Codex @codex  0

Write $s={\mu }_x+{\mu }_y$. The update can be rearranged asWhen $s\le 1$, the right side is a convex combination of old values: its coefficients are nonnegative and sum to one. Assume all old values lie in $[U_{\min },U_{\max }]$. A new maximum $v>U_{\max }$ is interior because boundary values are zero and $U_{\max }\ge 0$. At its node the left side is at least $(1+s)v-sv=v$, while the right side is at most $U_{\max }$, a contradiction. A new minimum $v<U_{\min }$ similarly makes the left side at most $v$ and the right side at least $U_{\min }$, again a contradiction.

The initial values satisfy the bounds by construction; mathematical induction provesThe new-time matrix is a strictly diagonally dominant matrix with positive diagonal and nonpositive off-diagonal entries, so the linear update is uniquely solvable. This Crank-Nicolson maximum principle under a step restriction concerns preservation of value bounds, which is a stronger requirement than scalar A-stability.

Using the semidiscrete Fourier transform from (b), the explicit scheme has amplification factorSet $r=a\mu$ and $s={\sin }^2(\theta /2)\in [0,1]$, so ${\sin }^2\theta =4s(1-s)$. Direct expansion givesThe expression in square brackets is an affine function of $s$. At the two endpoints its values areusing precisely ${\nu }^2\le 2r\le 1$. Every intermediate value is a convex combination of these endpoints, so $|G_E{|}^2\le 1$ for every frequency. The Parseval identity provesThus the scheme is practically stable under the given condition. For $a>0$ it is equivalent to $\Delta t\le 2a$ and $\Delta t\le (\Delta x{)}^2/(2a)$. This is the Forward Euler stability for centered advection-diffusion calculation with the diffusion coefficient retained. The proof does not require all three stencil weights to be nonnegative.

The mean of a shape-rate gamma distribution is shape divided by rate. Hence the posterior mean isBoth weights are positive and sum to one. This is a convex combination of the prior mean and the sample mean, provingFor $n\ge 1$ it lies strictly between them if the two means differ, and equals both if they agree. The rate $b$ acts as the prior weight relative to the $n$ observed exposure units.

The prior mean, posterior mean, and maximum-likelihood estimate areFor $0<S<N$, the last follows by maximizing $S\log \theta +(N-S)\log (1-\theta )$. At $S=0$ or $S=N$, use the usual binomial likelihood convention that admits the endpoints $0$ and $1$ as maximum-likelihood estimates; if the parameter domain is strictly $(0,1)$, these are boundary suprema rather than attained maxima. The posterior mean remains interior in all cases.

TakinggivesSince $N=nm>0$ and $\beta +1>0$, $0<\lambda <1$. Thus the posterior mean is a convex combination of the prior mean and observed binomial proportion, with larger prior weight when $\beta +1$ is large relative to $N$.

The posterior mean is a convex combination of the prior mean and the usual Poisson maximum-likelihood estimate:Both weights are positive and sum to one, so the mean lies between $r/\lambda$ and $\bar{x}$, strictly between them unless they coincide. If all counts are zero, $\bar{x}=0$ is the boundary estimate in the usual nonnegative Poisson parameter space; with a strictly positive parameter restriction it represents the likelihood supremum at that boundary. The displayed posterior-mean relation is valid in either interpretation.

For five points, partition the unit square into four squares of side $1/2$. Assign boundary points to just one neighbouring piece. By the pigeonhole principle, two of the points lie in one piece, whose diameter is its diagonal, $\sqrt{(1/2{)}^2+(1/2{)}^2}=\sqrt{2}/2$.

For eight points, use seven smaller-diameter pieces rather than eight rectangles. Start with the convex pentagonTake its four reflected copies at the corners of the square. The remaining pieces are the left triangle with vertices $(0,1/4),(1/4,1/2),(0,3/4)$, its right-hand reflection, and the central diamond with vertices $(1/4,1/2),(1/2,1/4),(3/4,1/2),(1/2,3/4)$. These seven polygons cover the square with disjoint interiors. To verify coverage directly, for $0\le y\le 1/4$ the two bottom pentagons split the horizontal section at $x=1/2$. For $1/4\le y\le 1/2$, the five successive sections have endpoints $0,y-1/4,3/4-y,y+1/4,5/4-y,1$ and belong respectively to the left triangle, bottom-left pentagon, diamond, bottom-right pentagon and right triangle. Reflection gives the top half.

The diameter of a finite convex hull is the greatest distance between its vertices: if two points are convex combinations of vertices, their difference is a convex combination of vertex differences, and the triangle inequality bounds its norm by the largest such distance. Checking the five vertices of $P$ gives diameter $\sqrt{(1/2{)}^2+(1/4{)}^2}=\sqrt{5}/4$; reflections preserve that value. Each remaining triangle and the diamond has diameter $1/2$, which is smaller. After assigning shared boundaries to one piece, eight points in these seven pieces force a pair in a common piece. Hence some pair is at distance at most $\sqrt{5}/4$.

Both arguments are instances of a pigeonhole distance bound from a partition.

Consider the advection equation $u_t+a{u}_x=0$ with constant speed $a>0$, no sources, and periodic boundary conditions on $[0,1]$. Use $N$ equal cells of width $h=1/N$, time step $k>0$, and indices modulo $N$. The exact solution is the periodic translation $u(x,t)=u_0(x-at)$. At a face, the Riemann problem transports the left state to the right, so the two-state upwind numerical flux is $F(L,R)=aL$, with $F(U,U)=aU$.^[ref]

The finite volume method with this numerical flux and an explicit Euler method step givesHere $U_j^n$ represents a cell average; the recurrence also occurs in the upwind finite difference scheme. Impose the specific CFL condition $0<\nu \le 1$. Each updated cell average is then a convex combination of two old ones. This proves preservation of their initial minimum and maximum, including nonnegativity for nonnegative data, and contraction of the maximum norm of the difference of two solutions on this periodic grid. These statements concern this constant-speed scalar update with this time step.

For an original synthetic example, take $N=8$, $a=1$, $h=1/8$, $k=1/16$, hence Courant number $\nu =1/2$. Choose a function constant on each successive cell, with the following values, so these are exact initial cell averages. One step givesFor example, cell zero receives from cell seven across the periodic face: $U_0^1=(0+1)/2=1/2$. All eight face transfers cancel in the total by shared face flux cancellation. More generally this periodic recurrence conserves the total algebraically for any $\nu$, even though the CFL condition is needed for the convex combination argument. With the same initial data and $\nu =3/2$, one step gives $U_2^1=-1$ and $U_4^1=5/2$, despite preserving the total.

After four steps, $t=1/4$, the exact translated profile has moved two cells to the right. Repeated averaging instead gives the explicit binomial coefficient formulawith all indices modulo eight. The diagram compares these computed cell averages with the exact translated cell averages. Their integrals agree, but the numerical jumps spread across cells. The finite volume method is useful for jump discontinuities because it retains the integral balance even at a jump; conservation alone does not guarantee a sharp profile. At $\nu =1$ this particular update is an exact one-cell shift, whereas for $0<\nu <1$ it mixes neighboring states. No accuracy conclusion at a jump follows just from the conserved total.

The function taking the greater value at every common-domain point. For finitely many affine functions it is a convex function, by bounding each affine branch by the convex combination of the endpoint maximum values. This is the finite-affine case of pointwise maximum of convex functions.

The function taking the lesser value at every common-domain point. It forms the lower envelope of two graphs. For finitely many affine functions it is a concave function: each affine branch is at least the corresponding convex combination of the two endpoint minimum values, so taking the minimum preserves that lower bound. A maximizing mixed strategy in a two-player zero-sum game maximizes such a lower envelope of opposing pure-choice expected payoffs.

For conditionally independent Poisson observations of mean $\theta$ and a gamma distribution prior of shape $r>0$ and rate $\lambda >0$, the likelihood function contributes ${\theta }^{{\sum }_i{x}_i}{e}^{-n\theta }$. Multiplying by the prior density gives a posterior distribution with shape $r+{\sum }_i{x}_i$ and rate $\lambda +n$. The single-observation update is shape $r+X$ and rate $\lambda +1$.

The posterior mean is the convex combinationThus the prior mean $r/\lambda$ and the sample maximum-likelihood estimate $\bar{x}$ bracket it. Integrating the next conditional Poisson distribution against this posterior distribution gives the Poisson-gamma mixture, a negative binomial distribution; its spread includes uncertainty in $\theta$ as well as the conditional count variation.

A nonconvex primal problem can have a positive duality gap even when both optima are attained. Choose the finite set $X=\{-2,0,2\}$, the objective function values $f(-2)=0$, $f(0)=3$, $f(2)=0$, and the equality optimization constraint $h(x)=x=0$. The only feasible point is $0$, so $p_{\ast }=3$.

With unrestricted equality Lagrange multiplier $\nu$, the three possible optimization Lagrangian values and their minimum areThe concave Lagrange dual function has unique maximizer ${\nu }_{\ast }=0$, with $d_{\ast }=0$. Thus the duality gap is $3$. At the optimal Lagrange multiplier, both inner minimizers $-2$ and $2$ violate the equality; the dual optimum does not recover a primal optimizer.

Why can no slope close the gap? Any affine function $a+bx$ lying below the two endpoint costs obeys $a-2b\le 0$ and $a+2b\le 0$. Adding gives $a\le 0$, so its value at residual zero cannot reach the feasible cost $3$. A convex combination of the endpoint cost-residual pairs with equal weights has residual zero and average cost zero. This averaged pair is unattainable: the actual decision $x=0$ costs $3$. This geometry explains precisely what the dual bounds fail to distinguish. Nonconvexity does not force a positive duality gap: changing only $f(0)$ to $0$ makes both optimal values zero on the same nonconvex domain.

Every set of at least $d+2$ points in ${\mathbb{R}}^d$ can be partitioned into two nonempty parts with intersecting convex hulls. To see this, take a nonzero affine dependence on $d+2$ of its points. Its positive and negative coefficient sums have the same magnitude $L>0$. Dividing by $L$ expresses one point as a convex combination of the positive-coefficient points and also as a convex combination of the negative-coefficient points. Distribute zero-coefficient and remaining points arbitrarily between the parts. This theorem obstructs arbitrary labels for affine linear classifiers.

The reachable set ${\mathcal{R}}_T(x_0)$ collects all endpoints attainable from $x_0$ at exactly $T$ by admissible control inputs. If arrival any time up to $T$ is allowed, use ${\mathcal{R}}_{\le T}(x_0)={\bigcup }_{0\le t\le T}{\mathcal{R}}_t(x_0)$, with ${\mathcal{R}}_0(x_0)=\{x_0\}$. These definitions differ: a general dynamical system may drift away after early arrival, so waiting need not preserve the target. The reachable set is always relative to a declared input class and any constraints on the entire path.

For the double integrator starting from $(0,0)$, now require measurable functions as control inputs with $|u(t)|\le 1$ almost everywhere. Write $(p,v)$ for its endpoint at $T>0$. Two integrals determine the endpoint:Our mathematical example has the exact reachable setTo see the upper boundary, fix $v$: put $u=1$ first and $u=-1$ last, switching at $s=(T+v)/2$. This maximizes the weighted integral for position because early commands have the largest weight $T-t$; the required net velocity fixes $s$. Formally, $(T-t)-(T-s)=s-t$ has the same sign as this maximizing command, so changing to any other admissible command with the same integral cannot increase $p$. Reversing the order gives the lower boundary. Pointwise convex combinations of these two commands preserve $v$ and attain every intermediate $p$, so the formula describes the whole region, not merely a bound. The extreme commands are bang-bang control.

In particular, the bounded double integrator can move from rest to position $d$ and rest again exactly when $|d|\le T^2/4$. Its minimum duration is therefore $2\sqrt{|d|}$ in these normalized units. With freely chosen time every rest-to-rest distance is attainable; with a fixed deadline many are excluded. Changing the amplitude bound to $|u|\le a$ with $a>0$ scales this condition to $|d|\le a{T}^2/4$. This shows why specifying the controls matters even when the Kalman rank criterion gives full rank of a matrix.

In closed quantum control, a density operator reachable set lies inside its unitary orbit of a density operator, since coherent evolution preserves the spectrum. That orbit need not be fully reachable under the available control inputs.

Suppose one row of a finite discrete memoryless channel is a convex combination of other rows: $W(\cdot \mid x_0)={\sum }_j{\alpha }_{j}W(\cdot \mid x_j)$. Deleting input $x_0$ leaves the channel capacity unchanged. Given any input law with mass $t$ on $x_0$, move that mass to the other symbols in proportions ${\alpha }_j$. The output law stays the same. By the concavity of information entropy, $H(W(\cdot \mid x_0))\ge {\sum }_j{\alpha }_{j}H(W(\cdot \mid x_j))$, so this replacement cannot increase conditional entropy $H(Y\mid X)$ and cannot decrease mutual information. Taking maxima proves the claim; the smaller channel is also trivially a restriction of the original one. If the entropy inequality is strict, every positive use of the redundant row is strictly suboptimal compared with its redistribution.

Let $K$ be a nonempty compact convex set in a Banach space and $T:K\to K$ a continuous map. For every $\epsilon >0$, there is a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ that approximates $T$ uniformly to within $\epsilon$ in norm.

Choose $y_1,\dots ,y_N\in T(K)$ whose open radius-$\epsilon$ balls cover the compact image $T(K)$. Set $w_i(x)=\max \{0,\epsilon -\Vert Tx-y_i\Vert \}$ andThe denominator is positive. This convex combination belongs to $K\cap \mathrm{span}\{y_1,\dots ,y_N\}$ and depends continuously on $x$. Every contributing $y_i$ has distance less than $\epsilon$ from $Tx$, so the triangle inequality gives $\Vert T_{\epsilon }(x)-Tx\Vert <\epsilon$.

A bipartite quantum state is separable when its density operator is a convex combination ${\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$ of product states; otherwise it has entanglement. Operationally, Alice and Bob can prepare a separable quantum state by sharing a classical random label $j$ and each preparing the corresponding local state.

Separable does not mean uncorrelated. The mixture in Bell correlations versus a classical mixture is separable with two terms of weight $1/2$, yet the parties' $Z$ records agree perfectly. A single product state would give independent local outcomes. Allowing a mixture of products captures shared classical randomness; excluding this possibility would incorrectly classify ordinary correlations as entanglement.

A subset $K$ of a normed space is sequentially weakly closed when every weakly convergent sequence in $K$ has its limit in $K$. Every weakly closed set has this property. For a convex set in a Banach space, it is equivalent to closure in the norm topology: Mazur lemma produces norm convergent convex combinations from any weakly convergent sequence, and conversely norm convergence implies weak convergence. For a Hilbert space, the weak Banach–Saks theorem in a Hilbert space supplies these convex combinations as subsequence averages. No equivalence with weak closedness is asserted here for arbitrary nonconvex sets.

Write $q=P(A)$, $r=P(B\mid A)$ and $s=P(B\mid A^c)$. The law of total probability gives the displayed convex combination. With $q,r$ fixed and $s$ free in $[0,1]$,With $r,s$ fixed and $q$ free in $[0,1]$,The joint cell masses $qr,q(1-r),(1-q)s,(1-q)(1-s)$ are nonnegative and sum to one, proving sharpness in the joint-model interpretation. For ordinary conditional probabilities, a conditioning event must have positive mass. If both $r$ and $s$ are specified by ratios, then $0<q<1$: for $r\ne s$ the last bounds are strict and their endpoints are the infimum and supremum. If $r=s$, the marginal equals that common value for every $q$. Endpoint mixtures remain meaningful if the unused zero-mass conditional is treated as a prescribed conditional model rather than a defined ratio.

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The reciprocal basis gives the coordinates of any position vector asThe point lies in the closed tetrahedron with vertices $0,\mathbf{a},\mathbf{b},\mathbf{c}$ exactly when $\alpha ,\beta ,\gamma \ge 0$ and $\alpha +\beta +\gamma \le 1$. Indeed, these conditions express it as a convex combination of the four vertices, with fourth weight $1-\alpha -\beta -\gamma$. They also describe the intersection of the four inward half-spaces bounded by its faces. Strict inequalities characterize its interior.

The weighted arithmetic mean combines values according to how much each contributes to the quantity being summarized. For a finite list of real numbers $x_i$ and nonnegative weights $w_i$ with $W={\sum }_i{w}_i>0$, defineZero weights ignore an entry; at least one weight must be positive. Multiplying every weight by the same positive number leaves the result unchanged. Writing $p_i=w_i/W$ makes the weighted mean a convex combination, since $p_i\ge 0$ and ${\sum }_i{p}_i=1$. It therefore lies between the smallest and largest input values. ^[ref]

For example, give the values $90$ and $60$ weights $2$ and $8$:The larger weight pulls the result closer to $60$. The balancing identityexplains a geometric interpretation: if the weights were masses placed at positions $x_i$, the weighted mean would be their center of mass. Here $8(66-60)=2(90-66)=48$, so the two sides balance even though their distances from the mean differ.

Weights need an interpretation. They can represent numbers of observations, masses, or probabilities in a finite probability distribution. These meanings lead to the same formula, while answering different questions. For instance, combining means of groups uses group sizes as weights to recover the mean over individual observations. For a function $f$ and nonnegative weight function $w$, the analogous integral is $\int wf/\int w$, provided $wf$ is integrable and the total weight $\int w$ is positive and finite.