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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

Here $T$ is a nonlinear compact operator: it is continuous and sends bounded subsets of $M$ to relatively compact subsets of $X$. Equivalently, the images of any bounded input sequence have a norm-convergent subsequence.

The Schauder fixed-point theorem states that if $M$ is a nonempty closed bounded convex set in a Banach space and $T:M\to M$ is continuous and compact, then $\boxed{T\text{ has a fixed point in }M}$. Linearity, a contraction estimate, and uniqueness are not assumptions or conclusions of this theorem.

Each closed ball is a convex set and contains the origin. Join any point to zero by a line segment inside its ball, then join zero to the other point. Concatenating the two segments gives a path in the union. It is therefore path-connected and connected.

On $X^{+}$, $z^2>x^2+y^2$, so $z\ne 0$ and therefore $a=-2z\ne 0$. Yet $a(I)=-2$ and $a(-I)=2$. The same continuous image of a connected space argument shows that $X^{+}$ is not connected.

Put $r=\sqrt{x^2+y^2}$. The required three disjoint nonempty open sets areTheir union is precisely $X$, the complement of the double cone $z^2=x^2+y^2$. The eigenvalues are $z\pm r$, so $X_1$ consists of positive-definite matrices above the upper sheet, $X_2$ of negative-definite matrices below the lower sheet, and $X_3$ of indefinite matrices outside the double cone. Examples are $I,-I,\mathrm{diag}(1,-1)$ respectively.

The two definite cones are convex sets. The indefinite region is path-connected: shrink $z$ continuously to zero while keeping $r>0$, then change the nonzero vector $(x,y)$ within the punctured plane. Thus these sets are also exactly the three connected components. This is the determinant cone for real symmetric two-by-two matrices.

Write $x(S)={\sum }_{i\in S}{x}_i$. The core of a cooperative game isTake any $x,y\in C(v)$ and $\alpha \in [0,1]$, and form the convex combination $z=\alpha x+(1-\alpha )y$. Its total payoff is $z(N)=\alpha v(N)+(1-\alpha )v(N)=v(N)$. For every coalition $S$,Hence $z$ remains in the core. The core is a convex set, even when the transferable utility game is not a convex cooperative game. An empty core is a convex set vacuously; the argument does not claim nonemptiness. Geometrically, the same conclusion follows because the constraints intersect an affine hyperplane with closed half-spaces.

Removing $1/2$ from the closed interval $I=[0,1]$ leaves the separation of a topological spaceThus $1/2$ is a cut point of $I$.

In contrast, removing any point $p$ from the square $Q=[0,1{]}^2$ leaves a path-connected space. To prove this, take distinct $u,v\in Q\setminus \{p\}$. Choose an interior point $w$ outside the two lines through $(u,p)$ and $(v,p)$. Such a choice exists: a small circle in the interior meets each of those lines in at most two points. Since $Q$ is a convex set, the line segments $[u,w]$ and $[w,v]$ stay in $Q$; our choice of $w$ ensures that both avoid $p$. Their concatenation is a continuous path from $u$ to $v$ in $Q\setminus \{p\}$.

The product topology on $I\times I$ agrees with its usual subspace topology in the Euclidean plane. If $h:Q\to I$ were a homeomorphism, its restriction would give a homeomorphismThe domain is connected, since path connectedness implies connectedness, while the target is not connected. This contradicts the preservation of connectedness by a homeomorphism.

Complete the square to write the objective asLet $z=(x,y)$ and $a=(1,2)$, so $\Vert a\Vert =\sqrt{5}$ while feasibility requires $\Vert z\Vert \ge \sqrt{20}=2\sqrt{5}$. The reverse triangle inequality givesThe objective is therefore at least zero. Equality is attained by $z=2a=(2,4)$, which lies on the boundary. Equality requires the vectors to have the same direction and $\Vert z\Vert =2\sqrt{5}$, so this minimizer is unique. HenceIt also satisfies the Karush-Kuhn-Tucker conditions with $\lambda =1/2$. The direct distance argument proves global optimality even though the exterior of a disk is not a convex set; this is nearest-point minimization outside a Euclidean ball.

For a compact convex set $K\subset \mathbb{C}$ and $\zeta \notin K$, strict separation gives a unit vector $u$ with $\mathrm{Re}(\bar{u}\zeta )>{\max }_K\mathrm{Re}(\bar{u}z)$. Taking $c=-Ru$ for large $R$ makes $q={\max }_K|z-c|/|\zeta -c|<1$. The polynomialhas uniform error at most $q^{N+1}/(|\zeta -c|(1-q))$. This elementary geometric series argument proves exterior-pole approximation without a general complex approximation theorem.

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.

A punctured ball is an open ball with its centre removed. In the real line it has two connected components. In Euclidean space of dimension $n\ge 2$ it is path-connected: for any two points, choose a third point inside the ball outside the two lines through the centre and those points. Such a point exists since two lines do not cover an open ball in dimension at least two. The two line segments joining through that point stay inside the ball, which is a convex set, and avoid its centre. They form a continuous path.