approximation of any unit vector by $x_k$ then gives convergence on all of $E$. The same finite-tail and approximation argument for neighborhoods proves equality with the weak-star topology, not just agreement of its convergent sequences.

Given a sequence $({\varphi }_n)$ in this ball, diagonal selection gives a subsequence converging at every $x_k$, since each coordinate lies in a compact disk. Uniform norm bounds then make it pointwise Cauchy at every $x\in E$. Define $\varphi (x)$ as the limit. Linearity passes to the limit, and $|\varphi (x)|\le \Vert x\Vert$, so $\varphi \in E_1^{\ast }$. The subsequence converges to it in the metric. Thus this metric space is sequentially compact, hence compact, proving

Use the convention that level $k$ counts the $k$ nonzero shifts, hence an arithmetic progression of length $k+1$. The SZ property at level $k$ means that every measurable set $A$ of positive measure satisfiesThis is the averaged multiple-recurrence property, stronger than merely finding one positive intersection. Shifting the indexing convention from $k+1$ terms to $k$ terms does not change the all-levels assertion. We work with probability measure-preserving systems.

Here is the Furstenberg correspondence principle construction proving the Szemerédi theorem. Let $D\subseteq \mathbb{N}$ have positive upper density $\delta ={\mathrm{limsup}}_N|D\cap \{1,\dots ,N\}|/N$. Put $z\in \{0,1{\}}^{\mathbb{Z}}$ equal to the indicator of $D$ on the positive coordinates and zero elsewhere, and let $S$ be the left shift. The binary full shift $\Omega =\{0,1{\}}^{\mathbb{Z}}$ is a compact metric space in the product topology, and $S$ is a homeomorphism. Choose $N_i\to \infty$ attaining the upper density and define the empirical measuresWe use the standard weak compactness fact that every sequence of Borel probability measures on a compact metric space has a weakly convergent subsequence, where weak convergence means convergence of integrals of continuous functions. Pass to a subsequence with limit $\mu$. For every continuous $F$, telescoping givesPassing to the limit proves $S_{\ast }\mu =\mu$. The cylinder set $A=\{w:w(0)=1\}$ is clopen, so its indicator is continuous andIf every measure-preserving system has the SZ property at level $k$, there is some $n\ge 1$ for which $\mu (C_n)>0$, where $C_n={\bigcap }_{j=0}^k{S}^{-jn}A$. This is also a clopen cylinder set, so ${\mu }_i(C_n)\to \mu (C_n)>0$. For some sufficiently large $i$, at least one term of the empirical average is positive: $S^{m}z\in C_n$ for some $1\le m\le N_i$. Unwinding the coordinates givesTherefore $D$ contains an arithmetic progression of length $k+1$ with positive common difference. Having the SZ property at every level gives the infinite-density form of the Szemerédi theorem.

The finite uniform form follows from the same argument. If, for fixed $\delta >0$ and $k$, arbitrarily large $N_i$ admitted sets $D_i\subseteq \{1,\dots ,N_i\}$ with $|D_i|\ge \delta N_i$ and no such arithmetic progression, replace $z$ above by the indicator $z_i$ of $D_i$, extended by zero outside that interval. Their empirical measures still have an invariant weak limit with $\mu (A)\ge \delta$. For every fixed positive $n$, however, ${\mu }_i(C_n)=0$, since a point contributing to that cylinder would be an actual progression in $D_i$. Thus $\mu (C_n)=0$ for all positive $n$, contradicting the SZ property. Consequently a sufficiently large $N$ depending only on $\delta ,k$ forces the required progression.

A compact measure-preserving system is one for which every observable $f\in L^2(\mu )$ has relatively compact forward orbit $\{U^{n}f:n\ge 0\}$ under the Koopman operator $Uf=f\circ T$. Equivalently, each of these orbits is totally bounded in the global $L^2$ norm; compactness refers to observables, not just to the underlying topological space. For an invertible system it is usual to use all integer times. One may also define compactness by density of almost periodic observables: the two definitions coincide because $\Vert U^n(f-g){\Vert }_2=\Vert f-g{\Vert }_2$, so a finite net for an approximating observable is a finite net, with slightly larger radius, for $f$.

To prove the SZ property, fix $A$ with $a=\mu (A)>0$, put $f=1_A$, and chooseCompactness gives a finite $\epsilon$-net for its orbit with centers $U^{r_1}f,\dots ,U^{r_s}f$, where $r_i\ge 0$. Let $M={\max }_i{r}_i$. For every integer $n\ge M+1$, some center satisfies $\Vert U^{n}f-U^{r_i}f{\Vert }_2<\epsilon$. Since $U$ is an isometry, cancellation givesThus the return set $R=\{r\ge 1:\Vert U^{r}f-f{\Vert }_2<\epsilon \}$ meets every interval $\{n-M,\dots ,n\}$ for $n\ge M+1$. In particular, $R$ is a syndetic set with lower density at least $1/(M+1)$. This proves the needed syndetic near returns of an almost periodic observable without any assumption of ergodicity.

For $r\in R$, telescoping and the isometry property giveAs $f$ is an indicator, $\Vert U^{jr}f-f{\Vert }_2^2=\mu (A△T^{-jr}A)$. The union bound therefore givesAll other multiple-intersection measures are nonnegative. Averaging over the bounded-gap set $R$ yieldsThus every compact measure-preserving system has the SZ property at every level, including nonergodic systems and, under the forward-orbit definition, noninvertible transformations.

Suppose $(x_n,f(x_n))$ converges to $(x,y)$. Continuity gives $f(x_n)\to f(x)$, and uniqueness of limits in a metric space gives $y=f(x)$. Hence every convergent sequence in the graph has its limit in the graph, proving ${\Gamma }_f$ is closed.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that some $0\le q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

For a proof, choose $x_0$ and iterate $x_{r+1}=T{x}_r$. Induction gives $d(x_{r+1},x_r)\le q^{r}d(x_1,x_0)$. For $m>n$, the triangle inequality and a geometric series giveThus the iterates form a Cauchy sequence, and completeness supplies a limit $a\in X$. A contraction mapping is continuous, so $Ta=\lim T{x}_n=\lim x_{n+1}=a$. If $b$ is another fixed point, $d(a,b)=d(Ta,Tb)\le qd(a,b)$, forcing $d(a,b)=0$. This proves existence and uniqueness.

For the exponential problem, use $h=f\circ f$ on the complete metric space $\mathbb{R}$. Its derivative isPutting $t=e^{-x}>0$, the function $t{e}^{-t}$ has maximum $1/e$ at $t=1$, as its derivative is $(1-t)e^{-t}$. By the mean value theorem, $|h(x)-h(y)|\le e^{-1}|x-y|$, so $h$ is a contraction mapping on all of $\mathbb{R}$. Let $a$ be its unique fixed point. The compositions commute, so $h(f(a))=f(h(a))=f(a)$; uniqueness of the fixed point of $h$ forces $f(a)=a$. Conversely every fixed point of $f$ is fixed by $h$. Hence there is exactly one real solution of $x=e^{-x}$. It lies in $(0,1)$ because $a=e^{-a}>0$ and then $e^{-a}<1$. This uses the iterated contraction argument without claiming that $f$ itself contracts on all of $\mathbb{R}$.

The Baire category theorem states that a complete metric space is a Baire space: the intersection of countably many open dense sets is dense. Equivalently, a nonempty open subset cannot be a countable union of nowhere dense sets.

Let $U_n$ be open dense and $O$ any nonempty open set. Choose a closed ball $B_1$ of positive radius at most $1/2$ inside $O\cap U_1$. Inductively choose $B_{n+1}$ inside the interior of $B_n$ intersected with $U_{n+1}$, with positive radius at most $2^{-(n+1)}$. Density makes this intersection nonempty, and openness supplies a sufficiently small closed ball. The centres form a Cauchy sequence because all later centres lie in $B_n$, whose diameter tends to zero. Completeness gives a limit $x$. Since each ball is closed and contains all later centres, $x\in B_n$ for every $n$. Thus $x\in O\cap {\bigcap }_n{U}_n$, proving density. Applying this to complements of the closures of nowhere dense sets gives the equivalent formulation.

Let $P_m$ be the linear subspace of polynomials of degree at most $m$. It has finite dimension $m+1$, is proper in $P$, and $P={\bigcup }_{m\ge 0}{P}_m$. For any norm on $P$, the restricted norm on $P_m$ is equivalent to its coefficient Euclidean norm: on the coefficient unit sphere the continuous norm has a positive minimum and a finite maximum. Thus $P_m$ is complete. A complete subspace of a metric space is closed, since any sequence converging to an ambient limit is Cauchy and has the same limit in the subspace.

Part (b) makes each $P_m$ a closed set with empty interior, hence nowhere dense. If $P$ were a Banach space, this countable union would contradict the Baire category theorem. No norm makes the space of all polynomials complete.

An extending absolute value on a field is non-Archimedean by Question 1(a). We prove finite-dimensional non-Archimedean norm equivalence over a complete field, using completeness rather than local compactness.

Fix a basis $e_1,\dots ,e_d$ of a finite-dimensional vector space over $K$, and let $\Vert \cdot \Vert$ be a norm homogeneous for the given absolute value on a field and satisfying the ultrametric inequality. Write $\Vert \sum a_i{e}_i{\Vert }_0={\max }_i|a_i|$. The upper boundis immediate. For the lower bound use induction on $d$. The one-dimensional case is homogeneity. For the induction step, the span $W$ of $e_1,\dots ,e_{d-1}$ is complete for its restricted norm by the induction comparison with the coordinate norm and completeness of $K$. A complete vector subspace of a metric space is closed, soFor $v=w+a{e}_d$ with $a\ne 0$, scaling by $a$ gives $\Vert v\Vert \ge |a|\delta$; the same coefficient bound is trivial when $a=0$. AlsoThe induction bound controls every coordinate of $w$ by a constant times $\Vert w\Vert$. Together with the bound on $a$, this proves $\Vert v{\Vert }_0\le C^{\prime }\Vert v\Vert$.

Two extending absolute values on a field on $L$ are norms of this kind on its finite-dimensional vector space over $K$. Hence there is $B>0$ with $|x{|}_2\le B|x{|}_1$ for every $x\in L$. Apply this to $x^N$ and use multiplicativity:Letting $N\to \infty$ gives $|x{|}_2\le |x{|}_1$. Interchanging the two norms gives equality. ThusThe proof covers the trivial base absolute value on a field too. It proves uniqueness, without assuming an existence theorem or compactness of the coordinate sphere $\{v:\Vert v{\Vert }_0=1\}$.

The relevant Banach-Alaoglu theorem says that for $1<p<\infty$ the closed bounded balls of $L^p({\mathbb{R}}^n)$ are compact in the weak topology. In particular, every bounded sequence has a weakly convergent subsequence whose limit has norm no larger than the common bound. Applying the preceding duality of Lp spaces also identifies this as weak-star compactness of $L^p=(L^q{)}^{\ast }$; its weak and weak-star test spaces are both $L^q$.

Here is a direct proof of both compactness and the subsequence assertion. Let $\Vert f_j{\Vert }_p\le M$ and choose a countable dense set $\{g_m\}$ in the separable Lp space $L^q$. Each scalar sequence $\int f_j{g}_m$ is bounded. Successive subsequence selection followed by the diagonal subsequence argument gives $f_{j_k}$ for which these pairings converge for every $m$. For arbitrary $g\in L^q$,First approximate $g$ by $g_m$ and then take $k,l$ large. This proves convergence of every pairing, and the limit $\Lambda (g)={\lim }_k\int f_{j_k}g$ is linear with $|\Lambda (g)|\le M\Vert g{\Vert }_q$. Apply the already proved duality of Lp spaces with $q$ in place of $p$: $\Lambda (g)=\int fg$ for an $f\in L^p$ with $\Vert f{\Vert }_p\le M$. ThusTo obtain compactness rather than only a sequential statement, on the radius-$M$ ball defineDensity and Hölder's inequality show that this is a metric space distance and that its topology is exactly the weak topology on this ball: every remaining test pairing is uniformly approximated there by a dense-set pairing. The subsequence argument proves sequential compactness for this metric. A sequentially compact metric space is compact: otherwise either an infinite separated sequence contradicts total boundedness or a Cauchy sequence without a limit contradicts sequential compactness; completeness and total boundedness give compactness by successive finite coverings. This completes the proof. The same argument is the Sequential Banach-Alaoglu theorem for a separable predual.

The Baire category theorem says that a countable intersection of open dense sets in a nonempty complete metric space is dense. Write $E=\{e_1,e_2,\dots \}$. Each singleton is closed in a metric space and has empty interior because there are no isolated points. Therefore every $X\setminus \{e_j\}$ is open and dense. Applying the theorem also to $G$ yieldswhich is dense in $X$. Thus every nonempty open subset meets the required difference, not merely the original dense set $G$.

Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal space admit a continuous $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. Using the usual convention, normal spaces are also $T_1$.

The Tietze extension theorem states that a continuous real-valued function on a closed subset of a normal space extends continuously to the whole space. If its range lies in a closed bounded interval, the extension may be chosen in that same interval.

Prove first the interval version for $f:S\to [-1,1]$. If a continuous residual $r$ on $S$ obeys $|r|\le M$, the sets $\{r\le -M/3\}$ and $\{r\ge M/3\}$ are disjoint closed subsets of $X$. Urysohn's lemma gives a continuous $h:X\to [-M/3,M/3]$ taking the endpoint values on those two sets. On each of the three ranges of $r$, it follows that $|r-h|\le 2M/3$ on $S$. Repeating with $M_j=(2/3{)}^j$ produces $h_j$ with $\Vert h_j{\Vert }_{\infty }\le M_j/3$ and a residual bounded by $M_{j+1}$. The series $\tilde{f}={\sum }_{j\ge 0}{h}_j$ converges uniformly, has absolute value at most $\sum M_j/3=1$, is continuous and restricts to $f$ as the residual tends to zero. Rescaling proves every closed bounded interval version.

For unbounded $f$, first extend $f/(1+|f|)$ to $g:X\to [-1,1]$. The closed set $B=g^{-1}(\{-1,1\})$ is disjoint from $S$. Urysohn's lemma gives $\psi :X\to [0,1]$ with $\psi {|}_S=1$ and $\psi {|}_B=0$. Then $h=\psi g$ takes values strictly between minus one and one everywhere and equals $f/(1+|f|)$ on $S$. Thus $h/(1-|h|)$ is a continuous real-valued extension of $f$, proving the full theorem.

Statement 1 is true: metric spaces are normal, and the bounded interval form applies. Statement 2 is true: compact Hausdorff spaces are normal, and the unrestricted real-valued form applies. Statement 3 is false. For a counterexample, let $X=\{a,b,c\}$ have open sets $\varnothing ,\{a\},\{a,b\},\{a,c\},X$. The subset $S=\{b,c\}$ is closed and discrete, so $f(b)=-1$, $f(c)=1$ is continuous on it. Every open neighborhood of $b$ or $c$ in $X$ contains $a$. A continuous extension would pull back disjoint neighborhoods of minus one and one to disjoint open sets containing $b$ and $c$, yet both would contain $a$. This is impossible.

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 Arzelà-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.

Use probability measure-preserving systems: $T:X\to X$ is measurable and $\mu (T^{-1}A)=\mu (A)$. Its invariant sigma-algebra isErgodicity means that every invariant measurable set has measure zero or one:Equivalently, every invariant measurable function is constant almost everywhere. Indeed all rational sublevel sets of a real invariant function belong to $\mathcal{I}$, so the zero-or-one alternative forces its distribution to concentrate at one value. Apply this to real and imaginary parts for complex functions. Conversely, an invariant set has an invariant indicator, so the function criterion implies the set criterion. This is the invariant-function characterization of ergodicity.

For an irrational rotation of the circle, $T(x)=x+\alpha (\mathrm{mod}1)$ with $\alpha \notin \mathbb{Q}$, an invariant $g\in L^2(m)$ has Fourier coefficients satisfyingFor $k\ne 0$ the multiplier is nonzero. Completeness of the Fourier series basis therefore makes $g$ constant. Apply this to an invariant indicator to prove the irrational rotation is ergodic. A second example is the cyclic permutation of $q$ points with equal masses: a nonempty invariant set contains the whole cycle and hence has measure one.

The identity on $[0,1]$ with Lebesgue measure is not ergodic, since $[0,1/2]$ is invariant and has measure $1/2$. Nor is rotation by $1/5$ on the circle: the setis invariant and has measure $1/2$. These examples also show why preservation of measure does not imply ergodicity.

Let $U=U_T$ be the Koopman operator, $Uf=f\circ T$, and setThe Von Neumann mean ergodic theorem says that, for $f\in L^2(\mu )$, these Cesàro averages converge in $L^2$ to the orthogonal projection onto the invariant functions:For an invertible system, $U$ is a unitary operator. The Hilbert-space theorem states more generally that the Cesàro averages of any unitary operator converge strongly to its fixed-space projection.

Here is the requested invertible-case proof. Since $U^{\ast }=U^{-1}$,Thus the orthogonal decomposition for unitary ergodic averages isOn the first summand, $A_{N}f=f$. On a vector $(I-U)g$, telescoping givesBecause $\Vert A_N\Vert \le 1$, approximation extends this convergence to the closure of the range. The orthogonal decomposition now proves the theorem for every $f$. An invariant $L^2$ function is $\mathcal{I}$-measurable, and an $\mathcal{I}$-measurable function is invariant by approximation with indicators; hence the projection is the stated conditional expectation.

In an ergodic system, invariant functions are constants by the criterion already proved. Their orthogonal projection is obtained by taking the inner product with $1$, whose norm is one. Therefore the ergodic limit is the space average:

The Birkhoff ergodic theorem states that, for $f\in L^1(\mu )$,The limit is invariant, integrable, and has the same integral as $f$. On a probability space convergence also holds in $L^1$. In an ergodic system the limit is $\int fd\mu$ almost everywhere.

For the topological sharpening, take a continuous map $T$ on a compact metric space. Unique ergodicity means there is exactly one invariant Borel probability measure $\mu$. The uniform ergodic convergence for uniquely ergodic systems says that every continuous real or complex $f$ satisfiesThus convergence holds at every point and uniformly in that point. The continuity, compactness and continuous-observable hypotheses belong to this theorem; it is not a claim of uniform convergence for all measurable $L^1$ functions.