Endpoint Inequalities and the Sign Test
A fixed point of a function \(f\) is an input \(x\) for which \(f(x)=x\). When \(f\) is continuous on an interval, the endpoint values can force such a point even if \(f\) does not map the interval into itself. The useful quantity is the difference between the output and input: \(g(x)=f(x)-x\). A zero of \(g\) is exactly a fixed point of \(f\).
Let \(a,b\in\mathbb{R}\) with \(a\le b\), give \([a,b]\) its usual subspace topology from \(\mathbb{R}\), and give \(\mathbb{R}\) its standard topology. Assume \(f:[a,b]\to\mathbb{R}\) is continuous. The inequalities of interest are \(f(a)\ge a\) and \(f(b)\le b\). They say that \(g(a)\ge0\) and \(g(b)\le0\), so the endpoint values of \(g\) have opposite weak signs.
Proof. First suppose \(a=b\). The hypotheses give \(f(a)\ge a\) and \(f(a)\le a\), hence \(f(a)=a\), so \(a\) is a fixed point. Now suppose \(a<b\), and define \(g:[a,b]\to\mathbb{R}\) by \(g(x)=f(x)-x\). The identity function and \(f\) are continuous, so their difference \(g\) is continuous. The endpoint assumptions give \(g(a)=f(a)-a\ge0\) and \(g(b)=f(b)-b\le0\). The Intermediate Value Theorem, applied to \(g\) on \([a,b]\), gives a point \(c\in[a,b]\) with \(g(c)=0\). Thus \(f(c)-c=0\), or \(f(c)=c\). If both endpoint inequalities are strict, then \(g(a)>0\) and \(g(b)<0\), so neither endpoint is a zero. The zero \(c\) therefore belongs to \((a,b)\). \(\square\)
The Intermediate Value Theorem used here states that a continuous real-valued function on a closed interval takes every value between its endpoint values. The endpoint test is precisely an application of that result to \(g=f-\operatorname{id}_{[a,b]}\). Weak inequalities suffice: if an endpoint value of \(g\) is already zero, that endpoint itself is a fixed point.
Examples: One, Several, and Infinitely Many Fixed Points
Worked Example: Exactly One Fixed Point
On \([0,1]\), define \(f(x)=\frac13\). This is a continuous function into \(\mathbb{R}\). At the endpoints, \(f(0)=\frac13\ge0\) and \(f(1)=\frac13\le1\), so the endpoint test guarantees a fixed point.
To find it, solve the fixed-point equation directly: \(f(x)=x\) becomes \(\frac13=x\). Thus the only candidate is \(x=\frac13\), and it belongs to \([0,1]\). Substitution checks it: \(f(\frac13)=\frac13\). There are no other solutions because the equation \(\frac13=x\) has exactly one solution.
Check: The difference function is \(g(x)=\frac13-x\). Its endpoint values are \(g(0)=\frac13-0=\frac13\) and \(g(1)=\frac13-1=-\frac23\), with the required signs. Its only zero is \(x=\frac13\), agreeing with the direct calculation.
Worked Example: Several Fixed Points
On \([0,1]\), define the polynomial function \(f(x)=x-(x-\frac14)(x-\frac12)(x-\frac34)\). Polynomials are continuous, so the endpoint test applies once its hypotheses are checked.
At \(x=0\), the product is \((-\frac14)(-\frac12)(-\frac34)=-\frac{3}{32}\). Therefore \(f(0)=0-(-\frac{3}{32})=\frac{3}{32}\ge0\). At \(x=1\), the product is \((\frac34)(\frac12)(\frac14)=\frac{3}{32}\). Therefore \(f(1)=1-\frac{3}{32}=\frac{29}{32}\le1\). Both endpoint inequalities are strict, so the theorem guarantees a fixed point in \((0,1)\).
For the complete set of fixed points, solve \(f(x)=x\). Subtracting \(x\) from each side gives \(-(x-\frac14)(x-\frac12)(x-\frac34)=0\). A product of real numbers is zero exactly when at least one factor is zero. Thus the solutions are \(x=\frac14\), \(x=\frac12\), and \(x=\frac34\), all in \([0,1]\).
Check: At each of these three values, one factor in the product is zero, so \(f(x)=x-0=x\). Conversely, the displayed equation requires one of those three factors to be zero, so there are no additional fixed points.
Worked Example: Infinitely Many Fixed Points
Define \(f:[0,1]\to\mathbb{R}\) by \(f(0)=0\) and, for \(x>0\), \(f(x)=x(1-\sin(1/x))\). For positive \(x\), this is continuous as a product and composition of continuous functions. At the endpoint \(0\), the bound \(|f(x)|\le2x\) follows from \(|\sin(1/x)|\le1\); since \(2x\to0\) as \(x\to0^+\), \(f(x)\to0=f(0)\). Thus \(f\) is continuous on \([0,1]\).
The endpoint values are \(f(0)=0\ge0\) and \(f(1)=1-\sin(1)\le1\), since \(\sin(1)\ge0\). The endpoint test guarantees a fixed point. In fact, \(0\) is already one. For \(x>0\), the equation \(f(x)=x\) is \(x(1-\sin(1/x))=x\). Subtracting \(x\) and dividing by the positive number \(x\) shows that this is equivalent to \(\sin(1/x)=0\).
The zeros of sine are the integer multiples of \(\pi\). Since \(1/x\ge1\), the solutions here are \(1/x=n\pi\) for positive integers \(n\), giving \(x=1/(n\pi)\). Each lies in \((0,1)\), because \(n\pi>1\) for \(n\ge1\). Together with \(x=0\), these give infinitely many distinct fixed points.
Check: For every positive integer \(n\), \(\sin(1/(1/(n\pi)))=\sin(n\pi)=0\), so \(f(1/(n\pi))=(1/(n\pi))(1-0)=1/(n\pi)\). The points are distinct because \(n\pi\) strictly increases with \(n\), so their reciprocals strictly decrease. This verifies the solutions and shows that they accumulate at \(0\).
What the Conditions Guarantee—and What They Do Not
The endpoint test guarantees existence, not uniqueness. The first example had one fixed point, while the polynomial example had three and the trigonometric example had infinitely many. The argument uses only continuity and the signs of \(f(x)-x\) at the endpoints; it gives no rule for counting the zeros between them.
Endpoint equality is also meaningful. If \(f(a)=a\), then \(a\) is a fixed point without needing the Intermediate Value Theorem. If \(f(b)=b\), the same is true of \(b\). If both endpoint inequalities are strict, the theorem instead forces a fixed point strictly between the endpoints. When \(a=b\), both endpoint conditions together force equality at that single point.
The assumptions do not require \(f\) to take values in \([a,b]\). For example, on \([0,1]\) the function \(f(x)=2-x\) satisfies \(f(0)=2\ge0\) and \(f(1)=1\le1\), but \(f(0)=2\) is outside \([0,1]\). It still has a fixed point: solving \(2-x=x\) gives \(x=1\), and \(f(1)=1\). The endpoint test applies because it is a statement about the two endpoint inequalities, not about the whole range of \(f\).
Conversely, the endpoint inequalities are sufficient but not necessary for a fixed point. On \([0,1]\), \(f(x)=2x\) has the fixed point \(0\), but \(f(1)=2\) does not satisfy \(f(1)\le1\). In this case the test’s hypothesis fails even though a fixed point exists. A failed hypothesis means this argument cannot guarantee a fixed point; it does not mean that no fixed point is present.
Common pitfall: reversing the sign test. The relevant difference is \(g(x)=f(x)-x\), not \(x-f(x)\). For \(f(x)=\frac13\) on \([0,1]\), the correct endpoint calculations are \(g(0)=\frac13-0=\frac13\) and \(g(1)=\frac13-1=-\frac23\). A mistaken calculation using \(x-f(x)\) gives \(-\frac13\) and \(\frac23\), then incorrectly compares those signs with the criterion for \(f(x)-x\). The reversed difference has the opposite signs, so one must reverse the sign condition as well. In either convention, the equation being solved must remain equivalent to the fixed-point equation; here it gives \(x=\frac13\).
Check Your Understanding
For each question, distinguish what the endpoint test guarantees from what must be checked by solving the fixed-point equation.
- On \([0,2]\), let \(f(x)=1\). Check the endpoint inequalities and find all fixed points.
- On \([0,1]\), let \(f(x)=x+(x-\frac14)(x-\frac34)\). Do the endpoint inequalities hold in the required direction? Find the fixed points directly.
- Explain why strict endpoint inequalities force a fixed point in \((a,b)\), rather than merely in \([a,b]\).
- Give an example of a continuous function with a fixed point for which at least one of the endpoint inequalities fails.
- For the function in the infinitely-many-fixed-points example, explain why the fixed points \(1/(n\pi)\) are distinct as \(n\) ranges over the positive integers.