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Brouwer's fixed-point theorem says that every continuous self-map of a nonempty closed Euclidean ball has a fixed point. It does not promise convergence of repeated application. For example $f(x)=1-x$ with starting point $x_1=0$ gives the nonconvergent alternating sequence $0,1,0,1,\dots$, although $1/2$ is a fixed point.

For an invariant ball from sublinear growth, chooseThen $A{R}^{\alpha }\le R/2$ and $B\le R/2$. For every $|x|\le R$ the given estimate yields $|f(x)|\le A{R}^{\alpha }+B\le R$. Thus $f$ restricts to a continuous self-map of the closed ball of this radius, and Brouwer gives a fixed point. The exponent $\alpha <1$ ensures that the linear radius eventually dominates its sublinear growth; no differentiability or contraction estimate is needed.

The Brouwer fixed-point theorem says that every continuous self-map of a closed ball in ${\mathbb{R}}^n$ has a fixed point.

Suppose $g$ has no zero in ${\overline{B}}_R$. Thenis a continuous map from ${\overline{B}}_R$ into its boundary, hence into ${\overline{B}}_R$. The Brouwer fixed-point theorem gives $T(x_0)=x_0$. This forces $|x_0|=R$ andcontradicting the assumed boundary inequality. Therefore $\boxed{g(x_0)=0\text{ for some }|x_0|\le R}$. This is the outward vector field zero theorem.

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.

For a strictly positive real matrix $A$, the map $x\mapsto Ax/{\sum }_i(Ax{)}_i$ is a continuous self-map of the probability simplex. A Brouwer fixed-point theorem fixed point satisfies $Ax=\lambda x$ with $\lambda >0$. The image lies in the interior of the simplex, so every entry of this eigenvector is strictly positive. Nonnegative entries and nonzero columns alone do not imply this strict conclusion.

Every continuous map from a nonempty compact convex set $K$ in a Banach space to itself has a fixed point. A useful equivalent form takes a nonempty closed bounded convex set $C$ and a continuous map $T:C\to C$ whose image is relatively compact. To obtain this version from the compact-domain version, restrict $T$ to $\overline{\mathrm{conv}}T(C)\subset C$ and use closed convex hull of a compact set.

For a proof on $K$, the Schauder approximation lemma gives continuous maps $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with finite-dimensional $L_{\epsilon }$ and $\Vert T-T_{\epsilon }{\Vert }_{\infty }\le \epsilon$. The Brouwer fixed-point theorem on the nonempty compact convex set $K\cap L_{\epsilon }$ gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, so $\Vert T{x}_{\epsilon }-x_{\epsilon }\Vert \le \epsilon$. A convergent subsequence as $\epsilon \downarrow 0$ and the continuity of $T$ give an exact fixed point.

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.