Fixed Points and the Interval Condition
A fixed point is an input that a map leaves unchanged. For a map from an interval to itself, the restriction on its values has an important consequence: at the left endpoint, the map cannot take a value below that endpoint, and at the right endpoint, it cannot take a value above that endpoint. Continuity then forces the map to agree with the identity somewhere between them.
We use \([a,b]\) with \(a\le b\), equipped with the subspace topology inherited from \(\mathbb{R}\) with its standard topology. A map \(f:[a,b]\to[a,b]\) is continuous with respect to these topologies. A fixed point of \(f\) is a point \(c\in[a,b]\) such that \(f(c)=c\).
The main tool is the standard Intermediate Value Theorem from real analysis. In the form needed here, if \(g:[a,b]\to\mathbb{R}\) is continuous, \(a<b\), and \(g(a)\ge0\ge g(b)\) (or \(g(a)\le0\le g(b)\)), then there is \(c\in[a,b]\) for which \(g(c)=0\). We use this standard theorem without reproving it here.
The Fixed-Point Theorem
Proof. First suppose \(a=b\). The domain and codomain each consist of the single point \(a\), so the condition \(f:[a,a]\to[a,a]\) forces \(f(a)=a\). Thus \(a\) is a fixed point.
Now suppose \(a<b\). Define \(g:[a,b]\to\mathbb{R}\) by \(g(x)=f(x)-x\). This function is continuous: \(f\) is continuous by hypothesis, the identity map \(x\mapsto x\) is continuous, and the difference of continuous real-valued functions is continuous.
Since \(f(a)\in[a,b]\), we have \(f(a)\ge a\), and therefore \(g(a)=f(a)-a\ge0\). Since \(f(b)\in[a,b]\), we have \(f(b)\le b\), and therefore \(g(b)=f(b)-b\le0\). By the Intermediate Value Theorem, there is \(c\in[a,b]\) such that \(g(c)=0\). The equation \(g(c)=0\) means \(f(c)-c=0\), or \(f(c)=c\). Hence \(c\) is a fixed point of \(f\). \(\square\)
The proof uses both hypotheses in distinct ways. Continuity makes the difference \(g\) continuous, while the requirement that \(f\) maps the interval into itself gives the endpoint inequalities. The conclusion is an existence statement: it guarantees at least one fixed point, but does not claim that the fixed point is unique.
Worked Example: A Reflection of the Unit Interval
Let \(f:[0,1]\to[0,1]\) be \(f(x)=1-x\). Find its fixed point using the theorem’s difference function.
First, the map really does take values in the interval. If \(0\le x\le1\), subtracting from \(1\) gives \(0\le1-x\le1\). The function is continuous because it is affine. Define \(g(x)=f(x)-x\). Substituting the formula for \(f\) gives
At the endpoints, \(g(0)=1-2(0)=1\) and \(g(1)=1-2(1)=1-2=-1\). The values have opposite signs, as required by the Intermediate Value Theorem. Solving \(g(x)=0\) gives \(1-2x=0\), so \(2x=1\) and \(x=\frac12\).
Check: substituting \(x=\frac12\) into the original map gives \(f(\frac12)=1-\frac12=\frac12\). Thus the computed point is indeed fixed.
Worked Example: A Map with Two Fixed Points
Let \(f:[0,1]\to[0,1]\) be \(f(x)=x^2\). Determine all its fixed points.
For \(0\le x\le1\), multiplication by \(x\ge0\) gives \(0\le x^2\le x\le1\), so \(f\) maps the interval into itself. It is continuous. The difference function is
The equation \(g(x)=0\) is \(x(x-1)=0\), whose solutions are \(x=0\) and \(x=1\). Both belong to the domain. The endpoint values are \(g(0)=0^2-0=0\) and \(g(1)=1^2-1=0\); here the endpoint inequalities hold with equality.
Check: \(f(0)=0^2=0\), and \(f(1)=1^2=1\). Both solutions satisfy the original fixed-point equation. Since the factored equation \(x(x-1)=0\) has no other real solutions, these are all the fixed points in \([0,1]\).
Worked Example: A Map on a Longer Interval
Let \(f:[-2,4]\to[-2,4]\) be \(f(x)=\frac{x}{2}+1\). Find its fixed point.
For \(x\in[-2,4]\), dividing the interval bounds by \(2\) gives \(-1\le x/2\le2\). Adding \(1\) gives \(0\le x/2+1\le3\), and \([0,3]\subseteq[-2,4]\). Thus the formula defines a self-map. It is continuous, and its difference function is
At the left endpoint, \(g(-2)=1-(-2)/2=1+1=2\). At the right endpoint, \(g(4)=1-4/2=1-2=-1\). The endpoint signs satisfy the theorem. Solving \(1-x/2=0\) gives \(x/2=1\), hence \(x=2\).
Check: the value \(2\) lies in \([-2,4]\), and \(f(2)=2/2+1=1+1=2\). It is a fixed point.
What the Theorem Does Not Say
A fixed point need not be unique. For example, the identity map \(f:[0,1]\to[0,1]\), defined by \(f(x)=x\), is continuous and maps the interval into itself. Its difference function is \(g(x)=x-x=0\) for every \(x\in[0,1]\). Consequently every point of the interval is fixed: for each \(x\in[0,1]\), substituting into the defining formula gives \(f(x)=x\). The theorem guarantees that the fixed-point set is nonempty; it does not specify how many points it contains.
There is a useful further property: for any continuous self-map of a closed interval, the set of its fixed points is closed in that interval. This statement concerns the topology on \([a,b]\), so “closed” here means closed in the subspace topology.
Proof. Let \(g(x)=f(x)-x\), which is continuous as in the theorem’s proof. Take any \(x\in[a,b]\) that is not in \(\operatorname{Fix}(f)\). Then \(g(x)\ne0\), so \(|g(x)|>0\). By continuity at \(x\), there is a neighborhood \(U\) of \(x\) in \([a,b]\) such that every \(y\in U\) satisfies \(|g(y)-g(x)|<|g(x)|/2\). The triangle inequality gives \(|g(y)|\ge |g(x)|-|g(y)-g(x)|>|g(x)|/2>0\). Thus \(g(y)\ne0\) for every \(y\in U\), and \(U\) contains no fixed point. Every point outside \(\operatorname{Fix}(f)\) therefore has a neighborhood contained in its complement, so that complement is open in \([a,b]\). Hence \(\operatorname{Fix}(f)\) is closed in \([a,b]\). \(\square\)
A Common Mistake
The fixed-point equation is \(f(x)=x\), not \(f(x)=0\). It is therefore not enough to apply the Intermediate Value Theorem directly to \(f\) and locate a zero of \(f\).
For a concrete example, use \(f(x)=1-x\) on \([0,1]\). A calculation of the zeros of \(f\) gives \(1-x=0\), hence \(x=1\). But \(f(1)=1-1=0\), which is not \(1\), so that zero is not a fixed point. The correction is to subtract the input: \(g(x)=f(x)-x=1-2x\). Solving \(g(x)=0\) gives \(x=\frac12\), and direct substitution confirms \(f(\frac12)=\frac12\).
The self-map condition matters as well. For instance, the constant function \(f:[0,1]\to\mathbb{R}\) given by \(f(x)=2\) is continuous, but it is not a map from \([0,1]\) to itself. Its difference \(g(x)=2-x\) has \(g(0)=2\) and \(g(1)=1\), so the endpoint sign change used in the proof does not occur. Indeed, the equation \(2=x\) has no solution in \([0,1]\). This does not contradict the theorem: the map fails its self-map hypothesis.
The argument is useful because it turns a fixed-point question into a zero-finding question for one continuous real-valued function. Its conclusion is robust in the sense that it does not require a formula for the fixed point, but it is limited in a precise way: without additional information, it gives existence rather than uniqueness or an explicit location.
Check Your Understanding
For each question, identify the difference function before using endpoint values or solving an equation.
- For \(f:[0,2]\to[0,2]\) defined by \(f(x)=2-x\), compute the difference function and find all fixed points.
- Explain why the condition \(f([a,b])\subseteq[a,b]\) implies \(f(a)-a\ge0\) and \(f(b)-b\le0\).
- For the identity map on \([0,1]\), determine the difference function and its zero set.
- Why does the theorem still hold when \(a=b\), even though the usual sign-change form of the Intermediate Value Theorem is stated for \(a<b\)?
- Give a continuous function on \([0,1]\) whose values do not all lie in \([0,1]\), and explain why the fixed-point theorem does not apply to it.