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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]

For each player in a finite bimatrix game, add the positive pure-strategy gains to its current mixed strategy and normalize. This gives a continuous self-map of the product of probability simplices. At a Brouwer fixed-point theorem fixed point, the gain coordinates equal the current strategy times their total gain. If that total were positive, every supported action would beat the current average payoff, contradicting that average. Thus every pure deviation has nonpositive gain for both players, proving a Nash equilibrium.

Let $V$ be a finite-dimensional real inner-product space and let $F:V\to V$ be continuous. If $⟨F(x),x⟩<0$ on the sphere $\Vert x\Vert =R$, then $F$ has a zero in the open ball. Otherwise the radial projection of $F$ gives a map of the ball that contradicts the Brouwer fixed-point theorem.

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.

A continuous Euclidean self-map with this growth has a fixed point. Choose $R\ge (2A{)}^{1/(1-\alpha )}$ and $R\ge 2B$. Then $|f(x)|\le R$ whenever $|x|\le R$, so the closed radius-$R$ ball is invariant and Brouwer applies. Existence does not imply convergence of iterating the original map: $x\mapsto 1-x$ on the unit interval has a fixed point but a two-cycle from either endpoint.

A Nash equilibrium is a strategy profile in which every player's strategy is a best response to the others. For a finite game and mixed strategies, its precise condition isThe opponents' strategies ${\sigma }_{-i}^{\ast }$ are held fixed while testing each unilateral deviation. Equality is allowed: a player may have other equally good best responses. Because expected payoff is a linear function of the player's own mixture, it suffices to test every pure strategy. Equivalently, every choice in the support of a mixed strategy must attain the largest pure-choice expected payoff, and no unused choice may exceed it.

The two worked examples illustrate different uses of this condition. Matching pennies has no pure strategy Nash equilibrium, but its unique mixed strategy Nash equilibrium removes each player's vulnerability to predictable play. The asymmetric coordination example has three Nash equilibria, and its mixed one fails Pareto efficiency. A Nash equilibrium tests individual incentives, whereas Pareto efficiency compares feasible outcomes that may require several players to change together. In the prisoner's dilemma, for example, mutual defection is a Nash equilibrium even though mutual cooperation gives both players a larger payoff because $R>P$.

Existence in a finite game is guaranteed by Nash's theorem. The theorem does not select one Nash equilibrium, promise Pareto efficiency, or assert convergence of a learning process. Nash's ^[ref] used the Kakutani fixed-point theorem; his ^[ref] gave a proof using the Brouwer fixed-point theorem. The connection to a fixed point captures the idea of choices that remain mutually optimal when considered together.

No closed disc retracts continuously onto its boundary. This is equivalent to Brouwer fixed-point theorem by the standard ray construction and the antipodal map.

A continuous vector field on a closed Euclidean ball whose boundary values have nonnegative outward component has a zero in the ball. Otherwise $T(x)=-Rg(x)/|g(x)|$ is a continuous self-map of the ball. Its fixed point, supplied by the Brouwer fixed-point theorem, lies on the boundary and satisfies $g(x)\cdot x=-R|g(x)|<0$, a contradiction. The argument turns a zero-finding problem into a fixed point problem without any differentiability assumption.

The Brouwer fixed-point theorem says that every continuous map $T:{\overline{B}}_R(0)\to {\overline{B}}_R(0)$ has a fixed point. To use the supplied outward vector field zero theorem, first extend the construction to all of ${\mathbb{R}}^n$. LetThe radial retraction onto a closed ball $P_R$ is continuous, so $g$ is a continuous vector field on the required whole space. On $|x|=R$, the Cauchy-Schwarz inequality givesThe supplied theorem gives $g(x_0)=0$ with $|x_0|\le R$. There $P_R(x_0)=x_0$, hence $\boxed{T(x_0)=x_0}$. This explicit extension avoids presuming that the original map $T$ was defined outside the ball.

A retraction from a topological space $X$ onto a subset $Y$ is a continuous map $r:X\to Y$ satisfying $r(y)=y$ for every $y\in Y$. Equivalently, $r\circ \iota ={\mathrm{id}}_Y$, where $\iota :Y\hookrightarrow X$ is the inclusion.

The no-retraction theorem for the finite-dimensional closed unit ball saysIndeed, if $r$ were such a retraction, $F(x)=-r(x)$ would be a continuous map from ${\mathbb{B}}^n$ into itself. The Brouwer fixed-point theorem would give $x=F(x)=-r(x)$. Since $|r(x)|=1$, $x$ lies on the boundary, where $r(x)=x$. Thus $x=-x$, contradicting $|x|=1$. Here the Brouwer fixed-point theorem asserts that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point.

Let $S_{+}^{n-1}$ be the nonnegative portion of the unit sphere specified in the paper. If $Ax=0$ for some $x\in S_{+}^{n-1}$, then this nonzero vector is an eigenvector with eigenvalue zero, which already proves the assertion.

Otherwise $Ax\ne 0$ for every $x\in S_{+}^{n-1}$, andis a continuous map from $S_{+}^{n-1}$ to itself. Indeed, every entry of $Ax$ is nonnegative because $A$ is a nonnegative matrix and every entry of $x$ is nonnegative. By the supplied homeomorphism $h:{\mathbb{B}}^{n-1}\to S_{+}^{n-1}$, the continuous map $h^{-1}Th$ maps the finite-dimensional closed unit ball into itself. The Brouwer fixed-point theorem supplies a fixed point, hence $T(x_{\ast })=x_{\ast }$ for some $x_{\ast }\in S_{+}^{n-1}$. ThereforeTogether with the zero case this proves that $A$ has a nonnegative eigenvalue, with a nonnegative eigenvector. For $n=1$, the auxiliary closed unit ball is the one-point space ${\mathbb{B}}^0$ and the argument remains valid. No claim of strict positivity of every entry of the eigenvector is needed.

We first prove the compact-domain Schauder fixed-point theorem. Let $K$ be a nonempty compact convex set in a Banach space $X$, and let $T:K\to K$ be a continuous map. The Brouwer fixed-point theorem states that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point. By the homeomorphism proved in part (a)(ii), it applies equally to any nonempty finite-dimensional compact convex set, including a singleton.

For $\epsilon >0$, use the supplied Schauder approximation lemma to obtain a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with $\Vert Tx-T_{\epsilon }x\Vert \le \epsilon$. The set $K_{\epsilon }=K\cap L_{\epsilon }$ is nonempty because it contains $T_{\epsilon }(K)$. It is a compact convex set, since a finite-dimensional linear subspace is closed. Restricting $T_{\epsilon }$ to $K_{\epsilon }$ and applying the Brouwer fixed-point theorem gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, and thusTake ${\epsilon }_j=1/j$. The compactness of $K$ yields a convergent subsequence $x_{{\epsilon }_{j_m}}\to x_{\ast }\in K$. The continuity of $T$ and the preceding estimate giveFor completeness, this also proves the version on a nonempty closed bounded convex set $C$ when $T(C)$ is relatively compact. Let $K=\overline{\mathrm{conv}}T(C)\subset C$. The closed convex hull of a compact set in a Banach space is compact. To verify the fact needed here, cover $T(C)$ by finitely many balls of radius $\delta$. Replacing every term of a convex combination by a centre changes it by at most $\delta$ in norm. The convex hull of those finitely many centres is compact in a finite-dimensional linear subspace, so it has a finite net of radius $\delta$. Thus $\mathrm{conv}T(C)$ is totally bounded. Its closure is complete and totally bounded, hence compact. Now $T(K)\subset T(C)\subset K$, so the compact-domain result supplies a fixed point in $C$.

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.

The triangular-grid form is Sperner's lemma: subdivide a triangle into small triangles meeting edge to edge, label its vertices $1,2,3$, and forbid label $i$ on the side opposite corner $i$. The corner labels are thereby forced. An odd number of small triangles, in particular at least one, carry all three labels.

Here is the parity proof. Count incidences of a small triangle with an edge whose endpoint labels are $1,2$. On the outer boundary these edges occur only on the side joining corners $1,2$; traversing that side begins with label one and ends with label two, so the number of changes of label is odd. Interior edges contribute twice to the incidence count. A small triangle with all three labels contributes once; a triangle having only labels one and two contributes twice if it has both, and all other triangles contribute zero. Reduction modulo two therefore proves the assertion.

To obtain the continuous Brouwer fixed-point theorem on the triangle, use barycentric coordinates and a continuous map $F$ from the triangle to itself. If a grid vertex is not already fixed, label it by an index $i$ with $x_i>F_i(x)$, which exists since both coordinate sums are one. A point on the side $x_i=0$ cannot receive label $i$, so the discrete lemma applies. On successively finer grids take a fully labelled small triangle. Compactness gives a subsequence converging to one point $x$; all three vertices converge to it because the mesh tends to zero. Passing to the limit in the three label inequalities gives $x_i\ge F_i(x)$ for every $i$. Equal coordinate sums force equality throughout, hence $F(x)=x$.

A closed disk is homeomorphic to a triangle: radial coordinates from an interior point, scaled by the distance to the boundary in each direction, give such a homeomorphism. Conjugation therefore gives the same fixed-point assertion on the disk. If $r$ were a continuous retraction from the unit disk to its boundary, define $F(x)=-r(x)$. A fixed point would have $|x|=1$, hence $r(x)=x$ and $x=-x$, impossible on the unit circle. Thus no continuous disk-to-boundary retraction exists. The continuous conclusion has been deduced from the grid parity argument, not used to prove it.

The Brouwer fixed-point theorem says that every continuous map from a closed Euclidean ball to itself has a fixed point. For the closed disc $D\subset {\mathbb{R}}^2$ its equivalent no-retraction theorem says that there is no continuous $r:D\to \partial D$ satisfying $r(z)=z$ on $\partial D$.

Assume the fixed-point theorem and suppose such a retraction exists. Then $F(x)=-r(x)$ maps $D$ continuously into itself. A fixed point must lie on the boundary, where $r(x)=x$, giving $x=-x$, impossible on the unit circle.

Conversely suppose $F:D\to D$ is continuous and has no fixed point. From $F(x)$ draw the ray through $x$ and let $r(x)$ be its exit point on the circle. To check continuity rather than merely relying on the picture, put $v=x-F(x)\ne 0$. The exit point is $F(x)+t(x)v$, whereThis is continuous, lies on the circle and has $t(x)\ge 1$. On the boundary the exit point is $x$, so $r$ is a retraction, contradicting the assumed theorem. Hence the two disc formulations are equivalent; the same argument works for a ball in any dimension.

Let $J$ be a finite collection of resources with positive integer capacities $C_j$, and let $R$ be a finite collection of nonempty routes. In the standard unit-resource loss network, a call on route $r$ needs one unit simultaneously at each resource in $r$; fixed routing means that these requirements do not change when a request is rejected. Suppose route $r$ has an independent Poisson process of arrivals of rate ${\nu }_r$ and independent holding times of mean $1/{\mu }_r$. Its offered traffic is ${\alpha }_r={\nu }_r/{\mu }_r$. The Erlang fixed point approximation approximates resource blocking events as independent, rather than claiming this independence for the exact loss network.

Write $b_j$ for the approximate blocking probability at resource $j$. The reduced-load approximation screens a call by every other resource that it needs, so the offered traffic seen by resource $j$ isFor a single resource with offered traffic $a$ and capacity $C$, the Erlang loss formula isThus the Erlang fixed point approximation is the systemThe resulting route acceptance probability is ${\prod }_{j\in r}(1-b_j)$. In particular, the screening product for $a_j$ excludes $j$ itself; including it would describe a different approximation.

For existence, define $T_j(b)=E(a_j(b),C_j)$ on $[0,1{]}^J$. Each $a_j$ is a continuous polynomial, and the denominator of the Erlang loss formula is at least one. Consequently $T$ is continuous and maps this nonempty compact convex set into itself. The Brouwer fixed-point theorem supplies a fixed point. At any such point $b_j<1$, since finite offered traffic and positive capacity give $E(a,C)<1$.

For uniqueness, we construct a strictly convex function whose minimizer is equivalent to an Erlang fixed point. Put $y_j=-\log (1-b_j)$. First establish the single-resource facts needed for this change of coordinates. For $a>0$, the inverse blocking probability has the expressionThe terms with $k<C$ decrease strictly with $a$, so $E(a,C)$ increases strictly from zero to one. Let $a_C(y)$ be its inverse parametrization defined by $E(a_C(y),C)=1-e^{-y}$, with $a_C(0)=0$. If $K$ has the truncated Poisson distribution with weights proportional to $a^k/k!$, then shifting the sum gives its mean, the carried load of an Erlang loss resource, asDifferentiating the normalized weights showsThe variance is positive because both $K=0$ and $K=1$ have positive probability. Moreover, $m_C(a)\to C$ as $a\to \infty$, since the largest term in the truncated Poisson distribution dominates. Henceis continuous, strictly increasing from zero to $C$.

Now minimize the following convex potential for the Erlang fixed point over the nonnegative orthant:Each exponential term is convex, and each integral is a strictly convex function of its coordinate because its derivative $U_{C_j}$ is strictly increasing. Their sum is therefore strictly convex. It is also a coercive function: once $z$ is large enough, $U_{C_j}(z)\ge C_j/2$, so each integral grows at least linearly in its coordinate, while the exponential terms are nonnegative. Thus $F$ attains exactly one minimum.

The coordinate derivative isAt an interior coordinate of a minimizer it vanishes. At $y_j=0$, it is nonpositive, whereas the one-sided minimum condition requires it to be nonnegative; it therefore vanishes there too. Conversely, a zero gradient is a global minimizer of this convex function. Multiplying ${\partial }_{j}F=0$ by $e^{y_j}$ givesApplying the Erlang loss formula converts this exactly to $b_j=E(a_j(b),C_j)$. The change of coordinates is one-to-one, proving uniqueness under fixed routing. A zero-capacity resource can instead be fixed as blocked with probability one, with all routes using it removed before applying the proof; an unused positive-capacity resource has blocking probability zero.

For alternative routing, consider a symmetric three-node loss network. Each of its three links has capacity $1000$; every node pair offers traffic $950$, tries its direct link first, and, if that link is blocked, tries the two-link path through the third node. Let all approximate link blocking probabilities equal $b$. Each link receives its direct offered traffic and the overflow from the other two node pairs. An overflow is generated with probability $b$ and reaches the link under consideration only when its other required link accepts, with probability $1-b$. The symmetric reduced-load approximation therefore givesFor $g(b)=E(950[1+2b(1-b)],1000)-b$, evaluate the Erlang loss formula by the stable recursion $E(a,0)=1$, $E(a,k)=aE(a,k-1)/(k+aE(a,k-1))$. It givesThese signs can also be certified with exact rational arithmetic, since all four loads are rational. Continuity and the intermediate value theorem give three distinct roots, one in each intervening interval. Thus the alternative-routing approximation need not be unique. This is multiplicity of approximate fixed points, not multiplicity of stationary distributions for the exact finite irreducible Markov chain.

The two-dimensional Brouwer fixed-point theorem states that every continuous map from a closed filled triangle to itself has a fixed point. Use the triangle $\Delta =\{x\in {\mathbb{R}}^3:x_i\ge 0,{\sum }_i{x}_i=1\}$, which lies in a two-dimensional affine plane. Strict positivity of the matrix entries ensures $(Ax{)}_i>0$ for every $x\in \Delta$. Thereforeis a continuous self-map of $\Delta$, in fact into its interior. Its fixed point $x$ satisfies $Ax=\lambda x$, where $\lambda ={\sum }_i(Ax{)}_i>0$. Since $x=T(x)$ lies in the interior, $A$ has an eigenvector all of whose entries are strictly positive. This is positive matrix eigenvector from simplex normalization; it uses strict positivity, not merely nonnegative entries.

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.

The planar Brouwer fixed-point theorem says that every continuous map from the closed disk $D$ to itself has a fixed point. If a retraction $r:D\to \partial D$ existed, $x\mapsto -r(x)$ would be a continuous self-map of $D$. At a fixed point $x=-r(x)$, $x$ lies on the boundary, where $r(x)=x$; this would give $x=-x$, impossible on the unit circle.

Conversely, suppose a continuous self-map $f:D\to D$ has no fixed point. For each $x$, take the ray starting at $f(x)$ through $x$, and let $r(x)$ be its exit point on $\partial D$. The denominator $|x-f(x)|$ never vanishes, and solving the quadratic intersection with the circle shows that the exit point depends continuously on $x$. More explicitly, with $v=x-f(x)$,The ray exits beyond or at $x$, so $\lambda \ge 1$, and when $x\in \partial D$ its exit point is $x$. Hence $r$ is a retraction. Nonexistence of a boundary retraction and the planar fixed-point theorem are equivalent.

Define the continuous map $T(z)=-(z^6-z^5+2z^2+1)/6$. For $|z|\le 1$, the triangle inequality gives $|T(z)|\le (1+1+2+1)/6=5/6$. Thus $T$ maps the closed unit disk into itself. By the Brouwer fixed-point theorem, some $z$ has $T(z)=z$, which rearranges toThis supplies the requested existence and even places a root strictly inside the unit disk.