What Is a Fixed-Point Problem?
A fixed-point problem asks whether a map sends some point to itself. This question connects an equation to the behavior of a function on its whole domain: instead of solving an arbitrary equation \(g(x)=0\), we seek a point \(x\) satisfying \(f(x)=x\). The general-topology results from the previous tutorial provide useful tools for studying such points, but existence depends on the map and the space. Even a continuous map on a compact space need not have a fixed point.
The condition \(f:X\to X\) matters: the value \(f(x)\) must lie in the same space as \(x\), so the equation \(f(x)=x\) makes sense. A fixed-point problem can ask whether \(\operatorname{Fix}(f)\) is empty, whether it contains exactly one point, or how to find or approximate its points. These are different questions. A theorem guaranteeing at least one fixed point does not, by itself, guarantee uniqueness or provide a formula for the point.
A Fixed-Point Theorem for Closed Intervals
On a closed interval, continuity and the requirement that the map take values back into that interval force a fixed point. The proof turns the fixed-point equation into a sign change. At the left endpoint, the image cannot fall to the left of the interval; at the right endpoint, it cannot rise to the right. The Intermediate Value Theorem then supplies a point where the difference between the map and the identity is zero.
Proof. If \(a=b\), then \([a,b]=\{a\}\). Since \(f\) maps this one-point set into itself, \(f(a)=a\), so \(a\) is a fixed point.
Now suppose \(a<b\). Define \(g:[a,b]\to\mathbb{R}\) by \(g(x)=f(x)-x\). The identity function and \(f\) are continuous, so \(g\) is continuous. Because \(f(a)\in[a,b]\), we have \(f(a)\geq a\), and therefore \(g(a)=f(a)-a\geq0\). Because \(f(b)\in[a,b]\), we have \(f(b)\leq b\), and therefore \(g(b)=f(b)-b\leq0\).
If \(g(a)=0\), then \(f(a)=a\). If \(g(b)=0\), then \(f(b)=b\). In either case there is a fixed point. Otherwise \(g(a)>0\) and \(g(b)<0\); the Intermediate Value Theorem gives a point \(c\in[a,b]\) such that \(g(c)=0\). Thus \(f(c)-c=0\), or \(f(c)=c\). This proves the theorem. \(\square\)
Worked Example: A Unique Fixed Point on an Interval
Consider \(f:[0,2]\to[0,2]\) defined by \(f(x)=(x+2)/3\). For \(0\leq x\leq2\), we have \(2/3\leq (x+2)/3\leq4/3\), so the formula does map the interval into itself. It is continuous, and the interval theorem guarantees at least one fixed point.
To find it, solve \(f(x)=x\):
Substitution verifies the solution: \(f(1)=(1+2)/3=1\). The equation has no other solution, so this map has exactly one fixed point. The theorem supplied existence; solving the equation established uniqueness in this example.
Worked Example: More Than One Fixed Point
Define \(f:[0,1]\to[0,1]\) by \(f(x)=x^2\). For every \(x\in[0,1]\), \(0\leq x^2\leq1\), and the map is continuous. The interval theorem applies. Solving the equation gives
Both candidates check directly: \(f(0)=0^2=0\) and \(f(1)=1^2=1\). Thus \(\operatorname{Fix}(f)=\{0,1\}\). This example shows why an existence theorem should not be read as a uniqueness theorem.
Worked Example: A Fixed Point Need Not Be the Only Candidate Initially
Let \(f:[-1,1]\to[-1,1]\) be given by \(f(x)=-x^3\). It maps into the interval because \(-1\leq x^3\leq1\) implies \(-1\leq -x^3\leq1\), and it is continuous. The interval theorem guarantees a fixed point. Solving \(f(x)=x\) gives
For real \(x\), \(x^2+1>0\), so the equation forces \(x=0\). Substitution confirms \(f(0)=0\). Hence the map has exactly one fixed point. In this case, checking the factor \(x^2+1\) rules out any additional real solutions.
Fixed-Point Sets in Hausdorff Spaces
Even when no theorem guarantees that a fixed point exists, the fixed points of a continuous map have a useful topological structure. In a Hausdorff space, the fixed-point set is closed. It may still be empty; being closed describes the set if it exists, not whether the set contains any points.
Proof. The diagonal of \(X\times X\) is \(\Delta=\{(x,x):x\in X\}\). By the diagonal characterization of Hausdorff spaces established earlier in this course, \(\Delta\) is closed in \(X\times X\).
Define \(F:X\to X\times X\) by \(F(x)=(x,f(x))\). Its two coordinate maps are the identity map on \(X\) and \(f\), both of which are continuous. The continuity criterion for maps into a product therefore shows that \(F\) is continuous. A point \(x\in X\) belongs to \(F^{-1}(\Delta)\) exactly when \((x,f(x))\in\Delta\), which holds exactly when \(f(x)=x\). Consequently, \(\operatorname{Fix}(f)=F^{-1}(\Delta)\). Since the preimage of a closed set under a continuous map is closed, \(\operatorname{Fix}(f)\) is closed in \(X\). \(\square\)
Worked Example: A Compact Continuous Map with No Fixed Point
Let \(X=\{z\in\mathbb{C}:|z|=1\}\), the unit circle, and define \(f:X\to X\) by \(f(z)=-z\). If \(|z|=1\), then \(|-z|=|z|=1\), so \(f\) maps \(X\) into itself. Multiplication by \(-1\) is continuous, so \(f\) is continuous. The unit circle is compact, but this map has no fixed point: the equation \(-z=z\) implies \(2z=0\), hence \(z=0\), which does not belong to \(X\).
This example rules out a tempting overgeneralization. Continuity and compactness alone do not guarantee a fixed point on every space. The interval theorem uses the order and endpoints of a real interval, not compactness alone. The closedness theorem is consistent with this example because the empty set is closed.
What a Fixed-Point Argument Must Establish
The examples suggest a useful discipline. First check that the map is a self-map, then check any regularity assumptions, and only then apply an existence result. A formula may be continuous on the real line without mapping a chosen interval into itself. For instance, \(x\mapsto x+1\) is a continuous self-map of \(\mathbb{R}\), but its fixed-point equation \(x+1=x\) has no solution. Thus continuity by itself is not enough, even when the domain and codomain agree.
A second common mistake is to use a fixed-point existence theorem to claim uniqueness. The map \(x\mapsto x^2\) on \([0,1]\) has two fixed points, while \(x\mapsto(x+2)/3\) on \([0,2]\) has one. These conclusions come from solving the equations in those examples; the interval theorem alone only guarantees at least one solution. In later work on contraction mappings, additional structure will provide a different route to existence and uniqueness.
For the interval theorem, the essential mechanism is the sign comparison at the endpoints. The image condition gives \(f(a)-a\geq0\) and \(f(b)-b\leq0\); continuity prevents the difference from changing from one sign to the other without attaining zero. For the closedness theorem, the mechanism is different: Hausdorffness makes the diagonal closed, and continuity makes its preimage closed. Keeping these mechanisms distinct helps identify which hypotheses are doing the work.
Confirm that \(f\) maps the space under consideration into itself, so \(f(x)=x\) is a meaningful equation there.
Decide whether the goal is existence, uniqueness, a description of all fixed points, or a structural property of the fixed-point set.
For a continuous self-map of a closed interval, use endpoint inequalities and the Intermediate Value Theorem; for closedness, use continuity and a Hausdorff target space.
Substitute each candidate into the original equation, and do not infer uniqueness from an existence theorem.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What is the difference between proving that a map has a fixed point and proving that it has exactly one?
- In the fixed-point theorem for a closed interval, why do the values of \(f(x)-x\) at the two endpoints have opposite weak signs?
- Why does the closedness theorem not imply that a continuous self-map of a Hausdorff space has any fixed points?
- Find all fixed points of \(f(x)=1-x\) on \([0,1]\), and verify them by substitution.
- Which hypothesis in the interval theorem fails to apply to the map \(f(x)=x+1\) on \(\mathbb{R}\), and what does its fixed-point equation show?