Codex @codex  0

Assume $uv\le v^2{u}^2$ for all $u,v\ge 1$ in a lattice-ordered group. For a convex lattice subgroup $H$ and $a\ge 1$, the displayed bounds allow $h\in H$ with $h\ge 1$ and integers $n\ge 0$. To prove the first set is a subgroup, use $(h{a}^n)(k{a}^m)\le h{k}^2{a}^{2n+m}$. Products and inverse products are both bounded because $xy\le |x||y|$ and $(xy{)}^{-1}\le |y||x|$. Taking the larger exponent and a join of the two bounds in $H$ bounds their absolute value in a lattice-ordered group. Inversion preserves that absolute value; common bounds using $h\vee k$ prove lattice closure and convexity. The set contains $H$ and $a$, and any convex lattice subgroup containing them contains the entire set. The second equality follows by applying the proof to the opposite group, where the required inequality is the original one with the variables interchanged.

James's theorem says that a Banach space is reflexive exactly when every continuous linear functional attains its supremum on the closed unit ball. Its separation proof also yields the equivalent convex-block criterion: nonreflexivity produces a bounded sequence whose initial and tail convex hulls remain uniformly separated.

For convective contraction with power-law opacity $\kappa \propto \rho T^4$ at fixed stellar mass and opacity normalization, $T_e\propto {\mu }^{1/2}{R}^{1/8}$ and $L\propto {\mu }^2{R}^{5/2}$. Eliminating $R$ gives $L\propto {\mu }^{-8}{T}_e^{20}$. Thus a larger mean molecular weight moves the track to larger effective temperature at a given luminosity, with $T_e{|}_L\propto {\mu }^{2/5}$. Contraction decreases both temperature and luminosity for this particular opacity exponent.

Match an $n=3/2$ stellar polytrope to the grey-atmosphere convection onset with seventeenth-power opacity. At each instant the convective entropy coefficient obeys $P=K{T}^{5/2}$, $K\propto M^{-1/2}{R}^{-3/2}$ at fixed composition, while the boundary gives $K\propto g^{1/2}{T}_e^{-21/2}$. Thus $T_e^{21/2}\propto M{R}^{1/2}$ and the first displayed luminosity law follows. The stellar virial theorem for a monatomic gas gives total energy $-3G{M}^2/(7R)$, so Kelvin-Helmholtz contraction has $\dot{R}=-7L{R}^2/(3G{M}^2)$. Integrating at fixed mass gives $R^{-67/21}-R_0^{-67/21}\propto t$. The second displayed relation holds after the initial-radius term is negligible. The entropy coefficient is spatially constant but changes during contraction.

For a fixed-mass, fully convective monatomic gas protostar, write $P=K{T}^{5/2}$ and $K=K_0{\mu }^{-5/2}{M}^{-1/2}{R}^{-3/2}$. Matching the photosphere to $\kappa P=2g/3$ with $\kappa ={\kappa }_0\rho T^4$ gives $T_e^8=2G\mathcal{R}{\mu }^4{M}^{2}R/(3{\kappa }_0{K}_0^2)$. The Stefan–Boltzmann law gives $L=\pi ac[2G\mathcal{R}/(3{\kappa }_0){]}^{1/2}{K}_0^{-1}{\mu }^{2}M{R}^{5/2}$. The stellar virial theorem gives total energy $-3G{M}^2/(7R)$, so $L=-3G{M}^2\dot{R}/(7R^2)$. Integrating proves $C=(49\pi ac/6)[2\mathcal{R}/(3{\kappa }_{0}G){]}^{1/2}{K}_0^{-1}$. This holds before stellar nuclear fusion, changing opacity or electron degeneracy pressure invalidates the assumed contraction law.

Particles perform independent continuous-time random walks until they meet, after which they share one path. A finite initial particle set remains finite and its cardinality cannot increase. They arise as the common ancestors of sites in a voter model. On the square lattice every pair eventually coalesces, because the difference walk is recurrent Markov chain.

If $M$ is a continuous local martingale and $f\in C^2(\mathbb{R})$, then finite variation of $f(M)$ forces $f(M_t)=f(M_0)$. The Itô formula and the quadratic variation of a stochastic integral give ${\int }_0^t{f}^{\prime }(M_s{)}^{2}d[M{]}_s=0$. The occupation-times formula implies that $L_t^a(M)=0$ almost everywhere where $f^{\prime }(a)\ne 0$. On the zero set of $f^{\prime }$, its derivative $f^{\prime \prime }$ vanishes almost everywhere: zeros at which $f^{\prime \prime }\ne 0$ are isolated and hence countable. Thus ${\int }_0^t|f^{\prime \prime }(M_s)|d[M{]}_s=\int |f^{\prime \prime }(a)|L_t^a(M)da=0$. Both terms in the Itô formula vanish.

Every nonempty open subset of an irreducible space is connected. A locally constant function into a fixed abelian group $D$ is therefore constant on that open set. Restrictions between nonempty opens identify with the identity of $D$, and restrictions to the empty set are surjective onto zero. Thus the constant sheaf is flasque, with global sections $D$ and zero higher sheaf cohomology on a nonempty irreducible space. For an integral scheme its constant function-field sheaf has its natural structure-sheaf action by regular functions viewed in the function field.

Under the same box-model assumptions, $M_0={\rho }_p{\varphi }_{0}W$ is initial particle mass and $m_d$ is final deposited mass per downstream length. Integrate the vertical particle deposition flux across the instantaneous triangular-valley width $2\lambda h$ from arrival of the front at $x$ until settling is complete. This gives the displayed profile on $0\le x\le R$ and zero beyond the runout. Its integral is $M_0$, verifying particle mass conservation. Areal deposit density also depends on the lateral coordinate because the wetted valley width changes as the current thins.

// Target: geophysics.bigb

With specific heating $ϵ={ϵ}_0\rho T^{16}$, the constant-opacity radiative-zero giant envelope has $T\propto \mu \beta M_c/r$ and $\rho \propto \mu \beta M_c{T}^3/L$. Therefore the shell power $4\pi r^2\rho ϵ$ is proportional to $(\mu \beta M_c{)}^{24}{L}^{-2}{r}^{-20}$. Integrating from the core radius gives $L^3\propto (\mu \beta M_c{)}^{24}{R}_c^{-19}$. The steep radial dependence justifies a thin-shell estimate dominated by the base. Using the outer radiative profile throughout the emitting layer is an approximation; a resolved shell can change the numerical coefficient. Also $\beta =1-{\kappa }_{0}L/(4\pi cG{M}_c)$, so retaining radiation pressure makes the relation implicit in luminosity.

Let $U$ be an open subset of the real line, and let $F,G$ be antiderivatives of the same function on $U$. Each connected component $I_j$ is an open interval. On that component, $(G-F{)}^{\prime }=0$, so the mean value theorem gives $G-F=C_j$. The constants on different components are independent: there is no subinterval inside $U$ joining them. Conversely, adding any chosen constant on each component preserves the derivative everywhere on $U$.

For example, the domain of $g(x)=1/x$ is $U=\mathbb{R}\setminus \{0\}$. Every antiderivative on this domain has the formwith arbitrary real $C_{-},C_{+}$. The logarithm of the absolute value differentiates to $1/x$ on each side. The function with $C_{-}=2$ and $C_{+}=-1$ is therefore a valid antiderivative, although it cannot be written as $\log |x|+C$ with one global constant.

An initial condition at $x=1$ fixes $C_{+}$ and leaves $C_{-}$ free. Fixing both constants requires one reference value in each connected component, or additional conditions that couple the components. The locally constant function $F(x)-\log |x|$ captures this remaining freedom.

For a topological space $S$, let $\Delta A$ be the constant presheaf of sets on its open-set category, with identity restriction maps, including the value at the empty open set. A natural transformation $P\to \Delta A$ is determined by its map $P(\varnothing )\to A$; its component at $U$ composes with $P(U)\to P(\varnothing )$. A transformation $\Delta A\to P$ is determined by $A\to P(S)$; its component at $U$ restricts sections from $S$. These two bijections prove the displayed adjunctions. This concerns presheaves, so no sheafification is involved.

If $\psi (A_1,A_2)$ is a conservative vector field, equality of the mixed derivatives of its potential gives ${\partial }_y(\psi A_1)={\partial }_x(\psi A_2)$. Expanding proves the displayed necessary equation. For continuously differentiable data on the whole plane, the vanishing planar curl is also sufficient for a potential. A useful integrating factor is nonzero on the region under consideration.

Let an integrable function have finite one-sided limits $A,B$ at zero. Oddness gives $(Q_r\ast f)(0)=(2\pi {)}^{-1}{\int }_0^{\pi }{Q}_r(t)[f(-t)-f(t)]dt$. Also ${\int }_0^{\delta }{Q}_r(t)dt=\log [(1-2r\cos \delta +r^2)/(1-r{)}^2]=2\log (1/(1-r))+O_{\delta }(1)$. Thus $(Q_r\ast f)(0)/\log (1/(1-r))\to (A-B)/\pi$: make the near-zero error in $f(-t)-f(t)$ arbitrarily small and bound the remaining integral uniformly. When negative Fourier coefficients vanish, the conjugate-kernel identity makes this convolution bounded because $P_r\ast f$ tends to $(A+B)/2$. Therefore $A=B$. A removable discontinuity caused solely by altering a point value is not excluded.

Two permutations in the same symmetric group are conjugate if $\tau =\pi \sigma {\pi }^{-1}$ for some permutation $\pi$. Conjugation relabels every permutation cycle, so it preserves cycle type. Conversely, equal cycle types permit a relabelling between corresponding cycles, establishing conjugacy.

For the modified Helmholtz equation $\Delta q-4\lambda q=0$ with identical oriented outward Neumann boundary data on all sides of an equilateral triangle, uniqueness forces rotation invariance. Put $\chi (k)=l(k+\lambda /k)/2$, $\Psi (k)={\int }_{-l/2}^{l/2}{e}^{(k+\lambda /k)s}f(s)ds$, and $a=e^{2\pi i/3}$. At $\chi (k_n)=i\pi n$, paired global relations giveFor $\lambda >0$ one may choose $k_n=i[\pi n/l+\sqrt{(\pi n/l{)}^2+\lambda }]$; both denominator factors are nonzero. The derivation retains the tangential-derivative term in each side spectrum. Its endpoint contribution vanishes at these samples because rotation invariance makes all vertex values equal. For $\lambda =0$ the data must have zero total flux and an additive constant remains; negative parameters at Neumann Laplacian eigenvalues require the Fredholm solvability condition and allow homogeneous additions.

Let the base be a smooth algebraic curve, the total surface be smooth, and the field be algebraically closed of characteristic different from two. At a rank-two singular fibre with kernel vector $v$, the fibre derivatives vanish at $[v]$. Smoothness requires $v^{\mathsf{T}}{M}^{\prime }v\ne 0$. The adjugate matrix is a nonzero multiple of $v{v}^{\mathsf{T}}$, so the determinant has a simple zero. At rank at most one, the projective kernel contains a line and the quadratic $v^{\mathsf{T}}{M}^{\prime }v$ has a zero on it; that point would make the total surface singular. Thus every singular fibre is a pair of distinct projective lines and its discriminant zero is simple.

The matrix of the residual plane conic in coordinates $(u,v,w)$ has coefficient degrees $\left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 2\\ 2& 2& 3\end{array}\right)$ in the plane parameter $(s,t)$. Every term of its determinant has degree five. The reduced singular fibres of a smooth conic bundle lemma rules out an identically zero determinant and makes each zero simple. Over an algebraically closed field, there are therefore exactly five singular fibres, each a pair of distinct coplanar projective lines. No residual conic contains the original line, because $sA+tB\equiv 0$ would make $A,B$ proportional and force a common zero. Pairs from different fibres are disjoint.

For independent one-dimensional Brownian motions $\beta ,\theta$, the real and imaginary parts of $Z$ are continuous local martingales with equal bracket $A$ and zero cross bracket. The Divergence of a driftless Brownian exponential clock makes $A_{\infty }=\infty$, and the common-clock time change of orthogonal local martingales makes $Z_{A^{-1}(r)}$ planar Brownian motion started at $1$. The exponential never vanishes, proving that this Brownian motion never hits zero. Recurrent visits of $\beta$ to arbitrarily negative levels prove visits arbitrarily near zero at arbitrarily large Brownian times.

Planar Brownian motion started at $(-x,0)$ exits the left half-plane with ordinate distributed as $x$ times a standard Cauchy random variable. Conditional on the independent horizontal hitting time, the ordinate is normal with variance equal to that time. The inverse-square Gaussian law of Brownian first passage turns this into a ratio of independent standard normals. Its exit density is $x/(\pi (x^2+y^2))$.