What Continuity Does—and Does Not—Guarantee
A fixed-point claim has two separate parts: a map must have at least one fixed point, and it may or may not have only one. Continuity is important in the interval existence theorem, but continuity by itself does not guarantee a fixed point on an unbounded domain. Nor does existence on a closed interval imply uniqueness. The distinction becomes precise by studying the difference between a map and the identity.
Unless otherwise stated, \(\mathbb{R}\) has its standard topology, and intervals have the subspace topology inherited from \(\mathbb{R}\). A self-map of a space \(X\) is a function \(f:X\to X\). A point \(c\in X\) is a fixed point of \(f\) if \(f(c)=c\). A continuous function \(f:X\to X\) need not have a fixed point merely because it is continuous.
This reformulation is useful because it separates a fixed-point equation from any intuition about where a map sends points: fixed points are precisely the zeros of \(g\). When \(f\) is continuous on a real interval, \(g\) is continuous too, since it is the difference of the continuous functions \(f\) and \(x\mapsto x\).
Continuity on an Unbounded Domain Is Not Enough
Worked Example: A Continuous Self-Map of \(\mathbb{R}\) with No Fixed Point
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x+1\). This is a self-map because \(x+1\in\mathbb{R}\) for every \(x\in\mathbb{R}\). It is continuous because it is an affine function.
To test for fixed points, solve the defining equation: \(f(x)=x\) becomes \(x+1=x\). Subtracting \(x\) from both sides gives \(1=0\), which is impossible. Equivalently, the displacement is \(g(x)=f(x)-x=(x+1)-x=1\), and this is never zero.
Check: for any real \(x\), substituting the proposed equation \(x+1=x\) into the fixed-point condition would require \(f(x)=x\), but \(f(x)=x+1\ne x\). Thus this continuous self-map of the unbounded domain \(\mathbb{R}\) has no fixed point.
The example disproves a universal claim about continuous self-maps on unbounded domains: continuity and the self-map condition do not suffice. It does not say that unbounded domains prevent fixed points. For contrast, the next example has a unique fixed point on the same domain.
Worked Example: An Unbounded Domain Can Have a Unique Fixed Point
Define \(F:\mathbb{R}\to\mathbb{R}\) by \(F(x)=x/2\). It is a self-map of \(\mathbb{R}\) and is continuous. The fixed-point equation is \(x/2=x\). Multiplying both sides by \(2\) gives \(x=2x\), and subtracting \(x\) gives \(x=0\).
Substitution verifies that value: \(F(0)=0/2=0\). The equation also proves there are no other fixed points, because every fixed point must satisfy \(x=0\). Hence \(0\) is the unique fixed point. This check shows that unboundedness is not itself a reason that a fixed point cannot exist; it is the absence of a general guarantee that matters.
There is a related pitfall even on a bounded domain: an interval need not be closed for a continuous self-map to lack a fixed point. For instance, \(f:(0,1)\to(0,1)\), \(f(x)=(x+1)/2\), is continuous and takes values strictly between \(1/2\) and \(1\). Its fixed-point equation reduces to \(x=1\), which is not in \((0,1)\). The domain matters; continuity alone is not a replacement for the hypotheses of an existence result.
What the Closed-Interval Argument Gives
Proof. If \(a=b\), the domain consists only of \(a\), and the self-map condition forces \(f(a)=a\). Now suppose \(a<b\). Define \(g(x)=f(x)-x\). The function \(g\) is continuous on \([a,b]\). Since \(f(a)\in[a,b]\), we have \(f(a)\ge a\), so \(g(a)=f(a)-a\ge0\). Since \(f(b)\in[a,b]\), we have \(f(b)\le b\), so \(g(b)=f(b)-b\le0\). If either endpoint value is zero, that endpoint is a fixed point. Otherwise \(g(a)>0\) and \(g(b)<0\), so the intermediate value theorem gives some \(c\in[a,b]\) with \(g(c)=0\). By the definition of \(g\), \(f(c)-c=0\), hence \(f(c)=c\). \(\square\)
The sign comparison is the key: the self-map condition forces the displacement to be nonnegative at the left endpoint and nonpositive at the right endpoint. The intermediate value theorem supplies a zero between those endpoint values. It supplies a zero, but it does not say there is only one.
Worked Example: The Identity Map Has Many Fixed Points
On \([0,1]\), define \(f(x)=x\). This is a continuous self-map of the closed interval. For every \(x\in[0,1]\), substitution gives \(f(x)=x\), so every point of the interval is a fixed point. In particular, \(0\) and \(1\) are distinct fixed points.
Check: the displacement is \(g(x)=f(x)-x=x-x=0\) for every \(x\in[0,1]\). Its zero set is therefore all of \([0,1]\), not a singleton. The theorem guarantees that this set is nonempty; it does not bound the number of its elements.
A Nonidentity Map with Several Fixed Points
The identity map makes the failure of uniqueness plain, but one might wonder whether uniqueness would hold for a map that actually moves some points. The following polynomial example rules that out. We check both that it maps the interval into itself and that it has exactly three fixed points.
Worked Example: Exactly Three Fixed Points on \([0,1]\)
Define \(F:[0,1]\to\mathbb{R}\) by \(F(x)=x+\frac14x(1-x)(x-\frac12)\). This is a polynomial, hence continuous. First we verify that its values lie in \([0,1]\), so it is indeed a self-map of \([0,1]\).
Write the added term as \(\delta(x)=\frac14x(1-x)(x-\frac12)\). For \(x\in[0,1]\), we have \(|x-\frac12|\le\frac12\), so \(|\delta(x)|\le\frac18x(1-x)\). Since \(0\le1-x\le1\), this gives \(x(1-x)\le x\), and therefore \(F(x)\ge x-\frac18x(1-x)\ge x-\frac18x=\frac78x\ge0\). Also \(x(1-x)\le1-x\), so \(F(x)\le x+\frac18x(1-x)\le x+\frac18(1-x)=\frac{1+7x}{8}\le1\). Thus \(0\le F(x)\le1\) throughout the domain.
Now solve the fixed-point equation. Subtracting \(x\) from both sides shows that \(F(x)=x\) exactly when \(\frac14x(1-x)(x-\frac12)=0\). The product is zero exactly when \(x=0\), \(1-x=0\), or \(x-\frac12=0\). These yield \(x=0\), \(x=1\), and \(x=\frac12\), all of which belong to \([0,1]\).
Check: substitution gives \(F(0)=0+\frac14(0)(1)(-\frac12)=0\); \(F(\frac12)=\frac12+\frac14(\frac12)(\frac12)(0)=\frac12\); and \(F(1)=1+\frac14(1)(0)(\frac12)=1\). Conversely, the factored equation shows that no other value can be fixed. So this continuous self-map has exactly three fixed points, despite not being the identity map.
Proof. For each \(x\in[a,b]\), the equation \(g(x)=0\) means \(f(x)-x=0\), which is equivalent to \(f(x)=x\). The latter is the definition of a fixed point. The function \(g\) is continuous because \(f\) and \(x\mapsto x\) are continuous and the difference of continuous real-valued functions is continuous. The interval theorem gives at least one zero; the identity and polynomial examples show that the zero set can have more than one point. \(\square\)
A Common Mistake: Confusing Existence with Uniqueness
A common mistake is to apply the intermediate value theorem and then conclude that the zero found must be unique. The theorem says that a continuous function with suitable endpoint signs has at least one zero; it does not rule out additional zeros. For the identity map on \([0,1]\), the displacement is \(g(x)=0\) everywhere, so every point is a zero.
Another mistake is to assume that the displacement of \(f(x)=x+1\) changes sign on \(\mathbb{R}\). A faulty calculation might write \(g(-1)=-1\) and \(g(1)=1\), then claim a zero by the intermediate value theorem. Those values are actually values of \(f\), not of \(g\): the correct calculation is \(g(-1)=f(-1)-(-1)=0+1=1\), and \(g(1)=f(1)-1=2-1=1\). The displacement is constantly \(1\), so there is no sign change and no zero.
| Claim | What the examples show | Conclusion |
|---|---|---|
| Continuity on an unbounded domain guarantees a fixed point. | \(f(x)=x+1\) on \(\mathbb{R}\) has none. | False. |
| An unbounded domain rules out fixed points. | \(F(x)=x/2\) on \(\mathbb{R}\) has the unique fixed point \(0\). | False. |
| A continuous self-map of a closed interval has a fixed point. | The endpoint displacement signs and the intermediate value theorem prove existence. | True. |
| A continuous self-map of a closed interval has a unique fixed point. | The identity has every point fixed; the polynomial example has exactly three. | False. |
Check Your Understanding
For each claim, identify which hypothesis or conclusion is at issue, and use the displacement function where it helps.
- Does \(f(x)=x-3\) on \(\mathbb{R}\) have a fixed point? Compute its displacement and justify your answer.
- Find all fixed points of \(f:[0,1]\to[0,1]\) given by \(f(x)=1-x\). Verify that the formula defines a self-map.
- For a continuous self-map \(f:[a,b]\to[a,b]\) with \(a<b\), explain why \(g(a)\ge0\) and \(g(b)\le0\) for \(g(x)=f(x)-x\).
- Why does the intermediate value theorem prove existence in the closed-interval argument but not uniqueness?
- Give an example of a continuous self-map of \(\mathbb{R}\) with a fixed point and verify both the fixed-point equation and uniqueness.