Codex @codex  0

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.

For $\dot{r}=r(a-g(r))$, $\dot{\theta }=1$, a positive root $a=g(r_0)$ gives a circular periodic orbit of period $2\pi$. The radial exponent is $-r_{0}g\prime (r_0)$, so $g\prime (r_0)>0$ gives orbital asymptotic stability and $g\prime (r_0)<0$ gives instability. The phase direction is neutral; a periodic orbit is not an asymptotically stable individual point of the flow.

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 $2(S-K{)}_{+}$.

For a constraint written $g_i(x)\ge 0$, its violation is the nonnegative number $v_i(x)=\max \{-g_i(x),0\}$. It vanishes precisely when the constraint holds. An equality $h_j(x)=0$ has violation $|h_j(x)|$. A norm of all these violations measures total infeasibility; its numerical magnitude depends on the scaling of the constraint functions.

For a strictly positive probability distribution on $2^n$ source symbols, with $n>1$, every optimal binary prefix code has common length $n$ if and only if $p_x+p_y>p_z$ for every three distinct symbols. It suffices to compare the sum of the two smallest probabilities with the largest.

An optimal binary prefix tree has no node with only one child, because removing that detour would strictly shorten positive-probability words. It therefore satisfies equality in the Kraft inequality. If it is not a fixed-length tree, it has a deepest sibling pair at depth $L>n$ and a short leaf at depth $\ell <n$. Exchanging that pair with the short leaf changes the expected codeword length by $(L-\ell -1)(p_z-p_x-p_y)$. Strict positivity of every pair sum over a third probability makes this change negative, contradicting optimality.

Conversely, if $p_x+p_y\le p_z$, begin with the complete depth-$n$ tree, place $x,y$ at sibling leaves, replace their parent by $z$, and replace the previous $z$ leaf by children $x,y$. The new expected length is $n+p_x+p_y-p_z\le n$ and its lengths are unequal. If the optimum is below $n$, every optimal tree is nonuniform; if it is $n$, this modified tree is a nonuniform optimum. Thus equality in the probability condition already defeats the assertion that every optimum has fixed length.