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For a planar nematic director $\mathbf{n}=(\cos \chi ,\sin \chi )$ and constant amplitude ${\lambda }_0$, the nematic order parameter isDifferentiation gives $\nabla \cdot Q={\lambda }_0(-\sin 2\chi {\chi }_x+\cos 2\chi {\chi }_y,\cos 2\chi {\chi }_x+\sin 2\chi {\chi }_y)$. The coefficient matrix is orthogonal, proving the identity. Hence $K|\nabla \cdot Q{|}^2/2$ is the angle-gradient energy with effective modulus $K{\lambda }_0^2$. The full tensor-gradient norm instead obeys ${\sum }_{ijk}({\partial }_k{Q}_{ij}{)}^2=2{\lambda }_0^2|\nabla \chi {|}^2$ in this normalization; the two local densities must not be confused.

On a bounded connected Lipschitz domain $U$,for every $u\in H^1(U)$. If this failed, the Rellich-Kondrachov compactness theorem would produce a normalized strong $L^2$ limit with zero gradient, hence a constant function; continuity of the Sobolev trace theorem would force that constant to vanish, contradicting its unit norm.

If $U$ is a bounded connected Lipschitz domain and a measurable boundary portion ${\Gamma }_D$ has positive surface measure, thenfor every $v\in H^1(U)$ whose trace vanishes on ${\Gamma }_D$.

Otherwise choose $\Vert v_j{\Vert }_2=1$ with $\Vert \nabla v_j{\Vert }_2\to 0$. The Rellich-Kondrachov compactness theorem for $H^1(U)$ gives a strongly $L^2$ convergent subsequence and a weakly $H^1$ convergent subsequence with the same limit. Its weak gradient is zero, so connectedness makes it constant. The trace-zero closed vector subspace is weakly closed; that constant has zero trace on a set of positive surface measure, so it vanishes, contradicting its unit norm. The whole-boundary $H_0^1$ compactness theorem alone would not apply to these partially constrained functions.

For a function space on a set $X$ and $x\in X$, the point evaluation functional is ${\delta }_x(f)=f(x)$. On $C(K)$ for a nonempty compact space $K$, it is a bounded linear functional for the supremum norm:The constant function one attains the bound. The same argument works for bounded continuous functions on any $X$.

Boundedness depends on the chosen norm. On $C([0,1])$ with the L2 norm, fix $x_0\in (0,1)$ and take $f_n(t)=\max (1-n|t-x_0|,0)$ for sufficiently large $n$. Then $f_n(x_0)=1$ but $\Vert f_n{\Vert }_2^2=2/(3n)$, so point evaluation is unbounded. On L2 space itself, it is not even well-defined on equivalence classes: changing a representative at the single point $x_0$ does not change its Lebesgue almost everywhere class.

On continuous functions on a compact interval, evaluation at an interior point is not continuous in the integral norm. Narrow triangular peaks have fixed point value and integral tending to zero. Subtracting such a peak from the constant function one gives functions vanishing at the point and converging in the integral norm to one. Thus the evaluation kernel is not closed, although it is closed in the uniform norm.

Here is a fully computed endpoint version, now using the L2 norm on the actual continuous functions $C([0,1])$. For each integer $n\ge 1$, set $f_n(x)=\max (1-nx,0)$. Its point evaluation at zero is $E(f_n)=f_n(0)=1$, whileConsequently $f_n\to 0$ in the L2 norm but $E(f_n)\nrightarrow 0$: point evaluation is discontinuous on this normed space. Quantitatively $|E(f_n)|/\Vert f_n{\Vert }_2=\sqrt{3n}$ is unbounded. At $n=100$, the L2 norm is approximately $0.057735$, the integral is $0.005$, and the point value is still $1$. In the supremum norm, $\Vert f_n{\Vert }_{\infty }=1$ and $|E(f)|\le \Vert f{\Vert }_{\infty }$, so $E$ has operator norm one instead.

Set $g_n=1-f_n$. Then $E(g_n)=0$ and $\Vert g_n-1{\Vert }_2=1/\sqrt{3n}\to 0$, although $E(1)=1$. Thus the kernel of $E$ fails to be a closed set. On the full L2 space, a further distinction is essential: its elements are equivalence classes under equality almost everywhere, and changing a representative at zero does not change its class. Arbitrary point evaluation is therefore not even well-defined there. Restricting first to continuous functions makes the discontinuity statement precise.^[ref]

In three dimensions let $f\in L^{6/5}({\mathbb{R}}^3)$ have compact support. Let $v_R\in H_0^1(B_R)$ solve $-\Delta v_R=f$ and extend it by zero. There is a unique $v\in {\dot{H}}^1({\mathbb{R}}^3)$ with $\int \nabla v\cdot \nabla \varphi =\int f\varphi$ for all $\varphi \in {\dot{H}}^1$. Indeed Holder inequality and the homogeneous Sobolev inequality make the right side a continuous linear functional in the gradient norm, and the Riesz representation theorem applies to this Hilbert space.

The weak formulation says that $v_R$ is the orthogonal projection of $v$ onto the closed subspace consisting of zero extensions of $H_0^1(B_R)$. These subspaces increase with $R$ and their union contains every test function, which is dense by the definition of the homogeneous Sobolev space. The best-approximation property of orthogonal projection therefore gives $\Vert \nabla (v_R-v){\Vert }_2\to 0$. The homogeneous Sobolev inequality also gives $\Vert v_R-v{\Vert }_6\to 0$. This convergence does not generally extend to the inhomogeneous Sobolev space $H^1$, as the spherical mean of a Poisson solution outside its source shows for a source of nonzero total mass.

For $m>0$, the three physical polarization vectors are transverse to $k$ and have norm $-1$ in metric $(+---)$. In the rest frame their sum is $\mathrm{diag}(0,1,1,1)$; Lorentz covariance then gives the displayed expression for $k^2=m^2$. This includes the longitudinal mode. It cannot be specialized to $m=0$ by setting its denominator to zero.

For a real or complex polynomial $p(t)={\sum }_{k\ge 0}{a}_k{t}^k$, this finite sum defines a norm: definiteness follows from uniqueness of polynomial coefficients, absolute homogeneity of a norm follows termwise, and the triangle inequality follows by summing $|a_k+b_k|\le |a_k|+|b_k|$. On $[0,1]$, its supremum norm satisfies $\Vert p{\Vert }_{\infty }\le \Vert p{\Vert }_{\mathrm{coef},1}$, since $|t^k|\le 1$. The reverse comparison has no constant independent of the degree of a polynomial: for $p_n(t)=(1-t{)}^n$, the two norms are $1$ and $2^n$ respectively, by the binomial theorem.

For a spherically symmetric bound state, write $\psi =e^{-br}\sum c_j{r}^j$. The radial Coulomb equation gives the displayed recurrence. Choosing $b=a/(2N)$ terminates it after degree $N-1$, producing a nonzero continuous polynomial times an exponentially decaying function. Its three-dimensional norm $4\pi {\int }_0^{\infty }{r}^2|\psi {|}^{2}dr$ is finite and can be normalized. The energy is $-{\hslash }^2{a}^2/(8m{N}^2)$. Continuity at the origin is compatible with the Coulomb cusp; no differentiability as a Cartesian radial function at the origin is asserted.

A real symmetric bilinear form is positive-definite when $B(v,v)>0$ for every nonzero vector. It defines an inner product and the associated norm $\sqrt{B(v,v)}$. Symmetry is part of this definition: positive diagonal values by themselves do not imply symmetry.^[ref]

The identity matrix $A=\mathrm{diag}(1,1)$ gives $B(v,w)=v_1{w}_1+v_2{w}_2$ and $B(v,v)=v_1^2+v_2^2$. For $v=(3,4)$ this is $25$, giving norm $5$. A weighted example, $A=\mathrm{diag}(4,1)$, gives norm $\sqrt{4x^2+y^2}$: the first coordinate costs twice as much in length. Such inner products let geometry assign different weights to different directions.

Every positive-definite bilinear form is nondegenerate: a vector in its radical of a bilinear form has $B(v,v)=0$, so positivity forces $v=0$. The converse fails, as $\mathrm{diag}(1,-1)$ shows. Even positivity of the associated quadratic form does not force a general bilinear form to be symmetric: $A=\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$ has $v^{T}Av=x^2+y^2$ but $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.

The complex Killing form is not positive-definite on the entire complex Cartan subalgebra; its restriction to the Euclidean subspace of a Cartan subalgebra is a real positive-definite form.

For real vectors $x_1,\dots ,x_n$ in a Euclidean space, their Gram matrix $G_{ij}=x_i^T{x}_j$ satisfiesThus $G$ is a positive semidefinite matrix. It is a positive-definite matrix exactly when the vectors are linearly independent: the squared norm vanishes precisely when the indicated linear combination vanishes.

For complex vectors, specify the convention. With the inner product conjugate-linear in its first argument, the same definition $G_{ij}=⟨x_i,x_j⟩$ givesIf the inner product is instead linear in its first argument, this definition of $G$ gives $c^{\ast }Gc=\Vert {\sum }_i\overline{c_i}{x}_i{\Vert }^2$. The positive definiteness criterion is unchanged, but the coefficient formula must respect the convention.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.

For a unit vector $u$ spanning the target eigenspace and a normalized iterate $x_k$, defineThis is the norm of the component in the orthogonal complement of $u$, equivalently the sine of the acute angle to its line. It ignores the arbitrary sign or complex unit factor of an eigenvector. Under power iteration convergence for a diagonalizable matrix, ${\epsilon }_k=O({\gamma }^k)$ in Big O notation. The orthogonal projection defines how we measure direction; it does not assume that the eigenbasis of $A$ is orthogonal. In computation, the projection form avoids subtracting nearly equal numbers inside the square root.

In the contraction convention for numerical evolution, practical stability means non-growth of the chosen norm for every datum and every permitted time level. This is stronger than finite-time mesh-uniform stability allowing a bound $C_T\Vert U^0\Vert$. For a scalar translation-invariant one-step scheme on the infinite grid, a Fourier amplification symbol $G$ with $|G|\le 1$ gives this contraction by the Parseval identity. If $|G|\ge 1+c>1$ on a frequency set of positive measure, choose a nonzero square-integrable transform supported there to obtain $\Vert U^m\Vert \ge (1+c{)}^m\Vert U^0\Vert$. The constant $c$ may depend on the mesh. Centered pure advection with forward time has $c=\sqrt{1+(\Delta t/\Delta x{)}^2/2}-1$ on the set ${\sin }^2\theta \ge 1/2$. The distinction from finite-time stability is quantified by centered-advection refinement-dependent stability.

For unit-modulus prime values, pretentious distance is the Euclidean metric on the finite vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$. On whole arithmetic functions it is only a pseudometric: it sees neither values on prime powers nor primes above the cutoff. Even for every cutoff, the functions one and ${\mu }^2$ have distance zero but differ on nonsquarefree integers. The norm representation proves the triangle inequality and identifies exactly the quotient on which separation holds.

For a bounded linear operator $A$ on a complex Banach space and $ϵ>0$, use the open conventionThus the pseudospectrum detects both noninvertibility and a large resolvent norm. A Neumann series shows that perturbations of operator norm less than $ϵ$ cannot create a new point of the spectrum where the displayed inverse norm is at most ${ϵ}^{-1}$.

For a square complex matrix, the same condition is $s_{\min }(A-zI)<ϵ$, where $s_{\min }$ is the least singular value. For general bounded linear operators, a small lower bound ${\inf }_{\Vert x\Vert =1}\Vert (A-zI)x\Vert$ alone need not characterize the entire spectrum of a bounded operator: the unilateral shift operator is bounded below but is not surjective. The definition retains these spectral points explicitly.

A pure state on a C-star algebra is an extreme point of its state space. For a nonzero unital C-star algebra, the state space is a nonempty compact convex set in the weak-star topology, so the Krein-Milman theorem guarantees pure states.

Every Positive element of a C-star algebra $a$ attains its norm at a pure state. Indeed, states satisfying $f(a)=\Vert a\Vert$ form a nonempty compact face of a convex set: the maximum property forces each term of any nontrivial convex decomposition to attain the same maximum. An extreme point of this face is an extreme point of the whole state space. The argument uses the unital compactness statement, rather than asserting compactness for a general nonunital algebra.

If vectors $u,v$ in a real or complex inner product space are orthogonal, thenIndeed, expanding the inner product gives $\Vert u+v{\Vert }^2=\Vert u{\Vert }^2+\Vert v{\Vert }^2+2\Re ⟨u,v⟩$, and orthogonality removes the last term. Repeating this argument yields the squared-norm identity for any finite orthogonal set.

Over the real numbers, the equality also implies orthogonality. Over the complex numbers it only implies $\Re ⟨u,v⟩=0$. For example, in $\mathbb{C}$ the vectors $1$ and $i$ satisfy $|1+i{|}^2=|1{|}^2+|i{|}^2=2$, but their inner product is nonzero. Thus the converse requires the real-space hypothesis.

A reduced QR decomposition of an $m\times k$ matrix over the real numbers with full column rank, $m\ge k$, writes $A=QR$, where $Q$ is $m\times k$ with orthonormal columns and $R$ is a $k\times k$ upper triangular matrix with positive diagonal. Applying the Gram-Schmidt process to the columns $v_j$ of $A$ constructs $Q=(q_1\cdots q_k)$ and records the coefficientsThis last column identity is exactly the matrix multiplication $A=QR$. The upper triangular matrix records that the $j$th input uses only the first $j$ new directions, while its positive diagonal records the norms of the nonzero orthogonal projection residuals. Since $Q^{T}Q=I$, one also has $R=Q^{T}A$; $Q^T$ denotes the transpose and $I$ the identity matrix. Over the complex numbers, use the conjugate transpose $Q^{\ast }$ in these identities.

For the three-dimensional Gram-Schmidt process example, the exact matrices areFor example, the last column reconstructs asThe diagonal entries are approximately $1.414214,1.224745,1.154701$. Each is a norm of the new direction before it was scaled to length one.

For a linear least-squares problem, the orthogonal projection of $b$ onto the column space of $A$ is $Q{Q}^{T}b$. Hence the minimizing coefficients solve the linear system $Rx=Q^{T}b$ by back substitution. Let $A_2=(v_1{v}_2)$ consist of the first two columns of the example; its reduced factors are $Q_2=(q_1{q}_2)$ and the leading $2\times 2$ block $R_2$ of $R$. For $b=(0,0,1{)}^T$, the linear system $R_{2}x=Q_2^{T}b$ isThus $A_{2}x=(1/3,-1/3,2/3{)}^T$, agreeing with the orthogonal projection computed above, and the minimum Euclidean norm of $A_{2}x-b$ is $1/\sqrt{3}$. The QR decomposition separates building an orthonormal basis from recovering coefficients in the original basis.^[ref]