The first condition is that $X$ is Markov with transition semigroup
$P(t)=e^{tQ}$. The second is that, for every function
$f:S\to \mathbb{R}$,is a martingale.

The first condition implies the second by conditioning over a short time interval, using
$P(h)f=f+hQf+o(h)$, summing increments, and passing to the limit. This is Dynkin's formula.

Conversely assume the martingale problem for a continuous-time Markov chain. Fix $T>0$ and a function $f$, and putThe backward equation gives ${\partial }_{t}u+Qu=0$. Applying the martingale identity to this time-dependent function shows that $u(t,X_t)$ is a martingale. Therefore, for $s\le T$,Taking indicator functions $f$ proves both the Markov property and the transition semigroup $e^{tQ}$.

Since $f_n\uparrow f$, the functionsare nonnegative and increase to $g=f-f_1$. The monotone convergence theorem gives
$\mu (g_n)\uparrow \mu (g)$. Because $f_1$ is integrable, adding the finite number
$\mu (f_1)$ yields

Integrability of $f_1$ is essential. On $\mathbb{R}$ with Lebesgue measure, letThen $f_n\uparrow 0$ pointwise, but every integral is $-\infty$, so
$\mu (f_n)$ does not increase to $\mu (0)=0$.

The monotone convergence theorem states that if
$0\le f_n\uparrow f$ pointwise, thenallowing the value $+\infty$.

Monotonicity of the integral shows that the increasing limit $L$ of the left side is at most $\int f$. Conversely, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. The setsincrease to the support of $s$. HenceLet $c\uparrow 1$ and take the supremum over all simple $s\le f$. This gives
$L\ge \int f$, proving equality.

Lift $f$ to a doubly periodic meromorphic function on $\mathbb{C}$. Integrate it around a fundamental parallelogram whose boundary avoids all poles. Integrals over opposite edges cancel by periodicity, while the residue theorem gives

Thus the principal-part map from part (b) takes values in the codimension-one hyperplane on which the sum of residue coefficients is zero. The Mittag--Leffler existence criterion on a compact Riemann surface says that prescribed principal parts occur precisely when their residues pair trivially with every holomorphic one-form. On a complex torus the holomorphic one-forms are the scalar multiples of $dz$, so the single condition is exactly the displayed residue sum. Hence the image has dimension
${\sum }_i{m}_i-1$, and the kernel of constants has dimension one. For $n\ge 1$,Equivalently, this is the genus-one case of the Riemann-Roch theorem for a positive divisor.

Choose a local coordinate $z_i$ at each $p_i$ and map $f\in V$ to all coefficients of the negative-power terms in its Laurent expansions:The kernel consists of globally holomorphic functions on the compact Riemann surface $R$. By the maximum modulus principle, every such function is constant, so the kernel has dimension one. Rank--nullity givesFor $n=0$, this simply says $\dim V=1$.

The sum of two meromorphic functions has no poles away from the union of their pole sets, and at each $p_i$ its pole order is at most the larger of the two orders, hence at most $m_i$. Scalar multiplication cannot increase a pole order. The zero function belongs to the set, so $V$ is a complex vector space.

For $x={\sum }_n{x}_n{e}_n$, Cauchy-Schwarz inequality givesThus every $T\in H(V,V)$ is bounded and
$\Vert T\Vert \le \Vert T{\Vert }_{\ast }$. The truncationshave finite rank and satisfyHence every such Hilbert-Schmidt operator is compact.

DefineThis is an inner product inducing $\Vert \cdot {\Vert }_{\ast }$. If $(T_j)$ is Cauchy in this norm, then each $T_j{e}_n$ converges to some $y_n$, and the usual Fatou/tail argument gives
${\sum }_n\Vert y_n{\Vert }^2<\infty$ and
${\sum }_n\Vert T_j{e}_n-y_n{\Vert }^2\to 0$. Defining
$Tx={\sum }_n{x}_n{y}_n$ produces a bounded operator by the preceding estimate and gives
$T_j\to T$ in $\Vert \cdot {\Vert }_{\ast }$. Thus $(H(V,V),\Vert \cdot {\Vert }_{\ast })$ is a Hilbert space.

The norms are not equivalent in infinite dimension. The rank-$N$ orthogonal projection hasAlthough $\Vert T\Vert \le \Vert T{\Vert }_{\ast }$, no uniform reverse inequality can hold.

A compact operator maps every bounded sequence to a sequence with a norm-convergent subsequence. A Hilbertian basis is a complete orthonormal sequence.

If $T$ is compact and ${\lambda }_n$ does not tend to zero, some subsequence has
$|{\lambda }_{n_j}|\ge \epsilon$. The vectors
$T{e}_{n_j}={\lambda }_{n_j}{e}_{n_j}$ are pairwise separated by at least $\sqrt{2}\epsilon$, so they have no convergent subsequence, a contradiction.

Conversely, if ${\lambda }_n\to 0$, define the finite-rank operatorThenAn operator-norm limit of compact operators is compact, so $T$ is compact.

Let $(T_m)$ be a Cauchy sequence in the operator norm. For each $v\in V$,so $(T_{m}v)$ is a Cauchy sequence. Completeness of the Banach space $V$ permits the definition
$Tv={\lim }_m{T}_{m}v$. Pointwise passage to the limit proves that $T$ is a linear. A norm-Cauchy sequence is a bounded sequence, say $\Vert T_m\Vert \le C$, and therefore
$\Vert Tv\Vert \le C\Vert v\Vert$, so $T$ is bounded. Finally, letting $m\to \infty$ in
$\Vert (T_n-T_m)v\Vert \le \epsilon \Vert v\Vert$ gives
$\Vert T_n-T\Vert \le \epsilon$. Thus $T_n\to T$ in operator norm and
$\mathcal{B}(V,V)$ is Banach.

The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

Distinct prime-power ideals are pairwise comaximal. For comaximal ideals, product equals intersection, so induction givesThe Chinese remainder theorem therefore gives the ring isomorphism

The ideal approximation theorem, proved by applying this Chinese-remainder map one prime power deeper, suppliesEquivalently, $v_{P_i}(\alpha )=m_i$ at every prime dividing $I$.

Since $(\alpha )\subseteq I$, the fractional idealis an integral ideal and $I{I}^{\prime }=(\alpha )$ is a principal ideal. At every $P_i\mid I$,
$v_{P_i}(I^{\prime })=v_{P_i}(\alpha )-m_i=0$, so $I+I^{\prime }={\mathcal{O}}_K$.

Choose $\beta \in I$ and $\gamma \in I^{\prime }$ with $\beta +\gamma =1$. Then
$\alpha ,\beta \in I$. For any $x\in I$,Thusso every ideal of ${\mathcal{O}}_K$ has a generating set of two elements.

An element $\alpha$ is separable over $K$ when its minimal polynomial has distinct roots in a splitting field. A polynomial has a repeated root exactly when it has a common root with its formal derivative. Hence
$\gcd (f,f^{\prime })=1$ implies that $f$ has no repeated root and $\alpha$ is separable.

Let $f$ be the minimal polynomial of $\beta$. Repeatedly factor through the Frobenius untilwhere $g^{\prime }\ne 0$. Irreducibility of $f$ makes $g$ irreducible, and
$\alpha ={\beta }^{p^h}$ has separable minimal polynomial $g$. PutThen $M/K$ is separable andEvery $\gamma \in L=K(\beta )$ is a polynomial in $\beta$ of degree below $[L:K]$. The freshman's-dream identity givesso $L/M$ is purely inseparable.

Finally, any intermediate field satisfying the stated conditions consists of elements separable over $K$, so it lies in the maximal separable subextension just constructed. Conversely, the condition that all relevant $p^h$th powers lie in that field puts ${\beta }^{p^h}$, and hence $M$, inside it. The two inclusions prove uniqueness.

If the minimal polynomial of $\gamma$ has degree $d$, then
${\mathbb{F}}_p(\gamma )$ has $p^d$ elements and is an intermediate field of
$L/{\mathbb{F}}_p$. The tower law for field extensions givesso $d\mid n$.

The multiplicative group of a finite field is cyclic. Choose a generator
$\gamma$ of $L^{\times }$, of order $p^n-1$. If $\gamma$ lay in a proper intermediate field of degree $d<n$, its order would divide $p^d-1<p^n-1$, a contradiction. Thus
${\mathbb{F}}_p(\gamma )=L$ and its minimal polynomial has degree $n$.

For arbitrary $r\ge 1$, let $E$ be the splitting field of
$X^{p^r}-X$ over ${\mathbb{F}}_p$. Its roots form a field: the Frobenius endomorphism shows they are closed under addition, subtraction, multiplication, and inversion. The derivative is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose minimal polynomial over ${\mathbb{F}}_p$ has degree $r$, proving that an irreducible polynomial of every positive degree exists.

Form the graph whose vertices are the points $x_i$, joining two when their Euclidean distance is one. Two distinct points have at most two common unit-distance neighbours, because two unit circles intersect in at most two points. The graph is therefore $K_{3,2}$-free. Part (c) bounds its unordered edges by $2n^{3/2}$, so the number of ordered unit-distance pairs is at most

If $G$ contains no $K_{3,2}$, every pair of vertices has at most two common neighbours. Double-counting a vertex together with an unordered pair of its neighbours givesConsequentlyBy Cauchy--Schwarz,This quadratic inequality implies $e<c{n}^{3/2}$ for an absolute constant $c$; for example $c=2$ works for every $n\ge 1$. Thus

Let $T$ be the number of triangles. Each edge $uv$ has at least
$d(u)+d(v)-n$ common neighbours, and summing common-neighbour counts over edges counts every triangle three times. HenceIf $e>(1+\delta )n^2/4$, this is greater than
$\delta en>\delta n^3/4$. Thereforeso one may take $\epsilon =\delta /12$.

The extremal number $\mathrm{ex}(n,H)$ is the maximum number of edges in an $n$-vertex graph containing no subgraph isomorphic to $H$.

Let $G$ be triangle-free with $e$ edges. For every edge $uv$, the neighbourhoods of $u$ and $v$ are disjoint apart from their endpoints, so
$d(u)+d(v)\le n$. Summing over edges and applying Cauchy-Schwarz inequality giveswhileThus $e\le n^2/4$, proving the required Mantel theorem bound.