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For the linear program maximizing $c^{T}x$ with $Ax\le b$, $x\ge 0$, a vector $y\ge 0$ with $A^{T}y\ge c$ proves $c^{T}x\le y^{T}b$ for every feasible $x$. If a feasible $x$ attains that bound, it is optimal. This is weak duality written as a directly checkable certificate; adding nonnegative multiples of constraints suffices to verify it. The equivalent reversed-inequality certificate applies to a minimization program.

The weak gradient is the vector of first-order weak derivatives of a function. Its $i$th component is specified by $\int u{\partial }_i\phi =-\int ({\partial }_{i}u)\phi$ for compactly supported test functions. For a differentiable function with appropriate local integrability it agrees with the classical gradient.

A sequence can converge as distributions off one point while retaining a nonconvergent multiple of a Dirac delta function at that point. If $u_j=v+c_j{\delta }_p+o_{{\mathcal{D}}^{\prime }}(1)$ with $c_j$ nonconvergent, restriction to the punctured domain tends to $v$, but a test nonzero at $p$ detects the obstruction to global convergence.

Every bounded sequence in a Hilbert space has a weakly convergent subsequence. The closed span of a sequence is separable; diagonal convergence of coordinates along an orthonormal basis gives a bounded limiting linear functional, represented by a vector of the Hilbert space. Applied to a bounded family of mollifications, this proves that their strong Lebesgue limit inherits the weak derivative bound.

A spin-zero parent decaying through a chiral weak charged current to a charged lepton and a massless neutrino cannot supply the helicities selected by a massless interaction. A charged-lepton mass insertion makes the amplitude proportional to $m_{\ell }$ and the decay width proportional to $m_{\ell }^2$. A spin-one W boson does not suffer this suppression.

The weak charged current pairs each neutrino with its corresponding charged lepton. Thus elastic muon-neutrino/electron scattering has only tree-level neutral-current exchange, while electron-neutrino and electron-antineutrino scattering also have crossed and annihilation charged-current diagrams. A contact term pairing a muon neutrino directly with an electron would violate this Standard Model flavour assignment.

For two outgoing massless leptons, ${\ell }_{\mu }=\bar{u}(q_1){\gamma }_{\mu }(1-{\gamma }^5)v(q_2)$ is transverse to $q=q_1+q_2$. This follows from the massless Dirac equation and anticommutation of the chirality matrix with the gamma matrices. Thus the $q^{\mu }{f}_{-}(q^2)$ term in a pseudoscalar-to-pseudoscalar form factor does not contribute to this scattering amplitude. A nonzero charged-lepton mass spoils exact transversality and restores a mass-suppressed scalar-form-factor contribution.

For a real wavefront error $h={\sum }_k{a}_k{s}_k$, its squared spatial root mean square is $⟨h^2⟩={\sum }_{k,l}{a}_k{a}_l⟨s_k{s}_l⟩$. Orthogonal unit-peak sine modes have mean square $1/2$, so $h_{\mathrm{rms}}^2=\frac{1}{2}{\sum }_k{a}_k^2$, whereas the coefficient norm is $\sqrt{{\sum }_k{a}_k^2}$. On a finite optical pupil orthogonality needs checking; independent uniform random wave phases give the same $1/2$ diagonal after wave phase averaging. This normalization matters when a speckle contrast of a sinusoidal wavefront error is specified by peak wave amplitude.

For the unit-speed wave equation, the integral of $(u_t^2+|\nabla u{|}^2)/2$ over $B_r(x_0)$ at time $T\ge 0$ is at most its initial integral over $B_{r+T}(x_0)$. On a shrinking ball the outward energy flux minus the loss from its moving boundary is $-((u_t-{\partial }_{n}u{)}^2+|{\nabla }_{\tan }u{|}^2)/2$, hence nonpositive. This estimate establishes the domain of dependence and finite propagation speed without any assumptions at spatial infinity.

For an individual monochromatic component in $c=c(Y)$, conserved $\omega ,k$ give $\ell =\sqrt{{\omega }^2/c^2-k^2}$. With amplitude independent of $X,T$, wave-action transport for a variable-speed wave conserves $c^2\ell A^2$ on a positive-$\ell$ branch, hence $A=\mu /(c\sqrt{\ell })$. The divergence at $\ell =0$ signals failure of the outer WKB approximation.

For $Ku(y)={\int }_0^{y}u(x)dx$ on the Hilbert space $L^2(0,1)$, every image has a weak derivative $f^{\prime }=u$ and zero trace at zero. Conversely every such Sobolev space element is this integral of its weak derivative. The range is dense because it contains smooth functions supported in $(0,1)$, but it is not closed: a step function can be approximated in $L^2$ by continuous ramps and is not itself in $H^1$. The Moore–Penrose inverse of an operator has this range as its domain and differentiates there.

On $(0,1/2)$ use the forward difference and on $(1/2,1)$ the backward difference, with $0<h<1/2$. This defines a bounded linear operator on $L^2(0,1)$ and $\Vert D_h\Vert \le \sqrt{6}/h$. For $f$ in the range of the Volterra integration operator, these differences are local averages of $f^{\prime }$ and converge to $f^{\prime }$ in $L^2$ by continuity of translations. Hence $D_h$ gives a linear regularization of differentiation. No pointwise evaluation of an arbitrary $L^2$ equivalence class is needed: translated functions are defined almost everywhere.

A lower-viscosity basal sediment layer changes the plate coupling in viscous mountain spreading over an underthrust plate. With mountain thickness $H$ and sediment depth $d$, equal densities give one hydrostatic pressure gradient $G=\rho g{\partial }_x(H+d)$. Fluxes are $q_s=Ud-G(d^{2}H/2+d^3/3)/{\mu }_s$ and $q_m=UH-G(d{H}^2+d^{2}H/2)/{\mu }_s-G{H}^3/(3\mu )$. Separate conservation of mass laws determine each layer. Steady zero mountain flux and sediment influx $U{d}_{\infty }$ give contact height $3d_{\infty }$ and surface slope $2{\mu }_{s}U/(9\rho g{d}_{\infty }^2)$.

For a Keplerian disk with evolution ${\partial }_t\Sigma =(2\pi r{)}^{-1}{\partial }_r\mathcal{F}+S$ and $\mathcal{F}=6\pi r^{1/2}{\partial }_r(r^{1/2}\bar{\nu }\Sigma )$, take zero viscous torque at $r_1>0$ and zero flux outside the source region at $r_2>r_1$. The steady state gives ${\mathcal{F}}^{\prime }=-2\pi rS$, and its outer condition gives the displayed flux. Integrating once more from the zero-torque inner boundary condition yieldsThe equation determines the viscosity-density product, so a thermodynamic viscosity closure can be imposed afterward. For nonnegative integrable $S$, it is nonnegative and the inner accretion rate is exactly the total supplied mass per unit time.

For a Keplerian accretion disk with $\bar{\nu }\propto r^{5/2}$, dimensionless $g=\bar{\nu }\Sigma /({\bar{\nu }}_0{\Sigma }_0)$ satisfies $g_{\tau }=4x^{5/2}{g}_{xx}+6x^{3/2}{g}_x$. In $y=x^{-1/2}$ this is $g_{\tau }=y{g}_{yy}$. The ansatz $g=G(y/\tau )$ gives $G=A+B{e}^{-y/\tau }$; finite disk mass selects $A=0$. Its nonzero profile has logarithmically infinite disk angular momentum on an infinite radial domain.

A Keplerian disk with surface density $\Sigma \sim r^{-p}$ at infinity has finite outer mass for $p>2$ and finite outer angular momentum for $p>5/2$, since the relevant integrals are $\int \Sigma rdr$ and $\int \Sigma r^{3/2}dr$. The borderline $p=5/2$ has logarithmically divergent angular momentum even though its mass converges.

The conduit equation evolves the positive cross-sectional area $A$ of a viscous buoyant conduit. Its scaling uses axial length $a_0/\sqrt{8\lambda }$ and velocity $\Delta \rho g{a}_0^2/(8\lambda {\mu }_o)$, where $\lambda ={\mu }_i/{\mu }_o$. Linearization at $A=1$ gives $\omega =2k/(1+k^2)$: phase velocity is upward but group velocity changes sign at $|k|=1$. Nonlinear elevation waves obey the solitary-wave amplitude-speed relation for the conduit equation.

A linear Maxwell fluid has constitutive equation $\mathbf{\sigma }+\lambda {\partial }_t\mathbf{\sigma }=\mu \dot{\mathbf{\gamma }}$, with $\dot{\mathbf{\gamma }}$ twice the rate-of-strain tensor. Here $\mu$ is the zero-frequency shear viscosity and $\lambda >0$ is the viscoelastic relaxation time. Stress at fixed strain decays as $e^{-t/\lambda }$; the corresponding series spring and dashpot have shear modulus $G=\mu /\lambda$. In small-amplitude simple shear from an isotropic relaxed state, the first-order forcing is off-diagonal, so both normal-stress differences vanish at linear order. This is not a statement about the nonlinear objective model at second order in shear rate.

Every virtually soluble group $K$ has a soluble characteristic subgroup of finite index. First take a soluble normal subgroup $K_0$ of finite index by the subgroup core construction. Choose a soluble normal subgroup $R\supseteq K_0$ maximizing its image size in the finite quotient $K/K_0$. For every soluble normal subgroup $S$, the product $RS$ is soluble, because its quotient by $R$ is a quotient of $S$. Maximality forces $S\le R$. Hence $R$ is the unique largest soluble normal subgroup and is invariant under all automorphisms. This avoids incorrectly intersecting infinitely many conjugates when proving closure under group extensions.

For halo mean density ${\Delta }_v$ times the critical density at its epoch, the circular virial velocity obeys $v_{\mathrm{vir}}^2=G{M}_h/R_h=({\Delta }_v/2)H^2{R}_h^2$. Thus $v_{\mathrm{vir}}=H{R}_h\sqrt{{\Delta }_v/2}$. Specifying whether overdensity is relative to mean matter or critical density is necessary for numerical inference.