Fixed point theorems

Question

It is surprising that fixed point theorems (FPTs) appear in so many different contexts throughout Mathematics: Applying Kakutani's FPT earned Nash a Nobel prize; I am aware of some uses in logic; and of course everyone should know Picard's Theorem in ODEs. There are also results about local and global structure OF the fixed points themselves, and quite some famous conjectures (also labeled FPT for the purpose of this question).

Many results are so far removed from my field that I am sure there are plenty of FPTs out there that I have never encountered. I know of several, and will post later if you do not beat me to them :)

Community wiki rules apply. One FPT per answer, preferably with an inspiring list of interesting applications.

Answer 1

The Lefschetz Fixed Point Theorem is wonderful. It generalizes the Fixed Point Theorem of Brouwer, and is an indispensable tool in topological analysis of dynamical systems.

The weakest form goes like this. For any continuous function $f:X \to X$ from a triangulable space $X$ to itself, let $H_\ast f:H_\ast X\to H_\ast X$ denote the induced endomorphism of the Rational homology groups. If the alternating sum (over dimension) of the traces

$$\Lambda(f) := \sum_{d \in \mathbb{N}}(-1)^d\text{ Tr}(H_df)$$

is non-zero, then $f$ has a fixed point! Since everything is defined in terms of homology, which is a homotopy invariant, one gets to add "for free" the conclusion that any other self-map of $X$ homotopic to $f$ also has a fixed point.

When $f$ is the identity map, $\Lambda(f)$ equals the Euler characteristic of $X$.

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Update: Here is a lively document written by James Heitsch as a tribute to Raoul Bott. Along with an outline of the standard proof of the LFPT, you can find a large list of interesting applications.

Answer 2

Lawvere's fixed point theorem. If $f \colon A \to Y^A$ is a surjective morphism in a Cartesian closed category, then any $t \colon Y \to Y$ has a fixed point.

(Surjectivity is a technical term, which basically means that any $g \colon A \to Y$ equals $f(a)$ on points for some point $a$ of $A$. See here)

Applications: Cantor's diagonal argument, Turing's halting problem, Russell's paradox, Gödel's incompleteness theorem, Tarski's incompleteness theorem, Rice's theorem, and many more, see here.

Answer 3

Knaster-Tarski's fixed point theorem: If $L$ is a complete lattice and $f:L \rightarrow L$ is order preserving, then the set of fixed points of $f$ form a (non-empty) complete lattice.

Answer 4

Let $p$ be a prime and let $G$ be a finite $p$-group which acts on a finite set $X$. Suppose that $p$ does not divide $|X|$ then this action has a fixed point.

This has many applications, e.g. the proof of the fact that Sylow subgroups are conjugated.

Answer 5

Euler's Theorem, that every non-trivial rotation $R$ of 3-space has a unique axis. It really just says that $R$ acting on the space of lines through the origin has a unique fixed point.

(Added April 11, 2013) I just received my copy of the latest issue of The Journal of Fixed Point Theory and its Applications (Vol.12, Nos. 1--2) and starting on page 27 there is an article with the title "Chasles' fixed point theorem for Euclidean motions". Chasles' theorem is a generalization of Euler's Theorem; it says that every orientation preserving Euclidean motion of 3-space that is not a pure translation is a "twist" or "screw motion", that is, a rotation about some unique line (NOT necessarily through the origin) called the axis followed by a translation that is parallel to the axis. I really should have given this as my example rather than Euler's Theorem, since as I said it is more general. And I have no excuse for not recalling it since the authors of that paper are myself and my son Bob.

Answer 6

Brouwer's FPT: Every continuous function from a closed ball in $\mathbb{R}^n$ to itself has a FP.

For applications see this question.

Answer 7

Ryll-Nardzewski FPT: If $K$ is a nonempty weakly compact convex subset of a Banach space, then every semigroup of affine isometries of $K$ has a common fixed point.

This implies the existence of Haar measures on compact groups.

Answer 8

Suppose that $S$ is a finite set with an odd number of elements. Then every involution $f:S\rightarrow S$ has a fixed point.

Application: Every prime of the form $p=4m+1$ may be written as a sum of two squares. The result above is used on p.20 here (Wayback Machine).

Also, although not of the usual fixed point theorem form, is something I call the fixed point factor theorem. If $f:\mathbb{C}\rightarrow \mathbb{C}$ is a polynomial and $n>1$, then $f(x)-x$ is a factor of $f^n (x)-x$ for natural $n$. This has a very obvious generalisation...

Answer 9

There is the Bruhat-Tits theorem that a group acting by isometries on a CAT(0) space with a bounded orbit has a fixed point. This is often applied to compact subgroups of groups acting on Euclidean buildings.

Answer 10

The Banach fixed-point theorem (or contraction mapping principle) was already mentioned by Rodrigo A. Pérez, but I would like to stress another application. The principle says that a contraction of a complete metric space $(X,d)$ (namely, a continuous function $f:X\to X$ such that $d\big(f(x),f(y)\big)\leq \rho d(x,y)$ for each $x,y\in X$ where $\rho<1$ is some positive constant depending on $f$ only) has a unique fixed point.

In his milestone 1981 paper Fractals and Self Similarities, (Indiana Univ. Math. J., vol. 30, n. 5) J. Hutchinson axiomatized the relation between fractals and collections of contractions of $\mathbb{R}^n$. He showed that for each set $\mathscr{S}=\{S_1,\dots,S_N\}$ of contractions $S_i\colon\mathbb{R}^n\to\mathbb{R}^n$, there exists a unique closed, bounded set $K$ such that $$ K=\bigcup_{i=1}^N S_i(K)\;. $$ Such fixed closed sets are "fractals" in a very natural way. For instance, the Koch curve can be obtained in $\mathbb{R}^2$ by using two contractions (see p. 729 of Hutchinson's work), as well as the Cantor set - for this, take $\mathscr{S}=\{S_1,S_2\}$ with $$ S_1(x)=\frac{x}{3}\quad\text{and}\quad S_2(x)=\frac{x}{3}+\frac{2}{3}\;. $$ The three-line proof of the existence of $K$ is an application of the contraction mapping principle (and is Theorem 1 on p. 728 of Hutchinsons's work) and goes as follows: let, as before, $n\geq 1$ and $\mathscr{S}=\{S_1,\dots,S_N\}$ be contractions of $\mathbb{R}^n$. Let $\mathscr{B}$ be the set of all closed bounded subsets of $\mathbb{R}^n$ and, for two bounded closed $A,B\in\mathscr{B}$, let $\delta(A,B)=\sup \{d(a,B),d(b,A):a\in A,b\in B\}$. This turns $(\mathscr{B},\delta)$ into a complete metric space for which $$ \mathscr{S}:A\mapsto \bigcup _{i=1}^{N}S_i(A) $$ is a contraction. Hence, there is a unique fixed point $K\in\mathscr{B}$. Needless to say, one can replace $\mathbb{R}^n$ with any other complete metric space without affecting the proof.

Answer 11

Kleene's Second Recursion Theorem If $F$ is a total computable function then there is an index $e$ such that $\{e\} \simeq \{F(e)\}$.

This has many applications such as effective transfinite recursion.

Answer 12

Here is a teeny tiny toy version of the Lefschetz fixed point theorem: let $f : S \to S$ be an endomorphism of a finite set and let $K[f] : K[S] \to K[S]$ be the induced linear map on free vector spaces. Then $\text{tr}(K[f])$ is the number of fixed points of $f$. This is one way to prove Burnside's lemma.

Answer 13

I forgot who proved it, but the statement is nice and very easy to prove: A function $f:X\to X$ is fixed point free if and only if there is a partition of $X$ into three subsets s.t. $f$ maps each of the three subsets into the union of the other two. An immediate application is that if $f$ is fixed point free on a set $X$ then so is its continuous extension to a function on $\beta X$.

Answer 14

The infinite dimensional version of Brouwer's FPT is Schauder's FPT. If $K$ is a non-void closed convex subset of a TVS, and $f:K\rightarrow K$ is compact ($f$ is continuous and $f(K)$ is compact), then $f$ has a fixed point.

It has numerous applications in nonlinear analysis. One of the earliest being the existence of a solution to the stationnary Navier-Stokes equations with Dirichlet boundary condition, proven by J. Leray.

Answer 15

One of the most awesome fixed-point theorems I know of is due to Pataraia:

• If $L$ is a poset with a bottom element and with joins of directed subsets, then every monotone function $f: L \to L$ has a (least) fixed point.

It is a strengthening of the Knaster-Tarski theorem, and is somewhat reminiscent of the Bourbaki-Witt theorem, but is entirely constructive. Related discussion at the n-Category Café here.

Answer 16

Allow me to mention another version of the Lefschetz fixed-point theorem. If $X$ is a (say smooth projective, though this works in greater generality) variety over $\mathbb F_q$ of dimension $d$, then \begin{equation} \left|X\left(\mathbb F_{q^n}\right)\right| = q^d \sum_i (-1)^i \mathrm{tr}\left(\Phi_{q^n} : H_{et}^i(\bar X,\mathbb Q_\ell)\right) \end{equation} where $\ell$ is prime to $q$ and $\Phi_{q^n}$ is the geometric Frobenius.

As a corollary one gets the rationality of the zeta-function of $X$.

(Note that this actually is a fixed-point theorem. $X(\mathbb F_{q^n})$ is just the set of fixed points under $\Phi_{q^n}$ applied to $X$.)

Answer 17

Kakutani's FPT: Let $S$ be a non-empty, compact, convex subset of $\mathbb{R}^n$, and $\varphi:S \longrightarrow 2^S$ a set-valued function with a closed graph and the property that $\varphi(x)$ is non-empty and convex for all $x \in S$. Then $\varphi$ has a fixed point.

Application: Consider a game with finitely many players and finitely many strategies. If players are allowed to choose mixed strategies, there is always a Nash equilibrium; that is, a set of strategy choices for all players such that no player can do better by unilaterally switching to a different strategy. This is the theorem that resulted in J. Nash getting the 1994 Nobel Prize in Economics.

Answer 18

The Fiber contraction theorem due to Hirsch and Pugh:

Let $F: E \to E$ be a mapping on the fiber bundle $\pi: E \to B$ covering $f: B \to B$, where $B$ is a topological space and the fibers $Y$ of $E$ are complete metric spaces. Let $f$ have a globally attractive fixed point $b \in B$ and the fiber mapping is a uniform contraction in a neighborhood $\pi^{-1}(U), b \in U \subset B$ (and thus there exists a unique fixed point $e = (b,y) \in \pi^{-1}(b)$), and $b \mapsto F(b,y)$ be continuous. Then $e$ is the unique, globally attracting fixed point of $F$.

This result is an extension of the Banach fixed point theorem that can be used to prove e.g. the existence of center manifolds and normally hyperbolic invariant manifolds. It is specifically useful when one cannot find a contraction on an space of $C^k$ functions, but can construct inductively a contraction on the $k$-th jet when the $k-1$ jets are known to converge to a fixed point.

Answer 19

I thought this result was a bit interesting. Mahlon M. Day in the paper [1] showed that the amenable groups are precisely the groups where there Markov-Kakutani theorem holds.

If $(X,\mathcal{M})$ is an algebra of sets, then a function $\mu:\mathcal{M}\rightarrow[0,1]$ is said to be a finitely additive probability measure if $\mu(\emptyset)=0,\mu(X)=1$ and $\mu(A\cup B)=\mu(A)+\mu(B)$ whenever $A,B\in\mathcal{M}$ and $A\cap B=\emptyset$. If $G$ is a group, then a finitely additive probability measure $\mu:P(G)\rightarrow G$ on the algebra of sets $(G,P(G))$ is said to be left-invariant if $\mu(aR)=\mu(R)$ for each $R\subseteq G$.

A group $G$ is said to be amenable if there exists a left-invariant finitely additive probability measure $\mu:P(G)\rightarrow[0,1]$. For example, every finite group is amenable, and every abelian group is amenable. Furthermore, the class of amenable groups is closed under taking quotients, subgroups, direct limits, and finite products.

Let $C$ be a convex subset of a real vector space. Then a function $f:C\rightarrow C$ is said to be an affine map if $f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)$ for each $\lambda\in[0,1]$ and $x,y\in C$.

$\textbf{Theorem}$(Day) Let $G$ be a group. Then the following are equivalent.

1. $G$ is amenable.

2. Let $X$ be a Hausdorff topological vector space and let $C\subseteq X$ be a compact convex subset. Let $\phi:G\rightarrow C^{C}$ be a group action such that each $\phi(g)$ is a continuous affine map. Then there is a point in $C$ fixed by every element of $G$.

3. Let $X$ be a locally convex topological vector space and let $C\subseteq X$ be a compact convex subset. Let $\phi:G\rightarrow C^{C}$ be a group action such that each $\phi(g)$ is a continuous affine map. Then there is a point in $C$ fixed by every element in $G$.

[1] Fixed-point theorems for compact convex sets. Mahlon M. Day.Illinois J. Math. Volume 5, Issue 4 (1961), 585-590.

[2] Ceccherini-Silberstein, Tullio, and M. Coornaert. Cellular Automata and Groups. Heidelberg: Springer, 2010.

Answer 20

Banach's FPT (or contraction FPT): Every contraction in a complete metric space has a unique FP.

Application: If $f(t,y(t))$ is a real-valued function, Lipschitz continuous in $ y$ and continuous in $t$, then the initial value problem $$y'(t) = f(t,y(t)),\quad y(t_0)=y_0,\quad t \in [t_0-\varepsilon,t_0+\varepsilon]$$ has a unique solution.

Answer 21

The Arithmetic fixed point theorem (see also MO/30874) states that if $F$ is a formula in number theory with only one free variable $v$, then there is a sentence $A$ such that number theory can prove $A \Leftrightarrow F_v(\underline{[A]})$. An immediate application is Gödel's Theorem.

Answer 22

Alexander Abian (1923-1999) proved around 1998 the following result he named "the most fundamental fixed-point theorem". "Let F be a mapping from a set A into itself. Let G(x,0)=x, G(1,x)=F(x), G(2,x)= F(F(x)) be the iterates values of the function F for the argument x in A. Then F has a fixed point if and only if: there exists an element x of A such that, for every ordinal v, G(v,x) is an element of A and if G(v) is not a fixed point of A then G(u,x)'s are all distincts elements of A for u∈v." Details can be found at http://us2.metamath.org:88//abianfp.html Gérard Lang

Answer 23

The main theorem of Smith theory asserts that if a $p$-group $G$ acts on a mod-$p$-acyclic space $X$ (which must also be 'finitistic', a fairly weak condition), then the fixed point set $X^G$ is also mod-$p$ acyclic; in particular, it is non-empty.

This is especially useful because $X$ is not assumed to be compact, as is the case for the Lefschetz fixed point theorem, say.

Answer 24

Arnold's Conjecture: A Hamiltonian map on a compact symplectic manifold $(M,\omega)$ has at least as many fixed points as a function on $M$ has critical points.

Answer 25

Another one, from MR0151632 Michel Hervé: Several complex variables. Local theory. Published for the Tata Institute of Fundamental Research, Bombay by Oxford University Press, London 1963 vii+134 pp.

Let $G$ be an open and connected set of the affine space $X$. If the image $f(G)$ under a holomorphic map $f: G \to G$ is relatively compact in $G$, then $f$ has a unique fixed point.

The proof uses Montel theorem and the fact that every analytic and compact subset of an affine space must be finite.

Answer 26

The Caristi fixed point theorem is a generalisation of the Banach fixed point theorem.

Theorem. Let $(X, d)$ be a complete metric space. Let $T : X \rightarrow X$ and $f : X → [0, +∞)$ be a lower semicontinuous function from $X$ into the non-negative real numbers. Suppose that, for all points $x$ in $X$,

$$d \big( x, T(x) \big) \leq f(x) - f \big( T(x) \big).$$

Then $T$ has a fixed point in $X$.

Take $f(x) = \sum_{k \in N} d(T^{k+1}(x),T^{k}(x))$ to recover the Banach fixed point theorem.

Answer 27

I really like the following result, which allows one to drop the usual compactness assumption.

Okhezin's theorem¹: For a polyhedron $K$ and a continouous map $f\colon K\to K$ at least one of the following conditions is true:

• $f$ has a fixed point;
• $f$ is not nullhomotopic;
• $K$ contains a closed subset homeomorphic to $[0,\infty)$ (a closed ray).

Since $[0,\infty)$ is an absolute retract without fixed point property, no polyhedron containing it as a closed subset has the fixed point property. This gives the following corollary.

Corollary (Okhezin): A contractible polyhedron has the fixed point property if and only if it is rayless, i.e. contains no closed subset homeomorphic to $[0,\infty)$.

This was not noticed by Okhezin, but the following stronger result is implied.

Corollary: An acyclic polyhedron has the fixed point property if and only if it is rayless.

Proof: As noted above, the "only if" part is obvious. For the "if" part, let $f\colon K\to K$ be a self-map of an acyclic, rayless polyhedron. The suspension $SK$ is a contractible, rayless polyhedron. Thus, by the results of Okhezin, the map $\tilde{f}\colon SK\to SK$ that extends $f$ and swaps the two added cones has a fixed point, which must also be a fixed point of $f$.

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Okhezin also proved some fixed point theorems that apply to other classes of rayless spaces, including some Lefschetz-type results.

¹Okhezin, Vladimir P., On the fixed-point theory for noncompact maps and spaces. I, Topol. Methods Nonlinear Anal. 5, No. 1, 83-100 (1995). DOI: 10.12775/TMNA.1995.005, projecteuclid; ZBL0917.54046, MR1350346.

Answer 28

Let $X$ be a nonempty compact Hausdorff space, and $f\colon X\to X$ be continuous. Denote by $\mathcal P(X)$ the powerset of $X$. Then the function $f^+\colon\mathcal P(X)\to\mathcal P(X)$ defined by $f^+(A)=f[A]$ has a fixed point $f^+(A)=A$, where $A\subseteq X$ is nonempty and closed.

Answer 29

It would be a pity not to mention the work of F. Browder, in particular his study of non linear pde's, the main tool being FPT's on Banach spaces. This is documented in many of his publications, perhaps most memorably in his "Nonlinear operators and nonlinear equations of evolution".

Answer 30

Let $p:E\rightarrow B$($B$ is locally path wise connected) be a covering map then every isomorphism $h:E\rightarrow E$(isomorphism between covering spaces)is called automorphism and the set of automorphisms of $E$ relative to $p$ has a group structure and is shown with $A(E,p)$,now if $f\in A(E,p)$ has a fixed point then $f=I_{E}$.