The softplus function $s(x)=\log (1+e^x)$ is convex because

For a finite convex function $f:\mathbb{R}\to \mathbb{R}$, its negative part $f^{-}=\max (-f,0)$ is globally Lipschitz continuous. The negative sublevel set is an interval. If bounded, local Lipschitz continuity of finite convex functions bounds the slopes on its closure. If it is a left ray, any negative slope there would force $f$ positive far to the left by the supporting-line inequality; all slopes on that ray are therefore nonnegative and bounded above near its finite endpoint. For a right ray the analogous slopes are nonpositive and bounded below. If $f$ is negative everywhere, convexity and boundedness above force it to be constant. These cases bound the slopes of $f^{-}$ globally, including its zero extension outside the interval. In particular, integrability of $f(Z)$ implies integrability of the negative part of every fixed translate $f(Z+x)$.

Let a real convex function satisfy $|f|\le M$ on the open unit ball of a normed space. For $4r=1-\Vert x\Vert$, points $u,v\in B(x,r)$ satisfy $|f(u)-f(v)|\le (M/r)\Vert u-v\Vert$. Extend the segment from $u$ through $v$ to length $2r$; its endpoint still lies in the domain, and convexity bounds the difference by $2M\Vert u-v\Vert /(2r)$. Reverse $u,v$ for the opposite inequality. Unlike finite-dimensional local continuity theorems, this argument applies in arbitrary dimension because boundedness on a full neighborhood is assumed.

A convex quadratic function on ${\mathbb{R}}^n$ has the form $q(x)=\frac{1}{2}{x}^{T}Hx+b^{T}x+c$ with a symmetric positive semidefinite matrix $H$. Its Hessian is $H$, and $q(y)-q(x)-\nabla q(x)\cdot (y-x)=\frac{1}{2}(y-x{)}^{T}H(y-x)\ge 0$, proving convexity directly. A stationary point is a global minimum. On a compact polytope, a minimum either is stationary in the interior or minimizes a restriction to a boundary face.

An extended-real convex function $\Lambda :{\mathbb{R}}^d\to \mathbb{R}\cup \{+\infty \}$ is essentially smooth if its effective domain has nonempty interior, it is differentiable throughout that interior, and $\Vert \nabla \Lambda ({\theta }_n)\Vert \to \infty$ whenever interior points converge to a finite boundary point of the domain. A differentiable convex function finite on all of ${\mathbb{R}}^d$ satisfies the boundary condition vacuously. The cumulant-generating function $\log (r/(r-\theta ))$ on $\theta <r$ has derivative $(r-\theta {)}^{-1}$ and is essentially smooth. Restricting its domain to $\theta <\lambda <r$ destroys this property. This distinction matters in the lower bounds of the Gärtner–Ellis theorem.

For a finite real convex function on an entire real vector space, apply the convex domination form of the Hahn-Banach theorem to $\varphi (y+z)-\varphi (y)$ and the zero functional on $\{0\}$. It supplies a linear functional ${\ell }_y$ dominated by this translated function. The affine function $a_y(x)=\varphi (y)+{\ell }_y(x-y)$ lies below $\varphi$ everywhere and equals it at $y$. Taking the supremum proves the formula. These are algebraic affine minorants; continuity requires additional topological hypotheses.

If the amounts in the budget example must be integers, the actual feasible region is $C\cap {\mathbb{Z}}^2$. Both $(0,0)$ and $(1,0)$ are feasible, but their midpoint $(1/2,0)$ is not. Thus the continuous feasible region is a convex set while the integer-restricted choices are not. Averaging two manufacturing plans may specify half an indivisible item. Integer programming must respect that restriction; using the surrounding convex set is a relaxation, not a proof that its averages are admissible integer plans.

The lineality space of a convex cone is its largest linear subspace, equal to $C\cap (-C)$. Every nonempty face contains it. A cone is pointed exactly when this space is zero.

// Target: mathematical-optimization.bigb

Every nonzero vector of $K$ is a nonnegative combination of at most as many support-minimal vectors as it has nonzero coordinates. Choose a nonnegative kernel vector $h$ of minimal nonempty support inside the support of $u$; subtract $\lambda h$, where $\lambda ={\min }_{h_i>0}{u}_i/h_i$. The remainder is nonnegative, remains in the kernel, and has strictly smaller support. Induction proves the decomposition. On a minimal support, the kernel is one-dimensional: an independent kernel direction could perturb a strictly positive vector to the boundary of that support without making it zero, contradicting minimality. For an integer matrix with entries bounded by $M$, cofactor determinants give an integer generator with coordinates bounded by $q!M^q$, where $q$ is the ambient number of coordinates.

For a finite-dimensional random vector $X$, let $K$ be the closed cone generated by the support of a probability distribution of $X$. Every integrable strictly positively weighted barycentre lies in the relative interior of $K$: every nonzero supporting functional on its span is positive on a set of positive probability. Conversely, put $w=(1+\Vert X\Vert {)}^{-1}$ and $D=\{\mathbb{E}[wfX]:0\le f\in L^{\infty }\}$. Its dual cone equals that of $K$, so $\overline{D}=K$ and $\mathrm{ri}K\subseteq D$. For $z\in \mathrm{ri}K$, subtract a sufficiently small positive multiple of $\mathbb{E}[wX]$ and represent the remainder using $D$; adding the constant positive weight back gives $\nu =w(f+\epsilon )>0$.

A closed coordinate-sign region of Euclidean space. The $2^n$ orthants cover the space. An orthant contains no nontrivial affine line: for a nonzero direction, some coordinate eventually violates its sign restriction in one direction. Adding orthant inequalities to a nonempty linear polyhedron therefore permits construction of a basic feasible solution even when the original polyhedron has no vertex.

Every allocation $(x,y)$ in the convex feasible region for a resource budget is a convex combination of its three extreme points:The condition $x+2y\le 6$ is exactly ${\lambda }_1\ge 0$; nonnegative $x,y$ give the other two nonnegative coefficients. Thus the half-space description and the convex hull description encode the same triangle. At $(2,1)$ all three coefficients are $1/3$, and at $(3,9/8)$ they are $1/8$, $1/2$, $3/8$. The green and orange points in the budget diagram therefore have two complementary interpretations: direct resource constraints and mixtures of extreme choices.

The three extreme points alone are not a convex set, because their midpoints are absent. Their convex hull fills the whole triangle. More generally, Caratheodory theorem bounds how many generating points are needed for one convex combination in a finite-dimensional real vector space; its geometric content is that a complicated collection can describe each attainable average with a short representation.

Two convex bodies are homothetic convex bodies if one is a translation of a positive uniform dilation of the other. Their lengths scale by $s$ and their volumes by $s^n$. Closed disks of positive radii are homothetic convex bodies; axis-aligned positive-sided rectangles are homothetic convex bodies exactly when their aspect ratios agree. The Brunn-Minkowski theorem characterizes this relation by equality of root-volumes in Minkowski addition.

Laws of large numbers give conditions under which sample means approach a deterministic expected value. For independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, the strong law of large numbers gives almost sure convergence, while the weak law of large numbers gives convergence in probability. These conclusions specify different modes of convergence of random variables, rather than a universal numerical accuracy guarantee. The central limit theorem is a separate result about the remaining fluctuations after rescaling.

An iteration converges at least quadratically to a limit if its errors eventually satisfy the displayed bound with a fixed finite $C>0$. If $C{e}_N<1$, induction gives $|e_{N+k}|\le C^{-1}(C|e_N|{)}^{2^k}$. Exact asymptotic quadratic convergence has $|e_{n+1}|/|e_n{|}^2\to c\in (0,\infty )$. Newton root-finding iteration at a simple scalar root has this order when the second derivative at the root is nonzero; if that derivative vanishes, convergence can be faster.

Let $\mathbb{N}=\{1,2,\dots \}$ and give each factor of ${\mathbb{R}}^{\mathbb{N}}$ the usual topology of the real line. The familiar standard basis vectors extend to unit sequences $e_n$ with $(e_n{)}_j=1$ when $j=n$ and zero otherwise. Their finite linear combinations are exactly the finitely supported sequences, rather than all of ${\mathbb{R}}^{\mathbb{N}}$.

For every fixed coordinate $j$, $(e_n{)}_j=0$ for all $n>j$. Therefore $e_n\to 0$ in the product topology. More directly, any basic cylinder set $B(F,U)$ containing zero contains every $e_n$ with $n>\max (F\cup \{0\})$: all of its tested coordinates are then zero. The moving unit entry escapes every fixed finite collection of observations.

In the l-infinity sequence space of bounded real sequences, however,Thus the same sequence does not converge to zero in the supremum norm. Here the comparison is between the subspace topology inherited from the product topology and the norm topology on ${\ell }^{\infty }$. The supremum norm is not a finite-valued norm on all of ${\mathbb{R}}^{\mathbb{N}}$, which also contains unbounded sequences. In the box topology on ${\mathbb{R}}^{\mathbb{N}}$, the neighborhood ${\prod }_{j\ge 1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a convergent sequence with limit zero there either.

A GHZ state is a multipartite entangled state with coherent amplitudes linking the alternatives in which all qubits are $0$ and all are $1$.

The three-qubit GHZ state is

Computational-basis measurements give only $000$ and $111$, each with probability one half. A classical mixture of those strings gives the same probabilities, but lacks the off-diagonal quantum coherence. For the displayed state, $⟨X\otimes X\otimes X⟩=1$, whereas it is zero for the mixture. Tracing out one qubit leaves a classically correlated mixture of $|00⟩$ and $|11⟩$. ^[ref]

A Markov policy uses only the current state and, in the nonstationary case, the current time. A stationary Markov policy has the form $u_t=u_{\ast }(X_t)$ for one measurable selector $u_{\ast }$. It is a restricted class of control policies, whose sufficiency for an optimization problem requires the appropriate dynamic-programming hypotheses.

For two contraction mappings of a nonempty complete metric space, their fixed points obey $d(x_f,x_g)\le q_{f}d(x_f,x_g)+d_{\infty }(f,g)$. Rearranging proves the displayed bound whenever the uniform metric is finite. It establishes continuity of the fixed-point map at each $f$ without assuming a contraction constant shared by all maps near $f$.

Let $A$ be differentiable on a neighbourhood of a closed radius-$r$ ball in a Banach space, with $\Vert DA\Vert \le \kappa <1$ on that ball. If $\Vert A(p_0)-p_0\Vert \le (1-\kappa )r$, the mean-value estimate gives $A(B)\subseteq B$ and $\Vert A(p)-A(q)\Vert \le \kappa \Vert p-q\Vert$. The contraction mapping theorem supplies exactly one fixed point in $B$, and its distance from $p_0$ is at most $\Vert A(p_0)-p_0\Vert /(1-\kappa )$.

// Target: dynamical-systems.bigb