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Given a partially ordered set $B$ and a set $A$, the pointwise order on the function space of functions $A\to B$ compares their values at every argument. It is a partial order: reflexivity and transitivity hold at every argument, and antisymmetry follows because two functions equal at every argument are equal. Even if $B$ has a total order, this function space need not: the two functions on $\{0,1\}$ with values $(0,1)$ and $(1,0)$ are incomparable when $B=\mathbb{R}$.

An irreflexive transitive comparison whose induced incomparability relation is transitive. Incomparability is then an equivalence relation, and its classes are totally ordered by the comparison. A comparison-based sorting algorithm can treat incomparable elements as ties; a merge algorithm that takes from the left input on ties preserves stable sorting. Ordinary ordering on real numbers is an example; machine NaN values require separate handling.

A subset $U$ of a partially ordered set is an upper set when $x\in U$ and $x\le y$ imply $y\in U$. The upward closure $\uparrow A=\{y:\exists a\in A,a\le y\}$ is the smallest upper set containing $A$. Both union and intersection preserve this property.

The partial order on nonempty closed integer intervals given by subset inclusion. The least upper bound of two such intervals is their interval hull $[\min (l,l^{\prime }),\max (u,u^{\prime })]$. The hull can contain points absent from both original intervals, so it need not equal their union.

In a partially ordered set, $x\ll y$ means that every nonempty directed set $D$ whose least upper bound exists and satisfies $y\le \sup D$ contains some $d$ with $x\le d$. Taking $D=\{y\}$ shows that $x\ll y$ implies $x\le y$; the converse need not hold. Some formulations quantify over the empty directed set as well; this changes the behaviour at the least element and must be specified.

A partially ordered set in which, for every $x$, the elements way-below $x$ form a directed set with least upper bound $x$. Directed least upper bounds other than these approximating ones need not exist.

Every subset having an upper bound has a least upper bound. If the empty subset is included, a nonempty such partially ordered set has a least element. This condition differs from a directed-complete partial order: a directed set need not have an upper bound.

A weakly linear partial function on the real numbers has a domain closed under rational scalar multiplication and addition, and it preserves addition on that domain. A nonempty domain is a vector space over the rational numbers, and additivity implies rational homogeneity of an additive function by integer multiplication and division. The empty function also satisfies the defining conditions vacuously. This terminology does not assert homogeneity under arbitrary real scalars.

The union of a chain of partial-function graphs is a partial function: two values at the same input occur together in a larger chain member and therefore agree. Any closure or algebraic identity involving finitely many inputs passes to the union if it holds in every chain member, because one member contains all those inputs.

A complex normed vector space has an inducing inner product exactly when its norm satisfies the parallelogram law. Define $B(u,v)=[\Vert u+v{\Vert }^2-\Vert u-v{\Vert }^2]/4$. Subtract two parallelogram identities to obtain $B(u+w,v)+B(u-w,v)=2B(u,v)$. Taking $w=u$ gives doubling, and then replacing $u,w$ by the half-sum and half-difference gives additivity. Integer scaling and division give rational homogeneity; continuity gives real homogeneity. Thus $B$ is a symmetric real-bilinear form. The complex formula $P(u,v)=B(u,v)+iB(u,iv)$ satisfies $P(iu,v)=iP(u,v)$, because $B(iu,v)=-B(u,iv)$ and $B(iu,iv)=B(u,v)$. Symmetry gives conjugate symmetry of $P$, and $P(u,u)=\Vert u{\Vert }^2$ gives positive definiteness. The reverse implication follows by expanding squared norms. Completeness is needed only for the resulting inner product space to be a Hilbert space.

An outer measure is regular in this sense when every set admits a measurable hull of equal outer measure. This is different from merely specifying inner or outer approximation of already measurable sets by compact or open sets. Lebesgue outer measure has a hull that is a Borel set: for finite outer measure choose open covers of total volumes within $1/j$ of its value and intersect their unions. Monotonicity gives equality of the hull's outer measure with that of the original set. For infinite outer measure the whole Euclidean space is a hull. This regularity is useful for turning numerical equality of inner and outer sizes into measurability.

An outer measure on a metric space is metric when positively separated sets satisfy the displayed equality. Every Borel set is then measurable under Caratheodory measurability criterion. Lebesgue outer measure is metric: refine every box of a covering into finitely many boxes of diameter smaller than the separation. Each refined box meets at most one of the two sets, splitting the cover without changing its total volume. Taking infima proves the additive lower bound; outer subadditivity gives equality.

A real $m\times k$ matrix whose columns are an orthonormal set, equivalently $Q^{T}Q=I_k$. In the complex case use the conjugate transpose instead. This requires $k\le m$; only in the square case does the transpose also provide a two-sided inverse, making $Q$ an orthogonal matrix.

An orthonormal basis of the polynomial space ${\Pi }_n$ need not be indexed by polynomial degree. Permuting the normalized Legendre polynomials, for example, preserves orthonormality but changes the degrees attached to the labels. Results about the number of polynomial roots of $p_k$ require the extra condition $\mathrm{deg}{p}_k=k$, or that the polynomials arise by applying the Gram-Schmidt process to $1,x,x^2,\dots$ in that order. Under this degree ordering, $p_k$ is orthogonal to every polynomial of lower polynomial degree because those earlier basis members span ${\Pi }_{k-1}$.

For a positive integration weight on a nondegenerate interval, this implies that $p_k$ has $k$ distinct polynomial roots in the open interval. If fewer than $k$ roots change sign there, multiply $p_k$ by the product of their linear factors. The product has constant nonzero sign except at its finitely many zeros, so its weighted integral cannot vanish, contradicting orthogonality to this lower-degree polynomial. The usual Gaussian quadrature construction therefore uses the degree-ordered family; arbitrary relabelling is not enough.

A finite collection of nonzero pairwise orthogonal vectors is linearly independent. Given ${\sum }_i{a}_i{q}_i=0$, take the inner product with $q_j$ to obtain $a_j\Vert q_j{\Vert }^2=0$. Positive definiteness and $q_j\ne 0$ force $a_j=0$ for every index. The nonzero condition matters: adjoining a zero vector preserves pairwise orthogonality but destroys linear independence.

In the weighted inner product $⟨f,g⟩={\int }_0^{1}xf(x)g(x)dx$, the best approximation to $x^2$ in the span of $1,x$ is the displayed polynomial. The normal equations are $1/4-A/2-B/3=0$ and $1/5-A/3-B/4=0$. Their solution gives $A=-3/10$, $B=6/5$. The residual is orthogonal to that span, so the Pythagorean theorem proves minimality and uniqueness. Its squared norm is $1/600$. A weight vanishing at a single endpoint still gives a positive-definite inner product on continuous functions: a nonzero continuous function remains nonzero on an interval where the weight is positive.

For a bilinear form $B$ on a finite-dimensional vector space $V$, let $\Phi :V\to V^{\ast }$ send $u$ to $B(u,\cdot )$. Under the bidual evaluation map, $U^{\perp }$ is the annihilator of a vector subspace $\mathrm{im}(\Phi {|}_U)$, so $\dim U^{\perp }=\dim V-\mathrm{rank}(\Phi {|}_U)$. The rank-nullity theorem on $U$ now gives the displayed equality. In particular the sum of dimensions is at least $\dim V$. If $U\cap U^{\perp }=0$, the dimension formula for a sum of subspaces forces $V=U\oplus U^{\perp }$. The exact formula records the extra dimensions contributed by the left radical $\ker \Phi$, and does not assume an inner product.

Forsuppose $r_0>0$ satisfies $g(r_0)=a$. Then $r=r_0$ is a circular periodic orbit of period $2\pi$. Linearizing its radial equation givesso its transverse Floquet multiplier is $\exp [-2\pi r_0{g}^{\prime }(r_0)]$. It attracts radially if $g^{\prime }(r_0)>0$ and repels if $g^{\prime }(r_0)<0$; a vanishing derivative requires nonlinear terms.

For example, $g(r)=r^2$ and $a>0$ give $r_0=\sqrt{a}$, radial exponent $-2a$, and an attracting limit cycle. The angular direction is neutral because shifting the initial angle changes the phase of the same orbit. Attraction is therefore orbital stability with convergence to the orbit, rather than asymptotic convergence to one moving point. The argument uses only differentiability of $g$ and does not assume that it is a polynomial.

A periodic orbit has orbital asymptotic stability if nearby trajectories remain close and their distance to the orbit tends to zero. A persistent phase shift along the orbit is allowed.

For contingent claim payoff $(S_T-K{)}_{+}^2$ in the non-dividend Black-Scholes model, differentiating the discounted expected value gives $\Delta =2e^{-r\tau }{S}^{-1}\mathbb{E}[(S_T^2-K{S}_T)1_{\{S_T\ge K\}}]$. Completing the square in the Gaussian exponent computes each truncated lognormal moment and yields the displayed formula, where $d_1=(\log (S/K)+(r+{\sigma }^2/2)\tau )/(\sigma \sqrt{\tau })$. Its terminal limit is $2(S-K{)}_{+}$.