Existence of Fixed Points

From Zeros to Fixed Points

The previous tutorial studied when a continuous function must take the value zero. A closely related question asks whether a function can take a value equal to its input. Such a point is called a fixed point. The connection to zeros is immediate: if \(f\) is a function, then \(f(x)=x\) precisely when \(f(x)-x=0\). The challenge is to find conditions that guarantee this equality somewhere.

For a continuous function that maps a closed interval into itself, the endpoint values provide those conditions. At the left endpoint, the image cannot lie to its left; at the right endpoint, the image cannot lie to its right. Thus the continuous function \(x\mapsto f(x)-x\) starts nonnegative and ends nonpositive. The Intermediate Value Theorem then forces a zero. This gives an existence theorem without requiring \(f\) to be increasing or to have any particular formula.

Fixed Points and Endpoint Displacement

The condition \(f:E\to E\) says that every value of \(f\) is still in its domain. A function with this property is called a self-map of \(E\). For the fixed-point theorem, the domain will be a closed interval and the function will be continuous. It is useful to record the difference between the output and the input in a single function:

The function \(g\) is continuous whenever \(f\) is continuous, because the identity function \(x\mapsto x\) is continuous and differences of continuous functions are continuous. Moreover, \(g(c)=0\) exactly when \(c\) is a fixed point of \(f\). The next criterion isolates the sign information that will be used.

Proof. Define \(g:[a,b]\to\mathbb{R}\) by \(g(x)=f(x)-x\). The function \(g\) is continuous. The hypotheses give \(g(a)=f(a)-a\geq0\) and \(g(b)=f(b)-b\leq0\). If either value is zero, the corresponding endpoint is already a zero of \(g\). Otherwise \(g(a)>0\) and \(g(b)<0\), so \(0\) lies between \(g(a)\) and \(g(b)\). By the Intermediate Value Theorem, there is \(c\in[a,b]\) with \(g(c)=0\). In either case, \(f(c)-c=0\), and hence \(f(c)=c\). \(\square\)

The criterion does not require \(f\) to map the entire interval into itself. It only requires the two endpoint inequalities. This slightly broader formulation is useful when a function has been defined on a larger set but its behavior at the endpoints is known. For a self-map of \([a,b]\), however, both inequalities follow automatically.

The Fixed-Point Theorem for a Closed Interval

Proof. First suppose \(a<b\). Since \(f\) maps into \([a,b]\), its endpoint values satisfy \(f(a)\in[a,b]\) and \(f(b)\in[a,b]\). In particular, \(f(a)\geq a\) and \(f(b)\leq b\). The Endpoint Displacement Criterion therefore gives \(c\in[a,b]\) such that \(f(c)=c\).

If \(a=b\), then the domain and codomain are both the singleton \(\{a\}\). Since \(f(a)\in\{a\}\), we have \(f(a)=a\), so \(a\) is a fixed point. This handles the degenerate interval as well as the case of positive length. \(\square\)

This is an existence statement: it guarantees at least one fixed point but does not promise exactly one. The hypotheses have distinct roles. The interval is closed and bounded, the image remains in that interval, and continuity prevents the displacement \(f(x)-x\) from changing sign without taking the value zero. The proof does not require compactness directly; it uses the Intermediate Value Theorem on an interval.

The Fixed-Point Set Is Compact

The existence theorem says that the fixed-point set is nonempty for a continuous self-map of a closed interval. There is also useful information about the set as a whole: it is compact. To see why, once again use the displacement function \(g(x)=f(x)-x\). The fixed points are exactly its zeros, so the earlier result that zero sets of continuous functions are relatively closed applies.

Proof. By the Existence of a Fixed Point Theorem, \(\operatorname{Fix}(f)\) is nonempty. Define \(g(x)=f(x)-x\), which is continuous on \([a,b]\). The identity \(\operatorname{Fix}(f)=\{x\in[a,b]:g(x)=0\}\) shows that the fixed-point set is the zero set of \(g\) relative to \([a,b]\). By the earlier theorem that zero sets of continuous functions are relatively closed, \(\operatorname{Fix}(f)\) is closed relative to \([a,b]\). The interval \([a,b]\) is compact by the Heine–Borel Theorem, and a relatively closed subset of a compact set is compact. Therefore \(\operatorname{Fix}(f)\) is compact. \(\square\)

Compactness gives more than boundedness. For example, by the earlier theorem that nonempty compact subsets of the real line have a minimum and a maximum, the fixed-point set has a smallest and a largest point. It need not consist of every point between those extremes: in the \(x^2\) example it is \(\{0,1\}\). The theorem describes its compactness, not its shape as an interval.

What the Theorem Does—and Does Not—Say

A common error is to conclude that continuity alone guarantees a fixed point. A continuous function need not map an interval into itself, and the endpoint inequalities can fail. For instance, the map \(f(x)=x+1\) is continuous on \([0,1]\), but it does not map \([0,1]\) into itself: \(f(1)=2\). In fact, solving \(x+1=x\) gives \(1=0\), so it has no fixed point. The self-map condition is essential to the theorem as stated.

The endpoint displacement formulation also clarifies what is doing the work. If \(f(a)\geq a\) and \(f(b)\leq b\), then \(g(a)\geq0\) and \(g(b)\leq0\), and the Intermediate Value Theorem applies. A sign change is sufficient, but equality at an endpoint is already enough. Reversing both inequalities works as well: if \(f(a)\leq a\) and \(f(b)\geq b\), apply the same argument to \(x-f(x)\), which has the same zeros as \(f(x)-x\).

Finally, do not infer uniqueness from the crossing argument. The displacement may have several zeros, as the example \(f(x)=x^2\) demonstrates. The main conclusion is precisely that at least one input agrees with its output. The compactness result then says that all such inputs together form a nonempty compact set.