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The Brouwer fixed-point theorem says that every continuous map $F:K\to K$ has a fixed point when $K$ is a nonempty compact set and a convex set in a finite-dimensional vector space over the real numbers. Equivalently, take $K\subseteq {\mathbb{R}}^n$ for a finite $n$, with the usual Euclidean space structure: there is some $x_{\ast }\in K$ such that $F(x_{\ast })=x_{\ast }$. The conclusion guarantees at least one fixed point, without asserting uniqueness or telling us how quickly to find one. ^[ref]

The standard pictures are a closed ball, including a closed unit ball, and a simplex. A filled disk qualifies; its bounding circle alone does not. A probability simplex also qualifies even though its dimension is smaller than that of its ambient Euclidean space. The hypotheses have concrete meanings: being a convex set keeps every line segment between allowed points inside $K$, and compactness prevents escape or loss of a limiting point. In Euclidean space, the Heine-Borel theorem identifies compact sets with closed sets that are bounded sets. Crucially, every output must belong to the same $K$ as the input.

For a fully computable example, let $D=\{(u,v):u^2+v^2\le 1\}$ be the filled unit disk. Apply a counterclockwise quarter-turn rotation, shrink by $0.4$, and add a translation:This affine map is a continuous map. Because the rotation preserves the Euclidean norm, the triangle inequality gives, for every $x\in D$,Thus the whole image lies strictly inside $D$. More precisely, it is a disk of radius $0.4$ centred at $(0.3,0.1)$. This containment check is what makes the Brouwer fixed-point theorem applicable; a formula being a continuous map by itself would not suffice.

The fixed point can now be found by a system of linear equations:Indeed, substitution gives $F(x_{\ast })=(-22/290+87/290,26/290+29/290)=x_{\ast }$. Its Euclidean norm is $\sqrt{290}/58\approx 0.293610$, safely inside $D$. The figure shows arrows from sample inputs to their outputs; an arrow disappears at the fixed point because the displacement $F(x)-x$ vanishes there.

Repeated iterations of a map happen to succeed for this example. Starting at $x_0=(1,0)$ and setting $x_{k+1}=F(x_k)$ gives $x_1=(0.3,0.5)$, $x_2=(0.1,0.22)$, and $x_3=(0.212,0.14)$. Subtracting the fixed point equation yields $x_{k+1}-x_{\ast }=0.4R_{90}(x_k-x_{\ast })$, so the Euclidean distance to $x_{\ast }$ shrinks by exactly $0.4$ each step. This extra fact follows from the contraction mapping property and the Banach fixed-point theorem. It is additional information beyond the Brouwer fixed-point theorem.

The distinction becomes visible with two simpler continuous maps of the same disk. The pure rotation $G(u,v)=(-v,u)$ has exactly one fixed point, $(0,0)$, since $u=-v$ and $v=u$ force both to vanish. Yet iterations of a map starting at $(1,0)$ here cycle through $(0,1)$, $(-1,0)$, and $(0,-1)$ forever. Their Euclidean distance from $(0,0)$ stays one, so there is no convergence in a metric space. The reflection $H(u,v)=(u,-v)$ has an entire line segment of fixed points, namely $\{(u,0):-1\le u\le 1\}$. Starting off that line segment produces an alternating pair. Existence, uniqueness, and convergence of iterations of a map are separate questions.

Removing hypotheses changes the answer. Restrict the same quarter-turn rotation to the circle $S^1=\{(u,v):u^2+v^2=1\}$. It remains a continuous map into that compact set, but its only possible fixed point, $(0,0)$, has been removed. The circle is not a convex set. Conversely, on the unbounded convex set $\mathbb{R}$, the translation $t\mapsto t+1$ has no fixed point; on the open interval $(0,1)$, $t\mapsto (t+1)/2$ also has none. Neither domain is a compact set. The finite-dimensional closed ball statement should likewise not be extended to arbitrary infinite-dimensional vector spaces, where a closed ball need not be a compact set.

The one-dimensional version explains the connection to the intermediate value theorem. For a continuous map $f:[a,b]\to [a,b]$, put $h(t)=f(t)-t$. Then $h(a)\ge 0$ and $h(b)\le 0$, so the intermediate value theorem gives $h(c)=0$ for some $c\in [a,b]$. In higher dimensions, the Brouwer fixed-point theorem supplies a corresponding existence principle when this scalar sign argument is unavailable. Its practical value is ensuring that a proposed equation $x=F(x)$ has a solution once a suitable domain is found; it supplies no general promise of a fast algorithm.

A useful application is positive matrix eigenvector from simplex normalization: rescale the output of a strictly positive matrix to make its entries sum to one. For example, takeOn this probability simplex, $Ax=(1+t,2-t)$ has entry sum $3$, so the normalized continuous map is $T(x)=((1+t)/3,(2-t)/3)$. Both entries are positive and sum to one, keeping the image inside the probability simplex. Its fixed point equation is $t=(1+t)/3$, hence $t=1/2$. Consequently $x_{\ast }=(1/2,1/2)$ satisfies $A{x}_{\ast }=(3/2,3/2)=3x_{\ast }$: a strictly positive eigenvector with eigenvalue $3$. In larger dimensions, the Brouwer fixed-point theorem guarantees the corresponding positive eigenvector without requiring us to solve a system of linear equations in advance. The fixed point expresses preservation of relative weights, rather than preservation of their unnormalized size.

Proofs can be approached through the no-retraction theorem or Sperner's lemma. In the latter approach, increasingly fine triangulations locate approximate fixed points; compactness supplies a convergent subsequence, and continuous maps turn its limit into an exact fixed point. ^[ref]

A primary historical reference is L. E. J. Brouwer's 1911 paper Über Abbildung von Mannigfaltigkeiten, whose digitized original is hosted by Zenodo under DOI 10.1007/BF01456812. This gives a direct route to the early literature behind the Brouwer fixed-point theorem. ^[ref]

On the probability simplex, ${\sum }_i{x}_i=1$. The Cauchy-Schwarz inequality gives $1\le \sqrt{n}\Vert x\Vert$, with equality only when all coordinates are equal. Thus the displayed centroid is the unique closest point to the origin. It is also equidistant from all coordinate vertices, since $\Vert x-e_i{\Vert }^2=\Vert x{\Vert }^2-2x_i+1$.

The convex hull of a set consists of every finite convex combination of its points. It collects all choices obtainable by mixing the original choices with nonnegative weights summing to one. For $S$ in a real vector space,Equivalently this is the smallest convex set containing $S$. Mixing two already mixed points produces another finite convex combination, so the displayed set is convex; every convex set containing $S$ must contain all these mixtures.^[ref]

In an original planar example, take four points $A=(0,0)$, $B=(4,0)$, $C=(3,3)$, $D=(0,2)$ and four additional samples $(1,1)$, $(2,1.7)$, $(3,0.5)$, $(1,0.5)$. Their convex hull is the filled convex polygon $ABCD$, including its boundary. None of the samples is $p=(2,1)$, butThe coefficients are nonnegative and sum to one, providing an explicit membership certificate. Geometrically, first form $U=(3/2,0)=\frac{5}{8}A+\frac{3}{8}B$ on the line segment $AB$, then mix $p=\frac{2}{3}U+\frac{1}{3}C$. The additional interior samples do not enlarge the convex hull.

For a finite nonempty $S\subset {\mathbb{R}}^d$, the convex hull is the image of a probability simplex of weights under the linear map $\lambda \mapsto {\sum }_i{\lambda }_i{s}_i$. The probability simplex is a compact set, and the map is continuous, so this convex hull is compact and closed. A one-point input stays one point; a collinear input stays in a line segment. Filling the convex hull does not necessarily produce a full-dimensional region.

The construction is useful whenever mixtures preserve the intended meaning of a choice. Mixing resource-allocation vectors, for example, interpolates their coordinate requirements. The convex hull describes the resulting feasible region. A coordinate-wise box can admit combinations that no mixture produces, as the outside-point certificate in linear optimization on a convex hull shows.

For a fixed probability mass function $p$ on a finite alphabet, relative entropy is a convex function of its first probability mass function. For $p_i>0$, the coordinate function $f_i(t)=t{\log }_2(t/p_i)$ has $f_i^{\prime \prime }(t)=1/(t\ln 2)>0$ on $(0,\infty )$ and extends continuously with $f_i(0)=0$. Applying scalar convexity coordinate by coordinate proves the inequality. If $p_i=0$, finite relative entropy requires $q_i=r_i=0$ for every positive-weight constituent; otherwise the right side is infinite. Thus the assertion also holds as an extended-valued convexity statement. On the probability simplex supported where $p>0$, the function is strictly convex: two distinct probability mass functions give a strict inequality for $0<\lambda <1$.

Evolutionary game theory studies how a population's composition changes when the success of a pure strategy depends on which other pure strategies it encounters. The central feedback is a frequency-dependent payoff: becoming common changes the opponents a strategy meets, which can change whether it does well. In a coordination game, common behaviour can reinforce itself; in the coexistence example below, being rare is an advantage. This makes evolutionary game theory useful for asking which behaviours can persist and which initial compositions lead to different outcomes, without assuming that each participant solves an optimization problem.

In the basic model, a large, well-mixed population has finitely many inherited pure strategies; interactions sample opponents according to their population frequencies. Assume that a strategy's expected payoff determines its per capita net reproductive growth rate, with a common baseline allowed. These are modelling assumptions: reproduction must transmit the strategy, encounters must match the proposed sampling rule, and the conversion from payoff to growth must be justified for an application. The replicator equation below describes the resulting continuous-time frequencies. Taylor and Jonker's ^[ref] explicitly derive this growth model.

For learned behaviour, reproduction can instead be replaced by an assumed copying rule: an individual samples another individual uniformly and copies a better-performing pure strategy at a rate proportional to the positive payoff difference. At frequencies $x_i,x_j$, the net flow from $j$ to $i$ is $x_i{x}_j(f_i-f_j)$, because the two directional positive parts subtract to $f_i-f_j$. Summing over $j$ gives $x_i(f_i-\bar{f})$, the replicator equation. This derivation states a particular mechanism; an arbitrary rule for learning or imitation need not give the same dynamical system.

Three questions must be distinguished. A symmetric Nash equilibrium tests whether an individual can improve its expected payoff against the resident composition. An evolutionarily stable strategy tests whether each distinct rare mutant is outperformed after it is introduced. Dynamical stability tests the response of an actual dynamical system to small changes of composition, and asymptotic stability additionally requires nearby trajectories to converge. For a finite symmetric bimatrix game under the continuous-time replicator equation, every evolutionarily stable strategy is a symmetric Nash equilibrium and a locally asymptotically stable rest point; the converse from asymptotic stability to an evolutionarily stable strategy fails in general. See Ross Cressman and Yi Tao, The replicator equation and other game dynamics (2014), Theorems 1–2 and Example 1. These statements refer to stability on the full probability simplex, including perturbations introducing currently absent pure strategies.

A concrete experiment makes the frequency feedback visible. In 2009, Gore, Youk and van Oudenaarden competed two yeast strains: a producer of the enzyme invertase, which enables sucrose hydrolysis, and a mutant lacking the corresponding SUC2 gene. Released sugars can feed neighbours, so avoiding production can be advantageous. They used shaken 5 mL cultures with 5% sucrose at $30{}^{\circ }\mathrm{C}$, diluted daily after 23 hours of growth. Different labelled proteins allowed the two strains to be counted separately by flow cytometry; selective plating periodically checked the counts. The cooperator required externally supplied histidine, so restricting that amino acid imposed an adjustable growth disadvantage. See Snowdrift game dynamics and facultative cheating in yeast, pp. 253 and 255–256, Methods Summary.

Starting with rare producers and, separately, rare nonproducers tested both invasion directions. Under the reported coexistence conditions each rare strain increased; changing histidine or added glucose shifted the eventual composition. Their explanation used nonlinear growth benefits, rather than a fixed pairwise payoff matrix. This motivates testing both rarity regimes before inferring an evolutionarily stable strategy or fitting a replicator equation; it does not identify the illustrative tables below with measured payoffs. See Gore, Youk and van Oudenaarden, Figures 1–3.

An early motivation was understanding why animal contests can involve restraint even when escalation might win an individual encounter. Maynard Smith and Price's ^[ref] introduced the evolutionarily stable strategy in this setting. Taylor and Jonker's ^[ref] then studied explicit continuous and discrete dynamics. Static resistance to invasion and the paths populations follow are related questions, but they require separate definitions.

For $0<c<1$ on the probability simplex, $x_i^c\ge x_i$ gives the lower bound, with equality exactly at its vertices. Strict concavity and the Jensen inequality give the upper bound, with equality exactly at the uniform point $x_i=1/n$. Thus the boundary can minimize a strictly concave objective, while its unique maximum lies at the symmetric interior point; a Lagrange-multiplier calculation alone would miss the boundary minimum.

For a symmetric finite game, pure-action expected payoffs $e(i,p)$ against independent identical opposing mixed strategies are continuous. The displayed map takes the probability simplex continuously into itself. At a Brouwer fixed-point theorem fixed point, put $r_i=[e(i,p)-e(p){]}_{+}$ and $S={\sum }_i{r}_i$. The fixed-point equations give $r_i=p_{i}S$. If $S>0$, every supported action has strictly positive excess payoff, contradicting ${\sum }_i{p}_i(e(i,p)-e(p))=0$. Hence $S=0$ and every supported action is a best response, proving a symmetric Nash equilibrium.

An information projection chooses the probability distribution that satisfies a constraint while departing least from a reference law in relative entropy. The constraint might specify an expected value, several marginal probability distributions, or another convex set of permissible laws. The reference is held fixed: the variable is the first argument of $D(q\Vert p)$. Reversing the arguments gives a different minimization problem.

For a finite alphabet $\mathcal{A}$, take a positive reference probability mass function $p(a)>0$ and a nonempty closed convex set $\mathcal{C}$ in the probability simplex. ThenThe minimum exists and its minimizing probability distribution is unique. In this article, natural logarithms give nats. Using base-two logarithms rescales the objective without changing the minimizer.^[ref]

If $p$ already meets the constraint, its information projection is $p$ itself by relative entropy nonnegativity. If not, a constraint can change the relative weights of outcomes without specifying every weight individually. Relative entropy minimization under a mean constraint gives an especially useful example: it tilts the reference law rather than choosing an arbitrary distribution with the right average.

A mixed strategy is a probability distribution over a player's pure strategies. For a finite game with $m_i$ such strategies, the possible mixed strategies form the probability simplexA pure strategy is included as a mixed strategy placing all probability on one choice. The support of a mixed strategy consists of the choices with positive probability. In a sequential game, sampling a complete pure strategy once is the definition; it is not automatically equivalent to fresh randomization at every move.

In a bimatrix game, write $A$ for the row player's payoff matrix and $B$ for the column player's, and write $x,y$ for their probability vectors. Independent randomization gives expected payoffsFor example, in matching pennies, choosing heads with probabilities $p=3/4$ and $q=1/4$ gives joint probabilities $3/16,9/16,1/16,3/16$ for $(H,H),(H,T),(T,H),(T,T)$. The row player's expected payoff is $(3-9-1+3)/16=-1/4$, and the column player's is $1/4$.

Randomization can prevent an opponent exploiting a predictable choice. It does not mean every choice is equally good against every opponent: only best responses belong in the support of a mixed strategy at a Nash equilibrium. A shared random signal can instead produce a lottery over joint action profiles whose joint probabilities do not factor into independent mixed strategies.

For independent and identically distributed observations from a probability mass function $p$ on a finite alphabet, expected relative entropy of the empirical distribution from $p$ is nonincreasing in sample size. The leave-one-out identity for empirical distributions and convexity of relative entropy in its first argument giveEvery deletion sample has the law of $n$ observations from $p$. Taking expected values proves the assertion without requiring independence between deletion samples. Zero-probability symbols are absent almost surely. Since $\mathbb{E}{\hat{p}}_n=p$, the identity $\mathbb{E}D({\hat{p}}_n\Vert p)=H(p)-\mathbb{E}H({\hat{p}}_n)$ also shows that expected empirical information entropy increases towards the source information entropy. Convergence follows from the strong law of large numbers for empirical frequencies and boundedness and continuity of information entropy on a finite probability simplex.

Let $B=A^T$ and choose any column probability vector $q$, for example the first coordinate unit vector. Because $A$ has nonnegative entries and each row sums to one, $B$ maps the probability simplexinto itself: $Bv\ge 0$ and $1^{T}Bv=(A1{)}^{T}v=1^{T}v=1$.

The iterates $B^{j}q$ need not converge, so take their arithmetic averages,A telescoping sum givesBoth vectors on the right before subtraction have L1 norm one, so $\Vert B{x}_N-x_N{\Vert }_1\le 2/N\to 0$.

The probability simplex is closed and bounded in finite-dimensional Euclidean space, hence compact by the Heine-Borel theorem. It follows that a subsequence $x_{N_j}$ converges to some $x\in \Delta$. Passing to the limit in the displayed identity, using continuity of the linear map $B$, gives $A^{T}x=Bx=x$, with $x\ge 0$ and ${\sum }_i{x}_i=1$.

This proves stationary distribution existence for a finite Markov chain without any Markov chain irreducibility or aperiodicity assumption. Those stronger hypotheses concern uniqueness and convergence, not the existence required here.

For a discrete memoryless channel with transition probabilities $M(y\mid x)$, the conditional probability of a length-$n$ output given a length-$n$ input factors asThe same transition probabilities apply at each use. Its channel capacity isThe unit is bits per channel use. For finite alphabets the maximum exists by continuity on the probability simplex; with more general alphabets one uses a supremum when appropriate.

An $(m,n)$ channel code consists of an encoder $c:{\mathcal{V}}_m\to {\mathcal{X}}^n$ and a decoder $d:{\mathcal{Y}}^n\to {\mathcal{V}}_m$. It represents each of $m$ messages by $n$ input symbols and recovers a message from the $n$ observed output symbols. Its code rate is $R=({\log }_{2}m)/n$ bits per use. For each message $w$, its conditional error probability is $\mathbb{P}(d(Y^n)\ne w\mid W=w)$; an average error also specifies the message probability distribution, while the maximum error takes the largest conditional error.

The basic reproduction number is the expected value of secondary infections caused by one typical infectious individual introduced into an otherwise susceptible population. For a heterogeneous model it is the dominant eigenvalue of the next-generation operator, so the appropriate infected-type distribution is part of “typical.” A small introduction grows when ${\mathcal{R}}_0>1$ and decays when ${\mathcal{R}}_0<1$, under the usual linear invasion assumptions.

The asymptotic susceptible level depends on the model. In a homogeneous SIS model with transmission rate $\beta$ and recovery rate $\gamma$, the positive endemic equilibrium has $S_{\ast }=1/{\mathcal{R}}_0$ when ${\mathcal{R}}_0=\beta /\gamma >1$. In a closed SIR model, however, the final size relation for an epidemic isFor an infinitesimal seed in a wholly susceptible population this becomes $-\log S_{\infty }={\mathcal{R}}_0(1-S_{\infty })$. The epidemic peak occurs at $S=1/{\mathcal{R}}_0$, and a nontrivial outbreak leaves $S_{\infty }<1/{\mathcal{R}}_0$. Thus $1/{\mathcal{R}}_0$ is an endemic or instantaneous invasion threshold, not a universal final susceptible fraction.

For cooperative infections, let $s,x,y,z$ be the fractions carrying neither infection, only $A$, only $B$, and both, with sum one. Let ${\lambda }_A={\beta }_A(x+z)$, ${\lambda }_B={\beta }_B(y+z)$, assume each infection independently recovers at rate $\gamma$, and multiply susceptibility to the second infection by $q>1$. One suitable pair of SIS models isRecovery of one infection from $z$ leads to the other single-infection class, not directly to $s$. The probability simplex is invariant. At the disease-free equilibrium the eigenvalues are ${\beta }_A-\gamma$, ${\beta }_B-\gamma$ and $-2\gamma$. Therefore failure of both isolated and low-density invasion requires ${\mathcal{R}}_A={\beta }_A/\gamma <1$, ${\mathcal{R}}_B={\beta }_B/\gamma <1$. Enhancement of secondary susceptibility does not alter these linear thresholds because coinfection is second order at introduction.

For completeness, a joint positive equilibrium must haveandTo obtain these, divide the equilibrium equations for total $A$ and total $B$ prevalence by their positive prevalences, yielding $s+qy={\mathcal{R}}_A^{-1}$ and $s+qx={\mathcal{R}}_B^{-1}$. Substitution into the $s$ balance gives the quadratic. Its discriminant must be nonnegative and the resulting $x,y,z$ positive. These conditions test equilibrium feasibility for unequal rates; stability must also be checked rather than inferred from the mere presence of a root.

A fully explicit Allee effect results in the symmetric permitted case ${\beta }_A={\beta }_B=\beta$, $\mathcal{R}=\beta /\gamma$. The joint equilibria haveTwo feasible positive equilibria coexist with the stable disease-free state exactly in the strict rangeIndeed at the fold the two roots coalesce, and at $\mathcal{R}=1$ the smaller-prevalence branch reaches the disease-free boundary. For $q\le 2$, roots at $\mathcal{R}<1$ do not give positive singly infected fractions. Strict inequalities exclude these marginal endpoints.

Here is the stability check underlying the cooperative SIS coinfection threshold. On the symmetric subspace put $v=x+z=y+z$ and $c=z$. ThenAt a positive equilibrium $c=q\mathcal{R}{v}^2/(1+q\mathcal{R}v)$. Its two-dimensional Jacobian matrix hasThe higher-prevalence branch has positive determinant and negative trace, while the lower-prevalence branch is a saddle equilibrium. The remaining antisymmetric mode, from $x-y$, has eigenvalue $-q\beta v<0$. Thus the smaller susceptible root $s_{-}$ is a stable joint endemic equilibrium, separated from the stable zero state by a finite seeding threshold. Neither infection can persist alone in this range. Strong cooperation can sustain both infections after a sufficiently large joint introduction even though neither invades from rarity. The numerical condition belongs to the specified secondary-susceptibility model; alternative cooperation mechanisms need their own threshold calculation.

Let $a,b,e$ be fractions of patches occupied by the two competitors and empty, with $a+b+e=1$. Assume well-mixed propagule dispersal, equal independent fire extinction rate $m>0$, colonization constants $c_A>c_B>0$, and net replacement advantage $q=k_B-k_A>0$ for the poorer colonizer. The Levins metapopulation model with replacement isReplacement fluxes $k_{B}ab$ and $k_{A}ab$ cancel in the total occupancy equation. At every face the outward flux is zero or points into the patch probability simplex, so these equations preserve valid fractions. The constants include both propagule arrival and establishment probability; the model does not require unequal fire susceptibility.

Put $\Delta c=c_A-c_B$. Solving the two occupied-patch balances and their sum constraint givesThus strict coexistence is equivalent toToo little disturbance lets the better competitor exclude the colonizer; too much disturbance favors colonization or destroys occupancy. A positive equilibrium automatically has $m<c_A$. If the poorer colonizer cannot survive alone, coexistence can nevertheless be possible because it repeatedly replaces established patches of the better colonizer.

These bounds also have an invasion interpretation. The $A$-only equilibrium is $a=1-m/c_A$ when $c_A>m$; the growth rate of rare $B$ there is $q-m(q+\Delta c)/c_A$, positive at the upper-bound condition above. At a feasible $B$-only equilibrium the rare-$A$ growth is $-q+m(q+\Delta c)/c_B$, giving the lower bound. At coexistence the occupied-patch Jacobian matrix has trace $-c_A{a}_{\ast }-c_B{b}_{\ast }<0$ and determinant $a_{\ast }{b}_{\ast }q(q+\Delta c)>0$, proving local asymptotic stability. The Bendixson-Dulac criterion with multiplier $1/(ab)$ gives divergence $-c_A/b-c_B/a<0$ in the interior, excluding periodic competition cycles.

To specify selection, a monotone tradeoff alone is insufficient: one must say how two different trait values determine replacement. A natural model uses a trait $x$ with increasing competitive ability $B(x)$ and decreasing colonization $c(x)$, and net replacement $B(y)-B(x)$ by a mutant $y$ against a resident $x$. The resident metapopulation occupies $1-m/c(x)$, requiring $c(x)>m$. The rare-mutant per-capita growth, or invasion fitness, isIt is zero at $y=x$. Differentiating with respect to the mutant trait gives the selective pressureHigher colonization is valuable in empty habitat; higher competitive ability is valuable where the resident already occupies patches. This is colonization-competition tradeoff invasion fitness.

At an interior candidate for a local evolutionarily stable strategy, $g(x_{\ast })=0$. A strict, nondegenerate local evolutionarily stable strategy follows ifsince nearby distinct mutants then have negative invasion fitness. For a strict global evolutionarily stable strategy, every feasible distinct mutant must satisfyThis is the supporting-line condition on the tradeoff curve. At a trait boundary replace the vanishing-gradient condition by the appropriate one-sided inequality. Equality against another mutant, or zero local curvature, needs the second condition of evolutionary stability or higher-order terms; the strict tests above are sufficient and are the usual nondegenerate conditions, not a claim that every tie is unstable.

For evolutionary approach to a singular point, the selection gradient must also point toward it, locally $g^{\prime }(x_{\ast })<0$. This is logically distinct from testing mutant fitness. In this particular additive-ability model with $c^{\prime }<0<B^{\prime }$, negative mutant curvature implies negative $g^{\prime }$ becauseIf replacement instead has a general kernel $\kappa (y,x)$, replace $B(y)-B(x)$ in $f$ by that kernel and differentiate it; monotonicity does not justify importing the displayed additive formula without that modeling assumption. The tradeoff shape, disturbance rate and replacement law together determine selection and local or global uninvadability.

Each coordinate lies in $[0,1]$. Since $0<c<1$, $x_i^c\ge x_i$, with equality exactly when $x_i=0$ or $1$. Summing gives $f(\mathbf{x})\ge 1$. Equality under the simplex constraint forces exactly one coordinate to be one. Thus the minimum is $1$, attained exactly at the vertices of the probability simplex.

For the maximum, $t\mapsto t^c$ is strictly concave on $[0,\infty )$. The Jensen inequality givesStrict concavity makes equality possible exactly when all coordinates agree. Hence the extrema of a concave power sum on a simplex areFor $n=1$ the simplex is a single point, and both descriptions coincide.

For a finite bimatrix game, let $A,B\in {\mathbb{R}}^{r\times s}$ be the two players' payoff matrices. A pair of mixed strategies $(p,q)\in {\Delta }_r\times {\Delta }_s$ is a Nash equilibrium ifHere ${\Delta }_r$ is the probability simplex on the first player's pure actions, and similarly for ${\Delta }_s$. The two payoff matrices need not be negatives of each other. Nash's theorem asserts that every finite two-person game has an equilibrium in mixed strategies.

We prove it using the Brouwer gain-map proof of bimatrix equilibrium. Put $u=p^{T}Aq$, $v=p^{T}Bq$ and define the positive pure-deviation gainsWrite $G={\sum }_i{g}_i$, $H={\sum }_j{h}_j$ and defineEvery coordinate is nonnegative and each player's coordinates sum to one. The map is continuous, since the payoffs and positive-part function are continuous and both denominators are at least one. Thus $T$ is a continuous self-map of the nonempty compact convex set ${\Delta }_r\times {\Delta }_s$. The Brouwer fixed-point theorem supplies a fixed point $(p,q)$.

At that fixed point the first coordinate equations give $g_i=G{p}_i$. Suppose $G>0$. For every $i$ with $p_i>0$, this makes $g_i>0$, so $(Aq{)}_i-u=g_i=G{p}_i$. But averaging the payoff excess over the support givesa contradiction. Therefore $G=0$, and every $g_i$ is zero. Each pure action has payoff at most $u$, so every mixed deviation, being a convex combination of pure actions, has payoff at most $u$ too. The same argument for $h_j=H{q}_j$ gives $H=0$ and excludes every unilateral deviation by the second player. The fixed point is therefore a Nash equilibrium, proving the stated theorem without a zero-sum assumption.

For $y\ne 0$, the squared Euclidean norm of the orthogonal projection residual is nonnegative:This proves the Cauchy-Schwarz inequality $|x\cdot y|\le |x||y|$. Equality holds exactly when $x$ is a scalar multiple of $y$. If $y=0$, equality is automatic. Equivalently, equality means the two vectors are linearly dependent, including the cases with a zero vector.

For a point $C$ equidistant from the coordinate vertices,must be independent of $i$. Thus all $C_i$ are equal, and their sum is one. The centroid is thereforeFor every $x$ in the base probability simplex, the Cauchy-Schwarz inequality gives $1=(1,\dots ,1)\cdot x\le \sqrt{n}|x|$. Equality forces the same coordinates as $C$. Thus $C$ is the unique closest point of a probability simplex to the origin.

The angle $\alpha$ between $OC$ and any edge from the origin to $e_i$ satisfiesConsequently

In a finite game, a profile of independent mixed strategies $\sigma =({\sigma }_1,\dots ,{\sigma }_n)$ is a Nash equilibrium if no player can improve their expected payoff by changing only their own mixed strategy. In symbols,Because expectation is linear in a player's own mixed strategy, checking this inequality for all their pure strategies suffices. Nothing requires the players' payoffs to sum to zero.

For the symmetric finite game, write $e_i(p)=e(i,p)$ and $\bar{e}(p)={\sum }_i{p}_i{e}_i(p)$. Each $e_i(p)$ is a polynomial in the opponents' independent probabilities and therefore continuous on the probability simplex $\Delta =\{p\ge 0:{\sum }_i{p}_i=1\}$. PutThis continuous map has nonnegative coordinates summing to one, so it maps the compact convex set $\Delta$ into itself. The Brouwer fixed-point theorem gives $p=F(p)$. At that fixed point $r_i=p_{i}S$. If $S>0$, every $i$ in the strategy support has $r_i>0$ and therefore $e_i(p)>\bar{e}(p)$. Butwhereas every supported term would be strictly positive, a contradiction. Thus $S=0$, all pure payoffs are at most $\bar{e}(p)$, and their weighted average equals $\bar{e}(p)$. Every supported pure strategy is consequently a best response to the common opposing mixed strategy. The profile in which every player uses $p$ is a symmetric Nash equilibrium. This proves the gain-map proof of symmetric equilibrium for any finite number of players, including boundary fixed points of the probability simplex.

In the least unique bid auction, let $p$ be the common probability of bid 1. Bidding 1 wins only when the other two bids are both 2, and its winning payoff is $3-1=2$. Bidding 2 wins only when the other two bids are both 1, and its winning payoff is $3-2=1$. Hence the two expected payoffs areNeither symmetric pure strategy profile is an equilibrium: against two bids of 1 a deviation to 2 earns 1 instead of 0, and against two bids of 2 a deviation to 1 earns 2 instead of 0. An interior symmetric equilibrium must make the two expected payoffs equal. Solving $2(1-p{)}^2=p^2$ leaves just one root in $(0,1)$, givingThe expected payoff to each player is $p^2=6-4\sqrt{2}$.

To count all nonsymmetric Nash equilibria, let $p_i$ be player $i$'s probability of bidding 1. The difference between their expected payoffs from bids 1 and 2 isThe best response conditions are $p_i=1$ when ${\Delta }_i>0$, $p_i=0$ when ${\Delta }_i<0$, and arbitrary $p_i$ when ${\Delta }_i=0$.

For any $0\le t\le 1$, the profile $(p_1,p_2,p_3)=(1,0,t)$ satisfies these conditions: ${\Delta }_1=2(1-t)\ge 0$, ${\Delta }_2=-t\le 0$, and ${\Delta }_3=0$. Thus every permutation of this profile is a Nash equilibrium. At $t=0,1$ these give the six pure profiles with at least one bid of each kind; for $0<t<1$, one player mixes freely and the other two use opposite pure strategies.

There are no other nonsymmetric equilibria. If all three probabilities are interior, every ${\Delta }_i$ must vanish. Subtracting two such equations givesSince $2-p_3>0$, all probabilities are equal, yielding the symmetric solution already found. If exactly two players mix and the third uses bid 2, either mixer strictly prefers bid 1 because $\Delta =2(1-p_j)>0$. If the third uses bid 1, either mixer strictly prefers bid 2 because $\Delta =-p_j<0$. Both cases are impossible. If exactly one player mixes, their indifference requires the other players to use opposite bids. Finally, with no mixing, the all-equal pure profiles fail and the remaining six satisfy the best response conditions.

Consequently the complete Nash equilibrium set consists of the single symmetric mixed profile together with the six closed one-parameter families given by permutations of $(1,0,t)$. Their endpoints overlap at the six pure equilibria. In particular,Counting only the six pure ones would miss the interior points of these families in the two-bid three-player least unique bid auction.

Take the sampled points to be independent with the uniform distribution on $T=\{(x,y):0\le y\le x\le 1\}$. Its area is $1/2$, so its probability density is two. With probability one there are no coincidences, equal relevant coordinates or collinear triples. We give an exact first moment method argument; no tangent construction is needed.

First count the probability that $k$ labelled points, together with the endpoints, form a convex chain. In their order along the lower boundary, write their coordinates as $(x_i,y_i)$, $1\le i\le k$, and put $(x_0,y_0)=(0,0)$, $(x_{k+1},y_{k+1})=(1,1)$. They necessarily satisfyIndeed the lower boundary of the convex polygon is the graph of a convex piecewise linear function. Its segment slopes strictly increase; the first is positive because the first point lies above the $x$ axis, making all later slopes positive too.

Set $u_i=x_i-x_{i-1}$ and $v_i=y_i-y_{i-1}$ for $1\le i\le k+1$. Both spacing vectors lie in the open probability simplex, with ${\sum }_i{u}_i={\sum }_i{v}_i=1$. The coordinate-monotone region has volume $1/(k!{)}^2$. The change to the first $k$ spacings in each vector has Jacobian one, and each simplex has uniform volume measure. Simultaneous permutation of all $k+1$ pairs $(u_i,v_i)$ preserves this product volume measure. Except on a null set of ties, the ratios $v_i/u_i$ therefore have all $(k+1)!$ possible orders with equal volume.

Strict increase of these ratios is exactly the convex chain condition. It also puts the intermediate points below the diagonal: the weighted mean of all slopes is ${\sum }_i{u}_i(v_i/u_i)=1$, and each proper initial weighted mean is strictly smaller than one, so $y_j<x_j$ for $j\le k$. Thus it suffices to impose the slope order on the whole coordinate-monotone region; no additional triangle restriction is needed.

The volume for one specified label order is consequently $1/((k!{)}^2(k+1)!)$. There are $k!$ disjoint orders of the labels, and the joint uniform-triangle density is $2^k$. Hence the convex-chain probability in a triangle isFor example $p_1=1$ and $p_2=1/3$.

Let $C_k$ count the $k$-element subsets of the $n$ sampled points that are convex chains. By linearity of expectation, $\mathbb{E}{C}_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p_k$. If $L_n\ge k$, an optimal chain contains a $k$-vertex subchain, since removing vertices from a convex polygon preserves convex position of the remaining vertices. Therefore, by the first moment method,We used $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le n^k/k!$, $(k+1)!\ge k!$, and $k!\ge (k/e{)}^k$, the last following from ${\sum }_{j=1}^k\log j\ge {\int }_1^k\log xdx$.

Take $k=\lceil 4e{n}^{1/3}\rceil$. If $k\le n$, the last bound is at most $32^{-k}<1/2$; if $k>n$, the event is empty. Every integer median $m_n$ must then satisfy $m_n<k$, and thereforeThe hint's curve $y=x^2$ is a parabola, not a hyperbola. The exact counting argument above proves the upper bound directly. The same chain probability is recorded in Ambrus and Bárány's analysis of longest convex chains.

Let $f(x)=-{\sum }_i\log ({\alpha }_i+x_i)$. The feasible probability simplex is a nonempty compact convex set, so the extreme value theorem guarantees a minimizer. Its Hessian matrix is diagonal with positive entries $({\alpha }_i+x_i{)}^{-2}$, making $f$ a strictly convex function. Hence the minimizer is unique.

Use the Lagrange multipliers $\lambda$ for the sum constraint and ${\nu }_i\ge 0$ for $-x_i\le 0$. The Lagrangian isThe KKT conditions areA strictly positive feasible allocation exists, so the Slater condition holds. These conditions are necessary and sufficient for this convex optimization. At least one coordinate is positive, giving $\lambda >0$. Put $\tau =1/\lambda$. For a positive coordinate, complementary slackness gives ${\alpha }_i+x_i=\tau$. At a zero coordinate, stationarity gives ${\nu }_i=1/\tau -1/{\alpha }_i\ge 0$, or ${\alpha }_i\ge \tau$. ThusHere $(u{)}_{+}=\max (u,0)$. This logarithmic water filling raises the smaller baseline values to a common level. The scalar left side is a continuous function and strictly increasing above ${\min }_i{\alpha }_i$, starts at zero, and tends to infinity, so exactly one $\tau$ solves it. One can find it by interval bisection, or sort the baseline values increasingly and find an index $k$ for whichusing ${\alpha }_{(n+1)}=+\infty$. Equality at the next baseline simply gives a zero allocation. Sorting and a running sum implement the water-filling algorithm in $O(n\log n)$ time.