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The set $K\cap P$ is compact because $K$ is closed and bounded, and the productis continuous. By the extreme value theorem, $G$ has a maximizer. Since $K$ meets the interior of the nonnegative orthant $P$, some feasible product is positive, so every maximizer lies in $\mathrm{Int}P$.

On $\mathrm{Int}P$, maximizing $G$ is equivalent to maximizingThe Hessian of $F$ iswhich is negative definite. Thus $F$ is a strictly concave function, and it has at most one maximizer on the convex set $K\cap P$. Denote this unique point by $x^{\ast }$.

For any $x\in K\cap P$, the line segmentlies in $K\cap \mathrm{Int}P$. Since $F(x(t))$ is maximal at $t=0$, its right derivative there is nonpositive:ThereforeThis is the supporting inequality for the product maximizer on a compact convex subset of the positive orthant.

Suppose now that $K$ is invariant under cyclic coordinate permutation. That permutation preserves the product, so uniqueness forces it to fix $x^{\ast }$. Hencefor some $r>0$. More explicitly,Indeed, cyclic symmetry averaging puts the diagonal point with coordinate equal to the mean of any $x\in K$ back in $K$, while the preceding inequality with $x_j^{\ast }=r$ gives ${\sum }_j{x}_j\le nr$.

Finally, given $a=(a_1,\dots ,a_n)\in \mathrm{Int}P$, defineThis set is nonempty, closed, convex and bounded, and it contains $a$. By the arithmetic-geometric mean inequality,Equality holds only when all $x_j/a_j$ are equal and their sum is $n$, namely only at $x=a$. Thus this $K$ has

The convex hull of $S\subseteq {\mathbb{R}}^d$ isThus $v$ is a convex combination of $v_1,\dots ,v_m$ when it has precisely this form. Equivalently, $\mathrm{conv}S$ is the smallest convex set containing $S$. The Caratheodory theorem sharpens the finite-combination description in ${\mathbb{R}}^d$: every point of $\mathrm{conv}S$ is a convex combination of at most $d+1$ points of $S$.

The origin belongs to $S$, so $S$ is nonempty. Each displayed inequality defines a closed half-space, and every half-space is a convex set. An intersection of convex sets is convex: if $x,y\in S$ and $0\le t\le 1$, each linear inequality remains valid for $(1-t)x+ty$. Thus $S$ is a nonempty convex polytope.

An extreme point of a convex set $C$ is a point $x\in C$ that cannot be written as $x=(1-t)y+tz$ with $y\ne z$ in $C$ and $0<t<1$.

Start with the eight vertices of the unit cube. The constraint $x_1+x_2+x_3\le 5/2$ removes $(1,1,1)$ and retains the other seven. The cutting plane meets the three cube edges incident to the removed vertex atThese are new vertices. No others occur: at a vertex in three dimensions, three linearly independent bounding planes are active, and choosing triples from the six cube faces and the cutting plane yields precisely the listed points. Hence the ten extreme points are

A function $f:D\to \mathbb{R}$ on a convex set $D$ is convex whenfor all $x,y\in D$ and $0\le t\le 1$.

If $f$ is once differentiable, this is equivalent to monotonicity of its gradient:Equivalently, every tangent hyperplane supports the graph:If $f$ is twice differentiable, convexity is equivalent to the Hessian matrix being positive semidefinite throughout $D$.

Forthe Hessian isA real symmetric two-by-two matrix is positive semidefinite exactly when its two diagonal entries and its determinant are nonnegative. ThusThe largest convexity domain is thereforeFor $A\ne 0$, its boundary is the hyperbola $y=A^2/(36x)$ in the first quadrant and the domain lies above it. For $A=0$, it is the closed first quadrant. This is the convexity domain of x cubed plus y cubed plus Axy.

A function $f:C\to \mathbb{R}$ is strictly convex when, for distinct $x,y\in C$ and $0<t<1$,

Suppose distinct points $x$ and $y$ were both minimizers, with minimum value $m$. Since $C$ is a convex set, their midpoint belongs to $C$, while strict convexity givesa contradiction. Thus any minimizer is unique, which is the uniqueness of a minimizer of a strictly convex function.

For $x\in {\mathbb{R}}^d$, the euclidean projection onto a convex set $C$ isThe squared distance $z\mapsto \Vert x-z{\Vert }_2^2$ is a strictly convex function. Its restriction to the convex set $C$ can therefore have at most one minimizer by part (b). Closedness, together with finite dimensionality, supplies existence, although existence was not required to be proved.

Minkowski's lemma states that if $\Lambda$ is a full lattice in ${\mathbb{R}}^n$ and $C\subset {\mathbb{R}}^n$ is measurable, convex, and centrally symmetric, thenimplies that $C$ contains a nonzero point of $\Lambda$.

Let ${\sigma }_1,{\sigma }_2:K\to \mathbb{R}$ be the two real embeddings. The Minkowski embedding of a real quadratic field identifies ${\mathcal{O}}_K$ with a latticeof covolume $\sqrt{|D_K|}$. Choose a constant $c$ satisfyingFor $R>0$, consider the closed diamondIt is convex and centrally symmetric, and its area isMinkowski's lemma supplies a nonzero ${\alpha }_R\in {\mathcal{O}}_K$ whose two embeddings lie in $C_R$.

For $(x,y)\in C_R$, the arithmetic-geometric mean inequality givesSince the coordinate product is the field norm,Finally, $|{\sigma }_2({\alpha }_R)|\le c/R\to 0$ as $R\to \infty$. No fixed nonzero algebraic integer can occur for arbitrarily large $R$, because its second embedding is nonzero. Consequently the elements ${\alpha }_R$ obtained along an unbounded sequence of $R$ contain infinitely many distinct values. Here $N(\alpha )$ denotes the absolute norm, as usual in this inequality.

Apply part (b). Since every $x_n$ belongs to the convex set $C$, every convex combination ${\tilde{x}}_k$ also belongs to $C$. Since $C$ is norm closed and ${\tilde{x}}_k\to x$ in norm, it follows that $x\in C$. Thus a norm-closed convex set is weakly closed.

The function $\varphi (x)={\max }_{v\in C}⟨x,v⟩$ is the support function ${\sigma }_C$. For a nonempty compact convex set,so its convex conjugate is the indicator function ${\iota }_C$. Applying the Moreau decomposition,Multiplication of an indicator function by a positive scalar does not change it, and its proximal operator is the Euclidean projection onto a convex set. Therefore

A set is convex when it contains every segment joining two of its points. A function on a convex set is convex whenfor all $x,y$ and $0\le t\le 1$.

LetThe minimum exists because the probability simplex is compact. Let $R^{\ast }$ be the information projection of $Q$ onto the closed convex set ${\mathcal{C}}_{\delta }$. Its Pythagorean inequality says that every $R\in {\mathcal{C}}_{\delta }$ satisfiesFor a string $x_1^n$ of type $R\in {\mathcal{C}}_{\delta }$, this givesSumming over the decision region $B_n$ proves the exact bound

Two real affine functionals on a convex set of lotteries that represent the same strict preferences differ by $V=aU+b$ with $a>0$. Equal utility values must first be shown to give equal values of the other functional. The induced function on the real interval of utility values preserves mixtures; anchor interpolation, including exterior points represented through an anchor, makes it linear throughout. Constant representations are handled separately and still admit a positive slope.

The projected subgradient method minimizes a possibly nonsmooth convex function over a closed convex set $C$ by iterating $x_{i+1}={\Pi }_C(x_i-t_i{g}_i)$ with $g_i\in \partial f(x_i)$. Nonexpansiveness of Euclidean projection onto a convex set and the subgradient inequality give its basic convergence bound.

The recession cone of a nonempty convex set $C$ consists of directions $d$ for which $x+td\in C$ for every $x\in C$ and $t\ge 0$. For a linear polyhedron $P=\{x:Ax\le b\}$ it is exactly $\{d:Ad\le 0\}$. A nonzero recession direction gives an unbounded ray, so a bounded nonempty polyhedron has recession cone $\{0\}$. This excludes spurious zero-scale feasible points in the Charnes-Cooper transformation.

Increasing marginal contributions imply $m_i^{\pi }\ge v((S\cap P_i)\cup \{i\})-v(S\cap P_i)$ for $i\in S$. Summing telescopes to ${\sum }_{i\in S}{m}_i^{\pi }\ge v(S)$, while efficiency follows by telescoping over the full ordering. Hence every marginal contribution vector is in the core of a cooperative game. The core is a convex set, so their average, the Shapley value, is in it too. This gives an elementary proof of nonemptiness and stability for a convex cooperative game.

Strict concavity is the displayed property for distinct $x,y$ in a convex set and $0<t<1$. It makes a stationary point of a differentiable function its unique global maximum, whenever such a stationary point exists.

A function $f$ on a convex set is strictly convex whenwhenever $x\ne y$ and $0<t<1$.

For a nonempty closed convex set $C$, the finite-valued support function has subgradients exactly at the maximizing points of $C$. This follows from Fenchel–Young inequality and the conjugacy between the support function and the indicator functional of a constraint set. For compact $C$, the exposed maximizing set is nonempty.

The utility ${\sum }_r{n}_r\log x_r$ sums logarithmic utilities over individual flows. Its first-order condition gives proportional fairness. Its negative is a strictly convex function on active flow rates, giving uniqueness of those rates on a convex set of feasible allocations.