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A norm-closed convex set in a real or complex Banach space is closed in the weak topology, so it contains limits of all weakly convergent nets as well as sequences. If $x$ lies outside the set, the Hahn-Banach separation theorem gives a bounded linear functional $f$ and a real $\alpha$ with $\mathrm{Re}f(c)\le \alpha <\mathrm{Re}f(x)$ for every point $c$ of the set. The weakly open half-space $\{y:\mathrm{Re}f(y)>\alpha \}$ contains $x$ and misses the set, proving weak closedness.

For sequences, Mazur lemma gives another useful proof: convex combinations of each tail converge in norm to the weak limit. They stay in the convex set, and norm closedness puts the limit there. Convexity is essential: the unit sphere of an infinite-dimensional Hilbert space is norm closed but contains an orthonormal sequence converging weakly to zero, which is outside that sphere.

Every orthonormal sequence in a Hilbert space converges weakly to zero, by Bessel inequality, but no subsequence converges strongly because distinct terms remain distance $\sqrt{2}$ apart.

In a Hilbert space, weak convergence $x_n\rightharpoonup x$ together with $\Vert x_n\Vert \to \Vert x\Vert$ implies strong convergence $x_n\to x$.

If $x_n\rightharpoonup x$ in a Hilbert space, thenFor $x\ne 0$, take the inner product with $x$ and apply Cauchy-Schwarz before passing to the lower limit.

Every bounded sequence in a separable Hilbert space has a weakly convergent subsequence. Successive subsequences make each basis coordinate converge, and a diagonal subsequence converges in every coordinate; the coordinate criterion for weak convergence in a separable Hilbert space finishes the proof.

For a Hilbertian basis $(e_i)$, a sequence $x_n$ converges weakly if and only if $(x_n)$ is norm bounded and every coordinate sequence $⟨x_n,e_i⟩$ converges.

For a slowly varying local dispersion relation $\omega =\Omega (\mathbf{k};\mathbf{x},t)$, wave-packet rays obey

On the unit interval $0<x<1$, a fixed-end string with unit wave speed and damping coefficient1 obeysExpanding in the Fourier sine series $y={\sum }_{n\ge 1}{T}_n(t)\sin (n\pi x)$ gives $T_n^{\prime \prime }+T_n^{\prime }+n^2{\pi }^2{T}_n=0$. ThusAll these modes are underdamped; their frequency depends on the endpoint geometry as well as the damping. Nonunit length, speed or damping changes this modal equation.

For a sufficiently regular finite-energy solution, the energy dissipation identity for a linearly damped string isMultiply the equation by $y_t$ and integrate by parts; the fixed endpoints remove the boundary term. This identifies exactly where mechanical energy is lost. The modal solution also separates decay of the amplitude envelope from oscillatory changes of the actual displacement.

For $\overline{X}\sim N(\mu ,1/n)$ under $H_0:\mu =0$, the two-sided level-$\alpha$ test rejects when $|\sqrt{n}\overline{X}|>{\Phi }^{-1}(1-\alpha /2)$, and the observed p-value is

In a regular scalar model under the null, Taylor expansion about the MLE givesA uniform law of large numbers for the observed information makes
$-{\ell }_n^{\prime \prime }(\tilde{\theta })/n$ and the Fisher information used in the Wald statistic converge to the same positive limit. Their ratio therefore converges in probability to one.

For a regular $p$-parameter model, the Wald statistic for a candidate $\theta$ isUnder the true parameter it converges in distribution to ${\chi }_p^2$.

For the recurrencea Fourier mode with $s={\sin }^2(\theta /2)$ has amplification polynomial of a multilevel finite difference schemeFor real $\rho$ and $\mu \ge 0$, both roots have unit modulus for every $s\in [0,1]$ precisely whenIndeed a real quadratic $G^2+aG+1$ has both roots on the unit circle exactly when $|a|\le 2$. Checking the two endpoints of the affine expression $a=4\mu s-2\rho$ gives these inequalities. For example $\rho =0$, $\mu =1/4$ satisfies this root condition, so unit-modulus roots do not force $\rho =1$.

With fixed $\rho$, the unmodified wave equation selects $\rho =1$ through its consistency requirements. At equality $|a|=2$, a repeated root may produce linear growth for arbitrary two-level starting perturbations. Thus this modal root classification alone is not a full uniform power bound for matrix Fourier symbols.

For the centered spatial second difference, the Crank-Nicolson diffusion scheme has amplification factorIt is stable for every $\mu \ge 0$.

For a schemethe Fourier ansatz gives the amplification polynomialIts roots are the amplification factors of that Fourier mode.

For a one-step translation-invariant recurrence, the amplification factor is the multiplier satisfyingThe Parseval identity converts the pointwise bound $|H|\le 1$ into non-growth of the discrete $2$-norm.

A decretion disk is supplied near its inner edge and transports matter outward. Internal stresses redistribute the supplied angular momentum through a nearly orbiting flow. Classical Be stars provide an example: their gaseous circumstellar disks are modeled as viscously spreading, approximately Keplerian disks. A disk carrying an outward torque is not automatically a decretion disk; it must actually have outward mass transport.^[ref]

The viscous torque in an accretion disk is the rate of outward angular momentum transport by an internal shear stress. For an axisymmetric thin disk rotating with angular velocity $\Omega (r)$, a local Newtonian viscous stress has $T_{r\varphi }=\rho \nu rd\Omega /dr$. Integrating that stress over a cylinder and multiplying by the lever arm givesHere $\bar{\nu }$ is the density-weighted mean kinematic viscosity, with units of length squared per time. The vertically integrated quantity is $\bar{\nu }\Sigma$. For decreasing $\Omega$, the sign convention makes $\mathcal{G}>0$, describing outward angular momentum transport. More general internal stresses can be represented by an effective $\bar{\nu }$ through the same integrated torque relation.^[ref]

For an integer $k\ge 2$, a finite simple graph that is $k$-connected graph has a cycle of length at least $\min \{|G|,2k\}$.

A closed vector subspace of a topological vector space is both a vector subspace and a closed set. If a continuous linear map has a Hausdorff target, its kernel of a linear map is a closed vector subspace, since it is the preimage of the closed singleton $\{0\}$. The target condition matters because our topological vector space convention permits non-Hausdorff spaces.

For a counterexample without the target hypothesis, the identity linear map of $\mathbb{R}$ with the indiscrete topology is continuous, but its kernel of a linear map $\{0\}$ is not closed.

The cross product $a\times b$ of two three-dimensional vectors is perpendicular to both, has magnitude $|a||b|\sin \theta$, and is oriented by the right-hand rule.