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Under the usual regularity conditions, consistency and asymptotic normality of a maximum likelihood estimator givewhere $I({\theta }_0)$ is the one-observation Fisher information matrix. In a scalar model this reads $N(0,I({\theta }_0{)}^{-1})$.

Use the convention $J_0=0$, so $A(t)=t$ when $N_t=0$. Conditional on $N_t=n\ge 1$, the last jump $J_n$ is the largest of $n$ independent uniform points on $[0,t]$. The stated order statistic formula givesand hence, for every $n\ge 0$,Putting $\mu =\lambda t$ and averaging over $N_t$,This agrees with age of a Poisson process.

Condition on $N_t=n$. By Poisson process conditional arrival times, the jump times are the ordered version of independent $U_1,\dots ,U_n$ uniformly distributed on $[0,t]$. The sum is symmetric in the marked pairs, and the marks are independent, soWrite the bracket as $a$. Averaging over the Poisson distribution of $N_t$ givesThis is the exponential formula for a marked Poisson sum.

For $0<j_1<\cdots <j_n<t$, the probability of one jump in each infinitesimal interval around $j_1,\dots ,j_n$ and no other jump by time $t$ gives the joint densityfor $(J_1,ldots,J_n)$ together with the event $N_t=n$. Sincethe conditional density ison that ordered simplex.

Now $n$ independent uniform points on $[0,t]$ have joint density $t^{-n}$ on the cube. Each ordered vector has $n!$ permutations, so their order statistics have density $n!/t^n$ on the same simplex. The densities agree, proving Poisson process conditional arrival times.

Superpose the two independent Poisson processes. The combined process has rate $9$, and each arrival is independently a Number 1 bus with probability $1/9$ and a Number 8 bus with probability $8/9$. Therefore the number of Number 8 buses before the first Number 1 has the geometric distribution on $\{0,1,2,\dots \}$, andThis is number of one Poisson type before another with rates $1$ and $8$.

Fix $A_1\in {\mathcal{A}}_1$ and letThis is a Dynkin system. It contains $\Omega$. If $B\in {\mathcal{D}}_{A_1}$, thenso it is closed under complements. Countable additivity gives closure under countable disjoint unions.

The hypothesis says ${\mathcal{A}}_2\subseteq {\mathcal{D}}_{A_1}$. Since ${\mathcal{A}}_2$ is a pi-system, Dynkin lemma yieldsThus the factorization holds for every $A_1\in {\mathcal{A}}_1$ and every $B\in \sigma ({\mathcal{A}}_2)$.

Now fix such a $B$ and defineThe identical calculation makes ${\mathcal{E}}_B$ a Dynkin system. The first step gives ${\mathcal{A}}_1\subseteq {\mathcal{E}}_B$, so a second application of Dynkin lemma givesTherefore the probability factorization holds for everywhich is precisely independence of the two sigma-algebras. This is independence extended from generating pi-systems.

Dynkin lemma, also called the pi-lambda theorem, states that if a Dynkin system $\mathcal{D}$ contains a pi-system $\mathcal{P}$, then

Let $\lambda (\mathcal{P})$ be the smallest Dynkin system containing $\mathcal{P}$. It is enough to prove that $\lambda (\mathcal{P})$ is a sigma-algebra, because then

First fix $A\in \mathcal{P}$ and defineThis is a Dynkin system. It contains $\Omega$ because $A\in \lambda (\mathcal{P})$; closure under relative complements follows fromand closure under disjoint unions follows by distributing $A\cap -$ over the union. Since $\mathcal{P}$ is closed under intersections, $\mathcal{P}\subseteq {\mathcal{G}}_A$. Minimality therefore givesThus $A\cap B\in \lambda (\mathcal{P})$ whenever $A\in \mathcal{P}$ and $B\in \lambda (\mathcal{P})$.

Now fix $B\in \lambda (\mathcal{P})$ and defineThe same argument shows that ${\mathcal{H}}_B$ is a Dynkin system, and the preceding paragraph shows that it contains $\mathcal{P}$. Hence ${\mathcal{H}}_B=\lambda (\mathcal{P})$. We have proved that $\lambda (\mathcal{P})$ is itself a pi-system.

A Dynkin system that is also a pi-system is a sigma-algebra: it is closed under arbitrary finite intersections, hence finite unions by complements, and any countable union can be disjointified before using closure under disjoint unions. Therefore $\lambda (\mathcal{P})=\sigma (\mathcal{P})$, proving the lemma.

The induced metric makes the compact surface $S$ a compact Riemannian manifold. By the Hopf-Rinow theorem it is geodesically complete, so every ${\exp }_p$ is defined on all of $T_{p}S$. In particular, ${\exp }_p(V(p))$ is defined for every $p$.

The global exponential mapis smooth by smooth dependence of geodesics on initial data. Since $p\mapsto (p,V(p))$ is a smooth section of $TS$, their compositionis smooth.

For $0\le t\le 1$, defineCompleteness makes this a well-defined smooth homotopy, with ${\varphi }_0={\mathrm{id}}_S$ and ${\varphi }_1=\varphi$. By homotopy invariance of degree modulo two,This is the exponential displacement map on a compact surface.

For $v\in T_{p}S$, let ${\gamma }_v$ be the maximal geodesic satisfyingThe exponential map ison the domain of the exponential mapThis is an open, star-shaped neighbourhood of zero. In local coordinates the geodesic equation is a smooth ordinary differential equation, so smooth dependence on its initial position and velocity proves that $(p,v)\mapsto {\exp }_p(v)$ is smooth on its domain.

For both requested phenomena consider the embedded surfacewith $p=(1,0,0)$. Its geodesics are Euclidean straight lines for as long as they remain in $S$. The initial vector $v=(-2,0,0)$ would givewhich reaches the missing origin at $t=1/2$. Thus $v\notin {\mathcal{D}}_p$, and the domain is not all of $T_{p}S$.

The point $q=(-1,0,0)$ is not in the image of ${\exp }_p$. Any geodesic from $p$ to $q$ would have to be their unique Euclidean straight line, which passes through the omitted origin. Hence this exponential map is also not surjective, as summarized by exponential map of the punctured plane.

Assume first that $Y$ is connected; otherwise the definition is made separately on each connected component. Choose a regular value $y\in Y$ and define the degree modulo two by

The following course results make this well-defined.

The same transverse-preimage argument applied to a smooth homotopy proves homotopy invariance. Consequently the definition depends only on $f$, rather than on the chosen regular value.

A point $y_0\in Y$ is a regular value of the smooth map $f:X\to Y$ if, for every $x\in f^{-1}(y_0)$, the derivativeis surjective. A value outside the image is regular vacuously.

For a nonvacuous example, takeThe value $1$ is regular: its fibre is the unit circle, and at every $(x,y)$ on that circle,is a nonzero linear functional and hence surjective.

Sard theorem states that the set of critical values of a smooth map between smooth manifolds has measure zero in the target. Equivalently, almost every target point is a regular value.

Compact support and the fundamental theorem of calculus give, for every $(x,y)$,whereApplying the same argument in the other coordinate gives

Multiplication and Fubini's theorem now yieldFinally, the Cauchy-Schwarz inequality givesThus the required estimate holds with $C=4$; it is the Ladyzhenskaya inequality in two dimensions.

For $1\le p<\infty$, choose nonzero $\varphi \in C_c({\mathbb{R}}^{n-1})$ and $\psi \in C_c(\mathbb{R})$ with $\psi (0)=1$. DefineThese functions are continuous and belong to the Lebesgue space $L^p({\mathbb{R}}^n)$, while their restrictions to the hyperplane are all the same:Changing variables in the normal coordinate givesIf the claimed bounded $T$ existed, thena contradiction. This scaling argument is the failure of an Lp hyperplane trace.

For $s>\frac{1}{2}$, the Sobolev trace theorem states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator

Take first $u$ in the Schwartz space. Up to the harmless constant determined by the Fourier transform convention,The Cauchy-Schwarz inequality, with weights $1+|{\xi }^{\prime }{|}^2+{\xi }_n^2$, givesAfter the substitution ${\xi }_n=(1+|{\xi }^{\prime }{|}^2{)}^{1/2}t$, the second factor iswhere finiteness is exactly the condition $s>1/2$. Multiplying by $(1+|{\xi }^{\prime }{|}^2{)}^{s-1/2}$ and integrating in ${\xi }^{\prime }$ yieldsDensity of the Schwartz space in the Sobolev space completes the unique bounded extension.

Write $T_v=T(v)$. We prove that $T$ has a closed graph. SupposePut $u_n=v_n-v$ and $g=f-T_v$. Then $u_n\to 0$ and $T_{u_n}\to g$ in $V^{\ast }$.

Fix $w\in V$ and $t\in \mathbb{R}$. Positivity and linearity giveBecause $(T_{u_n})$ converges in the continuous dual space, its norms are bounded, soAlso $T_{u_n}(w)\to g(w)$, while $T_w(u_n)\to 0$ because the fixed functional $T_w$ is continuous. Passing to the limit yieldsfor every real $t$. If $g(w)\ne 0$, a sufficiently small $t$ of the opposite sign makes the right-hand side negative. Hence $g(w)=0$ for every $w$, so $g=0$ and $f=T_v$.

The graph is therefore closed. The closed graph theorem now proves that $T:V\to V^{\ast }$ is continuous. This is precisely continuity of a positive linear map into a dual space.

The continuous dual space of a normed vector space $V$ iswith the operator norm

Let $(f_n)$ be a Cauchy sequence in this norm. For each $v\in V$,so $(f_n(v))$ is Cauchy in the scalar field. Define $f(v)={\lim }_n{f}_n(v)$. Passing to limits in the linearity identities shows that $f$ is linear.

A norm-Cauchy sequence is bounded, say $\Vert f_n\Vert \le M$. Hence $|f(v)|\le M\Vert v\Vert$, so $f\in V^{\ast }$. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$. Letting $m\to \infty$ givesfor every $v$, and therefore $\Vert f_n-f\Vert \le \epsilon$. Thus $f_n\to f$ in operator norm, proving completeness of the dual space: $V^{\ast }$ is Banach.

The open mapping theorem states that every surjective bounded linear map $S:X\to Y$ between Banach spaces maps open sets to open sets. Equivalently, there is $c>0$ such that

The closed graph theorem states that a linear map $T:X\to Y$ between Banach spaces is continuous if and only if its graphis closed in $X\times Y$.

Assume the open mapping theorem and suppose $\Gamma (T)$ is closed. As a closed subspace of the Banach space $X\times Y$, the graph is itself Banach. The coordinate projectionis a bounded linear bijection. By the open mapping theorem, $P^{-1}$ is bounded. The other coordinate projection $Q:\Gamma (T)\to Y$ is bounded, andThus $T$ is bounded and hence continuous. This is the closed graph theorem from the bounded inverse theorem argument.

Because $p\equiv 11(\mathrm{mod}12)$, we have $p\equiv 3(\mathrm{mod}4)$ andFor $\theta =(1+\sqrt{-p})/2$, the minimal polynomial isSince $p\equiv 2(\mathrm{mod}3)$, reduction modulo $3$ gives $f(T)\equiv T(T-1)$. Its two roots are distinct, sowith distinct prime ideals of norm $3$. Thus $3$ splits completely, as recorded in splitting of three in Q of square root minus p.

The ideal class group ${\mathrm{Cl}}_K$ is the group of nonzero fractional ideals of ${\mathcal{O}}_K$ modulo the subgroup of principal fractional ideals. Saying that ${\mathfrak{p}}_1$ has order $n$ means that $n$ is the least positive integer for which ${\mathfrak{p}}_1^n$ is principal.

Suppose for contradiction that $n$ is odd and ${\mathfrak{p}}_1$ has order $n$. Then ${\mathfrak{p}}_1^n=(\alpha )$ for some $\alpha \in {\mathcal{O}}_K$, and taking norms givesWrite $\alpha =(a+b\sqrt{-p})/2$, where $a,b\in \mathbb{Z}$ have the same parity. ThenIf $b\ne 0$, the right-hand side is at least $p$, contradicting $p>3^{n+2}>4\cdot 3^n$. Hence $b=0$, so $\alpha$ is a rational algebraic integer and therefore an integer. But then $3^n=N(\alpha )={\alpha }^2$, impossible when $n$ is odd. This proves the odd-order obstruction for a split prime in an imaginary quadratic field and shows that the order cannot equal $n$.