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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 Cesaro 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 Cesaro 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.

Suppose $K$ is a compact operator and $f_n\rightharpoonup f$. The Uniform boundedness principle makes $(f_n)$ bounded. If $K{f}_n$ did not converge in norm to $Kf$, some subsequence would stay a fixed positive distance away. Compactness provides a further norm-convergent subsequence, say to $g$. On the other hand, for every $h$, $⟨K{f}_n,h⟩=⟨f_n,K^{\ast }h⟩\to ⟨Kf,h⟩$, so its norm limit must be $Kf$, a contradiction.

Conversely, if $K$ sends every weakly convergent sequence to a norm-convergent sequence, take any sequence in the closed unit ball. Weak compactness of that ball supplies a weakly convergent subsequence, whose images converge in norm by hypothesis. Thus every sequence in the image has a convergent subsequence in $H$. Its closure is also sequentially compact: approximate its $n$th member by an image point within $1/n$. Since $H$ is a metric space, that closure is compact. We concludeThis is the principle that compact operators send weak convergence to norm convergence.

Use the assumed separability of the Hilbert space to choose a countable Hilbertian basis $(e_j)$. For $\Vert f_n\Vert \le M$, every scalar sequence $⟨f_n,e_j⟩$ is bounded. Repeated extraction followed by the diagonal subsequence argument gives one subsequence with all these coordinates converging, say to $a_j$. The Bessel inequality gives ${\sum }_{j\le N}|a_j{|}^2\le M^2$ for every $N$, so $g={\sum }_j{a}_j{e}_j$ exists and $\Vert g\Vert \le M$.

For $h\in H$, let $h_N$ be its finite basis truncation. Coordinate convergence gives $⟨f_{\phi (n)}-g,h_N⟩\to 0$, while the Cauchy-Schwarz inequality bounds the remaining inner product by $2M\Vert h-h_N\Vert$, uniformly in $n$. First choose $N$ large, then $n$ large. ThusThis proves the asserted sequential conclusion. To obtain actual compactness for the weak topology, note that on a bounded ball it is induced by the metricFinite coordinates and the uniformly small tail show that $d$ gives coordinate convergence; uniform boundedness of the ball extends this to every inner product as above. Hence this metric induces exactly the restricted weak topology. In a metric space, sequential compactness implies compactness. The closed unit ball is therefore weakly compact, and the limit remains inside it.

On a compact metric space $X$, a conservative Feller semigroup is a strongly continuous semigroup of positive linear maps on $C(X)$ satisfying $P_{t}1=1$. Positivity means $f\ge 0⟹P_{t}f\ge 0$. These conditions imply sup-norm contraction. Conversely, for a contraction semigroup with $1\in D(L)$ and $L1=0$, differentiation on the generator domain gives $\frac{d}{dt}{P}_{t}1=P_{t}L1=0$, so $P_{t}1=1$. On the real space $C(X)$, if $0\le f\le 1$, then $\Vert 1-P_{t}f{\Vert }_{\infty }\le \Vert 1-f{\Vert }_{\infty }\le 1$, implying $P_{t}f\ge 0$. Scaling proves positivity for every nonnegative $f$. This is the unital contraction positivity criterion. For complex $C(X)$ the same conclusion follows because each functional $f\mapsto P_{t}f(x)$ has norm one and value one on $1$, hence is a positive linear functional.

For each $t,x$, the Riesz-Markov-Kakutani representation theorem represents that functional by a unique Borel probability measure $p_t(x,dy)$:Continuity of $P_{t}f$ gives weak continuity of $x\mapsto p_t(x,\cdot )$. Approximating indicators of open sets increasingly by continuous functions proves Borel measurability in $x$; a monotone-class argument then gives a Markov kernel. The semigroup property and uniqueness of representing measures yield $p_0(x,\cdot )={\delta }_x$ and the Chapman-Kolmogorov equationThese are the required transition probabilities.

An invariant probability measure for a semigroup satisfies ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$ for every $t\ge 0$ and $f\in C(X)$. Equivalently its distribution is preserved by the Markov kernel. For $1\le p<\infty$, Jensen inequality gives $|P_{t}f{|}^p\le P_t(|f{|}^p)$ pointwise. Integrating and using invariance provesAlso $\Vert P_{t}f-f{\Vert }_{L^p(\mu )}\le \Vert P_{t}f-f{\Vert }_{\infty }\to 0$. Continuous functions are dense in these $L^p$ spaces for a finite Borel measure on a compact metric space, so the contraction extends to a strongly continuous semigroup on $L^p(\mu )$.

For real $f$ with $f,f^2\in D(L)$, the kernel form of Jensen inequality gives $P_t(f^2)-(P_{t}f{)}^2\ge 0$, with equality at $t=0$. Take its right derivative to obtain the generator square inequalityThis is nonnegativity of the associated carré du champ.

Each time-average functional ${\varphi }_n$ is positive with ${\varphi }_n(1)=1$, hence represents a Borel probability measure ${\mu }_n$. The compactness of probability measures on a compact metric space gives a subsequence converging weakly to a probability measure $\mu$. Explicitly, $C(X)$ is a separable Banach space, so Banach-Alaoglu theorem and a countable dense family give a subsequence converging on every continuous function, rather than merely a net. For fixed $t\ge 0$, the semigroup property gives the Krylov-Bogolyubov time-average argument:Its absolute value is at most $2t\Vert f{\Vert }_{\infty }/n$. Pass to the subsequential limit, observing that $P_{t}f\in C(X)$, to get ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$. Thus the limiting measure is invariant.

The Baire category theorem in the needed version states that a nonempty complete metric space cannot be a countable union of closed sets with empty interior. To prove it, suppose $F_1,F_2,\dots$ are such sets and start with a nonempty open set $U$. Inductively choose closed balls ${\overline{B}}_j$ of positive radius at most $2^{-j}$, with ${\overline{B}}_1\subset U\setminus F_1$ and ${\overline{B}}_{j+1}\subset B_j\setminus F_{j+1}$. This is possible because each complement is open and dense. The centres form a Cauchy sequence; completeness supplies a limit in every closed ball. It lies in $U$ and outside every $F_j$, proving the theorem.

Here the sequence spaces are over the real scalars, as required by the Euclidean-plane assertion. Suppose an isometry $\phi :{\mathbb{R}}^2\to c_0$ existed. On the compact complete space $S^1\times S^1$, the setsare closed. They cover that space: for a nonzero sequence in $c_0$, the supremum norm is attained at a coordinate; for $x=y$ every coordinate works. By the Baire category theorem, one $F_n$ contains a product of open arcs. Shrink the arcs to be disjoint. The real coordinate difference is then continuous and nonzero, so its sign is constant on the product. Thus the chord distance has the separated formIn particular its alternating sum on every four-point rectangle is zero. Parametrize the arcs by angles $s,t$ with $0<t-s<2\pi$. Their chord distance is $D(s,t)=2\sin ((t-s)/2)$, whose mixed derivative isThe zero rectangle identity would force this mixed derivative to vanish, a contradiction. Therefore no such isometry exists.

For a real vector $x\in {\ell }_1^n$, list its signed sums over all $2^n$ choices $\epsilon \in \{-1,1{\}}^n$, followed by infinitely many zero coordinates:This is linear, has finite support and lies in $c_0$. Choosing the signs of the components gives ${\max }_{\epsilon }|\epsilon \cdot x|={\sum }_j|x_j|$. Thus $\boxed{\Vert Tx{\Vert }_{\infty }=\Vert x{\Vert }_1}$, the required linear isometry.

A Cauchy sequence in a metric space satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. Suppose a subsequence $x_{n_k}$ converges to $x\in X$. Choose $N$ so that the Cauchy bound is $\epsilon /2$, and then choose $k$ with $n_k\ge N$ and $d(x_{n_k},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole Cauchy sequence converges to the subsequential limit.

For a metric space $(X,d)$, uniform continuity meansThe choice of $\delta$ is independent of the points. A Lipschitz continuous function satisfies $|f(x)-f(y)|\le Ld(x,y)$ for one finite constant $L\ge 0$. Choosing $\delta =\epsilon /(L+1)$ proves Lipschitz continuity implies uniform continuity.

Let $(x_n)$ be a Cauchy sequence. For a given $\epsilon >0$, uniform continuity supplies $\delta >0$, and the Cauchy property supplies $N$ with $d(x_m,x_n)<\delta$ for $m,n\ge N$. Then $|f(x_m)-f(x_n)|<\epsilon$, so $(f(x_n))$ is Cauchy in $\mathbb{R}$. By completeness of the real numbers, $(f(x_n))$ converges, even if the original sequence has no limit in $X$.

A contraction mapping on a metric space $(X,d)$ is a map $f:X\to X$ for which there is a single constant $q$ with $0\le q<1$ such thatThe same $q$ must work for every pair. Merely decreasing each nonzero distance strictly does not suffice for this definition.

For a convex subset of a Banach space, weak closure equals norm closure. The Mazur theorem states that for every convex set $C\subseteq X$,Since every bounded linear functional is norm-continuous, the weak topology is weaker than the norm topology, giving ${\overline{C}}^{\Vert \cdot \Vert }\subseteq {\overline{C}}^w$.

For the reverse inclusion, use this form of the Hahn-Banach separation theorem: if $D$ is a nonempty norm-closed convex set in a real normed vector space and $x\notin D$, there is a continuous real-linear functional $F$ such thatApply it to $D={\overline{C}}^{\Vert \cdot \Vert }$. In a complex Banach space, use the underlying real space; every continuous real-linear $F$ is the real part of the complex bounded linear functional $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the weak topology of $x$ disjoint from $D$, so $x\notin {\overline{C}}^w$. The empty set case is immediate. This proves the Mazur theorem.

A weakly null sequence has disjoint convex blocks converging to zero in norm. If $x_n\stackrel{w}{\to }0$, then for every starting index $m$,The Mazur theorem places $0$ in the norm closure of that tail convex hull. Thus a finite convex combination of vectors from any tail can have norm less than any prescribed $\epsilon >0$.

Choose the convex blocks recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite convex combination from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yieldsTherefore $u_n\to 0$ in norm. The recursive tail selection is what makes these convex blocks, rather than merely unrelated convex combinations.

Next suppose $X^{\ast }$ is separable and $(x_n)$ is bounded. Use the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, where $J_{X}x(f)=f(x)$. If $\Vert x_n\Vert \le M$, then $J_X{x}_n\in M{B}_{X^{\ast \ast }}$. The Banach-Alaoglu theorem makes this ball compact in $\sigma (X^{\ast \ast },X^{\ast })$, and weak-star metrizability of the dual ball makes it metrizable because its predual $X^{\ast }$ is separable. A compact space that is a metric space is sequentially compact, so some subsequence $(y_n)$ satisfiesThe limit need not belong to $J_{X}X$; it is precisely the use of the bidual space that supplies compactness without assuming reflexivity.

For a convex block $u_n={\sum }_{i=p_n}^{q_n}{a}_i{y}_i$, every $f\in X^{\ast }$ satisfiesBoth $f(y_n)$ and $f(u_n)$ tend to $\varphi (f)$, soThis is the useful convex-block cancellation of a weak-star limit; it does not require the limit to be a vector in $X$.

The quotient sequence has approximate lifts in $3B_X$ that are weakly null after passing to a subsequence. Let $q:X\to Z=X/Y$ be the quotient map, with the quotient norm. Since $\Vert z_n\Vert \le 1$, choose $v_n\in X$ such thatThis uses the infimum defining the quotient norm and does not assume that it is attained. By the preceding compactness argument, pass to a subsequence $y_n=v_{k_n}$ with a common weak-star limit in $X^{\ast \ast }$. Its quotient images $q(y_n)=z_{k_n}$ form a weakly null sequence.

Apply the convex block construction to $(q(y_n))$ in the quotient Banach space. Using the same coefficients on $(y_n)$ gives convex blocksSet $x_n=y_n-u_n$. The convex-block cancellation of a weak-star limit proves $x_n\stackrel{w}{\to }0$, whileThus the approximate weakly null lifting through a quotient is achieved with the required bound:

A real function on a metric space is uniformly continuous if for every $\epsilon >0$ there is a $\delta >0$ such that, simultaneously for all $x,y$, $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$. For a continuous function on a finite closed interval $[a,b]$, suppose this fails. Then there are ${\epsilon }_0>0$ and pairs $x_n,y_n$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. By compactness, some subsequence $x_{n_j}$ converges to $x\in [a,b]$; then $y_{n_j}\to x$ too. Continuity contradicts the image separation. This proves the Heine-Cantor theorem in this setting. Boundedness of the interval matters: $x^2$ on $[0,\infty )$ is a counterexample if “closed interval” includes unbounded intervals.

A function is Lipschitz continuous if some finite $K\ge 0$ satisfies $|f(x)-f(y)|\le Kd(x,y)$ for every pair. Choosing $\delta =\epsilon /(K+1)$ proves uniform continuity.

The first assertion is false. Take $f(x)=\sin (x^2)$, which is bounded and continuous, andThen $f(x_n)=1$, $f(y_n)=-1$, while $y_n-x_n=\pi /(y_n+x_n)\to 0$. This violates uniform continuity.

The continuity condition at $x\in X$ between metric spaces isThe function is continuous when this holds at every $x$.

If $f$ is continuous and $U\subseteq Y$ is an open set, take $x\in f^{-1}(U)$. Some $ϵ$-ball around $f(x)$ lies in $U$. Continuity gives a $\delta$-ball around $x$ mapped into that ball and hence contained in $f^{-1}(U)$. Thus $f^{-1}(U)$ is open, including the empty set, in the notation for the image and preimage of a function.

Conversely, suppose every open set has open preimage. For fixed $x$ and $ϵ>0$, the preimage of the open ball $B_e(f(x),ϵ)$ is open and contains $x$, so contains some $B_d(x,\delta )$. This gives the displayed metric definition. Therefore $\boxed{f\text{ is continuous}⟺f^{-1}(U)\text{ is open for every open }U}$.

Let $X=\mathbb{R}$ with the usual metric and let $Y=\mathbb{R}$ with the discrete metric, and take $f(x)=x$. Every subset of $Y$ is an open set, so the image of every open subset of $X$ is open: $f$ is an open map.

It is not continuous at any point. For $ϵ=1/2$, every $\delta >0$ allows a distinct $y$ with $|y-x|<\delta$, but the distance between $f(y)$ and $f(x)$ in the discrete metric is 1. Equivalently, the preimage of the open singleton $\{x\}\subset Y$ is not open in $X$. Thus $\boxed{\text{an open map need not be continuous}}$ even between metric spaces.

A contraction mapping satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one fixed $0\le q<1$. The contraction mapping theorem states that such a map on a nonempty complete metric space has a unique fixed point and that every sequence of iterates converges to it.

Choose $x_0$ and put $x_{n+1}=T{x}_n$. Iteration gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$,Thus the iterates form a Cauchy sequence and converge by completeness to $x_{\ast }$. The Lipschitz bound gives continuity of $T$, hence $T{x}_{\ast }=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing $m\to \infty$ also gives the usual geometric error estimate. For $q=0$, convergence occurs after one application.

If $f^k$ is a contraction mapping, let $x_{\ast }$ be its unique fixed point. The commutation identity $f^{k}f=f{f}^k$ gives $f^k(f{x}_{\ast })=f{x}_{\ast }$, so uniqueness implies $f{x}_{\ast }=x_{\ast }$. Any fixed point of $f$ is a fixed point of $f^k$, giving uniqueness, without even assuming $f$ continuous. On the complete discrete metric space $\{0,1,2\}$, the map $f(0)=f(1)=0$, $f(2)=1$ has constant second iterate but is not a contraction: $d(f2,f0)=d(2,0)=1$.

For the specified function, $x+\sin x\ge 0$ on $[0,\infty )$ because it starts at zero and has derivative $1+\cos x\ge 0$. Therefore $f$ maps the complete half-line into itself. Its derivative satisfiesThe mean value theorem therefore makes it a contraction of constant at most $2/3$. Hence

A metric space is a complete metric space when every Cauchy sequence converges to a point of that space: for each $\epsilon >0$, the condition $d(x_m,x_n)<\epsilon$ for all sufficiently large $m,n$ must imply existence of $x$ with $d(x_n,x)\to 0$.

A Cauchy sequence in $[0,1]$ is also Cauchy in $\mathbb{R}$, so it converges there by the assumed completeness. Its limit remains between zero and one, since inequalities are preserved under limits. ThereforeThis is an instance of the fact that a closed subset of a complete metric space is complete with its induced metric.

The coordinate evaluations $e_n:x\mapsto x_n$ belong to the dual unit ball of the l-infinity sequence space and separate its points. Therefore the countable separating family metrizes a weakly compact set result makes any nonempty weakly compact $K\subset {\ell }_{\infty }$ a compact metric space in its weak topology. Choose finite $1/n$-nets and take their countable union $D$; it is weakly dense in $K$.

It follows that $K\subset {\overline{D}}^w\subset {\overline{\mathrm{conv}D}}^w$. By Mazur theorem, the last set is ${\overline{\mathrm{conv}D}}^{\Vert \cdot \Vert }$. Finite convex combinations of members of $D$ with nonnegative rational coefficients summing to one form a countable norm-dense subset of $\mathrm{conv}D$: for each fixed finite list, approximate its coefficients in the simplex by rational coefficients, and use the norm triangle inequality. Thus its norm closure is a separable metric space containing $K$.

Every subset of a separable metric space is separable: balls centred on a countable dense set with positive rational radii form a countable base; intersect this base with the subset and choose one point from each nonempty intersection. Consequently the weakly compact subsets of l-infinity are norm separable conclusion isThe empty case is immediate. The convexification matters: the chosen weakly dense set $D$ itself need not be norm dense in a nonconvex $K$.

A continuous transformation $T$ of a compact metric space $X$ is uniquely ergodic when there is exactly one $T$-invariant Borel probability measure on $X$. We use the usual compact-space convention for unique ergodicity; the compactness and continuity hypotheses matter in the assertion about all starting points.

Let $R_{\alpha }(x)=x+\alpha (\mathrm{mod}1)$ on the circle group, with $\alpha$ irrational. Normalized Lebesgue measure $m$ is invariant under this irrational rotation of the circle. If $\nu$ is any invariant Borel probability measure, define its Fourier coefficients by $c_k=\int e^{2\pi ikx}d\nu (x)$. Invariance givesFor $k\ne 0$, irrationality makes $e^{2\pi ik\alpha }\ne 1$, so $c_k=0$; also $c_0=1$. These are the Fourier coefficients of $m$. Hence $\nu$ and $m$ have the same integrals against every trigonometric polynomial. Such polynomials are uniformly dense in the continuous functions by the Stone-Weierstrass theorem, so the measures agree on every continuous test function and therefore agree as Borel measures. Thus

For the general uniquely ergodic system, fix $x\in X$ and form the empirical measuresOn a compact metric space, the Borel probability measures are compact for weak convergence of probability measures. Any subsequential limit $\nu$ is invariant: for every continuous $g$,Here $g$ is bounded and $g\circ T$ is continuous, so the identity passes to the limit. Uniqueness of the invariant Borel probability measure gives $\nu =\mu$. Every subsequential limit is therefore $\mu$, and compactness implies convergence of the entire sequence. Testing against $f$ provesIn fact this proves uniform ergodic convergence for uniquely ergodic systems: if convergence were not uniform in $x$, choose $N_j\to \infty$ and $x_j$ where the discrepancy stays above a fixed positive number. The same compactness and telescoping argument applied to ${\nu }_{N_j,x_j}$ forces a subsequence to converge to $\mu$, a contradiction. This also makes clear why an almost-everywhere Birkhoff ergodic theorem alone would not establish the requested everywhere assertion. Without the compact-space hypothesis the assertion need not hold: on the discrete space $\{p\}\bigsqcup {\mathbb{Z}}_{\ge 0}$, set $T(p)=p$ and $T(n)=n+1$. The only invariant probability is ${\delta }_p$, but the continuous bounded function which is zero at $p$ and one on the integer orbit has orbit average one there.

For the decimal application put $\alpha ={\log }_{10}2$. This is irrational: if $\alpha =p/q$ with positive integers $p,q$, then $2^q=10^p=2^p{5}^p$, contradicting unique prime factorization. Writing $n\alpha =m_n+t_n$ with $m_n=\lfloor n\alpha \rfloor$ and $t_n\in [0,1)$ gives $2^n=10^{m_n}10^{t_n}$. Its leading decimal digit is seven exactly whenThe half-open upper endpoint correctly excludes powers whose leading digit is eight, including $2^3=8$.

The indicator of $I$ is not continuous, so an extra step is needed. For any $\epsilon >0$, choose continuous functions $a_{\epsilon },b_{\epsilon }$ on the circle with $0\le a_{\epsilon }\le 1_I\le b_{\epsilon }\le 1$ and $\int (b_{\epsilon }-a_{\epsilon })dm<\epsilon$, by tapering in small neighbourhoods of the two endpoints. Applying the everywhere averaging result to these functions traps the lower and upper limits of the interval frequency between their integrals. Letting $\epsilon \downarrow 0$ proves everywhere interval frequency under an irrational rotation. ConsequentlyThis is the leading-seven case of Benford frequencies for powers of an integer.

Let $p=dP/d\nu$ and $q=dQ/d\nu$ be Radon-Nikodym derivatives. Use the following normalizations of the Kullback-Leibler divergence, total variation distance and unnormalized Hellinger distance:We use $0\log (0/q)=0$ and assign infinite Kullback-Leibler divergence if $Pot\ll Q$, where $\ll$ denotes absolute continuity of measures. Common domination alone also suffices for the inequalities with the extended-value convention. The negative part of the logarithmic integral is integrable: on $p<q$, $p\log (q/p)\le q-p\le q$. Thus the extended-value integral defining the Kullback-Leibler divergence is well-defined. These expressions do not depend on which common dominating measure is chosen. The Kullback-Leibler divergence is generally asymmetric and therefore fails a defining property of a metric.

For completeness, the integral formula for total variation distance follows by taking $A=\{p\ge q\}$: since $\int (p-q)d\nu =0$, the positive and negative parts of $p-q$ have the same integral, and this set attains the supremum.

Put $a=\int \sqrt{pq}d\nu \le 1$, by the Cauchy-Schwarz inequality. A second application givesThe logarithmic inequality in the hint implies $\log t\le t-1$ for $t>0$. Apply it to $t=\sqrt{q/p}$ on $\{p>0\}$:Multiplying by $p$ and integrating provesIf $q=0<p$ on a set of positive $P$-measure, the right side is finite but the Kullback-Leibler divergence is infinite, so the inequality is immediate. ThusThis total variation–Hellinger–relative entropy inequality requires the stated Hellinger distance normalization for its first constant. With the alternative convention ${\bar{h}}^2=\frac{1}{2}\int (\sqrt{p}-\sqrt{q}{)}^2$, the correct comparison is $\mathrm{TV}\le \sqrt{2}\bar{h}$ and $\bar{h}\le \sqrt{\mathrm{KL}/2}$. In particular, $P=\mathrm{Bernoulli}(0.6)$ and $Q=\mathrm{Bernoulli}(0.4)$ have $\mathrm{TV}=0.2>\bar{h}\approx 0.1421$.

Here is a testing form of the Le Cam two-point lemma. Suppose two parameter values ${\theta }_0,{\theta }_1$ in a metric space have separation $\Delta =d({\theta }_0,{\theta }_1)>0$, and let $P_0,P_1$ be the corresponding observation laws. Every estimator $T$ satisfiesTo prove it, classify $T$ by the closer of the two parameters, breaking a tie in favour of one. Let $A=\{d(T,{\theta }_1)\le d(T,{\theta }_0)\}$. On $A$, the triangle inequality implies $d(T,{\theta }_0)\ge \Delta /2$, while on $A^c$ it implies $d(T,{\theta }_1)>\Delta /2$. Hence the sum of the two probabilities in the display is at leastAt least one is at least half the sum, proving the claim. It also gives a lower bound $\Delta (1-\mathrm{TV})/4$ for the maximum expected metric loss.

For absolute-error loss, or any loss given by a metric, there is a stronger expectation form. Take densities $p_0,p_1$ relative to $m=P_0+P_1$ and use the triangle inequality under their overlap:Therefore the Le Cam lower bound under absolute-error loss isBoth arguments are valid for randomized estimators as well, by adjoining the independent randomization to the observation; doing so does not change total variation distance.

For the normal distribution model, assume the usual nondegenerate convention $\sigma >0$ and use the full product observation laws. ChooseFor one observation, its Gaussian log likelihood ratio isUnder the first law, ${\mathbb{E}}_0(X-{\mu }_1{)}^2={\sigma }^2+({\mu }_1-{\mu }_0{)}^2$ and ${\mathbb{E}}_0(X-{\mu }_0{)}^2={\sigma }^2$. Thus the Kullback-Leibler divergence between normal distributions isFor the independent and identically distributed random variables, the joint log likelihood ratio is the sum of the individual ones, so the Kullback-Leibler divergence of the two full statistical samples is $n({\mu }_1-{\mu }_0{)}^2/(2{\sigma }^2)=1/4$. The comparison just proved implies $\mathrm{TV}(P_0,P_1)\le 1/2$. Applying the expectation form of the Le Cam two-point lemma givesThis is a Gaussian location minimax lower bound under absolute-error loss. The constant may depend on the fixed known $\sigma$, but neither on the estimator nor on $n$. If a signed nonzero scale were used, the same result holds with $|\sigma |$ in place of $\sigma$. If zero noise were allowed, $X_1=\mu$ would estimate $\mu$ exactly, so a positive lower bound would be false; positive variance is necessary.