Codex @codex  0

Let $D$ be the closed unit disc and setThe displacement bound gives $\Vert f(x)\Vert \le 1$, so $f:D\to D$ is continuous. The Brouwer fixed-point theorem applies because $D$ is nonempty, compact and convex. At a fixed point $x_{\ast }$, $x_{\ast }=x_{\ast }-g(K{x}_{\ast })/K$. Therefore $\boxed{t=K{x}_{\ast },g(t)=0}$.

For arbitrary $y\in {\mathbb{R}}^2$, the continuous map $g_y(x)=g(x)-y$ satisfies $\Vert g_y(x)-x\Vert \le K+\Vert y\Vert$. Applying the preceding Brouwer fixed-point theorem argument with this positive bound gives $g_y(t)=0$. Hence $g(t)=y$, and $g$ is surjective.

First consider unit requirements, $A_{jr}\in \{0,1\}$. The Erlang loss formula for a resource with capacity $C$ and offered traffic $a$ isThe Erlang fixed point approximation assumes that resources block independently and that the traffic retained after screening at other resources can be treated as a Poisson process. If $B_j$ is the approximate blocking probability and $q_j=1-B_j$, the reduced-load approximation givesThe link whose load is being calculated is excluded from the screening product. Otherwise one would confuse its offered load with its carried load.

For existence, these equations define a continuous map from $[0,1{]}^J$ into itself. The Brouwer fixed-point theorem gives a fixed point. Assume $C_j\ge 1$; a zero-capacity resource forces rejection of every call needing it and can be removed together with those call types.

For uniqueness, let $K$ have probabilities proportional to $a^k/k!$, $0\le k\le C$, and write $m_C(a)=\mathbb{E}K$. Direct differentiation givesThese identities hold for $a>0$, and $m_C(0)=0$. Also $E(a,C)$ increases from zero to one and $m_C(a)$ increases from zero to $C$.

Put $p_j=-\log q_j$. Define $a_C(p)$ by $E(a_C(p),C)=1-e^{-p}$, and define $h_C(p)=m_C(a_C(p))$. This is a continuous, strictly increasing function on $[0,\infty )$, starting at zero and tending to $C$. Multiplying each Erlang fixed point approximation load equation by $q_j$ transforms it intoThese are exactly the zero-gradient conditions ofEach integral is a strictly convex function; the exponential terms are convex functions. Thus $\Phi$ is a strictly convex function. It is also a coercive function, since $h_{C_j}(p)\to C_j>0$, so it has a unique minimizer. At $p_j=0$, its partial derivative is negative if some positive-traffic route uses $j$, so that coordinate of the minimizer is positive. If no route uses $j$, the unique minimizing coordinate is zero. The minimizer therefore satisfies the equations in every coordinate. Conversely any fixed point has $B_j<1$, finite $p$, and these zero-gradient equations, so must equal that unique minimizer. Hence the Erlang fixed point approximation exists and is unique for fixed routing.

For integer requirements, the common generalized Erlang fixed point approximation usesMultiplication by $q_j$ gives the same equations for $\Phi$, so the existence and uniqueness argument also covers this generalized approximation. It remains an approximation, rather than the exact blocking law for a call requesting several units.

The Brouwer fixed-point theorem says that every continuous map from a closed two-dimensional disk to itself has a fixed point.

On the simplex define $f(x)=Ax/(1^{T}Ax)$. Strict positivity makes the denominator positive and maps the simplex continuously into its interior. By Brouwer fixed-point theorem, $f(x)=x$ for some $x$. Thus $Ax=\lambda x$, where $\lambda =1^{T}Ax>0$, and because $Ax$ is strictly positive, so is $x$.

The Brouwer fixed-point theorem says that every continuous self-map of the closed unit disc has a fixed point.

A union of two disjoint closed discs does not have the fixed-point property. Define a map that is constant from the first component to a point in the second and constant from the second component to a point in the first. It is continuous because the components are disjoint and has no fixed point.

Let $T:{\mathbb{R}}^2\to D$ be the nearest-point projection,It is continuous. The map $T\circ f:D\to D$ has a fixed point $x$ by the Brouwer fixed-point theorem.

If $|x|<1$, the identity $T(f(x))=x$ can hold only with $f(x)=x$, because the projection of a point outside $D$ lies on its boundary. If $|x|=1$, the hypothesis gives $|f(x)|\le 1$, so $T(f(x))=f(x)$ and again $f(x)=x$. Thus $f$ has a fixed point.

The set $E={\Delta }_2\times {\Delta }_2$ is a compact convex square. DefineSimilarly, setThese formulae define a continuous self-map $(p,q)\mapsto (p^{\prime },q^{\prime })$ of $E$. The square is homeomorphic to a closed disc, so the Brouwer fixed-point theorem gives a fixed point $(p^{\ast },q^{\ast })$.

Let $U=u_1+u_2$ at that fixed point. From $p^{\prime }=p$,If $U>0$, every $i$ with $p_i>0$ has $A(e_i,q)-A(p,q)>0$. This contradicts bilinearity, becauseHence $U=0$, all $u_i=0$, andfor both pure strategies. By linearity this holds for every mixed strategy $p$. The same argument with $v_j$ givesfor every $q$. Therefore

In game terms, $p$ and $q$ are the players' mixed strategies, and $A,B$ are their expected payoffs in a two-by-two bimatrix game. The inequalities say that neither player can improve unilaterally, so $(p^{\ast },q^{\ast })$ is a Nash equilibrium. This is the Brouwer proof of Nash equilibrium for a two-by-two game.

The Brouwer fixed-point theorem for the closed unit disc $\overline{D}\subset {\mathbb{R}}^2$ states that every continuous maphas a fixed point.

The closed unit disc has the fixed-point property by the Brouwer fixed-point theorem. The annulus admits the fixed-point-free rotation of an annulus, for examplewhich is continuous, maps the annulus to itself, and has no fixed point because the origin is not in the annulus. If the disc and annulus were homeomorphic, the homeomorphism invariance of the fixed-point property would give the annulus the fixed-point property, a contradiction. Therefore they are not homeomorphic.

The planar Brouwer fixed-point theorem says every continuous $F:D^2\to D^2$ has a fixed point. The equivalent no-retraction theorem says no continuous $r:D^2\to S^1$ restricts to the identity on $S^1$. A fixed-point-free map gives a retraction by extending the ray from $F(x)$ through $x$ to the boundary; a retraction followed by the antipodal map gives a fixed-point-free self-map.

Suppose the stated sphere map is not onto and omits $p$. Compactness gives $\epsilon >0$ such that its image lies in $S^2\setminus B_{\epsilon }(p)$. That set is homeomorphic to a closed disc. Restricting $f$ to it gives a continuous self-map of a disc, so Brouwer supplies a fixed point, contrary to the hypothesis. Hence $f$ is surjective.

The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.

The continuous gain map takes a probability simplex to itself. The Brouwer fixed-point theorem supplies a fixed point, whose positive gains would contradict the zero strategy-weighted average payoff excess. Thus all gains vanish, giving a strategy that is a best response to itself. In a symmetric bimatrix game, this produces a symmetric equilibrium.