A nonzero weight-fourteen cusp form would have order at infinity at least one and order at $i$ at least one, since $i^{14}=-1$. The valence formula for the modular group would then give $1+1/2\le 14/12$, which is impossible.

The upcrossing count $U_N[a,b]$ counts completed upcrossings by time $N$, allowing the first purchase at time zero. For a martingale, the Doob upcrossing inequality bounds its expected value by $\mathbb{E}(M_N-a{)}^{-}/(b-a)$.

A universal sentence has the form $\forall \mathbf{x}\psi (\mathbf{x})$ with quantifier-free formula $\psi$, including an empty quantifier block. Its truth passes from a first-order structure to every substructure of a first-order structure: tuples from the smaller domain are also tuples of the larger domain, and quantifier-free truth agrees. Therefore a first-order theory of universal sentences has a first-order model class closed under substructures of a first-order structure.

A first-order theory has its class of first-order models closed under substructures of a first-order structure exactly when it is equivalent to its universal consequences $T_{\forall }$. If $A\models T_{\forall }$, then $T\cup \mathrm{Diag}(A)$ is consistent. Otherwise a finite logical conjunction $\delta$ of signed diagram of a structure sentences yields a universal consequence $\forall \mathbf{x}\neg \delta (\mathbf{x})$ false in $A$. The compactness theorem gives a first-order model of $T$ containing an embedded copy of $A$, and closure under substructures of a first-order structure implies $A\models T$. The converse is preservation of universal sentences. The diagram of a structure must include negated as well as positive atomic sentences so that the resulting map is an embedding, not just a homomorphism.

A biholomorphism of the unit disk to itself. Every such map has the form $e^{i\theta }(z-a)/(1-\overline{a}z)$ with $|a|<1$. These maps normalize an extremal univalent function in the proof of the Riemann mapping theorem.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension to every finite field extension. Any K-isomorphism $K({\alpha }_i)\to K({\alpha }_j)$ between conjugate roots pulls the absolute value back to an extension on the source field, which must be the given one. Thus $|h({\alpha }_i)|=|h({\alpha }_j)|$ for every $h\in K[X]$. This argument includes inseparable irreducible polynomials: no transitive Galois action or completeness assumption is needed.

Suppose a Non-Archimedean absolute value extends uniquely to every finite extension. If $f\in {\mathcal{O}}_K[X]$ is monic and irreducible, every root has absolute value at most one: otherwise its leading power dominates all lower terms, contradicting $f(\alpha )=0$ by the ultrametric inequality. In a splitting field, let $\phi$ be the minimal polynomial of an algebraic element of the residue of one root. Lift its coefficients to ${\mathcal{O}}_K$. The lift has absolute value less than one at that root and hence, by equal absolute values of algebraic conjugates, at every root. Consequently every root residue is a zero of $\phi$. Since reduction preserves the product of the linear factors with multiplicities, every irreducible factor of $\overline{f}$ is $\phi$, proving the displayed identity for some $m\ge 1$.

For a uniquely ergodic continuous map $T$ of a compact metric space, every continuous $f$ satisfies $N^{-1}{\sum }_{j=0}^{N-1}f(T^{j}x)\to \int fd\mu$ uniformly in $x$. Otherwise choose increasingly long orbit empirical measures with a fixed discrepancy. Compactness provides a weak limit; the telescoping identity makes it invariant, so it must be the unique measure $\mu$, contradicting that discrepancy. Compactness and continuity are essential hypotheses.

For the splitting field $M/{\mathbb{Q}}_3$ of $T^3-3T+3$, the cubic is Eisenstein and its discriminant is $-135$. The quadratic subfield is ${\mathbb{Q}}_3(\sqrt{3})$, and the degree-six extension is totally ramified. If $\alpha$ is a root and $s=\sqrt{3}$, then $\pi ={\alpha }^2/s$ is a uniformizer. Transpositions have $v_M(\sigma \pi -\pi )=1$, while three-cycles have value $2$, giving the displayed ramification groups. Its different exponent from ramification groups is $7$.

Let $K$ be an equal-characteristic-p local field and let $c\in K$ have $v_K(c)=-m<0$, with $p\nmid m$. If ${\alpha }^p-\alpha =c$, the Artin–Schreier extension $L=K(\alpha )$ has degree $p$ and is totally ramified: a root in $K$ would force the pole order to be divisible by $p$, and $p{v}_L(\alpha )=-me(L/K)$ forces $e=p$. Choose $1\le s<p$ and an integer $r$ with $rp-sm=1$. For a base uniformizer $t$, the element ${\pi }_L=t^r{\alpha }^s$ has valuation one. Under ${\sigma }_a(\alpha )=\alpha +a$, $a\in {\mathbb{F}}_p^{\times }$, the leading term in ${\sigma }_a({\pi }_L)-{\pi }_L$ has valuation $rp-(s-1)m=m+1$. The uniformizer criterion for lower ramification groups therefore gives the single lower break $m$. The Herbrand function is the identity up to that break, so the upper break is also $m$.

For $n\ge 3$, put $e_i=e(T_i(n))$, $t_3=k_3(T_3(n))$, and $c_n=(e_3-e_2)/t_3$. Every graph on $n$ vertices satisfiesThus an edge count $e_2+\theta (e_3-e_2)$ forces at least $\theta t_3$ triangles in a graph. This is an exact finite bound, with the rounding in the Turán graph retained. To prove it, use edge-triangle symmetrization. The inequality $c_n\ge 2/n$ allows merging the two smallest classes when there are at least four classes. With three classes, fixing the middle class makes the objective affine in the product of the other two sizes; either merge them or balance them. Only $T_2(n)$ and $T_3(n)$ need remain.

If $H{H}^2(A,A)=0$, every formal associative deformation on the adic completion of a module $A[[t]]$ is trivial. After killing lower coefficients, associativity makes the order-$r$ coefficient a Hochschild cocycle ${\mu }_r=\delta g_r$. Transport by $\mathrm{id}+t^r{g}_r$ subtracts $\delta g_r$ and kills it. The successive transformations stabilize modulo each $t^n$ and converge to an invertible change of coordinates. Completeness, and equivalence congruent to the identity modulo $t$, are essential to this argument.

On the ordered branch of tricritical crossover scaling, at zero field and $X\to +\infty$, the minimizing square is $a={\psi }^2=(\sqrt{X^2+4}-X)/2\sim 1/X$. Its stationarity relation $Xa+a^2=1$ gives ${\Phi }_{-}(X,0)=-a/4-a^3/12\sim -1/(4X)$. Thus the minimized free-energy density becomes $-r^2/(4q)$ for fixed $q>0$ as $r\uparrow 0$, recovering the ordinary quartic continuous phase transition and its heat-capacity critical exponent $\alpha =0$. Replacing the scaling function by a nonzero constant in this limit would give an incorrect exponent.

The trapezoidal rule is A-stable but is not B-stable. For the dissipative vector field $f(y)=-y^3$ at step size one, its uniquely defined step map $T$ obeys $T+T^3/2=y-y^3/2$. At $y=\sqrt{2}$, $T=0$ and implicit differentiation gives $T^{\prime }=-2$, so nearby states expand rather than contract. This is a concrete nonlinear obstruction, rather than merely failure of the sufficient algebraic stability criterion.

For standard three-dimensional Brownian motion started at $x$ with $r<|x|<R$,The harmonic function $x\mapsto 1/|x|$ gives a bounded stopped martingale in the annulus, so the optional stopping theorem gives the formula. Taking $R\to \infty$ yields ${\mathbb{P}}_x({\tau }_r<\infty )=r/|x|$ for $|x|>r$. This limit uses path continuity on the finite interval before a hit, rather than assuming transience to prove transience.

For a finite electrical network with positive electrical conductances $c_e$, choose a spanning tree with lawThe one-edge case of the transfer-current theorem gives $\mathrm{Pr}(e\in T)=c_e{R}_{\mathrm{eff}}(a,b)$ for $e=\{a,b\}$. A unit injected electric current produces voltage $R_{\mathrm{eff}}(a,b)$, and Ohm's law makes the electric current along $e$ equal to $c_e{R}_{\mathrm{eff}}(a,b)$.

In a triangle of unit resistors, one resistor is parallel to a series pair, so $R_{\mathrm{eff}}=1\cdot 2/(1+2)=2/3$. Each edge consequently has inclusion probability $2/3$, also clear from the three equally likely spanning trees. For a wired uniform spanning forest, take the finite wired limit and use the limiting wired effective resistance.

Let $V_T$ be total magma volume shared between a reservoir and an advancing toughness-controlled magma intrusion, $p_r=E(V_T-V)$, and conduit conductance $\beta >0$. With $A=\sqrt{8\pi /3}B{\kappa }_f^{3/2}$,Equilibria satisfy $\sqrt{V}(V_T-V)=A/E$. Two positive equilibria exist when $A/E<2V_T^{3/2}/(3\sqrt{3})$; the smaller is unstable and the larger is stable under this volume equation. An advancing solution between them approaches the larger. The singular pressure as $V\to 0$ prevents this late-time model from nucleating an intrusion by itself.

For a real quadratic form on an $n$-dimensional vector space, with rank $r$ and signature $s$ defined as positive minus negative index, the maximum dimension of a totally isotropic subspace is $n-(r+|s|)/2$. Quotienting by the radical of a bilinear form leaves a nondegenerate form of indices $p,q$. Projection of an isotropic subspace to each sign-coordinate space is injective, bounding its dimension by $\min (p,q)$. The radical and vectors pairing one positive with one negative unit coordinate attain the bound $n-r+\min (p,q)$.

For $a<b$ strictly inside a one-dimensional domain, the indicator $u=1_{[a,b]}$ has distributional derivative $Du={\delta }_a-{\delta }_b$, hence $\mathrm{TV}(u)=2$. Directly from the test-function definition, $\int u{\phi }^{\prime }=\phi (b)-\phi (a)\le 2$, and two disjoint normalized smooth bump functions attain the bound. Values at the endpoints themselves do not change the $L^1$ function.

For $g\in L^2$ on a bounded domain, total variation denoising has a unique minimizer. A bounded minimizing sequence has a weakly convergent $L^2$ subsequence; variation is a supremum of weakly continuous test-function pairings and the fidelity has weak lower semicontinuity. The strictly convex fidelity proves uniqueness. Bounded-variation contraction under clipping proves that data range bounds pass to the minimizer.